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 Post subject: How hard/difficult is Puzzle X?
PostPosted: Thu Oct 14, 2010 7:45 am 
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This thread brought up a topic I've thought of before. How can one quantify the difficulty of a twisty puzzle?

The number of states isn't the best number and the God's number of a puzzle (even if it can be calculated) isn't either. The 3x3x3 has a God's number of 20. Yet a 1x2x40 also has a God's number of 20 and is far "easier".

So if you have 2 puzzles X and Y. What is the best way to determine which is more difficult?

What about the product of the number of states and the God's number?

Curious if there are any good thoughts on this topic,
Carl

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Thu Oct 14, 2010 8:51 am 
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That would make the Megaminx much more difficult than the 3x3, but it's not.

The Bandaged Cube has far fewer states than most puzzles, but it's very difficult.

I think it's more a matter of what sorts of commutators are available.


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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Thu Oct 14, 2010 9:01 am 
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I'm not sure how to combine/weight these factors in a consistent formula, but here are some important factors that I think should be included:

1. No. of states (permutations)
2. Min. moves (God's number)
3. No. of potential positions per move (branching factor)
4. % of elements displaced per move
5. % of disallowed vs potential moves (degree of bandaging)
6. Pattern recognition

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Thu Oct 14, 2010 9:27 am 
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I think difficulty should be defined by average first-solve time among a big number of solvers.

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Thu Oct 14, 2010 9:27 am 
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Jared wrote:
That would make the Megaminx much more difficult than the 3x3, but it's not.

The Bandaged Cube has far fewer states than most puzzles, but it's very difficult.

I think it's more a matter of what sorts of commutators are available.


This sounds very reasonable. Anyone feel like looking at a few cases of how large G/[G,G] is for various puzzle groups G? (This is the abelianization, and roughly measures how non-commutative the group is)


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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Thu Oct 14, 2010 9:36 am 
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Jared wrote:
I think it's more a matter of what sorts of commutators are available.


Well said!!! This is an *excellent* criterion, which will suit both analytic *and* intuitive solvers.

The length of the *different* required algorithms is exactly what really separates a solved puzzle
and a scrambled puzzle. Therefore, the only challenge here is to find the minimum (optimal) algorithm
lengths for each puzzle. Shouldn't be very hard, should it?

Here is a break-down example. As known, the Rubik's cube needs four algorithms to completely
solve it (I am not talking about speed-solving here, as we are trying to succeed-solve!)
say, the lengths are r1, r2, r3, and r4.

Then we may use a number based on them, e.g. their sum, their product, or the square root of
the sum of their squares (my favourite! LOL), to determine the "difficulty number".

Things get interesting, as a 3x3x3 supercube for example, will require two more algorithms.
In any case, it seems that this commutator/algorithm-method is the correct way to go.

:D


Pantazis


PS. And my, this statement (dividing a puzzle into its basic piece-types and how they are connected
with the appropriate minimal algorithms), seems to be enough to even *classify* all twisty puzzles!

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Thu Oct 14, 2010 9:53 am 
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Jared wrote:
That would make the Megaminx much more difficult than the 3x3, but it's not.
Aggreed. I don't have the answer... just the question.
Kelvin Stott wrote:
I'm not sure how to combine/weight these factors in a consistent formula, but here are some important factors that I think should be included:

1. No. of states (permutations)
2. Min. moves (God's number)
3. No. of potential positions per move (branching factor)
4. % of elements displaced per move
5. % of disallowed vs potential moves (degree of bandaging)
6. Pattern recognition
I'm not sure how Pattern recognition is a factor. A measure of the quality of the solver maybe but the puzzle?

And a factor I'm inclined to add to this list would be the number of axis of rotation the puzzle has. Though that may already come into play in the branching factor. There is also the order (or number of cut planes per axis of rotation) that could play a role.
Iranon wrote:
This sounds very reasonable. Anyone feel like looking at a few cases of how large G/[G,G] is for various puzzle groups G? (This is the abelianization, and roughly measures how non-commutative the group is)
Looks like I need to brush up on my group theory.

http://mathworld.wolfram.com/Abelianization.html

So G/[G/G] is a group. Are you saying the number of elements in this group could directly relate to a puzzles difficulty? Interesting... So what does this set look like for a 3x3x3?

Carl

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Last edited by wwwmwww on Thu Oct 14, 2010 10:15 am, edited 2 times in total.

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Thu Oct 14, 2010 10:13 am 
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kastellorizo wrote:
Here is a break-down example. As known, the Rubik's cube needs four algorithms to completely
solve it (I am not talking about speed-solving here, as we are trying to succeed-solve!)
say, the lengths are r1, r2, r3, and r4.

Then we may use a number based on them, e.g. their sum, their product, or the square root of
the sum of their squares (my favourite! LOL), to determine the "difficulty number".

Things get interesting, as a 3x3x3 supercube for example, will require two more algorithms.
In any case, it seems that this commutator/algorithm-method is the correct way to go.

PS. And my, this statement (dividing a puzzle into its basic piece-types and how they are connected
with the appropriate minimal algorithms), seems to be enough to even *classify* all twisty puzzles!
Nice... this I could follow without knowing what G/[G,G] is. Still curious about the G/[G,G] method but this makes alot of sense.

As for the difficulty rating I think I'd be inclined to define it as follows:

a = number of algorithms needed to solve.
b = the standard deviation of those algorithms' lengths.

Difficulty = a*b

Just realized I think this just differs from Pantazis' "the square root of the sum of their squares" by the square root of a. And his number is small then mine so I think its just weighting the number of algorithms needed differently. As such his is probably better then mine.

Carl

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Last edited by wwwmwww on Thu Oct 14, 2010 10:17 am, edited 2 times in total.

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Thu Oct 14, 2010 10:15 am 
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Quote:
G/[G/G]


I am not sure this is a good criterion. I mean, puzzles like the Rubik's Clock, the Lights Out
or even the Brain-Chek, are abelian, *and* none-trivial.

And maybe the above are not clear-cut twisty puzzles, but I suspect that easier/surface
puzzles like the Dogic (which requires only two algorithms) may give a bigger quotient group.
In any case, such software to test those is readily available (e.g. GAP).

:)


Pantazis


PS. By the way Carl, I am glad you re-started this topic, and then Jared replied. I believe
we are on something good this time. :)

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Thu Oct 14, 2010 10:22 am 
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Pattern recognition I believe is about how hard it is to know which piece goes where. This is needed because otherwise the skewb would be the same difficulty as the golden cube.

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Thu Oct 14, 2010 10:30 am 
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I don't have enough time for a full response but I'll join the discussion by saying I have a lot of thoughts on what makes a puzzle hard.

We have already discussed this topic a bit on the Gelatinbrain solving thread.

See Julian's commutator measurement: http://twistypuzzles.com/forum/viewtopic.php?p=225288#p225288
I responded with a few ideas here: http://twistypuzzles.com/forum/viewtopic.php?p=225707#p225707.

Elwyn and Julian jumped in the conversation and there were a couple posts after those two about the subject.

In short, I think the minimum length of available commutators is a good measure but those lengths are influenced by factors such as depth of cuts, "extra room" available to move pieces for commutation, etc.

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Last edited by Brandon Enright on Fri Oct 15, 2010 10:34 am, edited 1 time in total.

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Thu Oct 14, 2010 10:58 am 
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bmenrigh wrote:
In short, I think the minimum length of available commutators is a good measure but those lengths are influenced by factors such as depth of cuts, "extra room" available to move pieces for commutation, etc.


I see, very interesting. :)

Still, the method seems to be really good. We first need to work out this concept with puzzles
which are not bandaged and/or not jumbable, and then we may proceed to the rest of the cases.
Something like defining the groups before the groupoids (the opposite way won't work!).

The length of the commutators based on the available generators is like counting steps
when going from one place to another. Why does depth of the cut has to come into the picture?

It doesn't matter which path you choose, as discovering an algorithm of length r is independent of
the structure of the puzzle. Simply put, we are deliberately avoiding the mechanism attributes and
we directly dive into the algebra. If there are more "non-trivial paths" (which arguably can make things
more difficult), then those will be already covered by the extraction of the rest of the algorithms.

:)


Pantazis

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Thu Oct 14, 2010 12:23 pm 
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I came up with a pretty nice formula, but I’m not sure how it would be applied to bandaging/jumbling puzzles. Here’s the formula:

# of Permutations / God’s Number / Pieces with multiple permutations / % Pieces moved per turn

Example: 43,252,003,274,489,856,000/20*20/40

Then, put this value into scientific notation.

4.3252003274489856 x 10^19 (simplified to 4.3)

Then, use the exponent as the difficulty level, followed by a decimal point and the first place of your number (rounded).

So, a 3x3 = 19.4

A 2x2x1 = -2.1 (Note zero is not the lower bound for difficulty.)

A 3x3x1 = -2.8

A Megaminx = 63.3*

*using an approximation for God’s Number.

I worked on this all study, does it make any sense to anyone? :lol:

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Thu Oct 14, 2010 12:28 pm 
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Maybe you should use the logarithm of the number of positions, instead of the number itself. If you just use the number, then the size of the number of positions will matter a LOT more than any of the other variables. For instance, your formula will say that the 5x5 is significantly harder than the Pentultimate.

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Thu Oct 14, 2010 3:36 pm 
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The only really meaningful measure of difficulty has to do with how long it takes people to solve it. That turns out to not be a simple measure though. Some puzzles take everyone about the same amount of time, some take some people a lot longer, some seem to randomly take some people a huge amount of time with very little correlation to how long those people take at other puzzles. Some are much easier if you've solved some others beforehand, etc. etc. etc.

