Zarankiewicz's problem

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Freywa
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Zarankiewicz's problem

Post by Freywa » January 5th, 2022, 5:08 am

All non-isomorphic solutions to certain small square instances of the 3×3 and 4×4 Zarankiewicz's problem, as verified through SAT solving:

Code: Select all

x = 176, y = 184, rule = LifeHistory
21.D5.3D4.3D5.D.3D4.D.3D5.3D.3D4.3D.3D5.D.D.3D6.3D.3D7.3D.D.D8.3D.3D
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.D5.D3.D8.D.D9.D.D.D8.D.D9.D.D3.D10.D3.D11.D.D12.D.D.D3.D$21.D5.3D4.
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Last edited by Freywa on January 5th, 2022, 4:33 pm, edited 1 time in total.
Princess of Science, Parcly Taxel

Code: Select all

x = 31, y = 5, rule = B2-a/S12
3bo23bo$2obo4bo13bo4bob2o$3bo4bo13bo4bo$2bo4bobo11bobo4bo$2bo25bo!

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dvgrn
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Re: Thread for your unsure discoveries

Post by dvgrn » January 5th, 2022, 8:39 am

Freywa wrote:
January 5th, 2022, 5:08 am
All non-isomorphic solutions to certain small square instances of the 3×3 and 4×4 Zarankiewicz's problem, as verified through SAT solving...
Is this a decent summary of Zarankiewicz's problem? It's a pretty sure sign that even the problem statement is over my head, when I can't even figure out what the definitions of "a" and "b" are.

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Freywa
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Location: Singapore
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Re: Thread for your unsure discoveries

Post by Freywa » January 5th, 2022, 4:37 pm

dvgrn wrote:
January 5th, 2022, 8:39 am
Is this a decent summary of Zarankiewicz's problem? It's a pretty sure sign that even the problem statement is over my head, when I can't even figure out what the definitions of "a" and "b" are.
Some of the results here (n >= 11 for 3×3, n >= 8 for 4×4) are genuinely new and proven by me. Indeed I've got an OEIS sequence with all the old and new data up in the 3×3 case.
Princess of Science, Parcly Taxel

Code: Select all

x = 31, y = 5, rule = B2-a/S12
3bo23bo$2obo4bo13bo4bob2o$3bo4bo13bo4bo$2bo4bobo11bobo4bo$2bo25bo!

FractalFusion
Posts: 52
Joined: March 27th, 2009, 2:07 pm

Re: Thread for your unsure discoveries

Post by FractalFusion » January 5th, 2022, 7:06 pm

dvgrn wrote:
January 5th, 2022, 8:39 am
Is this a decent summary of Zarankiewicz's problem? It's a pretty sure sign that even the problem statement is over my head, when I can't even figure out what the definitions of "a" and "b" are.
The Zarankiewicz problem actually has multiple manifestations, but the basic idea given on Wikipedia as well as the site you linked is:

Given positive integers m,n and positive integers a,b where a≤m and b≤n, what is the maximum number k such that there exists a {0,1}-matrix of size m×n with k 1s which does not contain the all-1s matrix of size a×b as a submatrix?

(However, there are related definitions, such as on OEIS, where it appears the definition used for most of them is the minimum number k such that every {0,1}-matrix of size m×n with k 1s contains somewhere the all-1s matrix of size a×b as a submatrix. This is just the number in the previous definition plus 1.)

It took me some time to figure out, but it looks like Freywa is using a third definition: the minimum number k such that there exists a {0,1}-matrix of size m×n with k 1s which does not contain the all-0s matrix of size a×b as a submatrix. In the case of this post, only the cases a=b=3, m=n≥3 and a=b=4, m=n≥4 are considered. This is n² minus the number in the first definition, plus 1. Also, not only is k given, but also all possible matrices that achieve such k.

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Freywa
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Re: Zarankiewicz's problem

Post by Freywa » January 7th, 2022, 3:33 am

Now with the 2×2 case:

