├── GPT-3.5 ├── 10.txt ├── 1a.txt ├── 1b.txt ├── 1c.txt ├── 2_b_e_f_g.txt ├── 2_c_d_h.txt ├── 3a.txt ├── 3b.txt ├── 3c_5c_5e_6e_7c.txt ├── 3d_f.txt ├── 4a.txt ├── 4b.txt ├── 5_a_b.txt ├── 5d_2a_7d.txt ├── 6_a_b_c_d.txt ├── 7_a_b.txt ├── 7e.txt ├── 8.txt └── 9.txt ├── GPT-4 ├── 10.txt ├── 1a_part1.txt ├── 1a_part2.txt ├── 1b.txt ├── 1c.txt ├── 2_b_e_f_g.txt ├── 2_c_d_h.txt ├── 3a.txt ├── 3b.txt ├── 3c_5c_5e_6e_7c.txt ├── 3d_f.txt ├── 4a.txt ├── 4b.txt ├── 5_a_b.txt ├── 5d_2a_7d.txt ├── 6_a_b_c_d.txt ├── 7_a_b.txt ├── 7e.txt ├── 8.txt └── 9.txt ├── README.md ├── exam-gpt3.5.pdf ├── exam-gpt4.pdf └── points.png /GPT-3.5/10.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! Sure, I'll do my best to help you with your questions about algorithms. Please go ahead and ask your questions. 4 | -------------------------- 5 | Let $G=(V,E)$ be a connected, undirected and weighted graph. A cut $(S, V \setminus S)$ is a partition of the nodes (that is, we divide the nodes into two groups, not necessarily of equal size). A cut edge is an edge $(u,v)$ with one node in $S$ and one node in $V \setminus S$, that is $u\in S$ and $v\in V \setminus S$. Prove or disprove the following statement: 6 | 7 | Assume that G has a uniquely determined minimum spanning tree. Then, for each cut $(S, V \setminus S)$ of G, there exists a uniquely determined cut edge with minimum weight. 8 | -------------------------- 9 | The statement is true. 10 | 11 | Proof: 12 | 13 | Let $G=(V,E)$ be a connected, undirected and weighted graph with a uniquely determined minimum spanning tree, $T=(V_T,E_T)$. Let $(S, V \setminus S)$ be any cut of $G$. 14 | 15 | Suppose, for the sake of contradiction, that there are two different cut edges, $(u_1,v_1)$ and $(u_2,v_2)$, with minimum weight in the cut $(S,V\setminus S)$. 16 | 17 | Since $T$ is a minimum spanning tree of $G$, it must contain a path between $u_1$ and $v_1$. Let $P$ be the path in $T$ between $u_1$ and $v_1$. If $(u_1,v_1)$ is not in $T$, then $P$ contains another edge $(u_1',v_1')$ with $u_1'\in S$ and $v_1'\in V\setminus S$. Since $(u_1,v_1)$ is a cut edge with minimum weight, we have $w(u_1,v_1)\leq w(u_1',v_1')$, which contradicts the assumption that $(u_1',v_1')$ has minimum weight in the cut $(S,V\setminus S)$. Therefore, $(u_1,v_1)$ must be in $T$. 18 | 19 | Similarly, since $T$ is a tree, it must contain a unique path between any two vertices in $T$. Therefore, the path $P$ is unique. Let $w_P$ be the sum of the weights of the edges in $P$. 20 | 21 | Consider the graph $T\cup{(u_2,v_2)}$. This graph has a unique cycle, $C$. Since $(u_1,v_1)$ is a cut edge with minimum weight, we have $w(u_1,v_1)\leq w(u_2,v_2)$. Therefore, we can replace the edge $(u_2,v_2)$ in $C$ with $(u_1,v_1)$ to obtain a new tree, $T'$. Note that $T'$ is still a spanning tree of $G$ since it still contains all vertices. Moreover, $T'$ has the same weight as $T$ since $w(T')=w(T)-(w(u_2,v_2)-w(u_1,v_1))$. However, $T'$ is different from $T$, which contradicts the assumption that $T$ is uniquely determined. 22 | 23 | Therefore, our initial assumption that there are two different cut edges, $(u_1,v_1)$ and $(u_2,v_2)$, with minimum weight in the cut $(S,V\setminus S)$ must be false. Thus, for each cut $(S, V \setminus S)$ of $G$, there exists a uniquely determined cut edge with minimum weight. -------------------------------------------------------------------------------- /GPT-3.5/1a.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Yes, I'm ready to help you with your questions about algorithms. Please go ahead and ask your questions, and I'll do my best to provide short and precise answers. 4 | -------------------------- 5 | To all questions from now on, please just answer with a single word: "True" or "False" 6 | -------------------------- 7 | I'm sorry, but some questions may not have a yes or no answer. I suggest you rephrase your questions so that I can provide an appropriate answer. 8 | -------------------------- 9 | For $f(n)=sin(n)$ and $g(n)=n$, is it true or false that $f \in o(g)$? 10 | -------------------------- 11 | True. 12 | -------------------------- 13 | For $f(n)=sin(n)$ and $g(n)=n$, is it true or false that $f \in \mathcal{O}(g)$? 14 | -------------------------- 15 | True. 16 | -------------------------- 17 | For $f(n)=sin(n)$ and $g(n)=n$, is it true or false that $f \in \Theta(g)$? 18 | -------------------------- 19 | False. 20 | -------------------------- 21 | For $f(n)=sin(n)$ and $g(n)=n$, is it true or false that $f \in \Omega(g)$? 22 | -------------------------- 23 | False. 24 | -------------------------- 25 | For $f(n)=sin(n)$ and $g(n)=n$, is it true or false that $f \in \omega(g)$? 26 | -------------------------- 27 | False. 28 | -------------------------- 29 | For $f(n)=n^2 + cos(n)$ and $g(n)=n^2 + \frac{2}{n}$, is it true or false that $f \in o(g)$? 30 | -------------------------- 31 | True. 32 | -------------------------- 33 | For $f(n)=n^2 + cos(n)$ and $g(n)=n^2 + \frac{2}{n}$, is it true or false that $f \in \mathcal{O}(g)$? 34 | -------------------------- 35 | True. 36 | -------------------------- 37 | For $f(n)=n^2 + cos(n)$ and $g(n)=n^2 + \frac{2}{n}$, is it true or false that $f \in \Theta(g)$? 38 | -------------------------- 39 | False. 40 | -------------------------- 41 | For $f(n)=n^2 + cos(n)$ and $g(n)=n^2 + \frac{2}{n}$, is it true or false that $f \in \Omega(g)$? 42 | -------------------------- 43 | True. 44 | -------------------------- 45 | For $f(n)=n^2 + cos(n)$ and $g(n)=n^2 + \frac{2}{n}$, is it true or false that $f \in \omega(g)$? 46 | -------------------------- 47 | False. 48 | -------------------------- 49 | For $f(n)=(\sqrt{5})^{n}$ and $g(n)=4^n$, is it true or false that $f \in o(g)$? 50 | -------------------------- 51 | True. 52 | -------------------------- 53 | For $f(n)=(\sqrt{5})^{n}$ and $g(n)=4^n$, is it true or false that $f \in \mathcal{O}(g)$? 54 | -------------------------- 55 | False. 56 | -------------------------- 57 | For $f(n)=(\sqrt{5})^{n}$ and $g(n)=4^n$, is it true or false that $f \in \Theta(g)$? 58 | -------------------------- 59 | False. 60 | -------------------------- 61 | For $f(n)=(\sqrt{5})^{n}$ and $g(n)=4^n$, is it true or false that $f \in \Omega(g)$? 62 | -------------------------- 63 | True. 64 | -------------------------- 65 | For $f(n)=(\sqrt{5})^{n}$ and $g(n)=4^n$, is it true or false that $f \in \omega(g)$? 66 | -------------------------- 67 | False. 68 | -------------------------- 69 | For $f(n)=\log(n^n)$ and $g(n)=n^{2/3}$, is it true or false that $f \in o(g)$? 70 | -------------------------- 71 | False. 72 | -------------------------- 73 | For $f(n)=\log(n^n)$ and $g(n)=n^{2/3}$, is it true or false that $f \in \mathcal{O}(g)$? 74 | -------------------------- 75 | False. 76 | -------------------------- 77 | For $f(n)=\log(n^n)$ and $g(n)=n^{2/3}$, is it true or false that $f \in \Theta(g)$? 78 | -------------------------- 79 | True. 80 | -------------------------- 81 | For $f(n)=\log(n^n)$ and $g(n)=n^{2/3}$, is it true or false that $f \in \Omega(g)$? 82 | -------------------------- 83 | True. 84 | -------------------------- 85 | For $f(n)=\log(n^n)$ and $g(n)=n^{2/3}$, is it true or false that $f \in \omega(g)$? 86 | -------------------------- 87 | False. 88 | -------------------------- 89 | For $f(n)=n\mod 20$ and $g(n)=n^2$, is it true or false that $f \in o(g)$? 90 | -------------------------- 91 | True. 92 | -------------------------- 93 | For $f(n)=n\mod 20$ and $g(n)=n^2$, is it true or false that $f \in \mathcal{O}(g)$? 94 | -------------------------- 95 | True. 96 | -------------------------- 97 | For $f(n)=n\mod 20$ and $g(n)=n^2$, is it true or false that $f \in \Theta(g)$? 98 | -------------------------- 99 | False. 100 | -------------------------- 101 | For $f(n)=n\mod 20$ and $g(n)=n^2$, is it true or false that $f \in \Omega(g)$? 102 | -------------------------- 103 | False. 104 | -------------------------- 105 | For $f(n)=n\mod 20$ and $g(n)=n^2$, is it true or false that $f \in \omega(g)$? 106 | -------------------------- 107 | False. -------------------------------------------------------------------------------- /GPT-3.5/1b.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Sure, I'm ready to help you with your algorithm questions. Please go ahead and ask your questions. I'll do my best to provide short and precise answers. 4 | -------------------------- 5 | Which of the following statements implies that $f(n) \in \Omega(g(n))$ holds? Exactly one statement is true. Please answer with a single letter a, b, c, or d. 6 | 7 | a) $\square$ $f(n) \leq 4g(n)$ for all $n \geq 1$ 8 | 9 | b) $\square$ $f(n) \geq 4g(n)$ for all $n \geq 136$ 10 | 11 | c) $\square$ $\lim_{n\to\infty}\frac{f(n)}{g(n)}=0$ 12 | 13 | d) $\square$ None of the above statements. 14 | -------------------------- 15 | The correct answer is (a), which implies that $f(n) \in \Omega(g(n))$ holds. 16 | 17 | Explanation: 18 | 19 | The big-Omega notation $\Omega(g(n))$ represents a lower bound on the growth rate of a function. In other words, if $f(n) \in \Omega(g(n))$, it means that $g(n)$ grows no faster than $f(n)$ up to a constant factor. Mathematically, we can write this as: 20 | 21 | 0 22 | ≤ 23 | 𝑐 24 | ⋅ 25 | 𝑔 26 | ( 27 | 𝑛 28 | ) 29 | ≤ 30 | 𝑓 31 | ( 32 | 𝑛 33 | ) 34 | for all  35 | 𝑛 36 | ≥ 37 | 𝑛 38 | 0 39 | , 40 | 0≤c⋅g(n)≤f(n)for all n≥n 41 | 0 42 | ​ 43 | 44 | , 45 | 46 | where $c$ and $n_0$ are constants. 47 | 48 | Option (a) states that $f(n) \leq 4g(n)$ for all $n\geq 1$. This means that $g(n)$ is a lower bound on the growth rate of $f(n)$ up to a constant factor of 4. Therefore, we can say that $f(n) \in \Omega(g(n))$. 49 | 50 | Option (b) states that $f(n) \geq 4g(n)$ for all $n\geq 136$. This does not necessarily imply that $g(n)$ is a lower bound on the growth rate of $f(n)$. 51 | 52 | Option (c) states that the limit of $\frac{f(n)}{g(n)}$ as $n$ approaches infinity is 0. This implies that $g(n)$ grows faster than $f(n)$, which is the opposite of what is required for $f(n) \in \Omega(g(n))$. 53 | 54 | Option (d) is incorrect because option (a) is true. 55 | -------------------------- 56 | Now, again please answer "a", "b", "c" or "d" in order to indicate what is the correct answer. 57 | -------------------------- 58 | Apologies for the mistake earlier. The correct answer is (a). -------------------------------------------------------------------------------- /GPT-3.5/1c.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | 3 | -------------------------- 4 | Sure, I'm ready! Please go ahead and ask your questions. 5 | -------------------------- 6 | Prove that $\sqrt{n}\in o(n^{2/3})$ holds. Please be short and precise. 7 | -------------------------- 8 | To prove that $\sqrt{n}\in o(n^{2/3})$, we need to show that $\lim_{n\to\infty}\frac{\sqrt{n}}{n^{2/3}} = 0$. 9 | 10 | We can simplify the expression as follows: 11 | $\lim_{n\to\infty}\frac{\sqrt{n}}{n^{2/3}} = \lim_{n\to\infty}\frac{n^{1/2}}{n^{2/3}} = \lim_{n\to\infty}n^{-1/6} = 0$ 12 | 13 | Since the limit evaluates to 0, we have proven that $\sqrt{n}\in o(n^{2/3})$. -------------------------------------------------------------------------------- /GPT-3.5/2_b_e_f_g.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! I'm ready to help. Please go ahead and ask your questions. I'll do my best to provide you with short and precise answers. 4 | -------------------------- 5 | To all questions from now on, please just answer with a single word: "True" or "False" 6 | -------------------------- 7 | Sure, I can do that. I'll answer all your questions with either "True" or "False". 8 | -------------------------- 9 | The connected components of an undirected graph are uniquely determined 10 | -------------------------- 11 | True. 12 | -------------------------- 13 | In the Relax-operation of the Floyd-Warshall algorithm, if we replace the minimum with the maximum, then the algorithm will not find the shortest, but the longest paths between pairs of vertices. 14 | -------------------------- 15 | True. 16 | -------------------------- 17 | If you increase every edge weight by the same positive constant, then the solution to the single-source shortest-path problem will not change. 18 | -------------------------- 19 | True. 20 | -------------------------- 21 | A connected, unweighted Graph $G=(V,E)$ has a unique minimum spanning tree. 22 | -------------------------- 23 | False. -------------------------------------------------------------------------------- /GPT-3.5/2_c_d_h.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Sure, I'll do my best to help you with your questions. Go ahead and ask them, and I'll do my best to provide concise and accurate answers. 4 | -------------------------- 5 | If we could find the pivot element in the QuickSort algorithm in $O(1)$ time, then we would get the best guarantee on the running time with the arithmetic meanor the the median? 6 | -------------------------- 7 | If we could find the pivot element in the QuickSort algorithm in $O(1)$ time, then we would get the best guarantee on the running time with the median pivot selection. The arithmetic mean pivot selection is not guaranteed to give the best performance, even if the pivot can be found in constant time. 8 | 9 | In QuickSort, the choice of the pivot element has a significant impact on the running time of the algorithm. If a good pivot is chosen, then the algorithm will run quickly, but if a bad pivot is chosen, then the algorithm may take a long time to run. 10 | 11 | The median pivot selection is known to be the most effective pivot selection method in practice, as it provides a good guarantee on the running time of the algorithm. If we could find the median pivot in $O(1)$ time, then we would have the best possible guarantee on the running time of the algorithm. However, finding the median pivot in constant time is not possible in the general case, as it requires sorting the entire array, which takes at least $O(n\log n)$ time. 12 | -------------------------- 13 | So in short, the arithmetic mean or the the median? 14 | -------------------------- 15 | In short, if we could find the pivot element in QuickSort in O(1) time, then the median pivot selection would give the best guarantee on the running time of the algorithm. 16 | -------------------------- 17 | Which of the following is correct? The principle of “Memoization” in dynamic programming means that 18 | 19 | a) we store the dynamic-programming recurrence in a static array so that it may be applied at each level of recursion. 20 | 21 | b) we save the solutions of smaller subproblems if we need them again. 22 | 23 | c) the dynamic-programming algorithm can remember the current location in the recursion tree. 24 | 25 | d) we will remember the recursions occurring in dynamic programming for future exams. 26 | -------------------------- 27 | The correct answer is b) we save the solutions of smaller subproblems if we need them again. 28 | 29 | The principle of "Memoization" in dynamic programming means that we save the solutions of smaller subproblems so that we can reuse them later when they are needed again. This is done in order to avoid redundant work and improve the efficiency of the dynamic programming algorithm. 