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Question: Prove that a 2^n x 2^n courtyard can be tiled using L-shaped tiles of size 3 with a statue in the middle. (Use the Induction axiom) Click here to view the original question

What I want to know is, can't we just nest a square of size 2^n x 2^n on which the inductive hypothesis has been applied in the centre of a square of size 2^(n+1) x 2^(n+1)? In that case, p(n) => p(n+1), right? Meaning every subsequent block up to n+1 will have a statue in the centre.

I'm sorry if the way I phrased the questions isn't right.

atomic-coder
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  • If you put four $4\times 4$ squares together, each with statue near the middle, you will have an $8\times8$ square with four separated statues when you only want one statue – Henry May 09 '22 at 21:19
  • [[# ,# ,# ,#] [# ,* ,S ,#] [# ,* ,*, #] [# ,# ,#, #]]

    This is what I meant. n = 2, which means the 2 x 2 square with a statue in its middle(i.e., the 2^0 x 2^0 square('S')) is in the middle of the 2^2 x 2^2 square. (Just view the nested list in 2d for clarification). I didn't mean 4 4x4 squares, I meant just one in the middle of the 8x8 square.

     – atomic-coder May 10 '22 at 06:15

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