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POFinding number of ways a piece on a chess board can move with number and type of moves given
 

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  1. POFinding number of ways a piece on a chess board can move with number and type of moves given
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    <p>This is the dynamic programming problem I encountered on a site. I solved the problem with the algorithm mentioned below. Though the answers I got were right, the evaluation shows that the Time limit exceeds.</p> <p>The problem is, on an N*N board, a piece is at a location (xp,yp).It can move to (xd,yd) if and only if abs(xd-xp)+abs(yd-yp) &lt;= S. The piece has to make M moves. The problem gives N,S, M and (xp,yp) and I have to find how many ways the piece can make M moves on this board. </p> <p>My solution was as follows. </p> <pre><code>Matrix m0 is initialized with all 0s except with 1 in the location (xp,yp) Make a N*N matrix m1 for one move from an initialized matrix m0 Make a N*N matrix m2 for two moves from matrix m1 ... carry on till m moves and the sum of elements of the final matrix mm gives me the number of ways the piece can make M moves on the N*N board with given distance constraint S. </code></pre> <p>My pseudocode for calculating a matrix is as follows</p> <pre><code>for each element (e) in matrix m, consider the elements (es) in S*S (S is the distance constraint given in the problem) box surrounding the elements. If (abs(es.x-e.x) + abs(es.y-e.y)) &lt;= S, then m[e.x][e.y] = m[e.x][e.y] + m[es.x][es.y]. (adding the # of ways that es can be reached with number of ways that e can be reached so that position of e in the matrix contains # of ways that e can be reached from es) In the S*S box, I considered all elements except the current element e. </code></pre> <p>As per my understanding, the above solution is (N^2)*(S^2) and runs slow though it gives the correct answer. Please suggest an idea of implementing this in N^2 time. Since it involves a 2D board, I presume it can't be done in O(N) time, but I may be wrong.</p>
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    1. CONote that the board is symmetric, so you only really need to work with 1/8th of it (not sure if this will result in a sufficient time reduction with your current algorithm though).
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