Tiling a 7x9 rectangle with 2x2 squares and L-shaped trominos
It's possible to cover a 7x9 rectangle using 0 2x2 squares and 21 L-shaped trominos, for example:
abbcddeef
aabccdeff
gghiijjkk
glhhimjnk
lloppmmnn
qoorpstuu
qqrrssttu
It's also possible to cover it with 3 2x2 squares and 17 L-shaped trominos, for example:
aabbccdee
fagbcddee
ffgghhijj
kkllhhiij
kklmmnnoo
ppqrmsnot
pqqrrsstt
However, it is not possible to use more than 3 2x2 squares, that is, the remaining four combinations don't work:
6 2x2 squares and 13 L-shaped trominos
9 2x2 squares and 9 L-shaped trominos
12 2x2 squares and 5 L-shaped trominos
15 2x2 squares and 1 L-shaped tromino
...at least this what I got using brute force (a simple C program based on backtracking). I tried to give a mathematical proof for this, but I couldn't.
Please help me find a proof which explains why more than 3 2x2 squares cannot be used. Thank you for your help in advance!
combinatorial-geometry rectangles tiling
add a comment |
It's possible to cover a 7x9 rectangle using 0 2x2 squares and 21 L-shaped trominos, for example:
abbcddeef
aabccdeff
gghiijjkk
glhhimjnk
lloppmmnn
qoorpstuu
qqrrssttu
It's also possible to cover it with 3 2x2 squares and 17 L-shaped trominos, for example:
aabbccdee
fagbcddee
ffgghhijj
kkllhhiij
kklmmnnoo
ppqrmsnot
pqqrrsstt
However, it is not possible to use more than 3 2x2 squares, that is, the remaining four combinations don't work:
6 2x2 squares and 13 L-shaped trominos
9 2x2 squares and 9 L-shaped trominos
12 2x2 squares and 5 L-shaped trominos
15 2x2 squares and 1 L-shaped tromino
...at least this what I got using brute force (a simple C program based on backtracking). I tried to give a mathematical proof for this, but I couldn't.
Please help me find a proof which explains why more than 3 2x2 squares cannot be used. Thank you for your help in advance!
combinatorial-geometry rectangles tiling
1
What's the question?
– Lord Shark the Unknown
Jan 4 at 19:29
@LordSharktheUnknown The question is, how can this claim be proved?
– kol
Jan 4 at 19:31
What's "the normal way" to prove this that you refer to?
– Michael Lugo
Jan 4 at 20:14
@MichaelLugo I would like to find a mathematical proof, because I don't like computer based proofs. My C code tried every possible way to find a tiling with a given number of 2x2 squares and L-trominos, but it does not tell anything about why one cannot use more than 3 2x2 squares.
– kol
Jan 4 at 20:20
1
Tile homology groups can be useful for these types of questions. cflmath.com/Research/Tilehomotopy/tilehomotopy.pdf
– Cheerful Parsnip
Jan 5 at 2:17
add a comment |
It's possible to cover a 7x9 rectangle using 0 2x2 squares and 21 L-shaped trominos, for example:
abbcddeef
aabccdeff
gghiijjkk
glhhimjnk
lloppmmnn
qoorpstuu
qqrrssttu
It's also possible to cover it with 3 2x2 squares and 17 L-shaped trominos, for example:
aabbccdee
fagbcddee
ffgghhijj
kkllhhiij
kklmmnnoo
ppqrmsnot
pqqrrsstt
However, it is not possible to use more than 3 2x2 squares, that is, the remaining four combinations don't work:
6 2x2 squares and 13 L-shaped trominos
9 2x2 squares and 9 L-shaped trominos
12 2x2 squares and 5 L-shaped trominos
15 2x2 squares and 1 L-shaped tromino
...at least this what I got using brute force (a simple C program based on backtracking). I tried to give a mathematical proof for this, but I couldn't.
Please help me find a proof which explains why more than 3 2x2 squares cannot be used. Thank you for your help in advance!
combinatorial-geometry rectangles tiling
It's possible to cover a 7x9 rectangle using 0 2x2 squares and 21 L-shaped trominos, for example:
abbcddeef
aabccdeff
gghiijjkk
glhhimjnk
lloppmmnn
qoorpstuu
qqrrssttu
It's also possible to cover it with 3 2x2 squares and 17 L-shaped trominos, for example:
aabbccdee
fagbcddee
ffgghhijj
kkllhhiij
kklmmnnoo
ppqrmsnot
pqqrrsstt
However, it is not possible to use more than 3 2x2 squares, that is, the remaining four combinations don't work:
6 2x2 squares and 13 L-shaped trominos
9 2x2 squares and 9 L-shaped trominos
12 2x2 squares and 5 L-shaped trominos
15 2x2 squares and 1 L-shaped tromino
...at least this what I got using brute force (a simple C program based on backtracking). I tried to give a mathematical proof for this, but I couldn't.
