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Handmade Puzzle!

1. Notation

Covarable - If a figure A can be covered with figure B, without double (or more)-covered areas, allowing rotations and flips, then A is coverable with B.

Non-Coverable - Not coverable.

2. Examples

  • Chessboards (8x8) are coverable with 2x1-sized dominos. (Maybe this could be a small hint - chessboard generally has some colors with specific pattern…)

  • 7x7 squares are non-coverable with 2x1-sized dominos.

3. Puzzle!

Given figure (figure B):

Figure B

Puzzles (figure A):

1.

Figure A - 1.

2.

Figure A - 2.

Some methods work here, I recommend finding / using the method(s).

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  • $\begingroup$ RDK: You were wondering if there are cases where the methods (coloring/numbering) used here don’t work. I suggest you ask that as a separate question. $\endgroup$ Commented Feb 24 at 5:49

3 Answers 3

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coverable

Covering 2 is not possible because each tile (figure B) covers two white squares and two grey squares, whereas figure A has 9 grey squares and 7 white squares.

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  • $\begingroup$ Wow, another coloring! Nice solution. (Surprisingly, my intended solution wasn’t this…) $\endgroup$ Commented Feb 24 at 2:14
  • $\begingroup$ @RDK What was it? Are you planning to post it? $\endgroup$ Commented Feb 24 at 2:23
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    $\begingroup$ Yes, check my solution! $\endgroup$ Commented Feb 24 at 2:35
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Here's another approach for number 2:

Denote the cell in the ith row and jth column as RiCj. Denote the fact that B does not have 3 cells in a single line as *. We want to prove that A is not coverable with B, so instead assume that A is coverable with B. Then R3C5 must be part of a copy of B. So R3C4 is forced to be joined to R3C5. R3C3 cannot be joined to them by *, so R4C4 and R4C3 must be joined to R3C4.

This forces R5C3 to be part of another copy of B, which forces R5C2 to be joined to R5C3. R5C1 cannot be joined by *, so R4C2 is forced to be joined. And R3C2 cannot be joined by * again, so R4C1 is forced to be joined. This leaves R5C1 as a single-cell group, so it is not part of a copy of B, contradicting our assumption. So R3C5 is in fact not part of a copy of B, again contradicting our assumption. Hence our assumption is wrong, and so A is not coverable with B.

Visual explanation:

Visual explanation of the answer

The square labeled "A" can only be covered by the red tetromino. Therefore, the square labeled "B" can only be covered by the orange tetromino. The bottom left square is then impossible to cover.

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    $\begingroup$ +1 for evidentsolution, now 1st text explanation not needed anymore. $\endgroup$ Commented Feb 23 at 22:35
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    $\begingroup$ This solution is great, but I think this method may not apply in other sophisticated cases. Still great explanation, thanks!(+1) $\endgroup$ Commented Feb 24 at 2:15
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Surprisingly, another method-using solutions appeared… Here is my intention.

Figure A-1.

Figure A-2.

Each figure B should contain all of 1~4, but this figure has 5 of 4s and 3 of 2s.

Anyway, @Pranay colored 1s and 4s as white, 2s and 3s as black. This may be simpler, I was surprised.

Well I believe @Pranay’s solution is as simple as (or simpler than) my intention…

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