CS302 --- Replacement Selection

Brad Vander Zanden

Replacement Selection


Perform the in-memory sort by passing the records through a large priority queue.

Detailed Strategy

  1. Choose as large a priority queue as possible, say of M elements
  2. Sort Step
  3. Merge Step: Same as before


The pseudo-code in this section shows how the heap code from earlier in the course must be modified to handle external sorting. It also shows how the create_initial_runs code must be modified from the balanced, multi-way sorting scheme. The other procedures from the balanced, multi-way sorting scheme remain the same.

Code for Create_Initial_Runs

create_initial_runs(input_file_name, run_size, num_ways) {

    allocate a dynamic array, a, large enough to accommodate runs of 
        size run_size
    open the input file using the fields package
    for i = 0 to NUM_WAYS-1 {
        open output_scratch_file i using fopen
    end_of_input = false
    next_output_file = 0
    num_runs_per_output_file = 0

  /* instantiate the heap */
  for i = 1 to run_size {
     read a record into a[i]
     if (end of input)
         end_of_input = true

     /* indicate the record belongs in the current run and insert the
        record into the heap */
     a[i]->mark = current_mark;
  /* initialize the heap size */
  N = i - 1;

  /* create the initial runs--get_item sets end_of_input to true only
      when there is no more input and the heap is exhausted */

  while (end_of_input == false) {
    /* keep outputting records to the current run until get_item returns
       a NULL record */
    for (record = get_item(input_file, &end_of_input); record != NULL; 
	 record = get_item(input_file, &end_of_input)) {
      write record to output_scratch_files[next_output_file]
    output the sentinel value to scratch_output_file[next_output_file]

    /* everytime we get back to the first output file, increment the
	number of runs per output file by 1 */
    if (next_output_file == 0)
	 num_runs_per_output_file = num_runs_per_output_file + 1
     next_output_file = (next_output_file + 1) % num_ways

  /* make sure the same number of runs are assigned to each scratch 
     output file */
  if (next_output_file != 0) {
      for i = next_output_file to (num_ways - 1) {
          output the sentinel value to scratch_output_file[i]
  for i = 0 to (num_ways -1)
      close scratch_output_file[i] using fclose
  close the input file using the fields package
  return num_runs_per_output_file

Heap Code

This code assumes there are two values in a record:

There is also a global variable:

current_mark: Indicates which of the two values a record's mark field must
              be set to in order to be included in this run.

Here is the pseudo-code:

/* if the two records both belong in the run or both don't belong in the
   run, compare the two values and return true if the first value is less
   than the first. Otherwise, return true if the first record belongs in
   the run, and false otherwise (in the latter case, the second record
   belongs in the run and hence it is "less than" the first record). */

int less_than_or_equal(record1, record2) {
  if (record1->mark == record2->mark)
    return (record1->key <= record2->key)
    return (record1->mark == current_mark)

downheap (int k) {
   int j;
   Record v;

   v = a[k];
   while (k <= N/2) {
       j = 2 * k;
       if (j < N && less_than_or_equal(a[j+1], a[j])) j++;
       if (less_than_or_equal(v, a[j])) break;
       a[k] = a[j];
       k = j;
   a[k] = v;

int heap_empty () {
  return (N == 0);

/* get_item returns the next item in the run. It returns
   NULL if the end of a run or end of the input is reached. The
   flag end_of_input allows get_item to indicate whether the reason for
   the NULL is due to the end of a run or the end of the input. 
   end_of_input must be a pointer to allow a value to be passed back
   through the end_of_input parameter. */

PERSONNEL get_item(input_file, *end_of_input) {
  static int more_input = true;

  /* determine if the heap is empty. If it is, then the end of the
     input has been reached, so set end_of_input to true */
  if (heap_empty()) {
    *end_of_input = true;
    return NULL;
  /* determine if the root element belongs in this run */
  if (a[1]->mark != current_mark) {
    /* reverse the mark so that all the elements in the heap will be
       available for the next run */
    current_mark = !(current_mark);
    *end_of_input = false;
    return NULL;
  if (more_input) {
    if (not end of file) {
      read a record into a[0]
      /* determine if the newly read record belongs in the current run
         or the next run. */
      if (a[0]->key >= a[1]->key)
          a[0]->mark = current_mark
          a[0]->mark = !(current_mark)
      /* insert the newly read record into the heap by executing a
         replace operation (i.e., the root of the heap will be returned
	 and the new value will be pushed onto the heap)
      return a[0];
    else { /* once the input is exhausted, start returning items from the
                heap */
      more_input = false;
      return remove_item();
  else {
    /* if the input has been exhausted, remove the top item on the heap
       and return it */
    return remove_item();


  1. For random keys, the runs produced by replacement selecton are about twice the size of the heap used
  2. Consequently, priority queues save approximately one merge pass. The number of passes is approximately 1 + log[P](N / 2M)

Practical Considerations

Systems Issues

The cost of I/O dominates the cost of computation so overlapping reading, writing, and computation is critical. A file is divided into blocks, each with a fixed number of bytes. When a file is read into memory or written out, the system maintains buffers that are equal to the block size of a file. Files are transferred to and from memory in chunks equal to the block size of a file, or equivalently, equal to the size of a buffer.
  1. Double Buffering: For each input file maintain two buffers--one that is used by the CPU and one that is used by the I/O processor. As the CPU uses its buffers the I/O processor reads its buffers.
  2. Forecasting: Double buffering's drawback is that it uses only half the available memory space. If there are P input units being used, 2P buffers are required. Forecasting can reduce this number to P+1. The idea is as follows:

Selecting P

  1. If N scratch tapes are available, use a N/2 - way sort (i.e. P = N/2)
  2. If disks are available, a good strategy is to choose P so that only 2 passes are required. This requires that:
    		log[P](N / 2M) < 2   
    	   ==>  N / 2M < 2**P
    	   ==>  sqrt(N / 2M) < P