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What is the minimum number of algorithms needed for 2 look last layer?

XTowncuber

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61

Was not optimised for low alg count, can be done.

No, you're right. It's actually unsolvable with this system.

There are around 10 cases like this. I'm thinking of either just generating algs for them for 1LLL when it comes up or doing some other thing (I have a few ideas). The probability of each appearing is 1/~4600, so not a huge deal at the moment.
Did you end up fixing this problem then? I just read through that thread, and I don't remember you saying that you dealt with those bad cases, so just wondering.

Glad you brought that thread up again, it's fun to read and I was too much of a nub to follow it when it was active.
 

Tempus

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I agree that with 32 being a lower bound.

Initial state: 1 state
AUF + 1 alg applied: 4*31 = 124 states
AUF + 1st alg + AUF + 2nd alg: 124*4*31 = 15376 states
Max total reached states: 1 + 124 + 15376 = 15501 < 15552
So 31 not sufficient:
For 32, I get 1 + 128 + 16384 = 16513 > 15552; so it seems 32 could work.
I see this as analogous to looking at the question of how many 1 inch discs you would need to fully cover a 5-inch disc and deciding that the lower bound is 25 because 25 1-inch discs together have the same surface area as 1 5-inch disc. In either case, common sense tells you that it is obviously impossible to satisfy the condition with a number of discs/algorithms anywhere near that low, but rigorously codifying exactly why is rather tricky, as is establishing a more realistic lower bound.

Meanwhile, I'm approaching it from the other end, trying to whittle down the upper bound I've already set at 67.
 

Kirjava

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Did you end up fixing this problem then? I just read through that thread, and I don't remember you saying that you dealt with those bad cases, so just wondering.

Glad you brought that thread up again, it's fun to read and I was too much of a nub to follow it when it was active.

61 is taking into account the x number of cases you'd need extra algs for. It would be 48 without.

I actually forgot that you could just invert the algs that were extra, so that would only require 55 because each has a pair.
 

Christopher Mowla

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It's also interesting to note that it's possible on the 4x4x4 last layer, at least, that you can have algs which, if you just invert some turns, creates an alternate case. For example, [r2 B2 U' l2 U: l2] and [r2 B2 U l2 U': l2]. (There are several other instances when this has happened).

Thus, it might be possible to create a list of algs (even though they might be longer than optimal) which are closely related as these are.

The result would be, of course, a list which has the fewest number of "base" algorithms (not using setup moves). If such a thing is possible, then the resulting set of such algs would be essentially smaller than the actual calculated/computed minimum number of algorithms.
 

Lucas Garron

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Yeah, if you want to draw from the same set, 32 sounds like a good lower bound. I'd love to see someone do this for PLL, to see if the numbers are more analogous to 62 or 32. I have no intuition for this.

The search space would almost certainly need to be restricted to PLLs, so this wouldn't be *that* hard.

Thus, it might be possible to create a list of algs (even though they might be longer than optimal) which are closely related as these are.

You could almost certainly make algs with small Hamming distances like this:

A (U*) B (U*) C (U*) D

... where A, B, C, D are fixed. This gives us lots of algs with minimal differences. I'm willing to bet that we can set this up so that you can convert a natural indexing scheme to U-directions.

I don't think that's very interesting, though.
 

Tempus

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61 is taking into account the x number of cases you'd need extra algs for. It would be 48 without.

I actually forgot that you could just invert the algs that were extra, so that would only require 55 because each has a pair.
I gather from this that the 48 also assumes that you are counting an algorithm and its inverse together as one algorithm. I, however, am not. By my reckoning, an algorithm is distinct from its inverse, its mirror, and it's mirror inverse. Together they would comprise 4 distinct algorithms. By this method of counting, I have now managed to lower the upper bound of n=67 to n=66, and I suspect it can go no lower than that.

Based on the output of my program so far, it is my gut instinct that 66 is the minimum number of algorithms required to solve the LL in 2 looks with AUF.
 

Tempus

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what is the minumum number of algorithms needed such that for any last layer case, you can apply 2 algorithms and solve the cube?
A few weeks have passed since my last message in this thread. During that time, I have run scenarios almost constantly, but the weather is getting too hot for non-stop processing. It burns about 100 extra watts to run, and that's heat I don't need right now, so I am discontinuing the work for the moment, but I think I've gathered some useful information with which to answer your question, and I've found some surprising results.

