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My Sixteen Friendly Students
Ernie's Clockwork CalculatorFriendly GolfersSleeping studentsTouching matchsticks with compass and straightedgeThe First Interplanetary Arithmetic SummitVegas Street Magician Math TrickCookie Box Mix-upA way to beat the system?What a weird final exam1984 - take the digits 1,9, 8 and 4 and make 123 - Part III
$begingroup$
I have sixteen students in my class who sit in four rows of four. Each week they sit in a different order.
After a number of weeks every student has sat next to every other student, next meaning side by side, one behind the other, or sitting diagonally together. What is the fewest number of weeks in which this can happen?
How many if my students were 25?
mathematics combinatorics
asked May 7 at 12:20
Bernardo Recamán SantosBernardo Recamán Santos
2,9351353
$endgroup$
add a comment |
$begingroup$
I have sixteen students in my class who sit in four rows of four. Each week they sit in a different order.
After a number of weeks every student has sat next to every other student, next meaning side by side, one behind the other, or sitting diagonally together. What is the fewest number of weeks in which this can happen?
How many if my students were 25?
mathematics combinatorics
asked May 7 at 12:20
Bernardo Recamán SantosBernardo Recamán Santos
2,9351353
$endgroup$
2
$begingroup$
Perhaps the second 25 student puzzle (which presumably has five rows of five) should be a new question after this is solved?
$endgroup$
– Weather Vane
May 7 at 18:27
2
$begingroup$
@WeatherVane: Following your suggestion, I have now posted the case for 25 students at: math.stackexchange.com/questions/3218384/…
$endgroup$
– Bernardo Recamán Santos
May 8 at 12:41
1
$begingroup$
I made a solution validator some may find useful.
$endgroup$
– TemporalWolf
May 8 at 23:30
2
$begingroup$
I found an optimal solution to the 25 student problem (5 weeks), or rather, my computer program did. The solution can be found over on Mathematics SE.
$endgroup$
– Jaap Scherphuis
May 9 at 15:59
add a comment |
$begingroup$
I have sixteen students in my class who sit in four rows of four. Each week they sit in a different order.
After a number of weeks every student has sat next to every other student, next meaning side by side, one behind the other, or sitting diagonally together. What is the fewest number of weeks in which this can happen?
How many if my students were 25?
mathematics combinatorics
asked May 7 at 12:20
Bernardo Recamán SantosBernardo Recamán Santos
2,9351353
$endgroup$
I have sixteen students in my class who sit in four rows of four. Each week they sit in a different order.
After a number of weeks every student has sat next to every other student, next meaning side by side, one behind the other, or sitting diagonally together. What is the fewest number of weeks in which this can happen?
How many if my students were 25?
mathematics combinatorics
mathematics combinatorics
asked May 7 at 12:20
Bernardo Recamán SantosBernardo Recamán Santos
2,9351353
asked May 7 at 12:20
Bernardo Recamán SantosBernardo Recamán Santos
2,9351353
edited May 7 at 18:24
Bernardo Recamán Santos
asked May 7 at 12:20
Bernardo Recamán SantosBernardo Recamán Santos
2,9351353
asked May 7 at 12:20
Bernardo Recamán SantosBernardo Recamán Santos
2,9351353
asked May 7 at 12:20
Bernardo Recamán SantosBernardo Recamán Santos
2,9351353
2,9351353
2
$begingroup$
Perhaps the second 25 student puzzle (which presumably has five rows of five) should be a new question after this is solved?
$endgroup$
– Weather Vane
May 7 at 18:27
2
$begingroup$
@WeatherVane: Following your suggestion, I have now posted the case for 25 students at: math.stackexchange.com/questions/3218384/…
$endgroup$
– Bernardo Recamán Santos
May 8 at 12:41
1
$begingroup$
I made a solution validator some may find useful.
$endgroup$
– TemporalWolf
May 8 at 23:30
2
$begingroup$
I found an optimal solution to the 25 student problem (5 weeks), or rather, my computer program did. The solution can be found over on Mathematics SE.
$endgroup$
– Jaap Scherphuis
May 9 at 15:59
add a comment |
2
$begingroup$
Perhaps the second 25 student puzzle (which presumably has five rows of five) should be a new question after this is solved?
$endgroup$
– Weather Vane
May 7 at 18:27
2
$begingroup$
@WeatherVane: Following your suggestion, I have now posted the case for 25 students at: math.stackexchange.com/questions/3218384/…
$endgroup$
– Bernardo Recamán Santos
May 8 at 12:41
1
$begingroup$
I made a solution validator some may find useful.
$endgroup$
– TemporalWolf
May 8 at 23:30
2
$begingroup$
I found an optimal solution to the 25 student problem (5 weeks), or rather, my computer program did. The solution can be found over on Mathematics SE.
$endgroup$
– Jaap Scherphuis
May 9 at 15:59
2
2
$begingroup$
Perhaps the second 25 student puzzle (which presumably has five rows of five) should be a new question after this is solved?
$endgroup$
– Weather Vane
May 7 at 18:27
$begingroup$
Perhaps the second 25 student puzzle (which presumably has five rows of five) should be a new question after this is solved?
$endgroup$
– Weather Vane
May 7 at 18:27
2
2
$begingroup$
@WeatherVane: Following your suggestion, I have now posted the case for 25 students at: math.stackexchange.com/questions/3218384/…
$endgroup$
– Bernardo Recamán Santos
May 8 at 12:41
$begingroup$
@WeatherVane: Following your suggestion, I have now posted the case for 25 students at: math.stackexchange.com/questions/3218384/…
$endgroup$
– Bernardo Recamán Santos
May 8 at 12:41
1
1
$begingroup$
I made a solution validator some may find useful.
$endgroup$
– TemporalWolf
May 8 at 23:30
$begingroup$
I made a solution validator some may find useful.
$endgroup$
– TemporalWolf
May 8 at 23:30
2
2
$begingroup$
I found an optimal solution to the 25 student problem (5 weeks), or rather, my computer program did. The solution can be found over on Mathematics SE.
$endgroup$
– Jaap Scherphuis
May 9 at 15:59
$begingroup$
I found an optimal solution to the 25 student problem (5 weeks), or rather, my computer program did. The solution can be found over on Mathematics SE.
