Talk:Sorting algorithms/Radix sort

From Rosetta Code


Beware negative number handling! See Wiki's python demo. dingowolf 13:25, 19 January 2011 (UTC)

An interesting problem; the easiest way to handle it seems to me to be to double the number of bins and put negative values in the first half and positive in the second. Or at least it produces correct results when I implemented it in the Tcl solution. (I suspect that the original algorithm simply didn't implement them, or sorted by printed digit instead of logical digit.) –Donal Fellows 13:22, 19 January 2011 (UTC)
The easiest way to handle negative numbers might be to find the minimum value in the list, subtract it from every item in the unsorted list and add it to every item in the sorted list. This approach is modular and can wrap any "non-negative integers only" implementation, and work well in a variety of circumstances. That said the "double the bins" approach might have an efficiency advantage when the the absolute value of the maximum equals the absolute value of the minimum. --Rdm 16:13, 19 January 2011 (UTC)
It was a smaller change to the code I already had working for the positive case. :-) –Donal Fellows 16:42, 19 January 2011 (UTC)
Yuppers, the negative integers were a small annoyance, all right (concerning the REXX example). -- Gerard Schildberger 22:03, 11 June 2012 (UTC)

C code[edit]

in the C code for radix sort, it seems to me that the condition ll < to after the "while (1)" loop is always fulfilled, and can thus be removed. Indeed in the "while (1)" loop we always have ll <= rr, thus since rr decreases ll cannot exceed the initial value of rr, which is to - 1. User:Paul Zimmermann 13:09, 30 October 2012


I made a shorter/simpler implementation of the java example (it also handles negatives)

public static int[] sort(int[] old) {
for(int shift = Integer.SIZE-1; shift > -1; shift--) { //Loop for every bit in the integers
int[] tmp = new int[old.length]; //the array to put the partially sorted array into
int j= 0; //The number of 0s
int i; //Iterator
for(i = 0; i < old.length; i++){ //Move the 0s to the new array, and the 1s to the old one
boolean move = old[i] << shift >= 0; //If there is a 1 in the bit we are testing, the number will be negative
if(shift == 0 ? !move : move) { //If this is the last bit, negative numbers are actually lower
tmp[j] = old[i];
} else { //It's a 1, so stick it in the old array for now
old[i-j] = old[i];
for(i = j; i < tmp.length; i++) { //Copy over the 1s from the old array
tmp[i] = old[i-j];
old = tmp; //And now the tmp array gets switched for another round of sorting
return old;

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