A committee consists of 12 members. Each pair of members of the committee either mutually like each other, or...
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A committee consists of 12 members. Each pair of members of the committee either mutually like each other, or mutually dislike each other. Each member likes exactly 6 other members. Is it necessarily possible to sit all 12 members around a round table so that any two adjacent members like each other?
combinatorics
asked Jan 19 at 5:45
Anson ChanAnson Chan
162
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add a comment |
$begingroup$
A committee consists of 12 members. Each pair of members of the committee either mutually like each other, or mutually dislike each other. Each member likes exactly 6 other members. Is it necessarily possible to sit all 12 members around a round table so that any two adjacent members like each other?
combinatorics
asked Jan 19 at 5:45
Anson ChanAnson Chan
162
$endgroup$
1
$begingroup$
Thoughts on this question? Have you tried with smaller numbers like a total of $4$ members with every member liking exactly $2$ others? This will give you an idea of whether or not such a situation is possible.
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– астон вілла олоф мэллбэрг
Jan 19 at 5:48
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Where does this problem come from?
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– Lord Shark the Unknown
Jan 19 at 5:48
$begingroup$
Yes, this can be shown to be true using Hall's marriage theorem. This is made easier by the fact that all likes are reciprocal and that, because the member is not counted among the 6 members they like, even a group of 6 has to like someone outside the group.
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– Old Pro
Jan 19 at 6:43
add a comment |
$begingroup$
A committee consists of 12 members. Each pair of members of the committee either mutually like each other, or mutually dislike each other. Each member likes exactly 6 other members. Is it necessarily possible to sit all 12 members around a round table so that any two adjacent members like each other?
combinatorics
asked Jan 19 at 5:45
Anson ChanAnson Chan
162
$endgroup$
A committee consists of 12 members. Each pair of members of the committee either mutually like each other, or mutually dislike each other. Each member likes exactly 6 other members. Is it necessarily possible to sit all 12 members around a round table so that any two adjacent members like each other?
combinatorics
combinatorics
asked Jan 19 at 5:45
Anson ChanAnson Chan
162
asked Jan 19 at 5:45
Anson ChanAnson Chan
162
asked Jan 19 at 5:45
Anson ChanAnson Chan
162
asked Jan 19 at 5:45
Anson ChanAnson Chan
162
asked Jan 19 at 5:45
Anson ChanAnson Chan
162
162
1
$begingroup$
Thoughts on this question? Have you tried with smaller numbers like a total of $4$ members with every member liking exactly $2$ others? This will give you an idea of whether or not such a situation is possible.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 5:48
$begingroup$
Where does this problem come from?
$endgroup$
– Lord Shark the Unknown
Jan 19 at 5:48
$begingroup$
Yes, this can be shown to be true using Hall's marriage theorem. This is made easier by the fact that all likes are reciprocal and that, because the member is not counted among the 6 members they like, even a group of 6 has to like someone outside the group.
$endgroup$
– Old Pro
Jan 19 at 6:43
add a comment |
1
$begingroup$
Thoughts on this question? Have you tried with smaller numbers like a total of $4$ members with every member liking exactly $2$ others? This will give you an idea of whether or not such a situation is possible.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 5:48
$begingroup$
Where does this problem come from?
$endgroup$
– Lord Shark the Unknown
Jan 19 at 5:48
$begingroup$
Yes, this can be shown to be true using Hall's marriage theorem. This is made easier by the fact that all likes are reciprocal and that, because the member is not counted among the 6 members they like, even a group of 6 has to like someone outside the group.
$endgroup$
– Old Pro
Jan 19 at 6:43
1
1
$begingroup$
Thoughts on this question? Have you tried with smaller numbers like a total of $4$ members with every member liking exactly $2$ others? This will give you an idea of whether or not such a situation is possible.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 5:48
$begingroup$
Thoughts on this question? Have you tried with smaller numbers like a total of $4$ members with every member liking exactly $2$ others? This will give you an idea of whether or not such a situation is possible.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 5:48
$begingroup$
Where does this problem come from?
