You are looking to partition the distance from 0 to 1. Choose n – 1 numbers from 0 to 1, sort them and determine the distances between each of them. This will partition the space 0 to 1, which should yield the occasional large result which you aren’t getting. Even so, for large values of … Read more
This is challenging to explain in a single StackOverflow post. In general, the simplest beat-detection algorithms work by locating peaks in sound energy, which is easy to detect. More sophisticated methods use comb filters and other statistical/waveform methods. For a detailed explication including code samples, check this GameDev article out.
Here’s how you can get explorer-like behaviour in Python: #!/usr/bin/env python “”” >>> items = u’a1 a003 b2 a2 a10 1 10 20 2 c100′.split() >>> items.sort(explorer_cmp) >>> for s in items: … print s, 1 2 10 20 a1 a2 a003 a10 b2 c100 >>> items.sort(key=natural_key, reverse=True) >>> for s in items: … print … Read more
If you need just constant color (not interpolated by used area of pixel) then use DDA: void line_DDA_subpixel(int x0,int y0,int x1,int y1,int col) // DDA subpixel -> thick { int kx,ky,c,i,xx,yy,dx,dy; x1-=x0; kx=0; if (x1>0) kx=+1; if (x1<0) { kx=-1; x1=-x1; } x1++; y1-=y0; ky=0; if (y1>0) ky=+1; if (y1<0) { ky=-1; y1=-y1; } y1++; … Read more
def total_subsets_matching_sum(numbers, sum): array = [1] + [0] * (sum) for current_number in numbers: for num in xrange(sum – current_number, -1, -1): if array[num]: array[num + current_number] += array[num] return array[sum] assert(total_subsets_matching_sum(range(1, 10), 9) == 8) assert(total_subsets_matching_sum({1, 3, 2, 5, 4, 9}, 9) == 4) Explanation This is one of the classic problems. The idea … Read more
DFS: list nodes_to_visit = {root}; while( nodes_to_visit isn’t empty ) { currentnode = nodes_to_visit.take_first(); nodes_to_visit.prepend( currentnode.children ); //do something } BFS: list nodes_to_visit = {root}; while( nodes_to_visit isn’t empty ) { currentnode = nodes_to_visit.take_first(); nodes_to_visit.append( currentnode.children ); //do something } The symmetry of the two is quite cool. Update: As pointed out, take_first() removes and … Read more
if it helps a bit here is my ancient C++ source for triangle in VCL/GDI: //————————————————————————— class gfx_main { public: Graphics::TBitmap *bmp; int **pyx,xs,ys; gfx_main(); ~gfx_main(); void resize(int _xs=-1,int _ys=-1); void troj(int x0,int y0,int x1,int y1,int x2,int y2,int col); // this is filled triangle void _troj_line(int *pl,int *pr,int x0,int y0,int x1,int y1); // this is … Read more
As other posters have said, there are potentially O(n^2) substrings for a given string, so printing them out cannot be done faster than that. However there exists an efficient representation of the set that can be constructed in linear time: the suffix tree.
One trick for analyzing the time complexity of Euclid’s algorithm is to follow what happens over two iterations: a’, b’ := a % b, b % (a % b) Now a and b will both decrease, instead of only one, which makes the analysis easier. You can divide it into cases: Tiny A: 2a <= … Read more
The typical way is to shift and subtract. This is basically pretty similar to long division as we learned it in school. The big difference is that in decimal division you need to estimate the next digit of the result. In binary, that’s trivial. The next digit is always either 0 or 1. If the … Read more