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How to colour the US map with Yellow, Green, Red and Blue to minimize the number of states with the colour of Green
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How to colour the US map with Yellow, Green, Red and Blue to minimize the number of states with the colour of Green
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Did the Appel/Haken graph colouring (four colour map) proof really not contribute to understanding?How many four-vertex graphs are there up to isomorphism;$Z_n backslash 0$ splits into octetsHow many different cubes can be obtained if four colours are used?Coloring the 6 vertices of a regular hexagon with a limited use per colorKempe's proof of the four colour theoremAlgorithm to solve this grid puzzle?Explain this proof of the 5-color theoremFinding a maximum connected planar graph to prove the four colour theoremGraph colouring problem
$begingroup$
I want to colour the US (only the states) map with Yellow, Green, Red and Blue. I was wondering what would be the lowest number of states with the colour of Green. We can of course use the other colours as much as we want. Please note that I want to follow the Four Color Theorem rules.
Motivation:
I am studying graph theory and I want to know if there is a way that we could limit the use of the fourth colour as much as possible. This is not a homework problem.
My attempt:
I have tried many variations and can limit it to 6 and it seems like the
minimum possible but there are many possibilities to try ($4^50$). Therefore I was wondering if there is a simpler method? Thank you in advance.
Clarification:
I am interested in only the mainland of USA. For states like Michigan that are split, I used the same colour for both parts (since they were not connected directly).
graph-theory recreational-mathematics coloring
asked 19 hours ago
Bor KariBor Kari
44910
$endgroup$
|
show 2 more comments
$begingroup$
I want to colour the US (only the states) map with Yellow, Green, Red and Blue. I was wondering what would be the lowest number of states with the colour of Green. We can of course use the other colours as much as we want. Please note that I want to follow the Four Color Theorem rules.
Motivation:
I am studying graph theory and I want to know if there is a way that we could limit the use of the fourth colour as much as possible. This is not a homework problem.
My attempt:
I have tried many variations and can limit it to 6 and it seems like the
minimum possible but there are many possibilities to try ($4^50$). Therefore I was wondering if there is a simpler method? Thank you in advance.
Clarification:
I am interested in only the mainland of USA. For states like Michigan that are split, I used the same colour for both parts (since they were not connected directly).
graph-theory recreational-mathematics coloring
asked 19 hours ago
Bor KariBor Kari
44910
$endgroup$
1
$begingroup$
you would need to agree on a favorite version of the graph. In the actual US, there are islands, states split into disconnected regions, other things forbidden
$endgroup$
– Will Jagy
19 hours ago
1
$begingroup$
blog.computationalcomplexity.org/2006/05/… They correctly point out that three colors cannot work, as Nevada has an odd number of neighbors
$endgroup$
– Will Jagy
19 hours ago
$begingroup$
thank you for your suggestion, I made a few clarifications.
$endgroup$
– Bor Kari
18 hours ago
$begingroup$
@WillJagy Maine has an odd number of neighbours too, but that's not much of a problem.
$endgroup$
– Marc van Leeuwen
5 hours ago
1
$begingroup$
There are not infinite possibilities. There is only $4^50$ ways of assigning colors to all the states even if we consider no other constraints or symmetries.
$endgroup$
– Derek Elkins
2 hours ago
|
show 2 more comments
$begingroup$
I want to colour the US (only the states) map with Yellow, Green, Red and Blue. I was wondering what would be the lowest number of states with the colour of Green. We can of course use the other colours as much as we want. Please note that I want to follow the Four Color Theorem rules.
Motivation:
I am studying graph theory and I want to know if there is a way that we could limit the use of the fourth colour as much as possible. This is not a homework problem.
My attempt:
I have tried many variations and can limit it to 6 and it seems like the
minimum possible but there are many possibilities to try ($4^50$). Therefore I was wondering if there is a simpler method? Thank you in advance.
Clarification:
I am interested in only the mainland of USA. For states like Michigan that are split, I used the same colour for both parts (since they were not connected directly).
graph-theory recreational-mathematics coloring
asked 19 hours ago
Bor KariBor Kari
44910
$endgroup$
I want to colour the US (only the states) map with Yellow, Green, Red and Blue. I was wondering what would be the lowest number of states with the colour of Green. We can of course use the other colours as much as we want. Please note that I want to follow the Four Color Theorem rules.
Motivation:
I am studying graph theory and I want to know if there is a way that we could limit the use of the fourth colour as much as possible. This is not a homework problem.
My attempt:
I have tried many variations and can limit it to 6 and it seems like the
minimum possible but there are many possibilities to try ($4^50$). Therefore I was wondering if there is a simpler method? Thank you in advance.