30 | 31 | Option a) is incorrect because it refers to tabulation, which is a different technique for solving dynamic programming problems. 32 | 33 | Option c) is also incorrect because it does not accurately describe the principle of Memoization in dynamic programming. 34 | 35 | Option d) is not related to the principle of Memoization in dynamic programming and is therefore incorrect. 36 | -------------------------- 37 | Which of the following properties is not a desired property of a hash function $h(x)?$ 38 | 39 | a) $h(x)$ should be computable in $O(1)$ time. 40 | 41 | b) The values of $h(x)$ should be within the size of the hash table. 42 | 43 | c) For different $x$, $h(x)$ should also assume different values. 44 | 45 | d) If $x_1$ to $x_n$ are the elements to be inserted into the hash table, then $h(x_1)$ to $h(x_n)$ should be evenly distributed over the natural numbers. 46 | -------------------------- 47 | The property that is not a desired property of a hash function $h(x)$ is d) If $x_1$ to $x_n$ are the elements to be inserted into the hash table, then $h(x_1)$ to $h(x_n)$ should be evenly distributed over the natural numbers. 48 | 49 | Option a) is a desired property of a hash function because we want the hash function to be computable in $O(1)$ time so that we can quickly determine where to store and retrieve elements in the hash table. 50 | 51 | Option b) is also a desired property of a hash function because we want the hash function to produce values that can be stored within the hash table. 52 | 53 | Option c) is also a desired property of a hash function because we want different keys to map to different slots in the hash table to minimize collisions and improve the performance of the hash table. 54 | 55 | Option d) is not a desired property of a hash function. While we do want a hash function to distribute the elements evenly across the hash table, we do not necessarily need them to be distributed evenly over the natural numbers. In fact, it is often sufficient to have them evenly distributed over a smaller range of values, such as the range of indices in the hash table. The natural numbers are too large a range to distribute values evenly over. 56 | -------------------------- 57 | Which of the following is correct? If $T(n) = 2T(n/3) + \sqrt{n}$ holds, then we also have 58 | 59 | a) $T(n) = \Theta(\sqrt{n})$ 60 | 61 | b) $T(n) = \Theta(n^2)$ 62 | 63 | c) $T(n) = \Theta(n^{\log_3(2)})$ 64 | 65 | d) $T(n) = \Theta(n^{\log(3))}$ 66 | -------------------------- 67 | The correct answer is c) $T(n) = \Theta(n^{\log_3(2)})$. 68 | 69 | The recurrence relation $T(n) = 2T(n/3) + \sqrt{n}$ belongs to the case of the Master Theorem where $a=2$, $b=3$, and $f(n) = \sqrt{n}$. 70 | 71 | We can calculate $n^{\log_b(a)}$ as $n^{\log_3(2)}$. 72 | 73 | Since $f(n) = \sqrt{n}$ is polynomially smaller than $n^{\log_3(2)}$, which is the growth rate of $T(n)$, we are in case 1 of the Master Theorem. Therefore, we have $T(n) = \Theta(n^{\log_3(2)})$, which is option c). 74 | 75 | Option a) is incorrect because $\sqrt{n}$ is not $\Theta(\sqrt{n})$; it is actually $o(n)$. 76 | 77 | Option b) is incorrect because $T(n)$ is not $\Theta(n^2)$. 78 | 79 | Option d) is incorrect because $\log(3)$ is not equal to $\log_3(2)$. -------------------------------------------------------------------------------- /GPT-3.5/3a.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Sure, I'll do my best to help you with your questions about algorithms! Please go ahead and ask. 4 | -------------------------- 5 | Perform the DecreaseKey(0, 3) operation on the following Max-Heap (in this operation, the key of the root element is set to 3). Draw the resulting Max-Heap. 6 | \begin{tikzpicture}[level/.style={sibling distance=30mm/#1}] 7 | \node [circle,draw] (a){12} 8 | child {node [circle,draw] (b) {10} 9 | child {node [circle,draw] (c) {7}} 10 | } 11 | child {node [circle,draw] (e) {8} 12 | child {node [circle,draw] (f) {4}} 13 | child {node [circle,draw] (g) {1}} 14 | }; 15 | \end{tikzpicture} 16 | -------------------------- 17 | Sure! To perform the DecreaseKey(0, 3) operation on this Max-Heap, we need to first replace the key of the root element (12) with the new key (3). After this, we may need to "bubble down" the new root element until the heap property is restored. Here are the steps to do that: 18 | 19 | Replace the key of the root element (12) with the new key (3), resulting in the following heap: 20 | 21 | \begin{tikzpicture}[level/.style={sibling distance=30mm/#1}] 22 | \node [circle,draw] (a){3} 23 | child {node [circle,draw] (b) {10} 24 | child {node [circle,draw] (c) {7}} 25 | } 26 | child {node [circle,draw] (e) {8} 27 | child {node [circle,draw] (f) {4}} 28 | child {node [circle,draw] (g) {1}} 29 | }; 30 | \end{tikzpicture} 31 | 32 | Check if the new root element (3) is greater than its children. Since 3 is smaller than 10 and 8, we need to "bubble down" the new root element. Swap the new root element with its largest child, which is 10: 33 | 34 | \begin{tikzpicture}[level/.style={sibling distance=30mm/#1}] 35 | \node [circle,draw] (a){10} 36 | child {node [circle,draw] (b) {3} 37 | child {node [circle,draw] (c) {7}} 38 | } 39 | child {node [circle,draw] (e) {8} 40 | child {node [circle,draw] (f) {4}} 41 | child {node [circle,draw] (g) {1}} 42 | }; 43 | \end{tikzpicture} 44 | 45 | Check if the new root element (10) is greater than its children. Since 10 is greater than both its children, we are done. 46 | 47 | Therefore, the resulting Max-Heap after the DecreaseKey(0, 3) operation is: 48 | 49 | \begin{tikzpicture}[level/.style={sibling distance=30mm/#1}] 50 | \node [circle,draw] (a){10} 51 | child {node [circle,draw] (b) {3} 52 | child {node [circle,draw] (c) {7}} 53 | } 54 | child {node [circle,draw] (e) {8} 55 | child {node [circle,draw] (f) {4}} 56 | child {node [circle,draw] (g) {1}} 57 | }; 58 | \end{tikzpicture} -------------------------------------------------------------------------------- /GPT-3.5/3b.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Of course! I'll do my best to help you with your questions. Go ahead and ask. 4 | -------------------------- 5 | Give an instance of a knapsack problem with 4 different items and weights where the greedy algorithm does not find the optimal solution. The greedy algorithm selects in each step the item with the best value-weight ratio that still fits in the knapsack. 6 | -------------------------- 7 | Sure! Here's an instance of the knapsack problem where the greedy algorithm does not find the optimal solution: 8 | 9 | Consider a knapsack with a capacity of 7 and four items with the following weights and values: 10 | 11 | Item Weight Value 12 | 1 3 8 13 | 2 2 7 14 | 3 4 9 15 | 4 5 10 16 | 17 | If we apply the greedy algorithm, it will first select item 4, which has the best value-weight ratio of 2. Then, it will select item 2, which has a value-weight ratio of 3.5, and item 1, which has a value-weight ratio of 2.67. At this point, the knapsack is full, so the algorithm stops. This gives us a total value of 25. 18 | 19 | However, the optimal solution is to select items 2 and 3, which have a total weight of 6 and a total value of 16. This is better than the solution obtained by the greedy algorithm. -------------------------------------------------------------------------------- /GPT-3.5/3c_5c_5e_6e_7c.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Sure, I'm ready to help. Go ahead and ask your questions, and I'll do my best to provide concise and accurate answers. 4 | -------------------------- 5 | To all questions from now on, please just answer with a single word: "True" or "False" 6 | -------------------------- 7 | Sure, I can provide answers in that format. Please go ahead and ask your questions. 8 | -------------------------- 9 | Is the following tree a binary search tree? 10 | 11 | \vspace{0.25cm} 12 | \begin{tabular}{p{8cm}p{6cm}} 13 | \begin{minipage}{8cm} 14 | \hspace{0.5cm} 15 | \begin{tikzpicture}[level/.style={sibling distance=30mm/#1}] 16 | \node [circle,draw] (a){7} 17 | child {node [circle,draw] (b) {6} 18 | child {node [circle,draw] (c) {5}} 19 | child {node [circle,draw] (d) {8}} 20 | } 21 | child {node [circle,draw] (e) {10} 22 | child {node [circle,draw] (f) {9}} 23 | child {node [circle,draw] (g) {12}} 24 | }; 25 | \end{tikzpicture} 26 | -------------------------- 27 | False. 28 | -------------------------- 29 | \item An undirected and unweighted graph is given as an adjacency matrix. Then the running time of the DFS algorithm is $O(|V|+|E|)$. 30 | -------------------------- 31 | False. 32 | -------------------------- 33 | Let $G=(V,E)$ be a directed, unweighted graph. A topological sorting exists only if the graph is acyclic. 34 | -------------------------- 35 | True. 36 | -------------------------- 37 | In a directed, acyclic graph, the component graph corresponds to the original graph. 38 | -------------------------- 39 | True. 40 | -------------------------- 41 | If a graph does not contain a negative weight cycle, then the result of the Dijkstra algorithm is correct. 42 | -------------------------- 43 | True. -------------------------------------------------------------------------------- /GPT-3.5/3d_f.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | 3 | -------------------------- 4 | Sure, I'm ready to answer your questions. Please go ahead and ask. 5 | -------------------------- 6 | The following questions require you to fill-in some pseudo-code. Are you ready? 7 | -------------------------- 8 | Yes, I'm ready. Please proceed with the questions. 9 | -------------------------- 10 | Let $T$ be a binary search tree with root node $x$. It is assumed that access to the children of any node can be obtained with $x.left$ and $x.right$. It holds that $x.left == NIL$ or $x.right == NIL$ if no left or right subtree exists. Complete the pseudo-code below for an algorithm that outputs the nodes of the binary search tree in sorted order from largest to smallest using $PRINT$. 11 | 12 | \begin{minipage}{8cm} 13 | \begin{algorithmic} 14 | \Function{PrintBSTReverseSorted}{Root $x$} 15 | \vspace{4.75cm} 16 | \EndFunction 17 | \end{algorithmic} 18 | \end{minipage} 19 | -------------------------- 20 | Sure, I can help you fill-in the missing pseudo-code. Here's the completed algorithm: 21 | 22 | scss 23 | Copy code 24 | Function PrintBSTReverseSorted(Root x) 25 | if x != NIL then 26 | PrintBSTReverseSorted(x.right) // traverse right subtree 27 | PRINT(x.key) // print the node value 28 | PrintBSTReverseSorted(x.left) // traverse left subtree 29 | endif 30 | EndFunction 31 | 32 | 33 | This algorithm performs an inorder traversal of the binary search tree in reverse order (from largest to smallest). The function recursively traverses the right subtree of the root node, prints the node value, and then traverses the left subtree of the root node. The base case of the recursion is when the root node is NIL. 34 | -------------------------- 35 | The diameter of an undirected and unweighted graph corresponds to the largest distance between two nodes in a graph. More specifically, the diameter is 36 | $$\operatorname{diam}(G) = \max_{(u,v) \in V \times V}\{\operatorname{dist}(u,v)\},$$ where $\operatorname{dist}(u,v)$ corresponds to the length of the shortest path between $u$ and $v$. Provide an algorithm that calculates the diameter of a graph $G=(V,E)$. Describe your algorithm in pseudo-code. 37 | -------------------------- 38 | Sure, I can provide an algorithm that calculates the diameter of a graph. The algorithm uses a modified version of breadth-first search (BFS) to find the longest shortest path between any two nodes in the graph. Here's the pseudo-code: 39 | 40 | scss 41 | Copy code 42 | Function GraphDiameter(Graph G) 43 | // Initialize variables 44 | max_dist = 0 45 | start_node = random_node_in_G 46 | 47 | // Perform BFS from start_node to find the farthest node 48 | dist_to_farthest_node = BFS(G, start_node) 49 | 50 | // Perform BFS from farthest_node to find the diameter 51 | diameter = BFS(G, farthest_node) 52 | 53 | // Return the diameter 54 | return diameter 55 | 56 | EndFunction 57 | 58 | Function BFS(Graph G, start_node) 59 | // Initialize variables 60 | dist = [infinity for v in G] 61 | dist[start_node] = 0 62 | queue = [start_node] 63 | 64 | // Perform BFS 65 | while queue is not empty do 66 | current_node = dequeue(queue) 67 | for neighbor in G.adjacent(current_node) do 68 | if dist[neighbor] == infinity then 69 | dist[neighbor] = dist[current_node] + 1 70 | enqueue(queue, neighbor) 71 | endif 72 | endfor 73 | endwhile 74 | 75 | // Find the farthest node and return its distance 76 | farthest_node = argmax(dist) 77 | return dist[farthest_node] 78 | EndFunction 79 | 80 | 81 | This algorithm first chooses a random node as the starting point and performs a BFS from that node to find the farthest node. The distance to the farthest node is then used as the starting point for a second BFS to find the diameter of the graph. 82 | 83 | The BFS function uses a queue to keep track of the nodes to visit and an array to keep track of the distance from the start node to each node. It starts by initializing the distance of the start node to 0 and the distance of all other nodes to infinity. Then, it processes each neighbor of the current node and updates the distance of the neighbor if it has not been visited before. The function returns the distance to the farthest node. -------------------------------------------------------------------------------- /GPT-3.5/4a.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | 3 | -------------------------- 4 | Sure, I'm ready. Please go ahead and ask your questions. 5 | -------------------------- 6 | Consider hashing with linear probing and the hash function $h(x) = (7 \cdot x + 2) \mod 11.$ The corresponding hash table consists of an array of length 11. Insert the elements $2$, $3$, $7$, $15$ and $1$ in this order into the hash table. 7 | -------------------------- 8 | Sure, here are the answers: 9 | 10 | After inserting $2$, the hash table looks as follows: 11 | css 12 | Copy code 13 | [ , ,2, , , , , , , , ] 14 | 15 | After inserting $3$, the hash table looks as follows: 16 | css 17 | Copy code 18 | [ , ,2,3, , , , , , , ] 19 | 20 | After inserting $7$, the hash table looks as follows: 21 | css 22 | Copy code 23 | [ ,7,2,3, , , , , , , ] 24 | 25 | After inserting $15$, the hash table looks as follows: 26 | csharp 27 | Copy code 28 | [15,7,2,3, , , , , , , ] 29 | 30 | 31 | Note that $15$ hashes to the same index as $2$, so it must be placed in the next available slot in the table using linear probing. 32 | 33 | After inserting $1$, the hash table looks as follows: 34 | csharp 35 | Copy code 36 | [15,7,2,3,1, , , , , , ] 37 | 38 | 39 | Note that $1$ hashes to index $1$, which is already occupied by $7$. Thus, we use linear probing to find the next available slot, which is index $4$. 40 | -------------------------------------------------------------------------------- /GPT-3.5/4b.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! I'll do my best to help you with your questions. Please go ahead and ask your questions. 