Firstly, I'm assuming that AUF is allowed both before and after algorithms. Secondly, I don't know whether you're counting the number of algorithms overall, or are you allowing an algorithm and it's inverse to count as one? Or are you perhaps allowing an algorithm and it's mirror to count as one? Maybe you're even counting an algorithm, its mirror, its inverse, and its mirror-inverse to all count as one.

Since I did not know the answer, I calculated all four, running 24 hours a day. On three separate occasions, I found ways to radically increase the speed of my program, each time making it several times faster than before, so it should now be faster than it started out by about two orders of magnitude.

Here are the results in short form:
  • If you count each algorithm separately, you can do 2-Look Last Layer (2LLL) using just 63 algorithms. I found this rather surprising, as at one time I thought that 66 was the best that could be done.
  • If you count an algorithm and it's mirror together as one, you can do 2LLL in just 33 algorithms.
  • If you count an algorithm and it's inverse together as one, you can do 2LLL in just 33 algorithms.
  • If you count an algorithm, its mirror, its inverse, AND its mirror-inverse all as one algorithm, you can do 2LLL in just 19 algorithms.
If anyone is interested in examples of sets of algorithms that display these properties, let me know.


what about 3 look?
This one has a very interesting result because it seems to defy what's written in the wiki. Here's what the wiki says about 3LLL:
The Speedsolving.com Wiki said:
The 3 Look Last Layer involves completing the last layer in 3 steps or looks. It usually consists of a 2-Look OLL, followed by a 1-Look PLL which requires knowledge of 31 algorithms in total (including mirrors) with an average of 31 moves. Another method is an LLEF followed by a OCLL-EPP and finally a CPLL which requires 26 algorithms in total (25 if excluding a possible reuse of an algorithm) with an average of 27 moves.
That would seem to indicate that the smallest set of algorithms anyone else has found that is sufficient to do 3-Look Last Layer is a set of 25. My program, however, has found that it is possible to do 3LLL using just ELEVEN.

Here is one such set:
  1. U F2 R' F' R F' U2 F R U2 R' F' (12f)
  2. U2 F U F' U' R' U F' R' F' R F2 U' F' R (15f)
  3. U' F' R2 F' U F2 U F2 U2 F2 U F' R2 F U (15f)
  4. U2 R' F' R U2 F R' F R F2 U2 R' F R U' (15f)
  5. R U2 R U2 R' U F R2 U' R' U R' U R2 U F' (16f)
  6. R U R2 F R2 U R' U' F2 U2 F (11f)
  7. U R U R' F2 U2 F U F' U F2 R U' R' U (15f)
  8. U R U2 R2 F R F' R U2 R2 F' U' F U R (15f)
  9. R U2 R2 F' U' F U R U R U R' F' U' F U' (16f)
  10. F' R' U' R U2 F2 R2 F' U' R U' F' U2 F R2 (15f)
  11. U' F U F' U F' U' F2 R U' R' F2 U F U2 (15f)
I imagine the number would be even smaller if one counted mirrors and/or inverses as being the same.
 

Dane man

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That would seem to indicate that the smallest set of algorithms anyone else has found that is sufficient to do 3-Look Last Layer is a set of 25. My program, however, has found that it is possible to do 3LLL using just ELEVEN.

Here is one such set:
U F2 R' F' R F' U2 F R U2 R' F' (12f)
U2 F U F' U' R' U F' R' F' R F2 U' F' R (15f)
U' F' R2 F' U F2 U F2 U2 F2 U F' R2 F U (15f)
U2 R' F' R U2 F R' F R F2 U2 R' F R U' (15f)
R U2 R U2 R' U F R2 U' R' U R' U R2 U F' (16f)
R U R2 F R2 U R' U' F2 U2 F (11f)
U R U R' F2 U2 F U F' U F2 R U' R' U (15f)
U R U2 R2 F R F' R U2 R2 F' U' F U R (15f)
R U2 R2 F' U' F U R U R U R' F' U' F U' (16f)
F' R' U' R U2 F2 R2 F' U' R U' F' U2 F R2 (15f)
U' F U F' U F' U' F2 R U' R' F2 U F U2 (15f)
So are those based on a set of steps used to achieve the solved state, or are they simply algs that any combination required to solve the last layer stays within 3 steps?
 

qqwref

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My program, however, has found that it is possible to do 3LLL using just ELEVEN.

