math problem using recursion?

[Gribouillis]

You could use recursion to build the tree. The key is to suppose that the problem is solved. Let func() be a function that takes a number as argument and returns None if it is not a semi-prime or the root of this semi-prime's tree if it is a semi-prime.

We need to define what the root of a tree will be. We'll take a tuple ((factor0, factor1), (subroot0, subroot1)). For this we could use a namedtuple. Here is an algorithm

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Root = namedtuple('Root', 'factor subroot')   def func(number):     try:         fac0, fac1 = decompose_semi_prime_in_factors(number)     except NotSemiPrime:         return None     root0 = func(int(f'{fac0}{fac1}'))     root1 = None     if fac0 != fac1:         root1 = func(int(f'{fac1}{fac0}'))     return Root((fac0, fac1), (root0, root1))

[SmartGrid]

The longest chain length could be millions , no ? Or is the start number limited?

[mnh001]

Ok, I got that alright. Here:

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try:     divisors(newList[passes])     passes += 1 except:     break

It seems to give the same results as yours. I have since added more code to make
it automatic. If I'm going to be searching millions of start numbers I can't do
it by hand.
I also tried to put in a counter for max chain length (inside the 'íf len(div) == 2:'
statement but it's not accurate. Since each 'level' of the tree can have anywhere
from 0 up to 2^(n-1) leaves, I may have to put in some conditionals to keep track,
although I'm not sure how that would work yet.

You could use recursion to build the tree.

Ok, I'll look at that also.

[DPaul]

Ok, I got that alright. Here:

Could not have done it better! :-)

Paul

[mnh001]

Thanks for the help. Sometimes the logic behind these things eludes me.
I think we can say this one is solved. :)