A polyomino puzzle
up vote
12
down vote
favorite
Is there a polyomino such that it can be glued to an I-shaped pentomino and to a X-shaped pentomino to obtain the same polyomino?
Or is there simple proof for non-existence of such polyomino? [Edit: See "I-shaped" and "X/(+)-shaped" pentominos below:]
puzzle
edited Jun 30 '11 at 23:32
amWhy
191k27223438
asked Jun 28 '11 at 21:50
bleh
30318
|
show 6 more comments
up vote
12
down vote
favorite
Is there a polyomino such that it can be glued to an I-shaped pentomino and to a X-shaped pentomino to obtain the same polyomino?
Or is there simple proof for non-existence of such polyomino? [Edit: See "I-shaped" and "X/(+)-shaped" pentominos below:]
puzzle
edited Jun 30 '11 at 23:32
amWhy
191k27223438
asked Jun 28 '11 at 21:50
bleh
30318
1
What do you mean by "the same polyomino" ? How could it stay the same if you had stuff to it ?
– Joel Cohen
Jun 28 '11 at 22:42
1
oh sorry, what i wanted to say was can the polyomino obtained after gluing some polyomino to the I shaped pentomino and the polyomino obtained after gluing the same polyomino to the X shaped one be the same
– bleh
Jun 28 '11 at 22:51
I'm wondering, whether we can use the fact that, on a chess board, the I-shape covers 3 squares of one color and 2 of the opposite, whereas the cross shape splits 4+1. This won't do it by itself, because translation by a single square switches the colors, but may be we can rule out some cases?
– Jyrki Lahtonen
Jun 29 '11 at 9:52
An immediate consequence of that is that the color imbalance of the third polyomino must be 1 or 2. I don't see this leading anywhere now.
– Jyrki Lahtonen
Jun 29 '11 at 10:13
@Jyrki: that was my first take on it - in fact, I think you can say that the color imbalance must be 2, since the equations aren't quite consistent otherwise, but simple parity doesn't say anything beyond that.
– Steven Stadnicki
Jun 29 '11 at 21:57
|
show 6 more comments
up vote
12
down vote
favorite
up vote
12
down vote
favorite
Is there a polyomino such that it can be glued to an I-shaped pentomino and to a X-shaped pentomino to obtain the same polyomino?
Or is there simple proof for non-existence of such polyomino? [Edit: See "I-shaped" and "X/(+)-shaped" pentominos below:]
puzzle
edited Jun 30 '11 at 23:32
amWhy
191k27223438
asked Jun 28 '11 at 21:50
bleh
30318
Is there a polyomino such that it can be glued to an I-shaped pentomino and to a X-shaped pentomino to obtain the same polyomino?
Or is there simple proof for non-existence of such polyomino? [Edit: See "I-shaped" and "X/(+)-shaped" pentominos below:]
puzzle
puzzle
edited Jun 30 '11 at 23:32
amWhy
191k27223438
asked Jun 28 '11 at 21:50
bleh
30318
edited Jun 30 '11 at 23:32
amWhy
191k27223438
asked Jun 28 '11 at 21:50
bleh
30318
edited Jun 30 '11 at 23:32
amWhy
191k27223438
edited Jun 30 '11 at 23:32
amWhy
191k27223438
edited Jun 30 '11 at 23:32
amWhy
191k27223438
191k27223438
asked Jun 28 '11 at 21:50
bleh
30318
asked Jun 28 '11 at 21:50
bleh
30318
asked Jun 28 '11 at 21:50
bleh
30318
30318
1
What do you mean by "the same polyomino" ? How could it stay the same if you had stuff to it ?
– Joel Cohen
Jun 28 '11 at 22:42
1
oh sorry, what i wanted to say was can the polyomino obtained after gluing some polyomino to the I shaped pentomino and the polyomino obtained after gluing the same polyomino to the X shaped one be the same
– bleh
Jun 28 '11 at 22:51
I'm wondering, whether we can use the fact that, on a chess board, the I-shape covers 3 squares of one color and 2 of the opposite, whereas the cross shape splits 4+1. This won't do it by itself, because translation by a single square switches the colors, but may be we can rule out some cases?
– Jyrki Lahtonen
Jun 29 '11 at 9:52
An immediate consequence of that is that the color imbalance of the third polyomino must be 1 or 2. I don't see this leading anywhere now.
