Recent content by noelhustler

djvu is a very good document compression format (much better than pdf). You can get viewers from here or here or here

strstr and strchr You don't need to use c++ to write a program like that. You can stick with plain c.

You don't need to use regular expressions in c in order to find substrings and characters. You can use the c functions strstr and strchr. By the way, the line in your perl program: elsif (/a$/ or /e$/ or /i$/ or /o$/ or /u$/ or /a\?$/ or /e\?$/ or /i\?$/ or /o\?$/ or /u\?$/ ) { would...

The area of math dealing with this kind of problem would be percolation theory. I wrote a program to run the problem experimentally and came up with a fairly consistent number of clumps. There were rougly 0.066A clumps where A is the total number of pixels. I also came across s-Clusters...

Well, I imagine this may be related to your clump query. Looking at Self-Avoiding Walks there seems to be some solutions for limited cases but these don't really help.

It is an interesting problem. I'm not sure there's an elegant way to solve it. The mean number of pixels changed on each fill is related to the distribution of black and white pixels in the image and this distribution changes with each fill. This is also ignoring distribution locality...

10 minutes and I find it hard to believe that 98% wouldn't be able to solve it.

Almost. There are 2^{2n+1} subsets in total. The number of subsets with k elements is actually \binom{2n+1}{k}. Now, because we want the number of subsets of length <= n we can use the fact that the binomial expansion is symmetrical (about n and n+1 since 2n+1 is always odd. This means...

Of course calculating the dot product is a much easier way to solve the problem. However, I think it is useful to show that there is more than one way to skin a cat, especially when they don't seem to grasp the first method.

Calculate the cross product of your first vector and A. You'll end up with 3 equations, one for each component. Set these equations to equal the respective components of your final vector. Then try and solve the equations simultaneously. If there is a solution, then the vector A exists.

A solution for this problem is: \binom{n}{k}-\sum_{i=2}^{x}\left[\binom{x}{i}\times\binom{n-x}{k-i}\right] I'm sure there's a more elegant formulation, but this one works. Of course, for x=2, this simplifies to: \binom{n}{k}-\binom{n-2}{k-2}

I don't think that's quite right. It ignores the combinations that don't have one of the x objects.

Ah yes, I misread the actual question. Must remember to read the post before hitting submit. Very important, that.

The series mod 7 is peculiar. the other moduli 2..10 have a single period if prime or a single period plus the periods of their prime factors if not. 7 however has two distinct periods depite being prime. Any ideas?

Yes, I digressed. I forgot the question simply asked whether or not it repeated. I was thinking of a derivation of the period. It is interesting to note that there are only 3 possible periods for any four starting digits. For the particular starting sequence a,0,0,b they are: If a = b =...