If I can solve Sudoku, can I solve the Travelling Salesman Problem (TSP)? If so, how?
$begingroup$
Let us say there is a program such that if you give a partially filled Sudoku of any size it gives you corresponding completed Sudoku.
Can you treat this program as a black box and use this to solve TSP? I mean is there a way to represent TSP problem as partially filled Sudoku, so that if I give you the answer of that Sudoku, you can tell the solution for TSP in polynomial time?
If yes, how? how do you represent TSP as a partially filled Sudoku and interpret corresponding filled Sudoku for the result.
algorithms np-complete reductions traveling-salesman sudoku
edited 14 hours ago
Rodrigo de Azevedo
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asked yesterday
Chakrapani N RaoChakrapani N Rao
7918
New contributor
Chakrapani N Rao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Let us say there is a program such that if you give a partially filled Sudoku of any size it gives you corresponding completed Sudoku.
Can you treat this program as a black box and use this to solve TSP? I mean is there a way to represent TSP problem as partially filled Sudoku, so that if I give you the answer of that Sudoku, you can tell the solution for TSP in polynomial time?
If yes, how? how do you represent TSP as a partially filled Sudoku and interpret corresponding filled Sudoku for the result.
algorithms np-complete reductions traveling-salesman sudoku
edited 14 hours ago
Rodrigo de Azevedo
700615
asked yesterday
Chakrapani N RaoChakrapani N Rao
7918
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Chakrapani N Rao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
1
$begingroup$
This paper claims to give a constructive reduction from Sudoku to Hamiltonian cycle problem: sciencedirect.com/science/article/pii/S097286001630038X
$endgroup$
– C. Windolf
yesterday
$begingroup$
@C.Windolf The question is asking for the other direction. (Indeed, there's a deleted answer that made the same mistake and cited the same paper.)
$endgroup$
– David Richerby
yesterday
add a comment |
$begingroup$
Let us say there is a program such that if you give a partially filled Sudoku of any size it gives you corresponding completed Sudoku.
Can you treat this program as a black box and use this to solve TSP? I mean is there a way to represent TSP problem as partially filled Sudoku, so that if I give you the answer of that Sudoku, you can tell the solution for TSP in polynomial time?
If yes, how? how do you represent TSP as a partially filled Sudoku and interpret corresponding filled Sudoku for the result.
algorithms np-complete reductions traveling-salesman sudoku
edited 14 hours ago
Rodrigo de Azevedo
700615
asked yesterday
Chakrapani N RaoChakrapani N Rao
7918
New contributor
Chakrapani N Rao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let us say there is a program such that if you give a partially filled Sudoku of any size it gives you corresponding completed Sudoku.
Can you treat this program as a black box and use this to solve TSP? I mean is there a way to represent TSP problem as partially filled Sudoku, so that if I give you the answer of that Sudoku, you can tell the solution for TSP in polynomial time?
If yes, how? how do you represent TSP as a partially filled Sudoku and interpret corresponding filled Sudoku for the result.
algorithms np-complete reductions traveling-salesman sudoku
algorithms np-complete reductions traveling-salesman sudoku
edited 14 hours ago
Rodrigo de Azevedo
700615
asked yesterday
Chakrapani N RaoChakrapani N Rao
7918
New contributor
Chakrapani N Rao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 14 hours ago
Rodrigo de Azevedo
700615
asked yesterday
Chakrapani N RaoChakrapani N Rao
7918
New contributor
Chakrapani N Rao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 14 hours ago
Rodrigo de Azevedo
700615
edited 14 hours ago
Rodrigo de Azevedo
700615
edited 14 hours ago
Rodrigo de Azevedo
700615
700615
asked yesterday
Chakrapani N RaoChakrapani N Rao
7918
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Chakrapani N Rao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Chakrapani N RaoChakrapani N Rao
7918
asked yesterday
Chakrapani N RaoChakrapani N Rao
7918
7918
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Chakrapani N Rao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Chakrapani N Rao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.
