Given a number n, write a function that returns true if n is divisible by 9, else false. The most simple way to check for n’s divisibility by 9 is to do n%9.
Another method is to sum the digits of n. If sum of digits is multiple of 9, then n is multiple of 9.
The above methods are not bitwise operators based methods and require use of % and /.
The bitwise operators are generally faster than modulo and division operators. Following is a bitwise operator based method to check divisibility by 9.
Another method is to sum the digits of n. If sum of digits is multiple of 9, then n is multiple of 9.
The above methods are not bitwise operators based methods and require use of % and /.
The bitwise operators are generally faster than modulo and division operators. Following is a bitwise operator based method to check divisibility by 9.
#include<iostream> using namespace std; // Bitwise operator based function to check divisibility by 9 bool isDivBy9( int n) { // Base cases if (n == 0 || n == 9) return true ; if (n < 9) return false ; // If n is greater than 9, then recur for [floor(n/9) - n%8] return isDivBy9(( int )(n>>3) - ( int )(n&7)); } // Driver program to test above function int main() { // Let us print all multiples of 9 from 0 to 100 // using above method for ( int i = 0; i < 100; i++) if (isDivBy9(i)) cout << i << " " ; return 0; } |
Output:
0 9 18 27 36 45 54 63 72 81 90 99
How does this work?
n/9 can be written in terms of n/8 using the following simple formula.
n/9 can be written in terms of n/8 using the following simple formula.
n/9 = n/8 - n/72
Since we need to use bitwise operators, we get the value of floor(n/8) using n>>3 and get value ofn%8 using n&7. We need to write above expression in terms of floor(n/8) and n%8.
n/8 is equal to “floor(n/8) + (n%8)/8″. Let us write the above expression in terms of floor(n/8) andn%8
n/8 is equal to “floor(n/8) + (n%8)/8″. Let us write the above expression in terms of floor(n/8) andn%8
n/9 = floor(n/8) + (n%8)/8 - [floor(n/8) + (n%8)/8]/9 n/9 = floor(n/8) - [floor(n/8) - 9(n%8)/8 + (n%8)/8]/9 n/9 = floor(n/8) - [floor(n/8) - n%8]/9
From above equation, n is a multiple of 9 only if the expression floor(n/8) – [floor(n/8) - n%8]/9 is an integer. This expression can only be an integer if the sub-expression [floor(n/8) - n%8]/9 is an integer. The subexpression can only be an integer if [floor(n/8) - n%8] is a multiple of 9. So the problem reduces to a smaller value which can be written in terms of bitwise operators.
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