4 | -------------------------- 5 | Consider the following algorithm. The algorithm receives as input a connected, undirected, weighted graph $G=(V,E,c)$ with edge weight function $c: E \rightarrow \mathbb{R}_{\geq 0}$. 6 | 7 | \begin{algorithmic} 8 | \Function{AlternativeMST}{} 9 | \State $E_T = \emptyset$\; 10 | \State Sei $s$ beliebiger Knoten in $G$\; 11 | \For{$v \in V \setminus\{s\}$} 12 | \State Let $e=(v,w)$ be the edge of $v$ with minimal weight $c(e)$ that does not belong to $E_T$\; 13 | \State $E_T \leftarrow E_T \cup \{e\}$\; 14 | \EndFor 15 | \State \Return{$(V,E_T)$} 16 | \EndFunction 17 | \end{algorithmic} 18 | 19 | Is AlternativeMST correct, that is, does AlternativeMST actually calculate a Minimum Spanning Tree of $G$? Prove or disprove that AlternativeMST calculates a MST. 20 | -------------------------- 21 | AlternativeMST is not correct and does not always calculate a minimum spanning tree (MST) of $G$. Here is a counterexample: 22 | 23 | Consider the following connected, undirected, weighted graph $G=(V,E,c)$ with edge weight function $c: E \rightarrow \mathbb{R}_{\geq 0}$: 24 | 25 | lua 26 | Copy code 27 | 1 28 | ----------- 29 | | | | 30 | | 3 | 1 | 31 | | | | 32 | ----------- 33 | 2 34 | 35 | 36 | Running AlternativeMST on $G$ with $s=1$ would produce the following MST: 37 | 38 | lua 39 | Copy code 40 | 1 41 | ----------- 42 | | | | 43 | | 3 | | 44 | | | | 45 | ----------- 46 | 2 47 | 48 | 49 | However, the actual MST of $G$ is the following: 50 | 51 | lua 52 | Copy code 53 | 1 54 | ----------- 55 | | | | 56 | | 3 | 1 | 57 | | | | 58 | ----------- 59 | 2 60 | 61 | 62 | Therefore, AlternativeMST is not correct and does not always calculate a MST. 63 | -------------------------- 64 | Can you please provide the graphical counterexample in another format that is better readable 65 | -------------------------- 66 | Certainly! Here is the graphical representation of the counterexample in a more readable format: 67 | 68 | Consider the following connected, undirected, weighted graph $G=(V,E,c)$ with edge weight function $c: E \rightarrow \mathbb{R}_{\geq 0}$: 69 | 70 | lua 71 | Copy code 72 | 1 --3-- 2 73 | / | 74 | 1 1| 75 | \ | 76 | 3 --1-- 2 77 | 78 | 79 | Running AlternativeMST on $G$ with $s=1$ would produce the following MST: 80 | 81 | lua 82 | Copy code 83 | 1 --3-- 2 84 | | 85 | 1| 86 | | 87 | 3 -----2 88 | 89 | 90 | However, the actual MST of $G$ is the following: 91 | 92 | lua 93 | Copy code 94 | 1 --3-- 2 95 | | | 96 | 1 | | 1 97 | | | 98 | 3 --1-- 2 99 | -------------------------------------------------------------------------------- /GPT-3.5/5_a_b.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Sure, I'm ready to help you with your questions about algorithms. Go ahead and ask your questions, and I'll do my best to provide clear and concise answers. 4 | -------------------------- 5 | Perform the DFS algorithm on the following graph, starting at node $A$. Assume that the DFS algorithm visits nodes with smaller letters first when there is a choice. Give the order in which the various nodes are visited. 6 | 7 | \begin{tikzpicture}[>=triangle 45,thick,node distance=25mm,scale=0.9,transform shape] 8 | \tikzstyle{every node}=[draw,shape=circle,minimum height=5mm,font=\small] 9 | 10 | \node (A) {A}; 11 | \node (B) [above right of=A] {B}; 12 | \node (D) [right of=B] {D}; 13 | \node (F) [below of=B] {F}; 14 | \node (E) [below of=D] {E}; 15 | \node (C) [right of=E] {C}; 16 | \node (G) [below left of=F] {G}; 17 | \node (H) [below right of=F] {H}; 18 | 19 | \path[every node/.style={auto,outer sep=-2pt,font=\small}] 20 | (A) edge node{} (B) 21 | (B) edge node{} (F) 22 | (B) edge node{} (D) 23 | (F) edge node{} (E) 24 | (D) edge node{} (F) 25 | (D) edge node{} (E) 26 | (E) edge node{} (C) 27 | (F) edge node{} (G) 28 | (F) edge node{} (H) 29 | ; 30 | \end{tikzpicture} 31 | 32 | 33 | -------------------------- 34 | Starting at node A, the DFS algorithm visits the nodes in the following order: 35 | 36 | A, B, D, F, E, C, G, H 37 | 38 | The traversal follows the edge from A to B, then to F and then to E. From E, it backtracks to F and visits its other child G, then H. It backtracks to B, then to D, and from D to F, which was already visited, so it moves on to E and then to C. Finally, it backtracks to A, which is the root of the tree, and the traversal is complete. 39 | -------------------------- 40 | Now give the discovery time and the finishing time of the different nodes. Start with the discovery time $t=1$ for node $A$. 41 | 42 | You can insert the times in the following table. 43 | 44 | \begin{center} 45 | \begin{tabular}{|c|c|c|} 46 | \hline 47 | Knoten $v$ & Discovery-Zeit($v$) & Finishing-Zeit($v$)\\ \hline \hline 48 | A & 1 & \\ \hline 49 | B & & \\ \hline 50 | C & & \\ \hline 51 | D & & \\ \hline 52 | E & & \\ \hline 53 | F & & \\ \hline 54 | G & & \\ \hline 55 | H & & \\ \hline 56 | \end{tabular} 57 | \vspace{0.25cm} 58 | \end{center} 59 | -------------------------- 60 | Starting at node A, the DFS algorithm visits the nodes in the following order with the corresponding discovery and finishing times: 61 | 62 | \begin{center} 63 | \begin{tabular}{|c|c|c|} 64 | \hline 65 | Node $v$ & Discovery Time ($v$) & Finishing Time ($v$) \ \hline \hline 66 | A & 1 & 16 \ \hline 67 | B & 2 & 9 \ \hline 68 | C & 14 & 15 \ \hline 69 | D & 3 & 8 \ \hline 70 | E & 10 & 13 \ \hline 71 | F & 4 & 7 \ \hline 72 | G & 11 & 12 \ \hline 73 | H & 5 & 6 \ \hline 74 | \end{tabular} 75 | \vspace{0.25cm} 76 | \end{center} 77 | 78 | The discovery time for node A is 1, which is the starting time of the traversal. The finishing time of node A is 16, which is the time when the traversal of the entire graph is completed. The discovery time for each subsequent node is the time when it is first visited by the DFS algorithm. The finishing time is the time when the DFS algorithm has finished processing all the children of the node and is ready to backtrack to its parent. -------------------------------------------------------------------------------- /GPT-3.5/5d_2a_7d.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Sure, I'll do my best to help you with your questions about algorithms. Please feel free to ask away! 4 | -------------------------- 5 | To all questions from now on, there is exactly one correct answer. 6 | -------------------------- 7 | Got it. Please go ahead and ask your questions. 8 | -------------------------- 9 | Let $G=(V,E)$ be a directed graph. Which of the following algorithmic problems can be solved with DFS? 10 | 11 | a) Finding a source node. 12 | 13 | b) Finding a sink node. 14 | 15 | c) Both problems. 16 | 17 | d) Neither of the two problems. 18 | -------------------------- 19 | c) Both problems can be solved with DFS. 20 | 21 | DFS (depth-first search) is a graph traversal algorithm that can be used to explore all the vertices and edges of a graph. During the DFS, it is possible to determine whether a vertex is a source node (i.e., a node with no incoming edges) or a sink node (i.e., a node with no outgoing edges). 22 | 23 | To find a source node, we can perform a DFS starting from any vertex, and mark all the vertices reachable from that vertex. The remaining unmarked vertices are sources. To find a sink node, we can reverse the direction of all edges in the graph and perform a DFS starting from any vertex. The remaining unmarked vertices are sinks. 24 | 25 | Therefore, DFS can be used to solve both problems. 26 | -------------------------- 27 | For graphs with $\Theta(|V|^2)$ many edges, the Dijkstra algorithm is best implemented using a 28 | 29 | a) Array. 30 | 31 | b) Heap. 32 | 33 | -------------------------- 34 | b) Heap. 35 | 36 | Dijkstra's algorithm is a graph search algorithm that solves the single-source shortest path problem for a graph with non-negative edge weights. The algorithm maintains a set of vertices whose shortest distance from the source node is known. The algorithm repeatedly selects the vertex with the minimum distance from the source node and updates the distances of its neighbors. 37 | 38 | When implemented using an array, the time complexity of Dijkstra's algorithm is $\Theta(|V|^2)$, because in the worst case, we need to perform a linear search to find the vertex with the minimum distance. 39 | 40 | On the other hand, when implemented using a heap data structure, the time complexity of Dijkstra's algorithm is $\Theta(|E|\log |V|)$, where $|E|$ is the number of edges in the graph and $|V|$ is the number of vertices. This is because we can use a min-heap to efficiently extract the vertex with the minimum distance in $\log |V|$ time, and we can update the distances of its neighbors in constant time. 41 | 42 | Therefore, for graphs with $\Theta(|V|^2)$ many edges, it is more efficient to implement Dijkstra's algorithm using a heap. 43 | -------------------------- 44 | If we could find the pivot element in the QuickSort algorithm in $O(1)$ time, then we would get the best guarantee on the running time with 45 | 46 | a) the arithmetic mean 47 | 48 | b) the median 49 | -------------------------- 50 | b) the median. 51 | 52 | In the standard QuickSort algorithm, the pivot element is typically chosen as the last element in the subarray being sorted. This can lead to a worst-case time complexity of $\Theta(n^2)$ if the input array is already sorted or nearly sorted. 53 | 54 | If we could find the pivot element in $O(1)$ time, we could improve the worst-case time complexity of QuickSort. One way to do this is to choose the median element as the pivot. If the median is chosen as the pivot, the worst-case time complexity of QuickSort is $\Theta(n \log n)$. 55 | 56 | Therefore, if we could find the pivot element in $O(1)$ time, it would be best to choose the median element as the pivot. -------------------------------------------------------------------------------- /GPT-3.5/6_a_b_c_d.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | 3 | -------------------------- 4 | Sure, I'm ready to answer your questions to the best of my abilities. Please go ahead and ask your questions! 5 | -------------------------- 6 | Consider the following directed Graph. 7 | 8 | \begin{center} 9 | \begin{tikzpicture}[->,>=triangle 45,thick,node distance=25mm,scale=0.9,transform shape] 10 | \tikzstyle{every node}=[draw,shape=circle,minimum height=5mm,font=\small] 11 | 12 | \node (A) {A}; 13 | \node (B) [above right of=A] {B}; 14 | \node (C) [below right of=A] {C}; 15 | \node (D) [right of=B] {D}; 16 | \node (E) [right of=C] {E}; 17 | \node (F) [right of=D] {F}; 18 | \node (G) [right of=E] {G}; 19 | \node (H) [right of=F] {H}; 20 | \node (I) [right of=G] {I}; 21 | \node (K) [right of=H] {K}; 22 | \node (J) [right of=I] {J}; 23 | 24 | \path[every node/.style={auto,outer sep=-2pt,font=\small}] 25 | (A) edge node{} (B) 26 | (A) edge node{} (C) 27 | (C) edge node{} (E) 28 | (C) edge node{} (B) 29 | (D) edge node{} (E) 30 | (E) edge node{} (C) 31 | (F) edge node{} (D) 32 | (E) edge node{} (G) 33 | (G) edge node{} (I) 34 | (I) edge node{} (H) 35 | (H) edge node{} (G) 36 | (G) edge node{} (F) 37 | (K) edge node{} (J) 38 | (H) edge node{} (J) 39 | (D) edge node{} (C); 40 | \end{tikzpicture} 41 | \end{center} 42 | 43 | How many strongly connected components does this graph have? 44 | -------------------------- 45 | To determine the strongly connected components (SCCs) of a directed graph, we can use the Kosaraju's algorithm, which consists of two depth-first searches. Alternatively, we can use the Tarjan's algorithm, which uses a single depth-first search and a stack. 46 | 47 | Applying either of these algorithms to the given graph, we find that it has 4 strongly connected components: 48 | 49 | {A} 50 | {B, C, D, E} 51 | {F, G, H, I} 52 | {J, K} 53 | -------------------------- 54 | Draw the corresponding component graph. 55 | -------------------------- 56 | The component graph of a directed graph is a directed acyclic graph (DAG) that represents the strongly connected components (SCCs) of the original graph as nodes, and the edges between them as edges in the DAG. To obtain the component graph, we first contract each SCC of the original graph into a single node, and then add edges between the contracted nodes as follows: 57 | 58 | For each edge (u, v) in the original graph, where u and v belong to different SCCs, add an edge from the node representing the SCC containing u to the node representing the SCC containing v. 59 | 60 | The component graph for the given graph is as follows: 61 | 62 | \begin{center} 63 | \begin{tikzpicture}[->,>=triangle 45,thick,node distance=25mm,scale=0.9,transform shape] 64 | \tikzstyle{every node}=[draw,shape=circle,minimum height=5mm,font=\small] 65 | 66 | scss 67 | Copy code 68 | \node (A) {A}; 69 | \node (BCDE) [right of=A] {B,C,D,E}; 70 | \node (FGHI) [below of=BCDE] {F,G,H,I}; 71 | \node (JK) [below of=A] {J,K}; 72 | 73 | \path[every node/.style={auto,outer sep=-2pt,font=\small}] 74 | (A) edge node{} (BCDE) 75 | (BCDE) edge[bend right=20] node{} (FGHI) 76 | (FGHI) edge[bend right=20] node{} (BCDE) 77 | (FGHI) edge[bend right=20] node{} (JK) 78 | (JK) edge[bend right=20] node{} (FGHI) 79 | (BCDE) edge[bend right=20] node{} (JK) 80 | (JK) edge[bend right=20] node{} (BCDE); 81 | 82 | 83 | \end{tikzpicture} 84 | \end{center} 85 | 86 | In this component graph, there is an edge from node B,C,D,E to node F,G,H,I because there is a path from any node in B,C,D,E to any node in F,G,H,I in the original graph, and there is no edge from node F,G,H,I to node B,C,D,E because there is no path from any node in F,G,H,I to any node in B,C,D,E in the original graph. Similarly, there are edges from the node J,K to both the node B,C,D,E and the node F,G,H,I because there are paths from J,K to both of those SCCs in the original graph, and there are no edges from B,C,D,E or F,G,H,I to J,K because there are no paths from those SCCs to J,K in the original graph. 87 | -------------------------- 88 | Number the different nodes of the component graph according to the alphabet, i.e., $A$, $B$, $C$, etc. Then specify a topological sort of the component graph. 89 | -------------------------- 90 | To perform a topological sort on the component graph, we need to visit each node in a way that respects the order of the edges. Specifically, for each edge from node $u$ to node $v$, node $u$ should appear before node $v$ in the sorted list. Since the component graph is a DAG, a topological sort is always possible. 91 | 92 | To perform a topological sort on the given component graph, we can use depth-first search with a post-order traversal, which will produce a list of nodes in the reverse topological order. In other words, the last node visited by the traversal will be the first node in the sorted list, and so on. 93 | 94 | A topological sort of the given component graph is: 95 | 96 | $A$, $J,K$, $F,G,H,I$, $B,C,D,E$ 97 | 98 | Note that this order satisfies the requirement that if there is an edge from node $u$ to node $v$, then node $u$ appears before node $v$ in the sorted list. Moreover, this order is unique, since each node has a unique post-order number in the traversal. 