Mixing
Edge colourings of an icosahedron
I'm referring to problem A6 of the 2017 Putnam competition -- the question is "How many ways exist to colour the labelled edges of an icosahedron such that every face has two edges of the same colour and one edge of another colour, where the colours are either red, white or blue?".
My solution is as follows: consider the planar representation of the icosahedron:
Note that:
There are 18 ways to colour the edges of a triangle such that it has two edges of the same colour and one edge of another colour (3 choices for the colour that appears twice, 2 choices for the colour that appears once, and 3 choices for the arrangement).
Given the colouring on any one edge of a triangle, there are 6 ways to colour the remaining edges in a way that satisfies the condition (WLOG suppose the given edge is white -- then either one other edge is white (and the other edge is red or blue), which has 4 possibilities, or both the other edges have the same colour, red or blue, which is possible in 2 ways.
Given the colouring on two edges of a triangle, there are 2 ways to colour the remaining edge in a way that satisfies the condition. WLOG, the given edges are coloured either "R R" or "R B". If it's "R R", the 2 ways to choose the other edge are "W" and "B" -- if it's "R B", the 2 ways to choose the other edge are "R" and "B".
So there are 18 ways to choose the colouring on the central triangle (the base case) of the planar representation, and you can write each the number of ways to colour each successive "containing triangle" as $6^32^3$ multiplied by the smaller triangle it contains, so the number of ways to colour the entire icosahedron should be:
$$18(6^32^3)^3=2^{19}3^{11}$$
Unfortunately, the official solution (p. 5) presents an answer of $2^{20}3^{10}$ -- I'm off by a factor of $2/3$! What's going on? What did I do wrong?
combinatorics coloring planar-graph
asked Nov 20 '18 at 12:47
Abhimanyu Pallavi Sudhir
876619
add a comment |
I'm referring to problem A6 of the 2017 Putnam competition -- the question is "How many ways exist to colour the labelled edges of an icosahedron such that every face has two edges of the same colour and one edge of another colour, where the colours are either red, white or blue?".
My solution is as follows: consider the planar representation of the icosahedron:
Note that:
There are 18 ways to colour the edges of a triangle such that it has two edges of the same colour and one edge of another colour (3 choices for the colour that appears twice, 2 choices for the colour that appears once, and 3 choices for the arrangement).
Given the colouring on any one edge of a triangle, there are 6 ways to colour the remaining edges in a way that satisfies the condition (WLOG suppose the given edge is white -- then either one other edge is white (and the other edge is red or blue), which has 4 possibilities, or both the other edges have the same colour, red or blue, which is possible in 2 ways.
Given the colouring on two edges of a triangle, there are 2 ways to colour the remaining edge in a way that satisfies the condition. WLOG, the given edges are coloured either "R R" or "R B". If it's "R R", the 2 ways to choose the other edge are "W" and "B" -- if it's "R B", the 2 ways to choose the other edge are "R" and "B".
So there are 18 ways to choose the colouring on the central triangle (the base case) of the planar representation, and you can write each the number of ways to colour each successive "containing triangle" as $6^32^3$ multiplied by the smaller triangle it contains, so the number of ways to colour the entire icosahedron should be:
$$18(6^32^3)^3=2^{19}3^{11}$$
Unfortunately, the official solution (p. 5) presents an answer of $2^{20}3^{10}$ -- I'm off by a factor of $2/3$! What's going on? What did I do wrong?
combinatorics coloring planar-graph
asked Nov 20 '18 at 12:47
Abhimanyu Pallavi Sudhir
876619
Check the update to my answer! I just spotted a serious flaw in your setup.
– Christian Blatter
Nov 22 '18 at 7:55
If I understand you correctly, your picture is supposed to be the icosahedron graph, where vertices & edges in your graph = vertices & edges in an icosahedron, right? If so, this is the wrong graph. Each vertex in an actual icosahedron has edge-degree 5. See en.wikipedia.org/wiki/Regular_icosahedron#Icosahedral_graph Your graph may actually represent a column of 3 octahedra glued together face to face...?
– antkam
Nov 22 '18 at 16:20
add a comment |
I'm referring to problem A6 of the 2017 Putnam competition -- the question is "How many ways exist to colour the labelled edges of an icosahedron such that every face has two edges of the same colour and one edge of another colour, where the colours are either red, white or blue?".
