Is Hamiltoncycle and Euler cycle NP-complete or not?Every $t$-coloring of $K_{2t+1}$ contains a monochromatic...
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Is Hamiltoncycle and Euler cycle NP-complete or not?
Every $t$-coloring of $K_{2t+1}$ contains a monochromatic cycleComplete graph-coloringColouring $n$ vertices with $(n-1)$ colours in a graph with chromatic number $n$Check existence of walk visiting every node in graph odd-timesUse of pigeonhole principle in ramsey-theorem about monochromatic triangles.Alternating cycle in a graphGraph Coloring : How to Think about Algorithms Exercise 1.6.2Complete graphs in the plane with colored edges where an edge don't cross edges with same colorIs it possible to start with a partially colored graph for a graph $G$ and complete it into a coloring with $chi(G)$ colors?A new graph invariant? The maximum number of non-equivalent colorings with $n$ colors.
$begingroup$
Question: Determine which of these computational problems are NP-complete.
- Determining whether a number with $n$ digits is a prime number.
- Determining whether a graph with $n$ nodes has a hamiltoncycle.
- Determining whether a graph with $n$ nodes has an eulercycle.
- Determining whether a graph with $n$ nodes can be coloured with $2$ colors.
- Determining whether a graph with $n$ nodes can be colored with $3$ colors.
My Attempts:
1) I'm not sure if this is correct because the question specifies the number of digits to be $n$ and not that the actual prime number is $n$. So in the latter case, we only need to check the division of $n$ with primes that are less than $sqrt{n}$. This reasoning doesnt feel correct.
2) So if we first choose one node we can choose among $n$ of them, then we can choose to go to $n-1$ other nodes, and then $n-2$ and so on. So we get $n!$ which is not in polynomial time, thus it is NP-complete.
3) This one is trickier, i can't really use a similar argument as for the hamiltoncycle. I don't know the number of edges. How should I think here?
4) I assume that a coloring here should be such that no neighbouring/connected nodes have the same colour. But I have no idea how these $n$ nodes are connected in order to figure this out?
5) Same question as 4) essentially.
discrete-mathematics graph-theory
asked Mar 18 at 13:38
ParsevalParseval
3,0721719
$endgroup$
add a comment |
$begingroup$
Question: Determine which of these computational problems are NP-complete.
- Determining whether a number with $n$ digits is a prime number.
- Determining whether a graph with $n$ nodes has a hamiltoncycle.
- Determining whether a graph with $n$ nodes has an eulercycle.
- Determining whether a graph with $n$ nodes can be coloured with $2$ colors.
- Determining whether a graph with $n$ nodes can be colored with $3$ colors.
My Attempts:
1) I'm not sure if this is correct because the question specifies the number of digits to be $n$ and not that the actual prime number is $n$. So in the latter case, we only need to check the division of $n$ with primes that are less than $sqrt{n}$. This reasoning doesnt feel correct.
2) So if we first choose one node we can choose among $n$ of them, then we can choose to go to $n-1$ other nodes, and then $n-2$ and so on. So we get $n!$ which is not in polynomial time, thus it is NP-complete.
3) This one is trickier, i can't really use a similar argument as for the hamiltoncycle. I don't know the number of edges. How should I think here?
4) I assume that a coloring here should be such that no neighbouring/connected nodes have the same colour. But I have no idea how these $n$ nodes are connected in order to figure this out?
5) Same question as 4) essentially.
discrete-mathematics graph-theory
asked Mar 18 at 13:38
ParsevalParseval
3,0721719
$endgroup$
$begingroup$
For (3), it is known that a graph has an eulerian cycle if and only if all the nodes have an even degree. That's linear on the number of nodes.
$endgroup$
– frabala
Mar 18 at 13:52
$begingroup$
What do you mean by "Thats linear on the number of nodes"? So the problem is then to check whether or not all of these $n$ nodes have an even degree. How is this a linear problem?
$endgroup$
– Parseval
Mar 18 at 14:21
1
$begingroup$
You're right. I falsely assumed constant time for checking the degree of a node.