$endgroup$
– Jaap Scherphuis
May 9 at 15:59
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
Continuing from Arnaud Mortier's observations,
I noticed that each student in the always-on-the-edge group will sit by one student from each rotating group while the rotating students are in the corners, and two more students from each rotating group while they are in the center. So each student in the always-on-the-edge group needs to sit by exactly one student in each rotating group while it is on the edge. Since each edge seat is adjacent to two other edge seats, this means that each student in the always-on-the-edge group will sit by one other student in the always-on-the-edge group each week.
Using that knowledge and a bit of trial and error,
I was able to find a solution for 3 weeks.
Week 1 Week 2 Week 3
1 2 3 4 9 11 10 12 10 2 14 6
5 6 7 8 2 4 1 3 16 8 12 4
9 10 11 12 15 13 16 14 1 9 5 13
13 14 15 16 8 6 7 5 7 15 3 11
Here's a table which show which weeks the students sit by each other.
Student 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
-----------------------------------------------------------------
1 | - 1 2 2 1 1 3 3 3 2 2 2 2 2 3 2,3
2 | 1 - 1 2 1 1 1 3 2 3 2 3 2 3 2 3
3 | 2 1 - 1 3 1 1 1 3 2 3 2 3 2 3 2
4 | 2 2 1 - 3 3 1 1 2 2 2 3 2,3 3 2 2
5 | 1 1 3 3 - 1 2 3 1,3 1 3 3 3 2 3 2
6 | 1 1 1 3 1 - 1,2 2 1 1 1 3 2 3 2 2
7 | 3 1 1 1 2 1,2 - 1 3 1 1 1 2 2 3 2
8 | 3 3 1 1 3 2 1 - 3 3 1 1,3 2 3 2 3
9 | 3 2 3 2 1,3 1 3 3 - 1 2 3 1 1 3 3
10 | 2 3 2 2 1 1 1 3 1 - 1,2 2 1 1 1 3
11 | 2 2 3 2 3 1 1 1 2 1,2 - 1 3 1 1 1
12 | 2 3 2 3 3 3 1 1,3 3 2 1 - 3 3 1 1
13 | 2 2 3 2,3 3 2 2 2 1 1 3 3 - 1 2 2
14 | 2 3 2 3 2 3 2 3 1 1 1 3 1 - 1 1
15 | 3 2 3 2 3 2 3 2 3 1 1 1 2 1 - 1
16 | 2,3 3 2 2 2 2 2 3 3 3 1 1 2 1 1 -
answered May 8 at 17:55
Ben JBen J
54618
$endgroup$
2
$begingroup$
Well done! I was waiting for this answer to be re-posted after it was deleted: beat me to the solution, but I too was working on the idea that 4 students rotate around two pairs of centre-edge seats, and each meets 5+5+5 others, and the other 12 students meet 3+5+8 others, having been at the corner-edge-middle positions.
$endgroup$
– Weather Vane
May 8 at 19:17
add a comment |
$begingroup$
Partial answer to Question 1.
First, the average number of friends a student gets per week is
$$frac4times 3 +8times 5+4times 816=5.25$$
(there are $4$ corners each of which has $3$ friends, and so on).
At the end of the day, we want each student to have had $15$ different friends, so this has to take at least $3$ weeks.
Moreover, in three weeks at most $12$ students can be in the center once:
Those who are never in the center will need to be on an edge all three times to get $3times 5$ friends.
The scheme above is therefore the only way to get 15 friends for everyone in 3 rounds. It is very rigid - there is no room for redundancy for the all-edgers, and there can be one redundancy for the others. Yet I can't prove yet that the constraints are too heavy to be met.
answered May 7 at 16:05
Arnaud MortierArnaud Mortier
3,066933
$endgroup$
add a comment |
$begingroup$
Here's 16 students in
4
weeks.
0123 A8B9 CAD0 135E
4567 2031 B179 A2BC
89AB ECFD 58E4 94D0
CDEF 6475 F263 7F68
answered May 7 at 15:56
noednenoedne
10.6k12976
$endgroup$
$begingroup$
I can confirm this works! Good job. How did you find it?
$endgroup$
– Dr Xorile
May 8 at 1:29
$begingroup$
@DrXorile Nothing special, just filled out the grids using some symmetries, trying to maximize new pairs. On my first attempt, the last grid fit together perfectly to finish off the pairs.
$endgroup$
– noedne
May 8 at 1:56
2
$begingroup$
It may be possible with 3. Tricky to find though.
$endgroup$
– Dr Xorile
May 8 at 1:58
add a comment |
$begingroup$
As a starting point for Question 2, here it is done in
7 weeks
Solution
01234 IG968 HCFA1 97HEP BA92H HDCE0 EB095
56789 B1HKJ 3P4JD MF8AK L3CJ8 8K1F2 GN46H
ABCDE P4307 E20BM N3LI4 KDGN7 GMBNI 3IACK
FGHIJ D5MAL 7698K J50D6 50461 P6LJA DF712
KLMNP NF2CE GILN5 1CG2B IEMPF 93475 JM8PL
answered May 8 at 16:35
hexominohexomino
50k4149239
$endgroup$
$begingroup$
Nice! I found one for nine weeks.
$endgroup$
– Freddy Barrera
May 8 at 19:20
add a comment |
$begingroup$
A quick intuitive answer:
For 16 students:
6 weeks
Reasoning (edited to account for the corner case, as pointed out by hexomino);
2nd week: swap 1st and 3rd columns. 3rd week: swap 2nd and 4th columns. Do the same operations with the rows over the next two weeks. Before swapping, either the first or the preceding week's formation can be taken as the starting point. 5th week: swap the two middle columns. 6th week: swap the middle rows. At the beginning of the 7th week, each student would have been next to each other student.
For 25 students:
8 weeks
Reasoning:
Follow the above reasoning, except that the rows/columns to be swapped would be 1st/4th, 2nd/5th, 2nd/3rd, and 3rd/4th.
answered May 7 at 15:12
Mazurka FahrMazurka Fahr
1114
$endgroup$
2
$begingroup$
Won't students who are originally in opposite corners stay in opposite corners?
$endgroup$
– hexomino
May 7 at 15:14
$begingroup$
hmm. Guess, I'd need to add a few extra steps for the corner cases. Not sure if this would be the optimum solution.