Clarification:
I am interested in only the mainland of USA. For states like Michigan that are split, I used the same colour for both parts (since they were not connected directly).
graph-theory recreational-mathematics coloring
graph-theory recreational-mathematics coloring
asked 19 hours ago
Bor KariBor Kari
44910
asked 19 hours ago
Bor KariBor Kari
44910
edited 2 hours ago
Bor Kari
asked 19 hours ago
Bor KariBor Kari
44910
asked 19 hours ago
Bor KariBor Kari
44910
asked 19 hours ago
Bor KariBor Kari
44910
44910
1
$begingroup$
you would need to agree on a favorite version of the graph. In the actual US, there are islands, states split into disconnected regions, other things forbidden
$endgroup$
– Will Jagy
19 hours ago
1
$begingroup$
blog.computationalcomplexity.org/2006/05/… They correctly point out that three colors cannot work, as Nevada has an odd number of neighbors
$endgroup$
– Will Jagy
19 hours ago
$begingroup$
thank you for your suggestion, I made a few clarifications.
$endgroup$
– Bor Kari
18 hours ago
$begingroup$
@WillJagy Maine has an odd number of neighbours too, but that's not much of a problem.
$endgroup$
– Marc van Leeuwen
5 hours ago
1
$begingroup$
There are not infinite possibilities. There is only $4^50$ ways of assigning colors to all the states even if we consider no other constraints or symmetries.
$endgroup$
– Derek Elkins
2 hours ago
|
show 2 more comments
1
$begingroup$
you would need to agree on a favorite version of the graph. In the actual US, there are islands, states split into disconnected regions, other things forbidden
$endgroup$
– Will Jagy
19 hours ago
1
$begingroup$
blog.computationalcomplexity.org/2006/05/… They correctly point out that three colors cannot work, as Nevada has an odd number of neighbors
$endgroup$
– Will Jagy
19 hours ago
$begingroup$
thank you for your suggestion, I made a few clarifications.
$endgroup$
– Bor Kari
18 hours ago
$begingroup$
@WillJagy Maine has an odd number of neighbours too, but that's not much of a problem.
$endgroup$
– Marc van Leeuwen
5 hours ago
1
$begingroup$
There are not infinite possibilities. There is only $4^50$ ways of assigning colors to all the states even if we consider no other constraints or symmetries.
$endgroup$
– Derek Elkins
2 hours ago
1
1
$begingroup$
you would need to agree on a favorite version of the graph. In the actual US, there are islands, states split into disconnected regions, other things forbidden
$endgroup$
– Will Jagy
19 hours ago
$begingroup$
you would need to agree on a favorite version of the graph. In the actual US, there are islands, states split into disconnected regions, other things forbidden
$endgroup$
– Will Jagy
19 hours ago
1
1
$begingroup$
blog.computationalcomplexity.org/2006/05/… They correctly point out that three colors cannot work, as Nevada has an odd number of neighbors
$endgroup$
– Will Jagy
19 hours ago
$begingroup$
blog.computationalcomplexity.org/2006/05/… They correctly point out that three colors cannot work, as Nevada has an odd number of neighbors
$endgroup$
– Will Jagy
19 hours ago
$begingroup$
thank you for your suggestion, I made a few clarifications.
$endgroup$
– Bor Kari
18 hours ago
$begingroup$
thank you for your suggestion, I made a few clarifications.
$endgroup$
– Bor Kari
18 hours ago
$begingroup$
@WillJagy Maine has an odd number of neighbours too, but that's not much of a problem.
$endgroup$
– Marc van Leeuwen
5 hours ago
$begingroup$
@WillJagy Maine has an odd number of neighbours too, but that's not much of a problem.
$endgroup$
– Marc van Leeuwen
5 hours ago
1
1
$begingroup$
There are not infinite possibilities. There is only $4^50$ ways of assigning colors to all the states even if we consider no other constraints or symmetries.
$endgroup$
– Derek Elkins
2 hours ago
$begingroup$
There are not infinite possibilities. There is only $4^50$ ways of assigning colors to all the states even if we consider no other constraints or symmetries.
$endgroup$
– Derek Elkins
2 hours ago
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The minimum is two states that use the fourth color. Nevada and its five neighbors cannot be colored with only three colors, and similarly West Virginia and its five neighbors cannot be colored with only three colors. In both cases, once you color the center state one color (say, red), you can't use it again on its neighbors: without using green, they'd have to alternate yellow-blue-yellow-blue, but because the number of neighbors is odd, you'd get stuck at the end.
