Polyomino question 2

[Dr. Gordon Hamilton]

Darn it - you're right after all - a 9-square solution does contain all the 
pentominoes. Amazing - the
school children usually pick up on my mistakes, but this time I'll have to 
tell them a real mathematician found it ;)



Here is another tough sequence (see "un-Babushka Squares" on 
www.galileo.org/math/puzzles.html)

A polyomino eats smaller polyominos that fit into it.  I want an infinite 
set of such polyominos none of which eat each other and built by maximizing 
the number of triominos, P(3), then by maximizing the number of tetrominos, 
P(4), then by maximizing the number of pentominos, P(5), etc.

All I have is that P(1) = P(2) = P(3) = 0 and P(4) = 2.

Gord!





> On Tue, 28 Feb 2006, Dr. Gordon Hamilton wrote:
>
>> The best that can be done is 10. There are 11 different solutions. The 
>> 9-solution below isn't right because it does not include the L polyomino. 
>> I have been presenting this problem to elementary school students in 
>> Calgary, Canada for the last year. You can see the problem at: 
>> www.galileo.org/math/puzzles.html under the name "Babushka Squares".
>
> Dear Gord - I hope they don't make you go back and visit all the primary 
> schools to show the kids the solution with 9 squares!
>
> I looked at the other problems you have for children at that site, and I 
> think it's great if you can get kids to puzzle over these kinds of 
> things - certainly more fun than anything I ever did at school in maths. 
> The extension to that polyomino problem at your site contains another good 
> sequence for the encyclopedia, if anyone would like to calculate it:
>
> start with the set S0 of all (i.e. both) trominoes. Then find the set S1 
> of all smallest n1-ominoes that contain any member of S0. Then find the 
> set S2 of all smallest n2-ominoes that contain any member of S1. Iterate. 
> This leads to two sequences: 3, n1, n2, n3...  and  2, #S1, #S2, #S3...
>
> (From Gord's hint it seems the sequences start 3,4,5,7... and 2,2,4...)
>
> I bet someone from the yahoo polyomino group can attack this with one of 
> their solving problems and give us quite an extended answer in no time.
>
> -Jon
>
>
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