$endgroup$
– frabala
Mar 18 at 14:29
add a comment |
$begingroup$
Question: Determine which of these computational problems are NP-complete.
- Determining whether a number with $n$ digits is a prime number.
- Determining whether a graph with $n$ nodes has a hamiltoncycle.
- Determining whether a graph with $n$ nodes has an eulercycle.
- Determining whether a graph with $n$ nodes can be coloured with $2$ colors.
- Determining whether a graph with $n$ nodes can be colored with $3$ colors.
My Attempts:
1) I'm not sure if this is correct because the question specifies the number of digits to be $n$ and not that the actual prime number is $n$. So in the latter case, we only need to check the division of $n$ with primes that are less than $sqrt{n}$. This reasoning doesnt feel correct.
2) So if we first choose one node we can choose among $n$ of them, then we can choose to go to $n-1$ other nodes, and then $n-2$ and so on. So we get $n!$ which is not in polynomial time, thus it is NP-complete.
3) This one is trickier, i can't really use a similar argument as for the hamiltoncycle. I don't know the number of edges. How should I think here?
4) I assume that a coloring here should be such that no neighbouring/connected nodes have the same colour. But I have no idea how these $n$ nodes are connected in order to figure this out?
5) Same question as 4) essentially.
discrete-mathematics graph-theory
asked Mar 18 at 13:38
ParsevalParseval
3,0721719
$endgroup$
Question: Determine which of these computational problems are NP-complete.
- Determining whether a number with $n$ digits is a prime number.
- Determining whether a graph with $n$ nodes has a hamiltoncycle.
- Determining whether a graph with $n$ nodes has an eulercycle.
- Determining whether a graph with $n$ nodes can be coloured with $2$ colors.
- Determining whether a graph with $n$ nodes can be colored with $3$ colors.
My Attempts:
1) I'm not sure if this is correct because the question specifies the number of digits to be $n$ and not that the actual prime number is $n$. So in the latter case, we only need to check the division of $n$ with primes that are less than $sqrt{n}$. This reasoning doesnt feel correct.
2) So if we first choose one node we can choose among $n$ of them, then we can choose to go to $n-1$ other nodes, and then $n-2$ and so on. So we get $n!$ which is not in polynomial time, thus it is NP-complete.
3) This one is trickier, i can't really use a similar argument as for the hamiltoncycle. I don't know the number of edges. How should I think here?
4) I assume that a coloring here should be such that no neighbouring/connected nodes have the same colour. But I have no idea how these $n$ nodes are connected in order to figure this out?
5) Same question as 4) essentially.
discrete-mathematics graph-theory
discrete-mathematics graph-theory
asked Mar 18 at 13:38
ParsevalParseval
3,0721719
asked Mar 18 at 13:38
ParsevalParseval
3,0721719
asked Mar 18 at 13:38
ParsevalParseval
3,0721719
asked Mar 18 at 13:38
ParsevalParseval
3,0721719
asked Mar 18 at 13:38
ParsevalParseval
3,0721719
3,0721719
$begingroup$
For (3), it is known that a graph has an eulerian cycle if and only if all the nodes have an even degree. That's linear on the number of nodes.
$endgroup$
– frabala
Mar 18 at 13:52
$begingroup$
What do you mean by "Thats linear on the number of nodes"? So the problem is then to check whether or not all of these $n$ nodes have an even degree. How is this a linear problem?
$endgroup$
– Parseval
Mar 18 at 14:21
1
$begingroup$
You're right. I falsely assumed constant time for checking the degree of a node.
$endgroup$
– frabala
Mar 18 at 14:29
add a comment |
$begingroup$
For (3), it is known that a graph has an eulerian cycle if and only if all the nodes have an even degree. That's linear on the number of nodes.
$endgroup$
– frabala
Mar 18 at 13:52
$begingroup$
What do you mean by "Thats linear on the number of nodes"? So the problem is then to check whether or not all of these $n$ nodes have an even degree. How is this a linear problem?
$endgroup$
– Parseval
Mar 18 at 14:21
1
$begingroup$
You're right. I falsely assumed constant time for checking the degree of a node.