1
$begingroup$
This paper claims to give a constructive reduction from Sudoku to Hamiltonian cycle problem: sciencedirect.com/science/article/pii/S097286001630038X
$endgroup$
– C. Windolf
yesterday
$begingroup$
@C.Windolf The question is asking for the other direction. (Indeed, there's a deleted answer that made the same mistake and cited the same paper.)
$endgroup$
– David Richerby
yesterday
add a comment |
1
$begingroup$
This paper claims to give a constructive reduction from Sudoku to Hamiltonian cycle problem: sciencedirect.com/science/article/pii/S097286001630038X
$endgroup$
– C. Windolf
yesterday
$begingroup$
@C.Windolf The question is asking for the other direction. (Indeed, there's a deleted answer that made the same mistake and cited the same paper.)
$endgroup$
– David Richerby
yesterday
1
1
$begingroup$
This paper claims to give a constructive reduction from Sudoku to Hamiltonian cycle problem: sciencedirect.com/science/article/pii/S097286001630038X
$endgroup$
– C. Windolf
yesterday
$begingroup$
This paper claims to give a constructive reduction from Sudoku to Hamiltonian cycle problem: sciencedirect.com/science/article/pii/S097286001630038X
$endgroup$
– C. Windolf
yesterday
$begingroup$
@C.Windolf The question is asking for the other direction. (Indeed, there's a deleted answer that made the same mistake and cited the same paper.)
$endgroup$
– David Richerby
yesterday
$begingroup$
@C.Windolf The question is asking for the other direction. (Indeed, there's a deleted answer that made the same mistake and cited the same paper.)
$endgroup$
– David Richerby
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For 9x9 Sudoku, no. It is finite so can be solved in $O(1)$ time.
But if you had a solver for $n^2 times n^2$ Sudoku, that worked for all $n$ and all possible partial boards, then yes, that could be used to solve TSP in polynomial time, as completing a $n^2 times n^2$ Sudoku is NP-complete.
The proof of NP-completeness works by reducing from some NP-complete problem R to Sudoku; then because R is NP-complete, you can reduce from TSP to R (that follows from the definition of NP-completeness); and chaining those reductions gives you a way to use the Sudoku solver to solve TSP.
answered yesterday
D.W.♦D.W.
101k12125290
$endgroup$
1
$begingroup$
Could you please explain how? Yes lets assume I have general sudoku solver which acts as a black box. So how can you use it? How do you represent TSP as a partially filled Sudoku
$endgroup$
– Chakrapani N Rao
yesterday
2
$begingroup$
@ChakrapaniNRao, see updated answer. Yes, I understand it is a black box. To work out the details, find the proof of NP-completeness for Sudoku and understand how the reduction works.
$endgroup$
– D.W.♦
yesterday
6
$begingroup$
@ChakrapaniNRao It doesn't answer the question completely but the full answer would be ridiculously long, be full of intricate details and wouldn't give you any more enlightenment than the sketch here. It's possible that a reduction of some NP-complete problem to $n^2times n^2$ sudoku might be interesting but, unless the proof that sudoku is NP-complete was actually by reduction from TSP (unlikely), the answer is still going to end "and then chain those two reductions together".
$endgroup$
– David Richerby
yesterday
$begingroup$
OK, so we do not have a working solution for the above question then?
$endgroup$
– Chakrapani N Rao
yesterday
7
$begingroup$
@ChakrapaniNRao You are asking how to solve problem X using a black box for problem Y. That is literally asking for a reduction. That's what "reduction" means. And, as this answer explains, the answer to your yes/no question is yes.
