Kirkman's schoolgirl problem is a problem in combinatorics proposed by Thomas Penyngton Kirkman in 1850 as Query VI in The Lady's and Gentleman's Diary (pg.48). The problem states:
Fifteen young ladies in a school walk out three abreast for seven days in succession: it is required to arrange them daily so that no two shall walk twice abreast.
A solution to this problem is an example of a Kirkman triple system, which is a Steiner triple system having a parallelism, that is, a partition of the blocks of the triple system into parallel classes which are themselves partitions of the points into disjoint blocks. Such Steiner systems that have a parallelism are also called resolvable.
There are exactly seven non-isomorphic solutions to the schoolgirl problem, as originally listed by Frank Nelson Cole in Kirkman Parades in 1922. The seven solutions are summarized in the table below, denoting the 15 girls with the letters A to O.
Solution class | Automorphism group | Day 1 | Day 2 | Day 3 | Day 4 | Day 5 | Day 6 | Day 7 | |||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
Solution I | align='center' | Order 168, generated by (A K G E I L B)(C H M J N O D) and (A M L K O C D)(B H N G I E J). Related to PG(3,2). | align='center' | ABC DEF GHI JKL MNO | align='center' | ADG BEH CJM FKN ILO | align='center' | AEO BIJ CDN FHL GKM | align='center' | AIM BDL CEK FGO HJN | align='center' | AFJ BKO CGL DHM EIN | align='center' | AHK BGN CFI DJO ELM | align='center' | ALN BFM CHO DIK EGJ | |
Solution II | align='center' | Order 168, generated by (A B I M F C J)(D N H K O L E) and (A J M I B F C)(D H G N K E O). Related to PG(3,2). | align='center' | ABC DEF GHI JKL MNO | align='center' | ADG BEH CJM FKN ILO | align='center' | AEO BIJ CDN FHL GKM | align='center' | AFJ BGN CHO DIK ELM | align='center' | AHK BFM CGL DJO EIN | align='center' | AIM BDL CEK FGO HJN | align='center' | ALN BKO CFI DHM EGJ | |
Solution III | align='center' | Order 24, generated by (A H E)(B O K)(C F I)(D J L)(G N M) and (A J B M)(D L E O)(F I)(G K H N) | align='center' | ABC DEF GHI JKL MNO | align='center' | ADG BEH CJM FKN ILO | align='center' | AEO BIM CDK FGL HJN | align='center' | AFM BGN CHL DJO EIK | align='center' | AHK BFJ CGO DIN ELM | align='center' | AIJ BDL CEN FHO GKM | align='center' | ALN BKO CFI DHM EGJ | |
Solution IV | align='center' | Order 24, generated by (A J M)(C F I)(D E K)(H O L) and (A L B O)(C I)(D K E N)(G J H M) | align='center' | ABC DEF GHI JKL MNO | align='center' | ADG BEH CJM FKN ILO | align='center' | AEO BIM CDK FGL HJN | align='center' | AFM BKO CHL DIN EGJ | align='center' | AHK BGN CFI DJO ELM | align='center' | AIJ BDL CEN FHO GKM | align='center' | ALN BFJ CGO DHM EIK | |
Solution V | align='center' | Tetrahedral group of order 12, generated by (A L)(B G)(E O)(F J)(H K)(I M) and (A B C)(D L G)(F J I)(E K H) | align='center' | ABC DEF GHI JKL MNO | align='center' | ADG BEJ CHM FKN ILO | align='center' | AEM BDL CIK FGO HJN | align='center' | AFH BKM CGL DJO EIN | align='center' | AIJ BGN CEO DHK FLM | align='center' | AKO BFI CDN EHL GJM | align='center' | ALN BHO CFJ DIM EGK | |
Solution VI | align='center' | Tetrahedral group of order 12, generated by (A L)(B G)(E O)(H K)(F J)(I M) and (A B C)(D L G)(E K H)(F J I) | align='center' | ABC DEF GHI JKL MNO | align='center' | ADG BEJ CHM FKN ILO | align='center' | AEM BDL CIK FGO HJN | align='center' | AFH BKM CGL DJO EIN | align='center' | AIJ BHO CDN EGK FLM | align='center' | AKO BGN CFJ DIM EHL | align='center' | ALN BFI CEO DHK GJM | |
Solution VII | align='center' | Order 21, generated by (A B L C G D N)(E H K I O J F) and (B G L)(C D N)(E F K)(H I O) | align='center' | ABC DEF GHI JKL MNO | align='center' | ADG BEJ CHM FKN ILO | align='center' | AEI BDN CJO FHL GKM | align='center' | AFO BIK CGN DHJ ELM | align='center' | AHK BFM CDL EGO IJN | align='center' | AJM BGL CFI DKO EHN | align='center' | ALN BHO CEK FGJ DIM |
