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 science_man_88 2019-01-05 12:46

here's another thing you all may have noticed neither A or B can have more than 2 values mod 4 within them.

 henryzz 2019-01-05 17:06

Have found plenty of 4+3 solutions but no 4+4 yet. Not sure I can search much further

 henryzz 2019-01-07 09:57

[QUOTE=henryzz;505034]Have found plenty of 4+3 solutions but no 4+4 yet. Not sure I can search much further[/QUOTE]

Turns out I had a bug. 4+4 solutions found.

Something I hadn't realized is that if A=[a,b,c,d] and B=[e,f,g,h] is a solution then A=[a-1,b-1,c-1,d-1] and B=[e+1,f+1,g+1,h+1] is a solution. I think this means that you can set a=1 and generate all solutions based on this.

 axn 2019-01-07 10:44

[QUOTE=henryzz;505189]I think this means that you can set a=1 and generate all solutions based on this.[/QUOTE]
Or set a=0. This way, all the entries in the other set will be squares.

 henryzz 2019-01-07 12:47

Unfortunately I have found a bug that invalidates my solutions.

@axn that is potentially a useful idea.

 Dieter 2019-01-13 07:46

[FONT=Calibri][SIZE=3]Eight days ago, I have found a 4+4 solution and a few hours later a 4+5 solution - sufficient for a "*". But the search for a 4+6 solution - [a1,a2,a3,a4] and [b1,b2,b3,b4,b5,b6] with ai+bj = square number [/SIZE][SIZE=3]– [/SIZE][SIZE=3]seems to be difficult. Until now I have only three “quasi-solutions” with 23 of 24 fulfilled conditions: [/SIZE][/FONT]
[FONT=Calibri][SIZE=3]cij: = ai +bj = a square number is correct [/SIZE][SIZE=3]for i = 1 to 4 and j = 1 to 6; only [/SIZE][/FONT]
[FONT=Calibri][SIZE=3]a4 + b6 is no square number. [/SIZE][/FONT]
[SIZE=3][FONT=Calibri]I would like to know which sort of solutions other people with “*” have sent to the puzzle master. [/FONT][/SIZE]

 uau 2019-01-13 17:04

[QUOTE=Dieter;505742][SIZE=3][FONT=Calibri]I would like to know which sort of solutions other people with “*” have sent to the puzzle master. [/FONT][/SIZE][/QUOTE]
I've found a 4+6 solution.

 petrw1 2019-01-13 20:20

The age old binary question: 1 or 0

Just curious if you are having more luck if a1=1 or a1=0?
I found a (4,3) solution quite easily with a1=1.

I guess the third option could be: a1 != {0,1}

 Dieter 2019-01-14 07:50

[QUOTE=petrw1;505800]Just curious if you are having more luck if a1=1 or a1=0?
I found a (4,3) solution quite easily with a1=1.

That's no question for me, because I don't use ai and bj as parameters. My parameters for the search are the square numbers cij = ai + bj.

I don't know, if I am allowed to write more here.

 henryzz 2019-01-14 11:44

[QUOTE=petrw1;505800]Just curious if you are having more luck if a1=1 or a1=0?
I found a (4,3) solution quite easily with a1=1.

I guess the third option could be: a1 != {0,1}[/QUOTE]

The value for a1 shouldn't make a difference. If you add 1 onto all As and subtract 1 from all Bs then you have a new solution. As such something needs to be fixed. This can be 0 or 1.

 petrw1 2019-01-14 16:03

[QUOTE=henryzz;505860]The value for a1 shouldn't make a difference. If you add 1 onto all As and subtract 1 from all Bs then you have a new solution. As such something needs to be fixed. This can be 0 or 1.[/QUOTE]

Good point .... or if you have a 4X4 solution put a zero in one list.

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