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Shouldn't 17 Scramble Moves be Enough?

SF

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OK I think I get it now. Not all scrambles are alike. With just random moves, you'd need a whole bunch to create a "random state" cube. I wasn't aware that some programs produce random state scrambles that aren't just a series of n random moves but are really a much longer series of random moves then partially or fully optimized into an "algorithm" of n moves.

I see that Cube Explorer (a very cool program btw) has an option to generate WCA scrambles. I'm assuming these are random state. So I had the program produce 5 WCA scrambles. They are all in the low 20s in terms of the length of the algorithm, but they still aren't fully optimized. So I'm wondering why that is? I had the Cube Explorer program optimize the 5 scrambles (which takes a while), and ended up with four 18 move scrambles and one 17 move scramble.View attachment 5777

Is the reason why the WCA scrambles aren't fully optimized because of a concern that some competitors could actually memorize the series? That seems far-fetched. Or is there a concern that the length of the scramble would give away the difficulty level of the solve? If that's the case, why not make all WCA scrambles exactly 20 moves in length? If the fully optimized algorithm is only 16 or 17 moves for example, I'm sure the computer could figure out a way to add additional cancelling moves to reach the same cube state (pretty sure anyway).

With random state scramble generators (which admittedly I only found out existed as of yesterday) & optimizer programs, I just don't understand why we would ever have to do practice scrambles of greater than 20 moves (since we know that any cube state can be reached with <=20 moves and we know that we can mimic & optimize random move scrambles of any length, even into the hundreds of moves, using optimized algorithms). I practice solving the cube sometimes 2 or 3 hours in a day. Being able to do all my scrambles at between 16 & 20 moves seems like a really big time-saver and hassle-saver vs. doing 22 to 25 moves or more. It makes practicing more fun and leaves more time for actually solving the cube.

Please tell me if I'm missing something and if anyone knows where to find these shorter random state scrambles without having to wait for Cube Explorer to fully optimize the WCA scrambles.
 

shadowslice e

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OK I think I get it now. Not all scrambles are alike. With just random moves, you'd need a whole bunch to create a "random state" cube. I wasn't aware that some programs produce random state scrambles that aren't just a series of n random moves but are really a much longer series of random moves then partially or fully optimized into an "algorithm" of n moves.

I see that Cube Explorer (a very cool program btw) has an option to generate WCA scrambles. I'm assuming these are random state. So I had the program produce 5 WCA scrambles. They are all in the low 20s in terms of the length of the algorithm, but they still aren't fully optimized. So I'm wondering why that is? I had the Cube Explorer program optimize the 5 scrambles (which takes a while), and ended up with four 18 move scrambles and one 17 move scramble.View attachment 5777

Is the reason why the WCA scrambles aren't fully optimized because of a concern that some competitors could actually memorize the series? That seems far-fetched. Or is there a concern that the length of the scramble would give away the difficulty level of the solve? If that's the case, why not make all WCA scrambles exactly 20 moves in length? If the fully optimized algorithm is only 16 or 17 moves for example, I'm sure the computer could figure out a way to add additional cancelling moves to reach the same cube state (pretty sure anyway).

With random state scramble generators (which admittedly I only found out existed as of yesterday) & optimizer programs, I just don't understand why we would ever have to do practice scrambles of greater than 20 moves (since we know that any cube state can be reached with <=20 moves and we know that we can mimic & optimize random move scrambles of any length, even into the hundreds of moves, using optimized algorithms). I practice solving the cube sometimes 2 or 3 hours in a day. Being able to do all my scrambles at between 16 & 20 moves seems like a really big time-saver and hassle-saver vs. doing 22 to 25 moves or more. It makes practicing more fun and leaves more time for actually solving the cube.

Please tell me if I'm missing something and if anyone knows where to find these shorter random state scrambles without having to wait for Cube Explorer to fully optimize the WCA scrambles.

