reduction from 3sat to 3 dimensional matching.












0












$begingroup$


I've been reading about the standard reduction from 3sat to 3DM and my question was regarding the 'clean up gadgets'. So suppose i take an instance of 3-Sat with $n$ variables and $k$ clauses. Once we have constructed the reduction it says we need to add an extra $(n-1)k$ clean up gadgets for the remaining uncovered tips.



My question is if we were actually trying to solve the problem using this polynommial time reduction to a 3-dimensional matching, surely we wouldn't know which tips are free until we have actually found a matching for the clause gadgets and so have used $k$ tips-one for each of the clauses, then allowing us to match the remaining $(n-1)k$ tips.






combinatorics logic graph-theory computational-complexity satisfiability







share|cite|improve this question









$endgroup$












  • $begingroup$
    Can you point us to or describe the specific version of the reduction you are referring to?
    $endgroup$
    – Clement C.
    Jun 4 '15 at 20:21










  • $begingroup$
    I'm referring to the reduction used in these notes i think it starts around page 47 courses.cs.vt.edu/cs5114/spring2009/lectures/…
    $endgroup$
    – Pavan Sangha
    Jun 4 '15 at 22:53










  • $begingroup$
    They mention using clean up gadgets just not how they use them.
    $endgroup$
    – Pavan Sangha
    Jun 4 '15 at 22:53
















0












$begingroup$


I've been reading about the standard reduction from 3sat to 3DM and my question was regarding the 'clean up gadgets'. So suppose i take an instance of 3-Sat with $n$ variables and $k$ clauses. Once we have constructed the reduction it says we need to add an extra $(n-1)k$ clean up gadgets for the remaining uncovered tips.



My question is if we were actually trying to solve the problem using this polynommial time reduction to a 3-dimensional matching, surely we wouldn't know which tips are free until we have actually found a matching for the clause gadgets and so have used $k$ tips-one for each of the clauses, then allowing us to match the remaining $(n-1)k$ tips.






combinatorics logic graph-theory computational-complexity satisfiability







share|cite|improve this question









$endgroup$












  • $begingroup$
    Can you point us to or describe the specific version of the reduction you are referring to?
    $endgroup$
    – Clement C.
    Jun 4 '15 at 20:21










  • $begingroup$
    I'm referring to the reduction used in these notes i think it starts around page 47 courses.cs.vt.edu/cs5114/spring2009/lectures/…
    $endgroup$
    – Pavan Sangha
    Jun 4 '15 at 22:53










  • $begingroup$
    They mention using clean up gadgets just not how they use them.
    $endgroup$
    – Pavan Sangha
    Jun 4 '15 at 22:53














0












0








0





$begingroup$


I've been reading about the standard reduction from 3sat to 3DM and my question was regarding the 'clean up gadgets'. So suppose i take an instance of 3-Sat with $n$ variables and $k$ clauses. Once we have constructed the reduction it says we need to add an extra $(n-1)k$ clean up gadgets for the remaining uncovered tips.



My question is if we were actually trying to solve the problem using this polynommial time reduction to a 3-dimensional matching, surely we wouldn't know which tips are free until we have actually found a matching for the clause gadgets and so have used $k$ tips-one for each of the clauses, then allowing us to match the remaining $(n-1)k$ tips.






combinatorics logic graph-theory computational-complexity satisfiability







share|cite|improve this question









$endgroup$




I've been reading about the standard reduction from 3sat to 3DM and my question was regarding the 'clean up gadgets'. So suppose i take an instance of 3-Sat with $n$ variables and $k$ clauses. Once we have constructed the reduction it says we need to add an extra $(n-1)k$ clean up gadgets for the remaining uncovered tips.