Please help me find a proof which explains why more than 3 2x2 squares cannot be used. Thank you for your help in advance!
combinatorial-geometry rectangles tiling
combinatorial-geometry rectangles tiling
edited Jan 4 at 20:30
kol
asked Jan 4 at 19:28
kolkol
24828
24828
1
What's the question?
– Lord Shark the Unknown
Jan 4 at 19:29
@LordSharktheUnknown The question is, how can this claim be proved?
– kol
Jan 4 at 19:31
What's "the normal way" to prove this that you refer to?
– Michael Lugo
Jan 4 at 20:14
@MichaelLugo I would like to find a mathematical proof, because I don't like computer based proofs. My C code tried every possible way to find a tiling with a given number of 2x2 squares and L-trominos, but it does not tell anything about why one cannot use more than 3 2x2 squares.
– kol
Jan 4 at 20:20
1
Tile homology groups can be useful for these types of questions. cflmath.com/Research/Tilehomotopy/tilehomotopy.pdf
– Cheerful Parsnip
Jan 5 at 2:17
add a comment |
1
What's the question?
– Lord Shark the Unknown
Jan 4 at 19:29
@LordSharktheUnknown The question is, how can this claim be proved?
– kol
Jan 4 at 19:31
What's "the normal way" to prove this that you refer to?
– Michael Lugo
Jan 4 at 20:14
@MichaelLugo I would like to find a mathematical proof, because I don't like computer based proofs. My C code tried every possible way to find a tiling with a given number of 2x2 squares and L-trominos, but it does not tell anything about why one cannot use more than 3 2x2 squares.
– kol
Jan 4 at 20:20
1
Tile homology groups can be useful for these types of questions. cflmath.com/Research/Tilehomotopy/tilehomotopy.pdf
– Cheerful Parsnip
Jan 5 at 2:17
1
1
What's the question?
– Lord Shark the Unknown
Jan 4 at 19:29
What's the question?
– Lord Shark the Unknown
Jan 4 at 19:29
@LordSharktheUnknown The question is, how can this claim be proved?
– kol
Jan 4 at 19:31
@LordSharktheUnknown The question is, how can this claim be proved?
– kol
Jan 4 at 19:31
What's "the normal way" to prove this that you refer to?
– Michael Lugo
Jan 4 at 20:14
What's "the normal way" to prove this that you refer to?
– Michael Lugo
Jan 4 at 20:14
@MichaelLugo I would like to find a mathematical proof, because I don't like computer based proofs. My C code tried every possible way to find a tiling with a given number of 2x2 squares and L-trominos, but it does not tell anything about why one cannot use more than 3 2x2 squares.
– kol
Jan 4 at 20:20
@MichaelLugo I would like to find a mathematical proof, because I don't like computer based proofs. My C code tried every possible way to find a tiling with a given number of 2x2 squares and L-trominos, but it does not tell anything about why one cannot use more than 3 2x2 squares.
– kol
Jan 4 at 20:20
1
1
Tile homology groups can be useful for these types of questions. cflmath.com/Research/Tilehomotopy/tilehomotopy.pdf
– Cheerful Parsnip
Jan 5 at 2:17
Tile homology groups can be useful for these types of questions. cflmath.com/Research/Tilehomotopy/tilehomotopy.pdf
– Cheerful Parsnip
Jan 5 at 2:17
add a comment |
1 Answer
1
active
oldest
votes
This answer (to a different tiling problem) gave me the idea to use more than 2 colors to color the 7x9 rectangle (I have already tried the chessboard coloring without any success).
I devised a coloring scheme with which every 2x2 square always covers the same number of fields for each color, independently of its position. This can be achieved using 3 colors (say, a
, b
and c
):
ababababa
bcbcbcbcb
ababababa
bcbcbcbcb
ababababa
bcbcbcbcb
ababababa
The number of fields per color: $a = 20, b = 31, c = 12$.
Every 2x2 square covers 1 a
-, 2 b
- and 1 c
-fields. Let $S$ denote the total number of 2x2 squares.
