Thank you. I left them as currency here to tie in with the Snipit thread. Exactly as you say, they are very properly “longs” since I no longer had a need to store large square values.

Have you seen the “Code Snipit to test” thread? That came about because of a need I didn’t really have in this problem. When Dick allowed me to author, I didn’t think I’d become an insectologist.

I also didn’t quite realize whose notice I’d attract.

Thank you.

…mrt

As a recursive kind of guy you no doubt have this, but for those who don’t, here’s the recursive GCD version, GCD1().

I’ve found GCD2() above to be faster when I swap ‘em out.

…mrt

I can’t believe I’d missed the stupid GCD function, by the way, although I didn’t use it for this problem: I just used the formulae to crank out primitives until they reached the maximum, and added the multiples with a loop.

I don’t think dictionaries have much of an advantage – if any at all – over arrays in this case. They’re very handy as a substitute for sparsely-populated arrays (vectors, really) of large size. When, as in this case, over 10% of the possible values are of interest (and many more are used but not really interesting) then the additional overhead of the hashing in the Dictionary lookup probably isn’t worth it.

Comparing Dictionaries against Collections:
On the one hand, you lose “For Each” when you move from a Collection.
On the other hand, you gain a lot of performance, plus the “.Exist” method.
On the gripping hand they both suck against Arrays when you load them up with non-trivial quantities of non-primitive objects and then destroy them.

Different ways to get to P. If one uses P = 2m(m+n), then you don’t need a and b. I used c only as a kick-out to shave some msecs that I might waste in calculating it. Taking m to 999 and n to 2 does give c > 1000000.

Significantly less loops your way, though. I put in a counter and I reach Next n an even 250000 times.

My only experience with dictionaries, actually, is with this problem, and I found stuffing a series/array faster.

What do you get for run time?

…mrt

It can be shown that for a .LT. b .LT. c, c – b = 1 or 2 only for primitive triples. When c – b = 1, this means the difference between sequential squares b2 and c2 must also be a square. Since n2 is the sum of the first n positive odd integers, the square of every odd number is in a primitive triple, so

a = 3, b = 4, c = 5, a + b + c = 12 = 3 * 4 = a(a + 1)
a = 5, b = c – 1 = 12, c = (a2 + 1)/2 = 13, a + b + c = 30 = 5 * 6 = a(a + 1)
a = 7, b = c – 1 = 24, c = (a2 + 1)/2 = 25, a + b + c = 56 = 7 * 8 = a(a + 1)
a = 9, b = c – 1 = 40, c = (a2 + 1)/2 = 41, a + b + c = 90 = 9 * 10 = a(a + 1)

So sufficient to iterate a = 3 to a = 1413 step 2. For c – b = 2, a can be every multiple of 4 starting with 8 (a = 4 corresponds to b = 3, but that violates a .LT. b).

a = 8, b = 15, c = 17, a + b + c = 40 = 8 * 5 = a(a/2 + 1)
a = 12, b = 35, c = 37, a + b + c = 84 = 12 * 7 = a(a/2 + 1)
a = 16, b = 63, c = 65, a + b + c = 144 = 16 * 9 = a(a/2 + 1)

So sufficient to iterare a = 8 to 1996 step 4.

The a(a + 1) and a(a/2 + 1) terms give the perimeters of primitive triples. Their multiples are perimeters of nonprimitive triples. Easiest to iterate from 1 to 2E6 primitive perimeter storing the multiples of the primitive perimeters as keys in a WSH Dictionary object with the count as the key’s values. Then get the result by counting the resulting .Item values equal to 1.