From the number of automorphisms for each solution and the definition of an automorphism group, the total number of solutions including isomorphic solutions is therefore:
15! x \left(
1 | |
168 |
+
1 | |
168 |
+
1 | |
24 |
+
1 | |
24 |
+
1 | |
12 |
+
1 | |
12 |
+
1 | |
21 |
\right)
=15! x
13 | |
42 |
=404,756,352,000
=210 x 35 x 53 x 7 x 11 x 132
The problem has a long and storied history. This section is based on historical work done at different times by Robin Wilson[1] and by Louise Duffield Cummings. The history is as follows:
455=13 x 35
James Joseph Sylvester in 1850 asked if 13 disjoint Kirkman systems of 35 triples each could be constructed to use all triples on 15 girls. No solution was found until 1974 when RHF Denniston at the University of Leicester constructed it with a computer. Denniston's insight was to create a single-week Kirkman solution in such a way that it could be permuted according to a specific permutation of cycle length 13 to create disjoint solutions for subsequent weeks; he chose a permutation with a single 13-cycle and two fixed points like (1 2 3 4 5 6 7 8 9 10 11 12 13)(14)(15). Under this permutation, a triple like 123 would map to 234, 345, ... (11, 12, 13), (12, 13, 1) and (13, 1, 2) before repeating. Denniston thus classified the 455 triples into 35 rows of 13 triples each, each row being the orbit of a given triple under the permutation. In order to construct a Sylvester solution, no single-week Kirkman solution could use two triples from the same row, otherwise they would eventually collide when the permutation was applied to one of them. Solving Sylvester's problem is equivalent to finding one triple from each of the 35 rows such that the 35 triples together make a Kirkman solution. He then asked an Elliott 4130 computer to do exactly that search, which took him 7 hours to find this first-week solution, labeling the 15 girls with the letters A to O:
Day 1 ABJ CEM FKL HIN DGO Day 2 ACH DEI FGM JLN BKO Day 3 ADL BHM GIK CFN EJO Day 4 AEG BIL CJK DMN FHO Day 5 AFI BCD GHJ EKN LMO Day 6 AKM DFJ EHL BGN CIO Day 7 BEF CGL DHK IJM ANO
He stopped the search at that point, not looking to establish uniqueness.
The American minimalist composer Tom Johnson composed a piece of music called Kirkman's Ladies based on Denniston's solution.[4] [5]
As of 2021, it is not known whether there are other non-isomorphic solutions to Sylvester's problem, or how many solutions there are.
The equivalent of the Kirkman problem for 9 schoolgirls results in S(2,3,9), an affine plane isomorphic to the following triples on each day:
Day 1: 123 456 789 Day 2: 147 258 369 Day 3: 159 267 348 Day 4: 168 249 357
The corresponding Sylvester problem asks for 7 different S(2,3,9) systems of 12 triples each, together covering all triples. This solution was known to Bays (1917) which was found again from a different direction by Earl Kramer and Dale Mesner in a 1974 paper titled Intersections Among Steiner Systems (J Combinatorial Theory, Vol 16 pp 273-285). There can indeed be 7 disjoint S(2,3,9) systems, and all such sets of 7 fall into two non-isomorphic categories of sizes 8640 and 6720, with 42 and 54 automorphisms respectively.
Solution 1: Day 1 Day 2 Day 3 Day 4 Week 1 ABC.DEF.GHI ADG.BEH.CFI AEI.BFG.CDH AFH.BDI.CEG Week 2 ABD.CEH.FGI ACF.BGH.DEI AEG.BCI.DFH AHI.BEF.CDG Week 3 ABE.CDI.FGH ACG.BDF.EHI ADH.BGI.CEF AFI.BCH.DEG Week 4 ABF.CEI.DGH ACD.BHI.EFG AEH.BCG.DFI AGI.BDE.CFH Week 5 ABG.CDE.FHI ACH.BEI.DFG ADI.BCF.EGH AEF.BDH.CGI Week 6 ABH.CDF.EGI ACI.BDG.EFH ADE.BFI.CGH AFG.BCE.DHI Week 7 ABI.CFG.DEH ACE.BFH.DGI ADF.BEG.CHI AGH.BCD.EFISolution 1 has 42 automorphisms, generated by the permutations (A I D C F H)(B G) and (C F D H E I)(B G). Applying the 9! = 362880 permutations of ABCDEFGHI, there are 362880/42 = 8640 different solutions all isomorphic to Solution 1.
