Backtracking - Filling a grid with coins - c++

Question

I was trying to do this question i came across while looking up interview questions. We are asked the number of ways of placing r coins on a n*m grid such that each row and col contain at least one coin.
I thought of a backtracking solution, processing each cell in the grid in a row major order, I have set up my recursion in this way. Seems my approach is faulty because it outputs 0 every time. Could someone please help me find the error in my approach. ? Thanks.
constraints. n , m < 200 and r < n*m;
Here is the code i came up with.
#include<cstdio>
#define N 201
int n, m , r;
int used[N][N];
int grid[N][N] ; // 1 is coin is placed . 0 otherwise. // -1 undecided.
bool isOk()
{
int rows[N];
int cols[N];
for(int i = 0 ; i < n ; i++) rows[i] = 0;
for(int i = 0 ; i < m ; i++) cols[i] = 0;
int sum = 0;
for(int i = 0 ; i < n ; i++)for(int j = 0; j < m ; j++)
{
if(grid[i][j]==1)
{
rows[i]++;
cols[j]++;
sum++;
}
}
for(int i = 0 ; i < n ; i++)
{
if(rows[i]==0) return false;
}
for(int j = 0 ; j < n ; j++)
{
if(cols[j]==0) return false;
}
if(sum==r) return true;
else return false;
}
int calc_ways(int row , int col, int coins)
{
if(row >= n) return 0;
if(col >= m) return 0;
if(coins > r) return 0;
if(coins == r)
{
bool res = isOk();
if(res) return 1;
else 0;
}
if(row == n - 1 and col== m- 1)
{
bool res = isOk();
if(res) return 1;
else return 0;
}
int nrow, ncol;
if(col + 1 >= m)
{
nrow = row + 1;
ncol = 0;
}
else
{
nrow = row;
ncol = col + 1;
}
if(used[row][col]) return calc_ways(nrow, ncol, coins);
int ans = 0;
used[row][col] = 1;
grid[row][col] = 0;
ans += calc_ways(nrow , ncol , coins);
grid[row][col] = 1;
ans += calc_ways(nrow , ncol , coins + 1);
return ans;
}
int main()
{
int t;
scanf("%d" , &t);
while(t--)
{
scanf("%d %d %d" , &n , &m , &r);
for(int i = 0 ; i <= n ; i++)
{
for(int j = 0; j <= m ; j++)
{
used[i][j] = 0;
grid[i][j] = -1;
}
}
printf("%d\n" , calc_ways(0 , 0 , 0 ));
}
return 0;
}

Answer 1

You barely need a program to solve this at all.
Without loss of generality, let m <= n.
To begin with, we must have n <= r, otherwise no solution is possible.
Then, we subdivide the problem into a square of size m x m, on to which we will place m coins along the major diagonal, and a remainder, on to which we will place n - m coins so as to fulfil the remaining condition.
There is one way to place the coins along the major diagonal of the square.
There are m^(n - m) possibilities for the remainder.
We can permute the total so far in n! ways, although some of those will be duplicates (how many is left as an exercise for the student).
Furthermore, there are r - n coins left to place and (m - 1)n places left to put them.
Putting these all together we have an upper bound of
1 x m^(n - m) x n! x C((m - 1)n, r - n)
solutions to the problem. Divide this number by the number of duplicate permutations and you're done.

Answer 2

Problem 1
The code will start by placing a coin on each square and marking each square as used.
It will then test the final position and decide that the final position does not meet the goal of r coins.
Next it will start backtracking, but will never actually try another choice because used[row][col] is set to 1 and this shortcircuits the code to place coins.
In other words, one problem is that entries in "used" are set, but never cleared during the recursion.
Problem 2
Another problem with the code is that if n,m are of size 200, then it will never complete.
The issue is that this backtracking code has complexity O(2^(n*m)) as it will try all possible combinations of placing coins (many universe lifetimes for n=m=200...).
I would recommend you look at a different approach. For example, you might want to consider dynamic programming to compute how many ways there are of placing "k" coins on the remaining "a" columns of the board such that we make sure that we place coins on the "b" rows of the board that currently have no coins.

Answer 3

It can be treated as total ways in which d grid can b filled with r coins -(total ways leaving a single row nd filling in d rest -total ways leaving a single column nd filling in d rest- total ways leaving a row nd column together nd filling d rest) which implies
p(n*m ,r) -( (p((n-1)*m , r) * c(n,1)) +(p((m-1)*n , r) * c(m,1))+(p((n-1)*(m-1) , r) * c(n,1)*c(m,1)) )
I just think so but not sure of it!