11.2 Counting Sort and Radix Sort

In this section we consider two sorting algorithms that are not comparison-based. These algorithms are specialized for sorting small integers. These algorithms get around the lower-bounds of Theorem 11.5 by using (parts of) the elements of $\mathtt{a}$ as indices into an array. Consider a statement of the form

$$\ensuremath{\mathtt{c[a[i]]}} = 1 \enspace .$$

This statement executes in constant time, but has $\mathtt{c.length}$ possible different outcomes, depending on the value of $\mathtt{a[i]}$. This means that the execution of an algorithm that makes such a statement can not be modelled as a binary tree. Ultimately, this is the reason that the algorithms in this section are able to sort faster than comparison-based algorithms.

11.2.1 Counting Sort

Suppose we have an input array $\mathtt{a}$ consisting of $\mathtt{n}$ integers, each in the range $0,\ldots,\ensuremath{\mathtt{k}}-1$. The counting-sort algorithm sorts $\mathtt{a}$ using an auxiliary array $\mathtt{c}$ of counters. It outputs a sorted version of $\mathtt{a}$ as an auxiliary array $\mathtt{b}$.

The idea behind counting-sort is simple: For each $\ensuremath{\mathtt{i}}\in\{0,\ldots,\ensuremath{\mathtt{k}}-1\}$, count the number of occurrences of $\mathtt{i}$ in $\mathtt{a}$ and store this in $\mathtt{c[i]}$. Now, after sorting, the output will look like $\mathtt{c[0]}$ occurrences of 0, followed by $\mathtt{c[1]}$ occurrences of 1, followed by $\mathtt{c[2]}$ occurrences of 2,..., followed by $\mathtt{c[k-1]}$ occurrences of $\mathtt{k-1}$. The code that does this is very slick, and its execution is illustrated in Figure 11.7:

    int[] countingSort(int[] a, int k) {
        int c[] = new int[k];
        for (int i = 0; i < a.length; i++)
            c[a[i]]++;
        for (int i = 1; i < k; i++)
            c[i] += c[i-1];
        int b[] = new int[a.length];
        for (int i = a.length-1; i >= 0; i--)
            b[--c[a[i]]] = a[i];
        return b;
    }

Figure 11.7: The operation of counting sort on an array of length $\ensuremath {\mathtt {n}}=20$ that stores integers $0,\ldots ,\ensuremath {\mathtt {k}}-1=9$.

The first $\mathtt{for}$ loop in this code sets each counter $\mathtt{c[i]}$ so that it counts the number of occurrences of $\mathtt{i}$ in $\mathtt{a}$. By using the values of $\mathtt{a}$ as indices, these counters can all be computed in $O(\ensuremath{\mathtt{n}})$ time with a single for loop. At this point, we could use $\mathtt{c}$ to fill in the output array $\mathtt{b}$ directly. However, this would not work if the elements of $\mathtt{a}$ have associated data. Therefore we spend a little extra effort to copy the elements of $\mathtt{a}$ into $\mathtt{b}$.

The next $\mathtt{for}$ loop, which takes $O(\ensuremath{\mathtt{k}})$ time, computes a running-sum of the counters so that $\mathtt{c[i]}$ becomes the number of elements in $\mathtt{a}$ that are less than or equal to $\mathtt{i}$. In particular, for every $\ensuremath{\mathtt{i}}\in\{0,\ldots,\ensuremath{\mathtt{k}}-1\}$, the output array, $\mathtt{b}$, will have

$$\ensuremath{\mathtt{b[c[i-1]]}}=\ensuremath{\mathtt{b[c[i-1]+1]=}}\cdots=\ensuremath{\mathtt{b[c[i]-1]}}=\ensuremath{\mathtt{i}} \enspace .$$

Finally, the algorithm scans $\mathtt{a}$ backwards to put its elements in order into an output array $\mathtt{b}$. When scanning, the element $\mathtt{a[i]=j}$ is placed at location $\mathtt{b[c[j]-1]}$ and the value $\mathtt{c[j]}$ is decremented.

Theorem 11..7   The $\mathtt{countingSort(a,k)}$ method can sort an array $\mathtt{a}$ containing $\mathtt{n}$ integers in the set $\{0,\ldots,\ensuremath{\mathtt{k}}-1\}$ in $O(\ensuremath{\mathtt{n}}+\ensuremath{\mathtt{k}})$ time.