See my post above. In essence it is dwon to computer power and speed. The program uses Kociemba and IDA* to create a realtively short scramble so that the state can be reached more quickly when a human scrambles.

It does not need to create optimal scrambles as this uses more processing power and hence takes longer with little (if any) benefit. The point is, finding more optimal solutions would take longer than it would for you to do maybe 2-6 more moves.

This is also why the WCA uses random move scrambles on larger cubes and megaminx because no one has developed a program or method which can efficiently get to each state quickly (or in the amount of time a person would want to wait for new scrambles in a session) without a large amount of processing power.
 
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SF

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Thanks shadowslice. I did read your earlier post and it was very helpful, especially in defining "random state." I just don't know if I can fully buy-in to the argument about computer processing speed. On my generic 2 year-old home pc, I'm sure I could generate a few hundred fully optimized random state scrambles per day using Cube Explorer. (Actually, strike that, I guess the website is doing the processing, not my computer.) At any rate, while I agree it isn't worth it for me to spend the time and energy to have Cube Explorer generate marginally shorter scrambles for my own use, why can't some bright kid write a program that "stockpiles" these scrambles? Or why doesn't the Cube Explorer guy do it? He could be generating these shorter algorithms 24 hours a day and always have a huge stockpile for users to draw upon online. It wouldn't matter to me that the scrambles aren't being generated in real time since, to quote Allen Iverson, "we're talking about practice here."

Am I the only one out there who sees value in making scrambles shorter? If we do thousands of practice scrambles per year, then wouldn't lowering the average length from, say, 25 to 18 be a huge benefit? That's a 28% reduction in scramble moves. Even for the fastest cubers, that's a significant percentage reduction in the time spent scrambling their cubes.
 

shadowslice e

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Thanks shadowslice. I did read your earlier post and it was very helpful, especially in defining "random state." I just don't know if I can fully buy-in to the argument about computer processing speed. On my generic 2 year-old home pc, I'm sure I could generate a few hundred fully optimized random state scrambles per day using Cube Explorer. (Actually, strike that, I guess the website is doing the processing, not my computer.) At any rate, while I agree it isn't worth it for me to spend the time and energy to have Cube Explorer generate marginally shorter scrambles for my own use, why can't some bright kid write a program that "stockpiles" these scrambles? Or why doesn't the Cube Explorer guy do it? He could be generating these shorter algorithms 24 hours a day and always have a huge stockpile for users to draw upon online. It wouldn't matter to me that the scrambles aren't being generated in real time since, to quote Allen Iverson, "we're talking about practice here."

Am I the only one out there who sees value in making scrambles shorter? If we do thousands of practice scrambles per year, then wouldn't lowering the average length from, say, 25 to 18 be a huge benefit? That's a 28% reduction in scramble moves. Even for the fastest cubers, that's a significant percentage reduction in the time spent scrambling their cubes.

Because if you want optimal scrambles for all states you would need a few petabytes of space.

There was a team (which incidentally included the "cube explorer guy") who proved God's number was 20 by generating solutions for all the scrambles and they had to use google's supercomputer.

And no your computer is actually doing the processing.

Also, a "few hundred optimal scrambles" is nothing to the quadrillions of possible permutations.
 
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mark49152

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I wasn't aware that some programs produce random state scrambles that aren't just a series of n random moves but are really a much longer series of random moves then partially or fully optimized into an "algorithm" of n moves.
No you still don't get it - perhaps go back and read the thread again.
 

mark49152

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He's since reread some of the other posts since the I think.
According to his post, despite all the explanations in this thread, he still thinks a random state scramble is when you take a really long sequence of say 50 moves for "better randomness" and use a solver to shorten it.

Here's another attempt to explain random state scrambles in even simpler terms.

1. Take your cube and pull it apart. Put all the pieces in a bag and shake to mix.

2. Now, decide which cubie position you're going to fill first. Let's say it's the FB edge. Put your hand in the bag, pull out some random edge, and put it in the FB position. Make sure you insert it in a random orientation too! Maybe flip a coin, or don't look at the colours until after you've inserted it.

