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enter image description here

In the above picture, there are 24 squares. Can you only use L trominos to fill the figure? If yes, give an example. Otherwise, please explain why.

An L tromino is like this:

enter image description here

ACB
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Culver Kwan
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  • It is quite hard. – Culver Kwan Oct 16 '19 at 13:52
  • Can you rotate or mirror the L tromino? – Wheat Wizard Oct 17 '19 at 14:11
  • The standard name for these is "triomino", not "tromino". – Paul Sinclair Oct 17 '19 at 17:10
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    @PaulSinclair Tromino is more correct in my experience, and seems to jive with https://en.wikipedia.org/wiki/Tromino. Triominoes appears to refer to triangular shaped pieces used to play a game similar to dominoes. – MassDefect Oct 17 '19 at 23:10
  • @MassDefect - well, then, I'll acquiesce. Thanks for the correction. – Paul Sinclair Oct 17 '19 at 23:40
  • @All you can only fill the figure with max of 7 L trimos and you will left with 3 empty square( and I have not tried all the ways but I just showed few but I can assume it will always be left with 3 blank squares which will either vertical in a row or horizontal in a row, you can try to prove me wrong with reasoning that it is not a solution), yes, this puzzle is a way simpler even a kid can do it but yes I do agree magma gave more elegant solution. it is just like everyone knows from the start that everything falls down to the earth but newton gave the gravity :) – Sayed Mohd Ali Oct 18 '19 at 06:03
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    @PaulSinclair I'd always heard them referred to as "triominoes" too, but it looks like Martin Gardner used "tromino". (I can't find an easy reference for what Golomb used, other than one article which just has "3-omino".) Thinking about it, "tromino" fits better with "domino". – Especially Lime Oct 18 '19 at 08:28

3 Answers3

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Answer:

No, it's not possible.

Reasoning:

Consider the 9 marked squares in the following image:
enter image description here
Each L-tromino can only fill 1 of these squares, so you need at least 9 L-trominos. However, those 9 trominos will have 27 squares in total, and there are only 24 free squares in the figure.

Magma
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Note:

This is not a popular way of approaching this, and I personally love @Magma's proof, yet I have to disagree.

Answer:

solution
The problem as stated is perfectly solveable, since nowhere in the rules it states, that you are limited by the borders of the 5x5 quare. So if you are to leave some hangover beyond the bounding box of the original shape, you are easily able to cover the original square, but you will have these ugly hangover parts left. I do know that this is not a very elagant solution, but it is a valid one.

But I do agree that the problem as it is (probably) meant to be solved is impossible.

ACB
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Chund
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    yes, it is not written you are limited to 5*5 but it is written there you need to fill the below figure with L trimos. if you change the square position the figure will also change. and yes, it is possible to be filled with L trimos if you change the figure. – Sayed Mohd Ali Oct 17 '19 at 13:09
  • As I stated. Not elagant at all but the original shape is filled. And there are pieces of the tiles which just hang over the border of the original shape – Chund Oct 17 '19 at 13:11
  • okay I got you now +1 – Sayed Mohd Ali Oct 17 '19 at 13:15
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    Sorry i hope my edit makes it more clear now – Chund Oct 17 '19 at 13:19
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    It's also not stated that the triominos can't overlap or that they can't cover the black square. You could just put down 24 of them, with each bend covering a square. Heck it doesn't even say what size triominos you ned. You could cover the whole thing with one giant triomnio, or you could use 1/6 size triominos and fill the figure with 288 of them with no overlap or hangover. – user3294068 Oct 17 '19 at 13:44
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    Yes that is exactly right, that's why I mentioned, that this is quite a ugly way to look at the problem – Chund Oct 17 '19 at 14:00
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I was going to post a different answer, only to realize it was the same as Magma's. So I had to find a new one:

If one of the squares in a tromino is at a corner, it being the tromino's middle square is the better option, because otherwise it would always force another tromino to close the gap, forming a 3x2 rectangle combined with the first (1s and 2s together):

enter image description here

With the corner being in the middle, this option is also available, but not necessarily the only one. Hovever, even this advantage isn't enough to reach our goal:

enter image description here

Nautilus
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  • Could someone please explain what's wrong? – Nautilus Oct 17 '19 at 17:12
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    Does it really make a difference? It applies to ANY corner of the grid, not just the upper-right one. I just started from the corners with 1, 2, 3 and 4s, then filled the rest. – Nautilus Oct 17 '19 at 18:14
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    Oh I see now what you are getting at. It would help if you put in a diagram with just one piece in each corner, and write that the corners can therefore be assumed to be filled in this way. This can be assumed "without loss of generality" as a mathematician would say. Once those corners are filled, there is hardly any choice in filling the rest until it fails. I'll delete my previous comment. – Jaap Scherphuis Oct 17 '19 at 18:29
  • @Nautilus +1 I think most of the people will not get it you should give a diagram – Sayed Mohd Ali Oct 18 '19 at 05:58
  • @SayedMohdAli Done. – Nautilus Oct 18 '19 at 08:10
  • I'm not 100% clear why that initial 2x3 is so bad? You seem to be taking it for granted that your first diagram is a clear fail but I think you need to explain why that is so because I certainly can't see it as something obvious. I think I may have worked it out in my head but a proof shouldn't leave me needing to do a bunch of reasoning! – Chris Oct 18 '19 at 11:45
  • This is roughly how I solved it myself as well, although I think this can be presented a bit clearer. "Better option" doesn't sound quite right to me. What I think you should say instead is that to cover both corners of a long side of a 3x2 rectangle, you must cover the entire rectangle with 2 trominos. That this can be done in two ways doesn't matter, because in any case the 3x2 rectangle is filled, and this "reduces" the figure to a 3x1 rectangle, which can't be filled. – Discrete lizard Oct 18 '19 at 13:33
  • @Chris It may not really be a "clear fail" per se, but it just offers nothing the other option can't. Starting from a corner, you could fill the 3x2 rectangle the same way in the second diagram if you like or even avoid completing one. The first diagram shows you're forced to do the former unlike the second. – Nautilus Oct 18 '19 at 14:48
  • You have made me realise in thinking that if you discussed the top right corner and the forced piece (1 in your picture) and then work from that instead of an empty 5x5 grid as in your first picture you can prove that you must have the 3x2 in there. as you marked in your first picture no matter how you place that first corner piece. You can then look at the remaining 3x5 and conclude again that a 2x3 is the only option leaving you with a 3x3 to solve which the same logic can then be applied leading to a 3x1 which is clearly impossible. – Chris Oct 21 '19 at 11:30