$endgroup$
– frabala
Mar 18 at 14:29
$begingroup$
For (3), it is known that a graph has an eulerian cycle if and only if all the nodes have an even degree. That's linear on the number of nodes.
$endgroup$
– frabala
Mar 18 at 13:52
$begingroup$
For (3), it is known that a graph has an eulerian cycle if and only if all the nodes have an even degree. That's linear on the number of nodes.
$endgroup$
– frabala
Mar 18 at 13:52
$begingroup$
What do you mean by "Thats linear on the number of nodes"? So the problem is then to check whether or not all of these $n$ nodes have an even degree. How is this a linear problem?
$endgroup$
– Parseval
Mar 18 at 14:21
$begingroup$
What do you mean by "Thats linear on the number of nodes"? So the problem is then to check whether or not all of these $n$ nodes have an even degree. How is this a linear problem?
$endgroup$
– Parseval
Mar 18 at 14:21
1
1
$begingroup$
You're right. I falsely assumed constant time for checking the degree of a node.
$endgroup$
– frabala
Mar 18 at 14:29
$begingroup$
You're right. I falsely assumed constant time for checking the degree of a node.
$endgroup$
– frabala
Mar 18 at 14:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
1):
It is known that primality testing is in P. But this is not at all straightforward. You can read about the algorithm here or consult the paper itself: https://en.wikipedia.org/wiki/AKS_primality_test
The problem with your argumentation is that when we refer to polynomial time we mean that with respect to the input length. Now to represent n we need $O(log(n))$ bits. So the input length is order $O(log(n))$. Your proposed algorithm of checking all the values up to $sqrt{n}$ would satisfy (for asymptotic complexity): $$Omega(sqrt{n}) = Omega(sqrt{exp(log(n))}) = Omega(exp(1/2log(n)))$$
Thus that is at least exponential in the input length.
2):
This is in fact NP-complete. In order to prove this you have to prove that it is in NP and that it is NP-hard. It is not too hard to prove that it is in NP: You could guess a circle consisting of all nodes and check in polynomial time that it is actually hamiltonian.
It is more difficult to prove that this is NP-hard. You might want to have a look at reductions for this.
Note that your argumentation is not valid. You cannot conclude that something is NP-complete just because the only algorithm you found is outside of P.
3):
It is even possible to find an Eulerian path in linear time (in the number of edges). Note that there are at most $n^2$ edges in any simple undirected graph (where n is the number of nodes). You can try to find a polynomial algorithm yourself or just look it up on wikipedia. Hint to find the algorithm: Be Greedy.
4):
Your assumption is correct. Note that a graph can be colored with 2 colors if and only if it is bipartite. This can be done in polynomial time. My hint is to look at BFS (breadth first search).
5):
This is NP-complete. In order to show that you have to show that it is in NP (just give a nondeterministic polynomial algorithm that solves it) and that it is NP-hard. Again this is not very straightforward and you'll need to know about reductions to do this.
Link to wikipedia's page about reduction: https://en.wikipedia.org/wiki/Reduction_(complexity)
Feel free to ask if anything I wrote is unclear or if you'd like more information on reductions.
answered Mar 18 at 20:23
araomisaraomis
4039
$endgroup$
add a comment |
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1 Answer
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$begingroup$
1):
It is known that primality testing is in P. But this is not at all straightforward. You can read about the algorithm here or consult the paper itself: https://en.wikipedia.org/wiki/AKS_primality_test
The problem with your argumentation is that when we refer to polynomial time we mean that with respect to the input length. Now to represent n we need $O(log(n))$ bits. So the input length is order $O(log(n))$. Your proposed algorithm of checking all the values up to $sqrt{n}$ would satisfy (for asymptotic complexity): $$Omega(sqrt{n}) = Omega(sqrt{exp(log(n))}) = Omega(exp(1/2log(n)))$$
Thus that is at least exponential in the input length.
