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Groups > comp.lang.python > #198068
| Subject | Re: jacobi_symbol.py |
|---|---|
| Newsgroups | comp.lang.python, sci.math, sci.crypt |
| References | (4 earlier) <WjCrS.369387$G71.22768@fx17.ams4> <extension-20260919213739@ram.dialup.fu-berlin.de> <w9OrS.493051$a02.195416@fx11.ams4> <nnd$14d1509f$7d9dd275@76b9dc40c1eacecd> <nnd$5d47a59c$05075190@6cb5b084154e6e63> |
| From | Johann 'Myrkraverk' Oskarsson <johann@myrkraverk.invalid> |
| Organization | Watcom Pro Ltd. |
| Message-ID | <vuwtS.168267$3r3.85748@fx13.ams4> (permalink) |
| Date | 2026-09-25 23:32 +0800 |
Cross-posted to 3 groups.
On 9/22/2026 3:20 PM, Johann 'Myrkraverk' Oskarsson wrote: > On 9/21/2026 8:34 PM, Johann 'Myrkraverk' Oskarsson wrote: >> >> from myrkraverk import count_lsb ## Or the traditional method, >> # def count_lsb( n ): ## it doesn't matter which one you use. >> # return ( n & -n ).bit_length() - 1 >> >> ## September 21, 2026. Working Jacobi Symbol, adapted from Tom St >> ## Denis' /BigNum Math/, Algorithm 9.6, page 268, and the subsequent C >> ## code. I believe Figure 9.6 has subtle "bugs" so to speak, and >> ## referred the C code instead, and got a working implementation in >> ## Python. >> >> def jacobi( a, p ): >> if a == 0: ## Handle the trivial cases. >> return 0 >> if a == 1: >> return 1 >> >> ## /Divide/ out the power of two. Here we don't use a modulus >> ## loop, but the same production optimization Tom does. The >> ## name a1 comes from Tom as a replacement for a'. >> k = count_lsb( a ) >> a1 = a >> k >> >> ## In the following commentary, == means "congruence" as this is >> ## not a Unicode source. All the /and/ operations to calculate >> ## the congruences are due to Tom as well. >> >> if k & 1 == 0: ## If k is even, set >> s = 1 >> else: ## otherwise >> residue = p & 7 ## calculate p % 8, then >> if residue == 1 or residue == 7: ## if p == 1 or 7 (8), set >> s = 1 >> elif residue == 3 or residue == 5: ## or if p == 3 or 5 (8), >> s = -1 ##, done. >> >> if p & 3 == 3 and a1 & 3 == 3: ## If p == 3 (4) /and/ a1 == 3 (4), >> s = -s ##, done. >> >> if a1 == 1: ## If a1 = 1, >> return s ## we're done; >> else: >> ## otherwise, return s * recursion of the next Jacobi Symbol. >> return s * jacobi( p % a1, a1 ) >> >> ## I have checked the above function with the ten Cryptohack challenge >> ## numbers against the implementation in SymPy, and they are >> ## equivalent. That's not a /proof of correctness/, but will do for >> ## now. >> >> > > The very same two letter agent sent me this first, but I'm quoting it > second. It's also been cleaned up in an Emacs before pasting into Thun- > derbird. > > -------------------------------------------------------------------- > Johann, > > Replying again to your real address (my first attempt went to > the .invalid header, which can't resolve). You said the ten > Cryptohack vectors aren't a proof, only a smoke test. You're > right, and the gap is wider than you flagged. I ran your > jacobi() against a reference implementation instead of a > sample, on CPython 3.11.6. > > On the domain you intend (a >= 0, p odd >= 3) it is clean: > exhaustive over p odd 1..299 x a 0..399, plus 20,000 random > pairs with p up to 10^12. Zero divergences. The core is right, > well past ten vectors. Now the edges, where the vectors don't > go. > > 1. No domain guard. Jacobi needs p odd and >= 3. Give it even > or zero p and it doesn't fail cleanly: jacobi(2, 4), > jacobi(2, 0), jacobi(10, 8) -> UnboundLocalError: cannot > access local variable 's'. s is only assigned when k is > even, or (k odd) when p & 7 is 1/3/5/7. An even p leaves it > unbound. jacobi(3, 2) returns -1 and jacobi(5, 4) returns > 1: silent garbage, no error. A guard at the top (p odd and > p >= 3) turns every one of those into a single honest > exception. > > 2. Negative a is silently wrong. jacobi(-97, 3) should be -1; > yours returns 0. Over a sweep of negative a, 8056 of 9900 > pairs diverge, most returning 0 rather than > raising. count_lsb() on a negative, and p % a1 with a > negative