Total number of states is NOT a measure of difficulty, it doesn't even vaguely correlate. The things which correlate: Number of different kinds of pieces and whether they have orientations. Whether there's any bandaging. Minimum number of pieces in the overlap between to slices.


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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Thu Oct 14, 2010 4:53 pm 
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Bram wrote:
The only really meaningful measure of difficulty has to do with how long it takes people to solve it. That turns out to not be a simple measure though.

Not a factor in my opinion. THe reason I have so much trouble breaking one hour on the Teraminx is because it has low tolerance to misalignment. I could probably do it in half that time if I didn't spend longer aligning than turning. But on the Gigaminx (same difficulty I found) I can easily get 20 minutes and on the 7x7x7 (similar, simplified from the Teraminx) I can solve 7:30 every time, with pops.

I believe Kapusta has the right approach in terms of states- only the exponent really matters. But as Kelvin Scott suggested,
Kelvin Scott wrote:
3. No. of potential positions per move (branching factor)
4. % of elements displaced per move
5. % of disallowed vs potential moves (degree of bandaging)
6. Pattern recognition

So how about this formula:

% Displaced per move * (Kapusta's Formula + Degree of Bandaging)

It makes sense here to distribute displacements per move because it affects the number of positions as well as the degree of bandaging significantly.


Before I go ahead and give an example, are the centers of a solid-color 3x3x3 considered displaced in a turn?

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Thu Oct 14, 2010 5:47 pm 
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Rentlix wrote:
Bram wrote:
The only really meaningful measure of difficulty has to do with how long it takes people to solve it.

Not a factor in my opinion. THe reason I have so much trouble breaking one hour on the Teraminx is because it has low tolerance to misalignment.


By 'solve' I meant in the figure-out-a-solution sense, not in the run-through-a-solution sense.


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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Thu Oct 14, 2010 7:01 pm 
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I needed that clarification. In that case, how does the Latch cube fit into this question?

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Thu Oct 14, 2010 7:23 pm 
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Rentlix wrote:
how does the Latch cube fit into this question?


The Latch Cube is a bandaged puzzle.


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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Thu Oct 14, 2010 8:06 pm 
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(average pieces moved in rotation x possible permutations)/(puzzle constant x solved states x average rotating axis) = puzzle rating

Puzzle constant is determined by how different one piece is from another. Puzzle constant is 0 if all pieces are 100% identical and gets larger with more unique pieces. This would make the icon cube harder than a normal rubik's cube.

For example, a 3x3x3 would be (9 x 4.3 E19) / (.33 x 1 x 6) =

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Thu Oct 14, 2010 9:48 pm 
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Bram wrote:
By 'solve' I meant in the figure-out-a-solution sense, not in the run-through-a-solution sense.
Well again that could go back to finding the number of algorithms needed to solve the puzzle and then finding those algorithms. With those the puzzle could be solved from any state and we aren't talking about the time needed to apply those algorithms or how many times you'd need to apply them.

So if we want to talk about the time needed to find those algorithms is there a general method for finding them that we could apply to any puzzle? So you could write code to apply to any twisty puzzle and it would follow a fixed set of rules to tell you the number of algorithms needed and then supply those algorithms.

I would expect the time such code would require to "solve" the puzzle would be a function of the number possible permutations AND the lengh of those algorithms. I.e. the bigger the puzzle space and the longer the algorithms the more of the puzzle space that must be explored.

So is such a set of "rules" available?

Pantazis says the 3x3x3 needs 4 algorithms and the 3x3x3 super cube needs 6. Can someone tell me how these numbers are obtained? And what are the minimal lengths of these algorithms?

Using the info here:
http://www.research.att.com/~njas/sequences/A080601

Let's say the longest algorithm is 8 moves. Then we at most need to explore 1332343288/43252003274489856000 or only 0.00000000308041983522604% of the puzzle space needs to be explored to find a solution.
In this case I think the number directly related to the difficulty of the puzzle is 1332343288.

Now let's say the longest algorithm is 8 moves but that there are 10,000 such 8 move algorthms that can accomplish this task. We only need to find one to "solve" the puzzle so could we say:

Difficulty = 1332343288/10000 or 133234.

In the process of finding the longest such algorthm this "code" should have also found all the needed shorter algorthms.

Sound fair? What is the optimal set of rules such "code" should follow?

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Fri Oct 15, 2010 12:04 am 
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wwwmwww wrote:
Pantazis says the 3x3x3 needs 4 algorithms and the 3x3x3 super cube needs 6. Can someone tell me how these numbers are obtained? And what are the minimal lengths of these algorithms?


Ok, here is an example of how I would do a difficulty calculation:


Step 1.
Identify the different types of pieces of a symmetric puzzle
(I am considering symmetric puzzles which are unbandaged, as we need
to start from the simplest cases before extending to the rest of the variations).

Step 2.
For each type of piece, identify the types of movement needed
(which involves the minimum possible number of such pieces), such that
we can exchange them, and leaving the rest of the pieces invariant.
(.e. interchange types of movement, orientation types of movement, etc.
(I know that sometimes we cannot exchange a minimum of two pieces, but
that number can also be three, four etc, as long as it corresponds to the minimum
possible *same-type pieces* that can be "swapped around").

Step 3. Find the minimum possible paths of those *independent* to each other
algorithms, and then use a proposed formula to combine them meaningfully.
(square root of sum of squares?)

:)


Pantazis


PS. Example for the Rubik's Cube: We have corners and edges. So finding the shortest
four algorithms for swapping around and orienting the minimum number of corners,
as well as swapping around and orienting the minimum number of edges is enough.
For the supercube, the centers should be included. And this case, we have two orientation
cases independent of each other: a simultaneous 90 degree rotation of two centres, and
a 180 degree rotation of one center (both are required).

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Last edited by kastellorizo on Fri Oct 15, 2010 12:26 am, edited 1 time in total.

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Fri Oct 15, 2010 12:19 am 
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kastellorizo wrote:
As known, the Rubik's cube needs four algorithms to completely solve it

Ummmm no it doesn't :?

I have a method that uses only 2 (a T-perm and a modified V-perm), i take it you are using one each for both orienting and permuting edges and corners. That's not necesary as orienting can be done simultaniously with permuting if you use different setup moves hence eliminating the need for two algorithms.

That brings me to a factor i consider makes a puzzle hard which is the average number of setup moves needed to position a piece for cycling which fits in nicely with the number/length/"findability" of commutators needed. The problem is finding set up moves isn't really part of the "finding a solution" step but i still feel it is a contributing factor to the intrinsic difficulty of a puzzle.

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Fri Oct 15, 2010 12:33 am 
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Elwyn wrote:
kastellorizo wrote:
As known, the Rubik's cube needs four algorithms to completely solve it

Ummmm no it doesn't :?

I have a method that uses only 2 (a T-perm and a modified V-perm), i take it you are using one each for both orienting and permuting edges and corners. That's not necesary as orienting can be done simultaniously with permuting if you use different setup moves hence eliminating the need for two algorithms.

That brings me to a factor i consider makes a puzzle hard which is the average number of setup moves needed to position a piece for cycling which fits in nicely with the number/length/"findability" of commutators needed. The problem is finding set up moves isn't really part of the "finding a solution" step but i still feel it is a contributing factor to the intrinsic difficulty of a puzzle.


As I stated before, we are not speedsolving here, but succeedsolving. We need to split the puzzle
into its basic components. Let me put it in another way. When the first Rubik's cube came out, no one knew
(how could they?) any efficient algorithms to solve many things at once. They just needed a way to solve it,
just like anyone who sees it the first time.

And finding methods to swap and orient those basic components, constitute the intuitive way of someone
trying to solve a new puzzle, i.e. the difficulty he will encounter, as longer and different algorithms for the
basic components will make things more difficult.

:)


Pantazis


PS. Allow me to add here, that combining two algorithms, *should give* some similar indication number
when using the square root of sum of squares. Doesn't this sound fantastic?

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Fri Oct 15, 2010 1:08 am 
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kastellorizo wrote:
we are not speedsolving here, but succeedsolving.
How on earth is a method with only 2 algorithms going to be speed solving? I'd say it is the most basic method you could think of is it not? obviously the two algorithms i stated are drawn from speed solving methods however the basic idea is you only need 2 algorithms, they do not need to be for specific cases those are just the two i used, a simple (3,1) commutator R U R' D R U' R' D' could be used for corners and so on.
kastellorizo wrote:
When the first Rubik's cube came out, no one knew
(how could they?) any efficient algorithms to solve many things at once.
and hence they used commutators and as few of them as possible... which is 2, why bother looking for more algorithms when they are all you need? Not exactly sure what you mean by succeed solving and that could be where we are misunderstanding each other but i thought you mean the simplest method and i'd say that would be the one with the least algorithms to find/learn.

On a different note i'm surprised no one has mentioned (perhaps they have and i missed it) that paritys such as the single twisted corner parity on 1.2.9 and the icosimate and so on add to the difficulty.

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Fri Oct 15, 2010 1:36 am 
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Elwyn wrote:
kastellorizo wrote:
we are not speedsolving here, but succeedsolving.
How on earth is a method with only 2 algorithms going to be speed solving? I'd say it is the most basic method you could think of is it not?


Thank God, we can continue from here. I am talking about the *simplest* way of *breaking down*
a solution, not the most efficient. I have said this at least three times already...

Succeed-solving is the entire method I have also described above...