2):
This is in fact NP-complete. In order to prove this you have to prove that it is in NP and that it is NP-hard. It is not too hard to prove that it is in NP: You could guess a circle consisting of all nodes and check in polynomial time that it is actually hamiltonian.
It is more difficult to prove that this is NP-hard. You might want to have a look at reductions for this.
Note that your argumentation is not valid. You cannot conclude that something is NP-complete just because the only algorithm you found is outside of P.
3):
It is even possible to find an Eulerian path in linear time (in the number of edges). Note that there are at most $n^2$ edges in any simple undirected graph (where n is the number of nodes). You can try to find a polynomial algorithm yourself or just look it up on wikipedia. Hint to find the algorithm: Be Greedy.
4):
Your assumption is correct. Note that a graph can be colored with 2 colors if and only if it is bipartite. This can be done in polynomial time. My hint is to look at BFS (breadth first search).
5):
This is NP-complete. In order to show that you have to show that it is in NP (just give a nondeterministic polynomial algorithm that solves it) and that it is NP-hard. Again this is not very straightforward and you'll need to know about reductions to do this.
Link to wikipedia's page about reduction: https://en.wikipedia.org/wiki/Reduction_(complexity)
Feel free to ask if anything I wrote is unclear or if you'd like more information on reductions.
answered Mar 18 at 20:23
araomisaraomis
4039
$endgroup$
add a comment |
$begingroup$
1):
It is known that primality testing is in P. But this is not at all straightforward. You can read about the algorithm here or consult the paper itself: https://en.wikipedia.org/wiki/AKS_primality_test
The problem with your argumentation is that when we refer to polynomial time we mean that with respect to the input length. Now to represent n we need $O(log(n))$ bits. So the input length is order $O(log(n))$. Your proposed algorithm of checking all the values up to $sqrt{n}$ would satisfy (for asymptotic complexity): $$Omega(sqrt{n}) = Omega(sqrt{exp(log(n))}) = Omega(exp(1/2log(n)))$$
Thus that is at least exponential in the input length.
2):
This is in fact NP-complete. In order to prove this you have to prove that it is in NP and that it is NP-hard. It is not too hard to prove that it is in NP: You could guess a circle consisting of all nodes and check in polynomial time that it is actually hamiltonian.
It is more difficult to prove that this is NP-hard. You might want to have a look at reductions for this.
Note that your argumentation is not valid. You cannot conclude that something is NP-complete just because the only algorithm you found is outside of P.
3):
It is even possible to find an Eulerian path in linear time (in the number of edges). Note that there are at most $n^2$ edges in any simple undirected graph (where n is the number of nodes). You can try to find a polynomial algorithm yourself or just look it up on wikipedia. Hint to find the algorithm: Be Greedy.
4):
Your assumption is correct. Note that a graph can be colored with 2 colors if and only if it is bipartite. This can be done in polynomial time. My hint is to look at BFS (breadth first search).
5):
This is NP-complete. In order to show that you have to show that it is in NP (just give a nondeterministic polynomial algorithm that solves it) and that it is NP-hard. Again this is not very straightforward and you'll need to know about reductions to do this.
Link to wikipedia's page about reduction: https://en.wikipedia.org/wiki/Reduction_(complexity)
Feel free to ask if anything I wrote is unclear or if you'd like more information on reductions.
answered Mar 18 at 20:23
araomisaraomis
4039
$endgroup$
add a comment |
$begingroup$
1):
It is known that primality testing is in P. But this is not at all straightforward. You can read about the algorithm here or consult the paper itself: https://en.wikipedia.org/wiki/AKS_primality_test
The problem with your argumentation is that when we refer to polynomial time we mean that with respect to the input length. Now to represent n we need $O(log(n))$ bits. So the input length is order $O(log(n))$. Your proposed algorithm of checking all the values up to $sqrt{n}$ would satisfy (for asymptotic complexity): $$Omega(sqrt{n}) = Omega(sqrt{exp(log(n))}) = Omega(exp(1/2log(n)))$$
Thus that is at least exponential in the input length.