3. Continue until you've fully reassembled the cube with all pieces in random positions and orientations.

4. Go to Cube Explorer and input the colour pattern of your randomised cube. Click solve. The resulting solution will solve your random cube state, and thus the inverse of that solution will scramble a solved cube back to your random state. This is what is called a "random state scramble".

5. Only about 1/12 of your randomly reassembled cubes will be solvable, due to parities and twists. If unsolvable, go back and start again. When you are expert (like scrambling software) you can take care when inserting the last three pieces to make sure the cube is solvable.
 

SF

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According to his post, despite all the explanations in this thread, he still thinks a random state scramble is when you take a really long sequence of say 50 moves for "better randomness" and use a solver to shorten it.

Here's another attempt to explain random state scrambles in even simpler terms.

1. Take your cube and pull it apart. Put all the pieces in a bag and shake to mix.

2. Now, decide which cubie position you're going to fill first. Let's say it's the FB edge. Put your hand in the bag, pull out some random edge, and put it in the FB position. Make sure you insert it in a random orientation too! Maybe flip a coin, or don't look at the colours until after you've inserted it.

3. Continue until you've fully reassembled the cube with all pieces in random positions and orientations.

4. Go to Cube Explorer and input the colour pattern of your randomised cube. Click solve. The resulting solution will solve your random cube state, and thus the inverse of that solution will scramble a solved cube back to your random state. This is what is called a "random state scramble".

5. Only about 1/12 of your randomly reassembled cubes will be solvable, due to parities and twists. If unsolvable, go back and start again. When you are expert (like scrambling software) you can take care when inserting the last three pieces to make sure the cube is solvable.

OK thanks. I do get that a random state scramble is one in which there is essentially an equal probability that any one of the 43 quintillion possible cube states will result, and your example of disassembling and reassembling the cube seems like a good intuitive way to get there (at least 1 in 12 times). What I'm not sure of is how many random turns on a solved cube is enough to get there. Maybe it's hundreds, maybe it's thousands, maybe many more than that. But as long as we have software programs that can virtually disassemble and reassemble the cube to a random state, it doesn't matter if I know the answer. Because I can rely on these programs to create algorithms to get me to true random states.

What I wasn't getting before is that the scrambler programs (at least the good ones) aren't just returning a series of random moves, they're returning an algorithm that's an efficient way to get to the random cube state they've already picked. So they don't just take the letters "LRUDFB" (adding a 2, a prime, or nothing) and pick one randomly for each step in the scramble, thereby getting to the cube state that would result. Instead, it's the reverse - they're starting with the random cube state and then arranging the moves to arrive at that cube state.

Big difference, and there are some interesting consequences to this. Like now I'm going to be more careful that I don't make careless errors in performing my scrambles. Before, I felt it really didn't matter if I accidentally did a B' instead of a B as long as I was doing the same number of moves. Now I realize that a single error will most likely invalidate the "random state" of the scrambled cube.

I guess another consequence is that I need to start with a solved cube before performing the scramble. In the past, if I screwed up a solve & didn't feel like completing it I would just do the scramble on a partially solved cube. Seems like that would also invalidate the random state.
 
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mark49152

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What I'm not sure of is how many random turns on a solved cube is enough to get there. Maybe it's hundreds, maybe it's thousands, maybe many more than that.
It's an interesting question anyway and I don't know the answer, although I would hazard a guess that you would never actually reach an even distribution, just asymptotically converge on it. You would also need to make sure that scramble lengths are evenly distributed between odd and even numbers of quarter turns. As you point out though, it doesn't really matter, because there are quicker and more accurate ways to ensure an even distribution of random states.

