Prove that a 4 x 11 rectangle cannot be tiled by L-shaped tetrominoes.
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$\begingroup$ I have no idea how to start the proof. $\endgroup$– Tyler LewandowskiCommented Jul 25, 2019 at 4:29
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$\begingroup$ I can't resist linking to this (mildly embarrassing for me) old thread. $\endgroup$– Jyrki LahtonenCommented Jul 25, 2019 at 4:48
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$\begingroup$ But the question is in need of a bit of context as outlined in our guide fore new askers :-( $\endgroup$– Jyrki LahtonenCommented Jul 25, 2019 at 4:50
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1 Answer
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I am considering a grid with $4$ column and $11$ rows. Now, color the first and the third row with black color and the second and fourth with white. Clearly, whenever you place a $L$-Shaped domino there, it will cover either $(\hbox{3 black position and 1 white})$ or $(\hbox{1 black and 3 white})$. However, the number of black and white position is same in the hole grid. Thus, $L$-Shaped domino must exist in pairs. To be more precise, for every domino covering $3$ black and $1$ white space, there must be another domino covering $1$ black and $3$ white positions. Thus, total number of $L$-Shaped domino must be even. But each $L$-Shaped domino has $4$ blocks in it. Thus total number of blocks that an even number of dominoes can cover will be a multiple of $4\times 2=8$. But total number of blocks in the grid is $44$ which is not a multiple of $8$. Hence, a $4\times 11$ board cannot be covered with $L$-shaped dominoes.
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1$\begingroup$ woah..Nice solution ! *and someone says he is bad in combi -_- * $\endgroup$ Commented Oct 3, 2020 at 3:45
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1$\begingroup$ @SunainaPati Well, I'm not as good as you :P Thanks though :D $\endgroup$– AnandCommented Oct 3, 2020 at 4:18