(In the comments, David K points out that Kentucky is a third state with the same problem: it has seven neighbors. But this doesn't force us to use a third green state, because Kentucky and West Virginia share a border and some common neighbors.)
Using only two green states is possible. If we color Arizona (dealing with the Nevada situation) and Ohio (dealing with West Virginia and Kentucky) both green, then the remainder of the map can be completed using only blue, red, and yellow:
Adjacencies between the states may be easier to see here.
answered 18 hours ago
Misha LavrovMisha Lavrov
49.5k758109
$endgroup$
$begingroup$
I need a better atlas. I'm looking at the Philadelphia area, I cannot tell what happens among Pennsylvania, New Jersey, Delaware, Maryland.
$endgroup$
– Will Jagy
18 hours ago
2
$begingroup$
just curious -- did you write code to do this, or did you do this by hand?
$endgroup$
– antkam
17 hours ago
3
$begingroup$
@antkam By hand. I found two subgraphs where a fourth color is forced, and chose a state from each of them to color green that seemed to be a good choice. Then I just tried to color the rest with three colors - and for that, once you color the first two states, most of the rest of the map is forced, except for a few states like Maine.
$endgroup$
– Misha Lavrov
16 hours ago
1
$begingroup$
@WillJagy NJ and MD don't touch because DE interposes. All the other edges exist.
$endgroup$
– hobbs
15 hours ago
2
$begingroup$
Kentucky has seven neighbors, but coloring Ohio green fixes that as well as WV. I also notice that you managed to use four colors for UT, CO, AZ, and NM, which makes the map a little clearer but I think is not strictly required by the usual rules, that is, under the usual rules (as far as I know) you could make AZ yellow and CA green.
$endgroup$
– David K
9 hours ago
|
show 6 more comments
Your Answer
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1 Answer
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$begingroup$
The minimum is two states that use the fourth color. Nevada and its five neighbors cannot be colored with only three colors, and similarly West Virginia and its five neighbors cannot be colored with only three colors. In both cases, once you color the center state one color (say, red), you can't use it again on its neighbors: without using green, they'd have to alternate yellow-blue-yellow-blue, but because the number of neighbors is odd, you'd get stuck at the end.
(In the comments, David K points out that Kentucky is a third state with the same problem: it has seven neighbors. But this doesn't force us to use a third green state, because Kentucky and West Virginia share a border and some common neighbors.)
Using only two green states is possible. If we color Arizona (dealing with the Nevada situation) and Ohio (dealing with West Virginia and Kentucky) both green, then the remainder of the map can be completed using only blue, red, and yellow:
Adjacencies between the states may be easier to see here.
answered 18 hours ago
Misha LavrovMisha Lavrov
49.5k758109
$endgroup$
$begingroup$
I need a better atlas. I'm looking at the Philadelphia area, I cannot tell what happens among Pennsylvania, New Jersey, Delaware, Maryland.
$endgroup$
– Will Jagy
18 hours ago
2
$begingroup$
just curious -- did you write code to do this, or did you do this by hand?
$endgroup$
– antkam
17 hours ago
3
$begingroup$
@antkam By hand. I found two subgraphs where a fourth color is forced, and chose a state from each of them to color green that seemed to be a good choice. Then I just tried to color the rest with three colors - and for that, once you color the first two states, most of the rest of the map is forced, except for a few states like Maine.
$endgroup$
– Misha Lavrov
16 hours ago
1
$begingroup$
@WillJagy NJ and MD don't touch because DE interposes. All the other edges exist.
$endgroup$
– hobbs
15 hours ago
2
$begingroup$
Kentucky has seven neighbors, but coloring Ohio green fixes that as well as WV. I also notice that you managed to use four colors for UT, CO, AZ, and NM, which makes the map a little clearer but I think is not strictly required by the usual rules, that is, under the usual rules (as far as I know) you could make AZ yellow and CA green.
$endgroup$
– David K
9 hours ago
|
show 6 more comments
$begingroup$
The minimum is two states that use the fourth color. Nevada and its five neighbors cannot be colored with only three colors, and similarly West Virginia and its five neighbors cannot be colored with only three colors. In both cases, once you color the center state one color (say, red), you can't use it again on its neighbors: without using green, they'd have to alternate yellow-blue-yellow-blue, but because the number of neighbors is odd, you'd get stuck at the end.
(In the comments, David K points out that Kentucky is a third state with the same problem: it has seven neighbors. But this doesn't force us to use a third green state, because Kentucky and West Virginia share a border and some common neighbors.)
