# of possible pyraminx states?

[Carson]

How many unique possible states are there for the pyraminx NOT including the tips and not including the same states with different orientations?

OR

Can anyone give me some advice on how I could calculate this myself?

 

[Mike Hughey]

Whenever you wonder about something like this, and aren't determined to figure it out by yourself, the first place to go is almost always
Jaap's Puzzle Page.

(He has your answer, plus a pretty good description of how to calculate it.)

 

[Kirjava]

I'm not sure, but I think it's def. a number higher than 1.

 

[Mike Hughey]

I'm not sure, but I think it's def. a number higher than 1.

Wow, you're in fine form today.

 

[ben1996123]

9xx,xxx

 

[Carson]

Whenever you wonder about something like this, and aren't determined to figure it out by yourself, the first place to go is almost always
Jaap's Puzzle Page.

(He has your answer, plus a pretty good description of how to calculate it.)

I should have thought of this. I don't do much with theory, so I always forget about Jaap's page.

 

[miniGOINGS]

The maximum number of twists required to solve the Pyraminx is 11. There are 933,120 different positions (disregarding rotation of the trivial tips), a number that is sufficiently small to allow a computer search for optimal solutions.

 

[trying-to-speedcube...]

Centers: 3^4
Edge permutation: 6!/2
Edge orientation: 2^5
3^4*6!*2^4=933120

Yay I got it

 

[cuBerBruce]

I took the question to mean that you don't count as separate positions those positions that are reached by applying the same alg but from a different angle (e.g. RUR and LRL).

This can be computed in a brute force manner by the pseudocode algorithm shown below.

Let G := Pyraminx group (ignoring tips)
Let T := rotational symmetry group of the tetrahedron
Let S := empty set
Let n := 0
foreach g in G
  if g not in S then
    n := n + 1
    foreach t in T
      Let x := t' g t
      S := S union x 
    endfor
  endif
endfor
// n now contains the result.

Using GAP, I got a result of 78012 positions.

If you want mirrors to be considered equivalent as well, the number I get is 39036.

 

[Carson]

I took the question to mean that you don't count as separate positions those positions that are reached by applying the same alg but from a different angle (e.g. RUR and LRL).

This can be computed in a brute force manner by the pseudocode algorithm shown below.

Let G := Pyraminx group (ignoring tips)
Let T := rotational symmetry group of the tetrahedron
Let S := empty set
Let n := 0
foreach g in G
  if g not in S then
    n := n + 1
    foreach t in T
      Let x := t' g t
      S := S union x 
    endfor
  endif
endfor
// n now contains the result.

Using GAP, I got a result of 78012 positions.

If you want mirrors to be considered equivalent as well, the number I get is 39036.

Yes, this is exactly what I meant. Thank you very much.

 

[riffz]

What are the odds that one or more edge pieces will be solved relative to its center/vertex?

 

[cuBerBruce]

What are the odds that one or more edge pieces will be solved relative to its center/vertex?

I assume the question is to find the probability that an edge is "paired" (adjacent stickers matching in color) with either of the adjacent pieces (whatever you call those pieces). I'll refer to such a pair as a "CE pair."

Using GAP to do a brute force counting, I get the following for the distribution of the number of CE pairs for the positions of the Pyraminx.

Number of   Count of
 CE pairs  positions
---------  ---------
   0         348053
   1         336408
   2         172752
   3          57248
   4          14739
   5           3192
   6            600
   7             96
   8             15
   9             16
  10              0
  11              0
  12              1

Counting just the number of edges that are paired with at least one adjacent piece, I get:

  Number of    Count of
paired edges  positions
------------  ---------
     0          348053
     1          357792
     2          169023
     3           48656
     4            8571
     5             960
     6              65

So the probability of at least one paired edge is (933120-348053)/933120 = 585067/933120, or approximately 62.7%.