The counting-sort algorithm has the nice property of being stable; it preserves the relative order of elements that are equal. If two elements $\mathtt{a[i]}$ and $\mathtt{a[j]}$ have the same value, and $\ensuremath{\mathtt{i}}<\ensuremath{\mathtt{j}}$ then $\mathtt{a[i]}$ will appear before $\mathtt{a[j]}$ in $\mathtt{b}$. This will be useful in the next section.

11.2.2 Radix-Sort

Counting-sort is very efficient for sorting an array of integers when the length, $\mathtt{n}$, of the array is not much smaller than the maximum value, $\ensuremath{\mathtt{k}}-1$, that appears in the array. The radix-sort algorithm, which we now describe, uses several passes of counting-sort to allow for a much greater range of maximum values.

Radix-sort sorts $\mathtt{w}$-bit integers by using $\ensuremath{\mathtt{w}}/\ensuremath{\mathtt{d}}$ passes of counting-sort to sort these integers $\mathtt{d}$ bits at a time.² More precisely, radix sort first sorts the integers by their least significant $\mathtt{d}$ bits, then their next significant $\mathtt{d}$ bits, and so on until, in the last pass, the integers are sorted by their most significant $\mathtt{d}$ bits.

    int[] radixSort(int[] a) {
        int[] b = null;
        for (int p = 0; p < w/d; p++) {
            int c[] = new int[1<<d];
            // the next three for loops implement counting-sort
            b = new int[a.length];
            for (int i = 0; i < a.length; i++)
                c[(a[i] >> d*p)&((1<<d)-1)]++;
            for (int i = 1; i < 1<<d; i++)
                c[i] += c[i-1];
            for (int i = a.length-1; i >= 0; i--)
                b[--c[(a[i] >> d*p)&((1<<d)-1)]] = a[i];
            a = b;
        }
        return b;
    }

(In this code, the expression $\mathtt{(a[i] \text{\ttfamily >>} d*p)\text{\ttfamily\&}((1\text{\ttfamily <<}d)-1)}$ extracts the integer whose binary representation is given by bits $(\ensuremath{\mathtt{p}}+1)\ensuremath{\mathtt{d}}-1,\ldots,\ensuremath{\mathtt{p}}\ensuremath{\mathtt{d}}$ of $\mathtt{a[i]}$.) An example of the steps of this algorithm is shown in Figure 11.8.

Figure 11.8: Using radixsort to sort $\ensuremath {\mathtt {w}}=8$-bit integers by using 4 passes of counting sort on $\ensuremath {\mathtt {d}}=2$-bit integers.

This remarkable algorithm sorts correctly because counting-sort is a stable sorting algorithm. If $\ensuremath{\mathtt{x}} < \ensuremath{\mathtt{y}}$ are two elements of $\mathtt{a}$ and the most significant bit at which $\mathtt{x}$ differs from $\mathtt{y}$ has index $r$, then $\mathtt{x}$ will be placed before $\mathtt{y}$ during pass $\lfloor r/\ensuremath{\mathtt{d}}\rfloor$ and subsequent passes will not change the relative order of $\mathtt{x}$ and $\mathtt{y}$.

Radix-sort performs $\mathtt{w/d}$ passes of counting-sort. Each pass requires $O(\ensuremath{\mathtt{n}}+2^{\ensuremath{\mathtt{d}}})$ time. Therefore, the performance of radix-sort is given by the following theorem.

Theorem 11..8   For any integer $\ensuremath{\mathtt{d}}>0$, the $\mathtt{radixSort(a,k)}$ method can sort an array $\mathtt{a}$ containing $\mathtt{n}$ $\mathtt{w}$-bit integers in $O((\ensuremath{\mathtt{w}}/\ensuremath{\mathtt{d}})(\ensuremath{\mathtt{n}}+2^{\ensuremath{\mathtt{d}}}))$ time.

If we think, instead, of the elements of the array being in the range $\{0,\ldots,\ensuremath{\mathtt{n}}^c-1\}$, and take $\ensuremath{\mathtt{d}}=\lceil\log\ensuremath{\mathtt{n}}\rceil$ we obtain the following version of Theorem 11.8.

Corollary 11..1   The $\mathtt{radixSort(a,k)}$ method can sort an array $\mathtt{a}$ containing $\mathtt{n}$ integer values in the range $\{0,\ldots,\ensuremath{\mathtt{n}}^c-1\}$ in $O(c\ensuremath{\mathtt{n}})$ time.

Footnotes

... time.²

We assume that $\mathtt{d}$ divides $\mathtt{w}$, otherwise we can always increase $\mathtt{w}$ to $\ensuremath{\mathtt{d}}\lceil \ensuremath{\mathtt{w}}/\ensuremath{\mathtt{d}}\rceil$.