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No Card Game I Know Of!

Posted by Karl Sharman on July 29, 2011 – 10:45 am

You have a hand of cards that depict each of the letters of the alphabet. You fan the cards and see they are in the following order:

F, W, X, Y, Z, K, L, A, T, U, J, N, O, S, D, E, P, V, C, B, G, H, I, Q, R, M

Your job is to put the cards in alphabetical order in the minimum number of moves. You may remove any number of consecutive cards and insert them anywhere else in the fan; this counts as one move.

Of course, you must preserve the order of the cards you are moving, and keep all cards facing the same way.

Describe how you rearrange the cards in this (example) format:
1. Move D-K to between S and T.

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7 Responds so far- Add one»

  1. 1. cazayoux Said:

    I was able to do it in 7 moves.

  2. 2. Karl Sharman Said:

    That’s a lot of moves cazayoux – it can be done in less than that!

  3. 3. cazayoux Said:

    I was able to do it in 6 moves various ways… however, still searching for an optimal way.
    I have in my head how to map the possible moves in a manner that reveals the quickest route… just need to get it on the whiteboard.
    fun one!

  4. 4. cazayoux Said:

    Some of the letters are already in a run, such as WXYZ.
    I would consider this one segment.
    In total, we begin with 16 segments
    F,WXYZ,KL,A,TU,J,NO,S,DE,P,V,C,B,GHI,QR,M

    The best case scenarios is if we move a series of segments and, after combining congruous segments, we can reduce the number of segments by three. Of the series of segments moving, each end would reduce the segment count. As well, the gap left behind by the series of segments closes would reduce the segment count as well.

    For example, when we move J-P between IQ, then we have
    F,WXYZ,KL,A,TU,V,C,B,GHI,J,NO,S,DE,P,QR,M
    we can combine at UV, IJ, and PQ
    F,WXYZ,KL,A,TUV,C,B,GHIJ,NO,S,DE,PQR,M

    We have reduced the segment count by 3 to now have 13 segments.

    If we reduce by 3 each time, then it will take four more moves to get to 1 (goal!).
    16 – 3 – 3 – 3 – 3 – 3 = 1 in five moves.

    So I’m looking for a way to reduce the segment count by three each time.

    However, I’m having trouble doing this.
    If any move results in a segment reduction of only two, then the best I can do is accomplish the goal in 6 moves.

    Now that I think about it, to get the “A” on the left or the “Z” on the right, a 3-segment-reduction is not possible.

    Does this mean 6 moves is the minimum possible to accomplish the task?

    ………F,WXYZ,KL,A,TU,V,C,B,GHI,J,NO,S,DE,P,QR,M

    1. Move J-P to between I and Q. (same as V-I to U..J) [-3]
    ………F,WXYZ,KL,A,TUV,C,B,GHIJ,NO,S,DE,PQR,M

    2. Move T-C to between S and D. (same as B-S to A..T) [-3]
    ………F,WXYZ,KL,AB,GHIJ,NO,STUV,CDE,PQR,M

    3. Move W-B to between V and C. (same as G-V to F..W) [-3]
    ………FGHIJ,NO,STUVWXYZ,KL,ABCDE,PQR,M

    4. Move F-O to between E and P (same as S-E to _..F) [-2]
    ………STUVWXYZ,KL,ABCDEFGHIJ,NOPQR,M

    5. Move S-L to between R and M (same as A-R to _..S) [-2]
    ………ABCDEFGHIJ,NOPQRSTUVWXYZ,KLM

    6. Move K-M to between J and N (same as N-Z to M.._) [-2]
    ………ABCDEFGHIJKLMNOPQRSTUVWXYZ

    There are many scenarios for the last three steps.
    (For the moment) I’m convinced it cannot be done in less than 6 moves.

    :)

  5. 5. Karl Sharman Said:

    Cazayoux – as you seem to be the only one trying this puzzle, I will tell you that you are correct!

    J-P between I and Q
    B-S between A and T
    G-V between F and W
    N-Z after M
    P-R between O and S
    F-L between E and M

    I like the logical approach you took!

  6. 6. cazayoux Said:

    Thanks.
    It may not have been the most popular, but I enjoyed it.

  7. 7. Scott Said:

    I was attempting it, but my methods were quite crude and could only achieve 10 moves…
    Nice work Cazayoux.

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