In this puzzle, we have a rectangular board of cells, with some cells having a coin (the red disks in the plot). A robot is initially in the lower-left cell of the board. The task of the robot is to pick up as many coins as possible and carry them to the upper-right cell. However, at each step the robot can only move either one cell right or one cell up. Each time the robot visits a cell with a coin, it picks up that coin. The blue line shows an optimal tour, that is, a tour that allows the robot to pick up the maximum number of coins. To solve the puzzle by yourself, hide the optimal path and the maximum number of coins that can be picked up.
This is puzzle number 62 in [1]. As in the book, the optimal path is found here by using dynamic programming. Let the centers of the cells be at points  ,  ,  . The robot starts at  and ends at  . Let  be 1 if there is a coin at  and 0 otherwise. Let  be the largest number of coins the robot can pick up and bring to  . The task is to maximize  . The robot can reach the point  either from below—that is, from  —or from the left,  . The largest number of coins that can be brought to these points are  and  , respectively. Thus, the largest number of coins the robot can bring to  is the maximum of these two numbers plus  . So,  ,  ,  ,  . In addition,  ,  ,  , and  ,  . After we have coded these formulas, we simply ask for the value of  and we know the maximum number of coins that can be picked up! Mathematica does all the recursive calculations for us. To get the optimal path, we also write down from where it is optimal to come to each  . Note that often the optimal path is not unique. For more about dynamic programming with Mathematica, see Section 23.5.2 in [2]. [1] A. Levitin and M. Levitin, Algorithmic Puzzles, New York: Oxford University Press, 2011. [2] H. Ruskeepää, Mathematica Navigator: Mathematics, Statistics, and Graphics, 3rd ed., San Diego: Elsevier Academic Press, 2009.
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