Math Is Fun Forum
You are not logged in.
- Topics: Active | Unanswered
Pages: 1
#1 2025-02-19 02:16:59
- Nick79$$$
- Guest
Dinner puzzle....Please help!
I am hosting a progressive dinner with 28 guests. They need to rotate to 4 different homes for 4 different courses. I would like everyone to meet at least once. I cannot seem to think of a schedule that works. Any ideas?
#2 2025-02-19 02:40:21
- Bob
- Administrator
- Registered: 2010-06-20
- Posts: 10,702
Re: Dinner puzzle....Please help!
hi Nick
Welcome to the forum.
I would like everyone to meet at least once.
I'm thinking you'd want 7 at each venue, for each course. So person A would meet 6 others each time, that's only 24. So that won't work.
Can I do it with more folk at some venues and less elsewhere? I suspect not but I'll go back to my spreadsheet and try.
Trivial solution that you won't want: Everyone goes to venue 1, then all on to venue 2 etc etc.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob 
Offline
#3 2025-02-19 06:32:46
- Nick79$$$
- Novice
- Registered: 2025-02-19
- Posts: 1
Re: Dinner puzzle....Please help!
Yes.. they do not have to be all exact numbers at each spot. There can be some variation. Thanks for any insights you can provide !
Offline
#4 2025-02-25 10:53:42
- John E. Franklin(2)
- Member
- Registered: 2019-05-17
- Posts: 26
Re: Dinner puzzle....Please help!
I don't understand
what you guys are talking
about.
What is the number of times
they must meet at each home?
How many homes?
Last edited by John E. Franklin(2) (2025-02-25 11:04:32)
Offline
#6 2025-02-25 21:06:00
- Bob
- Administrator
- Registered: 2010-06-20
- Posts: 10,702
Re: Dinner puzzle....Please help!
There are 28 people who are going to share a four course meal. But not all in the same place. Some will go first to house 1, while others are in house 2 etc. There are 4 houses. Then they shuffle who goes where for the next course.
So the challenge is to come up with a rotation plan so that, over the 4 courses, everyone meets everyone else for at least one of the courses.
eg if abcdefgh meet in house 1 for the first course, you might try aijklmno in house 2 for the second course.
One trivial solution is for everyone to start in house 1 for course 1, then all move on to house 2 for the next and so on.
I'm assuming that 28 in one venue is too crowded, so I'm trying to spread people out more but still meet the criteria.
As 28/4 = 7, the first thing I tried is 7 in each house. But that must fail as 'a' only meets 4 x 6 = 24 of the friends. So at least 8 in some venues is needed.
So far I haven't come up with a plan using 8 people at each venue. I think that there may be some underlying maths that either helps to a solution or proves it cannot be done but I don't know what it is. Maybe you could set up a program that runs through the possibilities???
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
Offline
#7 2025-02-26 00:35:46
Re: Dinner puzzle....Please help!
Interesting. I will surely try it.
Why don't we use sudoku and abit algebra to solve it? 
I'll try doing in paper and if I don't get the answer, I'll try python or JS.
"Talent hits the target no one else can hit. Genius hits the target no one else can see." - Arthur Schopenhauer
Offline
#8 2025-02-26 03:12:41
- phrontister
- Real Member
- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,904
Re: Dinner puzzle....Please help!
I've done the following in Excel (spreadsheet...by hand, coz I don't know how to code progressive dinners): 
Guests are numbered 1-28, and are colour-coded green (1-7), peach (8-14), blue (15-21), pink (22-28).
All guests attend all 4 courses and all 4 houses (the image shows all guests 1-28 in each of the courses and in each of the houses). They do this in groups of 7 (of which there are 16, all unique).
No one attends...
(a) the same house more than once, nor
(b) the same course more than once.
Everyone...
(a) fails to meet many of the other guests, and
(b) meets several guests more than once (the colour-coding helps identify that).
...at least 8 in some venues is needed.
So far I haven't come up with a plan using 8 people at each venue.
Same here... 
Last edited by phrontister (2025-02-27 02:36:59)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
Offline
#9 2025-02-27 11:19:03
- KerimF
- Member
- From: Aleppo-Syria
- Registered: 2018-08-10
- Posts: 280
Re: Dinner puzzle....Please help!
This is what I did:
I started with the Appetizer as done by others.
I sent guest 1 to Soup at H1 where he shouldn't meet 2 to 7, so he was joined by 8 to 13.
Then I sent him to Main at H1 where he shouldn't meet 2 to 7 and 8 to 13, so he was joined by 14 to 19.
Again, I sent him to Dessert at H1 where he shouldn't meet 2 to 7, 8 to 13 and 14 to 19, so he was joined by 20 to 25.