Using only two green states is possible. If we color Arizona (dealing with the Nevada situation) and Ohio (dealing with West Virginia and Kentucky) both green, then the remainder of the map can be completed using only blue, red, and yellow:
Adjacencies between the states may be easier to see here.
answered 18 hours ago
Misha LavrovMisha Lavrov
49.5k758109
$endgroup$
$begingroup$
I need a better atlas. I'm looking at the Philadelphia area, I cannot tell what happens among Pennsylvania, New Jersey, Delaware, Maryland.
$endgroup$
– Will Jagy
18 hours ago
2
$begingroup$
just curious -- did you write code to do this, or did you do this by hand?
$endgroup$
– antkam
17 hours ago
3
$begingroup$
@antkam By hand. I found two subgraphs where a fourth color is forced, and chose a state from each of them to color green that seemed to be a good choice. Then I just tried to color the rest with three colors - and for that, once you color the first two states, most of the rest of the map is forced, except for a few states like Maine.
$endgroup$
– Misha Lavrov
16 hours ago
1
$begingroup$
@WillJagy NJ and MD don't touch because DE interposes. All the other edges exist.
$endgroup$
– hobbs
15 hours ago
2
$begingroup$
Kentucky has seven neighbors, but coloring Ohio green fixes that as well as WV. I also notice that you managed to use four colors for UT, CO, AZ, and NM, which makes the map a little clearer but I think is not strictly required by the usual rules, that is, under the usual rules (as far as I know) you could make AZ yellow and CA green.
$endgroup$
– David K
9 hours ago
|
show 6 more comments
$begingroup$
The minimum is two states that use the fourth color. Nevada and its five neighbors cannot be colored with only three colors, and similarly West Virginia and its five neighbors cannot be colored with only three colors. In both cases, once you color the center state one color (say, red), you can't use it again on its neighbors: without using green, they'd have to alternate yellow-blue-yellow-blue, but because the number of neighbors is odd, you'd get stuck at the end.
(In the comments, David K points out that Kentucky is a third state with the same problem: it has seven neighbors. But this doesn't force us to use a third green state, because Kentucky and West Virginia share a border and some common neighbors.)
Using only two green states is possible. If we color Arizona (dealing with the Nevada situation) and Ohio (dealing with West Virginia and Kentucky) both green, then the remainder of the map can be completed using only blue, red, and yellow:
Adjacencies between the states may be easier to see here.
answered 18 hours ago
Misha LavrovMisha Lavrov
49.5k758109
$endgroup$
The minimum is two states that use the fourth color. Nevada and its five neighbors cannot be colored with only three colors, and similarly West Virginia and its five neighbors cannot be colored with only three colors. In both cases, once you color the center state one color (say, red), you can't use it again on its neighbors: without using green, they'd have to alternate yellow-blue-yellow-blue, but because the number of neighbors is odd, you'd get stuck at the end.
(In the comments, David K points out that Kentucky is a third state with the same problem: it has seven neighbors. But this doesn't force us to use a third green state, because Kentucky and West Virginia share a border and some common neighbors.)
Using only two green states is possible. If we color Arizona (dealing with the Nevada situation) and Ohio (dealing with West Virginia and Kentucky) both green, then the remainder of the map can be completed using only blue, red, and yellow:
Adjacencies between the states may be easier to see here.
answered 18 hours ago
Misha LavrovMisha Lavrov
49.5k758109
edited 8 hours ago
answered 18 hours ago
Misha LavrovMisha Lavrov
49.5k758109
answered 18 hours ago
Misha LavrovMisha Lavrov
49.5k758109
answered 18 hours ago
Misha LavrovMisha Lavrov
49.5k758109
49.5k758109
$begingroup$
I need a better atlas. I'm looking at the Philadelphia area, I cannot tell what happens among Pennsylvania, New Jersey, Delaware, Maryland.
$endgroup$
– Will Jagy
18 hours ago
2
$begingroup$
just curious -- did you write code to do this, or did you do this by hand?
$endgroup$
– antkam
17 hours ago
3
$begingroup$
@antkam By hand. I found two subgraphs where a fourth color is forced, and chose a state from each of them to color green that seemed to be a good choice. Then I just tried to color the rest with three colors - and for that, once you color the first two states, most of the rest of the map is forced, except for a few states like Maine.
$endgroup$
– Misha Lavrov
16 hours ago
1
$begingroup$
@WillJagy NJ and MD don't touch because DE interposes. All the other edges exist.
$endgroup$
– hobbs
15 hours ago
2
$begingroup$
Kentucky has seven neighbors, but coloring Ohio green fixes that as well as WV. I also notice that you managed to use four colors for UT, CO, AZ, and NM, which makes the map a little clearer but I think is not strictly required by the usual rules, that is, under the usual rules (as far as I know) you could make AZ yellow and CA green.
