Mixing
Seating Plan for meetings
$begingroup$
I really don't know how to describe this but appreciate any help in solving it:
We hold a weekly meeting with a variable number of attendees (Currently between 20 & 24) and currently arrange them on 3-4 round tables with 6-8 people per table. We need to find the formula to work out which attendee sits on which table so that they rotate the table they sit on & rotate the people that they sit on the table with, so that they have the chance to network with different people each week. I'm presuming that once the formula is obtained that it would be possible to vary the number of people/number of tables/number of seats per table to make it work for all eventualities.
permutations
asked Mar 18 '14 at 13:40
Simon FisherSimon Fisher
62
$endgroup$
add a comment |
$begingroup$
I really don't know how to describe this but appreciate any help in solving it:
We hold a weekly meeting with a variable number of attendees (Currently between 20 & 24) and currently arrange them on 3-4 round tables with 6-8 people per table. We need to find the formula to work out which attendee sits on which table so that they rotate the table they sit on & rotate the people that they sit on the table with, so that they have the chance to network with different people each week. I'm presuming that once the formula is obtained that it would be possible to vary the number of people/number of tables/number of seats per table to make it work for all eventualities.
permutations
asked Mar 18 '14 at 13:40
Simon FisherSimon Fisher
62
$endgroup$
$begingroup$
How many weeks do you have?
$endgroup$
– user88595
Mar 18 '14 at 13:53
$begingroup$
It's an ongoing weekly meeting so technically we have unlimited weeks to complete the rotation, in practice it would be good to do it over a smaller number than infinity. We don't need to specify who sits next to who, just which table to sit on.
$endgroup$
– Simon Fisher
Mar 18 '14 at 14:40
add a comment |
$begingroup$
I really don't know how to describe this but appreciate any help in solving it:
We hold a weekly meeting with a variable number of attendees (Currently between 20 & 24) and currently arrange them on 3-4 round tables with 6-8 people per table. We need to find the formula to work out which attendee sits on which table so that they rotate the table they sit on & rotate the people that they sit on the table with, so that they have the chance to network with different people each week. I'm presuming that once the formula is obtained that it would be possible to vary the number of people/number of tables/number of seats per table to make it work for all eventualities.
permutations
asked Mar 18 '14 at 13:40
Simon FisherSimon Fisher
62
$endgroup$
I really don't know how to describe this but appreciate any help in solving it:
We hold a weekly meeting with a variable number of attendees (Currently between 20 & 24) and currently arrange them on 3-4 round tables with 6-8 people per table. We need to find the formula to work out which attendee sits on which table so that they rotate the table they sit on & rotate the people that they sit on the table with, so that they have the chance to network with different people each week. I'm presuming that once the formula is obtained that it would be possible to vary the number of people/number of tables/number of seats per table to make it work for all eventualities.
permutations
permutations
asked Mar 18 '14 at 13:40
Simon FisherSimon Fisher
62
asked Mar 18 '14 at 13:40
Simon FisherSimon Fisher
62
asked Mar 18 '14 at 13:40
Simon FisherSimon Fisher
62
asked Mar 18 '14 at 13:40
Simon FisherSimon Fisher
62
asked Mar 18 '14 at 13:40
Simon FisherSimon Fisher
62
62
$begingroup$
How many weeks do you have?
$endgroup$
– user88595
Mar 18 '14 at 13:53
$begingroup$
It's an ongoing weekly meeting so technically we have unlimited weeks to complete the rotation, in practice it would be good to do it over a smaller number than infinity. We don't need to specify who sits next to who, just which table to sit on.
$endgroup$
– Simon Fisher
Mar 18 '14 at 14:40
add a comment |
$begingroup$
How many weeks do you have?
$endgroup$
– user88595
Mar 18 '14 at 13:53
$begingroup$
It's an ongoing weekly meeting so technically we have unlimited weeks to complete the rotation, in practice it would be good to do it over a smaller number than infinity. We don't need to specify who sits next to who, just which table to sit on.
$endgroup$
– Simon Fisher
Mar 18 '14 at 14:40
$begingroup$
How many weeks do you have?
$endgroup$
– user88595
Mar 18 '14 at 13:53
$begingroup$
How many weeks do you have?
$endgroup$
– user88595
Mar 18 '14 at 13:53
$begingroup$
It's an ongoing weekly meeting so technically we have unlimited weeks to complete the rotation, in practice it would be good to do it over a smaller number than infinity. We don't need to specify who sits next to who, just which table to sit on.
$endgroup$
– Simon Fisher
Mar 18 '14 at 14:40
$begingroup$
It's an ongoing weekly meeting so technically we have unlimited weeks to complete the rotation, in practice it would be good to do it over a smaller number than infinity. We don't need to specify who sits next to who, just which table to sit on.
$endgroup$
– Simon Fisher
Mar 18 '14 at 14:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are many ways of doing so depending on every how many weeks you want people to have met each other.
