Blocking Polyominoes

Blocking Polyominoes
PUZZLE FUN Future Edition.

Unrestricted Problem: by Rodolfo Kurchan

In my bulletin of Los Acertijeros 74 of December 1995 I asked which is the smallest polyomino that can block itself using 2 replicas, and then the same for 3,4,5 and N replicas.
The only condition is that if we take out 1 piece there could not be part of the figure still blocked. The solutions are:

N = 2: an Octomino:	N = 3: a 10-omino:
N = 4: an hexomino (I think there are 3 different):	N = 5: a 9-omino: by Erich Friedman
N = 6: a 9-omino:	N = 7: a 7-omino: by Erich Friedman
N = 8: a 6-omino: by Erich Friedman	N = 9: a 7-omino: by Erich Friedman
N >= 10: a 6-omino: by Erich Friedman:
Example, N = 11:

The best solution we have are:

N	2	3	4	5	6	7	8	9	>
Pieces	8	10	6	9	9	7	6	7	6

New Pentomino Problem: by Rodolfo Kurchan

It is impossible to block a pentomino using replicas of itself. Find a solution using the least quantity of pentominoes (of course you have to use at least 1 pentomino different).

Restricted Problem: by Junk Kato and Patrick Hamlyn

In this variation we look for the least number of replicas that we need to block each polyomino.
The hexominoes of N=4 would not be solution because we can move together the 2 blue pieces to the left:

So to be a good solution to this puzzle we cannot move 1 or more pieces of the figure.
The best solutions for this problem are:

N = 2: an Octomino:	N = 3: a 10-omino:
N = 4: a 11-omino:	N = 5: a 12-omino: by Junk Kato
N = 6: a 11-omino:	N = 7: a 11-omino: by Junk Kato
N = 8: a 10-omino:	N = 9: a 11-omino:
N = 10: a 21-omino: by Erich Friedman

Erich found a generalization from his solution by fitting in more stacked vertical polyominoes on the right side.
The best solution we have are:

N	2	3	4	5	6	7	8	9	>
Pieces	8	10	11	12	11	11	10	11	3n-9

Please any comment or improvement write me to:

Rodolfo Marcelo Kurchan / Lavalle 3340 21"3" / (1190) Buenos Aires - Argentina
eMail: rodolfo@kurchan.com.ar
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