HoustonHokie

Well, I guess the last time I was wrong was just a few minutes ago...

Well, I might venture a new guess, although I'm unsure how to express it eloquently. From what I can see, if you are trying to come up with n divisions, you end up breaking the original square into predictable groupings of 2 smaller rectangles, which have to be further divided. I still postulate that rect(n)=2*square(n), and based on that, here's what I'm seeing: for 7 divisions: 1 rectangle divided 4 times, 1 rectangle divided 3 times 1 rectangle divided 5 times, 1 rectangle divided 2 times 1 rectangle divided 6 times, 1 rectangle divided 1 time for 6 divisions: 1 rectangle divided 3 times, 1 rectangle divided 3 times 1 rectangle divided 4 times, 1 rectangle divided 2 times 1 rectangle divided 5 times, 1 rectangle divided 1 time for 5 divisions: 1 rectangle divided 3 times, 1 rectangle divided 2 times 1 rectangle divided 4 times, 1 rectangle divided 1 time for 4 divisions: 1 rectangle divided 2 times, 1 rectangle divided 2 times 1 rectangle divided 3 times, 1 rectangle divided 1 time Then, if rect(n)=2*square(n), I think you can express: square(4)=rect(3)+rect(2)-1 =4+2-1=5 square(5)=rect(4)+rect(3)-3 =10+4-3=11 square(6)=rect(5)+rect(4)+rect(3)-5=22+10+4-5=29 for square(7), the answer would be 83 - not sure I want to draw that out to confirm! As far as an expression goes, I guess this might do: square(n)=rect(n-1)+rect(n-2)+...+rect(n/2)-[(n-3)*2-1] where rect(n) = 2*square(n) Of course, rect(n/2) is only defined for even values of n, so the recursion would have to stop with rect(n/2+0.5) for odd numbers. I can't figure out if my last term is correct - the subtraction of an ever-increasing odd number - because I can't understand a geometric reason for it. Also, the formula doesn't work for square(3), which makes me more suspicious of that term. But the geometry seems good to me for the recursion. And if I'm wrong, well - I've been wrong before. I think the last time was yesterday!

I agree. I edited the picture and it looks like this. Now f(6) = 29.

Chuck, I think you're on to something that I started to see yesterday. If you'll see my last post, I postulated that rect(n)=2*square(n), for precisely the reason that you mentioned - you get one solution for dividing it lengthwise and another for dividing it widthwise, but I think the solutions should be equal in terms of the actual number of different solutions. I'm trying to confirm that you'd never end up with the same solution if your initial cuts were along two axes of different lengths, but there might be some specific geometric proportions of the rectangle where that occurs (not that we'd necessarily find them in this problem - that's the next step if I find proportions that give the same results).

Well, I'm not going to have much more time for this today, but here's my next guess: If I'm wrong, you guys will have to fix it for me. Another interesting question comes out of this. Suppose the original shape were a rectangle instead of a square. I think rect(n)=2*square(n), but I'm not sure. Can anyone confirm or deny? I think it would actually help in the recursion of the original problem if we knew.

I'm trying to draw this bugger out. This is what I've got through 5. Am I missing anything?

jack0, I think your solution double-counts. Your equation gives f(4) = 9, I think we've established f(4) = 5. Not that I'm doing much better right now

forgot about the 4 1/2x1/2 squares. Gotta look harder. Its definitely recursive - the trick is figuring out how to recurse it properly.

I get:

All I have time for now is #1. So here goes:

At first, I was trying to figure out if the "/" was part of the palindrome, or if it was ignored. Rather than ask questions, I answered it both ways. If "/" is part of the palindrome, I get: If it's just the number, I get: