# cuts         # pieces
      0                1 start with 1 piece for 0 cuts
      1                2
      2                4
      3                7

For every cut, the maximum lines you can intersect with is the previous number of cuts. So the maximum number of pieces bisected is (previous cuts 1), which is the current number of cuts. To get to the total number of pieces you have to subtract the pieces you didn't bisect, which is your previous total number of pieces minus current bisected pieces.

in formula that would be current cuts * 2 (prev total - current cuts) which is current cuts previous total

The 'previous total' is (x-1) (x-2) (x-3) .... 1 (add the one because of the original piece for 0 cuts)

This is how I got to x (x-1) (x-2) (x-3) ..... 1

I think this is correct, but since my math skills are quite rusty, and my niece's math teacher (who gave her the problem in the first place) appears to disagree with this, I thought I'd get the advice of an experienced mathematician.

Thanks a lot for your time, and I hope my explanation was not too
confusing.

Paul.

>Msg=What is the maximum number of pieces you can
>get cutting a circle x times.

>I came up with x + (x-1) + (x-2).... + 1
>but I'm not sure this is correct.

Check this. For x = 0, 1, 2, 3, 4 you must get 1, 2, 4, 7, 11, respectively. Your formula simplifies to (just by induction)

F(x) = x + (x-1) + (x-2).... + 1 = x·(x + 1)/2

with

F(0) = 0
F(1) = 1
F(2) = 3
F(3) = 6
F(4) = 10

1 short every time. It appears that the correct formula may rather be

G(x) = x·(x + 1)/2 + 1

The reasoning is very much yours. The max number of crossings is the previous number of lines. The number of added regions is 1 more than the number of new crossings. Let G(x) be the max number of regions after x cuts. Then

G(x) = G(x-1) + (x-1) + 1 = G(x-1) + x

Use this recursively as you did:

G(x) = G(x-2) + x + (x-1)
G(x) = G(x-3) + x + (x-1) + (x-2)
...
G(x) = G(0) + x + (x-1) + (x-2) + ... + 1
= 1 + x + (x-1) + (x-2) + ... + 1
= 1 + x·(x + 1)/2

That's all.

All the best,
Alexander Bogomolny

Just a clarification, the + 1 in my formula was the + 1 you thought I didn't have, I should have written:

x + x-1 + x-2 + x-3 + ... + x-x + 1

Thanks for pointing out the x·(x+1)/2 simplification. It'seems to be high time I start reviewing some math before I forget everything. I have your site bookmarked and have the best intentions of reading and understanding previous postings.

Thanks again for your time,

Paul.