Elwyn wrote:
obviously the two algorithms i stated are drawn from speed solving methods


Yeap. And they are unecessary when breaking a puzzle into basic components,
because we are trying to start from the very beginning, not go up and then come back...


Elwyn wrote:
however the basic idea is you only need 2 algorithms


No you don't if you use a breakdown. You have two types of pieces in the 3x3x3.
Unless you want to consider an edge and a corner as being "one piece".
But that is not true!


Elwyn wrote:
and hence they used commutators and as few of them as possible... which is 2, why bother looking for more algorithms when they are all you need?


Wrong. I am not looking for "more" algorithms, but the *basic* ones. So basic, that each one focuses on only
one type of piece, and then I am trying to find the most efficient sequence for each one type of piece.
You insist in using some which are efficient, but *not* basic ones. Remember, my steps first identify unique pieces
of a puzzle, then the algorithms for them. Yours are focusing on many pieces at the same time. By using your way,
how would you compare difficulty with other puzzles? Will you introduce a "combination piece" notion? I am curious...
Your way will become overcomplicated in no time when going to the 4x4x4 and 5x5x5.


Elwyn wrote:
i thought you mean the simplest method and i'd say that would be the one with the least algorithms to find/learn.


I repeat, simplest (in terms of analysing a puzzle into its basic components) and efficient (in terms of combining
many different types of pieces together) are two different things. Very different. Your algorithms are the most
efficient to speedsolve (or since you don't want me to use the term "speedsolve", I would say "fast enough solve"),
but they are not the simplest ones (in a fundamental way) to solve.

In fact, anyone who had first solved the cube (I am not talking about anyone specific here, just in general)
based on another person's solution instead of their own, will never be able to understand what I am talking about.
And obviously, the difficulty of a puzzle does not depend at all on reading the solution made by others.


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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Fri Oct 15, 2010 3:00 am 
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kastellorizo wrote:
As I stated before, we are not speedsolving here, but succeedsolving. We need to split the puzzle
into its basic components. Let me put it in another way. When the first Rubik's cube came out, no one knew
(how could they?) any efficient algorithms to solve many things at once. They just needed a way to solve it,
just like anyone who sees it the first time.

And finding methods to swap and orient those basic components, constitute the intuitive way of someone
trying to solve a new puzzle, i.e. the difficulty he will encounter, as longer and different algorithms for the
basic components will make things more difficult.
Hi Pantazis, my apologies for saying so but I think all your experience with the Rubik's cube is blinding you from from seeing it in simplistic terms. I agree with everything I have quoted above but what you're saying in the quote doesn't match up well with your proposed 4-algorithm solving order. Specifically, you propose:
kastellorizo wrote:
Example for the Rubik's Cube: We have corners and edges. So finding the shortest four algorithms for swapping around and orienting the minimum number of corners, as well as swapping around and orienting the minimum number of edges is enough.

I agree that developing an intuitive method means treating the corners and edges as separate, independent pieces That means developing sequences that can cycle corners or cycle edges without moving the other. As you know, the shortest pure cycle that can be found is a 3-cycle, and the easiest way to find cycles is via commutation. There is a certain intuitiveness to commutation and a fresh solver who has never twisted a Rubik's cube should already know the basic techniques even if they don't recognize any math or theory behind them. Furthermore, a solver doesn't have to find a short sequence, they have to find a sequence that is easy to understand and apply. That usually involves making use of symmetries.

For example, to 3-cycle corners it's easy to find something like [R',F,R,F'],D,[F,R',F',R],D'. This sequence isn't the shortest but it sure is symmetrical. I argue that it is one of the easiest pure sequences to find and understand on the Rubik's cube.

To apply this or an edge-cycle sequence to place all the pieces on the cube will require another intuitive technique -- conjugation. If you want to move corners other than what your sequence touches, use setup moves, then apply the sequence, and then undo the setup moves. Never mind that conjugation is backed by strong mathematical theorem, on twisty puzzles its an obvious technique that doesn't warrant a fancy name.

My point to all this is that if you've developed two sequences, one for corners and one for edges, your still going to need to apply setup moves to actually place all the pieces. Everyone is going to recognize that depending on the setup moves you use, sometimes the edge or corner will be cycled into place with the correct orientation and sometimes it wont. Orientation is going to seem like a daunting task on its own and most reasonable solvers would choose to handle orientation with the setup moves alone. Why search for and learn sequences to handle the orientation of pieces when you can adjust (even using trial-and-error!) your setup moves until the pieces are cycled into place with the correct orientation.

Even if a solver happened to wind up with the last few pieces in their correct position but wrong orientation, I doubt they would go searching for orientation-only sequences. It is much more likely that they would just cycle the pieces out of place. Then apply setup moves to them so that when they are cycled them back they go in with the correct orientations.

So to summarize this long winded post: I totally agree with Elwyn. Two short algorithms for the Rubik's cube and that's it. Breaking the puzzle into separate positioning and orienting phases is actually more complex and imposes unnecessary complications. Using two sequences is not fast and it won't be move-efficient. It is definitely simple though.

The 1-commutator-sequence-per-piece solving strategy happens to work very well for almost every twisty puzzle. If your goal is simplicity it just doesn't make sense to develop more than one sequence when you can handle all the other sequences using a 3-cycle and a few setup moves.

Edit: My apologies to Elwyn, I spelled his name wrong. It's kinda scary that I without studying each letter in sequence I can't see the difference between "Elwyn" and "Elywn".

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Last edited by Brandon Enright on Fri Oct 15, 2010 10:33 am, edited 1 time in total.

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Fri Oct 15, 2010 4:48 am 
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bmenrigh wrote:
Hi Pantazis, my apologies for saying so but I think all your experience with the Rubik's cube is blinding you from from seeing it in simplistic terms.


Hi, but I think it is the other way around. And I will explain to you why. Most of the puzzlers here focus
too much on the 3x3x3, without considering how many other puzzles should be solved.
And I am not talking only about twisty puzzles here, but mathematical in general.


bmenrigh wrote:
I agree with everything I have quoted above but what you're saying in the quote doesn't match up well with your proposed 4-algorithm solving order.
My point to all this is that if you've developed two sequences, one for corners and one for edges, your still going to need to apply setup moves to actually place all the pieces.


Of course, we agree here. I am not talking about directly solving the cube,
but having the right tools (for each type of piece) to eventually solve the puzzle.


bmenrigh wrote:
Everyone is going to recognize that depending on the setup moves you use, sometimes the edge or corner will be cycled into place with the correct orientation and sometimes it wont.


If you ask many people who solve the cube on their own, they will all tell you
that those were the steps they took (i.e. the four moves). But as time went by,
people improved them more and more, creating many nice sequences.


bmenrigh wrote:
Orientation is going to seem like a daunting task on its own and most reasonable solvers would choose to handle orientation with the setup moves alone. Why search for and learn sequences to handle the orientation of pieces when you can adjust (even using trial-and-error!) your setup moves until the pieces are cycled into place with the correct orientation.


Yes, it can be a little bit more tedious, but daunting? In fact, your method will
become much more tedious for larger puzzles, while mine will remain simple enough.

Again, it is not about the efficiency, but breaking it into the simplest components.
We are not interested in solving it fast in terms of moves, but fast in terms of using
the most elemental sequences for each different piece separately.

If you want to propose something new, ok, but I am trying hard (as you can see)
to explain some conditions which will allow to tackle the difficulty task efficiently.

Even the trivial tips of the Pyraminx add something small, and this needs to be
counted (no orientation but there is exchange). Similarly for the centers of the 3x3x3
supercube (there is no exchange, but there is orientation).


Therefore, we *need* to differentiate swapping colors or pieces from their orientation,
and I hope you can see now my point more clearly.


bmenrigh wrote:
Even if a solver happened to wind up with the last few pieces in their correct position but wrong orientation, I doubt they would go searching for orientation-only sequences. It is much more likely that they would just cycle the pieces out of place. Then apply setup moves to them so that when they are cycled them back they go in with the correct orientations.


Again, we are *not* trying to simply solve a puzzle. We are trying to determine a nice
way and to give the required tools to solve *and* to calculate the difficulty of the puzzle.
A solver who wants to find the secrets of a puzzle on their own (like I have) will definitely go
and find those orientation sequences. Someone who only wants to solve, will not. Plain and simple.


bmenrigh wrote:
So to summarize this long winded post: I totally agree with Elywn. Two short algorithms for the Rubik's cube and that's it. Breaking the puzzle into separate positioning and orienting phases is actually more complex and imposes unnecessary complications. Using two sequences is not fast and it won't be move-efficient. It is definitely simple though.


I disagree as stated above. And I would love to see how you would do this for other larger puzzles.
The 3x3x3 and its variants are *not* the only puzzles around. From what I can see, this thinking
will create some sort of "levels" for the algorithms (e.g. should we use one which can orient a piece A
and another sequence which can orient and swap a piece B, or should we use a combination which can do both?
- I will not go into this thinking, as things will get messy very fast and many different types of pieces
can be combined. What then?).


bmenrigh wrote:
The 1-commutator-sequence-per-piece solving strategy happens to work very well for almost every twisty puzzle. If your goal is simplicity it just doesn't make sense to develop more than one sequence when you can handle all the other sequences using a 3-cycle and a few setup moves.


I explained all those above. Clearly, you disagree. But your method does not use elemental
pieces, and when a puzzle becomes complex, I have serious doubts it will work.
And do not underestimate the "set-up" moves. They won't be that simple.


Pantazis

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Last edited by kastellorizo on Fri Oct 15, 2010 6:20 am, edited 1 time in total.