$endgroup$
– Mazurka Fahr
May 7 at 15:17
1
$begingroup$
In the new version, students in the original first column never sit next to students in the original third column.
$endgroup$
– hexomino
May 7 at 15:31
$begingroup$
:(. Can't avoid the extra steps.
$endgroup$
– Mazurka Fahr
May 7 at 15:32
add a comment |
$begingroup$
The first question i believe it's
256
and the second one it's
625
I think the solution of this one is to multiply the two numbers given. An other example is:find all the combinations of these words PANEL(wich i'll use the same equation).
answered May 7 at 12:49
riki481riki481
339
$endgroup$
$begingroup$
it's a math logic that my teacher told me, we have 16 students (in the first question) and 16 chairs, 16x16 = 256. Like, we need to repeat the same process 16 times for all of the 16 students
$endgroup$
– riki481
May 7 at 13:03
$begingroup$
This seems more like the total number of combinations, rather than how quickly you can get each student to have sat next to every other 8 directions
$endgroup$
– Smock
May 7 at 13:56
$begingroup$
Guess there is another step i need to do, but i don't remember
$endgroup$
– riki481
May 7 at 14:00
1
$begingroup$
After actually looking into the question, these numbers are way too large.
$endgroup$
– Arnaud Mortier
May 7 at 14:28
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Continuing from Arnaud Mortier's observations,
I noticed that each student in the always-on-the-edge group will sit by one student from each rotating group while the rotating students are in the corners, and two more students from each rotating group while they are in the center. So each student in the always-on-the-edge group needs to sit by exactly one student in each rotating group while it is on the edge. Since each edge seat is adjacent to two other edge seats, this means that each student in the always-on-the-edge group will sit by one other student in the always-on-the-edge group each week.
Using that knowledge and a bit of trial and error,
I was able to find a solution for 3 weeks.
Week 1 Week 2 Week 3
1 2 3 4 9 11 10 12 10 2 14 6
5 6 7 8 2 4 1 3 16 8 12 4
9 10 11 12 15 13 16 14 1 9 5 13
13 14 15 16 8 6 7 5 7 15 3 11
Here's a table which show which weeks the students sit by each other.
Student 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
-----------------------------------------------------------------
1 | - 1 2 2 1 1 3 3 3 2 2 2 2 2 3 2,3
2 | 1 - 1 2 1 1 1 3 2 3 2 3 2 3 2 3
3 | 2 1 - 1 3 1 1 1 3 2 3 2 3 2 3 2
4 | 2 2 1 - 3 3 1 1 2 2 2 3 2,3 3 2 2
5 | 1 1 3 3 - 1 2 3 1,3 1 3 3 3 2 3 2
6 | 1 1 1 3 1 - 1,2 2 1 1 1 3 2 3 2 2
7 | 3 1 1 1 2 1,2 - 1 3 1 1 1 2 2 3 2
8 | 3 3 1 1 3 2 1 - 3 3 1 1,3 2 3 2 3
9 | 3 2 3 2 1,3 1 3 3 - 1 2 3 1 1 3 3
10 | 2 3 2 2 1 1 1 3 1 - 1,2 2 1 1 1 3
11 | 2 2 3 2 3 1 1 1 2 1,2 - 1 3 1 1 1
12 | 2 3 2 3 3 3 1 1,3 3 2 1 - 3 3 1 1
13 | 2 2 3 2,3 3 2 2 2 1 1 3 3 - 1 2 2
14 | 2 3 2 3 2 3 2 3 1 1 1 3 1 - 1 1
15 | 3 2 3 2 3 2 3 2 3 1 1 1 2 1 - 1
16 | 2,3 3 2 2 2 2 2 3 3 3 1 1 2 1 1 -
answered May 8 at 17:55
Ben JBen J
54618
$endgroup$
2
$begingroup$
Well done! I was waiting for this answer to be re-posted after it was deleted: beat me to the solution, but I too was working on the idea that 4 students rotate around two pairs of centre-edge seats, and each meets 5+5+5 others, and the other 12 students meet 3+5+8 others, having been at the corner-edge-middle positions.
$endgroup$
– Weather Vane
May 8 at 19:17
add a comment |
$begingroup$
Continuing from Arnaud Mortier's observations,
I noticed that each student in the always-on-the-edge group will sit by one student from each rotating group while the rotating students are in the corners, and two more students from each rotating group while they are in the center. So each student in the always-on-the-edge group needs to sit by exactly one student in each rotating group while it is on the edge. Since each edge seat is adjacent to two other edge seats, this means that each student in the always-on-the-edge group will sit by one other student in the always-on-the-edge group each week.
Using that knowledge and a bit of trial and error,
I was able to find a solution for 3 weeks.
Week 1 Week 2 Week 3
1 2 3 4 9 11 10 12 10 2 14 6
5 6 7 8 2 4 1 3 16 8 12 4
9 10 11 12 15 13 16 14 1 9 5 13
13 14 15 16 8 6 7 5 7 15 3 11
Here's a table which show which weeks the students sit by each other.
Student 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
-----------------------------------------------------------------
1 | - 1 2 2 1 1 3 3 3 2 2 2 2 2 3 2,3
2 | 1 - 1 2 1 1 1 3 2 3 2 3 2 3 2 3
3 | 2 1 - 1 3 1 1 1 3 2 3 2 3 2 3 2
4 | 2 2 1 - 3 3 1 1 2 2 2 3 2,3 3 2 2
5 | 1 1 3 3 - 1 2 3 1,3 1 3 3 3 2 3 2
6 | 1 1 1 3 1 - 1,2 2 1 1 1 3 2 3 2 2
7 | 3 1 1 1 2 1,2 - 1 3 1 1 1 2 2 3 2
8 | 3 3 1 1 3 2 1 - 3 3 1 1,3 2 3 2 3
9 | 3 2 3 2 1,3 1 3 3 - 1 2 3 1 1 3 3
10 | 2 3 2 2 1 1 1 3 1 - 1,2 2 1 1 1 3
11 | 2 2 3 2 3 1 1 1 2 1,2 - 1 3 1 1 1
12 | 2 3 2 3 3 3 1 1,3 3 2 1 - 3 3 1 1
13 | 2 2 3 2,3 3 2 2 2 1 1 3 3 - 1 2 2
14 | 2 3 2 3 2 3 2 3 1 1 1 3 1 - 1 1
15 | 3 2 3 2 3 2 3 2 3 1 1 1 2 1 - 1
16 | 2,3 3 2 2 2 2 2 3 3 3 1 1 2 1 1 -
answered May 8 at 17:55
Ben JBen J
54618
$endgroup$
2
$begingroup$
Well done! I was waiting for this answer to be re-posted after it was deleted: beat me to the solution, but I too was working on the idea that 4 students rotate around two pairs of centre-edge seats, and each meets 5+5+5 others, and the other 12 students meet 3+5+8 others, having been at the corner-edge-middle positions.