By applying the same logic:
I sent guest 2 to Soup at H2 where he shouldn't meet 1, 3 to 7, and 9 to 13 who were at H1 (with 1), so he was joined by 14 to 19.
Then I sent him to Main at H2 where he shouldn't meet 1, 3 to 7 and 9 to 13, also 14 to 19 who were at H1 (with 1), so he was joined by 20 to 25.
Again, I sent him to Dessert at H2 where he shouldn't meet 1, 3 to 7, 9 to 13 and 14 to 19, also 20 to 25 who were at H1 (with 1), so he was joined by 26 to 28 + 8 to 10.
Similarly:
I sent guest 3 to Soup at H3 where he shouldn't meet 1, 2 and 4 to 7, also 8 to 13 (with 1) and 14 to 19 (with 2), so he was joined by 20 to 25.
Then I sent 3 to Main at H3 where he shouldn't meet 1, 2 and 4 to 7 and 9 to 13, also 14 to 19 (with 1) and 20 to 25 (with 2), so he was joined by 26 to 28 + 8 to 10.
Again, I sent 3 to Dessert at H3 where he shouldn't meet 1, 2 and 4 to 7, 20 to 25 and 26 to 28 + 8 to 10, so he was joined by 11 to 16.
But this same reasoning couldn't be applied on the guest 4 as shown below. Perhaps, other methodic steps could work
=============================
Appetizer
H1: 1 2 3 4 5 6 7
H2: 8 9 10 11 12 13 14
H3: 15 16 17 18 19 20 21
H4: 22 23 24 25 26 27 28
Soup
H1: 1 8 9 10 11 12 13
H2: 2 14 15 16 17 18 19
H3: 3 20 21 22 23 24 25
H4: 4 26 27 28
Main
H1: 1 14 15 16 17 18 19
H2: 2 20 21 22 23 24 25
H3: 3 26 27 28 8 9 10
H4: 4
Dessert
H1: 1 20 21 22 23 24 25
H2: 2 26 27 28 8 9 10
H3: 3 11 12 13 14 15 16
H4: 4
=================
Edit: Perhaps instead of starting with 1,2,3 and 4, using 1, 8, 15 and 22 may work.
Last edited by KerimF (2025-02-27 11:29:11)
Every living thing has no choice but to execute its pre-programmed instructions embedded in it (known as instincts).
But only a human may have the freedom and ability to oppose his natural robotic nature.
But, by opposing it, such a human becomes no more of this world.
Offline
#10 2025-02-27 14:22:57
Re: Dinner puzzle....Please help!
I tried 5 different ways to complete this but all were useless. At first I tried to match all guests with each other so thought of first making G2 meet with all. I made him meet with 16 other guests but later didn't get anything useful so left the method.
"Talent hits the target no one else can hit. Genius hits the target no one else can see." - Arthur Schopenhauer
Offline
#11 2025-02-27 21:34:04
- KerimF
- Member
- From: Aleppo-Syria
- Registered: 2018-08-10
- Posts: 280
Re: Dinner puzzle....Please help!
Isn’t it a solution?
Appetizer
H1: 01 02 03 04 05 06 07
H2: 08 09 10 11 12 13 14
H3: 15 16 17 18 19 20 21
H4: 22 23 24 25 26 27 28
Soup
H1: 01 09 10 11 12 13 14
H2: 08 02 03 04 05 06 07
H3: 15 23 24 25 26 27 28
H4: 22 16 17 18 19 20 21
Main
H1: 01 16 17 18 19 20 21
H2: 08 23 24 25 26 27 28
H3: 15 02 03 04 05 06 07
H4: 22 09 10 11 12 13 14
Dessert
H1: 01 23 24 25 26 27 28
H2: 08 16 17 18 19 20 21
H3: 15 09 10 11 12 13 14
H4: 22 02 03 04 05 06 07
Last edited by KerimF (2025-02-27 21:37:19)
Every living thing has no choice but to execute its pre-programmed instructions embedded in it (known as instincts).
But only a human may have the freedom and ability to oppose his natural robotic nature.
But, by opposing it, such a human becomes no more of this world.
Offline
#13 2025-02-27 23:11:20
- KerimF
- Member
- From: Aleppo-Syria
- Registered: 2018-08-10
- Posts: 280
Re: Dinner puzzle....Please help!
Why do 1, 8, 15 and 22 take the same course 4 times?
Do you mean this is not allowed?
Every living thing has no choice but to execute its pre-programmed instructions embedded in it (known as instincts).
But only a human may have the freedom and ability to oppose his natural robotic nature.
But, by opposing it, such a human becomes no more of this world.
