Talk:Kirkman's schoolgirl problem

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I would like to point out that a sentence in the Wikipedia page "Kirkman's schoolgirl problem" is incorrect. It is in the fourth paragraph of the "History" section: "In 1850, Kirkman posed the 15 schoolgirl problem, which would become much more famous than the 1847 paper he had already written. (...) However, in publishing his solutions to the schoolgirl problem, Kirkman neglected to refer readers to his own 1847 paper, and this omission would have serious consequences for invention and priority as seen below."

It is incorrect to say that Kirkman neglected to refer to his 1847 paper. Indeed, in his 1850 paper "On the triads made with fifteen things", where he published his first solution, he wrote: "I obtained this property of the triads made with fifteen things four years ago, by observing that, if you substitute in Q_15, at page 195 of the second volume, N. S. of the Cambridge and Dublin Mathematical Journal, (...)". Here Kirkman was referring precisely to his 1847 paper "On a problem in combinations", published in The Cambridge and Dublin Mathematical Journal, volume II. 147.163.24.107 (talk) 13:33, 25 July 2022 (UTC)[reply]

Minimalist Equivalent alternative problem for 7 schoolboys walking

Minimalist solution to Kirkman's 15 girls' walking problem：Equivalent simplified alternative problem for 7 boys' walking. Question 1: There is a math teacher Hansen in a boarding school. After dinner, Teacher Hansen takes 7 boys in the same dormitory for a walk every day. Teacher Hansen is thinking about a question. Seven boys walked in a row of three. Every day only three of them walked in a row, and the other four walked alone without a row. How to arrange walks within a week so that a certain boy and the other six boys have the opportunity to walk in the same row and only once? Question 2: Mathematics teacher Hansen takes 7 boys in the same dormitory for a walk every day after dinner. Hope the arrangement: 7 boys walk in two rows every day, arranged in three rows of two people, one person in a row, a total of four groups. How to arrange walks within a week so that a certain boy and the other six boys have the opportunity to walk in the same row and only once? We will replace the original Kirkman 15 girls walking problem with 7 boys walking problem one and problem two. Combining the two arrangements of 7 students each, there are 14 students, and together with the leader, Mr. Hansen, there are 15 people. That is, the 5 rows listed in the first question, excluding the four rows with one person in each row after the first row that already has three people, plus the first three rows with two people in each row listed in the second question, these three rows are 3 people in each row. , the last row is one person for the first question, the fifth row, plus one person for the second question, the fourth row, two students plus the leader, Mr. Hansen, also a row of three. Then all five rows have three people in each row. The conditions are consistent with the original question of Kirkman's 15 girls walking problem. Liang Haisheng in Tokyo.

Minimalist solution to Kirkman's schoolgirl problem

《Pithy Formula 124 》

———————————————

1 2 4, 2 3 5, 3 4 6, 4 5 7, 5 6 1, 6 7 2, 7 1 3 ：7

2 3 5, 3 4 6, 4 5 7, 5 6 1, 6 7 2, 7 1 3, 1 2 4 ：6

3 4 6, 4 5 7, 5 6 1, 6 7 2, 7 1 3, 1 2 4, 2 3 5 ：5

~~4 5 7, 5 6 1, 6 7 2, 7 1 3, 1 2 4, 2 3 5, 3 4 6~~ :4

5 6 1, 6 7 2, 7 1 3, 1 2 4, 2 3 5, 3 4 6, 4 5 7 ：3

~~6 7 2, 7 1 3, 1 2 4, 2 3 5, 3 4 6, 4 5 7,5 6 1~~ :2

~~7 1 3, 1 2 4, 2 3 5, 3 4 6, 4 5 7, 5 6 1, 6 7 2~~ :1

———————————————

      1   2  4.                   OOO
      2’ 3’ 5        =           XXO
      3  4’ 6‘                    OXX
      5’ 6  1’                    XOX

———————————————

  124,  235,  346,  457,  561,  672, 713     
2’3’5,3’4’6,4’5’7,5’6’1,6’7’2,7’1’3,1’2’4
34’6’,45’7’,56’1’,67’2’,71’3’,12’4’,23’5’
5’61’,6’72’,7’13’,1’24’,2’35’,3’46’,4’57’

  77’0,    11’0,    22’0,    33’0,  44’0,   55’0,    66’0

———————————————

   1’=8,  2’=9,  3’=A,  4’=B,  5’=C, 6’=D,  7’=E, 0=F

     124    235    346    457     561    672    713            
     9A5    AB6    BC7    CD1     DE2    E83    894
     3BD    4CE    5D8    6E9     78A    19B    2AC
     C68    D79    E1A    82B     93C    A4D    B5E
     7EF    18F    29F    3AF     4BF    5CF    6DF

——by Liang Haisheng 2019 183.177.128.238 (talk) 13:42, 28 November 2023 (UTC)[reply]