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Dingyu is playing a game defined on an n X n board. Each cell (i, j) of the board (1 < i, j < positive integer. He starts from (1, 1). In each iteration, he can move down (increasing i by 1) long as he stays in the board. (For example, when he is at (1, n) and n > 2, he may only go to a move from cell C to cell D is |value of cell C – value of cell D|. The game ends when he is the sum of the rewards for each move he makes. - 1 For example, if n = 2 and A = 3 the answer is 4 since he can visit (1, 1) → (1,2)

Dingyu is playing a game defined on an n X n board. Each cell (i, j) of the board (1 < i, j < positive integer. He starts from (1, 1). In each iteration, he can move down (increasing i by 1) long as he stays in the board. (For example, when he is at (1, n) and n > 2, he may only go to a move from cell C to cell D is |value of cell C – value of cell D|. The game ends when he is the sum of the rewards for each move he makes. - 1 For example, if n = 2 and A = 3 the answer is 4 since he can visit (1, 1) → (1,2)

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Dingyu is playing a game defined on an \( n \times n \) board. Each cell \( (i, j) \) of the board \( (1 \le i, j \le n) \) has a value \( A_{i,j} \), which is a positive integer. He starts from \( (1, 1) \). In each iteration, he can move down (increasing \( i \) by 1) or right (increasing \( j \) by 1) as long as he stays in the board. (For example, when he is at \( (1, n) \) and \( n \ge 2 \), he may only go to \( (2, n) \).) The reward he earns for a move from cell \( C \) to cell \( D \) is \( | \text{value of cell } C - \text{value of cell } D | \). The game ends when he reaches \( (n, n) \). The total reward is the sum of the rewards for each move he makes. For example, if \( n = 2 \) and \( A = \begin{pmatrix} 1 & 3 \\ 2 & 1 \end{pmatrix} \), the answer is 4 since he can visit \( (1, 1) \rightarrow (1, 2) \rightarrow (2, 2) \), and no other solution will get a higher reward. **A.** Write a recurrence relation to express the maximum possible reward Dingyu can achieve in traveling from cell \( (1, 1) \) to cell \( (n, n) \). Be sure to include any necessary base cases. **B.** State the asymptotic (big-O) running time, as a function of \( n \), of a bottom-up dynamic programming algorithm based on your answer from the previous part. Briefly justify your answer. (You do not need to write down the algorithm itself.)
Transcribed Image Text:Dingyu is playing a game defined on an \( n \times n \) board. Each cell \( (i, j) \) of the board \( (1 \le i, j \le n) \) has a value \( A_{i,j} \), which is a positive integer. He starts from \( (1, 1) \). In each iteration, he can move down (increasing \( i \) by 1) or right (increasing \( j \) by 1) as long as he stays in the board. (For example, when he is at \( (1, n) \) and \( n \ge 2 \), he may only go to \( (2, n) \).) The reward he earns for a move from cell \( C \) to cell \( D \) is \( | \text{value of cell } C - \text{value of cell } D | \). The game ends when he reaches \( (n, n) \). The total reward is the sum of the rewards for each move he makes. For example, if \( n = 2 \) and \( A = \begin{pmatrix} 1 & 3 \\ 2 & 1 \end{pmatrix} \), the answer is 4 since he can visit \( (1, 1) \rightarrow (1, 2) \rightarrow (2, 2) \), and no other solution will get a higher reward. **A.** Write a recurrence relation to express the maximum possible reward Dingyu can achieve in traveling from cell \( (1, 1) \) to cell \( (n, n) \). Be sure to include any necessary base cases. **B.** State the asymptotic (big-O) running time, as a function of \( n \), of a bottom-up dynamic programming algorithm based on your answer from the previous part. Briefly justify your answer. (You do not need to write down the algorithm itself.)
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