$endgroup$
– David Richerby
yesterday
|
show 3 more comments
$begingroup$
It is indeed possible to use a general Sudoku solver to solve instances of TSP, and if this solver takes polynomial time then the whole process will as well (in complexity terminology, there is a polynomial-time reduction from TSP to Sudoku). This is because Sudoku is NP-complete and TSP is in NP. But as is usually the case in this area, looking at the details of the reduction isn't particularly illuminating. If you want, you can piece it together by using the simple reduction from Latin square completion to Sudoku here, the reduction from triangulating uniform tripartite graphs to Latin square completion here, the reduction from 3SAT to triangulation here, and a formulation of TSP as a 3SAT problem. However, if you want to understand the idea behind reducing from Sudoku to TSP I think you would be better off studying Cook's theorem (showing that SAT is NP-complete) and a couple of simple reductions from 3SAT (e.g. to 3-dimensional matching) and being satisfied in the knowledge that the TSP-Sudoku reduction is just the same kind of thing but longer and more fiddly.
answered yesterday
rlmsrlms
28114
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rlms is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
$begingroup$
For 9x9 Sudoku, no. It is finite so can be solved in $O(1)$ time.
But if you had a solver for $n^2 times n^2$ Sudoku, that worked for all $n$ and all possible partial boards, then yes, that could be used to solve TSP in polynomial time, as completing a $n^2 times n^2$ Sudoku is NP-complete.
The proof of NP-completeness works by reducing from some NP-complete problem R to Sudoku; then because R is NP-complete, you can reduce from TSP to R (that follows from the definition of NP-completeness); and chaining those reductions gives you a way to use the Sudoku solver to solve TSP.
answered yesterday
D.W.♦D.W.
101k12125290
$endgroup$
1
$begingroup$
Could you please explain how? Yes lets assume I have general sudoku solver which acts as a black box. So how can you use it? How do you represent TSP as a partially filled Sudoku
$endgroup$
– Chakrapani N Rao
yesterday
2
$begingroup$
@ChakrapaniNRao, see updated answer. Yes, I understand it is a black box. To work out the details, find the proof of NP-completeness for Sudoku and understand how the reduction works.
$endgroup$
– D.W.♦
yesterday
6
$begingroup$
@ChakrapaniNRao It doesn't answer the question completely but the full answer would be ridiculously long, be full of intricate details and wouldn't give you any more enlightenment than the sketch here. It's possible that a reduction of some NP-complete problem to $n^2times n^2$ sudoku might be interesting but, unless the proof that sudoku is NP-complete was actually by reduction from TSP (unlikely), the answer is still going to end "and then chain those two reductions together".
$endgroup$
– David Richerby
yesterday
$begingroup$
OK, so we do not have a working solution for the above question then?
$endgroup$
– Chakrapani N Rao
yesterday
7
$begingroup$
@ChakrapaniNRao You are asking how to solve problem X using a black box for problem Y. That is literally asking for a reduction. That's what "reduction" means. And, as this answer explains, the answer to your yes/no question is yes.
$endgroup$
– David Richerby
yesterday
|
show 3 more comments
$begingroup$
For 9x9 Sudoku, no. It is finite so can be solved in $O(1)$ time.
But if you had a solver for $n^2 times n^2$ Sudoku, that worked for all $n$ and all possible partial boards, then yes, that could be used to solve TSP in polynomial time, as completing a $n^2 times n^2$ Sudoku is NP-complete.
The proof of NP-completeness works by reducing from some NP-complete problem R to Sudoku; then because R is NP-complete, you can reduce from TSP to R (that follows from the definition of NP-completeness); and chaining those reductions gives you a way to use the Sudoku solver to solve TSP.
answered yesterday
D.W.♦D.W.
101k12125290
$endgroup$
1
$begingroup$
Could you please explain how? Yes lets assume I have general sudoku solver which acts as a black box. So how can you use it? How do you represent TSP as a partially filled Sudoku
$endgroup$
– Chakrapani N Rao
yesterday
2
$begingroup$
@ChakrapaniNRao, see updated answer. Yes, I understand it is a black box. To work out the details, find the proof of NP-completeness for Sudoku and understand how the reduction works.
$endgroup$
– D.W.♦
yesterday
6
$begingroup$
@ChakrapaniNRao It doesn't answer the question completely but the full answer would be ridiculously long, be full of intricate details and wouldn't give you any more enlightenment than the sketch here. It's possible that a reduction of some NP-complete problem to $n^2times n^2$ sudoku might be interesting but, unless the proof that sudoku is NP-complete was actually by reduction from TSP (unlikely), the answer is still going to end "and then chain those two reductions together".
