13.13.1. Binsort¶

Imagine that for the past year, as you paid your various bills, you then simply piled all the paperwork into a corner somewhere. Now the year has ended and you have decided that it is time to sort all of these papers by what the bill was for (phone, electricity, rent, etc.) and date. A pretty natural approach is to make some space on the floor and, as you go through the pile of papers, put the phone bills into one pile, the electric bills into another pile, and so on. Once this initial assignment of bills to piles is done (in one pass), you can then sort each pile by date relatively quickly, because each pile is fairly small. This is the basic idea behind a Binsort.

Let's start with an especially easy situation. Consider the following code fragment to sort a permutation of the numbers 0 through \(n-1\).

  for (i=0; i<A.length; i++)
    B[A[i]] = A[i];

  for (i=0; i<A.length; i++)
    B[A[i]] = A[i];

  for (i=0; i<A.length; i++)
    B[A[i]] = A[i];

  for (i=0; i<A.length; i++)
    B[A[i]] = A[i];

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Here the key value is used to determine the position for a record in the final sorted array. This is the most basic example of a Binsort, where key values are used to assign records to bins. This algorithm is extremely efficient, always taking \(\Theta(n)\) time regardless of the initial ordering of the keys. This is far better than the performance of any sorting algorithm that we have seen so far. The problem is that this algorithm has limited use because it works only for a permutation of the numbers from 0 to \(n-1\).

We can extend this simple version of the Binsort algorithm to be more useful. Because Binsort must perform direct computation on the key value (as opposed to just asking which of two records comes first as our previous sorting algorithms did), we will assume that the records use an integer key type.

The simplest extension is to allow for duplicate values among the keys. This can be done by turning array slots into arbitrary-length bins by turning array B into an array of linked lists. In this way, all records with key value \(i\) can be placed in bin B[i]. A second extension allows for a key range greater than \(n\). For example, a set of \(n\) records might have keys in the range 1 to \(2n\). The only requirement is that each possible key value have a corresponding bin in B. We assume that we know that the range of possible keys is between 0 and MaxKeyValue. Here is the extended Binsort algorithm.

void binsort(Integer[] A) {
  List[] B = new LinkedList[MaxKeyValue+1];
  Object item;
  for (int i=0; i<=MaxKeyValue; i++)
    B[i] = new LinkedList();
  for (int i=0; i<A.length; i++) B[A[i]].append(new Integer(A[i]));
  int pos = 0;
  for (int i=0; i<=MaxKeyValue; i++)
    for (B[i].moveToStart(); (item = B[i].getValue()) != null; B[i].next())
      A[pos++] = (Integer)item;
}

void binsort(Integer[] A) {
  List[] B = new LinkedList[MaxKeyValue+1];
  Object item;
  for (int i=0; i<=MaxKeyValue; i++)
    B[i] = new LinkedList();
  for (int i=0; i<A.length; i++) B[A[i]].append(new Integer(A[i]));
  int pos = 0;
  for (int i=0; i<=MaxKeyValue; i++)
    for (B[i].moveToStart(); (item = B[i].getValue()) != null; B[i].next())
      A[pos++] = (Integer)item;
}

void binsort(Integer[] A) {
  List[] B = new LinkedList[MaxKeyValue+1];
  Object item;
  for (int i=0; i<=MaxKeyValue; i++)
    B[i] = new LinkedList();
  for (int i=0; i<A.length; i++) B[A[i]].append(new Integer(A[i]));
  int pos = 0;
  for (int i=0; i<=MaxKeyValue; i++)
    for (B[i].moveToStart(); (item = B[i].getValue()) != null; B[i].next())
      A[pos++] = (Integer)item;
}

void binsort(Integer[] A) {
  List[] B = new LinkedList[MaxKeyValue+1];
  Object item;
  for (int i=0; i<=MaxKeyValue; i++)
    B[i] = new LinkedList();
  for (int i=0; i<A.length; i++) B[A[i]].append(new Integer(A[i]));
  int pos = 0;
  for (int i=0; i<=MaxKeyValue; i++)
    for (B[i].moveToStart(); (item = B[i].getValue()) != null; B[i].next())
      A[pos++] = (Integer)item;
}

This version of Binsort can sort any collection of records whose key values fall in the range from 0 to MaxKeyValue.

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The total work required is simply that needed to place each record into the appropriate bin and then take all of the records out of the bins. Thus, we need to process each record twice, for \(\Theta(n)\) work.

Does that cost analysis really make sense? Actually, that last statement is wrong, because it neglects a crucial observation. Taking all of the records out of the bins requires Binsort to look at every bin to see if it contains a record. Thus, the algorithm must process MaxKeyValue bins, regardless of how many of them actually hold records. If MaxKeyValue is small compared to \(n\), then this is not a great expense. Suppose that MaxKeyValue \(= n^2\). In this case, the total amount of work done will be \(\Theta(n + n^2) = \Theta(n^2)\). This results in a poor sorting algorithm. And the algorithm becomes even worse as the disparity between \(n\) and MaxKeyValue increases. In addition, a large key range requires an unacceptably large array B. Thus, even the extended Binsort is useful only for a limited key range.

A further generalization to Binsort would yield a bucket sort. Here, each bin (now called a bucket) is associated with not just one key, but rather a range of key values. A bucket sort assigns records to buckets and then relies on some other sorting technique to sort the records within each bucket. The hope is that the relatively inexpensive bucketing process will put only a small number of records into each bucket, and that a "cleanup sort" to each bucket will then be relatively cheap. This is similar in spirit to the Radix Sort, which extends the concept of the Binsort in a practical way.