Autum Challenge, 42 is the Answer

[Transfinite Numbers]

Find x, y, z such that:

x^3 + y^3 + z^3 = 42

[Transfinite Numbers]

Corr.: Typo Autum -> Autumn

You can cheat here:
https://www.heise.de/newsticker/meldung/Mathematiker-knacken-Gleichung-mit-Kubikzahlen-und-der-Summe-42-4523603.html

But some Prolog to get it?

[Transfinite Numbers]

For some algorithmic ideas:

Newer sums of three cubes
Sander G. Huisman
(Submitted on 26 Apr 2016)
https://arxiv.org/abs/1604.07746

[Transfinite Numbers]

This was an article about the 33 break through:

Cracking the problem with 33
Andrew R. Booker
Research in Number Theory
September 2019, 5:26
https://link.springer.com/article/10.1007/s40993-019-0162-1

[j4n bur53]

There is a new unit for computing time "core years".
Sounds familer, like "light years". LoL

The Springer / Booker article contains a sentence:

"The running time was approximately 8 core-years per number tested."

Transfinite Numbers schrieb:

[Transfinite Numbers]

Robin Houston says:

"I think that means 114 is now the smallest unsolved case."

https://twitter.com/robinhouston/status/1169877007045296128

[Transfinite Numbers]

Algorithmic complexity matters. Bitcoin is
now in quantum shock? 7’918.21 US-Dollar low.

Looks like a call for quantum resistant bitcoins:

https://medium.com/the-quantum-resistant-ledger/quantum-supremacy-and-the-case-for-quantum-security-today-in-blockchain-390fe55daab5

[Transfinite Numbers]

This looks interesting:

Applications of Roth’s Theorem to Diophantine Equation
There are only finitely many integer solutions to
x^3−2y^3 =a.

ALGEBRAIC AND TRANSCENDENTAL NUMBERS
STEVEN J. MILLER AND RAMIN TAKLOO-BIGHASH
https://pdfs.semanticscholar.org/71e0/cc231449bbba94acc20ff9d95a86704fb771.pdf

If there are solutions, then there are finitely
many ones. So not every infinite CLP(FD) might
never terminate. There are some heuristic criteria.

But the paper also says that the finite can
be very large, like 10^456 many solutions.

[Transfinite Numbers]

Did anybody ever try LLL for CLP(FD)?

https://en.wikipedia.org/wiki/Lenstra%E2%80%93Lenstra%E2%80%93Lov%C3%A1sz_lattice_basis_reduction_algorithm

[Mostowski Collapse]

Any progress during Corona?

Can this be solved via ZZD and the axiom of extensionality?
Or maybe the new CLP(Z) from Scryer Prolog can do it?

Or are we in the mist of Hilbert’s 10-th problem. Maybe Poincaré
would have objected we simply need new problem specific rules?

[Mostowski Collapse]

Does SWI-Prolog have modpow/4, a function popular in cryptography?
Isn’t this already in GMP, so that a verify fast implementation
would be available?

modpow(B, E, M, R):
The predicate succeeds in R with B^E mod M.

I guess we have also:

(modpow(X,3,M)+modpow(Y,3,M)+42 mod M) mod M =:= modpow(Z,3,M)

But not sure whether this will lead to anywhere. From Fermats
Last Theorem we know, whenever 42 mod M is zero, we
don’t have a non-trivial tripple (X,Y,Z) anyway.

[Mostowski Collapse]

No was talking nonsense, FLT doesn’t exclude non-trivial solutions. For
example we have this solution modulo 6, where it is 42 module 6 = 0:

1^3 + 1^3 + 42 = 2^3 (mod 6)

[Mostowski Collapse]

Furtther question, does SWI-Prolog have modinv/3, also a function popular
in cryptography? It can be used for a Chinese Remainder algorithm.

But currently banging my head on a slow (^)/2 for smallints. In modular
algorithms the arguments for (^)/2 are small. So I now get:

/* SWI-Prolog 8.5.17 */
?- time((modular([11,7,5], X, Y, Z), fail; true)).
% 15,388,917 inferences, 3.969 CPU in 3.963 seconds (100% CPU, 3877522 Lips)
true.

/* Jekejeke Prolog 1.5.4 */
?- time((modular([11,7,5], X, Y, Z), fail; true)).
% Threads 2,078 ms, GC 10 ms, Up 2,122 ms (Current 09/20/22 11:03:10)
true.

