Make a Bit Continent Let's imagine we have a matrix of bits (which contains at least one 1):

0 1 0 1 1 0 1 0 0 1 0
0 1 0 1 0 0 1 0 1 1 0
0 0 1 0 1 1 0 1 0 1 0
1 1 0 0 1 0 0 1 1 0 1
0 0 0 1 0 1 1 0 0 1 0


We want to set some of the bits in this matrix such that it forms a contiguous blob of 1s, in which every 1 is directly or indirectly connected to every other 1 through orthogonal movement:

0 1 1 1 1 1 1 0 0 1 0
0 1 0 1 0 0 1 0 1 1 0
0 1 1 0 1 1 1 1 0 1 0
1 1 0 0 1 0 0 1 1 1 1
0 0 0 1 1 1 1 0 0 1 0


(You can see this more clearly by searching for 1 with your browser's "find" feature.)

However, we also want to minimize the number of bits that we set.

The Task

Given a matrix (or array of arrays) of bits or booleans, return the minimum number of bits that need to be set to create a contiguous continent of 1s. It should be possible to get from one set bit in the matrix to another by only traveling in an orthogonal direction to other set bits.

This is code-golf, so the shortest valid submission (measured in bytes) wins.

Test Cases

0 1 0 1 1 0 1 0 0 1 0
0 1 0 1 0 0 1 0 1 1 0
0 0 1 0 1 1 0 1 0 1 0
1 1 0 0 1 0 0 1 1 0 1
0 0 0 1 0 1 1 0 0 1 0
=&gt; 6

1 0 0 0 0 0 1 0 0
1 1 0 0 1 1 1 0 0
1 1 1 0 1 1 1 1 1
0 1 0 0 0 0 0 0 0
0 0 0 0 0 1 1 1 1
0 1 0 0 0 0 1 1 0
1 0 0 0 0 0 1 0 0
=&gt; 4

0 0 0 1 1 1 0 1 1
0 0 1 0 0 0 0 1 0
0 0 1 1 1 1 1 1 0
1 1 0 0 1 1 0 0 0
0 0 1 1 1 0 0 1 1
0 1 1 1 0 0 0 0 0
1 1 1 0 0 1 1 1 0
1 1 1 0 1 1 0 1 1
0 0 0 0 1 0 0 0 1
1 1 0 0 1 1 0 1 1
0 0 0 0 0 0 0 1 0
0 1 1 1 1 0 0 0 0
0 0 0 1 1 0 0 0 1
0 1 0 0 1 0 1 1 0
0 1 1 1 0 0 0 0 1
=&gt; 8

1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
=&gt; 0
C (gcc), 308 306 bytes

Function f receives (height, width, flattened array, pointer to ans), and returns answer by pointer.

If there's no 1 in the matrix, it will return 0.



#define v A[i]
N,M,K,R,C,T,i,*A;s(x,y){i=x*M+y;if(!(x&lt;0|y&lt;0|x&gt;=N|y&gt;=M|v^1))v=2,s(x,y+1),s(x,y-1),s(x+1,y),s(x-1,y);}g(i){if(C&lt;R){if(i^K){g(i+1);if(!v)C+=v=1,g(i+1),v=0,C--;}else{T=1;for(i=0;i&lt;K&amp;&amp;!v;i++);s(i/M,i%M);for(i=0;i&lt;K;i++)T&amp;=v^1,v=!!v;if(T)R=C;}}}f(n,m,a,b)int*a,*b;{K=R=(N=n)*(M=m),A=a;g(0);*b=R;}


Try it online!

Ungolfed:



N,M,R,C,T,i,*A; // height, width, result, recursion depth

s(x,y)
{ // depth first search: replace all 1 in the same connected component with 2
    i=x*M+y;
    if(!(x&lt;0|y&lt;0|x&gt;=N|y&gt;=M|A[i]^1)) { // check if out of boundary
        A[i]=2;
        s(x, y+1),s(x, y-1),s(x+1, y),s(x-1, y);
    }
}

g(i)
{ // enumerate all posible solutions
    if(C&lt;R) {
        if(i!=N*M) {
            g(i+1);      // nothing change for this entry
            if (!A[i]) { // set the entry to 1
                C++, A[i]=1;
                g(i+1);
                C--, A[i]=0;
            }
        }
        else {
            T=1;
            for (i=0; i&lt;N*M &amp;&amp; !A[i]; i++); // find first non-zero entry
            s(i/M, i%M);     // replace the connected component
            for (i=0; i&lt;N*M; i++) {
                T&amp;=A[i]!=1;   // check if no other components
                A[i]=!!A[i]; // change 2s back to 1
            }
            if (T) R=C;      // update answer
        }
    }
}

f(n,m,a,b)int*a,*b;{
    R=(N=n)*(M=m), A=a;
    g(0);
    *b=R;
}