Well, I was playing with the UFO and please allow me to share a solution that is good for those who don't want to remember too many routines or generators!

Rubik's UFO numbering:

Top side:

*** 6 *** 1 ***

5 ********* 2

*** 4 *** 3 ***

Bottom side (imaginary view from the top side):

***12 *** 7 ***

11 ********* 8

*** 10 *** 9 ***

There are four generators in total

a = vertical right flip

b = lower (diagonal) right flip

c = upper (diagonal) right flip

d = clockwise rotation of front side

I used the GAP program to find those sequences (don't get me wrong, there may be simpler ones, but those I already found are easy enough to remember).

The code shows more clearly how the generators a,b,c and d act on the puzzle.

Code for generating the group:

a:=(1,9)(2,8)(3,7);

b:=(2,10)(3,9)(4,8);

c:=(1,7)(6,8)(2,12);

d:=(1,2,3,4,5,6);

G:=Group(a,b,c,d);

Order(G);

The required algorithms (

which are the only things required to remember in order to solve the puzzle) are two:

ddda da dbc (must be applied

SEVEN times) which swaps 1 and 8 and keeps all the rest invariant.

ad bdddd cd (mus be applied

THREE times) which swaps 1 and 7 and keeps all the rest invariant.

corresponding GAP code:

(d^3*a*d*a*d*b*c*)^7; equals (1,8)

(a*d*b*d^4*c*d)^3; equals (1, 7)

It can be mathematically proved, that the above two algorithms are enough to bring the Rubik's UFO into a state where only flips (and no rotations) are required to completely solve it. It all depends on targeting the correct pieces you need to swap. And since there are only 96 combinations of them in that state, there shouldn't be any trouble... right? RIGHT???

I realise that there may be simpler formulas, but in a memory point of view (where, the less defined, the better) and if you don't mind applying a simple algorithm seven times (until your fingers get blisters), I think this is not a bad method!

I hope you like it!

Peter

PS In most of times I mixed it and solved it, I only used the first algorithm!!! (that swaps 1 and 8).