My question is if we were actually trying to solve the problem using this polynommial time reduction to a 3-dimensional matching, surely we wouldn't know which tips are free until we have actually found a matching for the clause gadgets and so have used $k$ tips-one for each of the clauses, then allowing us to match the remaining $(n-1)k$ tips.






combinatorics logic graph-theory computational-complexity satisfiability




combinatorics logic graph-theory computational-complexity satisfiability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jun 4 '15 at 20:03









Pavan SanghaPavan Sangha

564211




564211












  • $begingroup$
    Can you point us to or describe the specific version of the reduction you are referring to?
    $endgroup$
    – Clement C.
    Jun 4 '15 at 20:21










  • $begingroup$
    I'm referring to the reduction used in these notes i think it starts around page 47 courses.cs.vt.edu/cs5114/spring2009/lectures/…
    $endgroup$
    – Pavan Sangha
    Jun 4 '15 at 22:53










  • $begingroup$
    They mention using clean up gadgets just not how they use them.
    $endgroup$
    – Pavan Sangha
    Jun 4 '15 at 22:53


















  • $begingroup$
    Can you point us to or describe the specific version of the reduction you are referring to?
    $endgroup$
    – Clement C.
    Jun 4 '15 at 20:21










  • $begingroup$
    I'm referring to the reduction used in these notes i think it starts around page 47 courses.cs.vt.edu/cs5114/spring2009/lectures/…
    $endgroup$
    – Pavan Sangha
    Jun 4 '15 at 22:53










  • $begingroup$
    They mention using clean up gadgets just not how they use them.
    $endgroup$
    – Pavan Sangha
    Jun 4 '15 at 22:53
















$begingroup$
Can you point us to or describe the specific version of the reduction you are referring to?
$endgroup$
– Clement C.
Jun 4 '15 at 20:21




$begingroup$
Can you point us to or describe the specific version of the reduction you are referring to?
$endgroup$
– Clement C.
Jun 4 '15 at 20:21












$begingroup$
I'm referring to the reduction used in these notes i think it starts around page 47 courses.cs.vt.edu/cs5114/spring2009/lectures/…
$endgroup$
– Pavan Sangha
Jun 4 '15 at 22:53




$begingroup$
I'm referring to the reduction used in these notes i think it starts around page 47 courses.cs.vt.edu/cs5114/spring2009/lectures/…
$endgroup$
– Pavan Sangha
Jun 4 '15 at 22:53












$begingroup$
They mention using clean up gadgets just not how they use them.
$endgroup$
– Pavan Sangha
Jun 4 '15 at 22:53




$begingroup$
They mention using clean up gadgets just not how they use them.
$endgroup$
– Pavan Sangha
Jun 4 '15 at 22:53










1 Answer
1






active

oldest

votes


















0












$begingroup$

There are 6 tips associated with each clause $C = x_a lor x_b lor x_c$. They represent $x_a$, $overline{x_a}$, $x_b$, $overline{x_b}$, $x_c$, and $overline{x_c}$. 3 of them (one from each pair) will be covered by matches from the TF-ring for that variable. 1 of them will be covered by the clause match. That leaves 2 uncovered tips from the original 6 tips associated with the clause, and they require cleaning up.



This is easily done. You create 4 nodes, $q_1,q_2,q_3,q_4$. Create 6 potential matches, $(q_1,q_2,x_a)$, $(q_1,q_2,overline{x_a})$, $(q_1, q_2, x_b)$... and 6 more potential matches $(q_3,q_4,x_a)$, $(q_3,q_4,overline{x_a})$... Now it will always be possible to cover the two unused tips. You could actually get away with 8 matches instead of 12 if you're clever.



Or if you wanted to be stingy with the nodes you could just allocate one of them, $q$, and create 12 matches that cover all the possibilities for leftover nodes: $(q,x_a,x_b)$, $(q, x_a, overline{x_b})$, $(q, overline{x_a}, x_b)$, etc.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry is your explanation related to the reduction that is in the notes that i attached?
    $endgroup$
    – Pavan Sangha
    Jun 5 '15 at 14:13










  • $begingroup$
    You make $(n-1)k$ cleanup gadgets, but each of them can clean up an unused tip from any of the $n$ variables. The choice of which cleanup gadget cleans up which tip is part of solving the 3DM problem, not something done beforehand.
    $endgroup$
    – NovaDenizen
    Jun 5 '15 at 16:02










  • $begingroup$
    Ok thanks, that's what i thought
    $endgroup$
    – Pavan Sangha
    Jun 5 '15 at 21:17










  • $begingroup$
    Just want to add the general reasoning of the first paragraph as to how it's $(n-1)k$ tips. Since $k$ clauses introduce $2k$ tips and there are $n$ gadgets, there are a total of $2nk$ tips. Half of them will be covered by covering the cores(TF-triples), and each clause gadget covers one tip, so there are $frac{2nk}{2} - k = (n-1)k$ tips left that need cleaning up.
    $endgroup$
    – RexYuan
    Mar 6 '18 at 17:09