Here is one such set:
  1. U F2 R' F' R F' U2 F R U2 R' F' (12f)
  2. U2 F U F' U' R' U F' R' F' R F2 U' F' R (15f)
  3. U' F' R2 F' U F2 U F2 U2 F2 U F' R2 F U (15f)
  4. U2 R' F' R U2 F R' F R F2 U2 R' F R U' (15f)
  5. R U2 R U2 R' U F R2 U' R' U R' U R2 U F' (16f)
  6. R U R2 F R2 U R' U' F2 U2 F (11f)
  7. U R U R' F2 U2 F U F' U F2 R U' R' U (15f)
  8. U R U2 R2 F R F' R U2 R2 F' U' F U R (15f)
  9. R U2 R2 F' U' F U R U R U R' F' U' F U' (16f)
  10. F' R' U' R U2 F2 R2 F' U' R U' F' U2 F R2 (15f)
  11. U' F U F' U F' U' F2 R U' R' F2 U F U2 (15f)
Very cool. I'd also be interested in having more of this information posted (maybe in a spoiler?).
 

Tempus

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So are those based on a set of steps used to achieve the solved state, or are they simply algs that any combination required to solve the last layer stays within 3 steps?
The latter. They are simply a set of algorithms that, bracketed by AUFs as necessary, my computer has determined sufficient to get from any last layer state to any other last layer state in only three hops. Put in terms of graph theory, if you envision the 15,552 possible upper layer states as nodes of a graph, and these eleven algorithms as dictating the edges of that graph, then the resulting graph has a diameter of 3.

I do not personally know how one would apply them intelligently, but the computer would know just by being able to remember every possible scenario at once. Perhaps a set of guidelines or rules could be developed, but I have not tried to do so. I've just thrown processing power and tricky programming at it to try to answer the question of what's possible.