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Fri Oct 15, 2010 4:50 am 
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kastellorizo wrote:
No you don't if you use a breakdown. You have two types of pieces in the 3x3x3.
Unless you want to consider an edge and a corner as being "one piece".
But that is not true!
What? Where on earth did that come from, i'm solving edges then corners using 2 algorithms, one for corners, one for edges... not thinking of them as the same thing in any way shape or form. All i was saying is orienting corners can be done using the same algorithm as permuting them and same for edges.
kastellorizo wrote:
I am talking about the *simplest* way of *breaking down*
a solution
And the simplest method is that with the least steps and algorithms that are the simplest to find. What commutator are you using to permute corners? As both i and Brandon said (his is essentially the same as mine but viewing the cube from a different angle) R U R' D R U' R' D' is probably the most simple to find and understand commutator to find for corners this very simple easy to find commutator can be used for both permuting and orienting which saves you looking for a corner orienting alg and hence makes finding a solution simpler.
kastellorizo wrote:
Wrong. I am not looking for "more" algorithms, but the *basic* ones. So basic, that each one focuses on only
one type of piece
I never ever said i was focusing on more than one piece with either alg. I regret mentioning that i use a T-perm and V-perm because you obviously misunderstood, i solve all edges, not even looking at the corners, with a t perm as it swaps only 2 edges there's no reason the most simple edge cycle couldn't be used i am just used to performing T-perms. I then solve all corners using a 3 cycle.
kastellorizo wrote:
You insist in using some which are efficient, but *not* basic ones.
Elwyn wrote:
they do not need to be for specific cases those are just the two i used, a simple (3,1) commutator R U R' D R U' R' D' could be used for corners and so on.
Define "Insist" for me again... perhaps you missed that sentence.

I see this post might seem a little redundant because Brandon explained it quite well (better than i did) in his post but i didn't like being told i insisted on something when i quite clearly didn't.
bmenrigh wrote:
I totally agree with Elywn.
And yet spelt my name wrong twice in this thread :lol:

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Fri Oct 15, 2010 4:55 am 
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Elwyn wrote:
kastellorizo wrote:
No you don't if you use a breakdown. You have two types of pieces in the 3x3x3.
Unless you want to consider an edge and a corner as being "one piece".
But that is not true!
What? Where on earth did that come from, i'm solving edges then corners using 2 algorithms, one for corners, one for edges... not thinking of them as the same thing in any way shape or form. All i was saying is orienting corners can be done using the same algorithm as permuting them and same for edges


Yes, but we need to differentiate corners from edges.
We need to differentiate swapping from orientation.
(please see my previous comments for the Pyraminx tips and the 3x3x3 centers)

Proposing an efficient algorithm is not the answer here, but an "easy enough" algorithm is.


Elwyn wrote:
And the simplest method is that with the least steps and algorithms that are the simplest to find. What commutator are you using to permute corners? As both i and Brandon said (his is essentially the same as mine but viewing the cube from a different angle) R U R' D R U' R' D' is probably the most simple to find and understand commutator to find for corners this very simple easy to find commutator can be used for both permuting and orienting which saves you looking for a corner orienting alg and hence makes finding a solution simpler.


Please see above. ;)


Elwyn wrote:
I see this post might seem a little redundant because Brandon explained it quite well (better than i did) in his post but i didn't like being told i insisted on something when i quite clearly didn't.


Good, so I won't have to repeat myself and answer again LOL

It is fair enough if you mentioned something which I misunderstood, but you should also
at least try to see how my method works.

:mrgreen:

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Fri Oct 15, 2010 7:19 am 
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kastellorizo wrote:
Yes, but we need to differentiate corners from edges.
I agree totally and never stated i was doing otherwise, you on the other hand stated i was, I think due to a misunderstanding.
kastellorizo wrote:
We need to differentiate swapping from orientation.
This is the only point i was ever disagreeing with and to be fair i was only disagreeing with it being applied to the Rubik's cube and other 3D twisty puzzles though i'll admit i wasn't very specific about that but that is due to this being twistypuzzles.com not puzzles.com. I do kind of see where you are coming from but still think adding steps is making your solution needlessly complex. On a pyraminx crystal would you say the most "basic" method is to find a way to permute them all then find a way to orient them all? It seems rather needlessly complex to me, though i'll agree, on that puzzle knowing how to flip two edges at the end makes the end last few edges easier but that to me is like saying knowing full PLL makes permuting the last layer of a 3x3x3 easier. It can be solved with just 3 cycles and that seems the most "basic" method to me.

I look at orientation as part of permutation, a piece is in the right spot or not and you can use one algorithm to put it there, why look for another. I see it like this because i feel using a short commutator to move a piece to a single position (with orientation) using setup moves is more basic than having an algorithm to permute and an algorithm to orient. Though there is the exception of corner orientation parity if it occurs on a puzzle but surely you agree parity algorithms for such cases fall under a different category to normal orientation algorithms, or do you think if you can turn a single corner that solving the orientation of each corner one at a time is the basic solution no matter how long the algorithm because that kind of goes against the look for simple commutators thing? We really are looking at the term basic in a different way i suppose, you see it as finding and using two easy to use tools is more basic than finding and using one tool requiring slightly more logic.

Find me a twisty puzzle where you cannot use setups and a permutation algorithm to orient pieces and i'll agree solving orientation needs to be separate to solving permutation, but to me it certainly isn't like that on a 3x3x3. This does not include orientation parity but i thought i was worth mentioning Brandon's recent solution to 1.2.2 orientation parity and indeed the way most people "orient" the last corner (the non trivial ones) of a pyraminx such as [X, Y, X', Y] x2 actually can be seen as permuting three edges (or groups of pieces for 1.2.2) around the corner rather than actually twisting it. Forgot to mention that solving trivial tips or 3x3x3 super centres could also be looked at in the perspective of permuting other pieces around them rather than "orienting" them. The centre rotating alg [R, U, R', U] x5 is another example.



I got rather off topic... At least the mention of parity making a puzzle harder was on topic i suppose.

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Fri Oct 15, 2010 8:03 am 
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Elwyn wrote:
kastellorizo wrote:
Yes, but we need to differentiate corners from edges.
I agree totally and never stated i was doing otherwise, you on the other hand stated i was, I think due to a misunderstanding.


Yeap, glad this is clarified and sorry if I sounded too "direct" towards you. :)


Elwyn wrote:
On a pyraminx crystal would you say the most "basic" method is to find a way to permute them all then find a way to orient them all?


In this case, and all other cases in general, what we need is to identify the smallest sequences which
exhange and orient the minimum number of pieces at the same time. By knowing this, you have the
core information to manipulate all pieces the way you want to.


Elwyn wrote:
I look at orientation as part of permutation, a piece is in the right spot or not and you can use one algorithm to put it there, why look for another. I see it like this because i feel using a short commutator to move a piece to a single position (with orientation) using setup moves is more basic than having an algorithm to permute and an algorithm to orient. Though there is the exception of corner orientation parity if it occurs on a puzzle but surely you agree parity algorithms for such cases fall under a different category to normal orientation algorithms, or do you think if you can turn a single corner that solving the orientation of each corner one at a time is the basic solution no matter how long the algorithm because that kind of goes against the look for simple commutators thing? We really are looking at the term basic in a different way i suppose, you see it as finding and using two easy to use tools is more basic than finding and using one tool requiring slightly more logic.


Yes, we are looking at the same problem from a different angle, that's for sure! :)
If there is parity, we need to identify it, there can be orientation parity or exchange parity.
And yes, they can be slightly longer, but very powerful (not fast or elegant, but powerful
in the sense you can directly handle the smallest possible changes).


Elwyn wrote:
Find me a twisty puzzle where you cannot use setups and a permutation algorithm to orient pieces and i'll agree solving orientation needs to be separate to solving permutation, but to me it certainly isn't like that on a 3x3x3.

This does not include orientation parity but i thought i was worth mentioning Brandon's recent solution to 1.2.2 orientation parity and indeed the way most people "orient" the last corner (the non trivial ones) of a pyraminx such as [X, Y, X', Y] x2 actually can be seen as permuting three edges (or groups of pieces for 1.2.2) around the corner rather than actually twisting it.


Ok. Apart from the Pyraminx and 3x3x3 Supercube examples, there are puzzles which only have
pieces to exchange (Varikon 2x2x2) or only to orient (Bolygok). I find those examples as I had a fast
glance at my IKEA stand with puzzles. Your statement that orientation and exchange are permutations
(they are both permutations - the orientation is, as you correctly said more complex) is true.

But looking it from the perspective of the algorithm will hide its real power. Self rotation of a piece
around its own axis (i.e. orientation), and exchange of pieces which is a rotation around an axis
created by those three pieces, are two different things, visually and mathematically.

One of them *can* contain a mix of moves of the other one, and this is normal.
And this is where we differ, as combining them will not allow us to do direct moves
which is focusing to as less pieces as possible.

To put it in another way. Yes, you are able to use an entire exchange algorithm,
combine it with another "shifted" exchange algorithm, to create an orientation algorithm.

So, in algorithm terms, the orientation can be a linear combination of two exchange algorithms.
But that little "shifting" adds something non-linear. So unless you declare *both*
(in a separate way) the previous exchange algorithms as part of the difficulty list, the
connection cannot be regarded linear any more and the orientation needs to be re-defined.

(shifting/shifted may refer to any change before re-starting the exchange algorithm,
and it can be as simple as rotating the entire puzzle - I am sure you know what I mean!)


