Say I have
f(x,y)=(f(x-1,y)*f(x,y-1)+1)%2
where $f(x,y)$ can be only 0,1
we are given $f(0,i)$ and $f(j,0)$ for all $1\leq i,j\leq n$
Now I need to compute $f(x,y)$ at some $x,y$ without actually taking out the value at all $f(i,j)$ therefore with minimum computation how can I take out $f(i,j)$ .
Example :
1 1
0 1 0
1 0 1
Say
$f(0,1)=1$
$f(0,2)=1$
$f(1,0)=0$
$f(2,0)=1$
then
$f(1,1)=1$
$f(1,2)=0$
$f(2,1)=0$
$f(2,2)=1$