$endgroup$
– David Richerby
yesterday
$begingroup$
OK, so we do not have a working solution for the above question then?
$endgroup$
– Chakrapani N Rao
yesterday
7
$begingroup$
@ChakrapaniNRao You are asking how to solve problem X using a black box for problem Y. That is literally asking for a reduction. That's what "reduction" means. And, as this answer explains, the answer to your yes/no question is yes.
$endgroup$
– David Richerby
yesterday
|
show 3 more comments
$begingroup$
For 9x9 Sudoku, no. It is finite so can be solved in $O(1)$ time.
But if you had a solver for $n^2 times n^2$ Sudoku, that worked for all $n$ and all possible partial boards, then yes, that could be used to solve TSP in polynomial time, as completing a $n^2 times n^2$ Sudoku is NP-complete.
The proof of NP-completeness works by reducing from some NP-complete problem R to Sudoku; then because R is NP-complete, you can reduce from TSP to R (that follows from the definition of NP-completeness); and chaining those reductions gives you a way to use the Sudoku solver to solve TSP.
answered yesterday
D.W.♦D.W.
101k12125290
$endgroup$
For 9x9 Sudoku, no. It is finite so can be solved in $O(1)$ time.
But if you had a solver for $n^2 times n^2$ Sudoku, that worked for all $n$ and all possible partial boards, then yes, that could be used to solve TSP in polynomial time, as completing a $n^2 times n^2$ Sudoku is NP-complete.
The proof of NP-completeness works by reducing from some NP-complete problem R to Sudoku; then because R is NP-complete, you can reduce from TSP to R (that follows from the definition of NP-completeness); and chaining those reductions gives you a way to use the Sudoku solver to solve TSP.
answered yesterday
D.W.♦D.W.
101k12125290
edited yesterday
answered yesterday
D.W.♦D.W.
101k12125290
answered yesterday
D.W.♦D.W.
101k12125290
answered yesterday
D.W.♦D.W.
101k12125290
101k12125290
1
$begingroup$
Could you please explain how? Yes lets assume I have general sudoku solver which acts as a black box. So how can you use it? How do you represent TSP as a partially filled Sudoku
$endgroup$
– Chakrapani N Rao
yesterday
2
$begingroup$
@ChakrapaniNRao, see updated answer. Yes, I understand it is a black box. To work out the details, find the proof of NP-completeness for Sudoku and understand how the reduction works.
$endgroup$
– D.W.♦
yesterday
6
$begingroup$
@ChakrapaniNRao It doesn't answer the question completely but the full answer would be ridiculously long, be full of intricate details and wouldn't give you any more enlightenment than the sketch here. It's possible that a reduction of some NP-complete problem to $n^2times n^2$ sudoku might be interesting but, unless the proof that sudoku is NP-complete was actually by reduction from TSP (unlikely), the answer is still going to end "and then chain those two reductions together".
$endgroup$
– David Richerby
yesterday
$begingroup$
OK, so we do not have a working solution for the above question then?
$endgroup$
– Chakrapani N Rao
yesterday
7
$begingroup$
@ChakrapaniNRao You are asking how to solve problem X using a black box for problem Y. That is literally asking for a reduction. That's what "reduction" means. And, as this answer explains, the answer to your yes/no question is yes.
$endgroup$
– David Richerby
yesterday
|
show 3 more comments
1
$begingroup$
Could you please explain how? Yes lets assume I have general sudoku solver which acts as a black box. So how can you use it? How do you represent TSP as a partially filled Sudoku
$endgroup$
– Chakrapani N Rao
yesterday
2
$begingroup$
@ChakrapaniNRao, see updated answer. Yes, I understand it is a black box. To work out the details, find the proof of NP-completeness for Sudoku and understand how the reduction works.