$endgroup$
– Weather Vane
May 8 at 19:17
add a comment |
$begingroup$
Continuing from Arnaud Mortier's observations,
I noticed that each student in the always-on-the-edge group will sit by one student from each rotating group while the rotating students are in the corners, and two more students from each rotating group while they are in the center. So each student in the always-on-the-edge group needs to sit by exactly one student in each rotating group while it is on the edge. Since each edge seat is adjacent to two other edge seats, this means that each student in the always-on-the-edge group will sit by one other student in the always-on-the-edge group each week.
Using that knowledge and a bit of trial and error,
I was able to find a solution for 3 weeks.
Week 1 Week 2 Week 3
1 2 3 4 9 11 10 12 10 2 14 6
5 6 7 8 2 4 1 3 16 8 12 4
9 10 11 12 15 13 16 14 1 9 5 13
13 14 15 16 8 6 7 5 7 15 3 11
Here's a table which show which weeks the students sit by each other.
Student 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
-----------------------------------------------------------------
1 | - 1 2 2 1 1 3 3 3 2 2 2 2 2 3 2,3
2 | 1 - 1 2 1 1 1 3 2 3 2 3 2 3 2 3
3 | 2 1 - 1 3 1 1 1 3 2 3 2 3 2 3 2
4 | 2 2 1 - 3 3 1 1 2 2 2 3 2,3 3 2 2
5 | 1 1 3 3 - 1 2 3 1,3 1 3 3 3 2 3 2
6 | 1 1 1 3 1 - 1,2 2 1 1 1 3 2 3 2 2
7 | 3 1 1 1 2 1,2 - 1 3 1 1 1 2 2 3 2
8 | 3 3 1 1 3 2 1 - 3 3 1 1,3 2 3 2 3
9 | 3 2 3 2 1,3 1 3 3 - 1 2 3 1 1 3 3
10 | 2 3 2 2 1 1 1 3 1 - 1,2 2 1 1 1 3
11 | 2 2 3 2 3 1 1 1 2 1,2 - 1 3 1 1 1
12 | 2 3 2 3 3 3 1 1,3 3 2 1 - 3 3 1 1
13 | 2 2 3 2,3 3 2 2 2 1 1 3 3 - 1 2 2
14 | 2 3 2 3 2 3 2 3 1 1 1 3 1 - 1 1
15 | 3 2 3 2 3 2 3 2 3 1 1 1 2 1 - 1
16 | 2,3 3 2 2 2 2 2 3 3 3 1 1 2 1 1 -
answered May 8 at 17:55
Ben JBen J
54618
$endgroup$
Continuing from Arnaud Mortier's observations,
I noticed that each student in the always-on-the-edge group will sit by one student from each rotating group while the rotating students are in the corners, and two more students from each rotating group while they are in the center. So each student in the always-on-the-edge group needs to sit by exactly one student in each rotating group while it is on the edge. Since each edge seat is adjacent to two other edge seats, this means that each student in the always-on-the-edge group will sit by one other student in the always-on-the-edge group each week.
Using that knowledge and a bit of trial and error,
I was able to find a solution for 3 weeks.
Week 1 Week 2 Week 3
1 2 3 4 9 11 10 12 10 2 14 6
5 6 7 8 2 4 1 3 16 8 12 4
9 10 11 12 15 13 16 14 1 9 5 13
13 14 15 16 8 6 7 5 7 15 3 11
Here's a table which show which weeks the students sit by each other.
Student 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
-----------------------------------------------------------------
1 | - 1 2 2 1 1 3 3 3 2 2 2 2 2 3 2,3
2 | 1 - 1 2 1 1 1 3 2 3 2 3 2 3 2 3
3 | 2 1 - 1 3 1 1 1 3 2 3 2 3 2 3 2
4 | 2 2 1 - 3 3 1 1 2 2 2 3 2,3 3 2 2
5 | 1 1 3 3 - 1 2 3 1,3 1 3 3 3 2 3 2
6 | 1 1 1 3 1 - 1,2 2 1 1 1 3 2 3 2 2
7 | 3 1 1 1 2 1,2 - 1 3 1 1 1 2 2 3 2
8 | 3 3 1 1 3 2 1 - 3 3 1 1,3 2 3 2 3
9 | 3 2 3 2 1,3 1 3 3 - 1 2 3 1 1 3 3
10 | 2 3 2 2 1 1 1 3 1 - 1,2 2 1 1 1 3
11 | 2 2 3 2 3 1 1 1 2 1,2 - 1 3 1 1 1
12 | 2 3 2 3 3 3 1 1,3 3 2 1 - 3 3 1 1
13 | 2 2 3 2,3 3 2 2 2 1 1 3 3 - 1 2 2
14 | 2 3 2 3 2 3 2 3 1 1 1 3 1 - 1 1
15 | 3 2 3 2 3 2 3 2 3 1 1 1 2 1 - 1
16 | 2,3 3 2 2 2 2 2 3 3 3 1 1 2 1 1 -
answered May 8 at 17:55
Ben JBen J
54618
edited May 8 at 19:24
answered May 8 at 17:55
Ben JBen J
54618
answered May 8 at 17:55
Ben JBen J
54618
answered May 8 at 17:55
Ben JBen J
54618
54618
2
$begingroup$
Well done! I was waiting for this answer to be re-posted after it was deleted: beat me to the solution, but I too was working on the idea that 4 students rotate around two pairs of centre-edge seats, and each meets 5+5+5 others, and the other 12 students meet 3+5+8 others, having been at the corner-edge-middle positions.
