Data Structures and Algorithms 
7.5 Radix Sorting 
The bin sorting approach can be generalised in a technique that is known as radix sorting.
As an example, consider the list of integers:
36 9 0 25 1 49 64 16 81 4
Bin  0  1  2  3  4  5  6  7  8  9 

Content  0  1 81      64 4 
25  36 16 
    9 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:
Bin  0  1  2  3  4  5  6  7  8  9 

Content  0 1 4 9 
16  25  36  49    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.
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 using the least significant "digit" (where "digit" here means a field with a limited range), then using 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, f_{i}i=0..k1, and that each f_{i} has s_{i} discrete values, then a generalised radix sort procedure can be written:
radixsort( A, n ) { for(i=0;i<k;i++) { for(j=0;j<s_{i};j++) bin[j] = EMPTY; 
O(s_{i}) 
for(j=0;j<n;j++) { move A_{i} to the end of bin[A_{i}>f_{i}] } 
O(n) 
for(j=0;j<s_{i};j++) concatenate bin[j] onto the end of A; } } 
O(s_{i}) 
Total 
Now if, for example, the keys are integers in (0,b^{k}1),
for some constant k,
then the keys can be viewed as kdigit baseb integers.
Thus, s_{i} = b for all i and
the time complexity becomes
O(n+kb) or O(n).
This result depends on k being constant.
If k is allowed to increase with n,
then we have a different picture.
For example,
it takes
log_{2}n
binary digits to represent an integer <n.
If the key length were allowed to increase with n,
so that k = logn,
then we would have:
.
Another way of looking at this is to note that if the range of the key is restricted to (0,b^{k}1), then we will be able to use the radixsort approach effectively if we allow duplicate keys when n>b^{k}. However, if we need to have unique keys, then k must increase to at least log_{b}n. Thus, as n increases, we need to have logn phases, each taking O(n) time, and the radix sort is the same as quick sort!
Radix Sort Animation This animation was written by Woi Ang. 

Please email comments to: morris@ee.uwa.edu.au 
Continue on to Search Trees  Back to the Table of Contents 