– Jyrki Lahtonen
Jun 29 '11 at 10:13
@Jyrki: that was my first take on it - in fact, I think you can say that the color imbalance must be 2, since the equations aren't quite consistent otherwise, but simple parity doesn't say anything beyond that.
– Steven Stadnicki
Jun 29 '11 at 21:57
|
show 6 more comments
1
What do you mean by "the same polyomino" ? How could it stay the same if you had stuff to it ?
– Joel Cohen
Jun 28 '11 at 22:42
1
oh sorry, what i wanted to say was can the polyomino obtained after gluing some polyomino to the I shaped pentomino and the polyomino obtained after gluing the same polyomino to the X shaped one be the same
– bleh
Jun 28 '11 at 22:51
I'm wondering, whether we can use the fact that, on a chess board, the I-shape covers 3 squares of one color and 2 of the opposite, whereas the cross shape splits 4+1. This won't do it by itself, because translation by a single square switches the colors, but may be we can rule out some cases?
– Jyrki Lahtonen
Jun 29 '11 at 9:52
An immediate consequence of that is that the color imbalance of the third polyomino must be 1 or 2. I don't see this leading anywhere now.
– Jyrki Lahtonen
Jun 29 '11 at 10:13
@Jyrki: that was my first take on it - in fact, I think you can say that the color imbalance must be 2, since the equations aren't quite consistent otherwise, but simple parity doesn't say anything beyond that.
– Steven Stadnicki
Jun 29 '11 at 21:57
1
1
What do you mean by "the same polyomino" ? How could it stay the same if you had stuff to it ?
– Joel Cohen
Jun 28 '11 at 22:42
What do you mean by "the same polyomino" ? How could it stay the same if you had stuff to it ?
– Joel Cohen
Jun 28 '11 at 22:42
1
1
oh sorry, what i wanted to say was can the polyomino obtained after gluing some polyomino to the I shaped pentomino and the polyomino obtained after gluing the same polyomino to the X shaped one be the same
– bleh
Jun 28 '11 at 22:51
oh sorry, what i wanted to say was can the polyomino obtained after gluing some polyomino to the I shaped pentomino and the polyomino obtained after gluing the same polyomino to the X shaped one be the same
– bleh
Jun 28 '11 at 22:51
I'm wondering, whether we can use the fact that, on a chess board, the I-shape covers 3 squares of one color and 2 of the opposite, whereas the cross shape splits 4+1. This won't do it by itself, because translation by a single square switches the colors, but may be we can rule out some cases?
– Jyrki Lahtonen
Jun 29 '11 at 9:52
I'm wondering, whether we can use the fact that, on a chess board, the I-shape covers 3 squares of one color and 2 of the opposite, whereas the cross shape splits 4+1. This won't do it by itself, because translation by a single square switches the colors, but may be we can rule out some cases?
– Jyrki Lahtonen
Jun 29 '11 at 9:52
An immediate consequence of that is that the color imbalance of the third polyomino must be 1 or 2. I don't see this leading anywhere now.
– Jyrki Lahtonen
Jun 29 '11 at 10:13
An immediate consequence of that is that the color imbalance of the third polyomino must be 1 or 2. I don't see this leading anywhere now.
– Jyrki Lahtonen
Jun 29 '11 at 10:13
@Jyrki: that was my first take on it - in fact, I think you can say that the color imbalance must be 2, since the equations aren't quite consistent otherwise, but simple parity doesn't say anything beyond that.
– Steven Stadnicki
Jun 29 '11 at 21:57
@Jyrki: that was my first take on it - in fact, I think you can say that the color imbalance must be 2, since the equations aren't quite consistent otherwise, but simple parity doesn't say anything beyond that.
– Steven Stadnicki
Jun 29 '11 at 21:57
|
show 6 more comments
3 Answers
3
active
oldest
votes
up vote
16
down vote
accepted
${}{}{}{}{}{}{}{}{}{}{}{}{}{}$
answered Jul 1 '11 at 0:45
TonyK
40.5k352130
Nice. Is there anything to be learned from how you arrived at this?
– joriki
Jul 1 '11 at 0:51
3
I arrived at it by proving its impossibility. Then I started to write up my proof here, and saw my mistake.