There are 3 groups of L-shaped trominos:
- Group-1 trominos cover 1
a
-, 1b
- and 1c
-fields. - Group-2 trominos cover 1
a
-, 2b
- and 0c
-fields. - Group-3 trominos cover 0
a
-, 2b
- and 1c
-fields.
Let $L$, $L_1$, $L_2$ and $L_3$ denote the total number of L-trominos, and the number of trominos in each group, respectively. The following equations hold:
begin{align}
L_1 + L_2 & = a - S \
L_1 + 2L_2 + 2L_3 & = b - 2S \
L_1 + L_3 & = c - S
end{align}
(Note: The summation of these equations gives
begin{align}
3(L_1 + L_2 + L_3) & = a + b + c - 4S \
3L + 4S & = a + b + c \
3L + 4S & = 63 ; (= 7 cdot 9)
end{align}
as expected.)
The solution for the above system of equations:
begin{align}
L_1 & = frac{2a - b + 2c - 2S}{3} = 11 - frac{2S}{3} \
L_2 & = frac{a + b - 2c - S}{3} = 9 - frac{S}{3} \
L_3 & = frac{-2a + b + c - S}{3} = 1 - frac{S}{3}
end{align}
It's already mentioned in the question that the only possible values for $S$ are 0, 3, 6, 9, 12 and 15. Substituting these values into the expressions for $L_1$, $L_2$ and $L_3$ shows that while both $L_1$ and $L_2$ are positive for all $S$ values, $L_3$ becomes negative for $S > 3$, which means that our tiling problem cannot be solved using more than 3 2x2 squares.
The calculations are even simpler if 4 colors are used:abab...
on the 1st line,cdcd...
on the 2nd line, etc. In this case, there are 4 tromino groups, and similarly to the 3-color case, the size of one group becomes negative for $S > 3$.
– kol
Jan 5 at 9:40
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062003%2ftiling-a-7x9-rectangle-with-2x2-squares-and-l-shaped-trominos%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
This answer (to a different tiling problem) gave me the idea to use more than 2 colors to color the 7x9 rectangle (I have already tried the chessboard coloring without any success).
I devised a coloring scheme with which every 2x2 square always covers the same number of fields for each color, independently of its position. This can be achieved using 3 colors (say, a
, b
and c
):
ababababa
bcbcbcbcb
ababababa
bcbcbcbcb
ababababa
bcbcbcbcb
ababababa
The number of fields per color: $a = 20, b = 31, c = 12$.
Every 2x2 square covers 1 a
-, 2 b
- and 1 c
-fields. Let $S$ denote the total number of 2x2 squares.
There are 3 groups of L-shaped trominos:
- Group-1 trominos cover 1
a
-, 1b
- and 1c
-fields. - Group-2 trominos cover 1
a
-, 2b
- and 0c
-fields. - Group-3 trominos cover 0
a
-, 2b
- and 1c
-fields.
Let $L$, $L_1$, $L_2$ and $L_3$ denote the total number of L-trominos, and the number of trominos in each group, respectively. The following equations hold:
begin{align}
L_1 + L_2 & = a - S \
L_1 + 2L_2 + 2L_3 & = b - 2S \
L_1 + L_3 & = c - S
end{align}
(Note: The summation of these equations gives
begin{align}
3(L_1 + L_2 + L_3) & = a + b + c - 4S \
3L + 4S & = a + b + c \
3L + 4S & = 63 ; (= 7 cdot 9)
end{align}
as expected.)
The solution for the above system of equations:
begin{align}
L_1 & = frac{2a - b + 2c - 2S}{3} = 11 - frac{2S}{3} \
L_2 & = frac{a + b - 2c - S}{3} = 9 - frac{S}{3} \
L_3 & = frac{-2a + b + c - S}{3} = 1 - frac{S}{3}
end{align}
It's already mentioned in the question that the only possible values for $S$ are 0, 3, 6, 9, 12 and 15. Substituting these values into the expressions for $L_1$, $L_2$ and $L_3$ shows that while both $L_1$ and $L_2$ are positive for all $S$ values, $L_3$ becomes negative for $S > 3$, which means that our tiling problem cannot be solved using more than 3 2x2 squares.
The calculations are even simpler if 4 colors are used:abab...
on the 1st line,cdcd...
on the 2nd line, etc. In this case, there are 4 tromino groups, and similarly to the 3-color case, the size of one group becomes negative for $S > 3$.
