CODEWITHSUNDEEP
cmathoverflowmodular-arithmetic
Bhavesh Munot
Bhavesh Munot

Reputation: 675

How can I avoid overflow in modular multiplication?

I know that,

(a*b)%m = ((a%m)*(b%m))%m

But there is a possibility of overflow. For simplicity lets assume size of integer is 2 bits. If a = 2 (i.e. 102) and b = 2 (i.e. 102), m = 3 (i.e. 112), then a%m and b%m turn out to be 2 and after multiplication, answer is 4 (i.e. 100) which does not fit in integer size. Final answer will be 0 if 2-lsb's are considered from 4. But actual answer is 1.

What should I do to avoid this situation?

Upvotes: 7

Views: 2836

Answers (3)

tboma t
tboma t

Reputation: 1

/// (a * n) mod m
fn mod_mul(mut a: i32, mut n: i32, m: i32) -> i32 {
    while n % 2 == 0 {
        a = (a + a) % m;
        n /= 2;
    }
    if n == 1 {
        a
    } else {
        mod_mul_acc(a, (a + a) % m, n / 2, m)
    }
}

fn mod_mul_acc(mut r: i32, mut a: i32, mut n: i32, m: i32) -> i32 {
    loop {
        if n % 2 == 1 {
            r = (r + a) % m;
            if n == 1 {
                return r;
            }
        }
        n /= 2;
        a = (a + a) % m;
    }
}

Upvotes: -1

user3629249
user3629249

Reputation: 16540

many small processor C implementations can directly check the result of a math operation for overflow/underflow.

Another way is to use a receiving field that is twice the length of the underlying int size I.E. for an int size of 2, use a result field of 4 bytes. (perhaps by a long long int) or transfer both numbers to double fields and multiply them then convert back to int (however, some precision in the result (I.E. the least significant digit) may be inaccurate.

Another way is to use an appropriate function from the math.h library.

Another way is to use long multiplication using arrays: this was copied from http://www.cquestions.com/2010/08/multiplication-of-large-numbers-in-c.html

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#include<string.h>
#define MAX 10000

char * multiply(char [],char[]);
int main(){
    char a[MAX];
    char b[MAX];
    char *c;
    int la,lb;
    int i;
    printf("Enter the first number : ");
    scanf("%s",a);
    printf("Enter the second number : ");
    scanf("%s",b);
    printf("Multiplication of two numbers : ");
    c = multiply(a,b);
    printf("%s",c);
    return 0;
}

char * multiply(char a[],char b[]){
    static char mul[MAX];
    char c[MAX];
    char temp[MAX];
    int la,lb;
    int i,j,k=0,x=0,y;
    long int r=0;
    long sum = 0;
    la=strlen(a)-1;
    lb=strlen(b)-1;

    for(i=0;i<=la;i++){
            a[i] = a[i] - 48;
    }

    for(i=0;i<=lb;i++){
            b[i] = b[i] - 48;
    }

    for(i=lb;i>=0;i--){
        r=0;
        for(j=la;j>=0;j--){
            temp[k++] = (b[i]*a[j] + r)%10;
            r = (b[i]*a[j]+r)/10;
        }
        temp[k++] = r;
        x++;
        for(y = 0;y<x;y++){
            temp[k++] = 0;
        }
   }

   k=0;
   r=0;
   for(i=0;i<la+lb+2;i++){
        sum =0;
        y=0;
        for(j=1;j<=lb+1;j++){
            if(i <= la+j){
                sum = sum + temp[y+i];
            }
            y += j + la + 1;
        }
        c[k++] = (sum+r) %10;
        r = (sum+r)/10;
   }
   c[k] = r;
   j=0;
   for(i=k-1;i>=0;i--){
      mul[j++]=c[i] + 48;
   }
   mul[j]='\0';
   return mul;

}

Sample output of above code:

Enter the first number: 55555555

Enter the second number: 3333333333

Multiplication of two numbers:

185185183314814815

Logic for multiplication of large numbers

As we know in c there are not any such data types which can store a very large numbers. For example we want to solve the expression:

55555555 * 3333333333

Result of above expression is very big number which beyond the range of even long int or long double. Then question is how to store such a big numbers in c?

Solution is very simple i.e. using array. Above program has used same logic that is we are using as usual logic to multiply two numbers except instead of storing the data in the normal variables we are storing into the array.

Upvotes: 2

R.. GitHub STOP HELPING ICE
R.. GitHub STOP HELPING ICE

Reputation: 215173

If m-1 squared doesn't fit in your integer type, you need to perform long multiplication. For your two-bit example, that means breaking your two-bit numbers into pairs of one-bit numbers (a high bit and a low bit) and multiplying all four pairs (high by high, high by low, low by high, low by low) individually. You can then get the result mod m for each pair (noting the actual places they represent, i.e. fours, twos, or ones) and add the results mod m.

Upvotes: 5

Related Questions