A Bakers Dozen

Introduction

I am going to investigate the pattern of numbers created by the task given. I have been told that two types of bun are laid out like so, A being one type of bun and B being the other:

ABAB

I then need to investigate how many times two adjacent buns must be swapped in order to sort the alternate pattern into two separate types of bun, one at each end. This particular arrangement shown above requires one switch:

1. AB↔AB

AABB

I have investigated the number of switches needed for the first 5 in this sequence. Here is a table of my results:

I will now attempt to find a formula solve any value in this sequence.

0    1    3    6    10

   1    2    3    4

      1    1    1

I will refer to the formula as fn as it changes depending on the value of n. This means fn is equal to the number of swaps needed. In my previous example 2 of each bun were used and one swap was required. I will now show this with the new lettering.

f(n) = 0.5k2 – 0.5k

f(n) = 0.5 x 22 – 0.5 x 2

f(n) = 2 – 1

f(n) = 1

This gives the same result as the previous example, showing that the formula works when n = k. To fully prove my formula works with the induction theory I will now need to show that the formula works when n = k+1

If f(n) = 0.5(k+1)2 – 0.5(k+1) is true then when k  = 2, according to the table and formula, (fn) should equal 3.

f(n) = 0.5(k+1)2 – 0.5(k+1)

f(n) = 0.5(2+1)2 – 0.5(2+1)

f(n) = 0.5 x 32 – 0.5 x 3

1

0

2

6

3

18

4

36

5

60

The formula I have found for this sequence, using the same method as before, is

3n2 – 3n.

Finally I have made a table of how many switches are needed to sort 5 different types of bun.

The formula I have found for this sequence, using the same method as before, is

5n2 – 5n.

With this set of formulas I now intend to work out a master formula to work out formulas for the different numbers of buns of different types.

The sequence of numbers used to make up the formulas is:

0.5     1.5      3      5

     0.5     1.5     2

          0.5     0.5

This is also a quadratic sequence. I have used the same formula as before to work out the master formula. It comes to N = 0.25n2 – 0.25n (N being the number which is then inserted into the formula for working out how many switches are required). Nn2 – Nn.