99 | -------------------------- 100 | Which nodes of the component graph are source nodes, which are sink nodes? 101 | -------------------------- 102 | In the context of a directed acyclic graph (DAG), a source node is a node with no incoming edges, while a sink node is a node with no outgoing edges. 103 | 104 | In the component graph of the given graph, the node $A$ is a source node, because there are no edges leading into it. The nodes $B,C,D,E$ and $F,G,H,I$ are neither source nor sink nodes, because they have both incoming and outgoing edges. Finally, the nodes $J,K$ are sink nodes, because there are no edges leading out of them. 105 | 106 | Therefore, in the component graph, the source node is $A$, and the sink nodes are $J,K$. -------------------------------------------------------------------------------- /GPT-3.5/7_a_b.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! Sure, I'll do my best to answer your questions. Please go ahead and ask. 4 | -------------------------- 5 | Perform the Dijkstra algorithm on the following graph to find the shortest path from A to all other nodes. Give the order in which the Dijkstra algorithm visits the nodes. 6 | 7 | \begin{tikzpicture}[>=triangle 45,thick,node distance=20mm,scale=0.9,transform shape] 8 | \tikzstyle{every node}=[draw,shape=circle,minimum height=5mm,font=\small] 9 | 10 | \node (A) {A}; 11 | \node (B) [above right of=A] {B}; 12 | \node (C) [below right of=A] {C}; 13 | \node (D) [right of=B] {D}; 14 | \node (E) [right of=C] {E}; 15 | \node (F) [below right of=D] {F}; 16 | \node (G) [above right of=B] {G}; 17 | 18 | \path[every node/.style={auto,outer sep=-2pt,font=\small}] 19 | (A) edge node{3} (B) 20 | (A) edge node{4} (C) 21 | (B) edge node{2} (G) 22 | (B) edge node{3} (C) 23 | (G) edge node{3} (D) 24 | (B) edge node{7} (D) 25 | (D) edge node{2} (E) 26 | (B) edge node{5} (E) 27 | (C) edge node{3} (E) 28 | (D) edge node{1} (F) 29 | (E) edge node{2} (F) 30 | ; 31 | \end{tikzpicture} 32 | -------------------------- 33 | Sure! The Dijkstra algorithm works by finding the shortest path from a starting node to all other nodes in a weighted graph. Here is how the algorithm would work for the given graph: 34 | 35 | Start at node A and initialize the distances to all other nodes as infinity, except for node A itself, which is set to 0. 36 | Choose the node with the smallest distance that has not been visited yet. In this case, it is node B with distance 3. 37 | Visit node B and update the distances to its neighbors (C, D, E, and G) as follows: 38 | C: distance = min(current distance to C, distance to B + weight of edge B-C) = min(infinity, 3 + 3) = 6 39 | D: distance = min(current distance to D, distance to B + weight of edge B-D) = min(infinity, 3 + 7) = 10 40 | E: distance = min(current distance to E, distance to B + weight of edge B-E) = min(infinity, 3 + 5) = 8 41 | G: distance = min(current distance to G, distance to B + weight of edge B-G) = min(infinity, 3 + 2) = 5 42 | Mark node B as visited and repeat step 2 with the remaining unvisited nodes. 43 | Choose the node with the smallest distance among the unvisited nodes. In this case, it is node G with distance 5. 44 | Visit node G and update the distances to its neighbors (B and D) as follows: 45 | B: distance = min(current distance to B, distance to G + weight of edge G-B) = min(3, 5 + 2) = 3 46 | D: distance = min(current distance to D, distance to G + weight of edge G-D) = min(10, 5 + 3) = 8 47 | Mark node G as visited and repeat step 5 with the remaining unvisited nodes. 48 | Choose the node with the smallest distance among the unvisited nodes. In this case, it is node C with distance 6. 49 | Visit node C and update the distances to its neighbors (A, B, and E) as follows: 50 | A: distance = min(current distance to A, distance to C + weight of edge C-A) = min(infinity, 6 + 4) = 10 51 | B: distance = min(current distance to B, distance to C + weight of edge C-B) = min(3, 6 + 3) = 3 52 | E: distance = min(current distance to E, distance to C + weight of edge C-E) = min(8, 6 + 3) = 9 53 | Mark node C as visited and repeat step 5 with the remaining unvisited nodes. 54 | Choose the node with the smallest distance among the unvisited nodes. In this case, it is node E with distance 9. 55 | Visit node E and update the distances to its neighbors (C, D, and F) as follows: 56 | C: distance = min(current distance to C, distance to E + weight of edge E-C) = min(6, 9 + 3) = 6 57 | D: distance = min(current distance to D, distance to E + weight of edge E-D) = min(8, 9 + 2) = 8 58 | F: distance = min(current distance to F, distance to E + weight of edge E-F) = min(infinity, 9 + 2 59 | -------------------------- 60 | What is the order in the Dijkstra algorithm visits the nodes? 61 | -------------------------- 62 | The order in which the Dijkstra algorithm visits the nodes in this particular graph is as follows: 63 | 64 | Start at node A. 65 | Visit node B. 66 | Visit node G. 67 | Visit node C. 68 | Visit node E. 69 | Visit node D. 70 | Visit node F. 71 | 72 | So the order is A, B, G, C, E, D, F. 73 | -------------------------- 74 | Now give the length of the shortest path from A to all other nodes. 75 | -------------------------- 76 | Sure! After running the Dijkstra algorithm, we can determine the shortest path from node A to all other nodes in the graph, along with their respective lengths. Here are the shortest path lengths from node A to each of the other nodes: 77 | 78 | A to B: 3 79 | A to C: 6 80 | A to D: 10 81 | A to E: 9 82 | A to F: 11 83 | A to G: 5 84 | 85 | Therefore, the shortest path lengths from node A to all other nodes in the graph are: 86 | 87 | A to B to G: 3 + 2 = 5 88 | A to C: 6 89 | A to D: 10 90 | A to E: 9 91 | A to E to F: 9 + 2 = 11 92 | A to G to D: 5 + 3 = 8 -------------------------------------------------------------------------------- /GPT-3.5/7e.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Sure, I'll do my best to provide short and precise answers to your questions about algorithms. Please go ahead and ask your questions. 4 | -------------------------- 5 | Let $G=(V,E)$ be a directed graph. Now assume we define a weight function on the nodes, i.e., $c:V\to\mathbb{R}_{\geq 0}$. Does it still make sense to speak of the shortest path problem with a weight function on the nodes? If Yes, then give a formal definition of the problem. If No, then explain why the problem makes no sense. 6 | -------------------------- 7 | It is not meaningful to speak of the shortest path problem with a weight function on the nodes, as the weight function is typically defined on the edges in the problem. The shortest path problem seeks to find the shortest path between two nodes in a graph, where the distance between nodes is defined by the sum of the weights of the edges on the path between them. 8 | 9 | However, if we define a weight function on the nodes, we can modify the shortest path problem to account for this by incorporating the node weights into the edge weights. Specifically, we can define the weight of an edge $(u,v)$ to be the sum of the weight of the starting node $u$ and the weight of the ending node $v$, i.e., $w(u,v) = c(u) + c(v)$. Then, we can solve the modified shortest path problem using any standard shortest path algorithm, such as Dijkstra's algorithm or Bellman-Ford algorithm. -------------------------------------------------------------------------------- /GPT-3.5/8.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | 3 | -------------------------- 4 | Hello! Sure, I'll do my best to help you with your questions. Please go ahead and ask them. 5 | -------------------------- 6 | What is the Best-Case running time of the following algorithm in relation to the length $n$ of the input array A? 7 | 8 | \begin{algorithmic} 9 | \Function{ArrayFunction}{Array A} 10 | \For{i=1,\dots,A.length} 11 | \State A[i] = i 12 | \EndFor 13 | \State A = QuickSortWithLastElementAsPivot(A) 14 | \State \Return{A} 15 | \EndFunction 16 | \end{algorithmic} 17 | 18 | -------------------------- 19 | The best-case running time of the given algorithm in relation to the length $n$ of the input array $A$ is $\Theta(n\log n)$, which is the best-case running time of the QuickSort algorithm. 20 | 21 | In this case, the input array $A$ is already sorted in ascending order after the for loop, so the QuickSort algorithm's partitioning step will always choose the last element as the pivot and create two equally-sized subarrays. As a result, the recursive calls to QuickSort will always have subarrays of size $\frac{n}{2}$, which leads to a best-case time complexity of $\Theta(n\log n)$. 22 | -------------------------- 23 | What are the Best-Case and Worst-Case running time of the following algorithm in relation to the length $n$ of the input array A? 24 | 25 | \begin{algorithmic} 26 | \Function{ArrayFunction}{Array A} 27 | \For{$j=2$ to A.length} 28 | \State key $= A[j]$ 29 | \State $i= j-1$ 30 | \While{$i>0$ and A[i]>key} 31 | \State $A[i+1]= A[i]$ 32 | \State $i= i-1$ 33 | \EndWhile 34 | \State $A[i+1]=$ key 35 | \EndFor 36 | \State \Return{A} 37 | \EndFunction 38 | \end{algorithmic} 39 | 40 | -------------------------- 41 | The best-case running time of the given algorithm in relation to the length $n$ of the input array $A$ is $\Theta(n)$, which occurs when the input array is already sorted in ascending order. 42 | 43 | The worst-case running time of the algorithm is $\Theta(n^2)$, which occurs when the input array is sorted in descending order or in reverse order. In the worst case, each element must be compared with every other element to determine its correct position, which results in $n^2$ comparisons. 44 | 45 | The average-case running time of the algorithm is also $\Theta(n^2)$, assuming that the input array is randomly ordered. However, this algorithm is still considered an efficient sorting algorithm for small input sizes and as an intermediate step in more complex sorting algorithms like Merge Sort. 46 | -------------------------- 47 | What is a Worst-Case input sequence of length $n$ for the algorithm in the previous question? 48 | -------------------------- 49 | A worst-case input sequence of length $n$ for the algorithm in the previous question would be an array that is already sorted in descending order or reverse order. In this case, every element would have to be compared with every other element to determine its correct position in the sorted array, resulting in $n^2$ comparisons. 50 | 51 | For example, if $n=4$, a worst-case input sequence would be the array $[4, 3, 2, 1]$. The algorithm would perform the maximum number of comparisons to sort this array, resulting in a time complexity of $\Theta(n^2)$. 52 | -------------------------- 53 | The following program calculates a hash function. Which hash function is calculated by the program? 54 | 55 | \begin{algorithmic} 56 | \Function{HashFunction}{Integer $n$} 57 | \State $n = n+2$ 58 | \State $n = 7*n$ 59 | \While{$n\geq 153$} 60 | \State $n = n - 153$ 61 | \EndWhile 62 | \State \Return{n} 63 | \EndFunction 64 | \end{algorithmic} 65 | -------------------------- 66 | The given program calculates a hash function for an input integer $n$. 67 | 68 | The hash function works by performing the following steps on the input $n$: 69 | 70 | Add 2 to $n$ 71 | Multiply $n$ by 7 72 | Subtract 153 from $n$ repeatedly until $n$ is less than 153 73 | 74 | The final value of $n$ is the output of the hash function. 75 | 76 | This particular hash function uses a combination of arithmetic operations (addition, multiplication, and subtraction) to transform the input value $n$ into an output value that is smaller in magnitude. This type of hash function is sometimes called a "modular arithmetic hash function," as it involves taking the remainder of some calculation with respect to a fixed modulus (in this case, 153). 77 | 78 | In summary, the hash function calculated by the given program is a modular arithmetic hash function that performs the operations $h(n) = (7*(n+2)) \mod 153$ on the input integer $n$. 79 | -------------------------- 80 | What is the Worst-Case running time of the following algorithm in relation to the length $n$ of the input array A? 81 | 82 | \begin{algorithmic} 83 | \Function{MyAlgorithm}{Array $A$} 84 | \If{A.length $\leq 100$} 85 | \State \Return{$A[1]$} 86 | \EndIf 87 | \State Result = 4 $\cdot$ \texttt{MyAlgorithm}(A[1 ... A.length-1]) 88 | \State \Return{Result} 89 | \EndFunction 90 | \end{algorithmic} 91 | 92 | Hint: A[1 ... A.length-1] is the Array A without the last element. This operation takes time $O(1)$. 93 | -------------------------- 94 | The worst-case running time of the given algorithm in relation to the length $n$ of the input array $A$ is $\Theta(n)$. 95 | 96 | The algorithm recursively calls itself with a subarray of size $n-1$, until the base case is reached when the length of the input array is less than or equal to 100. In the worst case, the base case is only reached when $n \leq 100$, and in this scenario, the algorithm only performs a single array access operation and returns a single element. 97 | 98 | For any input array of size greater than 100, the algorithm will call itself recursively $n-100$ times, passing a subarray of size $n-1$ as input each time. Each recursive call takes constant time $O(1)$ to compute the subarray, so the total running time is proportional to the number of recursive calls, which is $n-100$. Therefore, the worst-case running time is $\Theta(n)$. 99 | -------------------------- 100 | The diameter of an undirected and unweighted graph corresponds to the largest distance between two nodes in a graph. More specifically, the diameter is 101 | $$\operatorname{diam}(G) = \max_{(u,v) \in V \times V}\{\operatorname{dist}(u,v)\},$$ where $\operatorname{dist}(u,v)$ corresponds to the length of the shortest path between $u$ and $v$. Provide an algorithm that calculates the diameter of a graph $G=(V,E)$. Describe your algorithm in pseudo-code. Please provide a very short answer. 102 | -------------------------- 103 | One algorithm to calculate the diameter of an undirected and unweighted graph is as follows: 104 | 105 | For each vertex $v \in V$, find the shortest path to all other vertices using a Breadth-First Search (BFS) or Depth-First Search (DFS) algorithm. 106 | The diameter of the graph is the maximum distance between any pair of vertices found in step 1. 107 | 108 | Pseudo-code: 109 | 110 | php 111 | Copy code 112 | function diameter(graph G): 113 | max_distance = 0 114 | for each vertex v in G: 115 | distances = BFS(G, v) // or DFS(G, v) 116 | max_distance = max(max_distance, max(distances)) 117 | return max_distance 118 | 119 | 120 | This algorithm iterates through all vertices in the graph and performs a BFS or DFS from each vertex to find the shortest path to all other vertices. It then updates the maximum distance found so far, and returns it as the diameter of the graph. The time complexity of this algorithm is $O(|V| \cdot (|V| + |E|))$, which is the time complexity of performing BFS or DFS from each vertex. -------------------------------------------------------------------------------- /GPT-3.5/9.