# Kirkman's schoolgirl problem

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Kirkman's schoolgirl problem is a problem in combinatorics proposed by Rev. 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.

## Solution

If the girls are numbered from 0 to 14, the following arrangement is one solution:

Sun. Mon. Tues. Wed. Thurs. Fri. Sat.
0,  5, 10  0,  1,  4  1,  2,  5  4,  5,  8  2,  4, 10  4,  6, 12 10, 12,  3
1,  6, 11  2,  3,  6  3,  4,  7  6,  7, 10  3,  5, 11  5,  7, 13 11, 13,  4
2,  7, 12  7,  8, 11  8,  9, 12 11, 12,  0  6,  8, 14  8, 10,  1 14,  1,  7
3,  8, 13  9, 10, 13 10, 11, 14 13, 14,  2  7,  9,  0  9, 11,  2  0,  2,  8
4,  9, 14 12, 14,  5 13,  0,  6  1,  3,  9 12, 13,  1 14,  0,  3  5,  6,  9

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.

There are seven non-isomorphic solutions to the schoolgirl problem. Two of these can be visualized as relations between a tetrahedron and its vertices, edges, and faces. An approach using projective geometry of three dimensions over GF(2) is given below.

## XOR Triple Solution

If the girls are numbered in binary numbers 0001 to 1111, the following arrangement is one solution such at for any three girls that form a group, the bitwise XOR of any two is equal to the third:

Sun. Mon. Tues. Wed. Thurs. Fri. Sat.
0001, 0010, 0011 0001, 0100, 0101 0001, 0110, 0111 0001, 1000, 1001 0001, 1010, 1011 0001, 1100, 1101 0001, 1110, 1111
0100, 1000, 1100 0010, 1000, 1010 0010, 1001, 1011 0010, 1100, 1110 0010, 1101, 1111 0010, 0100, 0110 0010, 0101, 0111
0101, 1010, 1111 0011, 1101, 1110 0011, 1100, 1111 0011, 0101, 0110 0011, 0100, 0111 0011, 1001, 1010 0011, 1000, 1011
0110, 1011, 1101 0110, 1001, 1111 0100, 1010, 1110 0100, 1011, 1111 0101, 1001, 1100 0101, 1011, 1110 0100, 1001, 1101
0111, 1001, 1110 0111, 1011, 1100 0101, 1000, 1101 0111, 1010, 1101 0110, 1000, 1110 0111, 1000, 1111 0110, 1010, 1100

This solution has a geometric interpretation in connection with Galois geometry and PG(3,2). Take a tetrahedron and label its vertices as 0001, 0010, 0100 and 1000. Label its six edge centers as the XOR of the vertices of that edge. Label the four face centers as the XOR of the three vertices of that face, and the body center gets the label 1111. Then the 35 triads of the XOR solution correspond exactly to the 35 lines of PG(3,2). Each day corresponds to a spread and each week to a packing.

## History

The first solution was published by Arthur Cayley. This was shortly followed by Kirkman's own solution which was given as a special case of his considerations on combinatorial arrangements published three years prior. J. J. Sylvester also investigated the problem and ended up declaring that Kirkman stole the idea from him. The puzzle appeared in several recreational mathematics books at the turn of the century by Lucas, Rouse Ball, Ahrens, and Dudeney.

Kirkman often complained about the fact that his substantial paper (Kirkman 1847) was totally eclipsed by the popular interest in the schoolgirl problem.

## Galois geometry

In 1910 the problem was addressed using Galois geometry by George Conwell

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.:67

As there are seven days in a week, the heptad is an important part of the solution:

When two points as A and B of the line ABC are chosen, each of the five other lines through A is met by only one of the five other lines through B. The five points determined by the intersections of these pairs of lines, together with the two points A and B we designate a "heptad".:68

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.:69

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.:74

## Spreads and packing

A partition of points into lines is called a spread, and the partition of the spreads of a geometry is called a packing.:66 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,:91 and he presented two of them.:75

## Generalization

The problem can be generalized to $n$ girls, where $n$ must be an odd multiple of 3 (that is $n\equiv 3{\pmod {6}}$ ), walking in triplets for ${\frac {1}{2}}(n-1)$ days, with the requirement, again, that no pair of girls walk in the same row twice. The solution to this generalisation is a Steiner triple system, an S(2, 3, 6t + 3) with parallelism (that is, one in which each of the 6t + 3 elements occurs exactly once in each block of 3-element sets), known as a Kirkman triple system. It is this generalization of the problem that Kirkman discussed first, while the famous special case $n=15$ was only proposed later. A complete solution to the general case was published by D. K. Ray-Chaudhuri and R. M. Wilson in 1968, though it had already been solved by Lu Jiaxi (Chinese: 陆家羲) in 1965, but had not been published at that time.

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. This for 10 days in a row.

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. However, this term is currently not commonly used and evidence suggests that there is no common name for the process.

The Resolvable Coverings problem considers the general $n$ girls, $g$ groups case where each pair of girls must be in the same group at some point, but we want to use as few days as possible. This can, for example, be used to schedule a rotating table plan, in which each pair of guests must at some point be at the same table.

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.