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:
$\begingroup$
$\endgroup$
7
asked Oct 16, 2019 at 13:45
-
$\begingroup$ It is quite hard. $\endgroup$– Culver KwanCommented Oct 16, 2019 at 13:52
-
$\begingroup$ Can you rotate or mirror the L tromino? $\endgroup$– Wheat WizardCommented Oct 17, 2019 at 14:11
-
$\begingroup$ The standard name for these is "triomino", not "tromino". $\endgroup$– Paul SinclairCommented Oct 17, 2019 at 17:10
-
2$\begingroup$ @PaulSinclair Tromino is more correct in my experience, and seems to jive with en.wikipedia.org/wiki/Tromino. Triominoes appears to refer to triangular shaped pieces used to play a game similar to dominoes. $\endgroup$– MassDefectCommented Oct 17, 2019 at 23:10
-
1$\begingroup$ @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". $\endgroup$– Especially LimeCommented Oct 18, 2019 at 8:28
|
Show 2 more comments
3 Answers
$\begingroup$
$\endgroup$
3
Answer:
No, it's not possible.
Reasoning:
Consider the 9 marked squares in the following image:
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.
-
13
-
10
-
3
$\begingroup$
$\endgroup$
6
Note:
This is not a popular way of approaching this, and I personally love @Magma's proof, yet I have to disagree.
Answer:
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.
-
1$\begingroup$ 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. $\endgroup$ Commented Oct 17, 2019 at 13:09
-
$\begingroup$ 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 $\endgroup$– ChundCommented Oct 17, 2019 at 13:11
-
1$\begingroup$ Sorry i hope my edit makes it more clear now $\endgroup$– ChundCommented Oct 17, 2019 at 13:19
-
4$\begingroup$ 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. $\endgroup$ Commented Oct 17, 2019 at 13:44
-
1$\begingroup$ Yes that is exactly right, that's why I mentioned, that this is quite a ugly way to look at the problem $\endgroup$– ChundCommented Oct 17, 2019 at 14:00
$\begingroup$
$\endgroup$
9
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):
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:
-
$\begingroup$ Could someone please explain what's wrong? $\endgroup$– NautilusCommented Oct 17, 2019 at 17:12
-
1$\begingroup$ 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. $\endgroup$– NautilusCommented Oct 17, 2019 at 18:14
-
1$\begingroup$ 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. $\endgroup$ Commented Oct 17, 2019 at 18:29
-
$\begingroup$ @Nautilus +1 I think most of the people will not get it you should give a diagram $\endgroup$ Commented Oct 18, 2019 at 5:58
-