$endgroup$
– Weather Vane
May 8 at 19:17
add a comment |
2
$begingroup$
Well done! I was waiting for this answer to be re-posted after it was deleted: beat me to the solution, but I too was working on the idea that 4 students rotate around two pairs of centre-edge seats, and each meets 5+5+5 others, and the other 12 students meet 3+5+8 others, having been at the corner-edge-middle positions.
$endgroup$
– Weather Vane
May 8 at 19:17
2
2
$begingroup$
Well done! I was waiting for this answer to be re-posted after it was deleted: beat me to the solution, but I too was working on the idea that 4 students rotate around two pairs of centre-edge seats, and each meets 5+5+5 others, and the other 12 students meet 3+5+8 others, having been at the corner-edge-middle positions.
$endgroup$
– Weather Vane
May 8 at 19:17
$begingroup$
Well done! I was waiting for this answer to be re-posted after it was deleted: beat me to the solution, but I too was working on the idea that 4 students rotate around two pairs of centre-edge seats, and each meets 5+5+5 others, and the other 12 students meet 3+5+8 others, having been at the corner-edge-middle positions.
$endgroup$
– Weather Vane
May 8 at 19:17
add a comment |
$begingroup$
Partial answer to Question 1.
First, the average number of friends a student gets per week is
$$frac4times 3 +8times 5+4times 816=5.25$$
(there are $4$ corners each of which has $3$ friends, and so on).
At the end of the day, we want each student to have had $15$ different friends, so this has to take at least $3$ weeks.
Moreover, in three weeks at most $12$ students can be in the center once:
Those who are never in the center will need to be on an edge all three times to get $3times 5$ friends.
The scheme above is therefore the only way to get 15 friends for everyone in 3 rounds. It is very rigid - there is no room for redundancy for the all-edgers, and there can be one redundancy for the others. Yet I can't prove yet that the constraints are too heavy to be met.
answered May 7 at 16:05
Arnaud MortierArnaud Mortier
3,066933
$endgroup$
add a comment |
$begingroup$
Partial answer to Question 1.
First, the average number of friends a student gets per week is
$$frac4times 3 +8times 5+4times 816=5.25$$
(there are $4$ corners each of which has $3$ friends, and so on).
At the end of the day, we want each student to have had $15$ different friends, so this has to take at least $3$ weeks.
Moreover, in three weeks at most $12$ students can be in the center once:
Those who are never in the center will need to be on an edge all three times to get $3times 5$ friends.
The scheme above is therefore the only way to get 15 friends for everyone in 3 rounds. It is very rigid - there is no room for redundancy for the all-edgers, and there can be one redundancy for the others. Yet I can't prove yet that the constraints are too heavy to be met.
answered May 7 at 16:05
Arnaud MortierArnaud Mortier
3,066933
$endgroup$
add a comment |
$begingroup$
Partial answer to Question 1.
First, the average number of friends a student gets per week is
$$frac4times 3 +8times 5+4times 816=5.25$$
(there are $4$ corners each of which has $3$ friends, and so on).
At the end of the day, we want each student to have had $15$ different friends, so this has to take at least $3$ weeks.
Moreover, in three weeks at most $12$ students can be in the center once:
Those who are never in the center will need to be on an edge all three times to get $3times 5$ friends.
The scheme above is therefore the only way to get 15 friends for everyone in 3 rounds. It is very rigid - there is no room for redundancy for the all-edgers, and there can be one redundancy for the others. Yet I can't prove yet that the constraints are too heavy to be met.
answered May 7 at 16:05
Arnaud MortierArnaud Mortier
3,066933
$endgroup$
Partial answer to Question 1.
First, the average number of friends a student gets per week is
$$frac4times 3 +8times 5+4times 816=5.25$$
(there are $4$ corners each of which has $3$ friends, and so on).
At the end of the day, we want each student to have had $15$ different friends, so this has to take at least $3$ weeks.
Moreover, in three weeks at most $12$ students can be in the center once:
Those who are never in the center will need to be on an edge all three times to get $3times 5$ friends.
The scheme above is therefore the only way to get 15 friends for everyone in 3 rounds. It is very rigid - there is no room for redundancy for the all-edgers, and there can be one redundancy for the others. Yet I can't prove yet that the constraints are too heavy to be met.
answered May 7 at 16:05
Arnaud MortierArnaud Mortier
3,066933
answered May 7 at 16:05
Arnaud MortierArnaud Mortier
3,066933
answered May 7 at 16:05
Arnaud MortierArnaud Mortier
3,066933
answered May 7 at 16:05
Arnaud MortierArnaud Mortier
3,066933
3,066933
add a comment |
add a comment |
$begingroup$
Here's 16 students in
4
weeks.
0123 A8B9 CAD0 135E
4567 2031 B179 A2BC
89AB ECFD 58E4 94D0
CDEF 6475 F263 7F68
answered May 7 at 15:56
noednenoedne
10.6k12976
$endgroup$
$begingroup$
I can confirm this works! Good job. How did you find it?
$endgroup$
– Dr Xorile
May 8 at 1:29
$begingroup$
@DrXorile Nothing special, just filled out the grids using some symmetries, trying to maximize new pairs. On my first attempt, the last grid fit together perfectly to finish off the pairs.
$endgroup$
– noedne
May 8 at 1:56
2
$begingroup$
It may be possible with 3. Tricky to find though.
$endgroup$
– Dr Xorile
May 8 at 1:58
add a comment |
$begingroup$
Here's 16 students in
4
weeks.
0123 A8B9 CAD0 135E
4567 2031 B179 A2BC
89AB ECFD 58E4 94D0
CDEF 6475 F263 7F68
answered May 7 at 15:56
noednenoedne
10.6k12976
$endgroup$
$begingroup$
I can confirm this works! Good job. How did you find it?
$endgroup$
– Dr Xorile
May 8 at 1:29
$begingroup$
@DrXorile Nothing special, just filled out the grids using some symmetries, trying to maximize new pairs. On my first attempt, the last grid fit together perfectly to finish off the pairs.
$endgroup$
– noedne
May 8 at 1:56
2
$begingroup$
It may be possible with 3. Tricky to find though.
$endgroup$
– Dr Xorile
May 8 at 1:58
add a comment |
$begingroup$
Here's 16 students in
4
weeks.
0123 A8B9 CAD0 135E
4567 2031 B179 A2BC
89AB ECFD 58E4 94D0
CDEF 6475 F263 7F68
answered May 7 at 15:56
noednenoedne
10.6k12976
$endgroup$
Here's 16 students in
4
weeks.
