Show that the 3-color problem is in P when the input graph is a tree.
$begingroup$
This one is from university assignment. I am completely stuck on this one and I searched the internet but couldn't find a explanation.
Show that the 3-color problem is in P when the input graph is a tree.
Any explanation would be appreciated.
graph-theory computational-complexity coloring np-complete
$endgroup$
add a comment |
$begingroup$
This one is from university assignment. I am completely stuck on this one and I searched the internet but couldn't find a explanation.
Show that the 3-color problem is in P when the input graph is a tree.
Any explanation would be appreciated.
graph-theory computational-complexity coloring np-complete
$endgroup$
2
$begingroup$
Every tree is $2$-colorable, so I'm not quite getting the question.
$endgroup$
– Cheerful Parsnip
Nov 29 '18 at 21:09
$begingroup$
yeah I know but thats what the question states. I have no clue.
$endgroup$
– r4dicalbuRn
Nov 29 '18 at 21:11
add a comment |
$begingroup$
This one is from university assignment. I am completely stuck on this one and I searched the internet but couldn't find a explanation.
Show that the 3-color problem is in P when the input graph is a tree.
Any explanation would be appreciated.
graph-theory computational-complexity coloring np-complete
$endgroup$
This one is from university assignment. I am completely stuck on this one and I searched the internet but couldn't find a explanation.
Show that the 3-color problem is in P when the input graph is a tree.
Any explanation would be appreciated.
graph-theory computational-complexity coloring np-complete
graph-theory computational-complexity coloring np-complete
edited Nov 29 '18 at 21:41
gt6989b
33.6k22453
33.6k22453
asked Nov 29 '18 at 21:06
r4dicalbuRnr4dicalbuRn
31
31
2
$begingroup$
Every tree is $2$-colorable, so I'm not quite getting the question.
$endgroup$
– Cheerful Parsnip
Nov 29 '18 at 21:09
$begingroup$
yeah I know but thats what the question states. I have no clue.
$endgroup$
– r4dicalbuRn
Nov 29 '18 at 21:11
add a comment |
2
$begingroup$
Every tree is $2$-colorable, so I'm not quite getting the question.
$endgroup$
– Cheerful Parsnip
Nov 29 '18 at 21:09
$begingroup$
yeah I know but thats what the question states. I have no clue.
$endgroup$
– r4dicalbuRn
Nov 29 '18 at 21:11
2
2
$begingroup$
Every tree is $2$-colorable, so I'm not quite getting the question.
$endgroup$
– Cheerful Parsnip
Nov 29 '18 at 21:09
$begingroup$
Every tree is $2$-colorable, so I'm not quite getting the question.
$endgroup$
– Cheerful Parsnip
Nov 29 '18 at 21:09
$begingroup$
yeah I know but thats what the question states. I have no clue.
$endgroup$
– r4dicalbuRn
Nov 29 '18 at 21:11
$begingroup$
yeah I know but thats what the question states. I have no clue.
$endgroup$
– r4dicalbuRn
Nov 29 '18 at 21:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
UPDATED following a suggestion in the comments.
- Validate that the input is a tree
- Answer "yes"
Both (1) and (2) are doable in polynomial time (how?) so this is in $P$
$endgroup$
$begingroup$
Does this prove that 3-color is in P? 2-coloring can be done by BFS and tree is a bipartite by definition. But what about 3-color?
$endgroup$
– r4dicalbuRn
Nov 29 '18 at 21:33
$begingroup$
@semal259 any 2-coloring is also a 3-coloring, so it does indeed prove that 3-coloring a tree is in $P$
$endgroup$
– gt6989b
Nov 29 '18 at 21:40
1
$begingroup$
One could argue that it's even simpler than this, since the $2$-color problem should have a yes/no answer. Just write a program that always produces the answer yes (possibly after verifying that the input is a tree.)
