March 1997 Presentation Topic (continued)

Here the points represent the varieties; the six line segments and one circle represent the fields. This solution can be interpreted as a geometry (the "lines" of this geometry are the six line segments and one circle). In this geometric interpretation, the "points" stand for varieties of wheat and the "lines" stand for fields. There is an interesting duality to this geometry: you could equally well interpret the "lines" as being the varieties of wheat and the "points" as being the fields, and the same properties are preserved.

We can also list the fields and the varieties planted in each one:

field # | first variety | second variety | third variety |

1 | 1 | 2 | 4 |

2 | 2 | 3 | 5 |

3 | 3 | 4 | 6 |

4 | 4 | 5 | 7 |

5 | 5 | 6 | 1 |

6 | 6 | 7 | 2 |

7 | 7 | 1 | 3 |

**Solution to the Schoolgirls' Walk Problem.**
The following diagram pictorially represents the answer:

The blocks, divided up into the four days of walks, are:

**Solution to the Tournament Scheduling Problem.**
If each of the *n* teams plays every day, then there must be *n*/2 games
each day (as each game involves two teams). In particular, *n*/2 must
be an integer, so no solution is possible if *n* is odd.

If *n*=2 the solution is obvious: there is just one game, taking place on
one day. If *n*=4 the following schedule works:

day 1 | day 2 | day 3 | |

game 1 | A vs. B | A vs. C | A vs. D |

game 2 | C vs. D | B vs. D | B vs. C |

For *n*=6 the following schedule works:

day 1 | day 2 | day 3 | day 4 | day 5 | |

game 1 | A vs. B | A vs. C | A vs. D | A vs. E | A vs. F |

game 2 | C vs. D | B vs. E | B vs. F | B vs. D | B vs. C |

game 3 | E vs. F | D vs. F | C vs. E | C vs. F | D vs. E |

Many other schedules are possible as well.
No pattern is immediately obvious from the particular schedules we've
chosen above. For the *n*=8 case, let's be more careful to choose
a schedule in a manner which exemplifies a clear pattern:

day 1 | day 2 | day 3 | day 4 | day 5 | day 6 | day 7 | |

game 1 | A vs. H | B vs. H | C vs. H | D vs. H | E vs. H | F vs. H | G vs. H |

game 2 | B vs. G | C vs. A | D vs. B | E vs. C | F vs. D | G vs. E | A vs. F |

game 3 | C vs. F | D vs. G | E vs. A | F vs. B | G vs. C | A vs. D | B vs. E |

game 4 | D vs. E | E vs. F | F vs. G | G vs. A | A vs. B | B vs. C | C vs. D |

To get a general solution for any even number *n*=2*k* of teams: on day
one, have team 1 playing team 2*k*, team 2 playing team 2*k*-1,
team 3 playing team 2*k*-2, and so on, up through team *k* playing
team *k*+1. For day two, rotate the teams' positions in the schedule,
according to the following rotation:
1->2->3->4->. . . ->2*k*-1->1, with team 2*k* staying put.
(For example, in the above schedule we replaced A with B, B with C,
C with D, D with E, E with F, F with G, and G with A, with H staying
put, to get the schedule for day two from the schedule for day one).
Similarly, perform another rotation to get the third day's schedule,
and so on. It is left to you to verify that this gives a valid schedule
in which each team plays every other team exactly once.

**Answer to Question 1.** Pick a point *P*.
Since there are *v*-1 other points besides *P*, there are *v*-1
pairs which involve *P*. None of these pairs can occur in
more than one block. A block that contains *P* has *k*-1 of these
pairs in it (since that block contains *k*-1 points besides *P*).
Therefore, if *r*
is the number of blocks in which *P* occurs, we must have
*r*(*k*-1)= *v*-1, so *r*=(*v*-1)/(*k*-1), proving that *r* is independent
of the point *P*.

**Answer to Question 2.** The answer to qustion 1
reveals that one condition is that
*r*=(*v*-1)/(*k*-1), which also implies that (*v*-1)/(*k*-1) has to be an
integer.

The other condition is as follows. There are *v*(*v*-1)/2 pairs in total. Each block has *k*(*k*-1)/2 pairs
in it. Since each pair occurs once and only once we get that

In particular, it is necessary that *v*(*v*-1)/*k*(*k*-1) be an integer.

