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Is there an efficient solution to the travelling salesman problem with binary edge weights?
The Next CEO of Stack OverflowHow can I verify a solution to Travelling Salesman Problem in polynomial time?Travelling salesman problem with detoursEvolutionary algorithm for the Physical Travelling Salesman ProblemTravelling Salesman Problem with unknown shortest paths between nodesWhat if the travelling salesman travelled by plane?Traveling salesman problem with disconnected cities / infinite length edgeTravelling salesman problem with small edge weightsWhy does Travelling Salesman Problem pose the restriction that each vertex can only be visited once?Travelling Salesman problem using Guided Local SearchIf I can solve Sudoku, can I solve the Travelling Salesman Problem (TSP)? If so, how?
$begingroup$
Is there a way to solve TSP in polynomial time if there are only two kinds of weights, 0 and 1?
traveling-salesman
edited Mar 24 at 11:13
Apass.Jack
13.7k1940
asked Mar 24 at 7:33
WiccanKarnakWiccanKarnak
1185
$endgroup$
add a comment |
$begingroup$
Is there a way to solve TSP in polynomial time if there are only two kinds of weights, 0 and 1?
traveling-salesman
edited Mar 24 at 11:13
Apass.Jack
13.7k1940
asked Mar 24 at 7:33
WiccanKarnakWiccanKarnak
1185
$endgroup$
add a comment |
$begingroup$
Is there a way to solve TSP in polynomial time if there are only two kinds of weights, 0 and 1?
traveling-salesman
edited Mar 24 at 11:13
Apass.Jack
13.7k1940
asked Mar 24 at 7:33
WiccanKarnakWiccanKarnak
1185
$endgroup$
Is there a way to solve TSP in polynomial time if there are only two kinds of weights, 0 and 1?
traveling-salesman
traveling-salesman
edited Mar 24 at 11:13
Apass.Jack
13.7k1940
asked Mar 24 at 7:33
WiccanKarnakWiccanKarnak
1185
edited Mar 24 at 11:13
Apass.Jack
13.7k1940
asked Mar 24 at 7:33
WiccanKarnakWiccanKarnak
1185
edited Mar 24 at 11:13
Apass.Jack
13.7k1940
edited Mar 24 at 11:13
Apass.Jack
13.7k1940
edited Mar 24 at 11:13
Apass.Jack
13.7k1940
13.7k1940
asked Mar 24 at 7:33
WiccanKarnakWiccanKarnak
1185
asked Mar 24 at 7:33
WiccanKarnakWiccanKarnak
1185
asked Mar 24 at 7:33
WiccanKarnakWiccanKarnak
1185
1185
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No, since if every edge has weight 1, there is still the question of whether any such tour exists, which is the Hamiltonian Cycle problem, and this is still NP-hard. (The link is to a Wikipedia page for Hamiltonian Path -- both the path and cycle versions of the problem are hard.)
answered Mar 24 at 8:39
j_random_hackerj_random_hacker
2,92211016
$endgroup$
3
$begingroup$
I initially read the question with the assumption it's asking about complete graphs - but then you can still get the Hamiltonian Cycle problem by asking if a zero-length Hamiltonian cycle exists. And if you allow retrace, the problem becomes trivial.
$endgroup$
– John Dvorak
Mar 24 at 11:16
$begingroup$
@JohnDvorak thanks a lot, is there a way if I guarantee no Hamiltonian Cycles?
$endgroup$
– WiccanKarnak
Mar 24 at 14:43
$begingroup$
Every complete graph has a Hamiltonian cycle. And if your graph doesn't have a Hamiltonian cycle ... then it definitely doesn't have a Hamiltonian cycle, so what was the question again?
$endgroup$
– John Dvorak
Mar 24 at 15:00
$begingroup$
@WiccanKarnak : (A TSP solution is a Hamiltonian cycle ... of minimal total weight.)
$endgroup$
– Eric Towers
Mar 24 at 19:39
$begingroup$
Aren't you allowed to use the same edge twice in TSP?
$endgroup$
– immibis
Mar 24 at 22:00
|
show 1 more comment
$begingroup$
The accepted answer isn't quite right. An instance of TSP consists of a distance between every pair of cities: that is, it consists of a weighted complete graph. Every complete graph has a Hamiltonian cycle.