But I only solved this k=9 equation, not yet the k=42 problem:

/* x^3 + y^3 + 9 = z^3 */
?- modular([11,7,5], X, Y, Z).
X = 216,
Y = 52,
Z = 217 ;
X = 52,
Y = 216,
Z = 217 ;
false.

[Mostowski Collapse]

Using modular operations again, you can make it a tick faster by using
only two modules instead of three modules. But it still depends
heavily on the performance of ar_pow():

/* SWI-Prolog 8.5.17 */
?- time((modular([15,16], X, Y, Z), fail; true)).
% 3,816,486 inferences, 1.766 CPU in 1.762 seconds (100% CPU, 2161550 Lips)

true.

/* Jekejeke Prolog 1.5.4 */

?- modular([15,16], X, Y, Z).

X = 216, Y = 52, Z = 217;
X = 52, Y = 216, Z = 217;

fail.

?- time((modular([15,16], X, Y, Z), fail; true)).
% Threads 594 ms, GC 5 ms, Up 591 ms (Current 09/20/22 20:21:30)
true.

[Mostowski Collapse]

Thats the source code, it is ultimately intended to beat CLP(Z)
from Scryer Prolog some time in the future into dust, it also uses
ar_pow(), maybe better would be ar_modpow(), but for the small

values ar_modpow() wouldn't give some bang:

modular([P|L], X, Y, Z) :-
find(P, A, B, C),
many(L, A-P, B-P, C-P, X-_, Y-_, Z-_),
X^3+Y^3+9-Z^3 =:= 0.

many([], A, B, C, A, B, C).
many([P|L], U, V, W, A, B, C) :-
find(P, X, Y, Z),
cra_combine(X-P, U, S),
cra_combine(Y-P, V, T),
cra_combine(Z-P, W, R),
many(L, S, T, R, A, B, C).

find(M, X, Y, Z) :-
N is M-1,
between(0, N, X), A is X^3,
between(0, N, Y), B is Y^3,
between(0, N, Z),
(A+B+9-Z^3) mod M =:= 0.

/**
* Chinese Remainder Algorithm
* https://en.wikipedia.org/wiki/Chinese_remainder_theorem#Existence_%28constructive_proof%29
*/
% cra_combine(+Pair, +Pair, -Pair)
cra_combine(A-N, B-M, C-P) :-
cra_egcd(N, M, X, Y),
P is N*M,
C is (A*M*Y+B*N*X) mod P.

% cra_egcd(+Integer, +Integer, -Integer, -Integer)
cra_egcd(_, 0, 1, 0) :- !.
cra_egcd(A, B, X, Y) :-
divmod(A, B, Q, R),
cra_egcd(B, R, S, X),
Y is S - Q*X.

[Mostowski Collapse]

(Credits for the acronym cra for Chinese Remainder Algorithm in the
context of Prolog go to Amzi! Prolog, I found the acronym here:
https://amzi.com/manuals/amzi/pro/ref_math.htm )

[Mostowski Collapse]

Thats the performance of traditional CLP(FD) and CLP(Z)
for this kind of problem:

/* SWI-Prolog CLP(FD) */
?- time(([X,Y,Z] ins 0..239, X^3+Y^3+9 #= Z^3, label([X,Y,Z]), fail; true)).
% 36,572,338 inferences, 5.109 CPU in 5.117 seconds (100% CPU, 7157889 Lips)
true.

/* Scryer Prolog CLP(Z) */
?- time(([X,Y,Z] ins 0..239, X^3+Y^3+9 #= Z^3, label([X,Y,Z]), fail; true)).
% CPU time: 75.667s
true.

Wasnt using any bisection or somesuch option. Just the
defaults of CLP(FD) and CLP(Z).

[Mostowski Collapse]

Even Dogelog player is speedier than CLP(FD) and CLP(Z),
using the modular approach. I find:

?- modular([15,16], X, Y, Z).
X = 216, Y = 52, Z = 217;
X = 52, Y = 216, Z = 217;
fail.

?- time((modular([15,16], X, Y, Z), fail; true)).