0123 A8B9 CAD0 135E
4567 2031 B179 A2BC
89AB ECFD 58E4 94D0
CDEF 6475 F263 7F68
answered May 7 at 15:56
noednenoedne
10.6k12976
answered May 7 at 15:56
noednenoedne
10.6k12976
answered May 7 at 15:56
noednenoedne
10.6k12976
answered May 7 at 15:56
noednenoedne
10.6k12976
10.6k12976
$begingroup$
I can confirm this works! Good job. How did you find it?
$endgroup$
– Dr Xorile
May 8 at 1:29
$begingroup$
@DrXorile Nothing special, just filled out the grids using some symmetries, trying to maximize new pairs. On my first attempt, the last grid fit together perfectly to finish off the pairs.
$endgroup$
– noedne
May 8 at 1:56
2
$begingroup$
It may be possible with 3. Tricky to find though.
$endgroup$
– Dr Xorile
May 8 at 1:58
add a comment |
$begingroup$
I can confirm this works! Good job. How did you find it?
$endgroup$
– Dr Xorile
May 8 at 1:29
$begingroup$
@DrXorile Nothing special, just filled out the grids using some symmetries, trying to maximize new pairs. On my first attempt, the last grid fit together perfectly to finish off the pairs.
$endgroup$
– noedne
May 8 at 1:56
2
$begingroup$
It may be possible with 3. Tricky to find though.
$endgroup$
– Dr Xorile
May 8 at 1:58
$begingroup$
I can confirm this works! Good job. How did you find it?
$endgroup$
– Dr Xorile
May 8 at 1:29
$begingroup$
I can confirm this works! Good job. How did you find it?
$endgroup$
– Dr Xorile
May 8 at 1:29
$begingroup$
@DrXorile Nothing special, just filled out the grids using some symmetries, trying to maximize new pairs. On my first attempt, the last grid fit together perfectly to finish off the pairs.
$endgroup$
– noedne
May 8 at 1:56
$begingroup$
@DrXorile Nothing special, just filled out the grids using some symmetries, trying to maximize new pairs. On my first attempt, the last grid fit together perfectly to finish off the pairs.
$endgroup$
– noedne
May 8 at 1:56
2
2
$begingroup$
It may be possible with 3. Tricky to find though.
$endgroup$
– Dr Xorile
May 8 at 1:58
$begingroup$
It may be possible with 3. Tricky to find though.
$endgroup$
– Dr Xorile
May 8 at 1:58
add a comment |
$begingroup$
As a starting point for Question 2, here it is done in
7 weeks
Solution
01234 IG968 HCFA1 97HEP BA92H HDCE0 EB095
56789 B1HKJ 3P4JD MF8AK L3CJ8 8K1F2 GN46H
ABCDE P4307 E20BM N3LI4 KDGN7 GMBNI 3IACK
FGHIJ D5MAL 7698K J50D6 50461 P6LJA DF712
KLMNP NF2CE GILN5 1CG2B IEMPF 93475 JM8PL
answered May 8 at 16:35
hexominohexomino
50k4149239
$endgroup$
$begingroup$
Nice! I found one for nine weeks.
$endgroup$
– Freddy Barrera
May 8 at 19:20
add a comment |
$begingroup$
As a starting point for Question 2, here it is done in
7 weeks
Solution
01234 IG968 HCFA1 97HEP BA92H HDCE0 EB095
56789 B1HKJ 3P4JD MF8AK L3CJ8 8K1F2 GN46H
ABCDE P4307 E20BM N3LI4 KDGN7 GMBNI 3IACK
FGHIJ D5MAL 7698K J50D6 50461 P6LJA DF712
KLMNP NF2CE GILN5 1CG2B IEMPF 93475 JM8PL
answered May 8 at 16:35
hexominohexomino
50k4149239
$endgroup$
$begingroup$
Nice! I found one for nine weeks.
$endgroup$
– Freddy Barrera
May 8 at 19:20
add a comment |
$begingroup$
As a starting point for Question 2, here it is done in
7 weeks
Solution
01234 IG968 HCFA1 97HEP BA92H HDCE0 EB095
56789 B1HKJ 3P4JD MF8AK L3CJ8 8K1F2 GN46H
ABCDE P4307 E20BM N3LI4 KDGN7 GMBNI 3IACK
FGHIJ D5MAL 7698K J50D6 50461 P6LJA DF712
KLMNP NF2CE GILN5 1CG2B IEMPF 93475 JM8PL
answered May 8 at 16:35
hexominohexomino
50k4149239
$endgroup$
As a starting point for Question 2, here it is done in
7 weeks
Solution
01234 IG968 HCFA1 97HEP BA92H HDCE0 EB095
56789 B1HKJ 3P4JD MF8AK L3CJ8 8K1F2 GN46H
ABCDE P4307 E20BM N3LI4 KDGN7 GMBNI 3IACK
FGHIJ D5MAL 7698K J50D6 50461 P6LJA DF712
KLMNP NF2CE GILN5 1CG2B IEMPF 93475 JM8PL
answered May 8 at 16:35
hexominohexomino
50k4149239
answered May 8 at 16:35
hexominohexomino
50k4149239
answered May 8 at 16:35
hexominohexomino
50k4149239
answered May 8 at 16:35
hexominohexomino
50k4149239
50k4149239
$begingroup$
Nice! I found one for nine weeks.
$endgroup$
– Freddy Barrera
May 8 at 19:20
add a comment |
$begingroup$
Nice! I found one for nine weeks.
$endgroup$
– Freddy Barrera
May 8 at 19:20
$begingroup$
Nice! I found one for nine weeks.
$endgroup$
– Freddy Barrera
May 8 at 19:20
$begingroup$
Nice! I found one for nine weeks.
$endgroup$
– Freddy Barrera
May 8 at 19:20
add a comment |
$begingroup$
A quick intuitive answer:
For 16 students:
6 weeks
Reasoning (edited to account for the corner case, as pointed out by hexomino);
2nd week: swap 1st and 3rd columns. 3rd week: swap 2nd and 4th columns. Do the same operations with the rows over the next two weeks. Before swapping, either the first or the preceding week's formation can be taken as the starting point. 5th week: swap the two middle columns. 6th week: swap the middle rows. At the beginning of the 7th week, each student would have been next to each other student.