However, it is simple to reduce HAMILTON-CYCLE to $0$–$1$ TSP. Given a graph $G$, create a TSP instance where the cities are the vertices and the distance is $0$ if there is an edge between the cities and $1$ if there is not. Then $G$ has a Hamiltonian cyle if, and only if, the TSP instance has a tour of weight zero. Therefore, $0$–$1$ TSP is NP-complete.
answered Mar 24 at 20:20
David RicherbyDavid Richerby
69.4k15106195
$endgroup$
$begingroup$
This is a good point, though the choice of whether to require the input graph to be complete or not never makes a practical difference (for the purpose of finding a distance-minimal tour, missing edges in a graph can be encoded as arbitrarily-distant edges in a complete graph). Interestingly, in looking for a definitively canonical definition of the TSP problem, I found that on p. 211 of Garey & Johnson (1979) they require the edge weights to be in $mathbb Z^+$ -- i.e., 0-length edges are forbidden, meaning that for them, the "0-1 TSP" described here is technically not a special case of TSP!
$endgroup$
– j_random_hacker
Mar 25 at 11:24
$begingroup$
@j_random_hacker It's a good job I'm only throwing small stones in my glass house! (Actually, you can reduce $0$-$1$ TSP to $1$-$2$ TSP by just adding one to every edge weight and adding $n$ to the length of the path you're looking for.)
$endgroup$
– David Richerby
Mar 25 at 11:29
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, since if every edge has weight 1, there is still the question of whether any such tour exists, which is the Hamiltonian Cycle problem, and this is still NP-hard. (The link is to a Wikipedia page for Hamiltonian Path -- both the path and cycle versions of the problem are hard.)
answered Mar 24 at 8:39
j_random_hackerj_random_hacker
2,92211016
$endgroup$
3
$begingroup$
I initially read the question with the assumption it's asking about complete graphs - but then you can still get the Hamiltonian Cycle problem by asking if a zero-length Hamiltonian cycle exists. And if you allow retrace, the problem becomes trivial.
$endgroup$
– John Dvorak
Mar 24 at 11:16
$begingroup$
@JohnDvorak thanks a lot, is there a way if I guarantee no Hamiltonian Cycles?
$endgroup$
– WiccanKarnak
Mar 24 at 14:43
$begingroup$
Every complete graph has a Hamiltonian cycle. And if your graph doesn't have a Hamiltonian cycle ... then it definitely doesn't have a Hamiltonian cycle, so what was the question again?
$endgroup$
– John Dvorak
Mar 24 at 15:00
$begingroup$
@WiccanKarnak : (A TSP solution is a Hamiltonian cycle ... of minimal total weight.)
$endgroup$
– Eric Towers
Mar 24 at 19:39
$begingroup$
Aren't you allowed to use the same edge twice in TSP?
$endgroup$
– immibis
Mar 24 at 22:00
|
show 1 more comment
$begingroup$
No, since if every edge has weight 1, there is still the question of whether any such tour exists, which is the Hamiltonian Cycle problem, and this is still NP-hard. (The link is to a Wikipedia page for Hamiltonian Path -- both the path and cycle versions of the problem are hard.)
answered Mar 24 at 8:39
j_random_hackerj_random_hacker
2,92211016
$endgroup$
3
$begingroup$
I initially read the question with the assumption it's asking about complete graphs - but then you can still get the Hamiltonian Cycle problem by asking if a zero-length Hamiltonian cycle exists. And if you allow retrace, the problem becomes trivial.
$endgroup$
– John Dvorak
Mar 24 at 11:16
$begingroup$
@JohnDvorak thanks a lot, is there a way if I guarantee no Hamiltonian Cycles?
$endgroup$
– WiccanKarnak
Mar 24 at 14:43
$begingroup$
Every complete graph has a Hamiltonian cycle. And if your graph doesn't have a Hamiltonian cycle ... then it definitely doesn't have a Hamiltonian cycle, so what was the question again?
$endgroup$
– John Dvorak
Mar 24 at 15:00
$begingroup$
@WiccanKarnak : (A TSP solution is a Hamiltonian cycle ... of minimal total weight.)
$endgroup$
– Eric Towers
Mar 24 at 19:39
$begingroup$
Aren't you allowed to use the same edge twice in TSP?