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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

There are 6 tips associated with each clause $C = x_a lor x_b lor x_c$. They represent $x_a$, $overline{x_a}$, $x_b$, $overline{x_b}$, $x_c$, and $overline{x_c}$. 3 of them (one from each pair) will be covered by matches from the TF-ring for that variable. 1 of them will be covered by the clause match. That leaves 2 uncovered tips from the original 6 tips associated with the clause, and they require cleaning up.



This is easily done. You create 4 nodes, $q_1,q_2,q_3,q_4$. Create 6 potential matches, $(q_1,q_2,x_a)$, $(q_1,q_2,overline{x_a})$, $(q_1, q_2, x_b)$... and 6 more potential matches $(q_3,q_4,x_a)$, $(q_3,q_4,overline{x_a})$... Now it will always be possible to cover the two unused tips. You could actually get away with 8 matches instead of 12 if you're clever.



Or if you wanted to be stingy with the nodes you could just allocate one of them, $q$, and create 12 matches that cover all the possibilities for leftover nodes: $(q,x_a,x_b)$, $(q, x_a, overline{x_b})$, $(q, overline{x_a}, x_b)$, etc.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry is your explanation related to the reduction that is in the notes that i attached?
    $endgroup$
    – Pavan Sangha
    Jun 5 '15 at 14:13










  • $begingroup$
    You make $(n-1)k$ cleanup gadgets, but each of them can clean up an unused tip from any of the $n$ variables. The choice of which cleanup gadget cleans up which tip is part of solving the 3DM problem, not something done beforehand.
    $endgroup$
    – NovaDenizen
    Jun 5 '15 at 16:02










  • $begingroup$
    Ok thanks, that's what i thought
    $endgroup$
    – Pavan Sangha
    Jun 5 '15 at 21:17










  • $begingroup$
    Just want to add the general reasoning of the first paragraph as to how it's $(n-1)k$ tips. Since $k$ clauses introduce $2k$ tips and there are $n$ gadgets, there are a total of $2nk$ tips. Half of them will be covered by covering the cores(TF-triples), and each clause gadget covers one tip, so there are $frac{2nk}{2} - k = (n-1)k$ tips left that need cleaning up.
    $endgroup$
    – RexYuan
    Mar 6 '18 at 17:09


















0












$begingroup$

There are 6 tips associated with each clause $C = x_a lor x_b lor x_c$. They represent $x_a$, $overline{x_a}$, $x_b$, $overline{x_b}$, $x_c$, and $overline{x_c}$. 3 of them (one from each pair) will be covered by matches from the TF-ring for that variable. 1 of them will be covered by the clause match. That leaves 2 uncovered tips from the original 6 tips associated with the clause, and they require cleaning up.



This is easily done. You create 4 nodes, $q_1,q_2,q_3,q_4$. Create 6 potential matches, $(q_1,q_2,x_a)$, $(q_1,q_2,overline{x_a})$, $(q_1, q_2, x_b)$... and 6 more potential matches $(q_3,q_4,x_a)$, $(q_3,q_4,overline{x_a})$... Now it will always be possible to cover the two unused tips. You could actually get away with 8 matches instead of 12 if you're clever.



Or if you wanted to be stingy with the nodes you could just allocate one of them, $q$, and create 12 matches that cover all the possibilities for leftover nodes: $(q,x_a,x_b)$, $(q, x_a, overline{x_b})$, $(q, overline{x_a}, x_b)$, etc.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry is your explanation related to the reduction that is in the notes that i attached?
    $endgroup$
    – Pavan Sangha
    Jun 5 '15 at 14:13










  • $begingroup$
    You make $(n-1)k$ cleanup gadgets, but each of them can clean up an unused tip from any of the $n$ variables. The choice of which cleanup gadget cleans up which tip is part of solving the 3DM problem, not something done beforehand.
    $endgroup$
    – NovaDenizen
    Jun 5 '15 at 16:02










  • $begingroup$
    Ok thanks, that's what i thought
    $endgroup$
    – Pavan Sangha
    Jun 5 '15 at 21:17










