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# CS3 Data Structures & Algorithms

## Chapter 8 Sorting

The major problem with Binsort is that it does not work so well for a large key range. Fortunately, there is a way to keep the number of bins small and the related processing relatively cheap while still using the idea of binning records that have similar key values. Consider a sequence of records with keys in the range 0 to 99. If we have ten bins available, we can first assign records to bins by taking their key value modulo 10. Thus, every key will be assigned to the bin matching its rightmost decimal digit. We can then take these records from the bins in order, and reassign them to the bins on the basis of their leftmost (10’s place) digit. We will define values in the range 0 to 9 to have a leftmost digit of 0. In other words, assign the $i$’th record from array A to a bin using the formula A[i]/10. If we now gather the values from the bins in order, the result is a sorted list. We can see this process in the following visualization.

In this example, we have $r=10$ bins and key values in the range 0 to $r^2-1$. The total computation is $\Theta(n)$, because we look at each record and each bin a constant number of times. This is a great improvement over the simple Binsort where the number of bins must be as large as the key range. Note that the example uses $r = 10$ so as to make the bin computations easy to visualize: Records were placed into bins based on the value of first the rightmost and then the leftmost decimal digits. Any number of bins would have worked if we interpret the key values in terms of the corresponding base. This is an example of a Radix Sort, so called because the bin computations are based on the radix or the base of the key values. This sorting algorithm can be extended to any number of keys in any key range. We simply assign records to bins based on the keys’ digit values working from the rightmost digit to the leftmost. If there are $k$ digits, then this requires that we assign keys to bins $k$ times.

Here is a practice exercise for placing keys into bins.

As with Mergesort, an efficient implementation of Radix Sort is somewhat difficult to achieve. In particular, we would prefer to sort an array of values and avoid processing linked lists. If we knew how many values would be in each bin, then an auxiliary array of size $r$ can be used to define these lengths and guide us to were each one starts in the output array. For example, if during the first pass the 0 bin will receive three records and the 1 bin will receive five records, then we could simply reserve the first three array positions for the 0 bin and the next five array positions for the 1 bin. Exactly this approach is taken by the following implementation. At the end of each pass, the records are copied back to the original array.

static void radix(Integer[] A, int k, int r) {
Integer[] B = new Integer[A.length];
int[] count = new int[r];     // Count[i] stores number of records with digit value i
int i, j, rtok;

for (i=0, rtok=1; i<k; i++, rtok*=r) { // For k digits
for (j=0; j<r; j++) { count[j] = 0; }    // Initialize count

// Count the number of records for each bin on this pass
for (j=0; j<A.length; j++) { count[(A[j]/rtok)%r]++; }

// count[j] will be index in B for last slot of bin j.
// First, reduce count[0] because indexing starts at 0, not 1
count[0] = count[0] - 1;
for (j=1; j<r; j++) { count[j] = count[j-1] + count[j]; }

// Put records into bins, working from bottom of bin
// Since bins fill from bottom, j counts downwards
for (j=A.length-1; j>=0; j--) {
B[count[(A[j]/rtok)%r]] = A[j];
count[(A[j]/rtok)%r] = count[(A[j]/rtok)%r] - 1;
}
for (j=0; j<A.length; j++) { A[j] = B[j]; } // Copy B back
}
}

static void radixsort(int A[], int k, int r, int n) {
int B[n];
int count[r];
int i, j, rtok;

for (i = 0, rtok = 1; i < k; i++, rtok *= r) {  // For k digits
for (j = 0; j < r; j++) count[j] = 0;  // Initialize count

// Count the number of records for each bin on this pass
for (j = 0; j < n; j++) count[(A[j]/rtok)%r]++;

// count[j] will be index in B for last slot of bin j.
// First, reduce count[0] because indexing starts at 0, not 1
count[0] = count[0] - 1;
for (j = 1; j < r; j++) count[j] = count[j-1] + count[j];

// Put records into bins, working from bottom of bin
// Since bins fill from bottom, j counts downwards
for (j  = n-1; j >= 0; j--) {
B[count[(A[j]/rtok)%r]] = A[j];
count[(A[j]/rtok)%r] = count[(A[j]/rtok)%r] - 1;
}
for (j = 0; j < n; j++) A[j] = B[j];  // Copy B back
}
}


The first inner for loop initializes array count. The second loop counts the number of records to be assigned to each bin. The third loop sets the values in count to their proper indices within array B. Note that the index stored in count[j] is the last index for bin j; bins are filled from high index to low index. The fourth loop assigns the records to the bins (within array B). The final loop simply copies the records back to array A to be ready for the next pass. Variable rtoi stores $r^i$ for use in bin computation on the $i$’th iteration.

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Is it really a reasonable assumption to treat $k$ as a constant? Or is there some relationship between $k$ and $n$? If the key range is limited and duplicate key values are common, there might be no relationship between $k$ and $n$. To make this distinction more clear, use $N$ to denote the number of distinct key values used by the $n$ records. Thus, $N \leq n$. Because it takes a minimum of $\log_r N$ base $r$ digits to represent $N$ distinct key values, we know that $k \geq \log_r N$.

Now, consider the situation in which no keys are duplicated. If there are $n$ unique keys then $n = N$. It would require $n$ distinct values to represent them. So now it takes a minimum of $\log_r n$ base $r$ digits to represent the $n$ distinct key values. This means that $k \geq \log_r n$. Because it requires at least $\log n$ digits to distinguish between the $n$ distinct keys (within a constant factor—meaning, the number of digits is $\Omega(\log n)$), $k$ is in $\Omega(\log n)$. This means that Radix Sort requires $\Omega(n \log n)$ time to process $n$ distinct key values.

Of course the key range could be much bigger $\log_r n$ bits is merely the best case possible for $n$ distinct values. Thus, the $\log_r n$ estimate for $k$ could be overly optimistic. The bottom line of this analysis is that, for the general case of $n$ distinct key values, Radix Sort is at best a $\Omega(n \log n)$ sorting algorithm.

Radix Sort’s running time can be much improved (by a constant factor) if we make base $r$ be as large as possible. This is simplest if we think about integer key values. Set $r = 2^i$ for some $i$. In other words, the value of $r$ is related to the number of bits of the key processed on each pass. Each time the number of bits is doubled, the number of passes is cut in half. When processing an integer key value, setting $r = 256$ allows the key to be processed one byte at a time. Processing a 32-bit integer key requires only four passes. It is not unreasonable on most computers to use $r = 2^{16} = 64\mbox{K}$, resulting in only two passes for a 32-bit key. Of course, this requires a count array of size 64K. Performance will be good only if the number of records is about 64K or greater. In other words, the number of records must be large compared to the key size for Radix Sort to be efficient. In many sorting applications, Radix Sort can be tuned in this way to give better performance.

Radix Sort depends on the ability to make a fixed number of multiway choices based on a digit value, as well as random access to the bins. Thus, Radix Sort might be difficult to implement for certain key types. For example, if the keys are real numbers or arbitrary length strings, then some care will be necessary in implementation. In particular, Radix Sort will need to be careful about deciding when the “last digit” has been found to distinguish among real numbers, or the last character in variable length strings. Implementing the concept of Radix Sort with the alphabet trie data structure is most appropriate for these situations.