Are there $3$ disjoint copies of $2K_{3,3} cup (K_{5,5} setminus C_{10})$ in $K_{11,11}$?
$begingroup$
Question: Are there $3$ edge disjoint copies of $H:=2K_{3,3} cup (K_{5,5} setminus C_{10})$ in $K_{11,11}$?
Here's a drawing of $H$:
I'm working on a Latin squares research problem and trying to get a construction to work. If it would work, it would give a solution to this problem. Only, I can't get it to work easily. Maybe the above doesn't exist, and my construction won't work in this case.
- We see $H$ is regular with degree $3$, and the degrees of vertices in $K_{11,11}$ is $11$, so no clash there.
- $H$ has $33$ edges while $K_{11,11}$ has $121$ edges, so no clash there.
graph-theory
edited Jan 13 '17 at 14:25
Ethan Bolker
46k553120
asked Jan 13 '17 at 14:21
Rebecca J. StonesRebecca J. Stones
21.1k22781
$endgroup$
add a comment |
$begingroup$
Question: Are there $3$ edge disjoint copies of $H:=2K_{3,3} cup (K_{5,5} setminus C_{10})$ in $K_{11,11}$?
Here's a drawing of $H$:
I'm working on a Latin squares research problem and trying to get a construction to work. If it would work, it would give a solution to this problem. Only, I can't get it to work easily. Maybe the above doesn't exist, and my construction won't work in this case.
- We see $H$ is regular with degree $3$, and the degrees of vertices in $K_{11,11}$ is $11$, so no clash there.
- $H$ has $33$ edges while $K_{11,11}$ has $121$ edges, so no clash there.
graph-theory
edited Jan 13 '17 at 14:25
Ethan Bolker
46k553120
asked Jan 13 '17 at 14:21
Rebecca J. StonesRebecca J. Stones
21.1k22781
$endgroup$
add a comment |
$begingroup$
Question: Are there $3$ edge disjoint copies of $H:=2K_{3,3} cup (K_{5,5} setminus C_{10})$ in $K_{11,11}$?
Here's a drawing of $H$:
I'm working on a Latin squares research problem and trying to get a construction to work. If it would work, it would give a solution to this problem. Only, I can't get it to work easily. Maybe the above doesn't exist, and my construction won't work in this case.
- We see $H$ is regular with degree $3$, and the degrees of vertices in $K_{11,11}$ is $11$, so no clash there.
- $H$ has $33$ edges while $K_{11,11}$ has $121$ edges, so no clash there.
graph-theory
edited Jan 13 '17 at 14:25
Ethan Bolker
46k553120
asked Jan 13 '17 at 14:21
Rebecca J. StonesRebecca J. Stones
21.1k22781
$endgroup$
Question: Are there $3$ edge disjoint copies of $H:=2K_{3,3} cup (K_{5,5} setminus C_{10})$ in $K_{11,11}$?
Here's a drawing of $H$:
I'm working on a Latin squares research problem and trying to get a construction to work. If it would work, it would give a solution to this problem. Only, I can't get it to work easily. Maybe the above doesn't exist, and my construction won't work in this case.
- We see $H$ is regular with degree $3$, and the degrees of vertices in $K_{11,11}$ is $11$, so no clash there.
- $H$ has $33$ edges while $K_{11,11}$ has $121$ edges, so no clash there.
graph-theory
graph-theory
edited Jan 13 '17 at 14:25
Ethan Bolker
46k553120
asked Jan 13 '17 at 14:21
Rebecca J. StonesRebecca J. Stones
21.1k22781
edited Jan 13 '17 at 14:25
Ethan Bolker
46k553120
asked Jan 13 '17 at 14:21
Rebecca J. StonesRebecca J. Stones
21.1k22781
edited Jan 13 '17 at 14:25
Ethan Bolker
46k553120
edited Jan 13 '17 at 14:25
Ethan Bolker
46k553120
edited Jan 13 '17 at 14:25
Ethan Bolker
46k553120
46k553120
asked Jan 13 '17 at 14:21
Rebecca J. StonesRebecca J. Stones
21.1k22781
asked Jan 13 '17 at 14:21
Rebecca J. StonesRebecca J. Stones
21.1k22781
asked Jan 13 '17 at 14:21
Rebecca J. StonesRebecca J. Stones
21.1k22781
21.1k22781
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Since I solved the other linked problem, I figured I might as well give simulated annealing a go on this one.