Offline
#14 2025-02-27 23:16:14
Re: Dinner puzzle....Please help!
I am hosting a progressive dinner with 28 guests. They need to rotate to 4 different homes for 4 different courses. I would like everyone to meet at least once.
So it means that each guest has to get a new course. So that isn't allowed.
"Talent hits the target no one else can hit. Genius hits the target no one else can see." - Arthur Schopenhauer
Offline
#15 2025-02-28 00:41:31
- KerimF
- Member
- From: Aleppo-Syria
- Registered: 2018-08-10
- Posts: 280
Re: Dinner puzzle....Please help!
I guess this is the solution:
Appetizer
H1: 01 02 03 04 05 06 07
H2: 08 09 10 11 12 13 14
H3: 15 16 17 18 19 20 21
H4: 22 23 24 25 26 27 28
Soup
H1: 15 23 24 25 26 27 28
H2: 22 16 17 18 19 20 21
H3: 01 09 10 11 12 13 14
H4: 08 02 03 04 05 06 07
Main
H1: 22 09 10 11 12 13 14
H2: 15 02 03 04 05 06 07
H3: 08 23 24 25 26 27 28
H4: 01 16 17 18 19 20 21
Dessert
H1: 08 16 17 18 19 20 21
H2: 01 23 24 25 26 27 28
H3: 22 02 03 04 05 06 07
H4: 15 09 10 11 12 13 14
Every living thing has no choice but to execute its pre-programmed instructions embedded in it (known as instincts).
But only a human may have the freedom and ability to oppose his natural robotic nature.
But, by opposing it, such a human becomes no more of this world.
Offline
#16 2025-02-28 01:18:56
- phrontister
- Real Member
- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,904
Re: Dinner puzzle....Please help!
Hi Kerim;
Isn’t it a solution?
Well, not really, unless the "I would like" in Nick's "I would like everyone to meet at least once" is only a desire, not a requirement.
01 doesn't meet 08, 15 & 22
08 doesn't meet 01, 15 & 22
15 doesn't meet 01, 08 & 22
22 doesn't meet 01, 08 & 15
02 doesn't meet 09, 16 & 22
09 doesn't meet 02, 16 & 22
etc, etc.
Ditto for 01, 08, 15 & 22 in your post #16.
Also, in post #16, guest 02 only meets 01, 03, 04, 05, 06, 07, 08, 15 & 22, and fails to meet any of the other 18 guests.
Just looking quickly, 03 does much the same as 02.
I didn't check further...
Last edited by phrontister (2025-02-28 01:40:18)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
Offline
#17 2025-02-28 11:16:00
- KerimF
- Member
- From: Aleppo-Syria
- Registered: 2018-08-10
- Posts: 280
Re: Dinner puzzle....Please help!
Yes.
In post #12, the solution was wrong as noted in post #15
In post #16, it is also wrong if everyone has to meet at least once.
Every living thing has no choice but to execute its pre-programmed instructions embedded in it (known as instincts).
But only a human may have the freedom and ability to oppose his natural robotic nature.
But, by opposing it, such a human becomes no more of this world.
Offline
#18 2025-02-28 14:46:31
Re: Dinner puzzle....Please help!
But mathematically it seems inmpossible to make the guests meet with all each other. Lets first check it mathematically.
"Talent hits the target no one else can hit. Genius hits the target no one else can see." - Arthur Schopenhauer
Offline
#19 2025-03-01 11:43:16
- Bob
- Administrator
- Registered: 2010-06-20
- Posts: 10,702
Re: Dinner puzzle....Please help!
Definitely possible. Let all 28 meet in one place. (They can move on to other houses for each course if you want.)
So I also think that somewhere between 7 meeting at a place and 28 meeting at a place there is the optimum solution which I take to be the minimum at each place subject to all meeting all at least once.
Trouble is I cannot find it.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
Offline
#20 2025-03-02 17:23:47
- KerimF
- Member
- From: Aleppo-Syria
- Registered: 2018-08-10
- Posts: 280
Re: Dinner puzzle....Please help!
It seems that, at best, a guest can meet 24 different guests.
In case the distribution is 7+7+7+7, a guest can meet 6 guests * 4 = 24 guests
If 8+8+8+4, a guest of the latter 4 can meet 3 + 3*7 = 24 guests
If 9+9+9+1, The latter guest can meet 0 + 3*8 = 24 guests
Is there other possibility?
Every living thing has no choice but to execute its pre-programmed instructions embedded in it (known as instincts).
But only a human may have the freedom and ability to oppose his natural robotic nature.
But, by opposing it, such a human becomes no more of this world.
Offline
Pages: 1