% Wall 3925 ms, gc 7 ms, 1544404 lips
true.

https://jsfiddle.net/Jean_Luc_Picard_2021/d2njehtp/3/

I don't have divmod/4 yet, so one needs to use the
following variant of Extended GCD:

cra_egcd(A, B, X, Y) :-

Q is A div B,
R is A mod B,
% divmod(A, B, Q, R),

cra_egcd(B, R, S, X),
Y is S - Q*X.

[Markus Triska]

Mostowski Collapse <burs...@gmail.com> writes:

> /* Scryer Prolog CLP(Z) */
> ?- time(([X,Y,Z] ins 0..239, X^3+Y^3+9 #= Z^3, label([X,Y,Z]), fail; true)).
> % CPU time: 75.667s
> true.
>

With the newly release Scryer Prolog version 0.9.2, I now get:

?- time(([X,Y,Z] ins 0..239, X^3+Y^3+9 #= Z^3, label([X,Y,Z]), fail; true)).

% CPU time: 41.658s
true.

On my machine, that's within a factor of 6 of SWI. That's quite
comparable to many other applications when it comes to time performance:
At the current state of Scryer Prolog development, its performance tends
to be within an order of magnitude of SWI's.

One neat feature of the Scryer Prolog toplevel is that we can press "a"
to obtain *all* solutions, one after the other. In this case:

?- time(([X,Y,Z] ins 0..239, X^3+Y^3+9 #= Z^3, label([X,Y,Z]))).
% CPU time: 16.080s

X = 52, Y = 216, Z = 217

; % CPU time: 25.942s

X = 216, Y = 52, Z = 217

; % CPU time: 0.250s
false.

It's often nice to get the Prolog system to enumerate all solutions
automatically. The GNU Prolog toplevel also has this feature, and I
highly recommend adding it in system where it is not yet available.

All the best,
Markus

--
comp.lang.prolog FAQ: http://www.logic.at/prolog/faq/
The Power of Prolog: https://www.metalevel.at/prolog

[Mild Shock]

One motivation to look at the problem, and to look at the
easier problem with 9 instead 42, is to elaborate a completely
new CLP(FD) solver, that is based on Chinese Remainder Theorem.

Actually I have already a prototype working, the timing was
also already published around 12 months ago. My system was much
faster than SWI-Prolog, since SWI-Prolog is slow with smallints and (^)/2:

Mostowski Collapse schrieb am Dienstag, 20. September 2022 um 20:28:33 UTC+2:
> /* Jekejeke Prolog 1.5.4 */
> ?- modular([15,16], X, Y, Z).

> X = 216, Y = 52, Z = 217;

> X = 52, Y = 216, Z = 217;

> fail.
>
> ?- time((modular([15,16], X, Y, Z), fail; true)).
> % Threads 594 ms, GC 5 ms, Up 591 ms (Current 09/20/22 20:21:30)
> true.

https://groups.google.com/g/comp.lang.prolog/c/mjpxkE3xVYk/m/cn0FICAQAAAJ

So these 594 ms are 10x times faster than the 5.117 seconds from
SWI-Prolog using ordinary CLP(FD). And around 100x times faster than
the 41.658s from Scryer Prolog. But I didn't yet publish this new

CLP(FD) solver, maybe I did some blogging and also some code
went into comp.lang.prolog here, but its not currently some officially
released module somewhere. Its also a solver which isn't based on

attributed variables per se. It requires that the constraint store is
re-evaluated with different moduli, so I envision something totally
new, that drops attributed variables, but nevertheless has a constraint

store. Attributed variables have become less important. The dis-
advantage of not having attributed variables would be that the
constraints cannot be used on ordinary predicates. I have no solution

for the later problem yet, but the figures in the former testing show,
that the method can be much much faster than ordinary CLP(FD).

[Mild Shock]

Actually its not that dim, one could integrate the Chinese
Reminder Algorithm (CRA) strategy into a CLP(FD) constraint
solver via a labeling variant. The labeling would

include not only running through values inside the current
modulo M, i.e. choosing values in the range 0..M-1, but also
choosing different moduli M1, .., Mk. But the question is

whether one should do that. The problem is that the solution
is obtained by backtracking over all moduli. So when E is the
equation, then its basically a large conjunction:

E_M1, ..., E_Mk

Where each element of the conjunction produces some bits,
and these bits are then combined via CRA. Currently when
E_Mj is a predicate, the equation sits in this predicate,

and each element of the conjunctions runs instances of
the clauses of the predicate, thus has a copy of the equation
available. This copying of the equation makes me think

that attribute variables don't work. But alternatively one
could also table the results of each E_Mj, and later do the
CRA combination. My current solution is not like that,

but this could have more affinity to attributed variables.