Elwyn wrote:
Forgot to mention that solving trivial tips or 3x3x3 super centres could also be looked at in the perspective of permuting other pieces around them rather than "orienting" them. The centre rotating alg [R, U, R', U] x5 is another example.


I hear you, and the above are remarkable algorithms, and I am sure many people can see that.

And the last algorithm ([R, U, R', U] x5 ) you mentioned is a lovely algorithm which I also (found and) use.
The goal here, is to be able to manipulate specific pieces. We should not think of going to the
other extreme and turning all other pieces, that definitely makes things more complex.


Elwyn wrote:
I got rather off topic... At least the mention of parity making a puzzle harder was on topic i suppose.


LOL yeap!


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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Fri Oct 15, 2010 10:15 am 
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kastellorizo wrote:
If there is parity, we need to identify it, there can be orientation parity or exchange parity.
I only mentioned orientation parity because i can't think of a permutation parity that isn't on a puzzle that people use reduction on and wouldn't occur if the puzzle was solved with only three cycles, feel free to show me one that can't be solved by the exact same alg you use to cycle the piece.
kastellorizo wrote:
Ok. Apart from the Pyraminx and 3x3x3 Supercube examples, there are puzzles which only have
pieces to exchange (Varikon 2x2x2) or only to orient (Bolygok). I find those examples as I had a fast
glance at my IKEA stand with puzzles. Your statement that orientation and exchange are permutations
(they are both permutations - the orientation is, as you correctly said more complex) is true.
First of all Varikon 2x2x2 isn't a twisty puzzle, i didn't know what the Bolygok was but from the description and pictures i found of it I wouldn't call it a true twisty puzzle either. The other examples such as super 3x3x3 centres and pyraminx corners are both of pieces that don't need permuting and can hence be solved simply by twisting them at the start in one move each, then solving with commutators wouldn't disturb them. I'd hardly call that adding to the difficulty (what this topic is actually about) and an algorithm of length one move isn't really an algorithm is it. Did you think of solving a super 3x3x3 centres first and then just using commutators? I hadn't till just then but to me that means you could solve a super cube with 2 simple commutators :lol: the centres add a step but no need for an algorithm.

(the thing about pieces just being turned individually into correct orientation goes for the Bolygok as well)

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Fri Oct 15, 2010 11:05 am 
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Elwyn wrote:
kastellorizo wrote:
If there is parity, we need to identify it, there can be orientation parity or exchange parity.
I only mentioned orientation parity because i can't think of a permutation parity that isn't on a puzzle that people use reduction on and wouldn't occur if the puzzle was solved with only three cycles, feel free to show me one that can't be solved by the exact same alg you use to cycle the piece.


Well this is true, we agree to this. But someone who doesn't know about cycles, won't agree.
And the shifting I mentioned, cannot help to constitute a a linear combination of the exchange move.
An experienced solver on a specific puzzle, yes, can see it easily. But if the solver is not experienced
or if the puzzle is not something new (and even if the solver is experienced), then finding an
orientation component may not be that simple. Yes, it can be found, but it will need some serious thought.

:)


Elwyn wrote:
kastellorizo wrote:
Ok. Apart from the Pyraminx and 3x3x3 Supercube examples, there are puzzles which only have
pieces to exchange (Varikon 2x2x2) or only to orient (Bolygok). I find those examples as I had a fast
glance at my IKEA stand with puzzles. Your statement that orientation and exchange are permutations
(they are both permutations - the orientation is, as you correctly said more complex) is true.
First of all Varikon 2x2x2 isn't a twisty puzzle, i didn't know what the Bolygok was but from the description and pictures i found of it I wouldn't call it a true twisty puzzle either. The other examples such as super 3x3x3 centres and pyraminx corners are both of pieces that don't need permuting and can hence be solved simply by twisting them at the start in one move each, then solving with commutators wouldn't disturb them. I'd hardly call that adding to the difficulty (what this topic is actually about) and an algorithm of length one move isn't really an algorithm is it. Did you think of solving a super 3x3x3 centres first and then just using commutators? I hadn't till just then but to me that means you could solve a super cube with 2 simple commutators :lol: the centres add a step but no need for an algorithm.

(the thing about pieces just being turned individually into correct orientation goes for the Bolygok as well)


Still, a Pyraminx *does* have a (tiny) different difficulty number when compared to the Tetraminx,
and this is why we need all those. The difficulty is not apparent to an experienced puzzler, but it
is still there, and we ought to count it if we want to create a reliable difficulty counting method.
Even if the length is 1, it must be counted. For the Boob Cube, the same length is 2. And so on.

There are plenty of simple examples as well as more complex examples, e.g. a 2x2x2
with the upper part all colored blue, and the bottom part all colored green, or even the 4D8.
You won't get any orientation trouble, but only exchanges. The more you search the more you find
such examples. And this is why I insist, because no ones knows what will be found in the future,
(or in the past!) so it is better to be properly prepared with the most powerful and elemental way.

As for twisty puzzles, the forum is for twisty puzzles, but it also covers many of the puzzles I mentioned.
In fact, the aim is to find the difficulty of all such mathematical puzzles, twisty or not. Therefore,
we need a more broad and global way, which breaks the components into the smallest parts.

:)


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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Fri Oct 15, 2010 2:18 pm 
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kastellorizo and Elwyn, I have read through much of the debate above, and have a request. Could each of you grade the following puzzles on a scale of 1 to 10 with 10 being the most difficult?

Rubik's Cube
Megaminx
Meffert's Bandage Cube
Helicopter Cube
Qubami
4x4x4 cube
Trajber's Octahedron 4x4x4
Crazy 4x4x4 II
Crazy 3x3x3 Plus Earth
Magic Octahedron
FTO

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Fri Oct 15, 2010 9:44 pm 
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robertpauljr wrote:
Could each of you grade the following puzzles on a scale of 1 to 10 with 10 being the most difficult?
Because i don't have a method of quantitatively measuring the difficulty of a puzzle that i actually think gives a good indication of true difficulty i suppose i'll just have to say how hard i thought they were to solve. This has some major problems though because The Rubik's cube was the first i solved and hence made the others seem a lot easier than they would have if i'd solved them first, i'll try and compensate for that. I don't know how i'll compensate for the fact i have only solved 8 of the 11 because i just haven't ever had the chance to try the others.

Rubik's cube I'd say about a 3, it does only have two types of pieces to solve, no bandaging or jumbling or difficulties in knowing what piece goes where. algs are simple to find in comparison to some of the other puzzles.

Megaminx 4, Very similar to the Rubik's cube but i never found a simple pure edge cycle on it when there are lots on the 3x3x3.

I haven't tried the bandaged cube, i do think a solve would involve a lot more thought than a normal cube and new algs for the joined pieces, i'll guess 7 but i'm not sure.

Helicopter cube non jumbled 3, introduces orbits which might be a little confusing but really corners can be solved by just trial and error and then there's just one alg to solve centres

Jumbled is a different story, there are one or two shapes that actually require a little thought to get it back to a cube and having to switch centres back to their orbits along with the possibility of having only two not solved makes it about 5 i'd say.

Quabami i haven't tried but i really see it (and other things like maze or sudoku cubes) as two puzzles. And i'm not sure if having to figure out where a piece goes is part of this thread's version of difficulty, that said i'm sure it would be hard to solve the puzzle of where things go but after that it's just another 3.

4x4x4 around a 5, It might be higher if reduction was the only possible method but there's always simple commutators, Oll parity is very annoying but pll parity isn't really difficult to see what's wrong and fix with the same simple commutator for cycling single edges so isn't really worth mentioning.

Trajber's 4x4x4 4, only three commutators, easy setups due to no orientation lots of identical pieces make parity a simple matter.

Crazy 4x4x4 II 7, yeah that's a bit high but having to solve a 2x2x2 and then solving a super 4x4x4 without mixing that 2x2x2 up is a bit more of a task than a normal 4x4x4 and i still don't know what to do if i get parity hahaha that said it like a lot of puzzles i have solve only in electronic form and haven't had a very good look at.

Crazy 3x3x3 plus earth... no idea, haven't got any crazy cubes but i wish i did, i'd guess very hard another 7 or 8.

Magic octahedron 2, Someone can solve the whole thing with R U R' U' and a small bit of logic hahaha the only puzzles i can think of that are easier are a pyraminx (though it's probably about the same) and a dino cube.

FTO 3, one more commutator needed than a 3x3x3 but the corners can be solved first with the easiest of commutators.

Sorry for the long winded explanation but i don't believe there is a good way of measuring difficulty so i thought i'd explain for each puzzle.

--------------------------------------------------------------------------------------------------------------------

For non jumbling or bandaged puzzles the number of commutators needed along with their length and difficulty to find (even some short ones are hard to find sometimes) affects the difficulty the most in my opinion, then there's average number of setup moves and possibility of parity, if there's a obvious simple reduction method that doesn't add parity that reduces the difficulty.
kastellorizo wrote:
Still, a Pyraminx *does* have a (tiny) different difficulty number when compared to the Tetraminx,
and this is why we need all those. The difficulty is not apparent to an experienced puzzler, but it
is still there, and we ought to count it if we want to create a reliable difficulty counting method.
Even if the length is 1, it must be counted.
I'd say that is just adding complexity not difficulty, If it does add some kind of difficulty it's so small i doubt it should be counted due to the fact if you want to put a "difficulty number" on a puzzle, say puzzle x gets a score of 5, puzzle x with trivial tips should get a score of 5.0000000001 so insignificant to me it's not worth mentioning.
kastellorizo wrote:
As for twisty puzzles, the forum is for twisty puzzles, but it also covers many of the puzzles I mentioned.
In fact, the aim is to find the difficulty of all such mathematical puzzles, twisty or not. Therefore,
we need a more broad and global way, which breaks the components into the smallest parts.
That seems like too big a goal for a simple thread like this, it's complex enough with just twisty puzzles for now, look at the second sentence of this thread
wwwmwww wrote:
How can one quantify the difficulty of a twisty puzzle?