My solution is as follows: consider the planar representation of the icosahedron:
Note that:
There are 18 ways to colour the edges of a triangle such that it has two edges of the same colour and one edge of another colour (3 choices for the colour that appears twice, 2 choices for the colour that appears once, and 3 choices for the arrangement).
Given the colouring on any one edge of a triangle, there are 6 ways to colour the remaining edges in a way that satisfies the condition (WLOG suppose the given edge is white -- then either one other edge is white (and the other edge is red or blue), which has 4 possibilities, or both the other edges have the same colour, red or blue, which is possible in 2 ways.
Given the colouring on two edges of a triangle, there are 2 ways to colour the remaining edge in a way that satisfies the condition. WLOG, the given edges are coloured either "R R" or "R B". If it's "R R", the 2 ways to choose the other edge are "W" and "B" -- if it's "R B", the 2 ways to choose the other edge are "R" and "B".
So there are 18 ways to choose the colouring on the central triangle (the base case) of the planar representation, and you can write each the number of ways to colour each successive "containing triangle" as $6^32^3$ multiplied by the smaller triangle it contains, so the number of ways to colour the entire icosahedron should be:
$$18(6^32^3)^3=2^{19}3^{11}$$
Unfortunately, the official solution (p. 5) presents an answer of $2^{20}3^{10}$ -- I'm off by a factor of $2/3$! What's going on? What did I do wrong?
combinatorics coloring planar-graph
asked Nov 20 '18 at 12:47
Abhimanyu Pallavi Sudhir
876619
I'm referring to problem A6 of the 2017 Putnam competition -- the question is "How many ways exist to colour the labelled edges of an icosahedron such that every face has two edges of the same colour and one edge of another colour, where the colours are either red, white or blue?".
My solution is as follows: consider the planar representation of the icosahedron:
Note that:
There are 18 ways to colour the edges of a triangle such that it has two edges of the same colour and one edge of another colour (3 choices for the colour that appears twice, 2 choices for the colour that appears once, and 3 choices for the arrangement).
Given the colouring on any one edge of a triangle, there are 6 ways to colour the remaining edges in a way that satisfies the condition (WLOG suppose the given edge is white -- then either one other edge is white (and the other edge is red or blue), which has 4 possibilities, or both the other edges have the same colour, red or blue, which is possible in 2 ways.
Given the colouring on two edges of a triangle, there are 2 ways to colour the remaining edge in a way that satisfies the condition. WLOG, the given edges are coloured either "R R" or "R B". If it's "R R", the 2 ways to choose the other edge are "W" and "B" -- if it's "R B", the 2 ways to choose the other edge are "R" and "B".
So there are 18 ways to choose the colouring on the central triangle (the base case) of the planar representation, and you can write each the number of ways to colour each successive "containing triangle" as $6^32^3$ multiplied by the smaller triangle it contains, so the number of ways to colour the entire icosahedron should be:
$$18(6^32^3)^3=2^{19}3^{11}$$
Unfortunately, the official solution (p. 5) presents an answer of $2^{20}3^{10}$ -- I'm off by a factor of $2/3$! What's going on? What did I do wrong?
combinatorics coloring planar-graph
combinatorics coloring planar-graph
asked Nov 20 '18 at 12:47
Abhimanyu Pallavi Sudhir
876619
asked Nov 20 '18 at 12:47
Abhimanyu Pallavi Sudhir
876619
edited Nov 20 '18 at 13:38
asked Nov 20 '18 at 12:47
Abhimanyu Pallavi Sudhir
876619
asked Nov 20 '18 at 12:47
Abhimanyu Pallavi Sudhir
876619
asked Nov 20 '18 at 12:47
Abhimanyu Pallavi Sudhir
876619
876619
Check the update to my answer! I just spotted a serious flaw in your setup.
– Christian Blatter
Nov 22 '18 at 7:55
If I understand you correctly, your picture is supposed to be the icosahedron graph, where vertices & edges in your graph = vertices & edges in an icosahedron, right? If so, this is the wrong graph. Each vertex in an actual icosahedron has edge-degree 5. See en.wikipedia.org/wiki/Regular_icosahedron#Icosahedral_graph Your graph may actually represent a column of 3 octahedra glued together face to face...?
– antkam
Nov 22 '18 at 16:20
add a comment |
Check the update to my answer! I just spotted a serious flaw in your setup.