The answer is also yes; in fact, here, we can find $3$ disjoint copies of $2K_{3,3} cup (K_{5,5} - P_{10})$. I added the extra edge in an attempt to make the problem a bit more constrained so that the solution would come out nicer, but I'm not sure how much of an effect it had.
The solution I found is below:
Here is my simulated annealing code (it might take a few tries before finding a zero-energy solution):
edges[{perm1_, perm2_}] :=
Join[
Tuples[{perm1[[1 ;; 3]], perm2[[1 ;; 3]]}],
Tuples[{perm1[[4 ;; 6]], perm2[[4 ;; 6]]}],
Complement[
Tuples[{perm1[[7 ;; 11]], perm2[[7 ;; 11]]}],
Table[{perm1[[i]], perm2[[i]]}, {i, 7, 11}],
Table[{perm1[[i]], perm2[[i + 1]]}, {i, 7, 10}]]];
value[state_] := 102 - Length[Union @@ (edges /@ state)];
randomPerm := {RandomSample[Range[11]], RandomSample[Range[11]]}
newState := {{Range[11], Range[11]}, randomPerm, randomPerm};
randomSwitch[state_] :=
Module[{h = RandomInteger[{2, 3}], i = RandomInteger[{1, 2}], j, k,
copy = state},
{j, k} = RandomSample[Range[11], 2];
copy[[h, i, {j, k}]] = Reverse[copy[[h, i, {j, k}]]];
Return[copy];
]
currentState = bestState = newState;
currentEnergy = bestEnergy = value[currentState];
temp = 1;
While[Exp[-1/temp] > 1/1000,
Do[
nextState = randomSwitch[currentState];
nextEnergy = value[nextState];
If[nextEnergy < bestEnergy, bestState = nextState;
bestEnergy = nextEnergy];
prob = Exp[-((nextEnergy - currentEnergy)/temp)];
If[RandomReal < prob, currentState = nextState;
currentEnergy = nextEnergy];
, {3000}];
If[bestEnergy == 0, Break];
temp *= 0.99; Print[{temp, currentEnergy}]
]
Print["Done ", bestEnergy];
answered Dec 14 '18 at 19:18
Misha LavrovMisha Lavrov
48.9k757107
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since I solved the other linked problem, I figured I might as well give simulated annealing a go on this one.
The answer is also yes; in fact, here, we can find $3$ disjoint copies of $2K_{3,3} cup (K_{5,5} - P_{10})$. I added the extra edge in an attempt to make the problem a bit more constrained so that the solution would come out nicer, but I'm not sure how much of an effect it had.
The solution I found is below:
Here is my simulated annealing code (it might take a few tries before finding a zero-energy solution):
edges[{perm1_, perm2_}] :=
Join[
Tuples[{perm1[[1 ;; 3]], perm2[[1 ;; 3]]}],
Tuples[{perm1[[4 ;; 6]], perm2[[4 ;; 6]]}],
Complement[
Tuples[{perm1[[7 ;; 11]], perm2[[7 ;; 11]]}],
Table[{perm1[[i]], perm2[[i]]}, {i, 7, 11}],
Table[{perm1[[i]], perm2[[i + 1]]}, {i, 7, 10}]]];
value[state_] := 102 - Length[Union @@ (edges /@ state)];
randomPerm := {RandomSample[Range[11]], RandomSample[Range[11]]}
newState := {{Range[11], Range[11]}, randomPerm, randomPerm};
randomSwitch[state_] :=
Module[{h = RandomInteger[{2, 3}], i = RandomInteger[{1, 2}], j, k,
copy = state},
{j, k} = RandomSample[Range[11], 2];
copy[[h, i, {j, k}]] = Reverse[copy[[h, i, {j, k}]]];
Return[copy];
]
currentState = bestState = newState;
currentEnergy = bestEnergy = value[currentState];
temp = 1;
While[Exp[-1/temp] > 1/1000,
Do[
nextState = randomSwitch[currentState];
nextEnergy = value[nextState];
If[nextEnergy < bestEnergy, bestState = nextState;
bestEnergy = nextEnergy];
prob = Exp[-((nextEnergy - currentEnergy)/temp)];
If[RandomReal < prob, currentState = nextState;
currentEnergy = nextEnergy];
, {3000}];
If[bestEnergy == 0, Break];
temp *= 0.99; Print[{temp, currentEnergy}]
]
Print["Done ", bestEnergy];
answered Dec 14 '18 at 19:18
Misha LavrovMisha Lavrov
48.9k757107
$endgroup$
add a comment |
$begingroup$
Since I solved the other linked problem, I figured I might as well give simulated annealing a go on this one.