In the Farmer's Wheat Problem, *v*=7. Since (*v*-1)/(*k*-1) must be an
integer, that means *k*-1 must be a factor of 6. Since the only factors
of 6 are 1, 2, 3, and 6, this means *k* must be 2, 3, 4,
or 7. The case *k*=4 is impossible since then *v*(*v*-1)/*k*(*k*-1) would
be 42/12 which is not an integer. This leaves only the cases
*k*=2 (the 21-field solution), *k*=3 (the Fano plane solution), or
*k*=7 (the plant-everything-in-one-field solution which the farmer
rejected).

**Answer to Question 3.** Since we can find a group of blocks
(which have *k* points each) that together represents every one of
the *v* points exactly once, *k* must divide *v*. That means
*v*/*k* must be an integer (it is the number of blocks per group).

**Answer to Question 4.** Although *v*/*k *= 12/3 = 4 is an integer and *v*(*v*-1)/*k*(*k*-1) = 132/6 =
22 is an integer, (*v*-1)/(*k*-1) = 11/2
is not an integer, so this situation fails one of
the necessary conditions. It is therefore not possible to find
such a schedule of walks.

**Answer to Question 5.**
*v*(*v*-1)/*k*(*k*-1) = 156/6 = 26 and (*v*-1)/(*k*-1) = 6 are both integers
so the necessary conditions for a combinatorial design are
met. *v*/*k *= 13/3 is not an integer so it cannot be resolvable.
A non-resolvable combinatorial design with these parameters has the
following
blocks. It is divided up into sets for convenience and also hoping
that you will notice a pattern (that also occurred in the Fano plane).

**Answer to Question 6.** We know that *k* has to be 3 and
*v*/*k* has to be an integer, so *v* must be a multiple of 3.
Also (*v*-1)/(*k*-1) must be an integer, so *v*-1 must be a multiple of
2, so *v* must be odd. Finally, *v*(*v*-1)/*k*(*k*-1) must be an integer,
so *v*(*v*-1) must be a multiple of 6; this condition will automatically
be satisfied if *v* is an odd multiple of 3.

Therefore, the next value of of *v* for which a schedule of walks
might be possible is the next odd multiple of three, namely 15
(which corresponds to six more students coming to the school).

**Answer to Question 7.**
Recall that our blocks are of two types, which we can call *Fano
blocks* (blocks from the Fano plane, consisting of numbers from
1 to 7) and *mixed blocks* (consisting of a tournament game,
which is two numbers from 8 to 15, together with one number from
1 to 7 representing the day on which that game was played).

Any pair of points in which both numbers are 7 or below occurs
exactly once among the Fano plane blocks, and never in the mixed
blocks. Any pair of points in which both numbers are 8 or above
occurs exactly once in the tournament schedule, and hence exactly
once in the mixed blocks, and never in the Fano plane blocks.
Any pair of points in which one number *a* is 7 or below and the other
number *b* is 8 or above occurs exactly once in the mixed
blocks (because team *b* plays exactly once on date *a*), and
never in the Fano plane blocks.

Therefore, each pair of points occurs exactly once amongst all the blocks.

**Answer to Question 8.**
Note the vertical structure to the Fano plane. The first block
(the first row of the table in the answer to the Farmer's Wheat
Problem) consists of the numbers 1, 2, and 4. If we add 1
to each number we get the second row. If we add 1 again we
get the third row. If we continue to add 1 each time (except that
when we are at the number 7 we go back down to 1 instead of up
to 8) we eventually generate all the blocks of the Fano plane.

Similarly, if we take the games from one day of the sports tournament and cycle the teams according to the rotation

8->9->10->11->12->13->14->8(with team 15 remaining fixed), we get the next day's games.

That means that if we take one of our "mixed" blocks of girls
(where two of the numbers, call them *a* and *b*, are from 8 to
15, representing
one of the sports games, and the third number is *c* is from 1 to 7
representing the day on which the *a* versus *b* game was played),
and cycle the numbers according to

1->2->3->4->5->6->7->1

8->9->10->11->12->13->14->8(with 15 remaining fixed),

Thus, if we start with any one of our 35 blocks (either one from the Fano plane, or one of the mixed blocks), and keep applying the above rotation, we get a cycle of seven blocks.

All we need to do to solve Kirkman's schoolgirl problem is to find a schedule of five blocks which works for day 1, making sure none of these five blocks are in the same cycle. Then, to get the schedule for the second day, simply apply the above rotation to the first day's schedule, and so on for each of the seven days.

The final solution is shown below.

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