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What is God's number on a 4x4 Rubik's cube?

Dacuba

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Remembers me of the surprise challenge thread. When I clicked this one, I first was like "oh no, not again"
 

cubernya

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It's been proven to be at least 33 (according to multiple sites), but as far as I know we don't even have an upper bound. What I want to know is how they got the lower bound
 

AbstractAlg

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It's been proven to be at least 33 (according to multiple sites), but as far as I know we don't even have an upper bound. What I want to know is how they got the lower bound

prove that you can't select all four aces in the deck of cards with only three selections.
now imagine that after you have selected one card from deck you return it and remove few non-aces cards from the deck depnding on last selected card... all the way to be left with only 4 cards, which are, of course 4 aces.
similar principle.

waiting to super-uber-computers to calculate all the states of 4x4 and fewest move solve for each one.
one can't simply calculate the 4God's number, but dozens can. :)
 

SenileGenXer

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We seem to have a lot of computing power now. I think it should be calculable.

How much computing power was needed to prove 20 on a 3x3 and how much time has gone by?
 

SenileGenXer

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That was 35 years of CPU time. That's a rounding error to projects like SETI at home.

Nonetheless a 4x4 would require an astronomical amont more. Probably more than all the distributed computing projects put together.
Plus interested qualified people doing the coding.

I would be interesting to know god's answers.
 

Christopher Mowla

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We seem to have a lot of computing power now. I think it should be calculable.
rokicki said:
Will the diameter of the 4x4x4 be known in my lifetime? I suspect it will be, despite the fact that just solving a single random 4x4x4 position optimally probably requires more CPU time using current techniques than were used to determine the diameter of the 3x3x3
Link.
 

qqwref

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Indeed... and there are about 10^26 million times as many 4x4 positions as there are 3x3 positions (that's a hundred million million million million). So even if it was as easy to solve a 4x4 scramble optimally as it is to solve a 3x3 scramble optimally, we'd still need a hundred million million million million times as long to get God's Algorithm as we took to do it for the 3x3...

Considering we're only a few orders of magnitude from the atomic computing limit, I feel like I can predict that this computation cannot be done using our current knowledge of optimal-solving techniques.
 

ben1996123

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Indeed... and there are about 10^26 million times as many 4x4 positions as there are 3x3 positions (that's a hundred million million million million). So even if it was as easy to solve a 4x4 scramble optimally as it is to solve a 3x3 scramble optimally, we'd still need a hundred million million million million times as long to get God's Algorithm as we took to do it for the 3x3...

Considering we're only a few orders of magnitude from the atomic computing limit, I feel like I can predict that this computation cannot be done using our current knowledge of optimal-solving techniques.

How do we actually know that Kociemba generates optimal solutions? I'm guessing some (probably confusing) group theory stuff.
 

Kirjava

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How do we actually know that Kociemba generates optimal solutions? I'm guessing some (probably confusing) group theory stuff.

Kociemba isn't used for generating optimal solutions - that is done by brute force. Simple to understand why they must be optimal.

IIRC when proving god's alg they didn't optimally solve every position, 20 moves is all they needed.
 

ben1996123

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Kociemba isn't used for generating optimal solutions - that is done by brute force. Simple to understand why they must be optimal.

oh ok.

IIRC when proving god's alg they didn't optimally solve every position, 20 moves is all they needed.

yeah, I read this on cube20.org. It says the computers they used could solve 0.36 positions/sec optimally, but 3900 positions/sec in <=20 moves.
 

Cool Frog

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prove that you can't select all four aces in the deck of cards with only three selections.
now imagine that after you have selected one card from deck you return it and remove few non-aces cards from the deck depnding on last selected card... all the way to be left with only 4 cards, which are, of course 4 aces.
similar principle.

waiting to super-uber-computers to calculate all the states of 4x4 and fewest move solve for each one.
one can't simply calculate the 4God's number, but dozens can. :)

Is it because you are only taking 3 cards total?

SURPRISE CHALLENGE IN THE MIDDLE OF A THREAD.

Do a 4x4 Linear FMC and post movecount (Don't have to type out all the moves)
 
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