# Can you solve a 3x3 puzzle in N moves, knowing...



## gurthbruins (Dec 14, 2018)

Can you solve a 3x3 puzzle in N moves, knowing that N moves have been used to scramble the Cube? For instance if I scramble a Cube by doing 17 moves, can you solve it in 17 moves? Or if 17 is too high a number to manage, what number can be managed?


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## AbsoRuud (Dec 14, 2018)

You can always solve it in X moves where X <= N. Whether you can figure out those moves is a whole other thing.


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## Billabob (Dec 14, 2018)

AbsoRuud said:


> You can always solve it in X moves where X <= N. Whether you can figure out those moves is a whole other thing.


Not entirely true. In QTM, a scramble with odd/even N will always take an odd/even number of moves to solve. This is because a single turn does an odd number of swaps on both corners and edges, so if the scramble has an odd permutation of corners it will always take an odd number of moves to solve. (At least I think so. Please correct me if I’m mistaken)

In HTM your statement is true because you can just do U2 U' instead of U to add 1 to the move count.


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## AbsoRuud (Dec 14, 2018)

I am not familiar with QTM or HTM.


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## cuber314159 (Dec 14, 2018)

AbsoRuud said:


> I am not familiar with QTM or HTM.


QTM: quarter turn metric: any 90° turn of any face is one move
HTM: half turn metric: any turn of any face is one move


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## AbsoRuud (Dec 14, 2018)

I still think my statement is true. The lowest number of moves needed to solve a cube is always lower or equal to the number of moves used to scramble. Whether it needs to be a multiple of 2 less than the number of moves to scramble is another discussion altogether.


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## cuber314159 (Dec 14, 2018)

AbsoRuud said:


> I still think my statement is true. The lowest number of moves needed to solve a cube is always lower or equal to the number of moves used to scramble. Whether it needs to be a multiple of 2 less than the number of moves to scramble is another discussion altogether.


Well obviously as long as the same way of counting the moves is used.


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## SenorJuan (Dec 14, 2018)

I've tried this 'solving challenge' quite a number of times, in fact we even did it back in the 80's. I would scramble, then put the puzzle down for an hour, so I didn't have any worthwhile memory of what moves I did.
A limit of 12 turns is the best I can manage, but I'm sure more competent people could do better, especially with practice. 17 turns is a lot, though. It would also be very time-consuming.
I recall Jessica Fridrich mentions it on her website, at an early competition she attended, another competitor was solving 10-move scrambles easily.


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## xyzzy (Dec 14, 2018)

SenorJuan said:


> 17 turns is a lot, though. It would also be very time-consuming.


At that point, you're really doing an FMC attempt, trying to look for the optimal solution. 17 moves is enough to cover around a quarter of all possible scrambles.

(Granted, a large proportion of 17-move scramble sequences will have optimal solutions that are 15 or 16 moves long, but that still doesn't mean it's easy to find a 17-move solution.)


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## SenorJuan (Dec 14, 2018)

Solving success depends on the 'rules' you choose. If you choose to only make 'counting' turns, and work it out entirely in your head, it's unsurprisingly harder than allowing 'trial moves' , to see where a two or three-move sequence gets you.

Maybe it's an idea for a Forum competition. 

It's obviously best if someone else executes the scramble for you, then you're totally unaware of the scramble, and the length of it.
If you're self-scrambling, one strategy is to scramble, say, 4 cubes with different scrambles. Mix them up, put them away for a few hours. When you come to the solve attempts, you'll have no idea which is which, and hopefully gain no advantage.


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## Klaudiusz Szyprocinski (Jan 7, 2019)

SenorJuan said:


> A limit of 12 turns is the best I can manage, but I'm sure more competent people could do better, especially with practice.



Do you have any proof?


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