The answer is also yes; in fact, here, we can find $3$ disjoint copies of $2K_{3,3} cup (K_{5,5} - P_{10})$. I added the extra edge in an attempt to make the problem a bit more constrained so that the solution would come out nicer, but I'm not sure how much of an effect it had.
The solution I found is below:
Here is my simulated annealing code (it might take a few tries before finding a zero-energy solution):
edges[{perm1_, perm2_}] :=
Join[
Tuples[{perm1[[1 ;; 3]], perm2[[1 ;; 3]]}],
Tuples[{perm1[[4 ;; 6]], perm2[[4 ;; 6]]}],
Complement[
Tuples[{perm1[[7 ;; 11]], perm2[[7 ;; 11]]}],
Table[{perm1[[i]], perm2[[i]]}, {i, 7, 11}],
Table[{perm1[[i]], perm2[[i + 1]]}, {i, 7, 10}]]];
value[state_] := 102 - Length[Union @@ (edges /@ state)];
randomPerm := {RandomSample[Range[11]], RandomSample[Range[11]]}
newState := {{Range[11], Range[11]}, randomPerm, randomPerm};
randomSwitch[state_] :=
Module[{h = RandomInteger[{2, 3}], i = RandomInteger[{1, 2}], j, k,
copy = state},
{j, k} = RandomSample[Range[11], 2];
copy[[h, i, {j, k}]] = Reverse[copy[[h, i, {j, k}]]];
Return[copy];
]
currentState = bestState = newState;
currentEnergy = bestEnergy = value[currentState];
temp = 1;
While[Exp[-1/temp] > 1/1000,
Do[
nextState = randomSwitch[currentState];
nextEnergy = value[nextState];
If[nextEnergy < bestEnergy, bestState = nextState;
bestEnergy = nextEnergy];
prob = Exp[-((nextEnergy - currentEnergy)/temp)];
If[RandomReal < prob, currentState = nextState;
currentEnergy = nextEnergy];
, {3000}];
If[bestEnergy == 0, Break];
temp *= 0.99; Print[{temp, currentEnergy}]
]
Print["Done ", bestEnergy];
answered Dec 14 '18 at 19:18
Misha LavrovMisha Lavrov
48.9k757107
$endgroup$
add a comment |
$begingroup$
Since I solved the other linked problem, I figured I might as well give simulated annealing a go on this one.
The answer is also yes; in fact, here, we can find $3$ disjoint copies of $2K_{3,3} cup (K_{5,5} - P_{10})$. I added the extra edge in an attempt to make the problem a bit more constrained so that the solution would come out nicer, but I'm not sure how much of an effect it had.