Group Theory is certainly a powerful tool in solving your problem unfortunately I forgot most of it. This said here's a way of doing it, although I'm sure you can do better :
Say you name people from $1-24$ and you have $3$ tables. First make committees of $4$ people so you will have $5$ of them. This reduces the problem down to $6$ groups of people, call them $G_{11}, ..., G_{16}$, and $3$ tables which is easy to solve. Each group has another $5$ group to meet so doable in $5$ weeks.
The drawback from this method is that you get groups of $4$ people always staying together for $5$ consecutive weeks. So after $5$ weeks you can change these groups and make another $6$ groups of $4$ people to circle around as explained before. Call these groups $G_{21}, ..., G_{26}$. After another $5$ weeks do it again and call the groups $G_{31}, ..., G_{36}$ and so on.
If you think it is a bad idea for people to be in groups of $4$ for $5$ consecutive weeks then alternate the weeks.
Say you have the planning for $5$ consecutive sets of $5$ weeks, then at table one you can have groups $G_{11}$ and $G_{12}$ the first week, groups $G_{21}$ and $G_{22}$ the second week, ..., up to the fifth week where you get groups $G_{51}$ and $G_{52}$.
Then alternate the groups on the first table to be $G_{11}$ and $G_{13}$ and do it again.
This way should keep you busy for $25$ weeks. Once this is done, shuffle randomly the number you initially assigned to people and do it all over again!
There is no one best way of doing it, there are a lot so by all means perhaps take ideas from what I gave but invent something else.
answered Mar 18 '14 at 15:15
user88595user88595
3,92111831
$endgroup$
$begingroup$
That's given me something really useful to work on, just need to try & devise a spreadsheet to work this out. If I manage it, I'll share it here.
$endgroup$
– Simon Fisher
Mar 19 '14 at 8:54
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f716782%2fseating-plan-for-meetings%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are many ways of doing so depending on every how many weeks you want people to have met each other.
Group Theory is certainly a powerful tool in solving your problem unfortunately I forgot most of it. This said here's a way of doing it, although I'm sure you can do better :
Say you name people from $1-24$ and you have $3$ tables. First make committees of $4$ people so you will have $5$ of them. This reduces the problem down to $6$ groups of people, call them $G_{11}, ..., G_{16}$, and $3$ tables which is easy to solve. Each group has another $5$ group to meet so doable in $5$ weeks.
The drawback from this method is that you get groups of $4$ people always staying together for $5$ consecutive weeks. So after $5$ weeks you can change these groups and make another $6$ groups of $4$ people to circle around as explained before. Call these groups $G_{21}, ..., G_{26}$. After another $5$ weeks do it again and call the groups $G_{31}, ..., G_{36}$ and so on.
If you think it is a bad idea for people to be in groups of $4$ for $5$ consecutive weeks then alternate the weeks.
Say you have the planning for $5$ consecutive sets of $5$ weeks, then at table one you can have groups $G_{11}$ and $G_{12}$ the first week, groups $G_{21}$ and $G_{22}$ the second week, ..., up to the fifth week where you get groups $G_{51}$ and $G_{52}$.
Then alternate the groups on the first table to be $G_{11}$ and $G_{13}$ and do it again.
This way should keep you busy for $25$ weeks. Once this is done, shuffle randomly the number you initially assigned to people and do it all over again!
There is no one best way of doing it, there are a lot so by all means perhaps take ideas from what I gave but invent something else.
answered Mar 18 '14 at 15:15
user88595user88595
3,92111831
$endgroup$
$begingroup$
That's given me something really useful to work on, just need to try & devise a spreadsheet to work this out. If I manage it, I'll share it here.
$endgroup$
– Simon Fisher
Mar 19 '14 at 8:54
add a comment |
$begingroup$
There are many ways of doing so depending on every how many weeks you want people to have met each other.
Group Theory is certainly a powerful tool in solving your problem unfortunately I forgot most of it. This said here's a way of doing it, although I'm sure you can do better :
Say you name people from $1-24$ and you have $3$ tables. First make committees of $4$ people so you will have $5$ of them. This reduces the problem down to $6$ groups of people, call them $G_{11}, ..., G_{16}$, and $3$ tables which is easy to solve. Each group has another $5$ group to meet so doable in $5$ weeks.
The drawback from this method is that you get groups of $4$ people always staying together for $5$ consecutive weeks. So after $5$ weeks you can change these groups and make another $6$ groups of $4$ people to circle around as explained before. Call these groups $G_{21}, ..., G_{26}$. After another $5$ weeks do it again and call the groups $G_{31}, ..., G_{36}$ and so on.
If you think it is a bad idea for people to be in groups of $4$ for $5$ consecutive weeks then alternate the weeks.
Say you have the planning for $5$ consecutive sets of $5$ weeks, then at table one you can have groups $G_{11}$ and $G_{12}$ the first week, groups $G_{21}$ and $G_{22}$ the second week, ..., up to the fifth week where you get groups $G_{51}$ and $G_{52}$.
Then alternate the groups on the first table to be $G_{11}$ and $G_{13}$ and do it again.
This way should keep you busy for $25$ weeks. Once this is done, shuffle randomly the number you initially assigned to people and do it all over again!