$endgroup$
– Cheerful Parsnip
Nov 29 '18 at 21:41
$begingroup$
@CheerfulParsnip :-) +1, funny, yes, definitely
$endgroup$
– gt6989b
Nov 29 '18 at 21:43
1
$begingroup$
@semal259 Indeed, a tree is $k$-colorable in polynomial time for all $k ge 2$.
$endgroup$
– gt6989b
Nov 29 '18 at 21:47
|
show 2 more comments
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
UPDATED following a suggestion in the comments.
- Validate that the input is a tree
- Answer "yes"
Both (1) and (2) are doable in polynomial time (how?) so this is in $P$
$endgroup$
$begingroup$
Does this prove that 3-color is in P? 2-coloring can be done by BFS and tree is a bipartite by definition. But what about 3-color?
$endgroup$
– r4dicalbuRn
Nov 29 '18 at 21:33
$begingroup$
@semal259 any 2-coloring is also a 3-coloring, so it does indeed prove that 3-coloring a tree is in $P$
$endgroup$
– gt6989b
Nov 29 '18 at 21:40
1
$begingroup$
One could argue that it's even simpler than this, since the $2$-color problem should have a yes/no answer. Just write a program that always produces the answer yes (possibly after verifying that the input is a tree.)
$endgroup$
– Cheerful Parsnip
Nov 29 '18 at 21:41
$begingroup$
@CheerfulParsnip :-) +1, funny, yes, definitely
$endgroup$
– gt6989b
Nov 29 '18 at 21:43
1
$begingroup$
@semal259 Indeed, a tree is $k$-colorable in polynomial time for all $k ge 2$.
$endgroup$
– gt6989b
Nov 29 '18 at 21:47
|
show 2 more comments
$begingroup$
UPDATED following a suggestion in the comments.
- Validate that the input is a tree
- Answer "yes"
Both (1) and (2) are doable in polynomial time (how?) so this is in $P$
$endgroup$
$begingroup$
Does this prove that 3-color is in P? 2-coloring can be done by BFS and tree is a bipartite by definition. But what about 3-color?
$endgroup$
– r4dicalbuRn
Nov 29 '18 at 21:33
$begingroup$
@semal259 any 2-coloring is also a 3-coloring, so it does indeed prove that 3-coloring a tree is in $P$
$endgroup$
– gt6989b
Nov 29 '18 at 21:40
1
$begingroup$
One could argue that it's even simpler than this, since the $2$-color problem should have a yes/no answer. Just write a program that always produces the answer yes (possibly after verifying that the input is a tree.)
$endgroup$
– Cheerful Parsnip
Nov 29 '18 at 21:41
$begingroup$
@CheerfulParsnip :-) +1, funny, yes, definitely
$endgroup$
– gt6989b
Nov 29 '18 at 21:43
1
$begingroup$
@semal259 Indeed, a tree is $k$-colorable in polynomial time for all $k ge 2$.
$endgroup$
– gt6989b
Nov 29 '18 at 21:47
|
show 2 more comments
$begingroup$
UPDATED following a suggestion in the comments.
- Validate that the input is a tree
- Answer "yes"
Both (1) and (2) are doable in polynomial time (how?) so this is in $P$
$endgroup$
UPDATED following a suggestion in the comments.
- Validate that the input is a tree
- Answer "yes"
Both (1) and (2) are doable in polynomial time (how?) so this is in $P$
edited Nov 29 '18 at 21:43
answered Nov 29 '18 at 21:27
gt6989bgt6989b
33.6k22453
33.6k22453
$begingroup$
Does this prove that 3-color is in P? 2-coloring can be done by BFS and tree is a bipartite by definition. But what about 3-color?
$endgroup$
– r4dicalbuRn
Nov 29 '18 at 21:33
$begingroup$
@semal259 any 2-coloring is also a 3-coloring, so it does indeed prove that 3-coloring a tree is in $P$
$endgroup$
– gt6989b
Nov 29 '18 at 21:40
1
$begingroup$
One could argue that it's even simpler than this, since the $2$-color problem should have a yes/no answer. Just write a program that always produces the answer yes (possibly after verifying that the input is a tree.)