$endgroup$
– immibis
Mar 24 at 22:00
|
show 1 more comment
$begingroup$
No, since if every edge has weight 1, there is still the question of whether any such tour exists, which is the Hamiltonian Cycle problem, and this is still NP-hard. (The link is to a Wikipedia page for Hamiltonian Path -- both the path and cycle versions of the problem are hard.)
answered Mar 24 at 8:39
j_random_hackerj_random_hacker
2,92211016
$endgroup$
No, since if every edge has weight 1, there is still the question of whether any such tour exists, which is the Hamiltonian Cycle problem, and this is still NP-hard. (The link is to a Wikipedia page for Hamiltonian Path -- both the path and cycle versions of the problem are hard.)
answered Mar 24 at 8:39
j_random_hackerj_random_hacker
2,92211016
answered Mar 24 at 8:39
j_random_hackerj_random_hacker
2,92211016
answered Mar 24 at 8:39
j_random_hackerj_random_hacker
2,92211016
answered Mar 24 at 8:39
j_random_hackerj_random_hacker
2,92211016
2,92211016
3
$begingroup$
I initially read the question with the assumption it's asking about complete graphs - but then you can still get the Hamiltonian Cycle problem by asking if a zero-length Hamiltonian cycle exists. And if you allow retrace, the problem becomes trivial.
$endgroup$
– John Dvorak
Mar 24 at 11:16
$begingroup$
@JohnDvorak thanks a lot, is there a way if I guarantee no Hamiltonian Cycles?
$endgroup$
– WiccanKarnak
Mar 24 at 14:43
$begingroup$
Every complete graph has a Hamiltonian cycle. And if your graph doesn't have a Hamiltonian cycle ... then it definitely doesn't have a Hamiltonian cycle, so what was the question again?
$endgroup$
– John Dvorak
Mar 24 at 15:00
$begingroup$
@WiccanKarnak : (A TSP solution is a Hamiltonian cycle ... of minimal total weight.)
$endgroup$
– Eric Towers
Mar 24 at 19:39
$begingroup$
Aren't you allowed to use the same edge twice in TSP?
$endgroup$
– immibis
Mar 24 at 22:00
|
show 1 more comment
3
$begingroup$
I initially read the question with the assumption it's asking about complete graphs - but then you can still get the Hamiltonian Cycle problem by asking if a zero-length Hamiltonian cycle exists. And if you allow retrace, the problem becomes trivial.
$endgroup$
– John Dvorak
Mar 24 at 11:16
$begingroup$
@JohnDvorak thanks a lot, is there a way if I guarantee no Hamiltonian Cycles?
$endgroup$
– WiccanKarnak
Mar 24 at 14:43
$begingroup$
Every complete graph has a Hamiltonian cycle. And if your graph doesn't have a Hamiltonian cycle ... then it definitely doesn't have a Hamiltonian cycle, so what was the question again?
$endgroup$
– John Dvorak
Mar 24 at 15:00
$begingroup$
@WiccanKarnak : (A TSP solution is a Hamiltonian cycle ... of minimal total weight.)
$endgroup$
– Eric Towers
Mar 24 at 19:39
$begingroup$
Aren't you allowed to use the same edge twice in TSP?
$endgroup$
– immibis
Mar 24 at 22:00
3
3
$begingroup$
I initially read the question with the assumption it's asking about complete graphs - but then you can still get the Hamiltonian Cycle problem by asking if a zero-length Hamiltonian cycle exists. And if you allow retrace, the problem becomes trivial.
$endgroup$
– John Dvorak
Mar 24 at 11:16
$begingroup$
I initially read the question with the assumption it's asking about complete graphs - but then you can still get the Hamiltonian Cycle problem by asking if a zero-length Hamiltonian cycle exists. And if you allow retrace, the problem becomes trivial.
$endgroup$
– John Dvorak
Mar 24 at 11:16
$begingroup$
@JohnDvorak thanks a lot, is there a way if I guarantee no Hamiltonian Cycles?
$endgroup$
– WiccanKarnak
Mar 24 at 14:43
$begingroup$
@JohnDvorak thanks a lot, is there a way if I guarantee no Hamiltonian Cycles?
$endgroup$
– WiccanKarnak
Mar 24 at 14:43
$begingroup$
Every complete graph has a Hamiltonian cycle. And if your graph doesn't have a Hamiltonian cycle ... then it definitely doesn't have a Hamiltonian cycle, so what was the question again?
$endgroup$
– John Dvorak
Mar 24 at 15:00
$begingroup$
Every complete graph has a Hamiltonian cycle. And if your graph doesn't have a Hamiltonian cycle ... then it definitely doesn't have a Hamiltonian cycle, so what was the question again?
$endgroup$
– John Dvorak
Mar 24 at 15:00
$begingroup$
@WiccanKarnak : (A TSP solution is a Hamiltonian cycle ... of minimal total weight.)
$endgroup$
– Eric Towers
Mar 24 at 19:39
$begingroup$
@WiccanKarnak : (A TSP solution is a Hamiltonian cycle ... of minimal total weight.)
