Converting a for loop to recursion in Python

Question

I am new to recursion and am trying to convert a for loop to recursion.

allProducts = []

for a in range(10):
    for b in range(10):
        for c in range(10):
            for d in range(10):
                if (a*b*c*d)%6==0:
                    allProducts.append(a*b*c*d)

I am not able to convert this to a recursive program, which means I am not able to scale up. The idea is this - define a recursive program in Python that takes input A (number of for loops) and B (number which is a divisor of the product).

Any help would be really helpful.

Answer 1

You should have a look at the builtin package itertools :

for a,b,c,d in itertools.product(range(10),range(10),range(10),range(10)):
    if (a*b*c*d)%6==0:
        allProducts.append(a*b*c*d)

Answer 2

You can use itertools.product and its repeat argument:

from operator import mul
import itertools


def myprod(n, div, repeat=4):
    # i is a list of factors
    for i in itertools.product(range(n), repeat=repeat):
        # calculate product of all elements of list            
        prod = reduce(mul, i, 1)
        if prod % div == 0:
            yield prod

print list(myprod(10, 6))

Changing the repeat argument of myprod will change the number of loops and factors you are calculating.

Also, since multiplication is commutative ( a * b == b * a ) you should eliminate repetitive computations using itertools.combinations_with_replacement :

from operator import mul
import itertools


def myprod_unique(n, div, repeat=4):
    for i in itertools.combinations_with_replacement(range(n), r=repeat):
        prod = reduce(mul, i, 1)
        if prod % div == 0:
            yield prod

print list(myprod_unique(10, 6))

If you remove duplicate results from myprod using set you will find that the two results are equal:

print set(myprod_unique(10, 6)) == set(myprod(10, 6))

but you have cut down the number of operations drastically from n ** r to (n+r-1)! / r! / (n-1)! (n+r-1)! / r! / (n-1)! . For example 92,378 instead of 10,000,000,000 for n=10, r=10 .

Answer 3

The other answers don't use recursion. For completeness, here is one that does.

Recursive functions usually have two parts.

• Base case: Handled directly. Corresponds to the body of the innermost loop in your code.

• Recursive case: Involves calling the function again, corresponds to one for loop in your code.

Sometimes extra parameters need to be introduced. In this example the variables a , b , etc. that have already been set need to be passed in. The natural way to do that is using a list.

allProducts = []

def product(vals):
    res = 1
    for val in vals:
        res *= val
    return res

# n is the number of for loops remaining
# mod is the divisor
# vals is a list where the values of a, b, c, etc. will be placed
# start this function off using run(4, 6, [])
def run(n, mod, vals):
    if n == 0:
        # base case
        if product(vals) % mod == 0:
            allProducts.append(product(vals))
    else:
        # recursive case
        for i in range(10):
            # i takes the role of a, b, c, etc.
            # add i to the vals
            vals.append(i)
            # recursively iterate over remaining n-1 variables
            run(n - 1, mod, vals)
            # remove i
            vals.pop()