Vedic Divisibility

July 8, 2011

We extract into separate functions the calculation of the last digit and the number remaining after stripping the last digit:

(define (last x) (car (reverse (digits x))))

(define (but-last x) (undigits (reverse (cdr (reverse (digits x))))))

For the calculation of the osculator we need to figure out the smallest multiple of the divisor that ends with 9. That’s easy. Since the divisor must be odd and not divisible by 5, it must end in 1, 3, 7 or 9. If it ends in 1, multiply by 9. If it ends in 3, multiply by 3. If it ends in 7, multiply by 7. And if it ends in 9, do nothing. Then the calculation of the osculator is simple:

(define (osculator n) (define (f x) (+ (but-last (* n x)) 1)) (let ((ds (digits n))) (case (car (reverse ds)) ((1) (f 9)) ((3) (f 3)) ((7) (f 7)) ((9) (f 1)) (else (error 'osculator "unsupported")))))

The divisibility test is recursive, reducing the size of the dividend n at each step:

(define (divisible? n d) (let ((osc (osculator d))) (let loop ((n n)) (let ((next (+ (but-last n) (* last n) osc)))) (if (< next n) (loop next) (zero? (modulo n d)))))))

Here are some examples:

> (divisible? 13174584 23) #t > (divisible? 175121 37) #t > (divisible? 134567 29) #f

We used digits and undigits from the Standard Prelude. You can run the program at http://programmingpraxis.codepad.org/qZrg7y54.

Posted by programmingpraxis

Filed in Exercises

5 Comments »

5 Responses to “Vedic Divisibility”

arturasl said

July 11, 2011 at 12:49 PM

Solution in Python:

def divides(dividend, divisor):
    endings = {1 : 9, 3 : 3, 7 : 7, 9 : 1}

    if divisor % 10 not in endings:
        raise Exception('divisor should end in 1, 3, 7 or 9')

    osculator = ((divisor * endings[divisor % 10]) // 10) + 1

    while True:
        prev_dividend = dividend
        dividend = dividend // 10 + (dividend % 10) * osculator

        if prev_dividend <= dividend:
            dividend = prev_dividend
            break

    return (dividend % divisor) == 0

print(divides(13174584, 23))

WillJ said

July 12, 2011 at 6:13 PM

#include <iostream>

using namespace std;

int findOsculator( int num )
{
    int temp, i = 1;
    do
    {
        temp = num * i;
        i++;
    }
    while( temp % 10 != 9 );
    temp = temp / 10;
    return temp + 1;
}

bool divisible(int dividend, int divisor)
{
    int remainder,prev_dividend, osc;
    osc = findOsculator( divisor );
    do
    {
        prev_dividend = dividend;
        remainder = dividend % 10;
        dividend = (dividend / 10) + (osc * remainder);
    }
    while( dividend < prev_dividend );
    return !( dividend % divisor );
}

int main()
{
    cout << divisible(13174584,23);
}

Graham said

July 13, 2011 at 12:31 AM

Python:

def is_divisible(n, d):
    osc = 1 + (d * {1: 9, 3: 3, 7: 7, 9: 1}[d % 10]) // 10
    old, new = float('inf'), n
    while new < old:
        a, b = divmod(new, 10)
        old, new = new, a + b * osc
    return new % d == 0

if __name__ == "__main__":
    for (n, d) in [(13174584, 23), (175121, 37), (134567, 29)]:
        print "Is %d divisible by %d?%7s" % (n, d, "Yes." if is_divisible(n, d)
                                             else "No.")

And since Remco hasn’t jumped on it yet, here’s my attempt in Haskell:

import Data.Map

osculator :: (Integral a) => a -> a
osculator d = f e where
          f x = 1 + d * x `div` 10
          e = fromList [(1,9), (3,3), (7,7), (9,1)] ! (d `mod` 10)

isDivisible :: (Integral a) => a -> a -> Bool
isDivisible n d = aux old new where
            old = n + 1
            new = n
            osc = osculator d
            aux x y | x <= y            = y `mod` d == 0
                    | otherwise         = aux y (a + b * osc) where (a,b) = divMod y 10

main :: IO ()
main = mapM_ (\(n,d) -> print $ "Is " ++ show n ++ " divisible by " ++ show d ++
     "? " ++ show (isDivisible n d)) [(13174584, 23), (175121, 37), (134567, 29)]

Mike said

July 14, 2011 at 12:02 AM

Just to be different, how about an F# version.


exception BadValue of string

let osculator n = 
    match (n / 10, n % 10) with
    | d,1 -> d*9 + 1
    | d,3 -> d*3 + 1
    | d,7 -> d*7 + 1
    | d,9 -> d   + 1
    | _,_ -> raise <| BadValue("divisor cannot be divisible by 2 or 5")

let isdivisible dividend divisor =
    let osc = osculator divisor
    let rec loop old_dividend =
        let new_dividend = old_dividend / 10 + old_dividend % 10 * osc
        if new_dividend >= old_dividend then
            old_dividend % divisor = 0
        else
            loop new_dividend
    loop dividend

Mike said
July 14, 2011 at 12:24 AM
Line 07 above should be:
```
    | d,7 -> d*7 + 5   // 7*7 = 49, so carry the 4
```

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