SEARCH HOME
 Math Central Quandaries & Queries
 Greetings Math Central! I searched the index, but could not find a similar quandary. Here's my predicament: I am seeking help in scheduling a 12 person fishing outing. We have four boats. I would like to schedule three fishermen to a boat. We fish 2 1/2 days, split into morning and afternoon sessions. Thursday afternoon, friday morning, Friday afternoon, Saturday morning, and Saturday afternoon We switch boats after lunch, with each fisherman moving to a different boat, with different partners. I would like to have each fisherman fish with as many others in the group as possible, with a minimum of partner duplication. I would like to have each fisherman fish in all of the four different boats over the course of the trip at least once. Here is the boat grid we use. Would you please suggest a boat / fishing partner schedule that fits my parameters as closely as possible. Thank you! Don

Don,

There isn't a perfect solution, but there is a pretty good one. The arithmetic does not work out to have every pair of people together at least once: there are 66 possible pairs and the schedule only has room for 60 pairs to be together (3 per boat per day).

In the schedule described below, each person is in each boat at least once, and every pair of people are together exactly once apart from very few exceptions. The people are 1, 2, ..., 9, a, b, c. There are 9 pairs missing, and three who are together twice. The missing pairs are 4 with each of 1, 2, 3; 8 with each of 5 6 7; and 9 with each of a b c. The only pairs together twice are 4 8, 4 9, and 8 9.

The boats are ordered left to right.

Time 1: 1 5 9; 2 6 a; 3 7 b; 4 8 c
Time 2: 3 8 a; 4 7 9; 1 6 c; 2 5 b
Time 3: 2 7 c; 1 8 b; 5 4 a; 3 6 9
Time 4: 4 6 b; 3 5 c; 2 8 9; 1 7 a
Time 5: 1 2 3; 5 6 7; a b c; 4 8 9

Hope this works for you.

--Victoria

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.