The Nine Schoolgirls Challenge

In this simpler variation of Thomas Kirkman’s famous fifteen schoolgirls problem, which he posed in 1850, nine girls in a school walk out in groups of three for four days in a row. You, the teacher, must arrange the girls in walking groups so that no pair of girls ever walks in the same row (group of three) more than once.

How to Play: Click and drag the girls’ first initials from the “Girls” column on the left to the “Groups” column on the right. Schoolgirl pairs that do not duplicate any pairs from other rows will appear in the “Completed Pairs” list at the far right. Pairs that duplicate previous pairs will turn red in the “Completed Pairs” list. When viable walking groups have been arranged for a given day, all nine girls for that day will turn green. The goal is to include all 36 possible pairings of schoolgirls, turning all of the walking groups green over the four days as fast as you can. The first day has been completed for you. Good luck!

Emily Fuhrman for Quanta Magazine, with design by Olena Shmahalo. Collage resources from The Graphics Fairy and and Clker.


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  • Trivial. 143 seconds for this. Just follow the following rule – a row in one iteration becomes the determinant calculation triple in the next one.

  • Agreed, there are several solutions based on diagonalisations, which work easily (with no editing). three design based ones so far which took 60-90s each. (Although these are all related – are there strictly seven solutions or seven classes of solutions?)

    I don’t think this article really makes clear why this problem is of such enduring mathematical interest.. Which I’m sure it is. The issue of solutions without apparent design seems to be key to the most recent work

  • I had a bit more trouble with it than the other commentors. This was my method.

    (1) I noticed that I could turn the figure 90 degrees to the right to get a valid solution

    (2) I noticed that I could turn it like a combination lock: the first column stayed in place, the second moved 1 place, the third moved 2 places.

    (3) However, for the third solution it ends up being a bit more unusual. I take the left-right diagonal and make it a row (makes sense). But then the other two rows are made from B, F, G and D, H, C.

    I don’t know why #3 worked.

  • @Campbell Hutcheson – Your permutation (3) is quite similar to (2), but you’ve moved the 2nd and 3rd rows by 2 and 4 (or 2 and 1) places respectively.

  • Finished in 107.0 seconds… First time actually hearing about this and I found that disappointing. This is really very interesting and should be included in school. It would make for a great game for all those to be nerds (:

  • Took me about 90 seconds. For the first blank i juts rotated the picture by 90 degrees. The second one i translated columns number 1 and 2 down by 1 and 2 spaces (respectively). The last one for the last one i put “ABC” in the first column like always and then tried letters in the rows. If there was no contradiction with the primary letter (the “A”, “B”, or “C”), it stayed. If the letter conflicted with a non-primary letter, I took them both out and tried something different. All in all an interesting puzzle

  • Here’s quick and simple logic for this. The sets are Rows, Columns, first-diagonal, other-diagonal.

    Rows: (The game starts out with this set already in place for the first day.)




    Gives ABC, DEF, GHI.


    A B C

    | | |

    D E F

    | | |

    C F I

    Gives ADC, BEF, CFI.

    First diagonal:

    A B C

    \ \

    D E F

    \ \

    G H I

    Gives AEI, BFG, DHC. (The BF diagonal is completed by wrapping around to G, which is the only letter in a different row&column from B+F. The DH diagonal is completed by wrapping around to C, which is the only letter in a different row&column from D+H.)

    Second diagonal:

    A B C

    / /

    D E F

    / /

    G H I

    Gives GEC, DBI, HFA. (The DB diagonal and FH diagonal are completed by the same logic as the other diagonal.)

  • I wonder how many of these fast solutions were obtained by people who had seen the puzzle before. One note: each row contains three pairs and in 4 days there are 3pairs x 3rows x 4days = 36, which is also the total number of ways of choosing 2 people from 9, so all possible pairings appear in each solution.

  • I did this in 104 seconds by randomly clicking and dragging, and I lost the first 15-20 seconds because I undid the first day and had to go back and re-do it. I can see how this would be far more irritating if you were doing it by pen & paper, though.

  • Write down A, B … in a three by three square i.e. first line A B C, right side C D E , etc
    And you have solved the problem.

  • I really still am not sure about the instructions. Are they pairs or triples? It seems no one from a triple can be in another triple with the same girls the next day, you can't just count ac and eg as different pairs. I used the diagnolization method but it took me 357 seconds because I couldn't figure out how to operate the puzzle or understand the instructions.

    I also don't understand how this scales up to the n, t, k problem n things with no repeating group of t nor do I understand what a design is after reading this article.

  • Got it in 134 sec, but how to solve the original, harder problem? I'm sure the solution is related to this one.

  • I guessed a solution that took me 65.3 seconds to implement. The underlying idea wasn't anything particularly deep.

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