| I found one reference, which has only this single example:
Find a set of 11 integral squares that can be assembled into a 13x13 square.
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Squares of primes are generally difficult to break up efficiently.
Squares with composite sides can usually be profitably reduced to a set of
squares whose sides are divisors plus the dissection of one of those
squares; i.e. 100^2 = 90^2 + 18*10^2 + (number of squares to make up a
10x10).
If the side N is a power of 2, the square can be broken up into 3log[2]N+1
smaller squares.
Lynn
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| To elaborate on what Lynn said in .1, you can set some upper limits
on the minumum that you are searching for as follows:
If [ N = (2^T) * K ] and (2^T) is the highest power of 2 that divides
N, : Then we can set some upper limit U(N):
(1) U(N) = 3T + 1 Where N = 2^T
(2) U(N) = 3T + N Where T > 0
(3) U(N) = N + 3 Where T = 0
These are probably far from optimal, except for (1), but they provide
a good starting point.
-- Barry
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