The solution I found is below:
Here is my simulated annealing code (it might take a few tries before finding a zero-energy solution):
edges[{perm1_, perm2_}] :=
Join[
Tuples[{perm1[[1 ;; 3]], perm2[[1 ;; 3]]}],
Tuples[{perm1[[4 ;; 6]], perm2[[4 ;; 6]]}],
Complement[
Tuples[{perm1[[7 ;; 11]], perm2[[7 ;; 11]]}],
Table[{perm1[[i]], perm2[[i]]}, {i, 7, 11}],
Table[{perm1[[i]], perm2[[i + 1]]}, {i, 7, 10}]]];
value[state_] := 102 - Length[Union @@ (edges /@ state)];
randomPerm := {RandomSample[Range[11]], RandomSample[Range[11]]}
newState := {{Range[11], Range[11]}, randomPerm, randomPerm};
randomSwitch[state_] :=
Module[{h = RandomInteger[{2, 3}], i = RandomInteger[{1, 2}], j, k,
copy = state},
{j, k} = RandomSample[Range[11], 2];
copy[[h, i, {j, k}]] = Reverse[copy[[h, i, {j, k}]]];
Return[copy];
]
currentState = bestState = newState;
currentEnergy = bestEnergy = value[currentState];
temp = 1;
While[Exp[-1/temp] > 1/1000,
Do[
nextState = randomSwitch[currentState];
nextEnergy = value[nextState];
If[nextEnergy < bestEnergy, bestState = nextState;
bestEnergy = nextEnergy];
prob = Exp[-((nextEnergy - currentEnergy)/temp)];
If[RandomReal < prob, currentState = nextState;
currentEnergy = nextEnergy];
, {3000}];
If[bestEnergy == 0, Break];
temp *= 0.99; Print[{temp, currentEnergy}]
]
Print["Done ", bestEnergy];
answered Dec 14 '18 at 19:18
Misha LavrovMisha Lavrov
48.9k757107
$endgroup$
Since I solved the other linked problem, I figured I might as well give simulated annealing a go on this one.
The answer is also yes; in fact, here, we can find $3$ disjoint copies of $2K_{3,3} cup (K_{5,5} - P_{10})$. I added the extra edge in an attempt to make the problem a bit more constrained so that the solution would come out nicer, but I'm not sure how much of an effect it had.
The solution I found is below:
Here is my simulated annealing code (it might take a few tries before finding a zero-energy solution):
edges[{perm1_, perm2_}] :=
Join[
Tuples[{perm1[[1 ;; 3]], perm2[[1 ;; 3]]}],
Tuples[{perm1[[4 ;; 6]], perm2[[4 ;; 6]]}],
Complement[
Tuples[{perm1[[7 ;; 11]], perm2[[7 ;; 11]]}],
Table[{perm1[[i]], perm2[[i]]}, {i, 7, 11}],
Table[{perm1[[i]], perm2[[i + 1]]}, {i, 7, 10}]]];
value[state_] := 102 - Length[Union @@ (edges /@ state)];
randomPerm := {RandomSample[Range[11]], RandomSample[Range[11]]}
newState := {{Range[11], Range[11]}, randomPerm, randomPerm};
randomSwitch[state_] :=
Module[{h = RandomInteger[{2, 3}], i = RandomInteger[{1, 2}], j, k,
copy = state},
{j, k} = RandomSample[Range[11], 2];
copy[[h, i, {j, k}]] = Reverse[copy[[h, i, {j, k}]]];
Return[copy];
]
currentState = bestState = newState;
currentEnergy = bestEnergy = value[currentState];
temp = 1;
While[Exp[-1/temp] > 1/1000,
Do[
nextState = randomSwitch[currentState];
nextEnergy = value[nextState];
If[nextEnergy < bestEnergy, bestState = nextState;
bestEnergy = nextEnergy];
prob = Exp[-((nextEnergy - currentEnergy)/temp)];
If[RandomReal < prob, currentState = nextState;
currentEnergy = nextEnergy];
, {3000}];
If[bestEnergy == 0, Break];
temp *= 0.99; Print[{temp, currentEnergy}]
]
Print["Done ", bestEnergy];
answered Dec 14 '18 at 19:18
Misha LavrovMisha Lavrov
48.9k757107
answered Dec 14 '18 at 19:18
Misha LavrovMisha Lavrov
48.9k757107
answered Dec 14 '18 at 19:18
Misha LavrovMisha Lavrov
48.9k757107
answered Dec 14 '18 at 19:18
Misha LavrovMisha Lavrov
48.9k757107
48.9k757107
add a comment |
add a comment |
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