How does Radix Sort work?

Question

I don't know why this is so hard for me to wrap my head around. I've looked through the wiki pages, and pseudo code (as well as actual code) trying to understand how radix sort algorithms work (with respect to buckets).

Am I looking into the wrong thing here? Should I be looking into bucket sort maybe? Can someone give me a dumbed down version of how it works? For reference, here is a codeblock that supposedly performs a radix sort:

// Sort 'size' number of integers starting at 'input' according to the 'digit'th digit // For the parameter 'digit', 0 denotes the least significant digit and increases as significance does void radixSort(int* input, int size, int digit) {     if (size == 0)         return;      int[10] buckets;    // assuming decimal numbers      // Sort the array in place while keeping track of bucket starting indices.     // If bucket[i] is meant to be empty (no numbers with i at the specified digit),     // then let bucket[i+1] = bucket[i]      for (int i = 0; i < 10; ++i)     {         radixSort(input + buckets[i], buckets[i+1] - buckets[i], digit+1);     } } 

And I've looked at non-recursive solutions also:

void radixsort(int *a, int arraySize) {     int i, bucket[sortsize], maxVal = 0, digitPosition =1 ;     for(i = 0; i < arraySize; i++) {         if(a[i] > maxVal) maxVal = a[i];     }      int pass = 1;      while(maxVal/digitPosition > 0) {         // reset counter          int digitCount[10] = {0};          // count pos-th digits (keys)          for(i = 0; i < arraySize; i++)             digitCount[a[i]/digitPosition%10]++;          // accumulated count          for(i = 1; i < 10; i++)             digitCount[i] += digitCount[i-1];          // To keep the order, start from back side         for(i = arraySize - 1; i >= 0; i--)             bucket[--digitCount[a[i]/digitPosition%10]] = a[i];          for(i = 0; i < arraySize; i++)             a[i] = bucket[i];          cout << "pass #" << pass++ << ": ";         digitPosition *= 10;     }   } 

Specifically, this line is giving me troubles. I've tried walking through it with pen and paper, but I still can't figure out what this is doing:

   // To keep the order, start from back side         for(i = arraySize - 1; i >= 0; i--)             bucket[--digitCount[a[i]/digitPosition%10]] = a[i]; 

Answer 1

In mathematics, radix means base, where decimal would be base 10. Imagine you have numbers some of which having more than one digits like

5, 213, 55, 21, 2334, 31, 20, 430 

For simplicity, say you want to use the decimal radix (=10) for sorting. Then you would start by separating the numbers by units and then putting them together again; next you would separate the numbers by tens and then put them together again; then by hundreds and so on until all the numbers are sorted. Each time you loop, just read the list from left to right. You can also imagine you are separating the numbers into buckets. Here is an illustration using 5, 213, 55, 21, 2334, 31, 20, 430

Separate by units:

• zeros: 20, 430

• ones: 21, 31

• twos:

• threes: 213

• fours: 2334

• fives: 5, 55

  Back together: 20, 430, 21, 31, 213, 2334, 5, 55

To put them back together, first read the zeroes bucket, then the ones bucket, then so on, until you read the nines bucket.

Separate by tens:

• zeros: 05

• ones: 213

• twos: 20, 21

• threes: 430, 31, 2334,

• fours:

• fives: 55

  Back together: 5, 213, 20, 21, 430, 31, 2334, 55

Separate by hundreds:

• zeros: 005, 020, 021, 031, 055

• ones:

• twos: 213

• threes: 2334

• fours: 430

• fives:

  Back together: 5, 20, 21, 31, 55, 213, 2334, 430

Separate by thousands:

• zeros: 0005, 0020, 0021, 0031, 0055, 0213, 0430

• ones:

• twos: 2334

• threes:

• fours:

• fives:

  Back together: 5, 20, 21, 31, 55, 213, 430, 2334

You are now done. I saw a nice code for this on Geekviewpoint both in Java and in python