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2x1 bricks solution
3x1 bricks solution
2x1 can be done with a 5x6 rectangle. 3x1 can be done with a 7x9 rectangle. I'm fairly sure these are the best solutions (though they are not unique). I do not think I could prove it.
Bonus points :-)
ReplyDeleteI have a sol'n for 4×1 blocks which is nicely rotationally symmetrical and can be extended to any n×1.
It fills a 9×12 rectangle which tallies nicely with the progression 5×7 -> 7×9 -> 9×12 but I don't know if it's minimal.
AAAABACBBBBC
BCBCBACAAAAC
BCBCBACBBBBC
BCBCBACAAAAC
BCBCDDDDCBCB
CAAAACABCBCB
CBBBBCABCBCB
CAAAACABCBCB
CBBBBCABAAAA
Actually, I think 9×12 is minimal for 4×1 blocks, and I think that I can show it. I'll use my sol'n for 2×1 and generalise from there.
ReplyDeleteHere is my 2×1 sol'n:
AAXFFH
BCXGGH
BCMMNQ
ZYYONQ
ZWWOPP
The smallest rectangle that can be made is 2×3, represented by
AA
BC
BC
There must be a block (XX) to cut off the fault line under AA, and there must be a block to cut off the potential new fault line down the left of XX. That gives us
AAX
BCX
BC-
-YY
(-YY could go above AAX - the same logic applies but then the sol'n reached is the one you gave).
At a minimum, there must also be blocks at ZZ and WW. This gives us a side of 5. There must also be a block at MM. So:
AAX
BCX
BCMM
ZYY
ZWW
In order to cut off the new fault line under MM, blocks must go at NN and then OO, PP, and QQ follow.
AAX
BCX
BCMMNQ
ZYYONQ
ZWWOPP
The top-right can be filled in without creating any more fault-lines so we have a length of 6 as the minimum for the other side.
Similar reasoning can be applied to creating sol'ns for 3×1 and 4×1 (and looks like it can be applied to all n×1).