$endgroup$
– Eric Towers
Mar 24 at 19:39
$begingroup$
Aren't you allowed to use the same edge twice in TSP?
$endgroup$
– immibis
Mar 24 at 22:00
$begingroup$
Aren't you allowed to use the same edge twice in TSP?
$endgroup$
– immibis
Mar 24 at 22:00
|
show 1 more comment
$begingroup$
The accepted answer isn't quite right. An instance of TSP consists of a distance between every pair of cities: that is, it consists of a weighted complete graph. Every complete graph has a Hamiltonian cycle.
However, it is simple to reduce HAMILTON-CYCLE to $0$–$1$ TSP. Given a graph $G$, create a TSP instance where the cities are the vertices and the distance is $0$ if there is an edge between the cities and $1$ if there is not. Then $G$ has a Hamiltonian cyle if, and only if, the TSP instance has a tour of weight zero. Therefore, $0$–$1$ TSP is NP-complete.
answered Mar 24 at 20:20
David RicherbyDavid Richerby
69.4k15106195
$endgroup$
$begingroup$
This is a good point, though the choice of whether to require the input graph to be complete or not never makes a practical difference (for the purpose of finding a distance-minimal tour, missing edges in a graph can be encoded as arbitrarily-distant edges in a complete graph). Interestingly, in looking for a definitively canonical definition of the TSP problem, I found that on p. 211 of Garey & Johnson (1979) they require the edge weights to be in $mathbb Z^+$ -- i.e., 0-length edges are forbidden, meaning that for them, the "0-1 TSP" described here is technically not a special case of TSP!
$endgroup$
– j_random_hacker
Mar 25 at 11:24
$begingroup$
@j_random_hacker It's a good job I'm only throwing small stones in my glass house! (Actually, you can reduce $0$-$1$ TSP to $1$-$2$ TSP by just adding one to every edge weight and adding $n$ to the length of the path you're looking for.)
$endgroup$
– David Richerby
Mar 25 at 11:29
add a comment |
$begingroup$
The accepted answer isn't quite right. An instance of TSP consists of a distance between every pair of cities: that is, it consists of a weighted complete graph. Every complete graph has a Hamiltonian cycle.
However, it is simple to reduce HAMILTON-CYCLE to $0$–$1$ TSP. Given a graph $G$, create a TSP instance where the cities are the vertices and the distance is $0$ if there is an edge between the cities and $1$ if there is not. Then $G$ has a Hamiltonian cyle if, and only if, the TSP instance has a tour of weight zero. Therefore, $0$–$1$ TSP is NP-complete.
answered Mar 24 at 20:20
David RicherbyDavid Richerby
69.4k15106195
$endgroup$
$begingroup$
This is a good point, though the choice of whether to require the input graph to be complete or not never makes a practical difference (for the purpose of finding a distance-minimal tour, missing edges in a graph can be encoded as arbitrarily-distant edges in a complete graph). Interestingly, in looking for a definitively canonical definition of the TSP problem, I found that on p. 211 of Garey & Johnson (1979) they require the edge weights to be in $mathbb Z^+$ -- i.e., 0-length edges are forbidden, meaning that for them, the "0-1 TSP" described here is technically not a special case of TSP!
$endgroup$
– j_random_hacker
Mar 25 at 11:24
$begingroup$
@j_random_hacker It's a good job I'm only throwing small stones in my glass house! (Actually, you can reduce $0$-$1$ TSP to $1$-$2$ TSP by just adding one to every edge weight and adding $n$ to the length of the path you're looking for.)
$endgroup$
– David Richerby
Mar 25 at 11:29
add a comment |
$begingroup$
The accepted answer isn't quite right. An instance of TSP consists of a distance between every pair of cities: that is, it consists of a weighted complete graph. Every complete graph has a Hamiltonian cycle.
However, it is simple to reduce HAMILTON-CYCLE to $0$–$1$ TSP. Given a graph $G$, create a TSP instance where the cities are the vertices and the distance is $0$ if there is an edge between the cities and $1$ if there is not. Then $G$ has a Hamiltonian cyle if, and only if, the TSP instance has a tour of weight zero. Therefore, $0$–$1$ TSP is NP-complete.
answered Mar 24 at 20:20
David RicherbyDavid Richerby
69.4k15106195
$endgroup$
The accepted answer isn't quite right. An instance of TSP consists of a distance between every pair of cities: that is, it consists of a weighted complete graph. Every complete graph has a Hamiltonian cycle.