– kol
Jan 5 at 9:40
add a comment |
This answer (to a different tiling problem) gave me the idea to use more than 2 colors to color the 7x9 rectangle (I have already tried the chessboard coloring without any success).
I devised a coloring scheme with which every 2x2 square always covers the same number of fields for each color, independently of its position. This can be achieved using 3 colors (say, a
, b
and c
):
ababababa
bcbcbcbcb
ababababa
bcbcbcbcb
ababababa
bcbcbcbcb
ababababa
The number of fields per color: $a = 20, b = 31, c = 12$.
Every 2x2 square covers 1 a
-, 2 b
- and 1 c
-fields. Let $S$ denote the total number of 2x2 squares.
There are 3 groups of L-shaped trominos:
- Group-1 trominos cover 1
a
-, 1b
- and 1c
-fields. - Group-2 trominos cover 1
a
-, 2b
- and 0c
-fields. - Group-3 trominos cover 0
a
-, 2b
- and 1c
-fields.
Let $L$, $L_1$, $L_2$ and $L_3$ denote the total number of L-trominos, and the number of trominos in each group, respectively. The following equations hold:
begin{align}
L_1 + L_2 & = a - S \
L_1 + 2L_2 + 2L_3 & = b - 2S \
L_1 + L_3 & = c - S
end{align}
(Note: The summation of these equations gives
begin{align}
3(L_1 + L_2 + L_3) & = a + b + c - 4S \
3L + 4S & = a + b + c \
3L + 4S & = 63 ; (= 7 cdot 9)
end{align}
as expected.)
The solution for the above system of equations:
begin{align}
L_1 & = frac{2a - b + 2c - 2S}{3} = 11 - frac{2S}{3} \
L_2 & = frac{a + b - 2c - S}{3} = 9 - frac{S}{3} \
L_3 & = frac{-2a + b + c - S}{3} = 1 - frac{S}{3}
end{align}
It's already mentioned in the question that the only possible values for $S$ are 0, 3, 6, 9, 12 and 15. Substituting these values into the expressions for $L_1$, $L_2$ and $L_3$ shows that while both $L_1$ and $L_2$ are positive for all $S$ values, $L_3$ becomes negative for $S > 3$, which means that our tiling problem cannot be solved using more than 3 2x2 squares.
The calculations are even simpler if 4 colors are used:abab...
on the 1st line,cdcd...
on the 2nd line, etc. In this case, there are 4 tromino groups, and similarly to the 3-color case, the size of one group becomes negative for $S > 3$.
– kol
Jan 5 at 9:40
add a comment |
This answer (to a different tiling problem) gave me the idea to use more than 2 colors to color the 7x9 rectangle (I have already tried the chessboard coloring without any success).
I devised a coloring scheme with which every 2x2 square always covers the same number of fields for each color, independently of its position. This can be achieved using 3 colors (say, a
, b
and c
):
ababababa
bcbcbcbcb
ababababa
bcbcbcbcb
ababababa
bcbcbcbcb
ababababa
The number of fields per color: $a = 20, b = 31, c = 12$.
Every 2x2 square covers 1 a
-, 2 b
- and 1 c
-fields. Let $S$ denote the total number of 2x2 squares.
There are 3 groups of L-shaped trominos:
- Group-1 trominos cover 1
a
-, 1b
- and 1c
-fields. - Group-2 trominos cover 1
a
-, 2b
- and 0c
-fields. - Group-3 trominos cover 0
a
-, 2b
- and 1c
-fields.
Let $L$, $L_1$, $L_2$ and $L_3$ denote the total number of L-trominos, and the number of trominos in each group, respectively. The following equations hold:
begin{align}
L_1 + L_2 & = a - S \
L_1 + 2L_2 + 2L_3 & = b - 2S \
L_1 + L_3 & = c - S
end{align}
(Note: The summation of these equations gives
begin{align}
3(L_1 + L_2 + L_3) & = a + b + c - 4S \
3L + 4S & = a + b + c \
3L + 4S & = 63 ; (= 7 cdot 9)
end{align}
as expected.)
The solution for the above system of equations:
begin{align}
L_1 & = frac{2a - b + 2c - 2S}{3} = 11 - frac{2S}{3} \
L_2 & = frac{a + b - 2c - S}{3} = 9 - frac{S}{3} \
L_3 & = frac{-2a + b + c - S}{3} = 1 - frac{S}{3}
end{align}
It's already mentioned in the question that the only possible values for $S$ are 0, 3, 6, 9, 12 and 15. Substituting these values into the expressions for $L_1$, $L_2$ and $L_3$ shows that while both $L_1$ and $L_2$ are positive for all $S$ values, $L_3$ becomes negative for $S > 3$, which means that our tiling problem cannot be solved using more than 3 2x2 squares.
