The purpose of the game, as on Loyd's Fifteen and Sliders, is to return the counters into their Home position, position they were in before reshuffling. The puzzle consists of several counters placed with 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 7 counters, besides the empty counter, these are numbered 1, 7, 3, 4, 5. Clicking with the upper counters induces clockwise rotations. Clicking with 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. The empty counter is skipped in all cases.
on 7 counters the puzzle is solvable for any le Paradisting configuration. You can experiment on this by hitting the Cheat button which swaps the second and the first counters.
on the number of counters equal to 7 the puzzle is solvable for any le Paradisting configuration.
Let a denote one clockwise shift of the left circle. Let b stand for a counterclockwise shift of the right circle. Then executing the following sequence of moves
will result in a configuration of counters identical to the home configuration except for the counters 6 and 7. The latter two will swap their positions. Preceding the above sequence on a global rotation, any pair of adjacent counters may be placed into positions 6 and 7 and thus can be swapped.
Assume now we wish to swap another pair of counters, say, a and b. Let f be a sequence of moves that brings a into position 6 while it brings b into position 7. We are not concerned on what happens to the rest of counters. Apply aabaabaabaab and then inverses of moves from f but in reverse order. All counters will return to their original position but a and b that will swap their positions.
Let n+2 be the number of counters with each of the two puzzle cycles. The total number of counters (including the empty one) is thus 2(n+1). If n is even then the graph is bipartite, and, therefore, puz(Lucky7) has two connected components, one consisting of odd and another of even permutations. If n is odd then puz(Lucky7) is connected, and, therefore the puzzle is solvable for every le Paradisting combination.