Is there an efficient solution to the travelling salesman problem with binary edge weights? ...
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Is there an efficient solution to the travelling salesman problem with binary edge weights?
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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
14.3k1940
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
14.3k1940
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
14.3k1940
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
14.3k1940
asked Mar 24 at 7:33
WiccanKarnakWiccanKarnak
1185
edited Mar 24 at 11:13
Apass.Jack
14.3k1940
asked Mar 24 at 7:33
WiccanKarnakWiccanKarnak
1185
edited Mar 24 at 11:13
Apass.Jack
14.3k1940
edited Mar 24 at 11:13
Apass.Jack
14.3k1940
edited Mar 24 at 11:13
Apass.Jack
14.3k1940
14.3k1940
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
3,02711016
$endgroup$
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
70.4k16107196
$endgroup$
$begingroup$
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
3,02711016
$endgroup$
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
3,02711016
$endgroup$
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
3,02711016
$endgroup$
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
3,02711016
answered Mar 24 at 8:39
j_random_hackerj_random_hacker
3,02711016
answered Mar 24 at 8:39
j_random_hackerj_random_hacker
3,02711016
answered Mar 24 at 8:39
j_random_hackerj_random_hacker
3,02711016
3,02711016
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
70.4k16107196
$endgroup$
$begingroup$
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
70.4k16107196
$endgroup$
$begingroup$
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
70.4k16107196
$endgroup$
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
70.4k16107196
answered Mar 24 at 20:20
David RicherbyDavid Richerby
70.4k16107196
answered Mar 24 at 20:20
David RicherbyDavid Richerby
70.4k16107196
answered Mar 24 at 20:20
David RicherbyDavid Richerby
70.4k16107196
70.4k16107196
$begingroup$
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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