[Mild Shock]

Whats the Scryer Performance of this Example?

/* GNU Prolog 1.5.0 */
?- between(1,200,N), length(L,N), nth_member(I,L,E), fail; true.
(94 ms) yes

/* SWI-Prolog 9.1.14 */
?- time((between(1,200,N), length(L,N), nth_member(I,L,E), fail; true)).
% 94,504,524 inferences, 4.766 CPU in 4.767 seconds (100% CPU, 19830457 Lips)
true.

That was measured on my machine,
and was a factor ca. 50x slower in SWI-Prolog.

Source Code:

nth_member(1, [M|_], M).
nth_member(N, [_|T], M) :-
N #> 1, N1 #= N - 1,
nth_member(N1, T, M).

[Markus Triska]

Mild Shock <burs...@gmail.com> writes:

> Whats the Scryer Performance of this Example?
>
> /* GNU Prolog 1.5.0 */
> ?- between(1,200,N), length(L,N), nth_member(I,L,E), fail; true.
> (94 ms) yes
>
> /* SWI-Prolog 9.1.14 */
> ?- time((between(1,200,N), length(L,N), nth_member(I,L,E), fail; true)).
> % 94,504,524 inferences, 4.766 CPU in 4.767 seconds (100% CPU, 19830457 Lips)
> true.

With Scryer Prolog 0.9.2 on my machine, I get:

?- time((between(1,200,N), length(L,N), nth_member(I,L,E), false; true)).
% CPU time: 84.182s
true.

On my machine, that's ca. 15 times slower than SWI 8.3.15, which is the
last SWI version I found installed on this machine. Newer versions may
well be faster! For best performance, I recommend to build Scryer Prolog
with the optional --release flag, i.e.:

$ cargo build --release

[Mild Shock]

I am also slower than SWI-Prolog with formerly Jekejeke Prolog,
but sandwhich between Scryer Prolog and SWI-Prolog for this example.
I never know whether my CLP(FD) still works, so first some sanity checks:

?- nth_member(X, [5,4,6], 4).
X = 2;
fail.
?- nth_member(2, [5,4,6], X).
X = 4;
fail.

Then the benchmark results:

?- time((between(1,200,N), length(L,N), nth_member(I,L,E), fail; true)).

% Time 48513 ms, GC 373 ms, Wall 04/09/2023 21:41
true.

But I don't know whether its comparable with Scryer Prolog since
I didn't test Scryer Prolog, and the Scryer Prolog figures are from
Markus Triskas machine.

I am currently phasing out formerly Jekejeke Prolog. It should receive
the same core as Dogelog Player. And then maybe can also re-add
CLP(FD) and see what happens. But this might take months. Especially

since I want to reinvent CLP(FD), adding some CRA Strategy, and I
don't know yet eactly how to do it. I could optimized the non-labeling
behaviour exactly towards this test case, where GNU Prolog performs well.

[Mild Shock]

Actually the test was done with JDK 21 as the Java Virtual
Machine. The older JDK 8 fares a little bit more favorable:

/* JDK 8 */

?- time((between(1,200,N), length(L,N), nth_member(I,L,E), fail; true)).

% Time 39146 ms, GC 254 ms, Wall 04/09/2023 21:49
true.

/* JDK 21 */

?- time((between(1,200,N), length(L,N), nth_member(I,L,E), fail; true)).

% Time 48513 ms, GC 373 ms, Wall 04/09/2023 21:41
true.

The formerly Jekejeke Prolog Minlog Extension, that has the
CLP(FD), isn't public anymore, because I am phasing it out.

[Mild Shock]

What speed-up could a N-qubit quantum computer give
to this problem, find a positive integer tripple (x,y,z) such that:

x^3 + y^3 + 42 = z^3

Mild Shock schrieb am Freitag, 1. September 2023 um 20:47:06 UTC+2:

[Mild Shock]

Lets say you don't trust what a quantum computer
finds as a tripple for (x,y,z) to satisfy:

x^3 + y^3 + 42 = z^3

You could then view it as a candidate triple
only, and still feed it into a normal computer

and see whether it is a solution. Kind of combining
approximate search with exact verification.