$endgroup$
– David K
9 hours ago
|
show 6 more comments
$begingroup$
I need a better atlas. I'm looking at the Philadelphia area, I cannot tell what happens among Pennsylvania, New Jersey, Delaware, Maryland.
$endgroup$
– Will Jagy
18 hours ago
2
$begingroup$
just curious -- did you write code to do this, or did you do this by hand?
$endgroup$
– antkam
17 hours ago
3
$begingroup$
@antkam By hand. I found two subgraphs where a fourth color is forced, and chose a state from each of them to color green that seemed to be a good choice. Then I just tried to color the rest with three colors - and for that, once you color the first two states, most of the rest of the map is forced, except for a few states like Maine.
$endgroup$
– Misha Lavrov
16 hours ago
1
$begingroup$
@WillJagy NJ and MD don't touch because DE interposes. All the other edges exist.
$endgroup$
– hobbs
15 hours ago
2
$begingroup$
Kentucky has seven neighbors, but coloring Ohio green fixes that as well as WV. I also notice that you managed to use four colors for UT, CO, AZ, and NM, which makes the map a little clearer but I think is not strictly required by the usual rules, that is, under the usual rules (as far as I know) you could make AZ yellow and CA green.
$endgroup$
– David K
9 hours ago
$begingroup$
I need a better atlas. I'm looking at the Philadelphia area, I cannot tell what happens among Pennsylvania, New Jersey, Delaware, Maryland.
$endgroup$
– Will Jagy
18 hours ago
$begingroup$
I need a better atlas. I'm looking at the Philadelphia area, I cannot tell what happens among Pennsylvania, New Jersey, Delaware, Maryland.
$endgroup$
– Will Jagy
18 hours ago
2
2
$begingroup$
just curious -- did you write code to do this, or did you do this by hand?
$endgroup$
– antkam
17 hours ago
$begingroup$
just curious -- did you write code to do this, or did you do this by hand?
$endgroup$
– antkam
17 hours ago
3
3
$begingroup$
@antkam By hand. I found two subgraphs where a fourth color is forced, and chose a state from each of them to color green that seemed to be a good choice. Then I just tried to color the rest with three colors - and for that, once you color the first two states, most of the rest of the map is forced, except for a few states like Maine.
$endgroup$
– Misha Lavrov
16 hours ago
$begingroup$
@antkam By hand. I found two subgraphs where a fourth color is forced, and chose a state from each of them to color green that seemed to be a good choice. Then I just tried to color the rest with three colors - and for that, once you color the first two states, most of the rest of the map is forced, except for a few states like Maine.
$endgroup$
– Misha Lavrov
16 hours ago
1
1
$begingroup$
@WillJagy NJ and MD don't touch because DE interposes. All the other edges exist.
$endgroup$
– hobbs
15 hours ago
$begingroup$
@WillJagy NJ and MD don't touch because DE interposes. All the other edges exist.
$endgroup$
– hobbs
15 hours ago
2
2
$begingroup$
Kentucky has seven neighbors, but coloring Ohio green fixes that as well as WV. I also notice that you managed to use four colors for UT, CO, AZ, and NM, which makes the map a little clearer but I think is not strictly required by the usual rules, that is, under the usual rules (as far as I know) you could make AZ yellow and CA green.
$endgroup$
– David K
9 hours ago
$begingroup$
Kentucky has seven neighbors, but coloring Ohio green fixes that as well as WV. I also notice that you managed to use four colors for UT, CO, AZ, and NM, which makes the map a little clearer but I think is not strictly required by the usual rules, that is, under the usual rules (as far as I know) you could make AZ yellow and CA green.
$endgroup$
– David K
9 hours ago
|
show 6 more comments
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-coloring, graph-theory, recreational-mathematics
1
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you would need to agree on a favorite version of the graph. In the actual US, there are islands, states split into disconnected regions, other things forbidden
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– Will Jagy
19 hours ago
1
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blog.computationalcomplexity.org/2006/05/… They correctly point out that three colors cannot work, as Nevada has an odd number of neighbors
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– Will Jagy
19 hours ago
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thank you for your suggestion, I made a few clarifications.
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– Bor Kari
18 hours ago
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@WillJagy Maine has an odd number of neighbours too, but that's not much of a problem.
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– Marc van Leeuwen
5 hours ago
1
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There are not infinite possibilities. There is only $4^50$ ways of assigning colors to all the states even if we consider no other constraints or symmetries.
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– Derek Elkins
2 hours ago