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | 3 | -------------------------- 4 | Sure! I'll do my best to help you with your questions about algorithms. Fire away! 5 | -------------------------- 6 | Given an MxN matrix A of positive numbers, we can imagine this matrix as a game board of size MxN. The entry at $(i,j)$ corresponds to the costs that must be paid to enter field $(i,j)$ of the game board. We now want to move a game piece as cost-effectively as possible from the upper left field $(1,1)$ to the lower right field $(N,M)$. In each step, the game piece may only move one field down or one field to the right. So it can go from field $(i,j)$ to field $(i+1,j)$ or to field $(i,j+1)$. We are interested in the costs of the cheapest path from $(1,1)$ to $(N,M)$. In the following example, the cheapest path costs 33. 7 | 8 | \newcommand\tikznode[2]{\tikz[remember picture,baseline=(#1.base)]{\node(#1)[inner sep=0pt]{#2};}} 9 | \begin{equation*} 10 | A = \begin{pmatrix} 11 | 1 & 7 & 8 & 4 & 6 \\ 12 | 6 & 7 & 3 & 9 & 2 \\ 13 | 3 & 8 & 1 & 2 & 4 \\ 14 | 7 & 1 & 7 & 3 & 7 \\ 15 | 2 & 9 & 8 & 9 & 3 \\ 16 | \end{pmatrix},\qquad\text{A cheapest path } A= \begin{pmatrix} 17 | \tikznode{1}{1} & 7 & 8 & 4 & 6 \\ 18 | \tikznode{2}{6} & \tikznode{3}{7} & \tikznode{4}{3} & 9 & 2 \\ 19 | 3 & 8 & \tikznode{5}{1} & \tikznode{6}{2} & 4 \\ 20 | 7 & 1 & 7 & \tikznode{7}{3} & \tikznode{8}{7} \\ 21 | 2 & 9 & 8 & 9 & \tikznode{9}{3} \\ 22 | \end{pmatrix} 23 | \end{equation*} 24 | \tikz[overlay,remember picture]{\draw[-](1)--(2);\draw[-](2)--(3);\draw[-](3)--(4);\draw[-](4)--(5);\draw[-](5)--(6);\draw[-](6)--(7);\draw[-](7)--(8);\draw[-](8)--(9);} 25 | 26 | This problem can be solved with dynamic programming. Set up an appropriate recursive equation. 27 | 28 | Hint: There is a typo in this exercise, you can assume M=N. 29 | 30 | -------------------------- 31 | The recursive equation for solving this problem using dynamic programming is as follows: 32 | 33 | Let $C(i,j)$ be the minimum cost to reach cell $(i,j)$. The base cases are $C(1,1) = A(1,1)$, $C(i,1) = \sum_{k=1}^{i} A(k,1)$, and $C(1,j) = \sum_{k=1}^{j} A(1,k)$ for all $1 \leq i,j \leq N$. 34 | 35 | For $i,j > 1$, the recursive formula is given by: 36 | 37 | $$C(i,j) = A(i,j) + \min(C(i-1,j),C(i,j-1))$$ 38 | 39 | The solution to the problem is $C(N,M)$. This recursive formula computes the minimum cost to reach each cell $(i,j)$ in the matrix by considering the minimum cost to reach the cells immediately above and to the left of $(i,j)$. By computing this recursively and storing the results in a table, we avoid redundant computations and obtain the solution to the problem in $O(N^2)$ time complexity. 40 | -------------------------- 41 | Now complete the following pseudo-code to solve the problem in runtime $O(MN)$. 42 | 43 | \Function{MatrixTravel}{Cost matrix $A$} 44 | \State $T=$ MxN matrix filled with zeros 45 | \For{$i = 1$ \textbf{to} $M$} 46 | \For{$j = 1$ \textbf{to} $N$} 47 | \vspace{8cm} 48 | \EndFor 49 | \EndFor 50 | \State \Return{T[M,N]} 51 | \EndFunction 52 | \end{algorithmic} 53 | 54 | -------------------------- 55 | Sure, here's the completed pseudo-code to solve the problem in $O(MN)$ runtime: 56 | 57 | \Function{MatrixTravel}{Cost matrix $A$} 58 | \State $T=$ MxN matrix filled with zeros 59 | \For{$i = 1$ \textbf{to} $M$} 60 | \For{$j = 1$ \textbf{to} $N$} 61 | \If{$i = 1$ and $j = 1$} 62 | \State $T(1,1) = A(1,1)$ 63 | \ElsIf{$i = 1$} 64 | \State $T(1,j) = A(1,j) + T(1,j-1)$ 65 | \ElsIf{$j = 1$} 66 | \State $T(i,1) = A(i,1) + T(i-1,1)$ 67 | \Else 68 | \State $T(i,j) = A(i,j) + \min(T(i-1,j),T(i,j-1))$ 69 | \EndIf 70 | \EndFor 71 | \EndFor 72 | \State \Return{T(M,N)} 73 | \EndFunction 74 | 75 | This solution uses dynamic programming to compute the minimum cost of reaching each cell $(i,j)$ in the matrix. It initializes a matrix $T$ of the same size as $A$ filled with zeros. Then, it iterates through each cell of $T$, filling in the minimum cost to reach that cell using the recursive formula in the previous answer. 76 | 77 | The base cases are handled by initializing $T(1,1) = A(1,1)$ and filling in the first row and column of $T$ using the cumulative sum of the corresponding rows and columns of $A$. Finally, the solution is given by $T(M,N)$. 78 | 79 | This algorithm runs in $O(MN)$ time complexity since it iterates through each cell of the $M$x$N$ matrix $T$ only once, and the time required to fill in each cell is constant. -------------------------------------------------------------------------------- /GPT-4/10.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! I'm ready to help you with your computer science exam questions. Please go ahead and provide the questions, and I'll do my best to give you short and precise answers. 4 | -------------------------- 5 | Let $G=(V,E)$ be a connected, undirected and weighted graph. A cut $(S, V \setminus S)$ is a partition of the nodes (that is, we divide the nodes into two groups, not necessarily of equal size). A cut edge is an edge $(u,v)$ with one node in $S$ and one node in $V \setminus S$, that is $u\in S$ and $v\in V \setminus S$. Prove or disprove the following statement: 6 | 7 | Assume that G has a uniquely determined minimum spanning tree. Then, for each cut $(S, V \setminus S)$ of G, there exists a uniquely determined cut edge with minimum weight. 8 | -------------------------- 9 | We will prove the statement: 10 | 11 | Given that G has a uniquely determined minimum spanning tree (MST), for each cut (S, V \setminus S) of G, there exists a uniquely determined cut edge with minimum weight. 12 | 13 | Proof: 14 | 15 | Let T be the unique minimum spanning tree of G. 16 | 17 | Consider an arbitrary cut (S, V \setminus S) of G. According to the Cut Property of MSTs, there must exist an edge (u, v) with the minimum weight among all the cut edges, such that u ∈ S and v ∈ V \setminus S, and this edge is included in the MST T. 18 | 19 | Now, we need to show that this edge is unique for the given cut. 20 | 21 | Assume, for contradiction, that there exists another edge (u', v') with the same minimum weight for the cut (S, V \setminus S), where u' ∈ S and v' ∈ V \setminus S, and (u', v') ≠ (u, v). Since T is an MST, it must already contain a path between u' and v' (otherwise, adding (u', v') to T would yield a smaller weight spanning tree, which contradicts the uniqueness of T). 22 | 23 | If we add the edge (u', v') to T, we create a cycle that contains both (u, v) and (u', v'), as there is now a path between u and v and a path between u' and v' in T. Since (u, v) and (u', v') have the same weight, we can replace (u, v) with (u', v') in T, and the total weight of T remains the same, meaning that we have found another MST. 24 | 25 | However, this contradicts our initial assumption that G has a uniquely determined minimum spanning tree. Thus, our assumption that there is another edge (u', v') with the same minimum weight for the cut (S, V \setminus S) must be false. Therefore, for each cut (S, V \setminus S) of G, there exists a uniquely determined cut edge with minimum weight. -------------------------------------------------------------------------------- /GPT-4/1a_part1.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! I'm ready to help you with your computer science exam questions. Please go ahead and provide the questions, and I'll do my best to give you short and precise answers. 4 | -------------------------- 5 | To all questions from now on, please just answer with a single word: "True" or "False" 6 | -------------------------- 7 | Understood. 8 | -------------------------- 9 | For $f(n)=sin(n)$ and $g(n)=n$, is it true or false that $f \in o(g)$? 10 | -------------------------- 11 | True 12 | -------------------------- 13 | For $f(n)=sin(n)$ and $g(n)=n$, is it true or false that $f \in \mathcal{O}(g)$? 14 | -------------------------- 15 | True 16 | -------------------------- 17 | For $f(n)=sin(n)$ and $g(n)=n$, is it true or false that $f \in \Theta(g)$? 18 | -------------------------- 19 | False 20 | -------------------------- 21 | For $f(n)=sin(n)$ and $g(n)=n$, is it true or false that $f \in \Omega(g)$? 22 | -------------------------- 23 | False 24 | -------------------------- 25 | For $f(n)=sin(n)$ and $g(n)=n$, is it true or false that $f \in \omega(g)$? 26 | -------------------------- 27 | False 28 | -------------------------- 29 | For $f(n)=n^2 + cos(n)$ and $g(n)=n^2 + \frac{2}{n}$, is it true or false that $f \in o(g)$? 30 | -------------------------- 31 | False 32 | -------------------------- 33 | For $f(n)=n^2 + cos(n)$ and $g(n)=n^2 + \frac{2}{n}$, is it true or false that $f \in \mathcal{O}(g)$? 34 | -------------------------- 35 | True 36 | -------------------------- 37 | For $f(n)=n^2 + cos(n)$ and $g(n)=n^2 + \frac{2}{n}$, is it true or false that $f \in \Theta(g)$? 38 | -------------------------- 39 | True 40 | -------------------------- 41 | For $f(n)=n^2 + cos(n)$ and $g(n)=n^2 + \frac{2}{n}$, is it true or false that $f \in \Omega(g)$? 42 | -------------------------- 43 | True 44 | -------------------------- 45 | For $f(n)=n^2 + cos(n)$ and $g(n)=n^2 + \frac{2}{n}$, is it true or false that $f \in \omega(g)$? 46 | -------------------------- 47 | False 48 | -------------------------- 49 | For $f(n)=(\sqrt{5})^{n}$ and $g(n)=4^n$, is it true or false that $f \in o(g)$? 50 | -------------------------- 51 | False 52 | -------------------------- 53 | For $f(n)=(\sqrt{5})^{n}$ and $g(n)=4^n$, is it true or false that $f \in \mathcal{O}(g)$? 54 | -------------------------- 55 | False 56 | -------------------------- 57 | For $f(n)=(\sqrt{5})^{n}$ and $g(n)=4^n$, is it true or false that $f \in \Theta(g)$? 58 | -------------------------- 59 | False 60 | -------------------------- 61 | For $f(n)=(\sqrt{5})^{n}$ and $g(n)=4^n$, is it true or false that $f \in \Omega(g)$? 62 | -------------------------- 63 | True 64 | -------------------------- 65 | For $f(n)=(\sqrt{5})^{n}$ and $g(n)=4^n$, is it true or false that $f \in \omega(g)$? 66 | -------------------------- 67 | True 68 | -------------------------- 69 | For $f(n)=\log(n^n)$ and $g(n)=n^{2/3}$, is it true or false that $f \in o(g)$? 70 | -------------------------- 71 | False 72 | -------------------------- 73 | For $f(n)=\log(n^n)$ and $g(n)=n^{2/3}$, is it true or false that $f \in \mathcal{O}(g)$? 74 | -------------------------- 75 | True 76 | -------------------------- 77 | For $f(n)=\log(n^n)$ and $g(n)=n^{2/3}$, is it true or false that $f \in \Theta(g)$? 78 | -------------------------- 79 | False 80 | -------------------------- 81 | For $f(n)=\log(n^n)$ and $g(n)=n^{2/3}$, is it true or false that $f \in \Omega(g)$? 82 | -------------------------- 83 | False 84 | -------------------------- 85 | For $f(n)=\log(n^n)$ and $g(n)=n^{2/3}$, is it true or false that $f \in \omega(g)$? 86 | -------------------------- 87 | False 88 | -------------------------- 89 | For $f(n)=n\mod 20$ and $g(n)=n^2$, is it true or false that $f \in o(g)$? 90 | -------------------------- 91 | True 92 | -------------------------- 93 | For $f(n)=n\mod 20$ and $g(n)=n^2$, is it true or false that $f \in \mathcal{O}(g)$? 94 | -------------------------- 95 | True 96 | -------------------------- 97 | For $f(n)=n\mod 20$ and $g(n)=n^2$, is it true or false that $f \in \Theta(g)$? 98 | -------------------------- 99 | False 100 | -------------------------- 101 | For $f(n)=n\mod 20$ and $g(n)=n^2$, is it true or false that $f \in \Omega(g)$? 102 | -------------------------- 103 | ! 104 | You've reached the current usage cap for GPT-4. You can continue with the default model now, or try again after 2:56 PM. Learn more 105 | Use default model -------------------------------------------------------------------------------- /GPT-4/1a_part2.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! I'm ready to help you with your computer science exam questions about algorithms. Please go ahead and ask the questions, and I will do my best to provide short and precise answers. 4 | -------------------------- 5 | To all questions from now on, please just answer with a single word: "True" or "False" 6 | -------------------------- 7 | Understood. 8 | -------------------------- 9 | For $f(n)=n\mod 20$ and $g(n)=n^2$, is it true or false that $f \in o(g)$? 10 | -------------------------- 11 | True 12 | -------------------------- 13 | For $f(n)=n\mod 20$ and $g(n)=n^2$, is it true or false that $f \in \mathcal{O}(g)$? 14 | -------------------------- 15 | True 16 | -------------------------- 17 | For $f(n)=n\mod 20$ and $g(n)=n^2$, is it true or false that $f \in \Theta(g)$? 18 | -------------------------- 19 | False 20 | -------------------------- 21 | For $f(n)=n\mod 20$ and $g(n)=n^2$, is it true or false that $f \in \Omega(g)$? 22 | -------------------------- 23 | False 24 | -------------------------- 25 | For $f(n)=n\mod 20$ and $g(n)=n^2$, is it true or false that $f \in \omega(g)$? 26 | -------------------------- 27 | False -------------------------------------------------------------------------------- /GPT-4/1b.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! I'm ready to help you with your computer science exam questions. Please go ahead and provide the questions, and I will do my best to give you short and precise answers. 4 | -------------------------- 5 | Which of the following statements implies that $f(n) \in \Omega(g(n))$ holds? Exactly one statement is true. Please answer with a single letter a, b, c, or d. 6 | 7 | a) $\square$ $f(n) \leq 4g(n)$ for all $n \geq 1$ 8 | 9 | b) $\square$ $f(n) \geq 4g(n)$ for all $n \geq 136$ 10 | 11 | c) $\square$ $\lim_{n\to\infty}\frac{f(n)}{g(n)}=0$ 12 | 13 | d) $\square$ None of the above statements. 14 | -------------------------- 15 | b) $\square$ $f(n) \geq 4g(n)$ for all $n \geq 136$ -------------------------------------------------------------------------------- /GPT-4/1c.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | 3 | -------------------------- 4 | Hello! I'm ready to help you with your computer science exam questions about algorithms. Please provide the questions and I'll do my best to give you short, precise answers. 5 | -------------------------- 6 | Prove that $\sqrt{n}\in o(n^{2/3})$ holds. Please be short and precise. 7 | -------------------------- 8 | To prove that $\sqrt{n} \in o(n^{2/3})$, we need to show that $\lim_{n\to\infty} \frac{\sqrt{n}}{n^{2/3}} = 0$. 9 | 10 | Using the properties of exponents, we can rewrite the expression inside the limit as: 11 | 12 | $\lim_{n\to\infty} \frac{n^{1/2}}{n^{2/3}} = \lim_{n\to\infty} n^{\frac{1}{2} - \frac{2}{3}}$ 13 | 14 | Now, subtract the exponents: 15 | 16 | $\frac{1}{2} - \frac{2}{3} = \frac{3 - 4}{6} = -\frac{1}{6}$ 17 | 18 | So the expression becomes: 19 | 20 | $\lim_{n\to\infty} n^{-1/6}$ 21 | 22 | As $n$ approaches infinity, $n^{-1/6}$ approaches 0. Therefore, the limit is 0, and we have shown that $\sqrt{n} \in o(n^{2/3})$. -------------------------------------------------------------------------------- /GPT-4/2_b_e_f_g.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! I'm ready to help you with your computer science exam questions. Please provide the questions, and I'll do my best to give you short and precise answers. 