  • $begingroup$
    Just want to add the general reasoning of the first paragraph as to how it's $(n-1)k$ tips. Since $k$ clauses introduce $2k$ tips and there are $n$ gadgets, there are a total of $2nk$ tips. Half of them will be covered by covering the cores(TF-triples), and each clause gadget covers one tip, so there are $frac{2nk}{2} - k = (n-1)k$ tips left that need cleaning up.
    $endgroup$
    – RexYuan
    Mar 6 '18 at 17:09
















0












0








0





$begingroup$

There are 6 tips associated with each clause $C = x_a lor x_b lor x_c$. They represent $x_a$, $overline{x_a}$, $x_b$, $overline{x_b}$, $x_c$, and $overline{x_c}$. 3 of them (one from each pair) will be covered by matches from the TF-ring for that variable. 1 of them will be covered by the clause match. That leaves 2 uncovered tips from the original 6 tips associated with the clause, and they require cleaning up.



This is easily done. You create 4 nodes, $q_1,q_2,q_3,q_4$. Create 6 potential matches, $(q_1,q_2,x_a)$, $(q_1,q_2,overline{x_a})$, $(q_1, q_2, x_b)$... and 6 more potential matches $(q_3,q_4,x_a)$, $(q_3,q_4,overline{x_a})$... Now it will always be possible to cover the two unused tips. You could actually get away with 8 matches instead of 12 if you're clever.



Or if you wanted to be stingy with the nodes you could just allocate one of them, $q$, and create 12 matches that cover all the possibilities for leftover nodes: $(q,x_a,x_b)$, $(q, x_a, overline{x_b})$, $(q, overline{x_a}, x_b)$, etc.






share|cite|improve this answer









$endgroup$



There are 6 tips associated with each clause $C = x_a lor x_b lor x_c$. They represent $x_a$, $overline{x_a}$, $x_b$, $overline{x_b}$, $x_c$, and $overline{x_c}$. 3 of them (one from each pair) will be covered by matches from the TF-ring for that variable. 1 of them will be covered by the clause match. That leaves 2 uncovered tips from the original 6 tips associated with the clause, and they require cleaning up.



This is easily done. You create 4 nodes, $q_1,q_2,q_3,q_4$. Create 6 potential matches, $(q_1,q_2,x_a)$, $(q_1,q_2,overline{x_a})$, $(q_1, q_2, x_b)$... and 6 more potential matches $(q_3,q_4,x_a)$, $(q_3,q_4,overline{x_a})$... Now it will always be possible to cover the two unused tips. You could actually get away with 8 matches instead of 12 if you're clever.



Or if you wanted to be stingy with the nodes you could just allocate one of them, $q$, and create 12 matches that cover all the possibilities for leftover nodes: $(q,x_a,x_b)$, $(q, x_a, overline{x_b})$, $(q, overline{x_a}, x_b)$, etc.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jun 4 '15 at 20:57









NovaDenizenNovaDenizen

3,417718




3,417718












  • $begingroup$
    Sorry is your explanation related to the reduction that is in the notes that i attached?
    $endgroup$
    – Pavan Sangha
    Jun 5 '15 at 14:13










  • $begingroup$
    You make $(n-1)k$ cleanup gadgets, but each of them can clean up an unused tip from any of the $n$ variables. The choice of which cleanup gadget cleans up which tip is part of solving the 3DM problem, not something done beforehand.
    $endgroup$
    – NovaDenizen
    Jun 5 '15 at 16:02










  • $begingroup$
    Ok thanks, that's what i thought
    $endgroup$
    – Pavan Sangha
    Jun 5 '15 at 21:17










  • $begingroup$
    Just want to add the general reasoning of the first paragraph as to how it's $(n-1)k$ tips. Since $k$ clauses introduce $2k$ tips and there are $n$ gadgets, there are a total of $2nk$ tips. Half of them will be covered by covering the cores(TF-triples), and each clause gadget covers one tip, so there are $frac{2nk}{2} - k = (n-1)k$ tips left that need cleaning up.
    $endgroup$
    – RexYuan
    Mar 6 '18 at 17:09




















  • $begingroup$
    Sorry is your explanation related to the reduction that is in the notes that i attached?
    $endgroup$
    – Pavan Sangha
    Jun 5 '15 at 14:13