– TonyK
Jul 1 '11 at 0:53
3
Specifically, I knew that the glued-together polyomino must have a column of five squares on the left, and a column of a single square on the right (or vice versa). I thought this led to a contradiction, but it turned out to lead to the solution.
– TonyK
Jul 1 '11 at 0:55
add a comment |
up vote
6
down vote
This isn't a real answer, but if you allow "infinite polyominos" then you can do it. Consider the following "infinite polyomino", without the yellow included:
Then adding a either "+" or an "I" pentomino where indicated in yellow will give you the same result up to translation. In fact, you can add on $n$ "+" pentominos or $n$ "I" pentominos to this so that they'll give you the same answer, for any $n in mathbb{Z}$. (Yes $mathbb{Z}$, as long as you consider "adding a negative pentomino" to mean what I think it should mean :) )
A similar construction works for any pair of finite polyominos.
answered Jul 1 '11 at 0:23
MartianInvader
5,1631225
add a comment |
up vote
3
down vote
By gluing, if you allow overlapping of a square: e.g. the top square of an "I" shaped "tri-omino" could be glued to the bottom square of the "+" pentomino (vertically aligned), while its center square (of I tri-omino) could be glued orthogonally to the 2nd square from the top of the I pentomino.) This would result in a matching 7-ominos (heptominos). (See my (pitiful) LaTeX attempt at constructing the figure I'm alluding to - just pretend there are no gaps vertically!).
If overlapping is not allowed, (i.e. gluing must occur from edge to edge of each respective polyomino, then I'm doubting the existence of a polyomino which can be attached to each separate figure with the result a match. But I've no proof, yet.
$quadsquare$
$squaresquaresquare$
$quadsquare$
$quadsquare$
$quadsquare$
answered Jun 29 '11 at 1:15
amWhy
191k27223438
The heptomino was all I could think of, too.
– Jack Henahan
Jun 29 '11 at 1:36
4
hi,thnx for the answer,but this puzzle is trivial if overlapping is allowed as we can glue a very large polyomino to the top of any pentomino to obtain the same large polyomino!
– bleh
Jun 29 '11 at 6:37
I suppose that putting the polyominoes on a cylinder of girth 3 (or other changes in the topology) would also amount to bending the rules :-)
– Jyrki Lahtonen
Jun 29 '11 at 18:14
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
16
down vote
accepted
${}{}{}{}{}{}{}{}{}{}{}{}{}{}$
answered Jul 1 '11 at 0:45
TonyK
40.5k352130
Nice. Is there anything to be learned from how you arrived at this?
– joriki
Jul 1 '11 at 0:51
3
I arrived at it by proving its impossibility. Then I started to write up my proof here, and saw my mistake.
– TonyK
Jul 1 '11 at 0:53
3
Specifically, I knew that the glued-together polyomino must have a column of five squares on the left, and a column of a single square on the right (or vice versa). I thought this led to a contradiction, but it turned out to lead to the solution.
– TonyK
Jul 1 '11 at 0:55
add a comment |
up vote
16
down vote
accepted
${}{}{}{}{}{}{}{}{}{}{}{}{}{}$
answered Jul 1 '11 at 0:45
TonyK
40.5k352130
Nice. Is there anything to be learned from how you arrived at this?
– joriki
Jul 1 '11 at 0:51
3
I arrived at it by proving its impossibility. Then I started to write up my proof here, and saw my mistake.
– TonyK
Jul 1 '11 at 0:53
3
Specifically, I knew that the glued-together polyomino must have a column of five squares on the left, and a column of a single square on the right (or vice versa). I thought this led to a contradiction, but it turned out to lead to the solution.
– TonyK
Jul 1 '11 at 0:55
add a comment |
up vote
16
down vote
accepted
up vote
16
down vote
accepted
${}{}{}{}{}{}{}{}{}{}{}{}{}{}$
answered Jul 1 '11 at 0:45
TonyK
40.5k352130
${}{}{}{}{}{}{}{}{}{}{}{}{}{}$
answered Jul 1 '11 at 0:45
TonyK
40.5k352130
edited yesterday
answered Jul 1 '11 at 0:45
TonyK
40.5k352130
answered Jul 1 '11 at 0:45
TonyK
40.5k352130
answered Jul 1 '11 at 0:45
TonyK
40.5k352130
40.5k352130
Nice. Is there anything to be learned from how you arrived at this?
