Well Ive been having a little bit of problems with my machine in the last week, with hardware problems, taking hours to get past the POST stage.

**Interesting problem number one**

Actually, Ive noticed something, it dosnt matter what base you work in (assuming that the differnece change in bit is that the bit can change to any other bit not just to short range of bits either side of it... hrm didnt explain that well).

Its reasonable to say that on a base two system:A b bit number will have b neighbours.

A b bit system of the nth base will have n^b combinations

So the maximum you could have if you ignore the blatently obvious fact that perhaps it just wont fit... is

max = (n^b) / (b + 1)

SOLUTIONS FOR THE FIRST FIVE BITS:

One bit

0 1

-O--+-

Two bits

0 0 1 1 0 1 0 1

-O--+--+--+-

Three Bits

0 0 1 1 0 1 0 1

| | | | 0 -O--+--+--+- | | | | 1 -+--+--O--+- | | | |

Four Bits

0 0 1 1 0 1 0 1

| | | | 00 -O--+--+--+- | | | | 01 -+--+--+--+- | | | | 10 -+--O--+--+- | | | | 11 -+--+--+--+- | | | |

Five Bits

(note that this is a 3d solution - the two graphs represent layers)

0 0 1 1 0 0 1 1 0 0 1 0 1 1 0 1 0 1

| | | | | | | | 00 -O--+--+--+- 00 -+--+--+--+- | | | | | | | | 01 -+--+--+--+- 01 -+--+--+--O- | | | | | | | | 10 -+--O--+--+- 10 -+--+--+--+- | | | | | | | | 11 -+--+--+--+- 11 -+--+--O--+- | | | | | | | |

Table of results:

+------+------+------------------+----------------+ | Base | Bits | Formula's answer | Found solution | +------+------+------------------+----------------+ | 2 | 1 | 1.00 | 1 | | 2 | 2 | 1.33 | 1 | | 2 | 3 | 2.00 | 2 | | 2 | 4 | 3.20 | 2 | | 2 | 5 | 5.33 | 4 | | | | | | +------+------+------------------+----------------+