– Christian Blatter
Nov 22 '18 at 7:55
If I understand you correctly, your picture is supposed to be the icosahedron graph, where vertices & edges in your graph = vertices & edges in an icosahedron, right? If so, this is the wrong graph. Each vertex in an actual icosahedron has edge-degree 5. See en.wikipedia.org/wiki/Regular_icosahedron#Icosahedral_graph Your graph may actually represent a column of 3 octahedra glued together face to face...?
– antkam
Nov 22 '18 at 16:20
Check the update to my answer! I just spotted a serious flaw in your setup.
– Christian Blatter
Nov 22 '18 at 7:55
Check the update to my answer! I just spotted a serious flaw in your setup.
– Christian Blatter
Nov 22 '18 at 7:55
If I understand you correctly, your picture is supposed to be the icosahedron graph, where vertices & edges in your graph = vertices & edges in an icosahedron, right? If so, this is the wrong graph. Each vertex in an actual icosahedron has edge-degree 5. See en.wikipedia.org/wiki/Regular_icosahedron#Icosahedral_graph Your graph may actually represent a column of 3 octahedra glued together face to face...?
– antkam
Nov 22 '18 at 16:20
If I understand you correctly, your picture is supposed to be the icosahedron graph, where vertices & edges in your graph = vertices & edges in an icosahedron, right? If so, this is the wrong graph. Each vertex in an actual icosahedron has edge-degree 5. See en.wikipedia.org/wiki/Regular_icosahedron#Icosahedral_graph Your graph may actually represent a column of 3 octahedra glued together face to face...?
– antkam
Nov 22 '18 at 16:20
add a comment |
1 Answer
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active
oldest
votes
Update
You have not checked whether the last triangle (the infinite outer triangle in the figure) is colored correctly. I don't know how to fix this. You cannot just say that with probability ${2over3}$ the last triangle is correctly colored.
Now for the biggest mistake: Your net is not the net of an icosahedron. In the icosahedron graph each vertex is of degree $5$, but the vertices in your graph are of degree $4$ or $6$.
answered Nov 20 '18 at 19:18
Christian Blatter
172k7112326
Oh, of course. I forgot the graph is actually a solid. Thanks.
– Abhimanyu Pallavi Sudhir
Nov 21 '18 at 11:38
@ChristianBlatter - I have always admired the quality of your answers (to many, many other questions), so when I saw your original answer I thought, "How can we say the final triangle has 2/3 prob of being OK? But this is Christian Blatter, so he must be right and I must be just missing some symmetry argument..." :) It took me quite a bit of thinking to convince myself you were wrong. Too bad the whole thing is moot because of using the wrong net.
– antkam
Nov 22 '18 at 16:29
add a comment |
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1 Answer
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1 Answer
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active
oldest
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Update
You have not checked whether the last triangle (the infinite outer triangle in the figure) is colored correctly. I don't know how to fix this. You cannot just say that with probability ${2over3}$ the last triangle is correctly colored.
Now for the biggest mistake: Your net is not the net of an icosahedron. In the icosahedron graph each vertex is of degree $5$, but the vertices in your graph are of degree $4$ or $6$.
answered Nov 20 '18 at 19:18
Christian Blatter
172k7112326
Oh, of course. I forgot the graph is actually a solid. Thanks.
– Abhimanyu Pallavi Sudhir
Nov 21 '18 at 11:38
@ChristianBlatter - I have always admired the quality of your answers (to many, many other questions), so when I saw your original answer I thought, "How can we say the final triangle has 2/3 prob of being OK? But this is Christian Blatter, so he must be right and I must be just missing some symmetry argument..." :) It took me quite a bit of thinking to convince myself you were wrong. Too bad the whole thing is moot because of using the wrong net.
– antkam
Nov 22 '18 at 16:29
add a comment |
Update
You have not checked whether the last triangle (the infinite outer triangle in the figure) is colored correctly. I don't know how to fix this. You cannot just say that with probability ${2over3}$ the last triangle is correctly colored.
Now for the biggest mistake: Your net is not the net of an icosahedron. In the icosahedron graph each vertex is of degree $5$, but the vertices in your graph are of degree $4$ or $6$.
answered Nov 20 '18 at 19:18
Christian Blatter
172k7112326
Oh, of course. I forgot the graph is actually a solid. Thanks.
