Writing bucket sort in c++

Question

A book I have says this:

a) Place each value of the one-dimensional array into a row of the bucket array based on the value's ones digit. For example, 97 is placed in row 7, 3 is placed in row 3, and 100 is placed in row 0. This is called a "distribution pass."

b) Loop through the bucket array row by row, and copy the values back to the original array. This is called a "gathering pass." The new order of the preceding values in the one-dimensional array is 100, 3, and 97.

c) Repeat this process for each subsequent digit position.

I am having a lot of trouble trying to understand and implement this. So far I have:

void b_sort(int sarray[], int array_size) {
    const int max = array_size;
    for(int i = 0; i < max; ++i)
        int array[i] = sarray[i];

    int bucket[10][max - 1];
}

I'm thinking that in order to sort them by ones, tens, hundreds, etc, I can use this:

for(int i = 0; i < max; ++i)
    insert = (array[i] / x) % 10;
    bucket[insert];

where x = 1, 10, 100, 1000, etc. I am totally lost on how to write this now.

Answer 1

Here's a bucket sort based on the info in the OP question.

void b_sort(int sarray[], int array_size) {
    const int max = array_size;
    // use bucket[x][max] to hold the current count
    int bucket[10][max+1];
    // init bucket counters
    for(var x=0;x<10;x++) bucket[x][max] = 0;
    // main loop for each digit position
    for(int digit = 1; digit <= 1000000000; digit *= 10) {
        // array to bucket
        for(int i = 0; i < max; i++) {
            // get the digit 0-9
            int dig = (sarray[i] / digit) % 10;
            // add to bucket and increment count
            bucket[dig][bucket[dig][max]++] = sarray[i];
        }
        // bucket to array
        int idx = 0;
        for(var x = 0; x < 10; x++) {
            for(var y = 0; y < bucket[x][max]; y++) {
                sarray[idx++] = bucket[x][y];
            }
            // reset the internal bucket counters
            bucket[x][max] = 0;
        }
    }
}

Notes Using a 2d array for the bucket wastes a lot of space... an array of queues/lists usually makes more sense.

I don't normally program in C++ and the above code was written inside the web browser, so syntax errors may exist.

Answer 2

The following code uses hex digits for a bucket sort (for BITS_PER_BUCKET=4). Ofcourse it is meant to be instructive, not productive.

#include <assert.h>
#include <stdio.h>

#define TEST_COUNT 100
#define BITS_PER_BUCKET 4
#define BUCKET_COUNT (1 << BITS_PER_BUCKET)
#define BUCKET_MASK (BUCKET_COUNT-1)
#define PASS_COUNT (8*sizeof(int)/BITS_PER_BUCKET)

int main(int argc, char** argv) {

  printf("Starting up ...");
  assert((PASS_COUNT*BITS_PER_BUCKET) == (8*sizeof(int)));
  printf("... OK\n");

  printf("Creating repeatable very-pseudo random test data ...");
  int data[TEST_COUNT];
  int x=13;
  int i;
  for (i=0;i<TEST_COUNT;i++) {
    x=(x*x+i*i) % (2*x+i);
    data[i]=x;
  }
  printf("... OK\nData is ");
  for (i=0;i<TEST_COUNT;i++) printf("%02x, ",data[i]);
  printf("\n");

  printf("Creating bucket arrays ...");
  int buckets[BUCKET_COUNT][TEST_COUNT];
  int bucketlevel[BUCKET_COUNT];
  for (i=0;i<BUCKET_COUNT;i++) bucketlevel[i]=0;
  printf("... OK\n");

  for (i=0;i<PASS_COUNT;i++) {

    int j,k,l;

    printf("Running distribution pass #%d/%d ...",i,PASS_COUNT);
    l=0;
    for (j=0;j<TEST_COUNT;j++) {
      k=(data[j]>>(BITS_PER_BUCKET*i)) & BUCKET_MASK;
      buckets[k][bucketlevel[k]++]=data[j];
      l|=k;
    }
    printf("... OK\n");

    if (!l) {
      printf("Only zero digits found, sort completed early\n");
      break;
    }

    printf("Running gathering pass #%d/%d ...",i,PASS_COUNT);
    l=0;
    for (j=0;j<BUCKET_COUNT;j++) {
      for (k=0;k<bucketlevel[j];k++) {
        data[l++]=buckets[j][k];
      }
      bucketlevel[j]=0;
    }
    printf("... OK\nData is ");
    for (l=0;l<TEST_COUNT;l++) printf("%02x, ",data[l]);
    printf("\n");

  }
}

Answer 3

a rewrite of Louis's code in C++11 with STL queues.

void bucket_sort(vector<int>& arr){
    queue<int> buckets[10];
    for(int digit = 1; digit <= 1e9; digit *= 10){
        for(int elem : arr){
            buckets[(elem/digit)%10].push(elem);
        }
        int idx = 0;
        for(queue<int>& bucket : buckets){
            while(!bucket.empty()){
                arr[idx++] = bucket.front();
                bucket.pop();
            }
        }
    } 
}