What I wasn't getting before is that the scrambler programs (at least the good ones) aren't just returning a series of random moves, they're returning an algorithm that's an efficient way to get to the random cube state they've already picked. So they don't just take the letters "LRUDFB" (adding a 2, a prime, or nothing) and pick one randomly for each step in the scramble, thereby getting to the cube state that would result. Instead, it's the reverse - they're starting with the random cube state and then arranging the moves to arrive at that cube state.
Correct, you got it now :D.

Big difference, and there are some interesting consequences to this. Like now I'm going to be more careful that I don't make careless errors in performing my scrambles. Before, I felt it really didn't matter if I accidentally did a B' instead of a B as long as I was doing the same number of moves. Now I realize that a single error will most likely invalidate the "random state" of the scrambled cube.

I guess another consequence is that I need to start with a solved cube before performing the scramble. In the past, if I screwed up a solve & didn't feel like completing it I would just do the scramble on a partially solved cube. Seems like that would also invalidate the random state.
That's another interesting question. I don't think this kid of error impacts on the randomness of the scramble. Think of it this way: a single quarter turn does a 4-cycle of corners and a 4-cycle of edges. So if you make a single quarter-turn error, what you end up with is a random cube state with 4 edges and 4 corners cycled (not necessarily on the same face any more, of course). Is that any less random? As far as I can see, no, as long as the error is genuinely random and not derived from the state of the cube at the time. If I take a random cube state, and apply two random 4-cycles, the result is still random.

Compare to this: I have a random number generator that gives me an even distribution of numbers between 1 and 10. Occasionally, randomly, I add 1 to the result (modulo 10). The result is still a random number, with the same distribution. The randomness would only be compromised if the act of adding 1 is derived somehow, for example by adding 1 iff the initial value is <5.

The same logic applies to your second example. The new scramble isn't designed to manipulate the specific unsolved cube state, so how could any of the randomness be invalidated?
 

SF

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Thanks Mark. Had to look up some of the words in your explanation, but it definitely makes sense! :)

Because if you want optimal scrambles for all states you would need a few petabytes of space.

There was a team (which incidentally included the "cube explorer guy") who proved God's number was 20 by generating solutions for all the scrambles and they had to use google's supercomputer.

And no your computer is actually doing the processing.

Also, a "few hundred optimal scrambles" is nothing to the quadrillions of possible permutations.

I may have overestimated the difference in length between Cube Explorer's instant scramble vs. their optimal scramble for a given random cube state. After using the program a little more, I'm seeing that most of the instant scrambles are averaging between 20 & 21 moves, while most of the optimized scrambles are between 17 & 18 moves. So I'm really only saving about 3 moves between the two (about 15% fewer moves). And it does take a long time to resolve the optimal scrambles, especially if I do a large block simultaneously.
 
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It takes at most 20 moves for solving the cube, therefore, some combinations can't be solved with 17 moves. This means that there would be no scramble for this positions. If you allow only 17 moves, you could get about 15 out of 43*10^18 possibilities. It's about 30%.
 

shadowslice e

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It takes at most 20 moves for solving the cube, therefore, some combinations can't be solved with 17 moves. This means that there would be no scramble for this positions. If you allow only 17 moves, you could get about 15 out of 43*10^18 possibilities. It's about 30%.

Yeah the OP gets it now. He was working on the assumption that most 17 move scrambles are no harder than the 20 move scrambles and there are far too many to memorise them all.

It was cleared up after the random state was explained and how scramblers work.
 

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That's another interesting question. I don't think this kid of error impacts on the randomness of the scramble. Think of it this way: a single quarter turn does a 4-cycle of corners and a 4-cycle of edges. So if you make a single quarter-turn error, what you end up with is a random cube state with 4 edges and 4 corners cycled (not necessarily on the same face any more, of course). Is that any less random?

Yes. Yes, it is less random. Suppose you're scrambling a 1x3, and the scrambles are: L, L R, L R2, L R', L2, L2 R, L2 R2, L2 R', L', L' R, L' R2, L' R', R, R2, R'

Now suppose you "accidentally" do an R turn instead of an L turn sometimes. In those cases, you'll never scramble the left layer!

