A191873 A problem of Zarankiewicz: maximal number of 1's in an n X n matrix of 0's and 1's with 0's on the main diagonal and no "rectangle" with 1's at the four corners.

0, 2, 6, 9, 12, 16, 21, 24, 29, 34, 39, 45

Offset: 1

Author

R. H. Hardin and N. J. A. Sloane, Jun 18 2011

Keywords

• nonn
• more

Comments

In other words, the pattern

1...1

.....

1...1

is forbidden.

From Don Knuth, Aug 13 2014: (Start)

Well, it is well known from A001197 that a(8)<25. A001197 is essentially the same problem, but increased by 1, and without restricting the diagonals. The diagonal restriction is however of little interest, because it's easy to permute rows and columns and get all 0's or all 1's or probably any of the 2^n possible settings of the diagonal. At least, this is true when n=8; hence a(8) in this sequence is 24.

Transposing cols 1<->4 and 5<->8 in the example by Guy 1967 page 130 as cited in A001197 gives a(8)=24:

01110000

10010100

00010011

01000101

10100010

11001000

00101001

00001110

But as stated above, I think this is quite trivial, and I believe this sequence should be downplayed. Readers should look at the sequence A001197 --- that's what Zarankiewicz's problem asked for in 1951, namely the min number that forces a rectangle, not the max number that doesn't exclude it. (End)

Conjecture: for n >= 3, a(n) = A072567(n) = A001197(n) - 1 (per above comment). - Max Alekseyev, Jan 29 2022

References

• B. Bollobas, Extremal Graph Theory, pp. 309ff.

Links

• D. Bienstock and E. Gyori, An extremal problem on sparse 0-1 matrices, SIAM J. Discrete Math. 4 (1991), 17-27.

Crossrefs

Cf. A072567, A191874, A191965, A191966.

Extensions

a(8) confirmed, a(9)-a(12) added by Max Alekseyev, Feb 07 2022