Groups | Search | Server Info | Keyboard shortcuts | Login | Register [http] [https] [nntp] [nntps]
| From | Ben Bacarisse <ben.usenet@bsb.me.uk> |
|---|---|
| Newsgroups | comp.theory |
| Subject | Re: Fiting squares under a curve. |
| Date | 2021-09-16 00:33 +0100 |
| Organization | A noiseless patient Spider |
| Message-ID | <87y27xe7nj.fsf@bsb.me.uk> (permalink) |
| References | <373e789b-4461-4ee2-8c7a-f4f82053dea1n@googlegroups.com> <87fsu5gby7.fsf@bsb.me.uk> <c2c7fbe3-a1db-4642-93a2-81b01042cdc4n@googlegroups.com> <875yv1frs1.fsf@bsb.me.uk> <d57c2f56-04fe-4e6d-b682-7472d477c01dn@googlegroups.com> |
Malcolm McLean <malcolm.arthur.mclean@gmail.com> writes: > OK, lets try a more formal approach. That's always my preference... Thanks. > We have a function f(x) which is defined over x = 0.0 to 1.0, and is > always non-negative. > > We have a set of values x_0 ... x_N, in ascending order. x_0 = 0.0, > x_N = 1.0. These define N-1 non-overlapping intervals. I think you mean you have an N and wish to find the x_i. If you have the x_i you can't change anything to minimise the "error" function you've defined. But I think you suggested that N was also up for grabs because you could, within reason, add cells (but I am not 100% sure I know what the cells correspond to in this version). > The integral f(x) from x=0 to x = 1.0 is known to be less than 1.0. > > The optimisation problem is to define the set of points such that the sum > > for each interval > > | integral f(x) from x_i to x_i+1 - (x_i+1 - x_i)^2 | > > is at a minimum. A curious problem. I can't visualise it well. I suspect it's hard. > Additionally, an interval cannot be less than delta_min, which will be > 1 pixel, so usually between 0.01 and 0.001, because the range x = 0 - > 1.0 will generally span 100 to 1000 pixels. So let me check a few cases. For f(x) = 1 the only perfect solution has N = 2 and the x_i being just [0,1]. If N is given and > 2, then the best solution is to pack the internal x_i (0 < i < N) as close to 0 or 1 as possible. And for f(x) = 1/2 the optimum solution has N = 3 and the x_i at [0, 1/2, 1]. Again, for N > 3, packing the extra points close to 0, 1/2 or 1 seems optimal. For f(x) = x, I don't know the best N, but for N = 3 the optimum solution is (I think) [0, 1/3, 1] with an error of 5/90. I don't think that can be beaten for larger N, but I've not played around with it much. Do you agree with these solutions? -- Ben.
Back to comp.theory | Previous | Next — Previous in thread | Next in thread | Find similar | Unroll thread
Fiting squares under a curve. Malcolm McLean <malcolm.arthur.mclean@gmail.com> - 2021-09-15 06:59 -0700
Re: Fiting squares under a curve. Ben Bacarisse <ben.usenet@bsb.me.uk> - 2021-09-15 15:17 +0100
Re: Fiting squares under a curve. Malcolm McLean <malcolm.arthur.mclean@gmail.com> - 2021-09-15 08:06 -0700
Re: Fiting squares under a curve. Ben Bacarisse <ben.usenet@bsb.me.uk> - 2021-09-15 22:33 +0100
Re: Fiting squares under a curve. Malcolm McLean <malcolm.arthur.mclean@gmail.com> - 2021-09-15 15:01 -0700
Re: Fiting squares under a curve. Ben Bacarisse <ben.usenet@bsb.me.uk> - 2021-09-16 00:33 +0100
Re: Fiting squares under a curve. Malcolm McLean <malcolm.arthur.mclean@gmail.com> - 2021-09-17 03:25 -0700
Re: Fiting squares under a curve. Ben Bacarisse <ben.usenet@bsb.me.uk> - 2021-09-17 14:22 +0100
Re: Fiting squares under a curve. Andy Walker <anw@cuboid.co.uk> - 2021-09-17 16:25 +0100
Re: Fiting squares under a curve. Ben Bacarisse <ben.usenet@bsb.me.uk> - 2021-09-17 20:59 +0100
Re: Fiting squares under a curve. Malcolm McLean <malcolm.arthur.mclean@gmail.com> - 2021-09-23 02:14 -0700
csiph-web