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 Rainbow Towers (Posted on 2012-02-23)

 No Solution Yet Submitted by brianjn No Rating

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 Part 1 plus a thought (spoiler) | Comment 1 of 4
Well, I'll start.  The pictured case can be done in 10 moves:

R1 --> 4 (ie, smallest red to spindle 4).
B2 --> 5
Y1 --> 5
R2 --> 3
R1 --> 3  (finishes the red tower)

B1 --> 4
Y2 --> 2
Y1 --> 2 (finishes the yellow tower)

B2 --> 1
B1 --> 1 (finishes the blue tower)

The question wasn't asked, but what about N towers of 3 discs and two empty spindles?

I believe that the lowest disc never needs to move.  Let's therefore designate each spindle as the home spindle for the color of its starting base disc.

Well, if they are set up just right, so that the middle disc in each is to the left of its home spindle, and the smallest disc is on its home spindle (like in the picture), then they can be rearranged using 5 moves for the first spindle to be fixed, 2 moves for the last spindle to be fixed, and 3 moves apiece for all the rest.
5 + 2 + 3(N-2) = 3N + 1 moves. 10 moves if N = 3 (as above), 13 moves if N = 4.

But initial starting position affects the count.
Consider if there are an even number of colors, where spindles are paired and the middle disk of each pair needs to be exchanged with the middle disk of its' paired spindle.  For instance:

Spindle          1          2        3       4         5       6
R1       B1       Y1     G1
B2       R2       G2     Y2
R3       B3       Y3      G3

It takes 7 moves to complete the R and B towers, and 7 more moves to complete the Y and G towers.  A total of 14 moves.  For N colors, a total of 7N/2 moves.

And of course, there are other possible configurations, so maybe 7N/2 is not even the maximum.

Maybe that is why this question wasn't asked, because the number of moves depends on the original configuration.  Or maybe this bodes ill for Part 2, 3, and 4.  I am scared off, and will let somebody else carry on from here.

Edited on February 23, 2012, 8:21 pm
 Posted by Steve Herman on 2012-02-23 20:13:46

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