# How to find number of recursive call from result value?

I have following recursive equation

    y(0) = 0

x(0) = 0

y(n+1) = If((x(n)<50) then (y(n)+1) else (y(n)-1))

x(n+1) = If((If((If((x(n)<50) then (y(n)+1) else (y(n)-1))<0) then 1
else 0)==0) then (x(n)+1) else x(n))


Where function x(),y() return integer and n natual number.

I know x(N)==100 is true. Want to find the value of N. In the paper, we can try to find x(N-1),x(N-2)... till we are able to find x(0)=0. Is there any other way to do that.Please suggest

• So I guess that the semantics of If(X,Y,Z) means "If 'X', then 'Y' else 'Z', right ? – Zubzub Oct 11 '17 at 9:49
• yes I have updated in question – Tom Oct 11 '17 at 10:04