4 | -------------------------- 5 | To all questions from now on, please just answer with a single word: "True" or "False" 6 | -------------------------- 7 | Understood. 8 | -------------------------- 9 | The connected components of an undirected graph are uniquely determined 10 | -------------------------- 11 | True 12 | -------------------------- 13 | In the Relax-operation of the Floyd-Warshall algorithm, if we replace the minimum with the maximum, then the algorithm will not find the shortest, but the longest paths between pairs of vertices. 14 | -------------------------- 15 | False 16 | -------------------------- 17 | If you increase every edge weight by the same positive constant, then the solution to the single-source shortest-path problem will not change. 18 | -------------------------- 19 | False 20 | -------------------------- 21 | A connected, unweighted Graph $G=(V,E)$ has a unique minimum spanning tree. 22 | -------------------------- 23 | False -------------------------------------------------------------------------------- /GPT-4/2_c_d_h.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! I'm ready to help you with your computer science exam questions. Please go ahead and provide the questions, and I'll do my best to give short and precise answers. 4 | -------------------------- 5 | If we could find the pivot element in the QuickSort algorithm in $O(1)$ time, then we would get the best guarantee on the running time with the arithmetic meanor the the median? 6 | -------------------------- 7 | If you could find the pivot element in O(1) time and always choose the median element as the pivot, then QuickSort would be guaranteed to have the best-case running time of O(n log n), where n is the number of elements being sorted. Choosing the median element as the pivot ensures that the array is consistently divided into two equal or almost equal parts, resulting in a balanced partitioning, which leads to the optimal O(n log n) average-case running time. 8 | -------------------------- 9 | So in short, the arithmetic mean or the the median? 10 | -------------------------- 11 | In short, the median. Choosing the median element as the pivot ensures the best-case running time for QuickSort. 12 | -------------------------- 13 | Which of the following is correct? The principle of “Memoization” in dynamic programming means that 14 | 15 | a) we store the dynamic-programming recurrence in a static array so that it may be applied at each level of recursion. 16 | 17 | b) we save the solutions of smaller subproblems if we need them again. 18 | 19 | c) the dynamic-programming algorithm can remember the current location in the recursion tree. 20 | 21 | d) we will remember the recursions occurring in dynamic programming for future exams. 22 | -------------------------- 23 | b) we save the solutions of smaller subproblems if we need them again. 24 | -------------------------- 25 | Which of the following properties is not a desired property of a hash function $h(x)?$ 26 | 27 | a) $h(x)$ should be computable in $O(1)$ time. 28 | 29 | b) The values of $h(x)$ should be within the size of the hash table. 30 | 31 | c) For different $x$, $h(x)$ should also assume different values. 32 | 33 | d) If $x_1$ to $x_n$ are the elements to be inserted into the hash table, then $h(x_1)$ to $h(x_n)$ should be evenly distributed over the natural numbers. 34 | -------------------------- 35 | d) If $x_1$ to $x_n$ are the elements to be inserted into the hash table, then $h(x_1)$ to $h(x_n)$ should be evenly distributed over the natural numbers. 36 | 37 | This statement is incorrect because the hash values should be evenly distributed over the range of the hash table, not over the natural numbers. The range of the hash table is generally much smaller than the set of all natural numbers. 38 | -------------------------- 39 | Which of the following is correct? If $T(n) = 2T(n/3) + \sqrt{n}$ holds, then we also have 40 | 41 | a) $T(n) = \Theta(\sqrt{n})$ 42 | 43 | b) $T(n) = \Theta(n^2)$ 44 | 45 | c) $T(n) = \Theta(n^{\log_3(2)})$ 46 | 47 | d) $T(n) = \Theta(n^{\log(3))}$ 48 | -------------------------- 49 | c) $T(n) = \Theta(n^{\log_3(2)})$ 50 | 51 | By applying the Master Theorem, this recurrence relation can be categorized as Case 2, where the work done at the root is asymptotically larger than the work done in the recursion. Therefore, the solution is $T(n) = \Theta(n^{\log_3(2)})$. -------------------------------------------------------------------------------- /GPT-4/3a.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! I'm ready to help you with your computer science exam questions about algorithms. Please provide the questions, and I will do my best to give short and precise answers. 4 | -------------------------- 5 | Perform the DecreaseKey(0, 3) operation on the following Max-Heap (in this operation, the key of the root element is set to 3). Draw the resulting Max-Heap. 6 | \begin{tikzpicture}[level/.style={sibling distance=30mm/#1}] 7 | \node [circle,draw] (a){12} 8 | child {node [circle,draw] (b) {10} 9 | child {node [circle,draw] (c) {7}} 10 | } 11 | child {node [circle,draw] (e) {8} 12 | child {node [circle,draw] (f) {4}} 13 | child {node [circle,draw] (g) {1}} 14 | }; 15 | \end{tikzpicture} 16 | -------------------------- 17 | To perform the DecreaseKey(0, 3) operation, we first change the root value from 12 to 3, and then we need to restore the Max-Heap property by moving the decreased key down the tree. 18 | 19 | Here's the resulting Max-Heap after performing the DecreaseKey operation: 20 | 21 | markdown 22 | Copy code 23 | 10 24 | / \ 25 | 7 8 26 | / / \ 27 | 3 4 1 28 | 29 | 30 | The DecreaseKey operation involved swapping the root value (3) with its largest child (10) and then swapping the new root's child value (3) with its largest child (7) to restore the Max-Heap property. -------------------------------------------------------------------------------- /GPT-4/3b.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! I'm ready to help you with your computer science exam questions. Please provide the questions, and I'll do my best to give short and precise answers. 4 | -------------------------- 5 | Give an instance of a knapsack problem with 4 different items and weights where the greedy algorithm does not find the optimal solution. The greedy algorithm selects in each step the item with the best value-weight ratio that still fits in the knapsack. 6 | -------------------------- 7 | Consider a knapsack with a capacity of 50 units and the following 4 items: 8 | 9 | Item A: weight = 40 units, value = 60 10 | Item B: weight = 25 units, value = 45 11 | Item C: weight = 25 units, value = 45 12 | Item D: weight = 10 units, value = 20 13 | 14 | Using the greedy algorithm, we first compute the value-weight ratios: 15 | 16 | Item A: 60/40 = 1.5 17 | Item B: 45/25 = 1.8 18 | Item C: 45/25 = 1.8 19 | Item D: 20/10 = 2.0 20 | 21 | The greedy algorithm selects items in the order D, B (or C), as they have the highest value-weight ratios. The total value is 20 + 45 = 65. 22 | 23 | However, the optimal solution is to select items A and C (or A and B), which results in a total value of 60 + 45 = 105. In this instance, the greedy algorithm does not find the optimal solution. -------------------------------------------------------------------------------- /GPT-4/3c_5c_5e_6e_7c.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! I'd be glad to help you with your computer science exam questions. Please go ahead and provide the questions, and I'll do my best to give you short and precise answers. 4 | -------------------------- 5 | To all questions from now on, please just answer with a single word: "True" or "False" 6 | -------------------------- 7 | Understood. I will provide single-word answers to your questions: "True" or "False." Please proceed with your questions. 8 | -------------------------- 9 | Is the following tree a binary search tree? 10 | 11 | \vspace{0.25cm} 12 | \begin{tabular}{p{8cm}p{6cm}} 13 | \begin{minipage}{8cm} 14 | \hspace{0.5cm} 15 | \begin{tikzpicture}[level/.style={sibling distance=30mm/#1}] 16 | \node [circle,draw] (a){7} 17 | child {node [circle,draw] (b) {6} 18 | child {node [circle,draw] (c) {5}} 19 | child {node [circle,draw] (d) {8}} 20 | } 21 | child {node [circle,draw] (e) {10} 22 | child {node [circle,draw] (f) {9}} 23 | child {node [circle,draw] (g) {12}} 24 | }; 25 | \end{tikzpicture} 26 | -------------------------- 27 | False 28 | -------------------------- 29 | \item An undirected and unweighted graph is given as an adjacency matrix. Then the running time of the DFS algorithm is $O(|V|+|E|)$. 30 | -------------------------- 31 | False 32 | -------------------------- 33 | Let $G=(V,E)$ be a directed, unweighted graph. A topological sorting exists only if the graph is acyclic. 34 | -------------------------- 35 | True 36 | -------------------------- 37 | In a directed, acyclic graph, the component graph corresponds to the original graph. 38 | -------------------------- 39 | False 40 | -------------------------- 41 | If a graph does not contain a negative weight cycle, then the result of the Dijkstra algorithm is correct. 42 | -------------------------- 43 | True -------------------------------------------------------------------------------- /GPT-4/3d_f.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! I'm ready to help you with your computer science exam questions about algorithms. Please go ahead and provide the questions, and I will do my best to give you short and precise answers. 4 | -------------------------- 5 | The following questions require you to fill-in some pseudo-code. Are you ready? 6 | -------------------------- 7 | Yes, I'm ready. Please provide the questions, and I will do my best to provide pseudo-code answers. 8 | -------------------------- 9 | Let $T$ be a binary search tree with root node $x$. It is assumed that access to the children of any node can be obtained with $x.left$ and $x.right$. It holds that $x.left == NIL$ or $x.right == NIL$ if no left or right subtree exists. Complete the pseudo-code below for an algorithm that outputs the nodes of the binary search tree in sorted order from largest to smallest using $PRINT$. 10 | 11 | \begin{minipage}{8cm} 12 | \begin{algorithmic} 13 | \Function{PrintBSTReverseSorted}{Root $x$} 14 | \vspace{4.75cm} 15 | \EndFunction 16 | \end{algorithmic} 17 | \end{minipage} 18 | -------------------------- 19 | Here's the pseudo-code for the algorithm that prints the nodes of a binary search tree in reverse sorted order (largest to smallest): 20 | 21 | \begin{minipage}{8cm} 22 | \begin{algorithmic} 23 | \Function{PrintBSTReverseSorted}{Root $x$} 24 | \If{$x \neq NIL$} 25 | \State \Call{PrintBSTReverseSorted}{$x.right$} 26 | \State $PRINT(x)$ 27 | \State \Call{PrintBSTReverseSorted}{$x.left$} 28 | \EndIf 29 | \EndFunction 30 | \end{algorithmic} 31 | \end{minipage} 32 | -------------------------- 33 | The diameter of an undirected and unweighted graph corresponds to the largest distance between two nodes in a graph. More specifically, the diameter is 34 | $$\operatorname{diam}(G) = \max_{(u,v) \in V \times V}\{\operatorname{dist}(u,v)\},$$ where $\operatorname{dist}(u,v)$ corresponds to the length of the shortest path between $u$ and $v$. Provide an algorithm that calculates the diameter of a graph $G=(V,E)$. Describe your algorithm in pseudo-code. 35 | -------------------------- 36 | To calculate the diameter of an undirected and unweighted graph, you can use a modified Breadth-First Search (BFS) algorithm. Here's the pseudo-code for the algorithm: 37 | 38 | scss 39 | Copy code 40 | Function Diameter(G) 41 | max_diameter = 0 42 | for each vertex u in V(G) 43 | dist = BFS(G, u) 44 | max_u = max(dist) 45 | if max_u > max_diameter 46 | max_diameter = max_u 47 | return max_diameter 48 | 49 | Function BFS(G, start) 50 | for each vertex u in V(G) 51 | dist[u] = infinity 52 | dist[start] = 0 53 | queue = empty queue 54 | enqueue(queue, start) 55 | while not isEmpty(queue) 56 | u = dequeue(queue) 57 | for each neighbor v of u 58 | if dist[v] == infinity 59 | dist[v] = dist[u] + 1 60 | enqueue(queue, v) 61 | return dist 62 | 63 | 64 | The Diameter function iterates over all vertices in the graph, calling the BFS function for each vertex. It calculates the maximum distance from the vertex to any other vertex in the graph and keeps track of the overall maximum distance (diameter). 65 | 66 | The BFS function is a standard Breadth-First Search algorithm that takes the graph and a starting vertex as input, and it calculates the shortest distances from the starting vertex to all other vertices in the graph. It returns an array of distances, which is used by the Diameter function to determine the maximum distance for each vertex. -------------------------------------------------------------------------------- /GPT-4/4a.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! I'm ready to help you with your computer science exam questions. Please provide the questions, and I will do my best to give you short and precise answers. 4 | -------------------------- 5 | Consider hashing with linear probing and the hash function $h(x) = (7 \cdot x + 2) \mod 11.$ The corresponding hash table consists of an array of length 11. Insert the elements $2$, $3$, $7$, $15$ and $1$ in this order into the hash table. 6 | -------------------------- 7 | To insert the elements 2, 3, 7, 15, and 1 into the hash table using the hash function h(x) = (7 * x + 2) mod 11 and linear probing, we follow these steps: 8 | 9 | Compute the hash for each element. 10 | Insert the element into the hash table at the corresponding index. If there's a collision, use linear probing to find the next available slot. 11 | 12 | Here's the process for each element: 13 | 14 | Element 2: 15 | h(2) = (7 * 2 + 2) mod 11 = 16 mod 11 = 5 16 | Insert 2 at index 5. 17 | 18 | Element 3: 19 | h(3) = (7 * 3 + 2) mod 11 = 23 mod 11 = 1 20 | Insert 3 at index 1. 21 | 22 | Element 7: 23 | h(7) = (7 * 7 + 2) mod 11 = 51 mod 11 = 6 24 | Insert 7 at index 6. 25 | 26 | Element 15: 27 | h(15) = (7 * 15 + 2) mod 11 = 107 mod 11 = 9 28 | Insert 15 at index 9. 29 | 30 | Element 1: 31 | h(1) = (7 * 1 + 2) mod 11 = 9 mod 11 = 9 32 | Collision at index 9. Using linear probing, we look for the next available slot. 33 | Index 10 is empty. 34 | Insert 1 at index 10. 35 | 36 | The final hash table looks like this: 37 | Index: 0 1 2 3 4 5 6 7 8 9 10 38 | Value: [ , 3, , , , 2, 7, , ,15, 1] -------------------------------------------------------------------------------- /GPT-4/4b.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! I'm ready to help you with your computer science exam questions about algorithms. Please go ahead and provide the questions, and I will do my best to give you short, precise answers. 4 | -------------------------- 5 | Consider the following algorithm. The algorithm receives as input a connected, undirected, weighted graph $G=(V,E,c)$ with edge weight function $c: E \rightarrow \mathbb{R}_{\geq 0}$. 