Send me everything.
Very cool. I'd also be interested in having more of this information posted (maybe in a spoiler?).
As you wish.
  1. U' R' U2 R U2 F R U R' U' F' R' U R (14f)
  2. U' F U R' U2 R2 U R2 U F' R U R U' R' U (16f)
  3. U2 F U R U' R' U F' R' F U' F' U R (14f)
  4. U R F U' R' U2 R' U' R2 U R2 F' R' U2 (14f)
  5. U R' U2 R U F R' F' U R U R U' R' (14f)
  6. U R U R2 F R F2 U' F U F' U F U' (14f)
  7. R' U F2 U' F R F' U R' F2 U' F' R U2 F2 (15f)
  8. U F U R U' R2 F' U2 R U R U' R2 U2 R U' (16f)
  9. U2 F R U R2 F' R' U2 R F R' U2 R2 U' F' U2 (16f)
  10. R' U' R' F' U' F U R2 U2 R' U' R U (13f)
  11. U' F R2 U F' R' F U' R2 F' U' R U R' (14f)
  12. U F' U2 R2 F R U' R U' R2 U' F' U R2 F U' (16f)
  13. R U2 R' F2 U' R U' R' F R U R' U F' R' F' R (17f)
  14. U' F U R' F' R F' U F2 U2 R' F' R F' U F (16f)
  15. R U2 R' F R' F R' F R' U2 R F' R F2 R (15f)
  16. U2 F' U2 F2 R' F' R' F' U2 F U2 F R2 F' (14f)
  17. U' R U2 R2 F U' R2 U' R2 U F' U R U2 (14f)
  18. U2 R' F R F U R' U F2 U F2 U' F' R U2 F2 U2 (17f)
  19. R' F R F2 U2 F2 R' F' R F' U F U2 F' U F (16f)
  20. U2 R U R' U2 F2 R U2 R' U2 R' F2 R2 U R' U2 (16f)
  21. U2 R2 F U F2 U F2 U' R2 U' R' F' U' R (14f)
  22. U2 R' U' F' U F2 U R U' R2 F' R2 U R' U' R U (17f)
  23. U' R' F' U' F2 R U' R' U2 R U R' F' R U (15f)
  24. R F' R' U2 F U2 R' U' R F' R' U R2 F R' (15f)
  25. U2 F' U R U R2 F R F2 U F2 U' F' U2 F U' (16f)
  26. U' F R U' R2 U2 R2 U R2 U F R F' U' F' U' (16f)
  27. R' U R' U' F R F' R' U R F' U F U' R (15f)
  28. F' U2 F2 R' F' R U2 F R' F2 U' F U R U' (15f)
  29. U F U R' F R F2 R' F U' F' U R (13f)
  30. R' U2 F' R F' U2 F2 U2 F R2 F2 R U2 F R (15f)
  31. U F U' R2 U' F2 U' F2 U2 R2 U F2 U2 F (14f)
  32. U R' U' F U R U' R' F2 U F U' F' U' F R U' (17f)
  33. U2 R' U2 R U2 F' U' F2 R' F' U R2 U2 R' U2 (15f)
  34. F' R' F R F R2 U R' F' R' U' F2 R U F R' (16f)
  35. F2 U' F2 U F U R' U' F2 R F R' U R (14f)
  36. U' R U R' U R F U' R' U R U F' U R' (15f)
  37. U' R' F2 U2 F R' U' F2 U F R U' F U' F2 R (16f)
  38. U R' U' F U' R2 F2 R2 U2 R2 F2 R' U' R' F' R (16f)
  39. U' F' R U' R F' R' U R F' U' F' U F2 R2 F (16f)
  40. U' F U2 F' R' U' F U F' R U2 F R' F R F2 (16f)
  41. R' F R U R' F' R F U' R U R' U' F' U (15f)
  42. U R F U' F' U' R' F R' F' R2 F U2 F' R' U2 (16f)
  43. U2 F U R U' R U R' U R F R' F' U2 R2 F' U' (17f)
  44. U' R U R' F2 U' F U' F' U2 F U F R U' R' (16f)
  45. R' U' F' U2 F2 U2 F2 U' F R U F U2 F' (14f)
  46. U2 R U R' U R' F2 U' R2 U' R2 U F2 U R (15f)
  47. U' F' R' U F U2 F U F2 U' F2 R F U' (14f)
  48. U F R' U' R2 U' R2 U2 F R F2 U F U2 F' (15f)
  49. U R' F2 U F R' F R F' U' F2 R U2 F U2 F' (16f)
  50. U R U' R2 F' U' F U R2 U' R' U R U R' U2 (16f)
  51. U R U2 R' U2 R' F R F' U2 R' F' U' F U R U (17f)
  52. U R' F' U' F2 R' F' R U' R U2 R' U F R' F' R2 (17f)
  53. U R' F R U R' U' F' R' F R F2 U F R U' (16f)
  54. R' U2 R U F2 R' F2 U2 F U' F' U' F2 R F2 (15f)
  55. U R2 F2 R2 F' U2 F R2 F2 R F R F2 U2 F (15f)
  56. R' F' R2 U R' F' U R' F R U F2 R U2 R' U2 (16f)
  57. U2 F R U2 R' F' U2 F R' F R F2 U (13f)
  58. R F' U2 F U' R U2 R F R2 F' U R U2 R2 U (16f)
  59. U R U R' U2 F' U2 F R U2 R' U F' U2 F (15f)
  60. R U2 R2 F' U2 R' U2 R U2 F R' U2 R2 U2 R (15f)
  61. U R' F2 U' F2 U F' R F' U2 F' R' F2 R U2 F (16f)
  62. F R' F' R U2 R U' R2 F R F2 U F (13f)
  63. R' F2 U2 R' F2 R U2 F2 U F' U2 F U' R U' (15f)
  1. U F U R' U' R' U R2 U' R2 F' U2 R' U2 R (15f)
  2. U' R U2 R U2 F R2 F' U2 R2 U' R U R' U (15f)
  3. U R' F' U' F R U R U2 F R' F' U2 R2 U2 R (16f)
  4. U' R' F R U2 F' U R U R' F' U2 F2 U' F' U2 (16f)
  5. U2 F' U2 F R2 F' U F2 R' F2 U' F R' U' R' (15f)
  6. U R' U' F U R U' R' F2 U F U' F' U' F R U' (17f)
  7. U2 R U2 R2 F R F2 U' F U R U2 R' F' U' F (16f)
  8. F' U2 F R' F R F' U2 F R' F' R U' (13f)
  9. F R' F' U F R U2 F' U' R2 F R2 U' R' F' R (16f)
  10. U2 F' U2 F2 U2 F' U' F' R' F2 R F2 U' F' U2 (15f)
  11. F' U F R' F R F' R U' R' U2 F' U2 F U (15f)
  12. U R F U' R' U' R U2 F' R2 F R U' R' F' R (16f)