– Abhimanyu Pallavi Sudhir
Nov 21 '18 at 11:38
@ChristianBlatter - I have always admired the quality of your answers (to many, many other questions), so when I saw your original answer I thought, "How can we say the final triangle has 2/3 prob of being OK? But this is Christian Blatter, so he must be right and I must be just missing some symmetry argument..." :) It took me quite a bit of thinking to convince myself you were wrong. Too bad the whole thing is moot because of using the wrong net.
– antkam
Nov 22 '18 at 16:29
add a comment |
Update
You have not checked whether the last triangle (the infinite outer triangle in the figure) is colored correctly. I don't know how to fix this. You cannot just say that with probability ${2over3}$ the last triangle is correctly colored.
Now for the biggest mistake: Your net is not the net of an icosahedron. In the icosahedron graph each vertex is of degree $5$, but the vertices in your graph are of degree $4$ or $6$.
answered Nov 20 '18 at 19:18
Christian Blatter
172k7112326
Update
You have not checked whether the last triangle (the infinite outer triangle in the figure) is colored correctly. I don't know how to fix this. You cannot just say that with probability ${2over3}$ the last triangle is correctly colored.
Now for the biggest mistake: Your net is not the net of an icosahedron. In the icosahedron graph each vertex is of degree $5$, but the vertices in your graph are of degree $4$ or $6$.
answered Nov 20 '18 at 19:18
Christian Blatter
172k7112326
edited Nov 22 '18 at 7:53
answered Nov 20 '18 at 19:18
Christian Blatter
172k7112326
answered Nov 20 '18 at 19:18
Christian Blatter
172k7112326
answered Nov 20 '18 at 19:18
Christian Blatter
172k7112326
172k7112326
Oh, of course. I forgot the graph is actually a solid. Thanks.
– Abhimanyu Pallavi Sudhir
Nov 21 '18 at 11:38
@ChristianBlatter - I have always admired the quality of your answers (to many, many other questions), so when I saw your original answer I thought, "How can we say the final triangle has 2/3 prob of being OK? But this is Christian Blatter, so he must be right and I must be just missing some symmetry argument..." :) It took me quite a bit of thinking to convince myself you were wrong. Too bad the whole thing is moot because of using the wrong net.
– antkam
Nov 22 '18 at 16:29
add a comment |
Oh, of course. I forgot the graph is actually a solid. Thanks.
– Abhimanyu Pallavi Sudhir
Nov 21 '18 at 11:38
@ChristianBlatter - I have always admired the quality of your answers (to many, many other questions), so when I saw your original answer I thought, "How can we say the final triangle has 2/3 prob of being OK? But this is Christian Blatter, so he must be right and I must be just missing some symmetry argument..." :) It took me quite a bit of thinking to convince myself you were wrong. Too bad the whole thing is moot because of using the wrong net.
– antkam
Nov 22 '18 at 16:29
Oh, of course. I forgot the graph is actually a solid. Thanks.
– Abhimanyu Pallavi Sudhir
Nov 21 '18 at 11:38
Oh, of course. I forgot the graph is actually a solid. Thanks.
– Abhimanyu Pallavi Sudhir
Nov 21 '18 at 11:38
@ChristianBlatter - I have always admired the quality of your answers (to many, many other questions), so when I saw your original answer I thought, "How can we say the final triangle has 2/3 prob of being OK? But this is Christian Blatter, so he must be right and I must be just missing some symmetry argument..." :) It took me quite a bit of thinking to convince myself you were wrong. Too bad the whole thing is moot because of using the wrong net.
– antkam
Nov 22 '18 at 16:29
@ChristianBlatter - I have always admired the quality of your answers (to many, many other questions), so when I saw your original answer I thought, "How can we say the final triangle has 2/3 prob of being OK? But this is Christian Blatter, so he must be right and I must be just missing some symmetry argument..." :) It took me quite a bit of thinking to convince myself you were wrong. Too bad the whole thing is moot because of using the wrong net.
– antkam
Nov 22 '18 at 16:29
add a comment |
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Check the update to my answer! I just spotted a serious flaw in your setup.
– Christian Blatter
Nov 22 '18 at 7:55
If I understand you correctly, your picture is supposed to be the icosahedron graph, where vertices & edges in your graph = vertices & edges in an icosahedron, right? If so, this is the wrong graph. Each vertex in an actual icosahedron has edge-degree 5. See en.wikipedia.org/wiki/Regular_icosahedron#Icosahedral_graph Your graph may actually represent a column of 3 octahedra glued together face to face...?
– antkam
Nov 22 '18 at 16:20