$endgroup$
– D.W.♦
yesterday
6
$begingroup$
@ChakrapaniNRao It doesn't answer the question completely but the full answer would be ridiculously long, be full of intricate details and wouldn't give you any more enlightenment than the sketch here. It's possible that a reduction of some NP-complete problem to $n^2times n^2$ sudoku might be interesting but, unless the proof that sudoku is NP-complete was actually by reduction from TSP (unlikely), the answer is still going to end "and then chain those two reductions together".
$endgroup$
– David Richerby
yesterday
$begingroup$
OK, so we do not have a working solution for the above question then?
$endgroup$
– Chakrapani N Rao
yesterday
7
$begingroup$
@ChakrapaniNRao You are asking how to solve problem X using a black box for problem Y. That is literally asking for a reduction. That's what "reduction" means. And, as this answer explains, the answer to your yes/no question is yes.
$endgroup$
– David Richerby
yesterday
1
1
$begingroup$
Could you please explain how? Yes lets assume I have general sudoku solver which acts as a black box. So how can you use it? How do you represent TSP as a partially filled Sudoku
$endgroup$
– Chakrapani N Rao
yesterday
$begingroup$
Could you please explain how? Yes lets assume I have general sudoku solver which acts as a black box. So how can you use it? How do you represent TSP as a partially filled Sudoku
$endgroup$
– Chakrapani N Rao
yesterday
2
2
$begingroup$
@ChakrapaniNRao, see updated answer. Yes, I understand it is a black box. To work out the details, find the proof of NP-completeness for Sudoku and understand how the reduction works.
$endgroup$
– D.W.♦
yesterday
$begingroup$
@ChakrapaniNRao, see updated answer. Yes, I understand it is a black box. To work out the details, find the proof of NP-completeness for Sudoku and understand how the reduction works.
$endgroup$
– D.W.♦
yesterday
6
6
$begingroup$
@ChakrapaniNRao It doesn't answer the question completely but the full answer would be ridiculously long, be full of intricate details and wouldn't give you any more enlightenment than the sketch here. It's possible that a reduction of some NP-complete problem to $n^2times n^2$ sudoku might be interesting but, unless the proof that sudoku is NP-complete was actually by reduction from TSP (unlikely), the answer is still going to end "and then chain those two reductions together".
$endgroup$
– David Richerby
yesterday
$begingroup$
@ChakrapaniNRao It doesn't answer the question completely but the full answer would be ridiculously long, be full of intricate details and wouldn't give you any more enlightenment than the sketch here. It's possible that a reduction of some NP-complete problem to $n^2times n^2$ sudoku might be interesting but, unless the proof that sudoku is NP-complete was actually by reduction from TSP (unlikely), the answer is still going to end "and then chain those two reductions together".
$endgroup$
– David Richerby
yesterday
$begingroup$
OK, so we do not have a working solution for the above question then?
$endgroup$
– Chakrapani N Rao
yesterday
$begingroup$
OK, so we do not have a working solution for the above question then?
$endgroup$
– Chakrapani N Rao
yesterday
7
7
$begingroup$
@ChakrapaniNRao You are asking how to solve problem X using a black box for problem Y. That is literally asking for a reduction. That's what "reduction" means. And, as this answer explains, the answer to your yes/no question is yes.
$endgroup$
– David Richerby
yesterday
$begingroup$
@ChakrapaniNRao You are asking how to solve problem X using a black box for problem Y. That is literally asking for a reduction. That's what "reduction" means. And, as this answer explains, the answer to your yes/no question is yes.