  13. R U R' U2 F' U F R U' R' U R U' R' U2 (15f)
  14. R' U' R' U' F U R2 U2 R' U R U R' F2 U F R (17f)
  15. R' U F2 R2 U' F2 U F2 R' F2 R2 U2 R' U2 R2 F2 (16f)
  16. R' U' R2 F' U2 F U2 F R2 F' U' F' U' F R (15f)
  17. U2 F U R' U' R F' R' U R U2 (11f)
  18. U R' U R' F R' F2 R' U' R F2 R F' R U' R U' (17f)
  19. R' F R U R' U' F' R U' R' U2 R U' (13f)
  20. R' F R2 U R' F2 U F2 U2 F' U2 R U' R' (14f)
  21. R U R' U R U' R2 F R2 U R' U' F2 U2 F U (16f)
  22. U2 F R U2 R2 U' F' U F R U2 F' (12f)
  23. F U F' R' F U' F' U F' U' F U R U (14f)
  24. F U' F' U R' F U F U F' U' F2 R U2 (14f)
  25. U F U2 F2 U' F2 U' F' R' F R F2 U2 F (14f)
  26. F R F' U2 F R' F2 U2 F2 R F' R' U F' U F (16f)
  27. R U2 R' U2 F' U' F R2 U F' U R' U' R F U' R2 (17f)
  28. U' F U2 F2 U' F2 U' F' U2 R U2 R' U' F' U' F (16f)
  29. U' F' R2 F2 R U' R' U' R U' R' U2 F2 R2 F (15f)
  30. U F' U2 F U F R U R' F2 U F2 U2 F' U' (15f)
  31. U2 F U R' F R F2 U F U' R U' R' F' U' (15f)
  32. U F R' F R F2 U F R U R' F' U2 (13f)
  33. R' F R F2 U2 F2 R F' U R' U R F' U2 F R' (16f)
  1. F' U2 F' U R' F U' F2 R U' R' U2 F2 R F2 U2 (16f)
  2. U R2 F R F' R U R' F' U' F U' R (13f)
  3. U F2 R F' U F R2 F' R2 F' U2 F U F R' F2 (16f)
  4. U F U R' F2 R F2 R F2 R2 F' R2 F' R' U' F' (16f)
  5. U F U2 F' R' U' F U R U2 F' U' F' U F U (16f)
  6. U' R U F R' U' R2 F R' F' R' U F' U2 R' U2 (16f)
  7. U R' U2 R2 U2 F R F' U2 R' U' R' F' U F R (16f)
  8. U2 R U R2 F R F R U2 R' U F U R' F2 R (16f)
  9. F U' R' U' R U F' U2 R' U R U' R' U2 R (15f)
  10. U2 R U' R2 F2 U' R2 U' R2 U F2 U R2 U R' U2 (16f)
  11. U2 R' F R2 U R' F' R' F R2 U' R2 F' R U (15f)
  12. U' R' F' U F2 R2 F' U2 F' U2 F R2 U R (14f)
  13. U F U R U' R' F' R' U2 R U R' U R (14f)
  14. U F2 R' U' F R F2 R' U F R F2 R' F R (15f)
  15. R U2 R' F2 R U2 R' U F' U' R U2 R' F U F (16f)
  16. U2 R2 F U F U F' U' F' R2 U' R' F' R U (15f)
  17. U' R2 F2 R' U2 R' U' R U' R F2 R2 U (13f)
  18. U F' U' F' U' R' F' R U2 F U R' F R (14f)
  19. U2 F R U' R2 F R F' U2 F' U2 F R U2 R' U' F' (17f)
  20. R' F' R U' F' U' F U' R U' R' U' F U (14f)
  21. F U R U R' U' F' U' R' U' F U F' R U2 (15f)
  22. U' F R U R2 F R F' R' F R F2 U F U2 F' (16f)
  23. R' U2 R2 U R2 U R U' F' U F R U' R' U (15f)
  24. U2 R U' R' U2 R F U R2 F R2 F' U' F' R' (15f)
  25. U R U F2 R' F2 U2 F U' R F' U' F U F' U' R' (17f)
  26. U2 F U R U' R' F' U R' U' F' U F R U (15f)
  27. U2 F U F2 U2 F R U' R' U' F U2 R' F' R U2 (16f)
  28. F' U R' F R U2 F' U2 F2 U F' U' F U2 F U F (17f)
  29. U' R F' R' F2 U2 F2 R F R2 F2 U2 F U2 F2 R (16f)
  30. U F' U2 F R' F R U F2 U F2 U2 F' U (14f)
  31. R F U' R2 U2 R U2 R U2 R U' F' R' (13f)
  32. U' F U2 F2 U2 F' U F' U' F2 U2 F U2 R' F' R (16f)
  33. U2 R F R' F' R U R' F' U' F R2 U' R2 U' (15f)
  1. R U2 R2 F R2 U R' U' F2 U2 F R' F R F' (15f)
  2. U' F R F' U2 F U R U' R' F' U F R' F' (15f)
  3. U' F2 R' F' U R2 U R U' R U R U2 F R F2 (16f)
  4. U2 R U2 R' F R' F' R2 U' R2 F R F2 U F U' (16f)
  5. U' F U R U' R' U2 F' R F R' U2 R F' R' (15f)
  6. R2 U' R' F R F2 U F2 R F' R U' (12f)
  7. U' F U R' U F2 U' F2 U' R F2 U2 F U' (14f)
  8. U2 R' U2 R U F R' F' U F R F' U2 (13f)
  9. R U' R' F R' F R2 F U F' R2 F2 R U (14f)
  10. U' R' F2 R2 U' F U' F' U2 R2 F2 R U (13f)
  11. U2 F' U2 R' U F' R' F' R F2 U' F' R U F (15f)
  12. R' U2 R2 U R F' U' F U R' U2 R' U R' U2 R (16f)
  13. U2 F' U2 F U' F U R' U' F' U R F' U2 F (15f)
  14. U2 R2 U2 R' U R U' R U2 R F' U2 F R (14f)
  15. U2 R' F2 U' R F R' U F2 R U F' U' F (14f)
  16. R U2 R' U R U' R2 U2 R2 U F R2 F' R2 U R (16f)
  17. F U R U F R' F' U F R F' U R' U F' (15f)
  18. U' F' U' F R U2 F' U' F U F R' F2 U F (15f)
  19. U' F2 R2 F2 U F U F R2 F2 U' F R2 F' U (15f)
For what it's worth, the program that generated these does not care about the average number of moves needed, only the number of algorithms, so these are probably not optimal for designing a practical solution system. Still, if anyone comes up with anything new using these, I'll be interested to see it.


