All Pentominoes in 5

December 13th, 2010 by munizao Leave a reply »

I’ve been thinking about variations on the problem of cycling through all twelve pentominoes by moving a single cell at a time. (I wrote about this in a previous post.) Constraining the way that the squares are allowed to move led to something almost like a chess problem.

The problem:

Starting with the above position, take five turns as follows:

A turn consists of moving one white knight, then moving one black knight, according to standard chess rules.

After each turn, the squares occupied by the ten knights must form two separate pentominoes.

After the fifth turn, all twelve pentominoes must have appeared exactly once. (This includes the two that are present in the starting position.)

[I may make a separate post discussing and spoiling the puzzle later.]


  1. Bryce Herdt says:

    First clue: the knights must mingle. Otherwise, the only legal moves to or from each of W and Z would be on P, so keeping the white and black knights separate is a no-go.

    By the way, where you say the knight must form two separate pentominoes, do you mean the pentominoes shouldn’t share an edge, or are you using a different definition?

  2. munizao says:

    do you mean the pentominoes shouldn’t share an edge, or are you using a different definition?

    Right. They can share a corner, but not an edge.

  3. Kate Jones says:

    Forming all 12 pentominoes by moving one square at a time is also the theme in my Fractured Fives puzzle, – hard! And our Leap set includes a series of doubled pentominoes and hexominoes formed of alternating black and white knights to be changed into any of the other doubled pentominoes and hexominoes in the minimum of knight’s moves – Interestingly and obviously, only those polyominoes that fit within a 3×3 square are used. If you like, I can mail you the entire Leap booklet. It includes the stats.

  4. munizao says:

    One thing I was informed of after I posted this is that G. P. Jelliss (in his Games and Puzzles Journal, issue 12, p. 189) posed the problem of starting with a pentomino and sequentially generating the other 11 via a series of king moves. The sequence of pentominoes formed in solving this is unique up to reversal. (I believe I considered this problem before I posted the above, but I missed one of the possible moves, so I thought that there was no solution.) And yes, I would be interested in seeing the Leap booklet.

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