The same is the case if you accidentally do an R' turn instead of an R turn sometimes. The effect is much, much smaller, but you're almost certainly going to bias the outcome.

As far as I can see, no, as long as the error is genuinely random and not derived from the state of the cube at the time. If I take a random cube state, and apply two random 4-cycles, the result is still random.

Only if the combined effect of the 4-cycles is independent of the cube state.

Doing a certain wrong turn is *not* independent of the scramble (which is clearly not independent of the cube state). Perhaps you mix up the direction of D sometimes. Maybe R turns. The frequency of D turns or of double turns now affects your chance of biasing the scramble. Same reasoning as above.

The randomness would only be compromised if the act of adding 1 is derived somehow, for example by adding 1 iff the initial value is <5.

This is mostly just reiteration, but this is the core of the problem. Doing certain scramble turns wrong is biased exactly like adding 1 iff the initial value is <5. It just has an effect that is not practical to measure concretely.
 

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Doing certain scramble turns wrong is biased exactly like adding 1 iff the initial value is <5. It just has an effect that is not practical to measure concretely.
True, but that's not a random error. I did say random error. You are correct that if you apply some rule, consistently, like always inverting B/B', you may well skew the distribution.

Actually scramblers could address that by choosing the solution to a given random state randomly rather than deterministically. I don't know if any do, since it's really far too tiny an issue to care about :).
 

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True, but that's not a random error. I did say random error. You are correct that if you apply some rule, consistently, like always inverting B/B', you may well skew the distribution.

My argument shows that i's problematic even if you do it consistently. If you make it a "random" error, that doesn't make it better.

Actually scramblers could address that by choosing the solution to a given random state randomly rather than deterministically. I don't know if any do, since it's really far too tiny an issue to care about :).

... read the rest of this thread? ;-)
 

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Since I've been using Cube Explorer to generate random state scrambles, my average solve time has gone up by about two seconds (about 43 vs. 41 seconds). We're only talking about 100 solves or so since I made the change from 17 moves on the jaapsch.net site to the CE site, so maybe the two things aren't related. But maybe they are, since I could be getting more flipped edges, etc.

The jaapsch site says in the help section, "The scrambles will not contain any moves that cancel each other, nor moves that simplify to a cube rotation." I wonder what that means. Is he saying he's generating something like random state scrambles, or just that he never repeats turning the same face twice in a row? Not that important that I find out the answer since I'm now using Cube Explorer, but my (unscientific) results so far are telling me that if I do true random state scrambles (and I'm careful not to make mistakes), the solutions are a little more difficult.
 

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Since I've been using Cube Explorer to generate random state scrambles, my average solve time has gone up by about two seconds (about 43 vs. 41 seconds). We're only talking about 100 solves or so since I made the change from 17 moves on the jaapsch.net site to the CE site, so maybe the two things aren't related. But maybe they are, since I could be getting more flipped edges, etc.

The jaapsch site says in the help section, "The scrambles will not contain any moves that cancel each other, nor moves that simplify to a cube rotation." I wonder what that means. Is he saying he's generating something like random state scrambles, or just that he never repeats turning the same face twice in a row? Not that important that I find out the answer since I'm now using Cube Explorer, but my (unscientific) results so far are telling me that if I do true random state scrambles (and I'm careful not to make mistakes), the solutions are a little more difficult.

Basically the jaapsch site is saying it won't do R R' or say you are doing 2x2 U D' etc.

e: to answer your question the solves may be harder but a 2 second jump when averaging sup 40 isn't too uncommon, plus these are the types of scrambles you will get in competitions so its best to practice with these
 

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My argument shows that i's problematic even if you do it consistently. If you make it a "random" error, that doesn't make it better.
Let's take scrambles for all 43 quintillion states. Now apply an error to each scramble. In effect, you are now selecting a new random state 43 quintillion times, and can expect some duplicates, meaning some other states are never reached. I don't have time to do the math so it is likely wrong but let's assume 30% of states are never reached.