28 or less if you count inverses and stuff with ZZ!
I'm not sure as to your meaning when you say "inverses and stuff". By "stuff", do you mean mirrors? If you are saying that ZZ can get from any LL state to any other in 2 looks using a set of 28 algorithms and their mirrors/inverses/mirror-inverses, then I suppose my reply would be that 19 is smaller than 28. If you meant something else, please clarify.
 
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Kirjava

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I do not personally know how one would apply them intelligently, but the computer would know just by being able to remember every possible scenario at once. Perhaps a set of guidelines or rules could be developed, but I have not tried to do so. I've just thrown processing power and tricky programming at it to try to answer the question of what's possible.

This is what I was attempting to do with my LL method. I have ideas and ways of getting it done but the sheer number of cases requires a huge time investment to finish it.

It's a shame that those 19 cases are not very nice to execute. Would you be able to generate other groups? Would you be able to force certain algs to appear in the lists?
 

Tempus

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It's a shame that those 19 cases are not very nice to execute. Would you be able to generate other groups? Would you be able to force certain algs to appear in the lists?
With some small changes to the code, that should be possible, but the more you micromanage the computer's decisions, the more likely it becomes that it will require more than 19 algorithms. I'll try to work up a version that allows the input of a list of forced algorithms, such that the machine will only try to decide the remaining ones.

Meanwhile, you can figure out a list of algorithms you'd like to force, and which one of the various questions you want them forced into. In other words, are we talking about 2LLL or 3LLL? Are we counting all algorithms as distinct, or are we allowing mirrors, inverses, or both? Please be precise; I hate ambiguity.