Code: Select all

x = 310, y = 228, rule = LifeHistory
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3.D.D3.A.A.A.A2.D$66.D2.A2.A.A.D14.D3.A.A.A3.D.D3.2A.A.A3.D$66.11D14.
D5.2A2.2AD.D2.A2.2A4.AD$66.D.A7.D14.D3.2A3.A.AD.D2.A.A2.A2.A.D$66.DA
8.D14.D.2A4.A2.AD.D.A3.A2.A.A.D$66.D3.A.A3.D14.D7.4AD.D.A2.A4.A.AD$
66.D2.A4.A.D14.13D.DA6.2A2.AD$66.D4.A.A2.D14.D.A.A.A5.D.DA5.A2.2A.D$
66.D2.A2.2A2.D14.D2A4.A4.D.14D$66.D4.2A2.AD14.D2.A2.A.A3.D$66.D3.A3.
2AD14.DA3.A2.A3.D$66.D6.3AD14.D3.2A3.A2.D$66.11D14.DA.A6.A.D$66.D.A7.
D14.D.A6.2A.D$66.DA8.D14.D2.2A6.AD$66.D4.A3.AD14.D4.A.A3.AD$66.D3.A3.
A.D14.D5.2A2.2AD$66.D2.A3.A2.D14.D7.4AD$66.D5.2A.AD14.13D$66.D4.2A.A.
D14.D.A6.3AD$66.D3.A2.A.AD14.DA4.3A3.D16.D.D.3D$66.D2.A2.A.A.D14.D4.A
2.A2.AD16.D.D3.D$66.11D14.D3.A2.A2.A.D16.3D.3D$66.D.A7.D14.D2.A2.A2.A
2.D18.D3.D$66.DA8.D14.D.A2.2A5.D18.D.3D$66.D4.A3.AD14.D.A.A3.A3.D$66.
D3.A3.A.D14.D.2A3.A4.D16.15D$66.D2.A3.A2.D14.DA3.A4.A.D16.DA12.D$66.D
6.3AD14.DA2.A4.A2.D16.D.A.A6.A.AD$66.D4.2A.A.D14.DA.A7.AD16.D.A2.A3.A
2.A.D$66.D3.A.2A2.D14.13D16.D.A3.A.A.A3.D$66.D2.A2.A2.AD43.D2.2A5.A.A
.D$66.11D43.D2.A.A2.A4.AD$66.DA8.D43.D2.A2.A2.A.A2.D$66.D5.A2.AD43.D
3.A2.3A4.D$66.D4.A2.A.D43.D4.A.A2.2A2.D$66.D3.A3.A.D43.D5.2A4.2AD$66.
D2.A3.A2.D43.D7.6AD$66.D.A4.A2.D43.15D$66.D4.3A2.D43.DA12.D$66.D2.2A
4.AD43.D.2A7.A.AD$66.D.A5.2AD43.D.A2.A3.A2.A.D$66.11D43.D.A3.A.A.A3.D
$66.DA8.D43.D2.2A5.A.A.D$66.D3.A2.A2.D43.D2.A3.3A4.D$66.D2.A3.A2.D43.
D3.2A2.A4.AD$66.D.A5.A.D43.D3.A.A2.A.A2.D$66.D4.A2.A.D43.D4.A.A2.2A2.
D$66.D5.A2.AD43.D5.2A4.2AD$66.D.2A5.AD43.D7.6AD$66.D3.2A3.AD43.15D$
66.D5.3A.D43.DA12.D$66.11D43.D.2A7.A.AD$66.DA8.D43.D.A.A5.A.A.D$66.D.
2A6.D43.D.A4.3A4.D$66.D.A4.A2.D43.D2.A.A3.A2.A.D$66.D4.A2.A.D43.D2.A
2.A.A.A3.D$66.D3.A4.AD43.D3.2A2.A4.AD$66.D5.A2.AD43.D3.A.A2.A.A2.D$
66.D2.A4.2AD43.D4.A.A2.2A2.D$66.D3.A2.2A.D43.D5.2A4.2AD$66.D4.3A2.D
43.D7.6AD$66.11D43.15D$66.DA8.D$66.D.A2.A4.D$66.D2.2A5.D$66.D2.A5.AD$
66.D.A5.A.D$66.D5.2A2.D$66.D5.A.2AD$66.D4.A.A.AD$66.D3.A2.2A.D$66.11D
!
Princess of Science, Parcly Taxel

Code: Select all

x = 31, y = 5, rule = B2-a/S12
3bo23bo$2obo4bo13bo4bob2o$3bo4bo13bo4bo$2bo4bobo11bobo4bo$2bo25bo!

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Freywa
Posts: 877
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Location: Singapore
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Re: Zarankiewicz's problem

Post by Freywa » January 18th, 2022, 1:29 pm

The values given by Guy in his 1967 and 1969 papers for z(2,2, 14,n) for n >= 24 are WRONG.

