regarding behaviour of modulo operation?

Question

I wrote the following program in python for the following codechef question http://www.codechef.com/problems/MOVES/

import sys

tokenizedInput = sys.stdin.read().split()
mod=1000000007
arr=[1]*5001
for i in range(1,5001):
    arr[i]=(arr[i-1]*i)%mod

def combo(r,n,mod):
    q=arr[n]
    print q
    r=(arr[r]*arr[n-r])
    print r
    return ((q/r)%mod)

elm=0

for i in range (0,5001):
    n=int(tokenizedInput[elm])
    elm=elm+1
    k=int(tokenizedInput[elm])
    elm=elm+1
    if(n==0 and k==0):
        break
    out=0
    if(((k-1)/2)!=(k/2)):
      out=(2*combo((k-1)/2,n-2,mod)*combo(k/2,n-2,mod))%mod
    else:
      out=(2*combo(k/2,n-2,mod)**2)%mod
    print out

but my modulo function is not working correctly for example for values n=498 and r=2 the answer returned by combo() is 0 because q=243293343 and r=1428355228 how to perform my modulo operation in arr[] to rectify this error ?

score 1 · Answer 1 · answered Apr 20 '14 at 12:28

The above power function calculates a^b in O( log(b) ) by using what is called as exponentiation by squaring. The idea is very simple:

       (a^2)^(b/2)          if b is even and b > 0
a^b =  a*(a^2)^((b-1)/2)    if b is odd
        1                  if b = 0

This idea can be implemented very easily as shown below:

/* This function calculates (a^b)%c */
int modulo(int a,int b,int c)
{
    long long x=1,y=a; 
    while(b > 0)
    {
        if(b%2 == 1)
        {
            x=(x*y)%c;
        }
        y = (y*y)%c; // squaring the base
        b /= 2;
    }
    return x%c;
}

The above power function is just a recursive way of doing this. and as you asked about this

but it will be of great help if anyone explains the mathematics behind using return(base*Power(base,expo-1)%mod)

this is the same as checking if expo is odd then multiplying base with base^(expo-1) so that the new expo i.e. (expo-1) will become even and repeated squaring can be done

For more info refer:

Topcoder Tutorial

wiki: expo by squaring

score 0 · Answer 2 · edited Jun 20 '20 at 09:12

Got the solution,Answering my own question , but search for an optimized version is encouraged the error was

return ((q/r)%mod)

is wrong for mod being 1000000007 ie a prime number , it has to be written as

r=(arr[r]*arr[n-r])%mod

return (q*Power(r,mod-2))%mod

where Power function is

def Power(base,expo):
if(expo==0):
  return 1
else:
    if(expo&1):
      return(base*Power(base,expo-1)%mod)
    else:
      root=Power(base,expo>>1)
      return(root*root%mod)

but it will be of great help if anyone explains the mathematics behind using return(base*Power(base,expo-1)%mod)

score 0 · Accepted Answer · answered Feb 04 '14 at 16:25

When we divider (a/b)%MOD , then we do something like this .

(a/b)%MOD

(a*inverse(b))%MOD // You have to find the inverse of b. To find the inverse of b use Fermat Theorem.

Note never divide a/b and then take MOD , first find the inverse of b and then do a*b and then MOD it .

regarding behaviour of modulo operation?

3 Answers3