EDIT: On second thought, the 19 you mentioned implies that you meant 2LLL and that you meant to include both mirrors and inverses.

EDIT #2: The new forced-algorithm version of the program is written, and I am currently testing it.

EDIT #3: As a test, I forced it to include these two algorithms:
  1. F R U R' U' F' (6f)
  2. R U R' U R U2 R' (7f)
It succeeded in finding an appropriate set of 19 that includes the two algorithms that I forced on it:
  1. F R U R' U' F' (6f)
  2. R U R' U R U2 R' (7f)
  3. U F R' F' R F R' F' R F' U' F U F' U' F (16f)
  4. U F R' F' U2 R' U2 F' U F R U F R2 F' U (16f)
  5. F R U R' U2 R' U R U F' U R' U R U' (15f)
  6. R' F R' U' R2 U' R2 U F2 U F2 R F' R U2 (15f)
  7. U' R U R2 F' U F U2 F' U' F R2 U2 R' (14f)
  8. R' U F' R U R2 F R2 U2 R' U' F2 U2 F2 R U2 (16f)
  9. U F2 U R' F' R U' F' U' F U F U' F2 U' (15f)
  10. F U R U' R' F' R' F' U' F U F' U' F U R (16f)
  11. U2 R' U2 F2 U2 F2 U2 F U2 F' U2 F U2 F2 R (15f)
  12. R2 U R' U2 F' U2 R' U R F2 R2 F' R' U2 R2 (15f)
  13. U' R U' F U' R F R F' U' R2 U R F' U2 R' (16f)
  14. U F' U F U R U2 R' U2 F' U F U' F' U2 F (16f)
  15. U R' U F' R' U2 R U2 F U' R' U R2 U2 (14f)
  16. U R U R2 F R F2 U' F U2 F R' F' R U2 (15f)
  17. U F U R2 U2 R F' R' U2 R F R U' F' (14f)
  18. U R' U F U' F R F2 U2 F' R' U F2 R F U2 (16f)
  19. F U R U' R2 F R F2 U F R U R' F' U2 (15f)
 
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Kirjava

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With some small changes to the code, that should be possible, but the more you micromanage the computer's decisions, the more likely it becomes that it will require more than 19 algorithms. I'll try to work up a version that allows the input of a list of forced algorithms, such that the machine will only try to decide the remaining ones.

More than 19 is not a problem at all. When I tried this I used about 30. We want to minimise the amount of 'extra' algs as much as possible (they don't look very good).

EDIT: On second thought, the 19 you mentioned implies that you meant 2LLL and that you meant to include both mirrors and inverses.

Yep.

How about;

RUR'URU2R'
RUR'URU'R'URU2R'
rUR'URU2r'
RUR2FRF2UF
R'U2R2UR2URU'RU'R'
FR'F'RU2RU2R'
FRUR'U'F'
FRUR'U'RUR'U'F'
RU'L'UR'U'L
RB'RF2R'BRF2R2
M2U'MU2M'U'M2
M2UM2U2M2UM2
RUR'U'R'FRF'
rUR'U'r'FRF'
R'FRUR'U'F'UR
R'U'R'FRF'UR
F'LFL'U'L'UL
RUR'U'M'URU'r'
M'UMU2M'UM
RU2R2'U'R2U'R2'U2'R
 

AvGalen

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I like this idea a lot.
* having a small list of, lets say, 15 of your favorite algs
* to which you only have to add, lets say 5-10 a few other algs
* to always have a 2 look last layer.
I think it would be doable to learn all combinations of those algs actually. The goal of course is
* to have the 15 algs be perfect, even when mirrored/inversed
* to have the 5-10 extra algs be really good as well (and closer to 5 than 10). Not one stinker

The problem would be of course the cases where you already know a 1 look alg that is bad (Let's say N-Perm for example). With this new system it might become a 2 look that is still slower than the 1 look bad alg that you already know
 

Carrot

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The problem would be of course the cases where you already know a 1 look alg that is bad (Let's say N-Perm for example). With this new system it might become a 2 look that is still slower than the 1 look bad alg that you already know

why would you unlearn 1-look cases you already know? that sounds like a stupid approach to me.
 
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