$endgroup$
– Cheerful Parsnip
Nov 29 '18 at 21:41
$begingroup$
@CheerfulParsnip :-) +1, funny, yes, definitely
$endgroup$
– gt6989b
Nov 29 '18 at 21:43
1
$begingroup$
@semal259 Indeed, a tree is $k$-colorable in polynomial time for all $k ge 2$.
$endgroup$
– gt6989b
Nov 29 '18 at 21:47
|
show 2 more comments
$begingroup$
Does this prove that 3-color is in P? 2-coloring can be done by BFS and tree is a bipartite by definition. But what about 3-color?
$endgroup$
– r4dicalbuRn
Nov 29 '18 at 21:33
$begingroup$
@semal259 any 2-coloring is also a 3-coloring, so it does indeed prove that 3-coloring a tree is in $P$
$endgroup$
– gt6989b
Nov 29 '18 at 21:40
1
$begingroup$
One could argue that it's even simpler than this, since the $2$-color problem should have a yes/no answer. Just write a program that always produces the answer yes (possibly after verifying that the input is a tree.)
$endgroup$
– Cheerful Parsnip
Nov 29 '18 at 21:41
$begingroup$
@CheerfulParsnip :-) +1, funny, yes, definitely
$endgroup$
– gt6989b
Nov 29 '18 at 21:43
1
$begingroup$
@semal259 Indeed, a tree is $k$-colorable in polynomial time for all $k ge 2$.
$endgroup$
– gt6989b
Nov 29 '18 at 21:47
$begingroup$
Does this prove that 3-color is in P? 2-coloring can be done by BFS and tree is a bipartite by definition. But what about 3-color?
$endgroup$
– r4dicalbuRn
Nov 29 '18 at 21:33
$begingroup$
Does this prove that 3-color is in P? 2-coloring can be done by BFS and tree is a bipartite by definition. But what about 3-color?
$endgroup$
– r4dicalbuRn
Nov 29 '18 at 21:33
$begingroup$
@semal259 any 2-coloring is also a 3-coloring, so it does indeed prove that 3-coloring a tree is in $P$
$endgroup$
– gt6989b
Nov 29 '18 at 21:40
$begingroup$
@semal259 any 2-coloring is also a 3-coloring, so it does indeed prove that 3-coloring a tree is in $P$
$endgroup$
– gt6989b
Nov 29 '18 at 21:40
1
1
$begingroup$
One could argue that it's even simpler than this, since the $2$-color problem should have a yes/no answer. Just write a program that always produces the answer yes (possibly after verifying that the input is a tree.)
$endgroup$
– Cheerful Parsnip
Nov 29 '18 at 21:41
$begingroup$
One could argue that it's even simpler than this, since the $2$-color problem should have a yes/no answer. Just write a program that always produces the answer yes (possibly after verifying that the input is a tree.)
$endgroup$
– Cheerful Parsnip
Nov 29 '18 at 21:41
$begingroup$
@CheerfulParsnip :-) +1, funny, yes, definitely
$endgroup$
– gt6989b
Nov 29 '18 at 21:43
$begingroup$
@CheerfulParsnip :-) +1, funny, yes, definitely
$endgroup$
– gt6989b
Nov 29 '18 at 21:43
1
1
$begingroup$
@semal259 Indeed, a tree is $k$-colorable in polynomial time for all $k ge 2$.
$endgroup$
– gt6989b
Nov 29 '18 at 21:47
$begingroup$
@semal259 Indeed, a tree is $k$-colorable in polynomial time for all $k ge 2$.
$endgroup$
– gt6989b
Nov 29 '18 at 21:47
|
show 2 more comments
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$begingroup$
Every tree is $2$-colorable, so I'm not quite getting the question.
$endgroup$
– Cheerful Parsnip
Nov 29 '18 at 21:09
$begingroup$
yeah I know but thats what the question states. I have no clue.
$endgroup$
– r4dicalbuRn
Nov 29 '18 at 21:11