CHALLENGE: Friedman Number

Question

A relative integer is a Friedman number if there is an expression using all of its digits equal to it. The mathematical operations allowed in this expression are:

The four basic arithmetic operations:
(+, -, /, *)

Eventually, other operations such as:
exponentiation, factorial, square root, root n-th

For example:
374 is a Friedman number in base 10 because it is possible to write it in the form 73 + 4.

Another example:
127 = -1 + 2 ^ 7

Parentheses are allowed, but only to change the order of operations: for example, 1024 = (4 - 2) 10. The use of parentheses without operators would result in trivial Friedman numbers like 24 = (24). Initial zeros are also not allowed for the same reason: 001729 = 1700 + 29.

So the challenge is :
Build a program that takes a number and display if this last one is a Friedman number or not.

Good luck.

Answer 1

@1of3

Yes, you can use it.

The allowed operations are +,/,-,* and exponentiation,factorial,square root and root n-th.

By the way: The second example define also what is an ordered Friedman Number, that means without changing the order of the digits.

25 is a Friedman Number, 25=5^2.

Good luck.

Answer 2

hm. but what if you can use all digits only once? otherwise any number can be build very easy. for example like 1+1+...+1 where 1 is n^(n-n) or something like that.

Answer 3

@VcC

Yes, yes !
It works.
Bright, bravo, you did it !

Answer 4

Not yet one liner but does the job. Easy to add any function
https://code.sololearn.com/cXHQF3Zz8M3i/?ref=app

Answer 5

Here is an ultrafast probabilistic algorithm:
f=lambda n: True

Because most number are Friedman. You should only allow ops on digits not using '73' for 374

Answer 6

I have an idea. But you used exponentiation in your second example, while you only said four basic arithmetic operations before. Is that in or out?

Answer 7

@Alex Gvozdyov, Yes, all digits are used only once I believe

Answer 8

My try in python use exhaustive permutations, and I cannot exclude the repetition...so it works but it reaches time limit for complicated numbers

Answer 9

https://code.sololearn.com/cyz0F0SuyTg2/?ref=app

Answer 10

Let me say again: Great challenge. I learned something about iterators.

Not done yet, those special rules for concatenation of digits are hard. Basically the tokens that can be operated on is any choice of digits from the original number with length strictly smaller than original number's.