$endgroup$
– David Richerby
yesterday
|
show 3 more comments
$begingroup$
It is indeed possible to use a general Sudoku solver to solve instances of TSP, and if this solver takes polynomial time then the whole process will as well (in complexity terminology, there is a polynomial-time reduction from TSP to Sudoku). This is because Sudoku is NP-complete and TSP is in NP. But as is usually the case in this area, looking at the details of the reduction isn't particularly illuminating. If you want, you can piece it together by using the simple reduction from Latin square completion to Sudoku here, the reduction from triangulating uniform tripartite graphs to Latin square completion here, the reduction from 3SAT to triangulation here, and a formulation of TSP as a 3SAT problem. However, if you want to understand the idea behind reducing from Sudoku to TSP I think you would be better off studying Cook's theorem (showing that SAT is NP-complete) and a couple of simple reductions from 3SAT (e.g. to 3-dimensional matching) and being satisfied in the knowledge that the TSP-Sudoku reduction is just the same kind of thing but longer and more fiddly.
answered yesterday
rlmsrlms
28114
New contributor
rlms is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
It is indeed possible to use a general Sudoku solver to solve instances of TSP, and if this solver takes polynomial time then the whole process will as well (in complexity terminology, there is a polynomial-time reduction from TSP to Sudoku). This is because Sudoku is NP-complete and TSP is in NP. But as is usually the case in this area, looking at the details of the reduction isn't particularly illuminating. If you want, you can piece it together by using the simple reduction from Latin square completion to Sudoku here, the reduction from triangulating uniform tripartite graphs to Latin square completion here, the reduction from 3SAT to triangulation here, and a formulation of TSP as a 3SAT problem. However, if you want to understand the idea behind reducing from Sudoku to TSP I think you would be better off studying Cook's theorem (showing that SAT is NP-complete) and a couple of simple reductions from 3SAT (e.g. to 3-dimensional matching) and being satisfied in the knowledge that the TSP-Sudoku reduction is just the same kind of thing but longer and more fiddly.
answered yesterday
rlmsrlms
28114
New contributor
rlms is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
It is indeed possible to use a general Sudoku solver to solve instances of TSP, and if this solver takes polynomial time then the whole process will as well (in complexity terminology, there is a polynomial-time reduction from TSP to Sudoku). This is because Sudoku is NP-complete and TSP is in NP. But as is usually the case in this area, looking at the details of the reduction isn't particularly illuminating. If you want, you can piece it together by using the simple reduction from Latin square completion to Sudoku here, the reduction from triangulating uniform tripartite graphs to Latin square completion here, the reduction from 3SAT to triangulation here, and a formulation of TSP as a 3SAT problem. However, if you want to understand the idea behind reducing from Sudoku to TSP I think you would be better off studying Cook's theorem (showing that SAT is NP-complete) and a couple of simple reductions from 3SAT (e.g. to 3-dimensional matching) and being satisfied in the knowledge that the TSP-Sudoku reduction is just the same kind of thing but longer and more fiddly.
answered yesterday
rlmsrlms
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It is indeed possible to use a general Sudoku solver to solve instances of TSP, and if this solver takes polynomial time then the whole process will as well (in complexity terminology, there is a polynomial-time reduction from TSP to Sudoku). This is because Sudoku is NP-complete and TSP is in NP. But as is usually the case in this area, looking at the details of the reduction isn't particularly illuminating. If you want, you can piece it together by using the simple reduction from Latin square completion to Sudoku here, the reduction from triangulating uniform tripartite graphs to Latin square completion here, the reduction from 3SAT to triangulation here, and a formulation of TSP as a 3SAT problem. However, if you want to understand the idea behind reducing from Sudoku to TSP I think you would be better off studying Cook's theorem (showing that SAT is NP-complete) and a couple of simple reductions from 3SAT (e.g. to 3-dimensional matching) and being satisfied in the knowledge that the TSP-Sudoku reduction is just the same kind of thing but longer and more fiddly.
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answered yesterday
rlmsrlms
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answered yesterday
rlmsrlms
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28114
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rlms is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Chakrapani N Rao is a new contributor. Be nice, and check out our Code of Conduct.
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This paper claims to give a constructive reduction from Sudoku to Hamiltonian cycle problem: sciencedirect.com/science/article/pii/S097286001630038X
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– C. Windolf
yesterday
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@C.Windolf The question is asking for the other direction. (Indeed, there's a deleted answer that made the same mistake and cited the same paper.)
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– David Richerby
yesterday