Code: Select all

x = 30, y = 76, rule = LifeHistory
26D$D3A3.2A16.D$DA2.2A3.2A14.D$DA9.5A9.D$DA14.5A4.D$D.A.A.A4.A4.A8.D$
D.A2.A6.A4.A3.A3.D$D.A6.A3.A4.A3.2A.D$D2.2A9.A4.A2.A2.D$D2.A.2A8.A2.A
6.D$D2.A6.A.A3.A6.2AD$D3.A2.A7.A4.2A.A.D$D4.A2.A2.A8.A.A.AD$D5.2A.A4.
A2.A6.AD$D5.A.A.A2.A5.A.A3.D$27D$D3A3.2A17.D$DA2.2A3.2A15.D$DA9.5A10.
D$DA14.5A5.D$D.A.A6.A4.A4.A3.AD$D.A2.2A5.A4.A8.D$D.A6.A3.A4.A3.2A2.D$
D2.2A.A6.A5.A6.D$D2.A.A8.A3.A5.2AD$D2.A6.A.A7.3A3.D$D3.A2.A7.A.A4.A.A
.D$D4.A2.A6.A3.A.A.A2.D$D5.2A2.A3.A.A6.A2.D$D5.A.2A.A8.A3.A.D$28D$D4A
22.D$DA3.5A17.D$DA8.5A12.D$DA13.5A7.D$D.A2.A4.A4.A4.2A5.D$D.A3.A4.A4.
A5.2A3.D$D.A4.A4.A4.A6.2A.D$D2.A3.A2.A7.A3.A3.AD$D2.A4.A4.A3.A2.A2.A
3.D$D2.A5.A.A7.A.A2.A2.D$D3.A.A6.A4.A2.A3.A.D$D3.A3.A5.2A6.A.A2.D$D3.
A4.A2.A3.A3.A5.AD$D4.A8.A4.A3.A.2AD$29D$DA2.5A19.D$DA7.5A14.D$DA12.5A
9.D$DA17.5A4.D$D.A.A4.A4.A4.A4.A3.D$D.A2.A4.A4.A4.A4.A2.D$D.A3.A4.A4.
A4.A4.A.D$D.A4.A4.A4.A4.A4.AD$D2.2A7.A2.A7.A2.A.D$D2.A.A7.2A6.A5.AD$D
2.A2.A3.A7.A3.A.A3.D$D2.A4.A2.A5.A.A5.A2.D$D6.A5.A2.A6.3A2.D$D7.2A8.A
.A5.2AD$30D$DA2.5A20.D$DA7.5A15.D$DA12.5A10.D$DA17.5A5.D$D.A.A4.A4.A
4.A4.A4.D$D.A2.A4.A4.A4.A4.A3.D$D.A3.A4.A4.A4.A4.A2.D$D.A4.A4.A4.A4.A
4.A.D$D2.2A5.A7.A3.A3.A2.D$D2.A.A7.A3.A3.A2.A4.D$D2.A2.A2.A5.A7.A3.A.
D$D2.A4.A3.A3.A2.A5.A3.D$D6.A5.2A8.A.2A.AD$D7.A2.A6.A.A3.A2.2AD$30D!
Princess of Science, Parcly Taxel

Code: Select all

x = 31, y = 5, rule = B2-a/S12
3bo23bo$2obo4bo13bo4bob2o$3bo4bo13bo4bo$2bo4bobo11bobo4bo$2bo25bo!

fluffykitty
Posts: 1175
Joined: June 14th, 2014, 5:03 pm
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Re: Zarankiewicz's problem

Post by fluffykitty » January 19th, 2022, 10:52 pm

Freywa wrote:
January 5th, 2022, 5:08 am
All non-isomorphic solutions to certain small square instances of the 3×3 and 4×4 Zarankiewicz's problem, as verified through SAT solving:

Code: Select all

hi
The 105 solution on the first row is constructible from the Fano plane. In this isomorphic form

Code: Select all

x = 15, y = 15, rule = b3s23
8b7o$4b7o$2b2o2b3o2b2o$2b4o2bo4b2o$bobobobobobobo$bob2obo2bo2bobo$b2o
2b2o3b2o2bo$b2obo2bo2bob2o$4o4b3o$2o2b2o2bo2b2o$2o4b3o4b2o$obobobo2bob
obo$obo2bobobo2bobo$o2b2o2bo2b2o2bo$o2bob2o3bob2o!
it can be describes as the block matrix [[0, 0s, 1s], [0s, 1s-(Fano plane incidence matrix), Fano], [1s, Fano, Fano]].

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Freywa
Posts: 877
Joined: June 23rd, 2011, 3:20 am
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Re: Zarankiewicz's problem

Post by Freywa » January 20th, 2022, 12:46 pm

fluffykitty wrote:
January 19th, 2022, 10:52 pm
The 105 solution on the first row is constructible from the Fano plane.
I already knew that – I have also proved that it is the unique extremal matrix (where graph isomorphism determines uniqueness) for z(3,3, 15,15).

The real challenge I'm facing now is to determine all non-isomorphic extremal matrices for z(2,2, 22,22) = 108. There are at least 10 of them.
Princess of Science, Parcly Taxel

Code: Select all

x = 31, y = 5, rule = B2-a/S12
3bo23bo$2obo4bo13bo4bob2o$3bo4bo13bo4bo$2bo4bobo11bobo4bo$2bo25bo!

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