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I'm trying to solve a math problem that hasn't been solved - to anyone's knowledge - in the community it's being used in. I am sure it is not difficult, but I am not smart enough to figure it out.

In England, when on a country shoot (part of Britain's heritage) there are 8 "pegs" (shooting position in a straight line numbered 1-8) and shoot four "drives" (45 minute period of shooting). People draw pegs blind and then there are several ways that people change pegs across the 4 drives. Move two up: 1 goes to 3 goes to 5 goes to 7. Move up three: 1 goes to 4 goes to 7 goes to 2. Odds up 3, evens down 3, etc. 4 and 5 are considered the best "pegs" and 1 and 8 are considered the worst.

The questions is this: How would you solve this problem trying to solve for two different parameters: 1) Everyone get an equal distribution of being at 4/5 and 1/8 or at least close to them such that no one is advantaged over the course of the four "drives" and everyone is equally in the center or on the ends. 2) People get to stand next to different people across the course of the day and not always next to the same people (the reason odds up and evens down was invented).

No one particularly likes the current numbering system and many are looking for an alternative where you draw a number sequence as opposed to a number. (IE, you draw a card that has the "peg" order pre-determined for the 8 people - eg 3,1,5,7)

Thanks for your help! :)

Rand

PS Someone tried to solve this problem previously and could only make it work with 9 "pegs" and not 8. See link - https://www.gunsonpegs.com/articles/shooting-talk/alternatives-to-moving-up-2-the-durnford-wheel

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If you want a wheel-like system, by my (corrected) calculation there are 4 possibilities (8 with reversals).

  • $1 \to 2 \to 4 \to 6 \to 8 \to 7 \to 5 \to 3 \to 1$ (almost evens-up odds-down)
  • $1 \to 2 \to 5 \to 3 \to 8 \to 7 \to 4 \to 6 \to 1$
  • $1 \to 3 \to 5 \to 7 \to 8 \to 6 \to 4 \to 2 \to 1$ (almost odds-up evens-down)
  • $1 \to 3 \to 4 \to 2 \to 8 \to 6 \to 5 \to 7 \to 1$
  • $1 \to 7 \to 4 \to 3 \to 8 \to 2 \to 5 \to 6 \to 1$
  • $1 \to 7 \to 5 \to 6 \to 8 \to 2 \to 4 \to 3 \to 1$
  • $1 \to 6 \to 5 \to 2 \to 8 \to 3 \to 4 \to 7 \to 1$
  • $1 \to 6 \to 4 \to 7 \to 8 \to 3 \to 5 \to 2 \to 1$

The first or third seem like they might be easiest to sell from a cultural standpoint.

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  • 3
    $\begingroup$ This is really elegant, and easy to remember. Evens move two up and odds two down, except that those on the ends move one place toward the middle. I wish I could understand CJam. $\endgroup$ – saulspatz Dec 14 '18 at 16:58
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    $\begingroup$ @saulspatz, thanks for the observation, which I have included in an edit I was working on. $\endgroup$ – Peter Taylor Dec 14 '18 at 17:01
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    $\begingroup$ You can imagine numbers 1 and 3 as having people going round a conveyor belt, with the evens on the “upper” side and the odds on the “lower” side. E.g. for number 3, this is a conveyor belt like /_⁻⁻_⁻⁻_⁻⁻/, going clockwise. $\endgroup$ – Peter LeFanu Lumsdaine Dec 14 '18 at 17:20
  • $\begingroup$ Thank you for your effort on this! I like these. What is the different between the first half of the sequences and the second half? When I look at something like number 5 on the list, what I like about it is that you get to move from middle to end to middle to end and it would appear get to stand next to the largest number of different people. $\endgroup$ – Rand Dec 18 '18 at 2:03
  • $\begingroup$ To be more clear, I was asking how you approached the second set of series from a calculation perspective when I said "What's the difference". $\endgroup$ – Rand Dec 18 '18 at 2:11
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How about this schedule? Everyone gets one drive on the end and one in the middle, and also one drive which is one from the middle and one which is one from the end. No two people stand together twice.

Letters are people, rows are drives, columns are pegs.

A B H C G D F E
B C A D H E G F
C D B E A F H G
D E C F B G A H
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    $\begingroup$ This happens to be the third of Peter Taylor's eight possibilities (i.e. his "almost odds-up evens-down"). $\endgroup$ – r.e.s. Dec 14 '18 at 18:20
  • $\begingroup$ It is implicit, but not only do no two people stand together twice, but also each pair of people are next to each other once $\endgroup$ – Henry Dec 14 '18 at 23:12
  • $\begingroup$ @Henry - If I'm not mistaken, those properties are also implicit in all eight of Peter Taylor's solutions. $\endgroup$ – r.e.s. Dec 16 '18 at 16:34
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    $\begingroup$ @r.e.s., that each pair stand together once is implicit in the requirement that no pair stand together twice. $\endgroup$ – Peter Taylor Dec 16 '18 at 18:51
  • $\begingroup$ Thanks Mike. Appreciate your time on this. Looks like you and Peter came up with a similar approach (Peter's number 3) on this particular one. $\endgroup$ – Rand Dec 18 '18 at 2:12

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