– joriki
Jul 1 '11 at 0:51
3
I arrived at it by proving its impossibility. Then I started to write up my proof here, and saw my mistake.
– TonyK
Jul 1 '11 at 0:53
3
Specifically, I knew that the glued-together polyomino must have a column of five squares on the left, and a column of a single square on the right (or vice versa). I thought this led to a contradiction, but it turned out to lead to the solution.
– TonyK
Jul 1 '11 at 0:55
add a comment |
Nice. Is there anything to be learned from how you arrived at this?
– joriki
Jul 1 '11 at 0:51
3
I arrived at it by proving its impossibility. Then I started to write up my proof here, and saw my mistake.
– TonyK
Jul 1 '11 at 0:53
3
Specifically, I knew that the glued-together polyomino must have a column of five squares on the left, and a column of a single square on the right (or vice versa). I thought this led to a contradiction, but it turned out to lead to the solution.
– TonyK
Jul 1 '11 at 0:55
Nice. Is there anything to be learned from how you arrived at this?
– joriki
Jul 1 '11 at 0:51
Nice. Is there anything to be learned from how you arrived at this?
– joriki
Jul 1 '11 at 0:51
3
3
I arrived at it by proving its impossibility. Then I started to write up my proof here, and saw my mistake.
– TonyK
Jul 1 '11 at 0:53
I arrived at it by proving its impossibility. Then I started to write up my proof here, and saw my mistake.
– TonyK
Jul 1 '11 at 0:53
3
3
Specifically, I knew that the glued-together polyomino must have a column of five squares on the left, and a column of a single square on the right (or vice versa). I thought this led to a contradiction, but it turned out to lead to the solution.
– TonyK
Jul 1 '11 at 0:55
Specifically, I knew that the glued-together polyomino must have a column of five squares on the left, and a column of a single square on the right (or vice versa). I thought this led to a contradiction, but it turned out to lead to the solution.
– TonyK
Jul 1 '11 at 0:55
add a comment |
up vote
6
down vote
This isn't a real answer, but if you allow "infinite polyominos" then you can do it. Consider the following "infinite polyomino", without the yellow included:
Then adding a either "+" or an "I" pentomino where indicated in yellow will give you the same result up to translation. In fact, you can add on $n$ "+" pentominos or $n$ "I" pentominos to this so that they'll give you the same answer, for any $n in mathbb{Z}$. (Yes $mathbb{Z}$, as long as you consider "adding a negative pentomino" to mean what I think it should mean :) )
A similar construction works for any pair of finite polyominos.
answered Jul 1 '11 at 0:23
MartianInvader
5,1631225
add a comment |
up vote
6
down vote
This isn't a real answer, but if you allow "infinite polyominos" then you can do it. Consider the following "infinite polyomino", without the yellow included:
Then adding a either "+" or an "I" pentomino where indicated in yellow will give you the same result up to translation. In fact, you can add on $n$ "+" pentominos or $n$ "I" pentominos to this so that they'll give you the same answer, for any $n in mathbb{Z}$. (Yes $mathbb{Z}$, as long as you consider "adding a negative pentomino" to mean what I think it should mean :) )
A similar construction works for any pair of finite polyominos.
answered Jul 1 '11 at 0:23
MartianInvader
5,1631225
add a comment |
up vote
6
down vote
up vote
6
down vote
This isn't a real answer, but if you allow "infinite polyominos" then you can do it. Consider the following "infinite polyomino", without the yellow included:
Then adding a either "+" or an "I" pentomino where indicated in yellow will give you the same result up to translation. In fact, you can add on $n$ "+" pentominos or $n$ "I" pentominos to this so that they'll give you the same answer, for any $n in mathbb{Z}$. (Yes $mathbb{Z}$, as long as you consider "adding a negative pentomino" to mean what I think it should mean :) )
A similar construction works for any pair of finite polyominos.
answered Jul 1 '11 at 0:23
MartianInvader
5,1631225
This isn't a real answer, but if you allow "infinite polyominos" then you can do it. Consider the following "infinite polyomino", without the yellow included:
Then adding a either "+" or an "I" pentomino where indicated in yellow will give you the same result up to translation. In fact, you can add on $n$ "+" pentominos or $n$ "I" pentominos to this so that they'll give you the same answer, for any $n in mathbb{Z}$. (Yes $mathbb{Z}$, as long as you consider "adding a negative pentomino" to mean what I think it should mean :) )
A similar construction works for any pair of finite polyominos.