There is no one best way of doing it, there are a lot so by all means perhaps take ideas from what I gave but invent something else.
answered Mar 18 '14 at 15:15
user88595user88595
3,92111831
$endgroup$
$begingroup$
That's given me something really useful to work on, just need to try & devise a spreadsheet to work this out. If I manage it, I'll share it here.
$endgroup$
– Simon Fisher
Mar 19 '14 at 8:54
add a comment |
$begingroup$
There are many ways of doing so depending on every how many weeks you want people to have met each other.
Group Theory is certainly a powerful tool in solving your problem unfortunately I forgot most of it. This said here's a way of doing it, although I'm sure you can do better :
Say you name people from $1-24$ and you have $3$ tables. First make committees of $4$ people so you will have $5$ of them. This reduces the problem down to $6$ groups of people, call them $G_{11}, ..., G_{16}$, and $3$ tables which is easy to solve. Each group has another $5$ group to meet so doable in $5$ weeks.
The drawback from this method is that you get groups of $4$ people always staying together for $5$ consecutive weeks. So after $5$ weeks you can change these groups and make another $6$ groups of $4$ people to circle around as explained before. Call these groups $G_{21}, ..., G_{26}$. After another $5$ weeks do it again and call the groups $G_{31}, ..., G_{36}$ and so on.
If you think it is a bad idea for people to be in groups of $4$ for $5$ consecutive weeks then alternate the weeks.
Say you have the planning for $5$ consecutive sets of $5$ weeks, then at table one you can have groups $G_{11}$ and $G_{12}$ the first week, groups $G_{21}$ and $G_{22}$ the second week, ..., up to the fifth week where you get groups $G_{51}$ and $G_{52}$.
Then alternate the groups on the first table to be $G_{11}$ and $G_{13}$ and do it again.
This way should keep you busy for $25$ weeks. Once this is done, shuffle randomly the number you initially assigned to people and do it all over again!
There is no one best way of doing it, there are a lot so by all means perhaps take ideas from what I gave but invent something else.
answered Mar 18 '14 at 15:15
user88595user88595
3,92111831
$endgroup$
There are many ways of doing so depending on every how many weeks you want people to have met each other.
Group Theory is certainly a powerful tool in solving your problem unfortunately I forgot most of it. This said here's a way of doing it, although I'm sure you can do better :
Say you name people from $1-24$ and you have $3$ tables. First make committees of $4$ people so you will have $5$ of them. This reduces the problem down to $6$ groups of people, call them $G_{11}, ..., G_{16}$, and $3$ tables which is easy to solve. Each group has another $5$ group to meet so doable in $5$ weeks.
The drawback from this method is that you get groups of $4$ people always staying together for $5$ consecutive weeks. So after $5$ weeks you can change these groups and make another $6$ groups of $4$ people to circle around as explained before. Call these groups $G_{21}, ..., G_{26}$. After another $5$ weeks do it again and call the groups $G_{31}, ..., G_{36}$ and so on.
If you think it is a bad idea for people to be in groups of $4$ for $5$ consecutive weeks then alternate the weeks.
Say you have the planning for $5$ consecutive sets of $5$ weeks, then at table one you can have groups $G_{11}$ and $G_{12}$ the first week, groups $G_{21}$ and $G_{22}$ the second week, ..., up to the fifth week where you get groups $G_{51}$ and $G_{52}$.
Then alternate the groups on the first table to be $G_{11}$ and $G_{13}$ and do it again.
This way should keep you busy for $25$ weeks. Once this is done, shuffle randomly the number you initially assigned to people and do it all over again!
There is no one best way of doing it, there are a lot so by all means perhaps take ideas from what I gave but invent something else.
answered Mar 18 '14 at 15:15
user88595user88595
3,92111831
answered Mar 18 '14 at 15:15
user88595user88595
3,92111831
answered Mar 18 '14 at 15:15
user88595user88595
3,92111831
answered Mar 18 '14 at 15:15
user88595user88595
3,92111831
3,92111831
$begingroup$
That's given me something really useful to work on, just need to try & devise a spreadsheet to work this out. If I manage it, I'll share it here.
$endgroup$
– Simon Fisher
Mar 19 '14 at 8:54
add a comment |
$begingroup$
That's given me something really useful to work on, just need to try & devise a spreadsheet to work this out. If I manage it, I'll share it here.
$endgroup$
– Simon Fisher
Mar 19 '14 at 8:54
$begingroup$
That's given me something really useful to work on, just need to try & devise a spreadsheet to work this out. If I manage it, I'll share it here.
$endgroup$
– Simon Fisher
Mar 19 '14 at 8:54
$begingroup$
That's given me something really useful to work on, just need to try & devise a spreadsheet to work this out. If I manage it, I'll share it here.
$endgroup$
– Simon Fisher
Mar 19 '14 at 8:54
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f716782%2fseating-plan-for-meetings%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
How many weeks do you have?
$endgroup$
– user88595
Mar 18 '14 at 13:53
$begingroup$
It's an ongoing weekly meeting so technically we have unlimited weeks to complete the rotation, in practice it would be good to do it over a smaller number than infinity. We don't need to specify who sits next to who, just which table to sit on.
$endgroup$
– Simon Fisher
Mar 18 '14 at 14:40