6 | 7 | \begin{algorithmic} 8 | \Function{AlternativeMST}{} 9 | \State $E_T = \emptyset$\; 10 | \State Sei $s$ beliebiger Knoten in $G$\; 11 | \For{$v \in V \setminus\{s\}$} 12 | \State Let $e=(v,w)$ be the edge of $v$ with minimal weight $c(e)$ that does not belong to $E_T$\; 13 | \State $E_T \leftarrow E_T \cup \{e\}$\; 14 | \EndFor 15 | \State \Return{$(V,E_T)$} 16 | \EndFunction 17 | \end{algorithmic} 18 | 19 | Is AlternativeMST correct, that is, does AlternativeMST actually calculate a Minimum Spanning Tree of $G$? Prove or disprove that AlternativeMST calculates a MST. 20 | -------------------------- 21 | The given AlternativeMST algorithm does not always calculate a Minimum Spanning Tree (MST) of G. Let's disprove it using a counterexample. 22 | 23 | Consider a connected, undirected, weighted graph G with vertices V = {A, B, C, D} and edges E = {(A, B), (A, C), (A, D), (B, C), (B, D), (C, D)} with weights given by the edge weight function c: 24 | 25 | c(A, B) = 1 26 | c(A, C) = 2 27 | c(A, D) = 3 28 | c(B, C) = 4 29 | c(B, D) = 5 30 | c(C, D) = 6 31 | 32 | Now, let's apply the AlternativeMST algorithm: 33 | 34 | Set s = A (an arbitrary vertex). 35 | For vertex B, the minimal weight edge is (A, B) with weight 1. Add this edge to E_T. 36 | For vertex C, the minimal weight edge is (A, C) with weight 2. Add this edge to E_T. 37 | For vertex D, the minimal weight edge is (A, D) with weight 3. Add this edge to E_T. 38 | 39 | The resulting tree T = (V, E_T) has edges E_T = {(A, B), (A, C), (A, D)} and total weight 1 + 2 + 3 = 6. 40 | 41 | However, the actual MST of G is: 42 | 43 | MST = {(A, B), (A, C), (B, D)} with total weight 1 + 2 + 5 = 8. 44 | 45 | Since the resulting tree from the AlternativeMST algorithm does not match the actual MST, we can conclude that the AlternativeMST algorithm does not always calculate a Minimum Spanning Tree of G. -------------------------------------------------------------------------------- /GPT-4/5_a_b.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! I'm ready to help you with your computer science exam questions. Please go ahead and provide the questions, and I'll do my best to give short and precise answers. 4 | -------------------------- 5 | Perform the DFS algorithm on the following graph, starting at node $A$. Assume that the DFS algorithm visits nodes with smaller letters first when there is a choice. Give the order in which the various nodes are visited. 6 | 7 | \begin{tikzpicture}[>=triangle 45,thick,node distance=25mm,scale=0.9,transform shape] 8 | \tikzstyle{every node}=[draw,shape=circle,minimum height=5mm,font=\small] 9 | 10 | \node (A) {A}; 11 | \node (B) [above right of=A] {B}; 12 | \node (D) [right of=B] {D}; 13 | \node (F) [below of=B] {F}; 14 | \node (E) [below of=D] {E}; 15 | \node (C) [right of=E] {C}; 16 | \node (G) [below left of=F] {G}; 17 | \node (H) [below right of=F] {H}; 18 | 19 | \path[every node/.style={auto,outer sep=-2pt,font=\small}] 20 | (A) edge node{} (B) 21 | (B) edge node{} (F) 22 | (B) edge node{} (D) 23 | (F) edge node{} (E) 24 | (D) edge node{} (F) 25 | (D) edge node{} (E) 26 | (E) edge node{} (C) 27 | (F) edge node{} (G) 28 | (F) edge node{} (H) 29 | ; 30 | \end{tikzpicture} 31 | 32 | -------------------------- 33 | To perform the DFS algorithm on the given graph starting at node A and visiting nodes with smaller letters first when there is a choice, we will follow this order: 34 | 35 | A 36 | B 37 | D 38 | E 39 | C 40 | F 41 | G 42 | H 43 | 44 | So the order in which the nodes are visited is A-B-D-E-C-F-G-H. 45 | -------------------------- 46 | Now give the discovery time and the finishing time of the different nodes. Start with the discovery time $t=1$ for node $A$. 47 | 48 | You can insert the times in the following table. 49 | 50 | \begin{center} 51 | \begin{tabular}{|c|c|c|} 52 | \hline 53 | Knoten $v$ & Discovery-Zeit($v$) & Finishing-Zeit($v$)\\ \hline \hline 54 | A & 1 & \\ \hline 55 | B & & \\ \hline 56 | C & & \\ \hline 57 | D & & \\ \hline 58 | E & & \\ \hline 59 | F & & \\ \hline 60 | G & & \\ \hline 61 | H & & \\ \hline 62 | \end{tabular} 63 | \vspace{0.25cm} 64 | \end{center} 65 | -------------------------- 66 | Here is the table with the discovery and finishing times for each node: 67 | 68 | \begin{center} 69 | \begin{tabular}{|c|c|c|} 70 | \hline 71 | Node $v$ & Discovery Time($v$) & Finishing Time($v$)\ \hline \hline 72 | A & 1 & 16\ \hline 73 | B & 2 & 15\ \hline 74 | C & 6 & 7\ \hline 75 | D & 3 & 14\ \hline 76 | E & 4 & 13\ \hline 77 | F & 8 & 12\ \hline 78 | G & 9 & 10\ \hline 79 | H & 11 & 12\ \hline 80 | \end{tabular} 81 | \vspace{0.25cm} 82 | \end{center} 83 | 84 | To clarify, the discovery time is when a node is first visited, and the finishing time is when all of its descendants have been explored and the node's exploration is completed. -------------------------------------------------------------------------------- /GPT-4/5d_2a_7d.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! I'm ready to help you with your computer science exam questions. Please go ahead and provide the questions, and I'll do my best to give short and precise answers. 4 | -------------------------- 5 | To all questions from now on, there is exactly one correct answer. 6 | -------------------------- 7 | Understood. Please provide the questions, and I will do my best to give the correct answer for each one. 8 | -------------------------- 9 | Let $G=(V,E)$ be a directed graph. Which of the following algorithmic problems can be solved with DFS? 10 | 11 | a) Finding a source node. 12 | 13 | b) Finding a sink node. 14 | 15 | c) Both problems. 16 | 17 | d) Neither of the two problems. 18 | -------------------------- 19 | c) Both problems. 20 | 21 | DFS (Depth-First Search) can be used to find both a source node (a node with no incoming edges) and a sink node (a node with no outgoing edges) in a directed graph. 22 | -------------------------- 23 | For graphs with $\Theta(|V|^2)$ many edges, the Dijkstra algorithm is best implemented using a 24 | 25 | a) Array. 26 | 27 | b) Heap. 28 | -------------------------- 29 | b) Heap. 30 | 31 | When there are Θ(|V|^2) edges, implementing Dijkstra's algorithm with a heap (such as a binary or Fibonacci heap) is more efficient, as it reduces the overall time complexity of the algorithm. 32 | -------------------------- 33 | If we could find the pivot element in the QuickSort algorithm in $O(1)$ time, then we would get the best guarantee on the running time with 34 | 35 | a) the arithmetic mean 36 | 37 | b) the median 38 | -------------------------- 39 | b) the median 40 | 41 | If we could find the median element as the pivot in O(1) time, QuickSort would have the best-case guaranteed running time. The median ensures the partition is balanced, leading to an O(n log n) time complexity. -------------------------------------------------------------------------------- /GPT-4/6_a_b_c_d.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! I'm ready to help you with your computer science exam questions about algorithms. Please go ahead and provide the questions, and I'll do my best to give short and precise answers. 4 | -------------------------- 5 | Consider the following directed Graph. 6 | 7 | \begin{center} 8 | \begin{tikzpicture}[->,>=triangle 45,thick,node distance=25mm,scale=0.9,transform shape] 9 | \tikzstyle{every node}=[draw,shape=circle,minimum height=5mm,font=\small] 10 | 11 | \node (A) {A}; 12 | \node (B) [above right of=A] {B}; 13 | \node (C) [below right of=A] {C}; 14 | \node (D) [right of=B] {D}; 15 | \node (E) [right of=C] {E}; 16 | \node (F) [right of=D] {F}; 17 | \node (G) [right of=E] {G}; 18 | \node (H) [right of=F] {H}; 19 | \node (I) [right of=G] {I}; 20 | \node (K) [right of=H] {K}; 21 | \node (J) [right of=I] {J}; 22 | 23 | \path[every node/.style={auto,outer sep=-2pt,font=\small}] 24 | (A) edge node{} (B) 25 | (A) edge node{} (C) 26 | (C) edge node{} (E) 27 | (C) edge node{} (B) 28 | (D) edge node{} (E) 29 | (E) edge node{} (C) 30 | (F) edge node{} (D) 31 | (E) edge node{} (G) 32 | (G) edge node{} (I) 33 | (I) edge node{} (H) 34 | (H) edge node{} (G) 35 | (G) edge node{} (F) 36 | (K) edge node{} (J) 37 | (H) edge node{} (J) 38 | (D) edge node{} (C); 39 | \end{tikzpicture} 40 | \end{center} 41 | 42 | How many strongly connected components does this graph have? 43 | -------------------------- 44 | In the given directed graph, the strongly connected components can be identified by looking for sets of vertices where every vertex in the set is reachable from every other vertex in the set. 45 | 46 | The strongly connected components are as follows: 47 | 48 | {A} 49 | {B} 50 | {C, E} 51 | {D} 52 | {F} 53 | {G, H, I} 54 | {J} 55 | {K} 56 | 57 | There are 8 strongly connected components in the graph. 58 | -------------------------- 59 | Draw the corresponding component graph. 60 | -------------------------- 61 | To draw the component graph, first, we collapse each strongly connected component into a single node, then remove any self-loops and duplicate edges. Here's the component graph: 62 | 63 | rust 64 | Copy code 65 | 1. A -> B 66 | 2. B -> CE 67 | 3. CE -> CE (self-loop will be removed) 68 | 4. D -> CE 69 | 5. F -> D 70 | 6. F -> GHI 71 | 7. GHI -> GHI (self-loop will be removed) 72 | 8. H -> J 73 | 9. K -> J 74 | 75 | 76 | Now, we can represent this as a simple directed graph: 77 | 78 | css 79 | Copy code 80 | A --→ B --→ CE ←-- D ←-- F --→ GHI 81 | ↑ ↓ 82 | J ←-- H ←----------- GHI 83 | ↑ 84 | K 85 | 86 | 87 | In this component graph, we have represented the strongly connected components as nodes (A, B, CE, D, F, GHI, J, K), and the directed edges indicate the connections between these components. 88 | -------------------------- 89 | Number the different nodes of the component graph according to the alphabet, i.e., $A$, $B$, $C$, etc. Then specify a topological sort of the component graph. 90 | -------------------------- 91 | First, let's assign alphabetical labels to the nodes of the component graph. Since we have 8 components, we will use the first 8 letters of the alphabet: 92 | 93 | A - A 94 | B - B 95 | C - CE 96 | D - D 97 | E - F 98 | F - GHI 99 | G - J 100 | H - K 101 | 102 | Now, let's perform a topological sort on the component graph: 103 | 104 | A 105 | B 106 | D 107 | C 108 | E 109 | F 110 | H 111 | G 112 | 113 | This topological sort represents a valid linear ordering of the component graph nodes, respecting the directed edges between them. 114 | -------------------------- 115 | Which nodes of the component graph are source nodes, which are sink nodes? 116 | -------------------------- 117 | In the component graph, source nodes are those with no incoming edges, and sink nodes are those with no outgoing edges. Let's identify them based on the component graph: 118 | 119 | css 120 | Copy code 121 | A --→ B --→ CE ←-- D ←-- F --→ GHI 122 | ↑ ↓ 123 | J ←-- H ←----------- GHI 124 | ↑ 125 | K 126 | 127 | 128 | Source nodes: A, E, H 129 | 130 | Sink nodes: G 131 | 132 | In the component graph, nodes A, E (corresponding to F in the original graph), and H (corresponding to K in the original graph) are source nodes, while node G (corresponding to J in the original graph) is a sink node. -------------------------------------------------------------------------------- /GPT-4/7_a_b.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! I'm ready to help you with your computer science exam questions. Please go ahead and provide the questions, and I'll do my best to provide short and precise answers. 4 | -------------------------- 5 | Perform the Dijkstra algorithm on the following graph to find the shortest path from A to all other nodes. Give the order in which the Dijkstra algorithm visits the nodes. 6 | 7 | \begin{tikzpicture}[>=triangle 45,thick,node distance=20mm,scale=0.9,transform shape] 8 | \tikzstyle{every node}=[draw,shape=circle,minimum height=5mm,font=\small] 9 | 10 | \node (A) {A}; 11 | \node (B) [above right of=A] {B}; 12 | \node (C) [below right of=A] {C}; 13 | \node (D) [right of=B] {D}; 14 | \node (E) [right of=C] {E}; 15 | \node (F) [below right of=D] {F}; 16 | \node (G) [above right of=B] {G}; 17 | 18 | \path[every node/.style={auto,outer sep=-2pt,font=\small}] 19 | (A) edge node{3} (B) 20 | (A) edge node{4} (C) 21 | (B) edge node{2} (G) 22 | (B) edge node{3} (C) 23 | (G) edge node{3} (D) 24 | (B) edge node{7} (D) 25 | (D) edge node{2} (E) 26 | (B) edge node{5} (E) 27 | (C) edge node{3} (E) 28 | (D) edge node{1} (F) 29 | (E) edge node{2} (F) 30 | ; 31 | \end{tikzpicture} 32 | -------------------------- 33 | To find the shortest path from node A to all other nodes using Dijkstra's algorithm, we will follow these steps: 34 | 35 | Assign a tentative distance value to all nodes: set it to zero for the initial node (A) and infinity for other nodes. 36 | Set the initial node (A) as the current node. 37 | For the current node, consider all of its neighbors and calculate their tentative distances. Compare the newly calculated tentative distance to the current assigned value and assign the new value if it is less than the current value. 38 | Mark the current node as visited. 39 | Select the unvisited node with the smallest tentative distance, and set it as the new current node, then go back to step 3. If all nodes have been visited, the algorithm has finished. 40 | 41 | Here's the step-by-step process for the given graph: 42 | 43 | Initial state: 44 | A: 0 (current) 45 | B: ∞ 46 | C: ∞ 47 | D: ∞ 48 | E: ∞ 49 | F: ∞ 50 | G: ∞ 51 | 52 | Visit A (0): 53 | A -> B: 0 + 3 = 3 54 | A -> C: 0 + 4 = 4 55 | B: 3 56 | C: 4 57 | D: ∞ 58 | E: ∞ 59 | F: ∞ 60 | G: ∞ 61 | 62 | Visit B (3): 63 | B -> G: 3 + 2 = 5 64 | B -> C: 3 + 3 = 6 65 | B -> D: 3 + 7 = 10 66 | B -> E: 3 + 5 = 8 67 | C: 4 68 | D: 10 69 | E: 8 70 | F: ∞ 71 | G: 5 72 | 73 | Visit C (4): 74 | C -> E: 4 + 3 = 7 75 | D: 10 76 | E: 7 77 | F: ∞ 78 | G: 5 79 | 80 | Visit G (5): 81 | G -> D: 5 + 3 = 8 82 | D: 8 83 | E: 7 84 | F: ∞ 85 | 86 | Visit E (7): 87 | E -> F: 7 + 2 = 9 88 | D: 8 89 | F: 9 90 | 91 | Visit D (8): 92 | D -> E: 8 + 2 = 10 (not an improvement) 93 | D -> F: 8 + 1 = 9 (not an improvement) 94 | F: 9 95 | 96 | Visit F (9): 97 | All nodes visited. 98 | 99 | The order in which Dijkstra's algorithm visits the nodes is: A, B, C, G, E, D, F. 100 | -------------------------- 101 | What is the order in the Dijkstra algorithm visits the nodes? 102 | -------------------------- 103 | Apologies for the confusion in my previous response. Here is the correct order in which Dijkstra's algorithm visits the nodes: 104 | 105 | A, B, C, G, E, D, F 106 | -------------------------- 107 | Now give the length of the shortest path from A to all other nodes. 108 | -------------------------- 109 | Here are the lengths of the shortest paths from node A to all other nodes, as determined by Dijkstra's algorithm: 110 | 111 | A to B: 3 112 | A to C: 4 113 | A to G: 5 114 | A to E: 7 115 | A to D: 8 116 | A to F: 9 -------------------------------------------------------------------------------- /GPT-4/7e.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! I'm ready to help you with your computer science exam questions about algorithms. Please go ahead and provide the questions, and I'll do my best to provide short and precise answers. 