Data Structures and Algorithms

Radix Sort

The bin sorting approach can be generalised in a technique that is known as radix sorting.

Assume that we have n integers in the range (0,n2) to be sorted. (For a bin sort, m = n2, and we would have an O(n+m) = O(n2) algorithm.) Sort them in two phases:

  1. Using n bins, place ai into bin ai mod n,
  2. Repeat the process using n bins, placing ai into bin floor(ai/n), being careful to append to the end of each bin.
This results in a sorted list.

As an example, consider the list of integers:

36 9 0 25 1 49 64 16 81 4
n is 10 and the numbers all lie in (0..99). After the first phase, we will have:
01 8164 42536 169 49
Note that in this phase, we placed each item in a bin indexed by the least significant decimal digit.

Repeating the process, will produce:

0 116253649 64 81

In this second phase, we used the leading decimal digit to allocate items to bins, being careful to add each item to the end of the bin.

We can apply this process to numbers of any size expressed to any suitable base or radix.

Generalised Radix Sorting

We can further observe that it's not necessary to use the same radix in each phase, suppose that the sorting key is a sequence of fields, each with bounded ranges, eg for a date using the structure: 
typedef struct t_date {
    int day;
    int month;
    int year; 
    } date;
If the ranges for day and month are limited in the obvious way, and the range for year is suitably constrained, eg 1900 < year <= 2000, then we can apply the same procedure except that we'll employ a different number of bins in each phase. In all cases, we'll sort first on the least significant "digit" (where "digit" means a field with a limited range), then on the next significant "digit", placing each item after all the items already in the bin, and so on.

Assume that the key of the item to be sorted has k fields, fi|i=0..k-1, and that each fi has si discrete values, then a generalised radix sort procedure might look like this: 
radixsort( A, n )
    for(i=0;i<k;i++)
        for(j=0;j<si;j++) bin[j] = EMPTY;
        for(j=0;j<n;j++)
            move Ai 
            to the end of bin[Ai->fi]
        for(j=0;j<si;j++) 
            concatenate bin[j] onto the end of A;
O(si)
O(si)
O(si)
Total

Now if, for example, the keys are integers in (0..nk-1), for some constant k, then the keys can be viewed as k-digit base-n integers. Thus, si = n| for all i and the time complexity becomes O(n + kn) or O(n). Note that this result depends on k being constant.

It takes logn binary digits to represent an integer <=n, so that if the key length were allowed to increase with n, so that k = logn, then:
.

Bin and radix sorting.
Table of Contents
© , 1996