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An exact whole power with a huge exponent hangs evaluation: 2^(10^9), 2^(3^20)

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Évaluation

Difficulté
5/5
Temps estimé
Plus d'une semaine
Accessibilité débutants
32/100
Type d'issue
Bug
Clarté
À clarifier
Activité
Active
Stack technique
csharp
Domaine
performance

Piste de recherche

Start by reproducing the two expressions through ToEntity().Evaled and trace the Number.Pow/BinaryIntPow path described in the issue. Review the quantifier witness-search path and related issue #1338, then establish the chosen behavior for oversized exact powers and confirm that witness searches remain bounded without silently losing exactness.

Rédigé par le modèle d'indexation à partir du texte de l'issue.

Description

"2^(10^9)".ToEntity().Evaled and "2^(3^20)".ToEntity().Evaled do not return (killed at 60 s each; 2^(3^12) returns in 1 s). The result of the first is a 10⁹-bit integer (125 MB) and of the second a 3.5·10⁹-bit one; Number.Pow's BinaryIntPow builds them by squaring in EInteger, and the last squarings are Toom-4 multiplications of numbers hundreds of megabytes long. Measured on 8fcfaf42 (a branch of 345914f6); the code path is unchanged since long before.

Found through a quantifier: forall n in ZZ* : 3^(n + 1) divides 2^(3^n) + 1 (Sullivan and Mackey's Prob 5.7.13) hangs because a witness search evaluates the body at n = 20 or so, where 2^(3^n) is the second number above. Any consumer that evaluates a tower at a large argument hits the same.

What the answer should be is the decision to make, and it belongs with #1338's numerics: an exact power whose result would exceed some size — say 2²⁴ bits, which is 2 MB and a fraction of a second — is either (a) left as written (2^(10^9) evaluates to itself, the way an unsolved statement stays a statement), or (b) refused with an exception naming the limit, or (c) evaluated approximately as Real at the working precision, which is what a decimal calculator does and loses exactness silently. (a) is the honest one under not answering is legitimate; answering wrongly is not, and it composes: a witness search that gets an unevaluated power back moves on. Until then, a quantifier's witness search should skip a sample it cannot evaluate in bounded time, which is the narrower fix I would take first.

Part of #1409.

🤖 Generated with Claude Code

https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura

Langage dominant
C#
Étoiles
831
Forks
79
Merge moyen
2 h 22 min
PR mergées (30 j)
507

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