Now let's take the same set of scrambles and apply the same errors, because we consistently forget which way round B/B' are on the first B turn, etc. You will end up with the same set of states never being reached. So you have indeed skewed the distribution. Those 30% of states are never reachable. Unless I misunderstood you, this is the point I thought you were trying to make above about errors biasing the outcome.

Now let's take the same set of scrambles and apply a different (random) set of errors. Again, 30% of states won't be reached, but this time, because they are different errors, it's not the same set of states. It's a different 30%. Repeat ad infinitum and observe that the distribution evens out. Nothing you have said so far has convinced me that random + random != random.

You could of course analyze every possible error to each of those 43 quintillion scrambles and plot the frequency at which each state occurs and show that they are no longer evenly distributed, but the margins of error are becoming increasingly irrelevant to practical cubing.

... read the rest of this thread? ;-)
I've been reading and posting since the start so maybe you can be more specific about what I've missed. However, I'll also try to explain better what I meant. A random state scrambler generates a random state and then finds a solution to that state, returning the inverse as its output. However, there are many possible solutions to each state, as you know. The effect described in my second para above, where we apply the same errors and get the same set of unreachable states, depends on the same set of scrambles being used. If the scrambler were to instead choose randomly from the set of possible solutions to each random state, then we would get different scrambles each time, even for the same random state, and the effect of the errors would therefore be different.


Anyway... I think this is getting somewhat distanced from the question, which was whether an error in scrambling can compromise the "randomness" of a scramble. Defining randomness in terms of distribution is fine from a theoretical viewpoint but I guess what most people are thinking when they ask that question is whether an error is likely to make the scramble easier to solve, meaning more blocks or pairs already formed, or whatever, which depends on event and method anyway. There's a tendency to think that each move in a scramble is performing some direct function in mixing up the cube and if that function is impaired the cube will not be properly mixed. That was what I was thinking of when I made my original comment about 4-cycles, and to me it's a more interesting and pragmatic question, although I have no better ideas on how to answer it.
 

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Anyway... I think this is getting somewhat distanced from the question, which was whether an error in scrambling can compromise the "randomness" of a scramble.

The only correct answer is "yes, it does".

Repeat ad infinitum and observe that the distribution evens out.

This is where your argument fails. Unless you do something specific and systematic (e.g. randomly add a U turn at the end half of the time after generating a scramble), there is absolutely no guarantee that the distribution "evens out" if you make different kinds of errors.

As I've pointed out above, the (possible, and therefore actual) errors depend on the scramble, which depends on the state.
I don't think you can come up with a good way to characterize the random errors humans make while scrambling, but you cannot assume that the distribution evens out until a) you do come up with a way to characterize the errors, and b) prove that it the distribution evens out for the scramble programs you care about.

You could of course analyze every possible error to each of those 43 quintillion scrambles and plot the frequency at which each state occurs and show that they are no longer evenly distributed, but the margins of error are becoming increasingly irrelevant to practical cubing.

Ah, yes. The margins may be very small. But can you prove it?

(I suspect not. And if not, you should not assume they're small.)
 

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The margins may be very small. But can you prove it?
As I said, I'm not inclined to, because I don't think it's practically relevant. The question of whether errors skew scramble distribution is purely academic because what most people care about is one scramble and one error at a time. Oops, I think I did a B instead of a B' there, does that mean there is now something inherently unfair about the state of my cube such that I should exclude this solve from my average? The fact that if I deliberately made that same error on every scramble I might only be seeing a possible 30 quintillion cube states instead of all 43 is nothing more than a mathematical curiosity.

The only correct answer is "yes, it does".
Maybe the question should be whether it affects it in a meaningful way, and indeed how we define randomness for the purpose of the question.
 
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