In python The following iterative sequence is defined for the set of positive in

Question

In python

The following iterative sequence is defined for the set of positive integers:

n ? n/2 (n is even)
n ? 3n + 1 (n is odd)

Using the rule above and starting with 13, we generate the following sequence:

13 ? 40 ? 20 ? 10 ? 5 ? 16 ? 8 ? 4 ? 2 ? 1

It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.

Which starting number, under one million, produces the longest chain?

NOTE: Once the chain starts the terms are allowed to go above one million.

Explanation / Answer

# return the number of elements in the chain produced

def getCount(n):

    count = 0

    while n != 1:

        # if n is even

        if int( n % 2 ) == 0:

            n = int( n / 2 )

        # if n is odd

        else:

            n = 3 * n + 1

        count += 1

    return count

def main():

    max = 0

    max_count = 0

    # loop from 1 to 1 million

    for i in range( 1 , 1000001 ):

        # get the number of elements in chain of i

        count = getCount(i)

        if count > max_count:

            max_count = count

            max = i

    print(max, ' produces the longest chain and has', max_count, ' elements.')

main()

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