Trying to Solve Math Problem for Real World Use - Combinatorics
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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
combinatorics combinatorial-designs
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add a comment |
up vote
6
down vote
favorite
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
combinatorics combinatorial-designs
New contributor
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
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
combinatorics combinatorial-designs
New contributor
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
combinatorics combinatorial-designs
combinatorics combinatorial-designs
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New contributor
edited 10 hours ago
Acccumulation
6,6412616
6,6412616
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asked 16 hours ago
Rand
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311
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2 Answers
2
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oldest
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up vote
8
down vote
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.
2
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.
– saulspatz
14 hours ago
1
@saulspatz, thanks for the observation, which I have included in an edit I was working on.
– Peter Taylor
14 hours ago
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.
– Peter LeFanu Lumsdaine
14 hours ago
add a comment |
up vote
5
down vote
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
1
This happens to be the third of Peter Taylor's eight possibilities (i.e. his "almost odds-up evens-down").
– r.e.s.
13 hours ago
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
– Henry
8 hours ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
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.
2
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.
– saulspatz
14 hours ago
1
@saulspatz, thanks for the observation, which I have included in an edit I was working on.
– Peter Taylor
14 hours ago
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.
– Peter LeFanu Lumsdaine
14 hours ago
add a comment |
up vote
8
down vote
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.
2
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.
– saulspatz
14 hours ago
1
@saulspatz, thanks for the observation, which I have included in an edit I was working on.
– Peter Taylor
14 hours ago
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.
– Peter LeFanu Lumsdaine
14 hours ago
add a comment |
up vote
8
down vote
up vote
8
down vote
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.
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.
edited 14 hours ago
answered 15 hours ago
Peter Taylor
8,59712240
8,59712240
2
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.
– saulspatz
14 hours ago
1
@saulspatz, thanks for the observation, which I have included in an edit I was working on.
– Peter Taylor
14 hours ago
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.
– Peter LeFanu Lumsdaine
14 hours ago
add a comment |
2
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.
– saulspatz
14 hours ago
1
@saulspatz, thanks for the observation, which I have included in an edit I was working on.
– Peter Taylor
14 hours ago
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.
– Peter LeFanu Lumsdaine
14 hours ago
2
2
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.
– saulspatz
14 hours ago
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.
– saulspatz
14 hours ago
1
1
@saulspatz, thanks for the observation, which I have included in an edit I was working on.
– Peter Taylor
14 hours ago
@saulspatz, thanks for the observation, which I have included in an edit I was working on.
– Peter Taylor
14 hours ago
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.– Peter LeFanu Lumsdaine
14 hours ago
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.– Peter LeFanu Lumsdaine
14 hours ago
add a comment |
up vote
5
down vote
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
1
This happens to be the third of Peter Taylor's eight possibilities (i.e. his "almost odds-up evens-down").
– r.e.s.
13 hours ago
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
– Henry
8 hours ago
add a comment |
up vote
5
down vote
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
1
This happens to be the third of Peter Taylor's eight possibilities (i.e. his "almost odds-up evens-down").
– r.e.s.
13 hours ago
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
– Henry
8 hours ago
add a comment |
up vote
5
down vote
up vote
5
down vote
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
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
answered 15 hours ago
Mike Earnest
19.8k11950
19.8k11950
1
This happens to be the third of Peter Taylor's eight possibilities (i.e. his "almost odds-up evens-down").
– r.e.s.
13 hours ago
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
– Henry
8 hours ago
add a comment |
1
This happens to be the third of Peter Taylor's eight possibilities (i.e. his "almost odds-up evens-down").
– r.e.s.
13 hours ago
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
– Henry
8 hours ago
1
1
This happens to be the third of Peter Taylor's eight possibilities (i.e. his "almost odds-up evens-down").
– r.e.s.
13 hours ago
This happens to be the third of Peter Taylor's eight possibilities (i.e. his "almost odds-up evens-down").
– r.e.s.
13 hours ago
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
– Henry
8 hours ago
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
– Henry
8 hours ago
add a comment |
Rand is a new contributor. Be nice, and check out our Code of Conduct.
Rand is a new contributor. Be nice, and check out our Code of Conduct.
Rand is a new contributor. Be nice, and check out our Code of Conduct.
Rand is a new contributor. Be nice, and check out our Code of Conduct.
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