Chessboard game...but a different one [closed] - c

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https://www.hackerrank.com/challenges/chessboard-game-again-1
I have tried the above question in the following manner but the answers are evaluated as wrong.(I am not asking for a solution, but I am asking for the defects in the approach);
my code (please ignore the c99 errors)
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int numofmov = 0;
int issafe(int a,int b){
if(a>=0 && a<15 && b>=0 && b<15)
return 1;
return 0;
}
void move(int board[][15]){
for(int i=0;i<15;i++){
for(int j=0;j<15;j++){
if(board[i][j]>0){
board[i][j]--;
if(issafe(board,i-2,j+1)==1) {
numofmov++;
board[i-2][j+1]++;
}
if(issafe(board,i-2,j-1)==1) {
numofmov++;
board[i-2][j-1]++;
}
if(issafe(board,i+1,j-2)==1) {
numofmov++;
board[i+1][j-2]++;
}
if(issafe(board,i-1,j-2)==1) {
numofmov++;
board[i-1][j-2]++;
}
}
}
}
}
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int t;
scanf("%d",&t);
while(t--){
int k;
scanf("%d",&k);
int board[15][15];
for(int j=0;j<15;j++){
for(int h=0;h<15;h++){
board[j][h]=0;
}
}
for(int i=0;i<k;i++){
int x,y;
scanf("%d %d",&x,&y);
board[x-1][y-1]++;
}
int bro=0,mov=numofmov;
while(bro==0){
move(board);
if(numofmov==mov){
bro++;
printf("Second\n");
break;
}
mov = numofmov;
move(board);
if(numofmov==mov){
bro++;
printf("First\n");
break;
}
mov = numofmov;
}
}
return 0;
}
My approach is to go on making all the moves possible for all the coins until we come to a point when no moves are possible. But this is giving wrong answers in some test cases.

Answer 1

You are asking what is wrong with this approach ?
"My approach is to go on making all the moves possible for all the
coins until we come to a point when no moves are possible. But this is
giving wrong answers in some test cases."
I didn't read your code, but I can say that the main issue is your approach itself. You are thinking of this problem as a brute-force (make all possible move paths, and see who is winning). The number of possible moves can be arbitrarily large, checking is moves lead to a win is infinitely slow. In reality it is either dynamic-programming, or even more relevant game-theory problem.
Think about it this way. Does the starting positions uniquely identifies the winner of this game? What if I change the initial position of a single coin, will the winner change too?
Best way to approach this kind of problems is to simplify it. Assume that there is only a single board with a single coin, positioned at (x,y). Now notice, that after each move of a coin from position (x,y) to position (a,b), the following is true a+b<x+y. So if (x,y) is one of (1,1),(1,2),(2,1),(2,2), player making the move already lost. So my goal is to make sure my opponent will be making a move from already lost position, and if I can do it, I am in winning position. If u follow the same logic, you will realise that this approach will uniquely identify if position is winning or losing. So for any position we can answer if it is winning or losing simply building the grid of answers by going backwards from (1,1) to (15,15).
Now what would u do if the number of boards is more than one? You need to dig into the game theory, in particular Grundy numbers and how they relate to Nim games. I would suggest you to check the following links for further information:
https://en.wikipedia.org/wiki/Nim
https://en.wikipedia.org/wiki/Nimber
https://www.topcoder.com/community/data-science/data-science-tutorials/algorithm-games/