Design #1309
SmoothFactor: allow limit = 1?
Description
The SmoothFactor function tries to factorize integers by dividing by all primes up to a specified limit.
Currently that limit must be at least 2, otherwise an error is reported.
Should the limit be lowered to 1?
One motivation: let p be a prime, then one would expect SmoothFactor(p, FloorSqrt(p)) to return an empty factorization (and the RemainingFactor would be just p). With the current limit of 2 this will not work for the primes 2 and 3 because the FloorSqrt is below 2.
History
#1
Updated by John Abbott over 4 years ago
Updated by - Status changed from New to In Progress
- % Done changed from 0 to 10
The original reason for imposing the limit was to help users detect "obviously silly" function calls. Now I think that SmoothFactor(n,1) is not always "obviously silly".
I encountered the problem in another program I have written. In this program I wanted to compute something like SmoothFactor(n,FloorSqrt(n)); just occasionally n can be small, so the program triggers a BadArg error.
Of course, I could change my program so that it detects if n is smaller than 4 and then handles that case specially; but that makes my program more complicated and less readable.
Comments? Opinions?
#2
Updated by John Abbott over 4 years ago
There are two separate impls for SmoothFactor (one for MachineInt and one for BigInt). One allowed the limit 1, the other did not.
#3
Updated by John Abbott over 4 years ago
I have just changed the code to allow limit=1. All tests and examples pass (which suggests that the tests were not very thorough... oh).
#4
Updated by John Abbott over 4 years ago
- Status changed from In Progress to Closed
- Assignee set to John Abbott
- % Done changed from 10 to 100
#5
Updated by Anna Maria Bigatti over 4 years ago
Updated by - Target version changed from CoCoALib-0.99700 to CoCoALib-0.99650 November 2019
- Estimated time set to 1.01 h