To show the number of possible seams grows exponential.

Consider an array A for the compressing the picture so it requires to remove the pixels in the two adjacent rows in the same or adjacent column and the pixels are removed forms a seam from the top row to the bottom row.

Suppose n>1 for the seam algorithm the choice of pixel at any row have at least 2 choices of the pixels to the next row added to the seam.

The total area enclosed by the seams is order of m and it has two choices for every area of m so the total probability of the total number of seams can be varies from 2^m to 2^m -1.

Hence, the total number of possibilities of seams is bounded by O(2m) so the grow of number of seam will be exponentially.

To give an algorithm that finds a seam with lowest disruption measure.

The algorithm to find a seam with the lowest disruption is given below:

Seam(A)

Initialize tables D[1.....m,1.....n] of zeros and S[1....m,1....n] of empty lists.

For i=1 to n do

  S[1,i]=(1,i).D[1,i]=d1i.

End for.

for i=2 to m do

for j=1 to n do

If j==1 then

If D[i,j]<D[i−1,j+1] then

  D[i,j]=D[i−1,j]+d.S[i,j]=S[i−1,j].insert(i,j).

End if.

Else

  D[i,j]=D[i−1,j+1]+d.S[i,j]=S[i−1,j+1].insert(i,j).

End if.

Else if( j==n ) then

If() then

  D[i,j]=D[i−1,j−1]+d.S[i,j]=S[i−1,j−1].insert(i,j).

End if.

End if.

  x=MIN(D[i−1,j−1],D[i−1,j],D[i−1,j+1])

  D[i,j]=D[i−1,j+x].S[i,j]=S[i−1,j+x].insert(i,j).

End for.

End for.

  q=1.

For j=1 to n do

If D[m,j]<D[m,q] then

  q=j .

End if.

End for.

Print the list S[m,q] .

End.

The algorithm compressing the picture so it requires to removes the pixels in the two adjacent rows in the same or adjacent column and the pixels are removed forms a seam from the top row to the bottom row.

This algorithm takes the array A as input and the cases to insert pixel according to the defined condition and return the list of seam as S[m,q] .