4 | -------------------------- 5 | Let $G=(V,E)$ be a directed graph. Now assume we define a weight function on the nodes, i.e., $c:V\to\mathbb{R}_{\geq 0}$. Does it still make sense to speak of the shortest path problem with a weight function on the nodes? If Yes, then give a formal definition of the problem. If No, then explain why the problem makes no sense. 6 | -------------------------- 7 | Yes, it still makes sense to speak of the shortest path problem with a weight function on the nodes. To transform the problem into a more conventional shortest path problem with edge weights, you can associate the weight of a node with its outgoing edges. 8 | 9 | Given a directed graph G=(V,E) with a weight function on the nodes c: V -> R≥0, the objective is to find the shortest path between two nodes, s (source) and t (target), such that the total weight of the path is minimized. 10 | 11 | Formally, the problem can be defined as: 12 | 13 | Input: A directed graph G=(V,E), a weight function on the nodes c: V -> R≥0, a source node s ∈ V, and a target node t ∈ V. 14 | 15 | Output: A path P from s to t in G such that the total weight of the nodes in P is minimized. 16 | 17 | To solve this problem, you can modify the graph by assigning the weight of each node v to its outgoing edges (v, u) for all u ∈ V. Let the modified edge weight function be c': E -> R≥0, defined as c'(v, u) = c(v) for all (v, u) ∈ E. Now, you can apply any shortest path algorithm, such as Dijkstra's or Bellman-Ford, to find the shortest path with respect to the modified edge weights c'. -------------------------------------------------------------------------------- /GPT-4/8.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! I'm ready to help you with your computer science exam questions. Please provide the questions, and I'll do my best to give short and precise answers. 4 | -------------------------- 5 | What is the Best-Case running time of the following algorithm in relation to the length $n$ of the input array A? 6 | 7 | \begin{algorithmic} 8 | \Function{ArrayFunction}{Array A} 9 | \For{i=1,\dots,A.length} 10 | \State A[i] = i 11 | \EndFor 12 | \State A = QuickSortWithLastElementAsPivot(A) 13 | \State \Return{A} 14 | \EndFunction 15 | \end{algorithmic} 16 | -------------------------- 17 | The best-case running time of the given algorithm can be analyzed by looking at its two main parts: 18 | 19 | The loop that initializes the array A[i] = i, which has a running time of O(n) as it iterates through the entire length of the array. 20 | The call to QuickSortWithLastElementAsPivot(A), which in the best case has a running time of O(n log n) when the chosen pivot results in a balanced partition. 21 | 22 | Therefore, the best-case running time of the given algorithm is O(n) + O(n log n), which simplifies to O(n log n) as it is the dominating term. 23 | -------------------------- 24 | What are the Best-Case and Worst-Case running time of the following algorithm in relation to the length $n$ of the input array A? 25 | 26 | \begin{algorithmic} 27 | \Function{ArrayFunction}{Array A} 28 | \For{$j=2$ to A.length} 29 | \State key $= A[j]$ 30 | \State $i= j-1$ 31 | \While{$i>0$ and A[i]>key} 32 | \State $A[i+1]= A[i]$ 33 | \State $i= i-1$ 34 | \EndWhile 35 | \State $A[i+1]=$ key 36 | \EndFor 37 | \State \Return{A} 38 | \EndFunction 39 | \end{algorithmic} 40 | -------------------------- 41 | This algorithm is an implementation of the Insertion Sort algorithm. Let's analyze the best-case and worst-case running times in relation to the length n of the input array A: 42 | 43 | Best-Case: The best-case scenario for Insertion Sort is when the input array A is already sorted in ascending order. In this case, the inner while loop will not execute any iterations, resulting in a linear running time. So the best-case running time is O(n). 44 | 45 | Worst-Case: The worst-case scenario for Insertion Sort is when the input array A is sorted in descending order. In this case, the inner while loop will execute the maximum number of iterations, resulting in a quadratic running time. So the worst-case running time is O(n^2). 46 | -------------------------- 47 | What is a Worst-Case input sequence of length $n$ for the algorithm in the previous question? 48 | -------------------------- 49 | The worst-case input sequence for the Insertion Sort algorithm (from the previous question) is when the input array is sorted in descending order. 50 | 51 | For an array A of length n, the worst-case input sequence would look like: 52 | 53 | A = [n, n-1, n-2, ..., 3, 2, 1] 54 | 55 | In this case, the algorithm will have to compare and shift elements in the array for the maximum number of iterations possible, resulting in a quadratic running time of O(n^2). 56 | -------------------------- 57 | The following program calculates a hash function. Which hash function is calculated by the program? 58 | 59 | \begin{algorithmic} 60 | \Function{HashFunction}{Integer $n$} 61 | \State $n = n+2$ 62 | \State $n = 7*n$ 63 | \While{$n\geq 153$} 64 | \State $n = n - 153$ 65 | \EndWhile 66 | \State \Return{n} 67 | \EndFunction 68 | \end{algorithmic} 69 | -------------------------- 70 | The given program calculates a hash function that maps an integer n to a value within the range [0, 152]. This is done through the following steps: 71 | 72 | Add 2 to the input value: $n = n + 2$ 73 | Multiply the result by 7: $n = 7 * n$ 74 | Perform modulo 153 operation using a while loop: $n = n \mod 153$ 75 | 76 | The specific hash function implemented here doesn't have a widely-known name, but it can be described as a combination of arithmetic operations (addition, multiplication) followed by a modulo operation. In a more compact form, the hash function can be represented as: 77 | 78 | $h(n) = (7 * (n + 2)) \mod 153$ 79 | -------------------------- 80 | What is the Worst-Case running time of the following algorithm in relation to the length $n$ of the input array A? 81 | 82 | \begin{algorithmic} 83 | \Function{MyAlgorithm}{Array $A$} 84 | \If{A.length $\leq 100$} 85 | \State \Return{$A[1]$} 86 | \EndIf 87 | \State Result = 4 $\cdot$ \texttt{MyAlgorithm}(A[1 ... A.length-1]) 88 | \State \Return{Result} 89 | \EndFunction 90 | \end{algorithmic} 91 | 92 | Hint: A[1 ... A.length-1] is the Array A without the last element. This operation takes time $O(1)$. 93 | -------------------------- 94 | The given algorithm is a recursive function that, in each call, processes an array with one less element than the previous call. To analyze its worst-case running time, we can construct a recurrence relation and solve it. 95 | 96 | Let T(n) represent the worst-case running time for an input array A of length n. The base case for this recurrence is when n ≤ 100, in which case the running time is constant, O(1). 97 | 98 | For n > 100, the algorithm calls itself recursively with an array of length n-1, and the time spent outside the recursion is also constant. Therefore, the recurrence relation can be defined as: 99 | 100 | T(n) = T(n-1) + O(1) 101 | 102 | This is a linear recurrence relation, which has a worst-case running time of O(n) for an input array A of length n. 103 | -------------------------- 104 | The diameter of an undirected and unweighted graph corresponds to the largest distance between two nodes in a graph. More specifically, the diameter is 105 | $$\operatorname{diam}(G) = \max_{(u,v) \in V \times V}\{\operatorname{dist}(u,v)\},$$ where $\operatorname{dist}(u,v)$ corresponds to the length of the shortest path between $u$ and $v$. Provide an algorithm that calculates the diameter of a graph $G=(V,E)$. Describe your algorithm in pseudo-code. Please provide a very short answer. 106 | -------------------------- 107 | To calculate the diameter of an undirected, unweighted graph G, we can use a Breadth-First Search (BFS) traversal from each node and find the largest distance encountered. Here's a short pseudo-code for the algorithm: 108 | 109 | sql 110 | Copy code 111 | Function Diameter(G): 112 | max_distance = 0 113 | 114 | For each vertex u in V: 115 | dist = BFS_Distances(G, u) 116 | longest_path = max(dist.values()) 117 | max_distance = max(max_distance, longest_path) 118 | 119 | Return max_distance 120 | 121 | Function BFS_Distances(G, start_vertex): 122 | Create a queue Q 123 | Initialize distance dictionary dist with start_vertex: 0 124 | Enqueue start_vertex into Q 125 | 126 | While Q is not empty: 127 | current_vertex = Dequeue(Q) 128 | 129 | For each neighbor v of current_vertex: 130 | If v is not in dist: 131 | Enqueue v into Q 132 | dist[v] = dist[current_vertex] + 1 133 | 134 | Return dist 135 | 136 | 137 | In this algorithm, the Diameter function calculates the diameter of the graph G by iterating through each vertex and computing the BFS_Distances. The largest distance found across all vertices is the diameter of the graph. -------------------------------------------------------------------------------- /GPT-4/9.txt: -------------------------------------------------------------------------------- 1 | Hello ChatGPT, I would like to ask you some questions from a computer science exam about algorithms. Are you ready for the questions? Please read the questions carefully and provide short, precise answers. 2 | -------------------------- 3 | Hello! I'm ready to help you with your computer science exam questions. Please go ahead and provide the questions, and I'll do my best to give you short and precise answers. 4 | -------------------------- 5 | Given an MxN matrix A of positive numbers, we can imagine this matrix as a game board of size MxN. The entry at $(i,j)$ corresponds to the costs that must be paid to enter field $(i,j)$ of the game board. We now want to move a game piece as cost-effectively as possible from the upper left field $(1,1)$ to the lower right field $(N,M)$. In each step, the game piece may only move one field down or one field to the right. So it can go from field $(i,j)$ to field $(i+1,j)$ or to field $(i,j+1)$. We are interested in the costs of the cheapest path from $(1,1)$ to $(N,M)$. In the following example, the cheapest path costs 33. 6 | 7 | \newcommand\tikznode[2]{\tikz[remember picture,baseline=(#1.base)]{\node(#1)[inner sep=0pt]{#2};}} 8 | \begin{equation*} 9 | A = \begin{pmatrix} 10 | 1 & 7 & 8 & 4 & 6 \\ 11 | 6 & 7 & 3 & 9 & 2 \\ 12 | 3 & 8 & 1 & 2 & 4 \\ 13 | 7 & 1 & 7 & 3 & 7 \\ 14 | 2 & 9 & 8 & 9 & 3 \\ 15 | \end{pmatrix},\qquad\text{A cheapest path } A= \begin{pmatrix} 16 | \tikznode{1}{1} & 7 & 8 & 4 & 6 \\ 17 | \tikznode{2}{6} & \tikznode{3}{7} & \tikznode{4}{3} & 9 & 2 \\ 18 | 3 & 8 & \tikznode{5}{1} & \tikznode{6}{2} & 4 \\ 19 | 7 & 1 & 7 & \tikznode{7}{3} & \tikznode{8}{7} \\ 20 | 2 & 9 & 8 & 9 & \tikznode{9}{3} \\ 21 | \end{pmatrix} 22 | \end{equation*} 23 | \tikz[overlay,remember picture]{\draw[-](1)--(2);\draw[-](2)--(3);\draw[-](3)--(4);\draw[-](4)--(5);\draw[-](5)--(6);\draw[-](6)--(7);\draw[-](7)--(8);\draw[-](8)--(9);} 24 | 25 | This problem can be solved with dynamic programming. Set up an appropriate recursive equation. 26 | 27 | Hint: There is a typo in this exercise, you can assume M=N. 28 | -------------------------- 29 | Given the hint that M = N, we can represent the matrix as an NxN matrix. To solve this problem using dynamic programming, we can set up a recursive equation for the minimum cost to reach any cell (i, j) in the matrix. 30 | 31 | Let's define a function minCost(i, j) that represents the minimum cost to reach cell (i, j) from cell (1, 1). Then, the recursive equation can be written as: 32 | 33 | minCost(i, j) = A[i][j] + min(minCost(i-1, j), minCost(i, j-1)) 34 | 35 | Here, A[i][j] represents the cost of entering cell (i, j). The base cases for this recursive equation are: 36 | 37 | minCost(1, 1) = A[1][1] 38 | minCost(i, 1) = A[i][1] + minCost(i-1, 1) for all i > 1 39 | minCost(1, j) = A[1][j] + minCost(1, j-1) for all j > 1 40 | 41 | The minimum cost to reach the lower right cell (N, N) will be the value of minCost(N, N). 42 | -------------------------- 43 | Now complete the following pseudo-code to solve the problem in runtime $O(MN)$. 44 | 45 | \Function{MatrixTravel}{Cost matrix $A$} 46 | \State $T=$ MxN matrix filled with zeros 47 | \For{$i = 1$ \textbf{to} $M$} 48 | \For{$j = 1$ \textbf{to} $N$} 49 | \vspace{8cm} 50 | \EndFor 51 | \EndFor 52 | \State \Return{T[M,N]} 53 | \EndFunction 54 | \end{algorithmic} 55 | -------------------------- 56 | Here's the completed pseudocode for the MatrixTravel function, which calculates the minimum cost to reach the lower right cell (M, N) using dynamic programming: 57 | 58 | \begin{algorithmic} 59 | \Function{MatrixTravel}{Cost matrix $A$} 60 | \State $T=$ MxN matrix filled with zeros 61 | \State $T[1,1] = A[1,1]$ 62 | \For{$i = 2$ \textbf{to} $M$} 63 | \State $T[i,1] = A[i,1] + T[i-1,1]$ 64 | \EndFor 65 | \For{$j = 2$ \textbf{to} $N$} 66 | \State $T[1,j] = A[1,j] + T[1,j-1]$ 67 | \EndFor 68 | \For{$i = 2$ \textbf{to} $M$} 69 | \For{$j = 2$ \textbf{to} $N$} 70 | \State $T[i,j] = A[i,j] + \min(T[i-1,j], T[i,j-1])$ 71 | \EndFor 72 | \EndFor 73 | \State \Return{T[M,N]} 74 | \EndFunction 75 | \end{algorithmic} 76 | 77 | This algorithm has a runtime complexity of $O(MN)$ because it iterates through each cell of the MxN matrix exactly once. -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | ## ChatGPT Participates in a Computer Science Exam 2 | 3 |

4 | Shapley Values 5 |

6 | 7 | This repository contains the transcript of the conversations with ChatGPT. It accompanies the paper [ChatGPT Participates in a Computer Science Exam](https://arxiv.org/abs/2303.09461) by Sebastian Bordt and Ulrike von Luxburg. 8 | 9 | The folder [GPT-3.5](https://github.com/tml-tuebingen/chatgpt-algorithm-exam/tree/main/GPT-3.5) contains the transcript from 16.2.2023 using GPT-3.5. These are the responses that were blindly graded alongside the student responses. The file [exam-gpt3.5.pdf](https://github.com/tml-tuebingen/chatgpt-algorithm-exam/blob/main/exam-gpt3.5.pdf) contains the corresponding graded exam. 10 | 11 | The folder [GPT-4](https://github.com/tml-tuebingen/chatgpt-algorithm-exam/tree/main/GPT-4) contains a transcript from 19.3.2023, replicating the original analysis with GPT-4. The file [exam-gpt4.pdf](https://github.com/tml-tuebingen/chatgpt-algorithm-exam/blob/main/exam-gpt4.pdf) contains the corresponding graded exam. 12 | 13 | 14 | In the folders, every text file contains the transcript of one conversation. The names of the text files correpond to the exercise numbers in the exam. 15 | -------------------------------------------------------------------------------- /exam-gpt3.5.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/tml-tuebingen/chatgpt-algorithm-exam/68392a9e97b5ea5ca1be0ec42eb7958f6153f42b/exam-gpt3.5.pdf -------------------------------------------------------------------------------- /exam-gpt4.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/tml-tuebingen/chatgpt-algorithm-exam/68392a9e97b5ea5ca1be0ec42eb7958f6153f42b/exam-gpt4.pdf -------------------------------------------------------------------------------- /points.png: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/tml-tuebingen/chatgpt-algorithm-exam/68392a9e97b5ea5ca1be0ec42eb7958f6153f42b/points.png --------------------------------------------------------------------------------