# Hamiltonian circuit for the entire 2x2x2 cube group

Discussion in 'Puzzle Theory' started by cuBerBruce, Dec 26, 2011.

1. ### StefanMember

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But if the algorithm is very long?

2. ### qqwrefMember

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If you only check the position after each run of the algorithm, it doesn't matter how long it is.

3. ### StefanMember

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Ah yes, I misunderstood, sorry. Probably didn't realize what you meant (and what I now see he had said) because I already knew it's impossible

Last edited: Dec 30, 2011
4. ### Lucas GarronSuper-Duper ModeratorStaff Member

A couple of letters in the algorithm are "G", but only "g" is defined. Are these intended to be the same?

5. ### qqwrefMember

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g=ncabVR

6. ### Lucas GarronSuper-Duper ModeratorStaff Member

Oh, I'm silly. I was looking at the algs so hard that I forgot the {U, V, R, S, F, G} definitions at the top (which I only read the first time I saw this thread).

Mm-hmm. Hence my confusion, because I only saw that, but not the definition of G.

7. ### qqwrefMember

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Oh, sorry. I misread that pretty hard, and assumed that G must have been the one you'd seen, since it was right there at the top. Never mind. (It WAS kinda weird to use S, V, and G - considering that other sequences are inverted throughout the definitions...)

Any feedback on my bounds and/or idea for generating a full 3x3 Hamiltonian path? Is the idea wrong, or right, or potentially useful?

PS, a little rewriting to (hopefully) make things slightly easier to understand:
Code:
P=UR
Q=U'R

a=P^5
b=P^3
c=Q P^14 Q
d=Q P^5 Q P^5
e=Q P^5 c P
f=P c P^5 Q
g=Q P^5 c P^8 Q
h=Q P^8 c P^5 Q
i=P^7
j=cc P^5 Q
k=P^6 Q
l=Q P^5 cc
m=Q P^6
n=Q P^5
o=Q P

r=adcPefknbhaodcPfccncabdodcPjabQdccfcgdccfccQicaPdnQdljPPejaocdccfcglccadQddPciQ
cjPcecmcPcadQdgaboPnbeccad
s=nbgaQdblinQeckdbcanbQmcPmcidoP
t=PPnUUaPdQihkhcecncabdQdcPfPPhaQdlfabnbQdeckcfQigkPdPcfccaPdQdejhhejaoPdcPPefaQe
gkglPcadQddPciQchejaoliQQigmcidQnbePePcadQdcabQhPefaQdhcjPgmefPmciddUUaPdQicknbef
aodcPjabnodlinboPePnbcadQdgciQcgncgcaPdQddPPckPjPcePncabdQccdcPfPPlPQicaPdnQdmccQ
ijaQhcQijaQnbefaodPPglfaQeckdcPfabQmhidnUUaPdeefknbhaodcPfccncabdodcPjabQdccfcgdc
cfccQicaPdnQdljPPejaocdccfcglccadQddPciQcjPcecmcPcadQdgaboPnbeccadQnbgaQdblinQeck
dbcanbQmcPmciddUUaPdQihkhcecncabdQdcPfPPhaQdlfabnbQdeckcfQigkPdPcfccaPdQdejhhejao
PdcPPefaQegkglPcadQddPciQchejaoliQQigmcidQnbePePcadQdcabQhPefaQdhcjPgmefPmciddUUa
PdQicknbefaodcPjabnodlinboPePnbcadQdgciQcgncgcaPdQddPPckPjPcePncabdQccdcPfPPlPQic
aPdnQdmccQijaQhcQijaQnbefaodPPglfaQeckdcPfabQmhidnUUaPdeefknbhaodcPfccncabdodcPja
bQdccfcgdccfccQicaPdnQdljPPejaocdccfcglccadQddPciQcjPcecmcPcadQdgaboPnbeccadQnbga
QdblinQeckdbcanbQmcPmciddUUaPdQihkhcecncabdQdcPfPPhaQdlfabnbQdeckcfQigkPdPcfccaPd
QdejhhejaoPdcPPefaQegkglPcadQddPciQchejaoliQQigmcidQnbePePcadQdcabQhPefaQdhcjPgme
fPmciddUUaPdQicknbefaodcPjabnodlinboPePnbcadQdgciQcgncgcaPdQddPPckPjPcePncabdQccd
cPfPPlPQicaPdnQdmc
w=cQijaQhcQijaQnbefaodPPglfaQeckdcPfabQmhidnU

u=krQsPtwF
v=u^8 aQrQsPtwUUF'
p=FU'w't'R'U's'RRRUR'R'UR'U'b'R'FRbPQrU'R'R'R'sPtwUF'
q=F'U'w't'R'U's'R'UR'R'Ua'R'FU'U'w't'R'U's'R'UR'R'Ua'FRrQsUR'R'R'twUaF'
x=krQsPtcQijaQhcQijaQnbefaodPPglfaQnQUpbPUpbPUpbockdcPfabQQkaboPUpbPUpbPUpPdPPdnU
FkrQsPtQUpaabQQijaQhcQicnaUpbdQnbefaodPPgnoPUpbPUpbPUpPQcfaQeckU'paQUpbPcPPcPUpbn
bQQkabnbPUpbPdPPdnUFaQrQsPtQUpaabQQijaQhcQijaQnbePoUpaidodPPglfaQenbPUqbPQknQUpbP
nabPU'pPfabQQkPUpUpbUpUpcUpPUpUpPQbUpbQbPUpQUpbPoUpbUUUF'

z=v^6 u^6 x

Last edited: Dec 30, 2011
8. ### cuBerBruceMember

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I thought (as far as my description is concerned) I would reserve capital letters for single moves and lower case letters for sequences. To try to make parsing it by a computer program as simple as I could, I avoided using exponentiation or other form of repetition counts. However, using an inverse operator avoided having to define a lot more variables/sequences, and I had used almost all the lower case letters already.

Last edited: Dec 30, 2011
9. ### stannicMember

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Jun 16, 2010
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If each of two undirected graphs G1 and G2 have a Hamiltonian path then their Cartesian product also has a Hamiltonian path. We can draw |G2| copies of G1 and add edges between copies to get Cartesian product. On all copies of G1, we will mark the same Hamiltonian path. Then we can connect all these paths into one Hamiltonian path.

Is it true that the cube group is the Cartesian product of G and H?

Let A=<U,D,R,L,F,B>, B=<U,D,R2,L2,F2,B2>. Let C be the group that arises when we merge positions in A that differ by permutation in B.
Then |A| = 4.32*10^19, |B| = 1.95*10^10, |C|=2.217*10^9.
Suppose there is a Hamiltonian path on graph {B, <U,D,R2,L2,F2,B2>}, and there is a Hamiltonian path on {C, <R,L,F,B>}. Can we conclude that there is a Hamiltonian path on {A, <U,D,R,L,F,B>}?
Is this interpretation correct?

10. ### qqwrefMember

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Yeah, that's what I was thinking too.

That's what I was going for. I'm not completely sure it's mathematically valid since it's been a while since I did any serious group theory, but if it does work, the relative sizes of these subgroups should make it a lot easier to find a Hamiltonian path.

11. ### stannicMember

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Jun 16, 2010
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