  • $begingroup$
    You make $(n-1)k$ cleanup gadgets, but each of them can clean up an unused tip from any of the $n$ variables. The choice of which cleanup gadget cleans up which tip is part of solving the 3DM problem, not something done beforehand.
    $endgroup$
    – NovaDenizen
    Jun 5 '15 at 16:02










  • $begingroup$
    Ok thanks, that's what i thought
    $endgroup$
    – Pavan Sangha
    Jun 5 '15 at 21:17










  • $begingroup$
    Just want to add the general reasoning of the first paragraph as to how it's $(n-1)k$ tips. Since $k$ clauses introduce $2k$ tips and there are $n$ gadgets, there are a total of $2nk$ tips. Half of them will be covered by covering the cores(TF-triples), and each clause gadget covers one tip, so there are $frac{2nk}{2} - k = (n-1)k$ tips left that need cleaning up.
    $endgroup$
    – RexYuan
    Mar 6 '18 at 17:09


















$begingroup$
Sorry is your explanation related to the reduction that is in the notes that i attached?
$endgroup$
– Pavan Sangha
Jun 5 '15 at 14:13




$begingroup$
Sorry is your explanation related to the reduction that is in the notes that i attached?
$endgroup$
– Pavan Sangha
Jun 5 '15 at 14:13












$begingroup$
You make $(n-1)k$ cleanup gadgets, but each of them can clean up an unused tip from any of the $n$ variables. The choice of which cleanup gadget cleans up which tip is part of solving the 3DM problem, not something done beforehand.
$endgroup$
– NovaDenizen
Jun 5 '15 at 16:02




$begingroup$
You make $(n-1)k$ cleanup gadgets, but each of them can clean up an unused tip from any of the $n$ variables. The choice of which cleanup gadget cleans up which tip is part of solving the 3DM problem, not something done beforehand.
$endgroup$
– NovaDenizen
Jun 5 '15 at 16:02












$begingroup$
Ok thanks, that's what i thought
$endgroup$
– Pavan Sangha
Jun 5 '15 at 21:17




$begingroup$
Ok thanks, that's what i thought
$endgroup$
– Pavan Sangha
Jun 5 '15 at 21:17












$begingroup$
Just want to add the general reasoning of the first paragraph as to how it's $(n-1)k$ tips. Since $k$ clauses introduce $2k$ tips and there are $n$ gadgets, there are a total of $2nk$ tips. Half of them will be covered by covering the cores(TF-triples), and each clause gadget covers one tip, so there are $frac{2nk}{2} - k = (n-1)k$ tips left that need cleaning up.
$endgroup$
– RexYuan
Mar 6 '18 at 17:09






$begingroup$
Just want to add the general reasoning of the first paragraph as to how it's $(n-1)k$ tips. Since $k$ clauses introduce $2k$ tips and there are $n$ gadgets, there are a total of $2nk$ tips. Half of them will be covered by covering the cores(TF-triples), and each clause gadget covers one tip, so there are $frac{2nk}{2} - k = (n-1)k$ tips left that need cleaning up.
$endgroup$
– RexYuan
Mar 6 '18 at 17:09




















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Cáceres

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Para otros usos de este término, véase Cáceres (desambiguación). Cáceres Municipio y ciudad Bandera Escudo Collage de Cáceres Cáceres Ubicación de Cáceres en España. Cáceres Ubicación de Cáceres en la provincia de Cáceres. País  España • Com. autónoma  Extremadura • Provincia  Cáceres • Partido judicial Cáceres Ubicación 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 Coordenadas: 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 • Altitud 457 [ 1 ] ​ msnm (mín.:211 [ 2 ] ​, máx.:708 [ 2 ] ​) Superficie 1750,33 km² Núcleos de población Cáceres Estación Arroyo-Malpartida Rincón de Ballesteros Valdesalor Fundación 34 a. C. Población 96 068 hab. (2018) • Densidad 54,8 hab./km² Gentilicio Cacereño/a [ 3 ] ​ Código postal 10001-10005, 10195 Alcaldesa (2015) María Elena Nevado del Campo [ 4 ] ​ Presupuesto 69.045.004€ [ 5 ] ​  (año ...
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