$endgroup$
– Lord Shark the Unknown
Jan 19 at 5:48
$begingroup$
Where does this problem come from?
$endgroup$
– Lord Shark the Unknown
Jan 19 at 5:48
$begingroup$
Yes, this can be shown to be true using Hall's marriage theorem. This is made easier by the fact that all likes are reciprocal and that, because the member is not counted among the 6 members they like, even a group of 6 has to like someone outside the group.
$endgroup$
– Old Pro
Jan 19 at 6:43
$begingroup$
Yes, this can be shown to be true using Hall's marriage theorem. This is made easier by the fact that all likes are reciprocal and that, because the member is not counted among the 6 members they like, even a group of 6 has to like someone outside the group.
$endgroup$
– Old Pro
Jan 19 at 6:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$12$-gon construction">
this is an example, every line connect two members mutually like another, and for two members not connected by a line, they mutually dislike others.
This is quite symmetric, but I'm not sure if there are other symmetric or non-symmetric solutions.
Edit: This table is constructed before we actually know who likes which people, if the relation between likes and dislikes are given first, then it should apply more effort to ensure the existence of the permutation.
answered Jan 19 at 6:07
kelvin hong 方kelvin hong 方
62018
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add a comment |
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1 Answer
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$begingroup$
$12$-gon construction">
this is an example, every line connect two members mutually like another, and for two members not connected by a line, they mutually dislike others.
This is quite symmetric, but I'm not sure if there are other symmetric or non-symmetric solutions.
Edit: This table is constructed before we actually know who likes which people, if the relation between likes and dislikes are given first, then it should apply more effort to ensure the existence of the permutation.
answered Jan 19 at 6:07
kelvin hong 方kelvin hong 方
62018
$endgroup$
add a comment |
$begingroup$
$12$-gon construction">
this is an example, every line connect two members mutually like another, and for two members not connected by a line, they mutually dislike others.
This is quite symmetric, but I'm not sure if there are other symmetric or non-symmetric solutions.
Edit: This table is constructed before we actually know who likes which people, if the relation between likes and dislikes are given first, then it should apply more effort to ensure the existence of the permutation.
answered Jan 19 at 6:07
kelvin hong 方kelvin hong 方
62018
$endgroup$
add a comment |
$begingroup$
$12$-gon construction">
this is an example, every line connect two members mutually like another, and for two members not connected by a line, they mutually dislike others.
This is quite symmetric, but I'm not sure if there are other symmetric or non-symmetric solutions.
Edit: This table is constructed before we actually know who likes which people, if the relation between likes and dislikes are given first, then it should apply more effort to ensure the existence of the permutation.
answered Jan 19 at 6:07
kelvin hong 方kelvin hong 方
62018
$endgroup$
$12$-gon construction">
this is an example, every line connect two members mutually like another, and for two members not connected by a line, they mutually dislike others.
This is quite symmetric, but I'm not sure if there are other symmetric or non-symmetric solutions.
Edit: This table is constructed before we actually know who likes which people, if the relation between likes and dislikes are given first, then it should apply more effort to ensure the existence of the permutation.
answered Jan 19 at 6:07
kelvin hong 方kelvin hong 方
62018
answered Jan 19 at 6:07
kelvin hong 方kelvin hong 方
62018
answered Jan 19 at 6:07
kelvin hong 方kelvin hong 方
62018
answered Jan 19 at 6:07
kelvin hong 方kelvin hong 方
62018
62018
add a comment |
add a comment |
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1
$begingroup$
Thoughts on this question? Have you tried with smaller numbers like a total of $4$ members with every member liking exactly $2$ others? This will give you an idea of whether or not such a situation is possible.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 5:48
$begingroup$
Where does this problem come from?
$endgroup$
– Lord Shark the Unknown
Jan 19 at 5:48
$begingroup$
Yes, this can be shown to be true using Hall's marriage theorem. This is made easier by the fact that all likes are reciprocal and that, because the member is not counted among the 6 members they like, even a group of 6 has to like someone outside the group.
$endgroup$
– Old Pro
Jan 19 at 6:43