Mail from Patrick Hamlyn

Sorry I didn't do your competition before.

It's easy to prove the best solution.

The maximum contribution to perimeter for each piece is: (if 2 figures are
given, the first is for placing the piece in a top corner, and only two of
these can be chosen)

I 5

L 5,4

V 5,3

Y 4

P 4,3

N 3

U 3

T 3

W 2

Z 2

F 2

X 1

So maximum 'perimeter squares' is 5+5+5+4+3+3 or 5+4+5+4+4+3 = 25. Since two
corner squares contribute to width as well as height, we have a real maximum
perimeter of 27.

This gives a maximum area when width is closest to double the height, as can
be seen by the maximum area for each of:

Height Width Area

5 17 85

6 15 90

7 13 91

8 11 88

9 9 81

So the best we can hope for is area 91 (less area of 6 pentominos, equals 61)

My program found 792 ways of tiling the 7*13. Choosing a likely solution:

Patrick Hamlyn

Perth, Western Australia

Multiprogramming P/L