A way of comparing difficulty...?

As a method for comparing the difficulty of the various versions of  Dioctipoid, expressing
the number of possible legal solution permutations against the total number of permutations
as a ratio is maybe as good a way as any.

The total number of possible permutations of the segments on a Dioctipoid is going to be
really large. If I have analysed the mechanism correctly this number K   can be calculated
as follows.

There are 6 central square segments that can be arranged in 6! (factorial) ways, each of
which can be arranged in 2 orientations, giving 2   possibilities for each permutation of the
centre pieces.

Only ½ * of the permutations have the rotations of the central squares in possible

There are 24 triangle segments that behave in quite a counter-intuitive manner.
A triangle with the acute apex touching "a pole" or rotation axis will not migrate to the
opposing pole for instance and thus these segments form two distinct sets of 12 for
analysis purposes.

These segments have a single possible orientation so each set has 12! possible  
permutations. If we discount the first set to 11! positions this provides for a global  
reference frame by holding one of the pieces fixed.

There are 12 diamond segments which can be arranged in 12! ways. By observation, it
can be shown that each diamond  has a single possible orientation.

But in the Dioctipoid, the permutations of the pieces are dependent on the positions of
related pieces and only ½ * of these have the correct segment arrangement parity; we can
not manipulate a single piece in isolation for instance.

This gives:                                                                              
( 6! · 2   · 11! · 12! · 12!  )
(2 · 2)


or   1.055 x 10     possible permutations.

* This parity figure of ½ applied to the whole device as per the MIT/Rubik                  

By way of comparison, a similar calculation** for Rubik’s Cube (K  ) can also be
performed :                                                                       

(8! ·  3    · 12! · 2    · 4    )
(3 · 2 · 2 · 2)

=  88,580,102,706,155.200,000,000  

or   8.86 x 10     possible permutations.

** factoring in the numbers for the orientation of the centre squares in order to compare
like with like, irrelevant in Rubik‘s design.




This article has drawn heavily on a paper produced by the
Massachusetts Institute of Technology
describing the mathematics of Rubik’s Cube. This original work can be found at :

Moreover, I made a complete hash of my first attempt at this page. As I still have not
solved my own puzzle I took a far too simplistic approach in my analysis of the

Thanks to bmenrigh, Konrad et al. at the                     
for putting me straight on a number of points.

Still not certain that I have this right. The math involved to prove this to an academic
standard is beyond my capability but I'll throw it out for comment.

                               Thank you,      
                                                            (Any mistakes herein are
still mine).
                                                                                      26th March 2013
Designersaurus Rex Ltd.
Moulding Innovations Ltd.
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