This answer (to a different tiling problem) gave me the idea to use more than 2 colors to color the 7x9 rectangle (I have already tried the chessboard coloring without any success).
I devised a coloring scheme with which every 2x2 square always covers the same number of fields for each color, independently of its position. This can be achieved using 3 colors (say, a
, b
and c
):
ababababa
bcbcbcbcb
ababababa
bcbcbcbcb
ababababa
bcbcbcbcb
ababababa
The number of fields per color: $a = 20, b = 31, c = 12$.
Every 2x2 square covers 1 a
-, 2 b
- and 1 c
-fields. Let $S$ denote the total number of 2x2 squares.
There are 3 groups of L-shaped trominos:
- Group-1 trominos cover 1
a
-, 1b
- and 1c
-fields. - Group-2 trominos cover 1
a
-, 2b
- and 0c
-fields. - Group-3 trominos cover 0
a
-, 2b
- and 1c
-fields.
Let $L$, $L_1$, $L_2$ and $L_3$ denote the total number of L-trominos, and the number of trominos in each group, respectively. The following equations hold:
begin{align}
L_1 + L_2 & = a - S \
L_1 + 2L_2 + 2L_3 & = b - 2S \
L_1 + L_3 & = c - S
end{align}
(Note: The summation of these equations gives
begin{align}
3(L_1 + L_2 + L_3) & = a + b + c - 4S \
3L + 4S & = a + b + c \
3L + 4S & = 63 ; (= 7 cdot 9)
end{align}
as expected.)
The solution for the above system of equations:
begin{align}
L_1 & = frac{2a - b + 2c - 2S}{3} = 11 - frac{2S}{3} \
L_2 & = frac{a + b - 2c - S}{3} = 9 - frac{S}{3} \
L_3 & = frac{-2a + b + c - S}{3} = 1 - frac{S}{3}
end{align}
It's already mentioned in the question that the only possible values for $S$ are 0, 3, 6, 9, 12 and 15. Substituting these values into the expressions for $L_1$, $L_2$ and $L_3$ shows that while both $L_1$ and $L_2$ are positive for all $S$ values, $L_3$ becomes negative for $S > 3$, which means that our tiling problem cannot be solved using more than 3 2x2 squares.
edited Jan 5 at 2:07
answered Jan 5 at 2:01
kolkol
24828
24828
The calculations are even simpler if 4 colors are used:abab...
on the 1st line,cdcd...
on the 2nd line, etc. In this case, there are 4 tromino groups, and similarly to the 3-color case, the size of one group becomes negative for $S > 3$.
– kol
Jan 5 at 9:40
add a comment |
The calculations are even simpler if 4 colors are used:abab...
on the 1st line,cdcd...
on the 2nd line, etc. In this case, there are 4 tromino groups, and similarly to the 3-color case, the size of one group becomes negative for $S > 3$.
– kol
Jan 5 at 9:40
The calculations are even simpler if 4 colors are used:
abab...
on the 1st line, cdcd...
on the 2nd line, etc. In this case, there are 4 tromino groups, and similarly to the 3-color case, the size of one group becomes negative for $S > 3$.– kol
Jan 5 at 9:40
The calculations are even simpler if 4 colors are used:
abab...
on the 1st line, cdcd...
on the 2nd line, etc. In this case, there are 4 tromino groups, and similarly to the 3-color case, the size of one group becomes negative for $S > 3$.– kol
Jan 5 at 9:40
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062003%2ftiling-a-7x9-rectangle-with-2x2-squares-and-l-shaped-trominos%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
What's the question?
– Lord Shark the Unknown
Jan 4 at 19:29
@LordSharktheUnknown The question is, how can this claim be proved?
– kol
Jan 4 at 19:31
What's "the normal way" to prove this that you refer to?
– Michael Lugo
Jan 4 at 20:14
@MichaelLugo I would like to find a mathematical proof, because I don't like computer based proofs. My C code tried every possible way to find a tiling with a given number of 2x2 squares and L-trominos, but it does not tell anything about why one cannot use more than 3 2x2 squares.
– kol
Jan 4 at 20:20
1
Tile homology groups can be useful for these types of questions. cflmath.com/Research/Tilehomotopy/tilehomotopy.pdf
– Cheerful Parsnip
Jan 5 at 2:17