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Treasure Arrow


July, 2005: Treasure Arrow

Dr. Ecco Solution

Solution to "The Pirates' Cantilever," DDJ, June 2005. These solutions are all due to Mike Birken.

  1. (See Figure 1.) You might think that you want the lightest on top of the heaviest, but it turns out you want just the opposite. You put the 1-meter plank on the bottom, then the 2, then the 3, all the way up to the 10. The 10-meter plank extends out a little more than 10.06 meters. Here are the left starting positions relative to the dock edge. Negative values are above the dock. Zero is the edge.
  2. 10: 0.06106185941435823
    9: -3.938938140585642
    8: -5.070517087954062
    7: -5.2557022731392475
    6: -4.976290508433366
    5: -4.426290508433366
    4: -3.704068286211143
    3: -2.8673335923335923
    2: -1.953872053872054
    1: -0.9909090909090909

  3. (See Figure 2.) When several planks can lie on a single one, you can extend out to nearly 13 meters.
  4. 7: -6.216239067837533
    6: -8.716239067837533
    4: -8.100854452452918
    3: -7.5714426877470355
    2: -6.796442687747036
    1: -5.887351778656127
    10: -4.909090909090909
    8: 4.977272727272727
    9: -0.022727272727273373
    5: -2.5227272727272734

Mike explains his solution to this second problem as follows: "My goal was to use the three longest planks to reach out as far as possible from the table edge. I started by treating plank 10 like a balance scale with planks 1-7 on the left side and planks 8 and 9 on the right. The largest value that can be moved from the left stack to the right stack while keeping the left stack heavier than the right is 5. I used plank 5 to counterbalance the weight of plank 8 on 9. The final stacks enabled me to push plank 10 a little more than halfway off the table. To make the solution more interesting, the stacking order of the left stack was intentionally selected to appear as precarious as possible, though mathematically stable."


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