Thank you to anyone who read this entire post. I know there are more interesting things you could have read instead :lol: .

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Sat Oct 16, 2010 1:24 am 
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robertpauljr wrote:
Could each of you grade the following puzzles on a scale of 1 to 10 with 10 being the most difficult?


There is still some work needed to complete my definitions. If someone could provide me with the
shortest algorithms for exhanging and orienting each individual type of pieces for specific puzzles,
then it could be easy. I am not good in finding such shortest paths, so any expert help would be welcome
(Also, we need to finalise a last factor - see below).


Elwyn wrote:
kastellorizo wrote:
Still, a Pyraminx *does* have a (tiny) different difficulty number when compared to the Tetraminx,
and this is why we need all those. The difficulty is not apparent to an experienced puzzler, but it
is still there, and we ought to count it if we want to create a reliable difficulty counting method.
Even if the length is 1, it must be counted.

I'd say that is just adding complexity not difficulty, If it does add some kind of difficulty it's so small i doubt it should be counted due to the fact if you want to put a "difficulty number" on a puzzle, say puzzle x gets a score of 5, puzzle x with trivial tips should get a score of 5.0000000001 so insignificant to me it's not worth mentioning.


I agree it adds too little, but it is part of solving the puzzle, even it is too easy for an expert.

As an example, let's say a puzzle needs three moves to be solved, of length 7, 4, and 3.
And also assume it has four tips of "solving length" 1.

Then, the difficulty number would be sqrt(7^2 +4^2 +3^2 +1^2)=8.6603,
compared to the sqrt(7^2 +4^2 +3^2)=8.6023 when the tip is ignored. As you may see, it is a fair difference.

A good and reasonable argument would now be: "What if we had a puzzle which had only one
very long cyclic generator?". There exists an answer here , but we need to consider a multiplying
factor of how many different types of pieces are shared by the puzzle generators. This will not
make it hard to calculate, but it needs to be defined properly. My suggestion is to rate the
generators by using a fraction of their types of pieces. Then, multiply each sequence (inside the
square root) by all the different multiplying actors which are used in that sequence. of course,
if there is a trivial cyclic move with no interaction, it will be rated even lower than before!

And after that, I don't think there is anything else to add in order to define the difficulty of not just a twisty,
but any type of mathematical puzzle. Please give me some time, and I will come up with an updated formula.

:)


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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Sat Oct 16, 2010 2:09 am 
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OK, I'm probably wrong in saying this, and I'm not going to be getting myself into these absolutely massive posts, but couldn't an idea of measuring the shortest possible algorithms for each possible combination of pieces work, and adding the results work?

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Sat Oct 16, 2010 2:58 am 
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OK, I have came up with a nice formula which does not require any difficult calculations
and will post it after I come back from training! (give 3-4 hours LOL)

Surprisingly, it does not involve length algorithms, but attributes related to them!
(I will also include results).

:)


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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Sat Oct 16, 2010 7:56 am 
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Considering the thread that got me started on this topic I'd like to see a measure that compares these:

Comboctahedron
Comboctahedron (without trivial tips)
FTO
VTO
VTO (without trivial tips)

Carl

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Sat Oct 16, 2010 8:07 am 
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I'm not sure if you can get away comparing puzzles like that. The Combohedron is far more difficult than a VTO/FTO but ignoring the special moves, a Helicopter Skewb is not much more difficult than a Helicopter Cube.

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Sat Oct 16, 2010 9:40 am 
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ok, some time ago, I attempted to categorise twisty puzzles using a cutting number.
Well, it seems that the old information could be used for the "difficulty" purpose.

Here is my take, and this time it is a good one (i.e. elegant and simple).
That said, each person is *different* and the formula I am presenting only
gives some sort of difficulty indication.

First let us assume that a puzzle has i different generators g_i, and each such
generator is repeated n_i times (Generators are considered to be repeated,
if their action results in a similar change). For example the 3x3x3 has one type of generator,
but it can be found six times. The 2x2x2 has one type, repeated three times. The Pyraminx
has two types, the trivial tip generator (repeated four times) and another deeper cut generator
(also repeated four times).

Now let us pick up a generator and define it by its cutting movement. It divides the puzzle
into two parts, and each part has a number of pieces presented as stickers or tiles (for
most puzzles that is). So we get two numbers, L_i is the low and H_i is the high number
(when compared). Finally, we check how many "different type" of pieces each part has,
and we define them as l_i and h_i respectively for the parts with low number and high number.
(for example, the 3x3x3 has a total of three different type of pieces, the 2x2x2 has one,
the Pyraminx has three, and so on).

Then my formula is:

Σ n_i * ( L_i * l_i )/( H_i * h_i )

The number it gives for some puzzles is:

1x1x1: 0 (no generators)
1x1x2 (Boob Cube): 1*(5*2)/(5*2) = 1
Pyraminx: 4*(3*1)/(33*3) + 4*(12*3)/(24*3) = 2.1212
Magic Octahedron: 8*(4*1)/(68*3) + 8*(16*3)/(56*3) = 2.4426
Megaminx: 12*(26*3)/(106*3) = 2.9434
2x2x2: 3*(12*1)/(12*1) = 3
Cubedron: 12*(5*2)/(20*2) = 3
3x3x3: 6*(21*3)/(33*3) = 3.8182
Skewb/Ultimate Skewb = 4*(15*2)/(15*2) = 4
Void Cube: 6*(20*2)/(28*2) = 4.2857
Dogic: 12*(5*1)/(75*2) + 12*(20*2)/(60*2) = 4.4
FT Octahedron: 8*(27*3)/(45*3) = 4.8
Pyraminx Crystal: 12*(35*2)/(85*2) = 4.9412
24 Cube: 6*(12*1)/(12*1) = 6
4x4x4: 6*(32*4)/(64*4) + 3*(48*4)/(48*4) = 6
5x5x5: 6*(45*7)/(105*7) + 6*(65*7)/(85*7) = 7.1597


I hope I haven't made any arithmetic mistake (shame on me if I did LOL)
And now you may also calculate the puzzles of your own choice. :)

In other words, deeper cut puzzles and more axis, are given more power,
which makes sense, as they make things more confusing for the typical solver.
And the sequence lengths required to solve a puzzle, always depend (heavily)
on the ratio between the pieces, as well as the type of pieces which interact,
and this is why I mixed them in a logical way.

Some may argue there must be another way (if yes, please elaborate?), but
I must remind you that I just made a reasonable suggestion based on logical arguments,
and I believe this formula gives some nice indication, while it is SUPER easy to calculate.

Any constructive comments are welcome.

;)


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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Sat Oct 16, 2010 9:52 am 
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kastellorizo wrote:
Megaminx: 12*(26*3)/(106*3) = 2.9434
2x2x2: 3*(12*1)/(12*1) = 3

Pantazis

A megaminx is less difficult than a 2x2x2? I don't think so.

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Sat Oct 16, 2010 10:17 am 
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robertpauljr wrote:
kastellorizo wrote:
Megaminx: 12*(26*3)/(106*3) = 2.9434
2x2x2: 3*(12*1)/(12*1) = 3
Pantazis

A megaminx is less difficult than a 2x2x2? I don't think so.


This is the only result which seems "interesting", and it is because as stated:

kastellorizo wrote:
deeper cut puzzles and more axis, are given more power,
which makes sense, as they make things more confusing for the typical solver.
And the sequence lengths required to solve a puzzle, always depend (heavily)
on the ratio between the pieces, as well as the type of pieces which interact,
and this is why I mixed them in a logical way.


In this case [deeper cut] > [more axis] I guess LOL.

I did notice that, and have attempted to modify the formula, but stretching one bit,
affects badly some other bit, and in the end the formula becomes too overcomplicated.
(I have a version where it deals with the "concentration of types of pieces" in each L_i and H_i).

So I decided to stick with the original one. Still, deeper cut *is* more confusing than
having more axis (and when I first solved them I had similar trouble solving the 2x2x2
and the Megaminx - maybe it is just me, and... my formula!)

Personally, I am quite happy with the indications, especially
as no other formula (that I know) has come any close to my results.

:)


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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Sun Oct 17, 2010 10:27 am 
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kastellorizo wrote:
Any constructive comments are welcome.
I like the simplicity of your formula, and it has a huge advantage that one can analyse the properties of a puzzle and not play with it or try to solve it yet still come up quickly with a numeric measure of its difficulty.

I offer an alternative way of estimating difficulty, as a supplement to yours, not a replacement. It has the disadvantage that one must first experiment with the puzzle to find optimal algorithm lengths, but then the calculation is quite easy. It also follows naturally from what has interested me for some time, and is compatible with what Jared and you said earlier in this thread:

kastellorizo wrote:
Jared wrote:
I think it's more a matter of what sorts of commutators are available.
Well said!!! This is an *excellent* criterion, which will suit both analytic *and* intuitive solvers.