However, it is simple to reduce HAMILTON-CYCLE to $0$–$1$ TSP. Given a graph $G$, create a TSP instance where the cities are the vertices and the distance is $0$ if there is an edge between the cities and $1$ if there is not. Then $G$ has a Hamiltonian cyle if, and only if, the TSP instance has a tour of weight zero. Therefore, $0$–$1$ TSP is NP-complete.
answered Mar 24 at 20:20
David RicherbyDavid Richerby
69.4k15106195
answered Mar 24 at 20:20
David RicherbyDavid Richerby
69.4k15106195
answered Mar 24 at 20:20
David RicherbyDavid Richerby
69.4k15106195
answered Mar 24 at 20:20
David RicherbyDavid Richerby
69.4k15106195
69.4k15106195
$begingroup$
This is a good point, though the choice of whether to require the input graph to be complete or not never makes a practical difference (for the purpose of finding a distance-minimal tour, missing edges in a graph can be encoded as arbitrarily-distant edges in a complete graph). Interestingly, in looking for a definitively canonical definition of the TSP problem, I found that on p. 211 of Garey & Johnson (1979) they require the edge weights to be in $mathbb Z^+$ -- i.e., 0-length edges are forbidden, meaning that for them, the "0-1 TSP" described here is technically not a special case of TSP!
$endgroup$
– j_random_hacker
Mar 25 at 11:24
$begingroup$
@j_random_hacker It's a good job I'm only throwing small stones in my glass house! (Actually, you can reduce $0$-$1$ TSP to $1$-$2$ TSP by just adding one to every edge weight and adding $n$ to the length of the path you're looking for.)
$endgroup$
– David Richerby
Mar 25 at 11:29
add a comment |
$begingroup$
This is a good point, though the choice of whether to require the input graph to be complete or not never makes a practical difference (for the purpose of finding a distance-minimal tour, missing edges in a graph can be encoded as arbitrarily-distant edges in a complete graph). Interestingly, in looking for a definitively canonical definition of the TSP problem, I found that on p. 211 of Garey & Johnson (1979) they require the edge weights to be in $mathbb Z^+$ -- i.e., 0-length edges are forbidden, meaning that for them, the "0-1 TSP" described here is technically not a special case of TSP!
$endgroup$
– j_random_hacker
Mar 25 at 11:24
$begingroup$
@j_random_hacker It's a good job I'm only throwing small stones in my glass house! (Actually, you can reduce $0$-$1$ TSP to $1$-$2$ TSP by just adding one to every edge weight and adding $n$ to the length of the path you're looking for.)
$endgroup$
– David Richerby
Mar 25 at 11:29
$begingroup$
This is a good point, though the choice of whether to require the input graph to be complete or not never makes a practical difference (for the purpose of finding a distance-minimal tour, missing edges in a graph can be encoded as arbitrarily-distant edges in a complete graph). Interestingly, in looking for a definitively canonical definition of the TSP problem, I found that on p. 211 of Garey & Johnson (1979) they require the edge weights to be in $mathbb Z^+$ -- i.e., 0-length edges are forbidden, meaning that for them, the "0-1 TSP" described here is technically not a special case of TSP!
$endgroup$
– j_random_hacker
Mar 25 at 11:24
$begingroup$
This is a good point, though the choice of whether to require the input graph to be complete or not never makes a practical difference (for the purpose of finding a distance-minimal tour, missing edges in a graph can be encoded as arbitrarily-distant edges in a complete graph). Interestingly, in looking for a definitively canonical definition of the TSP problem, I found that on p. 211 of Garey & Johnson (1979) they require the edge weights to be in $mathbb Z^+$ -- i.e., 0-length edges are forbidden, meaning that for them, the "0-1 TSP" described here is technically not a special case of TSP!
$endgroup$
– j_random_hacker
Mar 25 at 11:24
$begingroup$
@j_random_hacker It's a good job I'm only throwing small stones in my glass house! (Actually, you can reduce $0$-$1$ TSP to $1$-$2$ TSP by just adding one to every edge weight and adding $n$ to the length of the path you're looking for.)
$endgroup$
– David Richerby
Mar 25 at 11:29
$begingroup$
@j_random_hacker It's a good job I'm only throwing small stones in my glass house! (Actually, you can reduce $0$-$1$ TSP to $1$-$2$ TSP by just adding one to every edge weight and adding $n$ to the length of the path you're looking for.)
$endgroup$
– David Richerby
Mar 25 at 11:29
add a comment |
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