Polyomino question

[Dr. Gordon Hamilton]

Jon,

Corerect, starting with the trominoes you get:

The 2 trominos:
OO
O

OOO

The 2 tetrominos that hold them:
O
OO
O

OO
O
O

The 4 pentominos that hold these:
OO
OO
O

OO
   OO
   O

O
OOO
O

O
OO
O
O

The 5 heptominos that hold these:

O
OOO
OO
O

OO
OOO
O
O

   O
OOO
   OO
   O

OOO
   OO
   O
   O

   OO
OOO
   O
   O

Some of these type of sequences end with repetition because there is a 
single n-omino which holds the previous set. Does the sequence above end in 
repetition? What about a sequence beginning with all the k-ominos?

Do tell me more about the yahoo polyomino group... a google search brought 
up a motley mix of hits. Who would I approach?

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