For 25 students:
8 weeks
Reasoning:
Follow the above reasoning, except that the rows/columns to be swapped would be 1st/4th, 2nd/5th, 2nd/3rd, and 3rd/4th.
answered May 7 at 15:12
Mazurka FahrMazurka Fahr
1114
$endgroup$
2
$begingroup$
Won't students who are originally in opposite corners stay in opposite corners?
$endgroup$
– hexomino
May 7 at 15:14
$begingroup$
hmm. Guess, I'd need to add a few extra steps for the corner cases. Not sure if this would be the optimum solution.
$endgroup$
– Mazurka Fahr
May 7 at 15:17
1
$begingroup$
In the new version, students in the original first column never sit next to students in the original third column.
$endgroup$
– hexomino
May 7 at 15:31
$begingroup$
:(. Can't avoid the extra steps.
$endgroup$
– Mazurka Fahr
May 7 at 15:32
add a comment |
$begingroup$
A quick intuitive answer:
For 16 students:
6 weeks
Reasoning (edited to account for the corner case, as pointed out by hexomino);
2nd week: swap 1st and 3rd columns. 3rd week: swap 2nd and 4th columns. Do the same operations with the rows over the next two weeks. Before swapping, either the first or the preceding week's formation can be taken as the starting point. 5th week: swap the two middle columns. 6th week: swap the middle rows. At the beginning of the 7th week, each student would have been next to each other student.
For 25 students:
8 weeks
Reasoning:
Follow the above reasoning, except that the rows/columns to be swapped would be 1st/4th, 2nd/5th, 2nd/3rd, and 3rd/4th.
answered May 7 at 15:12
Mazurka FahrMazurka Fahr
1114
$endgroup$
2
$begingroup$
Won't students who are originally in opposite corners stay in opposite corners?
$endgroup$
– hexomino
May 7 at 15:14
$begingroup$
hmm. Guess, I'd need to add a few extra steps for the corner cases. Not sure if this would be the optimum solution.
$endgroup$
– Mazurka Fahr
May 7 at 15:17
1
$begingroup$
In the new version, students in the original first column never sit next to students in the original third column.
$endgroup$
– hexomino
May 7 at 15:31
$begingroup$
:(. Can't avoid the extra steps.
$endgroup$
– Mazurka Fahr
May 7 at 15:32
add a comment |
$begingroup$
A quick intuitive answer:
For 16 students:
6 weeks
Reasoning (edited to account for the corner case, as pointed out by hexomino);
2nd week: swap 1st and 3rd columns. 3rd week: swap 2nd and 4th columns. Do the same operations with the rows over the next two weeks. Before swapping, either the first or the preceding week's formation can be taken as the starting point. 5th week: swap the two middle columns. 6th week: swap the middle rows. At the beginning of the 7th week, each student would have been next to each other student.
For 25 students:
8 weeks
Reasoning:
Follow the above reasoning, except that the rows/columns to be swapped would be 1st/4th, 2nd/5th, 2nd/3rd, and 3rd/4th.
answered May 7 at 15:12
Mazurka FahrMazurka Fahr
1114
$endgroup$
A quick intuitive answer:
For 16 students:
6 weeks
Reasoning (edited to account for the corner case, as pointed out by hexomino);
2nd week: swap 1st and 3rd columns. 3rd week: swap 2nd and 4th columns. Do the same operations with the rows over the next two weeks. Before swapping, either the first or the preceding week's formation can be taken as the starting point. 5th week: swap the two middle columns. 6th week: swap the middle rows. At the beginning of the 7th week, each student would have been next to each other student.
For 25 students:
8 weeks
Reasoning:
Follow the above reasoning, except that the rows/columns to be swapped would be 1st/4th, 2nd/5th, 2nd/3rd, and 3rd/4th.
answered May 7 at 15:12
Mazurka FahrMazurka Fahr
1114
edited May 7 at 15:34
answered May 7 at 15:12
Mazurka FahrMazurka Fahr
1114
answered May 7 at 15:12
Mazurka FahrMazurka Fahr
1114
answered May 7 at 15:12
Mazurka FahrMazurka Fahr
1114
1114
2
$begingroup$
Won't students who are originally in opposite corners stay in opposite corners?
$endgroup$
– hexomino
May 7 at 15:14
$begingroup$
hmm. Guess, I'd need to add a few extra steps for the corner cases. Not sure if this would be the optimum solution.
$endgroup$
– Mazurka Fahr
May 7 at 15:17
1
$begingroup$
In the new version, students in the original first column never sit next to students in the original third column.
$endgroup$
– hexomino
May 7 at 15:31
$begingroup$
:(. Can't avoid the extra steps.
$endgroup$
– Mazurka Fahr
May 7 at 15:32
add a comment |
2
$begingroup$
Won't students who are originally in opposite corners stay in opposite corners?
$endgroup$
– hexomino
May 7 at 15:14
$begingroup$
hmm. Guess, I'd need to add a few extra steps for the corner cases. Not sure if this would be the optimum solution.
$endgroup$
– Mazurka Fahr
May 7 at 15:17
1
$begingroup$
In the new version, students in the original first column never sit next to students in the original third column.
$endgroup$
– hexomino
May 7 at 15:31
$begingroup$
:(. Can't avoid the extra steps.
$endgroup$
– Mazurka Fahr
May 7 at 15:32
2
2
$begingroup$
Won't students who are originally in opposite corners stay in opposite corners?
$endgroup$
– hexomino
May 7 at 15:14
$begingroup$
Won't students who are originally in opposite corners stay in opposite corners?
$endgroup$
– hexomino
May 7 at 15:14
$begingroup$
hmm. Guess, I'd need to add a few extra steps for the corner cases. Not sure if this would be the optimum solution.
$endgroup$
– Mazurka Fahr
May 7 at 15:17
$begingroup$
hmm. Guess, I'd need to add a few extra steps for the corner cases. Not sure if this would be the optimum solution.
$endgroup$
– Mazurka Fahr
May 7 at 15:17
1
1
$begingroup$
In the new version, students in the original first column never sit next to students in the original third column.
$endgroup$
– hexomino
May 7 at 15:31
$begingroup$
In the new version, students in the original first column never sit next to students in the original third column.
