# Happy 8

The purpose of the game, as with the Lucky 7 puzzle, is to return
the counters into their *Home position*, position they were in before reshuffling. The puzzle consists of several counters
placed on two intersecting circles. There are six points of control: the two central counters and the two pairs of counters adjacent to the central ones.
In the case of 8 counters, these are numbered 1, 8, 7, 3, 4, 5. Clicking on the upper counters induces clockwise rotations.
Clicking on the lower counters results in counterclockwise rotations. Left counters rotate the left circle. Right counters rotate the right circle.
The middle ones handle the whole set.

If the "Cycles" box is checked, a permutation of counters needed to solve the puzzle is displayed. The permutation is presented as a product of cycles.

What if applet does not run? |

In the general framework of Graph Theory, R.Wilson gave a complete treatment of puzzles on graphs of which the Lucky 7 is a particular case. His results do not directly apply to the Happy 8 puzzles. However, combining some of his auxiliary Lemmas I can claim the following

## Theorem

Let's denote the three available (*counterclockwise*) moves *r* (right cycle), *l* (left cycle), and *w* (whole set cycle).

Let *n+2* be the number of counters on each of the two puzzle cycles. The total number of counters is thus
*2(n+1)*. If *n* is odd then every even permutation can be obtained using *r* and *l* cycles.
If, in addition, *w* is also used, then every starting configuration becomes solvable.
If *n* is even, the puzzle is solvable for any starting configuration even
if only *r* and *l* cycles are used.

## Proof

The proof hinges on the ability to demonstrate that the two conditions of Lemma 5 are satisfied.
The fact that the group <*r*, *l* > generated by the two basic moves is 2-transitive
is tedious but otherwise straightforward (see a similar statement for Blithe 12.)
Less trivial but no less tedious is a demonstration that <*r*, *l* > contains a 3-cycle.

Counters are numbered 1,2,...,n+1,n+2,...,2n+1,2(n+1). The proof consists in constructing a
sequence of permutations (mostly applying Lemma 1) such that the last
permutation in the sequence is a 3-cycle. To make the derivations more intuitive, let's use the following numbering
of counters: a_{1},a_{2},...,a_{n+1},b_{1},b_{2},...,b_{n},b_{n+1}, where a_{1} = 1 and so on counterclockwise.

In these notations

*l*= (a

_{1},a

_{2},...,a

_{n+1},b

_{n+1})

*r*= (b

_{1},b

_{2},...,b

_{n+1},a

_{n+1})

Define *p* = *l**r* and *q* = *r**l*. We find that

*p*= (a

_{1}a

_{2}... a

_{n}b

_{1}b

_{2}... b

_{n}b

_{n+1})

*q*= (b

_{1}b

_{2}... b

_{n}a

_{1}a

_{2}... a

_{n}a

_{n+1})

Now let's start forming our sequence of permutations:

s_{1} | = p^{-1}q | = (a_{1} b_{n+1})(b_{1} a_{n+1}) |

s_{2} | = q^{-1}s_{1} q | = (a_{2} b_{n+1})(b_{1} b_{2}) |

s_{3} | = q^{-1}s_{2} q | = (a_{3} b_{n+1})(b_{2} b_{3}) |

s_{4} | = p^{n}s_{1} p^{-n} | = (a_{1} a_{n+1})(b_{1} b_{2}) |

s_{5} | = s_{2} s_{4} | = (a_{1} a_{n+1})(a_{2} b_{n+1}) |

s_{6} | = q^{-1}s_{5} q | = (a_{2} b_{1})(a_{3} b_{n+1}) |

s_{7} | = p^{n-1}s_{1} p^{-(n-1)} | = (a_{2} a_{n+1})(b_{2} b_{3}) |

s_{8} | = s_{3} s_{7} | = (a_{2} a_{n+1})(a_{3} b_{n+1}) |

s_{9} | = s_{6} s_{8} | = (a_{2} a_{n+1})(a_{2} b_{1}) |

The latter is a 3-cycle.

## Related material
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## Sliders and Puzzles on Graphs | |

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