answered Jul 1 '11 at 0:23
MartianInvader
5,1631225
answered Jul 1 '11 at 0:23
MartianInvader
5,1631225
answered Jul 1 '11 at 0:23
MartianInvader
5,1631225
answered Jul 1 '11 at 0:23
MartianInvader
5,1631225
5,1631225
add a comment |
add a comment |
up vote
3
down vote
By gluing, if you allow overlapping of a square: e.g. the top square of an "I" shaped "tri-omino" could be glued to the bottom square of the "+" pentomino (vertically aligned), while its center square (of I tri-omino) could be glued orthogonally to the 2nd square from the top of the I pentomino.) This would result in a matching 7-ominos (heptominos). (See my (pitiful) LaTeX attempt at constructing the figure I'm alluding to - just pretend there are no gaps vertically!).
If overlapping is not allowed, (i.e. gluing must occur from edge to edge of each respective polyomino, then I'm doubting the existence of a polyomino which can be attached to each separate figure with the result a match. But I've no proof, yet.
$quadsquare$
$squaresquaresquare$
$quadsquare$
$quadsquare$
$quadsquare$
answered Jun 29 '11 at 1:15
amWhy
191k27223438
The heptomino was all I could think of, too.
– Jack Henahan
Jun 29 '11 at 1:36
4
hi,thnx for the answer,but this puzzle is trivial if overlapping is allowed as we can glue a very large polyomino to the top of any pentomino to obtain the same large polyomino!
– bleh
Jun 29 '11 at 6:37
I suppose that putting the polyominoes on a cylinder of girth 3 (or other changes in the topology) would also amount to bending the rules :-)
– Jyrki Lahtonen
Jun 29 '11 at 18:14
add a comment |
up vote
3
down vote
By gluing, if you allow overlapping of a square: e.g. the top square of an "I" shaped "tri-omino" could be glued to the bottom square of the "+" pentomino (vertically aligned), while its center square (of I tri-omino) could be glued orthogonally to the 2nd square from the top of the I pentomino.) This would result in a matching 7-ominos (heptominos). (See my (pitiful) LaTeX attempt at constructing the figure I'm alluding to - just pretend there are no gaps vertically!).
If overlapping is not allowed, (i.e. gluing must occur from edge to edge of each respective polyomino, then I'm doubting the existence of a polyomino which can be attached to each separate figure with the result a match. But I've no proof, yet.
$quadsquare$
$squaresquaresquare$
$quadsquare$
$quadsquare$
$quadsquare$
answered Jun 29 '11 at 1:15
amWhy
191k27223438
The heptomino was all I could think of, too.
– Jack Henahan
Jun 29 '11 at 1:36
4
hi,thnx for the answer,but this puzzle is trivial if overlapping is allowed as we can glue a very large polyomino to the top of any pentomino to obtain the same large polyomino!
– bleh
Jun 29 '11 at 6:37
I suppose that putting the polyominoes on a cylinder of girth 3 (or other changes in the topology) would also amount to bending the rules :-)
– Jyrki Lahtonen
Jun 29 '11 at 18:14
add a comment |
up vote
3
down vote
up vote
3
down vote
By gluing, if you allow overlapping of a square: e.g. the top square of an "I" shaped "tri-omino" could be glued to the bottom square of the "+" pentomino (vertically aligned), while its center square (of I tri-omino) could be glued orthogonally to the 2nd square from the top of the I pentomino.) This would result in a matching 7-ominos (heptominos). (See my (pitiful) LaTeX attempt at constructing the figure I'm alluding to - just pretend there are no gaps vertically!).
If overlapping is not allowed, (i.e. gluing must occur from edge to edge of each respective polyomino, then I'm doubting the existence of a polyomino which can be attached to each separate figure with the result a match. But I've no proof, yet.
