There is a mistake with such expressions as mod(23^12,24). MATLAB gives an answer of 0 which, given that 23 and 24 are relative primes,

1 vue (au cours des 30 derniers jours)
Clearly, 23 = -1 mod 24. So, mod(23^12,24) should result in 1.

Réponse acceptée

Walter Roberson
Walter Roberson le 12 Juin 2019

Plus de réponses (3)

Rik
Rik le 12 Juin 2019
The number you're trying to process is larger than flintmax(), so you can rely on single digit precision operations.

AdamE
AdamE le 12 Juin 2019
I the function mod could recognize -1 = 23 mod 24, then act on -1, the number would not be too large. This logic should be employed in the function.
  3 commentaires
Walter Roberson
Walter Roberson le 12 Juin 2019
No, it should not be. Doing that would require rewriting the entire way that MATLAB operates on parameters.
MATLAB currently always evaluates all parameters before the function is called, so when you call mod(23^12, 24) then 23^12 and 24 are evaluated and their full evaluated results are passed to mod() .
In order for MATLAB to do what you are suggesting automatically, it would have to somehow record the formula 23^12 and 24 and pass that into mod() and then mod would have to know how to decompose formulas in terms of modular arithmetic. It would have to be pretty deep, such as knowing that
mod( (23^12)^3 + 5*23^12 + 7, 24)
could be reduced in modular arithmetic.
AdamE
AdamE le 12 Juin 2019
I completely agree. I meant that there should be, and apparently are, scripts to handle this situation.

Connectez-vous pour commenter.


AdamE
AdamE le 12 Juin 2019
John,
Firstly, calm down. I've been using MATLAB since the late 1980s and have a PhD in Math. I wasn't implying that MATLAB performed erroneously; I was implying that there should be a modulo script which can employ some basic number theory to reduce incorrect results. Now that Walter provided the links, I see that there are such scripts.

Catégories

En savoir plus sur Loops and Conditional Statements dans Help Center et File Exchange

Produits

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by