The length of the *different* required algorithms is exactly what really separates a solved puzzle and a scrambled puzzle. Therefore, the only challenge here is to find the minimum (optimal) algorithm lengths for each puzzle.
My primary measure is the longest sequence of moves that must be found to solve the puzzle. Sequences will mostly be commutators, but not always. Where the q of (p,q) is a single move (counting a slice move where applicable as a single move), I count p as this length. For non-commutators I just count the whole length of the sequence. All of these accomplish simple operations such as cycling a piece type, or successfully isolating a single swapped piece in a turning layer to enable the sequence to be commutated to cycle a piece type. If we call the longest required sequence x, and we devise a solution to minimize x, what is the value of x?

My second measure, to separate puzzles with the same value of x, is a summation of the separate processes required to solve the puzzle. Because the difficulty of tracking pieces and groups increases greatly as the length of a sequence grows, I decided it would be a good idea to square the length of each sequence. If we are permuting and orienting pieces at the same time, I add 1 to the sequence length before squaring it, to reflect a slight increase in complexity (because more precise setups are needed when solving). If l is the length of a sequence and o is 1 if pieces are being oriented and permuted simultaneously and 0 otherwise, the formula for this second measure is:

∑ (l+o

The first measure is a good guide to the greatest difficulty in solving the puzzle, while the second measure is more the overall "quantity" of difficulty in solving the puzzle. The solution order is given after the name of the puzzle, with the length l of each part. Where pieces are permuted and oriented simultaneously (o=1) the piece type is marked with *. Then my first measure, the highest value of l, is shown in bold, followed by the second measure in brackets. I have not played with a Cubedron so I could not include it in my list, but I have slipped in a couple of other interesting puzzles instead:

1x1x1: 0 (0)
1x1x2: halves 1 => 1 (1)
Pyraminx: tips 1, edges* 1 => 1 (5)
Magic Octahedron: corners 1, edges* 1 => 1 (5)
FT Octahedron: edges 1, corners* 1, centers 3 => 3 (14)
2x2x2: corners* 3 => 3 (16)
Helicopter Cube (non-jumbling): permute corners 1, orient corners 2, centers 3, center double swap 2 => 3 (18)
3x3x3: edges* 1, corners* 3 => 3 (20)
Megaminx: edges* 1, corners* 3 => 3 (20)
Pyraminx Crystal: corners* 3, edges* 1 => 3 (20)
Void Cube: edges* 1, corners* 3, parity fix 1 => 3 (21)
Dogic: corners* 3, centers 3 => 3 (25)
4x4x4: centers 3, edges 3, corners* 3, parity fix 1 => 3 (35)
5x5x5: X centers 3, T centers 3, central edges* 1, outer edges 3, corners* 3, parity fix 1 => 3 (48)
6x6x6: X centers 3, oblique centers 3, inner edges 3, outer edges 3, corners* 3, parity fix 1 => 3 (53)
7x7x7: X centers 3, T centers 3, oblique centers 3, central edges* 1, inner edges 3, outer edges 3, corners* 3, parity fix 1 => 3 (66)
Skewb: permute corners 1, permute centers 3, orient corners 4 => 4 (26)
Skewb Ultimate: permute corners 1, permute centers 3, orient corners 4, orient centers 4 => 4 (42)
Deeper cut FT Icosahedron: flattish triangles 1, corners* 3, centers 3, edges 5 => 5 (51)
Pentultimate: centers 1, corners* 6 => 6 (50)
24 Cube (non-jumbling): pieces 10 => 10 (100)


--------------------------------------------------------------------------------------------------------


General notes:

In all cases I assume that no algorithm is purer than it needs to be, and I also make an important assumption that where pieces have visible orientations, we generally orient and permute them at the same time. The Skewb and Skewb Ultimate are exceptions, due to the restrictive orbits their corners are in. It takes more advanced and complex routines to simultaneously permute and orient them than it does to orient them as separate operations. With the Skewb Ultimate, when we find a (4,1) algo that twists two corners in opposite directions and also flips two large pieces, we have a two-in-one algo. If we perform this commutator twice, the flips cancel out and we have a pure corner twist algo. If we perform the commutator three times, the twists cancel out and we have a pure big piece flip algo. It's a fairly easy observation to make and we have a complete solution without ever having to notice the effect on the puzzle of a sequence longer than 4 moves - which is why I place emphasis on this maximum number. Whether or not we should really count the same algo twice when used two different ways, my instinct says yes, because having to do something in multiple ways does add to the difficulty of the puzzle.

Regarding the algorithm lengths, many of these seem to have been commonly known and agreed for some time, at least I think all of the <=4 ones are. I have not seen anyone offer a way of solving the DeFTI that does not use at least (5,1) commutators, and Brandon has verified by a computer search that the shortest commutator to solve the Pentultimate corners pure is (6,1). So the only questionable one is (10,1) to cycle three 24 Cube pieces pure, and I am fairly confident that a computer search will not yield a shorter commutator.

Regarding the parity fixes, I count them as 1 because a single slice move and re-solving using commutators only is a successful method. Admittedly it isn't very obvious to someone new to a 4x4x4 that a single 90 degree slice move will shift the edges to an even permutation and thus allow the puzzle to be successfully completed after re-solving the displaced centers and edges again exclusively with commutators and setup moves. It takes a certain amount of insight and I think it adds more to the difficulty than is reflected by adding just 1 to the total, but I wanted to keep my calculations as objective as possible, so there we are. And of course, although it is nice to find sequences such as r U2 r U2 r U2 r U2 r for big cubes, or a longer and more complex one that completely solves the odd edge perm with no side effects, such a sequence is not a necessity to solve the puzzle.


Notes regarding orientation:

Strange things happen to the list if we make it a requirement to include separate sequences for orienting pieces for all puzzles with orientable pieces: the primary measure for the Pyraminx jumps from 1 to 3, putting it in the same category as the 3x3x3; for all the face turning cubes it jumps from 3 to 6 (the moves to re-orient a corner in a layer), putting the cubes in the same category as the Pentultimate; for the Megaminx it jumps up from 3 to 8 (the moves to flip an edge in a layer); and for the Pentultimate it jumps up from 6 to 12 (the fastest routine I've seen to twist 2 corners pure is the (12,2) posted by Danny Devitt).

It is my observation, based on solving a wide variety of virtual puzzles including some quite complex ones, that algorithms to orient pieces in place are usually harder to find and nearly always longer than algorithms to permute pieces. When creating these commutators, it is a stricter requirement mathematically and practically to re-orient a piece type in a turning region than to swap a piece in a turning region with any other of the same type. So from the success-solving point of view (I love that word by the way!), when doing a measure based on algo lengths, as I do here, I think it is better only to separate orientation processes when absolutely necessary.


Edit #1: Fixed a couple of typos and corrected an error in the Skewb and Skewb Ultimate figures -- just realized that the centers can be permuted (3,1) instead of (4,1)!

Edit #2: Corrected 4x4x4 and 5x5x5 edge calculations (non-central edges of higher cubes only permute!) and added a few puzzles to the list.

Edit #3: Pulled the following from the list because I have realized that I missed out the important initial stage of restoring the cube shape from a jumbled state:
Helicopter Cube (jumbling): center orbital swap 3, permute corners 1, orient corners 2, centers 3, center double swap 2 => 3 (27)


Last edited by Julian on Mon Oct 18, 2010 2:28 am, edited 3 times in total.

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Sun Oct 17, 2010 12:58 pm 
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Julian, that was a magnificent post, it is nice to see your approach, as it is very interesting.


Julian wrote:
kastellorizo wrote:
Any constructive comments are welcome.
I like the simplicity of your formula, and it has a huge advantage that one can analyse the properties of a puzzle and not play with it or try to solve it yet still come up quickly with a numeric measure of its difficulty.


Yes, it is very simple, included my final update (I show it below). :)


Julian wrote:
My primary measure is the longest sequence of moves that must be found to solve the puzzle. Sequences will mostly be commutators, but not always.


Indeed, this is a very acceptable way. Using the algorithms can be extremely efficient.



Julian wrote:
Whether or not we should really count the same algo twice when used two different ways, my instinct says yes, because having to do something in multiple ways does add to the difficulty of the puzzle.


Yes! This is exactly what I meant before when I mentioned the "independent sequences". :D
Totally agreed!

**********************************************


And after Julian's alternative proposal, I have added a fix to my previous formula. I knew I was
missing something which had to do with the "concentration" of the puzzle, which would help
to avoid silly cases. In the beginning, I thought a sticker by pieces ratio would do, but that
was already included! Then I realised that the shape of a puzzle plays a huge role. But since
we have so many different shapes, what could we do? Well, the solution was actually simpler
than I thought: To find the maximum number of pieces which are adjacent (with edges) to one
piece. And this ladies and gentlemen is the concentration coefficient "cc" of the puzzle!


For example, the cc for the Dogic, the Pyraminx, is equal to 3, for the 3x3x3 it is equal to 4,
and for the Megaminx, it is 5! For the Boob Cube it is 1, and for the 2x2x1 it is 2. Therefore,
by multiplying the entire previous formula by that number we get some excellent indication
which has also been calibrated to accommodate all past experiences from puzzle solvers,
and which can give an excellent indication for future cases or when comparing overcomplicated cases.
In fact, after checking some of the rest of the results, the concentration coefficient also fixed
other ambiguous cases!