$quadsquare$
$squaresquaresquare$
$quadsquare$
$quadsquare$
$quadsquare$
answered Jun 29 '11 at 1:15
amWhy
191k27223438
By gluing, if you allow overlapping of a square: e.g. the top square of an "I" shaped "tri-omino" could be glued to the bottom square of the "+" pentomino (vertically aligned), while its center square (of I tri-omino) could be glued orthogonally to the 2nd square from the top of the I pentomino.) This would result in a matching 7-ominos (heptominos). (See my (pitiful) LaTeX attempt at constructing the figure I'm alluding to - just pretend there are no gaps vertically!).
If overlapping is not allowed, (i.e. gluing must occur from edge to edge of each respective polyomino, then I'm doubting the existence of a polyomino which can be attached to each separate figure with the result a match. But I've no proof, yet.
$quadsquare$
$squaresquaresquare$
$quadsquare$
$quadsquare$
$quadsquare$
answered Jun 29 '11 at 1:15
amWhy
191k27223438
edited Jun 29 '11 at 1:22
answered Jun 29 '11 at 1:15
amWhy
191k27223438
answered Jun 29 '11 at 1:15
amWhy
191k27223438
answered Jun 29 '11 at 1:15
amWhy
191k27223438
191k27223438
The heptomino was all I could think of, too.
– Jack Henahan
Jun 29 '11 at 1:36
4
hi,thnx for the answer,but this puzzle is trivial if overlapping is allowed as we can glue a very large polyomino to the top of any pentomino to obtain the same large polyomino!
– bleh
Jun 29 '11 at 6:37
I suppose that putting the polyominoes on a cylinder of girth 3 (or other changes in the topology) would also amount to bending the rules :-)
– Jyrki Lahtonen
Jun 29 '11 at 18:14
add a comment |
The heptomino was all I could think of, too.
– Jack Henahan
Jun 29 '11 at 1:36
4
hi,thnx for the answer,but this puzzle is trivial if overlapping is allowed as we can glue a very large polyomino to the top of any pentomino to obtain the same large polyomino!
– bleh
Jun 29 '11 at 6:37
I suppose that putting the polyominoes on a cylinder of girth 3 (or other changes in the topology) would also amount to bending the rules :-)
– Jyrki Lahtonen
Jun 29 '11 at 18:14
The heptomino was all I could think of, too.
– Jack Henahan
Jun 29 '11 at 1:36
The heptomino was all I could think of, too.
– Jack Henahan
Jun 29 '11 at 1:36
4
4
hi,thnx for the answer,but this puzzle is trivial if overlapping is allowed as we can glue a very large polyomino to the top of any pentomino to obtain the same large polyomino!
– bleh
Jun 29 '11 at 6:37
hi,thnx for the answer,but this puzzle is trivial if overlapping is allowed as we can glue a very large polyomino to the top of any pentomino to obtain the same large polyomino!
– bleh
Jun 29 '11 at 6:37
I suppose that putting the polyominoes on a cylinder of girth 3 (or other changes in the topology) would also amount to bending the rules :-)
– Jyrki Lahtonen
Jun 29 '11 at 18:14
I suppose that putting the polyominoes on a cylinder of girth 3 (or other changes in the topology) would also amount to bending the rules :-)
– Jyrki Lahtonen
Jun 29 '11 at 18:14
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f48315%2fa-polyomino-puzzle%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 do you mean by "the same polyomino" ? How could it stay the same if you had stuff to it ?
– Joel Cohen
Jun 28 '11 at 22:42
1
oh sorry, what i wanted to say was can the polyomino obtained after gluing some polyomino to the I shaped pentomino and the polyomino obtained after gluing the same polyomino to the X shaped one be the same
– bleh
Jun 28 '11 at 22:51
I'm wondering, whether we can use the fact that, on a chess board, the I-shape covers 3 squares of one color and 2 of the opposite, whereas the cross shape splits 4+1. This won't do it by itself, because translation by a single square switches the colors, but may be we can rule out some cases?
– Jyrki Lahtonen
Jun 29 '11 at 9:52
An immediate consequence of that is that the color imbalance of the third polyomino must be 1 or 2. I don't see this leading anywhere now.
– Jyrki Lahtonen
Jun 29 '11 at 10:13
@Jyrki: that was my first take on it - in fact, I think you can say that the color imbalance must be 2, since the equations aren't quite consistent otherwise, but simple parity doesn't say anything beyond that.
– Steven Stadnicki
Jun 29 '11 at 21:57