2):
This is in fact NP-complete. In order to prove this you have to prove that it is in NP and that it is NP-hard. It is not too hard to prove that it is in NP: You could guess a circle consisting of all nodes and check in polynomial time that it is actually hamiltonian.
It is more difficult to prove that this is NP-hard. You might want to have a look at reductions for this.
Note that your argumentation is not valid. You cannot conclude that something is NP-complete just because the only algorithm you found is outside of P.
3):
It is even possible to find an Eulerian path in linear time (in the number of edges). Note that there are at most $n^2$ edges in any simple undirected graph (where n is the number of nodes). You can try to find a polynomial algorithm yourself or just look it up on wikipedia. Hint to find the algorithm: Be Greedy.
4):
Your assumption is correct. Note that a graph can be colored with 2 colors if and only if it is bipartite. This can be done in polynomial time. My hint is to look at BFS (breadth first search).
5):
This is NP-complete. In order to show that you have to show that it is in NP (just give a nondeterministic polynomial algorithm that solves it) and that it is NP-hard. Again this is not very straightforward and you'll need to know about reductions to do this.
Link to wikipedia's page about reduction: https://en.wikipedia.org/wiki/Reduction_(complexity)
Feel free to ask if anything I wrote is unclear or if you'd like more information on reductions.
answered Mar 18 at 20:23
araomisaraomis
4039
$endgroup$
1):
It is known that primality testing is in P. But this is not at all straightforward. You can read about the algorithm here or consult the paper itself: https://en.wikipedia.org/wiki/AKS_primality_test
The problem with your argumentation is that when we refer to polynomial time we mean that with respect to the input length. Now to represent n we need $O(log(n))$ bits. So the input length is order $O(log(n))$. Your proposed algorithm of checking all the values up to $sqrt{n}$ would satisfy (for asymptotic complexity): $$Omega(sqrt{n}) = Omega(sqrt{exp(log(n))}) = Omega(exp(1/2log(n)))$$
Thus that is at least exponential in the input length.
2):
This is in fact NP-complete. In order to prove this you have to prove that it is in NP and that it is NP-hard. It is not too hard to prove that it is in NP: You could guess a circle consisting of all nodes and check in polynomial time that it is actually hamiltonian.
It is more difficult to prove that this is NP-hard. You might want to have a look at reductions for this.
Note that your argumentation is not valid. You cannot conclude that something is NP-complete just because the only algorithm you found is outside of P.
3):
It is even possible to find an Eulerian path in linear time (in the number of edges). Note that there are at most $n^2$ edges in any simple undirected graph (where n is the number of nodes). You can try to find a polynomial algorithm yourself or just look it up on wikipedia. Hint to find the algorithm: Be Greedy.
4):
Your assumption is correct. Note that a graph can be colored with 2 colors if and only if it is bipartite. This can be done in polynomial time. My hint is to look at BFS (breadth first search).
5):
This is NP-complete. In order to show that you have to show that it is in NP (just give a nondeterministic polynomial algorithm that solves it) and that it is NP-hard. Again this is not very straightforward and you'll need to know about reductions to do this.
Link to wikipedia's page about reduction: https://en.wikipedia.org/wiki/Reduction_(complexity)
Feel free to ask if anything I wrote is unclear or if you'd like more information on reductions.
answered Mar 18 at 20:23
araomisaraomis
4039
answered Mar 18 at 20:23
araomisaraomis
4039
answered Mar 18 at 20:23
araomisaraomis
4039
answered Mar 18 at 20:23
araomisaraomis
4039
4039
add a comment |
add a comment |
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$begingroup$
For (3), it is known that a graph has an eulerian cycle if and only if all the nodes have an even degree. That's linear on the number of nodes.
$endgroup$
– frabala
Mar 18 at 13:52
$begingroup$
What do you mean by "Thats linear on the number of nodes"? So the problem is then to check whether or not all of these $n$ nodes have an even degree. How is this a linear problem?
$endgroup$
– Parseval
Mar 18 at 14:21
1
$begingroup$
You're right. I falsely assumed constant time for checking the degree of a node.
$endgroup$
– frabala
Mar 18 at 14:29