$4$ Square of Ones

A classical problem in image processing is to find in a binary image $A$ of size $nm$ the largest square of

ones. The following code shows the main computational bottleneck of this computation.

void square$\_$of$\_$ones$($int $n,$ int $m,$

int $(*$A$)[$n$][$m$],//$Yes$,$ this works in $C$

int $(**t)[n][m]\{//$assume $**t$ and $**A$ are separate arrays

for $($int $(**t)[i][m-1]=(**A)[i][m-1](**t)[n-1][j]=(**A)[n-1][j]k=n-2$;$k0$;$--k1=m-2$;$10$;$--1((A**)[k][1]==0)(**t)[k][l]=0p=$min$((**t)[k][1+1],(**t)[k+1][1])p=$min$(p,(**t)[k+1][1+1])(**t)[k][l]=1+$pij$(k,l)nmj=0$;$ji=0$;$i$

$(**t)[i][m-1]=(**A)[i][m-1]$;