So, my updated formula is:

cc * Σ [ n_i * ( L_i * l_i )/( H_i * h_i ) ]

The number it gives for some puzzles is (I added a couple more):

1x1x1: 0 (no generators)
1x1x2 (Boob Cube): 1*1*(5*2)/(5*2) = 1
1x2x2: 2*2*(8*1)/(8*1) = 4
Pyraminx: 3*[4*(3*1)/(33*3) + 4*(12*3)/(24*3)] = 6.3636
Magic Octahedron: 3*[8*(4*1)/(68*3) + 8*(16*3)/(56*3)] = 7.3278
Helicopter Cube: 3*12*(10*2)/(48*2) = 7.5 (jumbling is not counted!)
2x2x2: 3*3*(12*1)/(12*1) = 9
Cubedron: 4*12*(5*2)/(20*2) = 12
Dogic: 3*[12*(5*1)/(75*2) + 12*(20*2)/(60*2)] = 13.2
Megaminx: 5*12*(26*3)/(106*3) = 14.7170
Pyraminx Crystal: 3*12*(35*2)/(85*2) = 14.8236
3x3x3: 4*6*(21*3)/(33*3) = 15.2728
Skewb = 4*4*(15*2)/(15*2) = 16
Void Cube: 4*6*(20*2)/(28*2) = 17.1428
FT Octahedron: 4*8*(27*3)/(45*3) = 19.2
24 Cube: 3*6*(12*1)/(12*1) = 18
4x4x4: 4*[6*(32*4)/(64*4) + 3*(48*4)/(48*4)] = 24
5x5x5: 4*[6*(45*7)/(105*7) + 6*(65*7)/(85*7)] = 28.6388
7x7x7: 4*[6*(77*12)/(217*12) + 6*(105*12)/(189*12) + 6*(133*12)/(161*12)] = 41.376

To summarise, the method favors deeper cut puzzles and puzzles which have
more "social" pieces. Which is exactly what intuition tells us!

:)


Pantazis

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Sun Oct 17, 2010 8:29 pm 
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Pantazis, prompted by your new list, I have added the Helicopter Cube, 6x6x6, and 7x7x7 to the list in my previous post. By my estimate the non-jumbling Helicopter Cube is between the 2x2x2 and 3x3x3 in difficulty. As a couple of my Helicopter Cube sequence lengths might seem surprising, I will explain how I obtained them:

Helicopter Cube (non-jumbling): permute corners 1, orient corners 2, centers 3, center double swap 2.

Observation #1: A single move swaps two corners, which gives us "permute corners 1".

Observation #2: Turning two parallel edges, such as FU BU, moves corners URF and URB to new adjacent positions with different relative orientations, so if we then swap them with LU, bring them back with BU FU and swap them again with UR, we have twisted them without affecting any other corners. It is not a "literal" commutator because the 3rd and 6th moves are in different positions, but it is what I consider a "conceptual" (2,1) commutator. As it is a process made by observing the effect of a sequence of 2 moves, it gives us "orient corners 2".

Observation #3: FU UR FU leaves the FR region unaltered apart from a single swapped center, which gives us "centers 3".

Observation #4: Turning two adjacent edges, such as FU UR, permutes 3 corners, permutes 3 centers, and does 2 center swaps. So if we repeat three times, everything cancels apart from the 2 center swaps. That gives us "center double swap 2".


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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Mon Oct 18, 2010 3:08 pm 
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Thanks Julian, I was also preparing a nice list covering many puzzles,
and will update it whenever I have more time (and requests!)

It seems that the definition covers a wide range of puzzles, so here we go:

1x1x1: 0.0000 (no generators)
1x1x2 (Boob Cube): 1*1*(5*2)/(5*2) = 1.0000
1x2x2: 2*2*(8*1)/(8*1) = 4.0000
Floppy Cube 1x3x3: 4*(4*(11*1)/(19*2)) = 4.6316
Chronos: 4*(6*(1*1)/(5*1)) = 4.8000
Missing Link: 4*(8*(1*1)/(14*1)+2*(4*1)/(11*1)) = 5.1948
Braintwist: 2*(8*(6*1)/(18*1)) = 5.3333
Rubik's Magic (4-tiles): 2*(2*(8*2)/(8*2)+1*(8*2)/(8*2)) = 6.0000
Orbik: 2*(12*(2*1)/(12*1)) = 6.0000
Pyraminx: 3*[4*(3*1)/(33*3) + 4*(12*3)/(24*3)] = 6.3636
Rubik's Shells (no buttons pushed): 4*(4*(8*1)/(20*1)) = 6.4000
15-Puzzle: 4*(24*(1*1)/(14*1)) = 6.8571
Dino Cube: 3*(8*(6*1)/(21*1)) = 6.8571
Rubik's Rings: 4*(2*(16*1)/(18*1)) = 7.1111
Rubik's Shells (one button pushed): 4*(2*(8*1)/(20*1)+1*(14*1)/(14*1)) = 7.2000
Magic Octahedron: 3*[8*(4*1)/(68*3) + 8*(16*3)/(56*3)] = 7.3278
Helicopter Cube: 3*12*(10*2)/(48*2) = 7.5000 (jumbling is not counted!)
Rubik's Shells (both buttons pushed): 4*(2*(14*1)/(14*1)) = 8.0000
Octo: 4*(1*(16*1)/(16*1)+8*(4*1)/(28*1)) = 8.5714
Orbit: 2*(1*(28*1)/(28*1)+2*(8*1)/(48*1)+2*(20*1)/(36*1)+2*(28*1)/(28*1)) = 8.8889
2x2x2: 3*3*(12*1)/(12*1) = 9.0000
2x2x3: 4*(2*(12*1)/(20*2)+2*(16*2)/(16*2)) = 10.4000
Impossiball: 3*(12*(5*1)/(15*1) = 12.0000
Puck: 2*(6*(6*1)/(6*1)) = 12.0000
Cubedron: 4*12*(5*2)/(20*2) = 12.0000
2x2x4: 4*(2*(12*1)/(28*2)+2*(20*2)/(20*2)+1*(20*2)/(20*2)) = 12.8571
Dogic: 3*[12*(5*1)/(75*2) + 12*(20*2)/(60*2)] = 13.2000
Columbus Egg: 2*(10*(20*1)/(30*1)) = 13.3333
Rubik's Clock: 1*(8*(5*4/13*6)+8*(11*4)/(7*6)+4*(9*4)/(9*6)+2*(5*4)/(13*6)) = 13.6117
Rubik's Magic (8-tiles): 2*(2*(16*2)/(16*2)+1*(16*2)/(16*2)+4*(16*2)/(16*2)) = 14.0000
Megaminx: 5*12*(26*3)/(106*3) = 14.7170
Pyraminx Crystal: 3*12*(35*2)/(85*2) = 14.8236
3x3x3: 4*6*(21*3)/(33*3) = 15.2728
Skewb = 4*4*(15*2)/(15*2) = 16.0000
Astrolabacus: 2*(6*(18*1)/(18*1)+6*(6*1)/(18*1)) = 16.0000
Void Cube: 4*6*(20*2)/(28*2) = 17.1428
24 Cube: 3*6*(12*1)/(12*1) = 18.0000
FT Octahedron: 4*8*(27*3)/(45*3) = 19.0000
Masterball: 4*(2*(8*1)/(24*2)+1*(16*1)/(16*1)+4*(16*2)/(16*2)) = 21.3333
4x4x4: 4*[6*(32*4)/(64*4) + 3*(48*4)/(48*4)] = 24.0000
5x5x5: 4*[6*(45*7)/(105*7) + 6*(65*7)/(85*7)] = 28.6388
7x7x7: 4*[6*(77*12)/(217*12) + 6*(105*12)/(189*12) + 6*(133*12)/(161*12)] = 41.3760


:)


Pantazis

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Educational R&D, Gravity, 4D Symmetry, Puzzle Ninja, Matrix Mech, Alien Technology.


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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Mon Oct 18, 2010 4:29 pm 
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kastellorizo wrote:
Thanks Julian, I was also preparing a nice list covering many puzzles,
and will update it whenever I have more time (and requests!)

How about the Crazy 3x3 Plus series? And the Crazy 4x4x4 series?

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 Post subject: Re: How hard/difficult is Puzzle X?
PostPosted: Tue Oct 19, 2010 3:19 am 
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Location: Greece, Australia, Thailand, India, Singapore.
robertpauljr wrote:
kastellorizo wrote:
Thanks Julian, I was also preparing a nice list covering many puzzles,
and will update it whenever I have more time (and requests!)

How about the Crazy 3x3 Plus series? And the Crazy 4x4x4 series?


There you go:

Circle Cube: 6*(6*(32*4)/(70*5)) = 13.1657

Mercury: 6*(1*(41*5)/(61*5)+5*(32*4)/(70*5)) = 15.0042
Venus/Earth: 6*(2*(41*5)/(61*5)+4*(32*4)/(70*5)) = 16.8427
Mars/Neptune: 6*(3*(41*5)/(61*5)+3*(32*4)/(70*5)) = 18.6812
Saturn/Uranus: 6*(4*(41*5)/(61*5)+2*(32*4)/(70*5)) = 20.5197
Jupiter: 6*(5*(41*5)/(61*5)+1*(32*4)/(70*5)) = 22.3582

Crazy 4x4x4 Type I: 5*(6*(32*5)/(88*5)+3*(56*4)/(64*5)) = 21.4091
Crazy 4x4x4 Type II: 5*(6*(25*5)/(44*6)+3*(64*6)/(80*6)) = 26.2045
Crazy 4x4x4 Type III: 6*(6*(48*5)/(90*6)+3*(72*6)/(86*6)) = 31.0698

Bonus calculation:

Varikon 2x2x2: 3*(12*(1*1)/(7*1)) = 5.1429
Varikon 3x3x3: 6*(30*(1*1)/(25*1)) = 7.2000

:)


Pantazis

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