Solution 2: Day 1 Day 2 Day 3 Day 4 Week 1 ABC.DEF.GHI ADG.BEH.CFI AEI.BFG.CDH AFH.BDI.CEG Week 2 ABD.CEH.FGI ACF.BGH.DEI AEG.BCI.DFH AHI.BEF.CDG Week 3 ABE.CGH.DFI ACI.BFH.DEG ADH.BGI.CEF AFG.BCD.EHI Week 4 ABF.CGI.DEH ACE.BDG.FHI ADI.BCH.EFG AGH.BEI.CDF Week 5 ABG.CDI.EFH ACH.BDF.EGI ADE.BHI.CFG AFI.BCE.DGH Week 6 ABH.CEI.DFG ACD.BFI.EGH AEF.BCG.DHI AGI.BDE.CFH Week 7 ABI.CDE.FGH ACG.BDH.EFI ADF.BEG.CHI AEH.BCF.DGISolution 2 has 54 automorphisms, generated by the permutations (A B D)(C H E)(F G I) and (A I F D E H)(B G). Applying the 9! = 362880 permutations of ABCDEFGHI, there are 362880/54 = 6720 different solutions all isomorphic to Solution 2.
Thus there are 8640 + 6720 = 15360 solutions in total, falling into two non-isomorphic categories.
In addition to S(2,3,9), Kramer and Mesner examined other systems that could be derived from S(5,6,12) and found that there could be up to 2 disjoint S(5,6,12) systems, up to 2 disjoint S(4,5,11) systems, and up to 5 disjoint S(3,4,10) systems. All such sets of 2 or 5 are respectively isomorphic to each other.
In the 21st century, analogues of Sylvester's problem have been visited by other authors under terms like "Disjoint Steiner systems" or "Disjoint Kirkman systems" or "LKTS" (Large Sets of Kirkman Triple Systems), for n > 15.[6] Similar sets of disjoint Steiner systems have also been investigated for the S(5,8,24) Steiner system in addition to triple systems.[7]
In 1910 the problem was addressed using Galois geometry by George Conwell.[8]
The Galois field GF(2) with two elements is used with four homogeneous coordinates to form PG(3,2) which has 15 points, 3 points to a line, 7 points and 7 lines in a plane. A plane can be considered a complete quadrilateral together with the line through its diagonal points. Each point is on 7 lines, and there are 35 lines in all.
The lines of PG(3,2) are identified by their Plücker coordinates in PG(5,2) with 63 points, 35 of which represent lines of PG(3,2). These 35 points form the surface S known as the Klein quadric. For each of the 28 points off S there are 6 lines through it which do not intersect S.[8]
As there are seven days in a week, the heptad is an important part of the solution:A heptad is determined by any two of its points. Each of the 28 points off S lies in two heptads. There are 8 heptads. The projective linear group PGL(3,2) is isomorphic the alternating group on the 8 heptads.[8]
The schoolgirl problem consists in finding seven lines in the 5-space which do not intersect and such that any two lines always have a heptad in common.[8]
In PG(3,2), a partition of the points into lines is called a spread, and a partition of the lines into spreads is called a or . There are 56 spreads and 240 packings. When Hirschfeld considered the problem in his Finite Projective Spaces of Three Dimensions (1985), he noted that some solutions correspond to packings of PG(3,2), essentially as described by Conwell above, and he presented two of them.
The problem can be generalized to
n
n
n\equiv3\pmod{6}
n=15
Many variations of the basic problem can be considered. Alan Hartman solves a problem of this type with the requirement that no trio walks in a row of four more than once using Steiner quadruple systems.
More recently a similar problem known as the Social Golfer Problem has gained interest that deals with 32 golfers who want to get to play with different people each day in groups of 4, over the course of 10 days.
As this is a regrouping strategy where all groups are orthogonal, this process within the problem of organising a large group into smaller groups where no two people share the same group twice can be referred to as orthogonal regrouping. [9]
The Resolvable Coverings problem considers the general
n
g
The Oberwolfach problem, of decomposing a complete graph into edge-disjoint copies of a given 2-regular graph, also generalizes Kirkman's schoolgirl problem. Kirkman's problem is the special case of the Oberwolfach problem in which the 2-regular graph consists of five disjoint triangles.