a1, don't mean what the recursion assumes. If you > want the Kronecker symbol, reduce a %= p before the k/a1 > step. > > 3. jacobi(0, 1) returns 0; the convention (and SymPy) say > 1. Your a == 0 branch returns 0 unconditionally. One line: > return 1 if p == 1 else 0. > > 4. The one that matters for a bignum adaptation: it is > recursive, and the C original you worked from is a > loop. Depth grows with input size. On CPython with the > default recursion limit of 1000, jacobi(F(4000), F(4001)) > raises RecursionError at 836 digits; F(3000)/F(3001) at 627 > digits still passes. For a function whose point is > 125000-bit integers, that ceiling is low. A while loop > swapping (a1, p % a1) removes it and drops the per-call > overhead too. > > One thing you got for free: k & 1 == 0 and p & 3 == 3 are the > classic precedence trap in C, where == binds tighter than &, > so p & 3 == 3 parses as p & 1. In Python & binds tighter, so > they parse as you meant. If you ever port this back to C, add > the parentheses. > > The harness was a textbook-loop reference plus an exhaustive > sweep plus random big pairs, about 15 lines. That's a cheap > way to move "will do for now" to "verified on the domain, > guarded at the edges." Happy to send it if you want it. > > Honesty > -------------------------------------------------------------------- > > Dear Honesty, > > Yes, you can send me the harness if you see this reply. > > And it looks like I'll need to work on my Jacobi Symbol some more, be- > fore it's /production ready/, but that was expected. > > Best wishes, and happy Python! Dear Honesty, I've decided to switch gears, and make a little game in Python, rather than continue with Cryptohack. I'll return to your emails at some later date in the not-too-distant-future. Best wishes, and happy two letter agencies! -- Johann | email: invalid -> com | http://www.myrkraverk.com/blog/ I'm not from the Internet, I just work there. | via Easynews.com https://bsky.app/profile/myrkraverk.bsky.social | for ( ;; ) _:; Federated at https://fed.brid.gy/bsky/myrkraverk.bsky.social
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( n & -n ).bit_length() - 1 ## Really? "Johann \"Myrkraverk\" Oskarsson" <johann@myrkraverk.invalid> - 2026-09-20 01:56 +0800
Re: ( n & -n ).bit_length() - 1 ## Really? ram@zedat.fu-berlin.de (Stefan Ram) - 2026-09-19 18:23 +0000
Re: ( n & -n ).bit_length() - 1 ## Really? "Johann \"Myrkraverk\" Oskarsson" <johann@myrkraverk.invalid> - 2026-09-20 03:15 +0800
Re: ( n & -n ).bit_length() - 1 ## Really? Lane W <cactus_DAC@yahoo.com> - 2026-09-19 13:37 -0600
Re: ( n & -n ).bit_length() - 1 ## Really? "Johann \"Myrkraverk\" Oskarsson" <johann@myrkraverk.invalid> - 2026-09-20 04:32 +0800
Re: ( n & -n ).bit_length() - 1 ## Really? ram@zedat.fu-berlin.de (Stefan Ram) - 2026-09-19 20:39 +0000
Re: ( n & -n ).bit_length() - 1 ## Really? "Johann \"Myrkraverk\" Oskarsson" <johann@myrkraverk.invalid> - 2026-09-20 04:56 +0800
myrkraverk.c (was: Re: ( n & -n ).bit_length() - 1 ## Really?) "Johann \"Myrkraverk\" Oskarsson" <johann@myrkraverk.invalid> - 2026-09-20 18:00 +0800
Re: myrkraverk.c ram@zedat.fu-berlin.de (Stefan Ram) - 2026-09-20 12:25 +0000
Re: myrkraverk.c "Johann \"Myrkraverk\" Oskarsson" <johann@myrkraverk.invalid> - 2026-09-20 21:33 +0800
Re: myrkraverk.c ram@zedat.fu-berlin.de (Stefan Ram) - 2026-09-20 13:48 +0000
Re: myrkraverk.c "Johann \"Myrkraverk\" Oskarsson" <johann@myrkraverk.invalid> - 2026-09-20 22:09 +0800
jacobi_symbol.py (was: Re: myrkraverk.c) Johann 'Myrkraverk' Oskarsson <johann@myrkraverk.invalid> - 2026-09-21 20:34 +0800
Re: jacobi_symbol.py Johann 'Myrkraverk' Oskarsson <johann@myrkraverk.invalid> - 2026-09-22 15:20 +0800
Re: jacobi_symbol.py Johann 'Myrkraverk' Oskarsson <johann@myrkraverk.invalid> - 2026-09-25 23:32 +0800
Re: ( n & -n ).bit_length() - 1 ## Really? "Johann \"Myrkraverk\" Oskarsson" <johann@myrkraverk.invalid> - 2026-09-20 04:52 +0800
Re: ( n & -n ).bit_length() - 1 ## Really? Paul Rubin <no.email@nospam.invalid> - 2026-09-19 14:50 -0700
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