$endgroup$
– hexomino
May 7 at 15:31
$begingroup$
:(. Can't avoid the extra steps.
$endgroup$
– Mazurka Fahr
May 7 at 15:32
$begingroup$
:(. Can't avoid the extra steps.
$endgroup$
– Mazurka Fahr
May 7 at 15:32
add a comment |
$begingroup$
The first question i believe it's
256
and the second one it's
625
I think the solution of this one is to multiply the two numbers given. An other example is:find all the combinations of these words PANEL(wich i'll use the same equation).
answered May 7 at 12:49
riki481riki481
339
$endgroup$
$begingroup$
it's a math logic that my teacher told me, we have 16 students (in the first question) and 16 chairs, 16x16 = 256. Like, we need to repeat the same process 16 times for all of the 16 students
$endgroup$
– riki481
May 7 at 13:03
$begingroup$
This seems more like the total number of combinations, rather than how quickly you can get each student to have sat next to every other 8 directions
$endgroup$
– Smock
May 7 at 13:56
$begingroup$
Guess there is another step i need to do, but i don't remember
$endgroup$
– riki481
May 7 at 14:00
1
$begingroup$
After actually looking into the question, these numbers are way too large.
$endgroup$
– Arnaud Mortier
May 7 at 14:28
add a comment |
$begingroup$
The first question i believe it's
256
and the second one it's
625
I think the solution of this one is to multiply the two numbers given. An other example is:find all the combinations of these words PANEL(wich i'll use the same equation).
answered May 7 at 12:49
riki481riki481
339
$endgroup$
$begingroup$
it's a math logic that my teacher told me, we have 16 students (in the first question) and 16 chairs, 16x16 = 256. Like, we need to repeat the same process 16 times for all of the 16 students
$endgroup$
– riki481
May 7 at 13:03
$begingroup$
This seems more like the total number of combinations, rather than how quickly you can get each student to have sat next to every other 8 directions
$endgroup$
– Smock
May 7 at 13:56
$begingroup$
Guess there is another step i need to do, but i don't remember
$endgroup$
– riki481
May 7 at 14:00
1
$begingroup$
After actually looking into the question, these numbers are way too large.
$endgroup$
– Arnaud Mortier
May 7 at 14:28
add a comment |
$begingroup$
The first question i believe it's
256
and the second one it's
625
I think the solution of this one is to multiply the two numbers given. An other example is:find all the combinations of these words PANEL(wich i'll use the same equation).
answered May 7 at 12:49
riki481riki481
339
$endgroup$
The first question i believe it's
256
and the second one it's
625
I think the solution of this one is to multiply the two numbers given. An other example is:find all the combinations of these words PANEL(wich i'll use the same equation).
answered May 7 at 12:49
riki481riki481
339
edited May 8 at 10:10
answered May 7 at 12:49
riki481riki481
339
answered May 7 at 12:49
riki481riki481
339
answered May 7 at 12:49
riki481riki481
339
339
$begingroup$
it's a math logic that my teacher told me, we have 16 students (in the first question) and 16 chairs, 16x16 = 256. Like, we need to repeat the same process 16 times for all of the 16 students
$endgroup$
– riki481
May 7 at 13:03
$begingroup$
This seems more like the total number of combinations, rather than how quickly you can get each student to have sat next to every other 8 directions
$endgroup$
– Smock
May 7 at 13:56
$begingroup$
Guess there is another step i need to do, but i don't remember
$endgroup$
– riki481
May 7 at 14:00
1
$begingroup$
After actually looking into the question, these numbers are way too large.
$endgroup$
– Arnaud Mortier
May 7 at 14:28
add a comment |
$begingroup$
it's a math logic that my teacher told me, we have 16 students (in the first question) and 16 chairs, 16x16 = 256. Like, we need to repeat the same process 16 times for all of the 16 students
$endgroup$
– riki481
May 7 at 13:03
$begingroup$
This seems more like the total number of combinations, rather than how quickly you can get each student to have sat next to every other 8 directions
$endgroup$
– Smock
May 7 at 13:56
$begingroup$
Guess there is another step i need to do, but i don't remember
$endgroup$
– riki481
May 7 at 14:00
1
$begingroup$
After actually looking into the question, these numbers are way too large.
$endgroup$
– Arnaud Mortier
May 7 at 14:28
$begingroup$
it's a math logic that my teacher told me, we have 16 students (in the first question) and 16 chairs, 16x16 = 256. Like, we need to repeat the same process 16 times for all of the 16 students
$endgroup$
– riki481
May 7 at 13:03
$begingroup$
it's a math logic that my teacher told me, we have 16 students (in the first question) and 16 chairs, 16x16 = 256. Like, we need to repeat the same process 16 times for all of the 16 students
$endgroup$
– riki481
May 7 at 13:03
$begingroup$
This seems more like the total number of combinations, rather than how quickly you can get each student to have sat next to every other 8 directions
$endgroup$
– Smock
May 7 at 13:56
$begingroup$
This seems more like the total number of combinations, rather than how quickly you can get each student to have sat next to every other 8 directions
$endgroup$
– Smock
May 7 at 13:56
$begingroup$
Guess there is another step i need to do, but i don't remember
$endgroup$
– riki481
May 7 at 14:00
$begingroup$
Guess there is another step i need to do, but i don't remember
$endgroup$
– riki481
May 7 at 14:00
1
1
$begingroup$
After actually looking into the question, these numbers are way too large.
$endgroup$
– Arnaud Mortier
May 7 at 14:28
$begingroup$
After actually looking into the question, these numbers are way too large.
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– Arnaud Mortier
May 7 at 14:28
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2
$begingroup$
Perhaps the second 25 student puzzle (which presumably has five rows of five) should be a new question after this is solved?
$endgroup$
– Weather Vane
May 7 at 18:27
2
$begingroup$
@WeatherVane: Following your suggestion, I have now posted the case for 25 students at: math.stackexchange.com/questions/3218384/…
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– Bernardo Recamán Santos
May 8 at 12:41
1
$begingroup$
I made a solution validator some may find useful.
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– TemporalWolf
May 8 at 23:30
2
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I found an optimal solution to the 25 student problem (5 weeks), or rather, my computer program did. The solution can be found over on Mathematics SE.
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– Jaap Scherphuis
May 9 at 15:59