Geek Answers Handbook

Fast byte allocation in bytes

I have a little problem and I can not find a satisfactory solution. There's an array of bytes, and I need those bytes sorted by high 7 bits, while keeping the order of low bits.

Initially, it looked like this:

// sort buf[N] to tmp[N]
uint offs[128+1]; uint c,i,s;
for( i=0; i<128; i++ ) offs[i]=0;
for( i=0; i<l; i++ ) offs[buf[i]>>1]++;
for( i=0,s=0; i<128; i++ ) c=offs[i], offs[i]=s, s+=c; offs[i]=s;

byte* tmp = new byte[N];
for( i=0; i<N; i++ ) c=buf[i], tmp[offs[c>>1]++]=c; // sort

But these blocks are large enough (currently 8M), and I want to use multiple streams, and an extra 8M per stream is noticeable.

So, I tried using a simple radix sort:

void radix( byte* buf, uint h, uint l, uint mask ) {
  uint p = (h+l)>>1, q = h; 
  uint i = offs[h], j = offs[l]-1; h = offs[p]; 
  if( (i<h) && (j>=h) ) {
    byte c = buf[i], d = buf[j];
    while( (i<h) && (j>=h) ) {
      while( (c&mask)==0 ) c = buf[++i]; // find value with bit 1
      while( (d&mask)!=0 ) d = buf[--j]; // find value with bit 0
      buf[i]=d; buf[j]=c; // swap 1-0 -> 0-1
      c = buf[++i]; d = buf[--j];
    }
    if( mask>=4 ) {
      radix( buf, q,p, mask>>1 );
      radix( buf, p,l, mask>>1 );
    }
  }
}

But he changes the order of these least significant bits and becomes unusable.

In fact, some simpler methods like bubblesort just do what I want, but they are much slower and speed is also a problem.

, , :

struct tmpsort {

  enum{ blocksize = (1<<16)-1 };

  unsigned short ofs[(max_quants+blocksize-1)/blocksize][probN];

  tmpsort( byte* buf, uint f_len ) {
    uint i,j,k;
    uint freq[2*probN]; // prob freqs
    byte tmp[blocksize+1];

    for( k=0,j=0; k<f_len; k+=blocksize,j++ ) {
      uint l = Min(k+blocksize,f_len)-k;
      byte* p = &buf[k];

      // compute offsets of sorted chunks
      for( i=0; i<2*probN; i++ ) freq[i]=0;
      for( i=0; i<l; i++ ) freq[p[i]]++;
      for( i=0; i<probN; i++ ) freq[i+1]=freq[2*i+0]+freq[2*i+1]; // 1=0+1, 2=2+3, 3=4+5
      freq[0] = 0;
      for( i=0; i<probN; i++ ) freq[i+1]+=freq[i];
      for( i=0; i<probN; i++ ) ofs[j][i]=freq[i+1];

      // sort the block via tmp
      for( i=0; i<l; i++ ) { byte c=p[i]; tmp[freq[c>>1]++]=c; }
      for( i=0; i<l; i++ ) p[i]=tmp[i];
    }
  }

};

[...]

tmpsort ts( buf, f_len );
for( i=0; i<probN; i++ ) {
  for( k=0,j=0; k<f_len; k+=ts.blocksize,j++ ) {
    uint x = i>0 ? ts.ofs[j][i-1] : 0;
    for(; x<ts.ofs[j][i]; x++ ) putc( buf[k+x],g );
  }
}

tmp [] ofs [] , , , .

: http://nishi.dreamhosters.com/u/tmpsort_v0.rar

+5
4

64 , ( ) 512 ( ) ( ). , , .

( 7M). .

O (N), 3 , .

0

, , . ?

+1

, O (n log n), radix, 7 , . - , , .

. O (lg n) ( , , for divide and conquer):

// sort array x from i to j by bit b
sort(x, i, j, b) {
  if (i >= j - 1) return;
  mid = (i + j) / 2;
  sort(x, i, mid, b);
  sort(x, mid, j, b);
  first1 = -1;
  last0 = -1;
  for (k = i; k < j; k++) {
    if (first1 < 0 && isSet(x[k], b)) first1 = k;
    if (!isSet(x[k], b)) last0 = k;
  }
  if (last0 < first1) return;

  // the sequence of bit b generally looks something like 0000011100000111111
  // so we reverse from the first 1 to the last 0
  reverse(x, first1, last0afterfirst1);
  newlast0 = first1;
  while (!isSet(x[++newlast0], b));
  newlast0--;

  // the elements in the range first1..last0 are in the wrong order, so reverse
  reverse(x, first1, newlast0);
  reverse(x, newlast0 + 1, last0);
}

isSet , , reverse . ( radix):

sort(x) {
  for (b = 1; b < 8; b++) {
    sort(x, 0, n, b);
  }
}

: "O (7 * n log n)". 7 , .

+1

quicksort . big-O, , , . 6 8, , , .

Actually ... presumably there is such a thing as a place, a stable merger. In terms of ideal theoretical characteristics, this is the holy grail of sorting — in place, true O(n log n)and stable, all at the same time. But I suspect that this is a huge pain to implement and has fairly large permanent conditions to go with this big-O.

0
source

More articles:


All Articles