[seqfan] Re: Unique Products Regarding Binary Matrices

hv at crypt.org hv at crypt.org
Fri Jun 11 18:54:26 CEST 2010

```Leroy Quet <q1qq2qqq3qqqq at yahoo.com> wrote:
:For those of you who are confused about what I am asking, here is a sample 9-by-9 grid with just 16 distinct values that occur. (14 products don't occur elsewhere.)
:There seem to be a number of 16-unique-products solutions, perhaps.
:
:1 0 1 0 0 0 1 1 0
:1 0 0 1 1 1 1 1 0
:0 1 1 1 0 0 0 0 0
:1 0 0 1 1 0 0 0 0
:0 0 0 0 0 0 0 0 0
:0 0 0 1 1 0 0 0 1
:0 0 0 0 0 0 0 1 1
:0 0 0 1 1 0 0 1 1
:1 1 1 0 0 0 1 1 1
:
:The row products are:
:1*1*1*3*2*1 = 6, 1*2*5*1 = 10, 1*3*5 = 15, continuing:
:16, 9, 18, 14, 24, 27.
:
:And the column products:
:2*1*1*4*1 = 8, 2*1*5*1 = 10, 1*1*1*5*1 = 5, continuing:
:3, 1, 7, 12, 24, 20
:
:So, the products 2 and 4 don't occur in my partial "solution".

Ah, I understand.

You originally wrote:

:Say we have an n-by-n binary matrix (all elements either 0 or 1).
:
:For a given row of the matrix, take the lengths of the runs of 0's and 1's and multiply these lengths.
:(By "run", it is meant a string of consecutive elements in the row (or column) all of the same value b, bounded by the value 1-b or by the edge of the row (or column).)
:
:Do this for all rows and all columns to get 2n products.
:
:Let a(n) = the number of such n-by-n binary matrices such that the 2n products are all unique.

If you replace "all unique" at the end with "distinct", I no longer suffer
any confusion.

(I thought you were looking for nxn matrices whose 2n products were not
the products of any other nxn matrix.)

Hugo

```