Balanced merge is a type of sorting algorithm where data is stored on multiple tapes (or other forms of storage). The tapes are divided into two groups: odd-numbered tapes and even-numbered tapes. The data on each tape is initially sorted, and the algorithm’s goal is to merge the data from all of the tapes into a single sorted output.
Approach: Here’s an approach for the above algorithm:
The algorithm works by repeatedly making passes over the data, merging pairs of tapes at a time. On each pass, the tapes are divided into pairs, and each pair’s data is merged into a temporary output tape. After all, pairs have been processed, the data on the temporary output tapes are copied back to the original tapes, and the process is repeated with a new set of tape pairs until all of the data is on a single tape.
Illustration: Here is an illustration with the approach used:
Suppose we have the following list of data that we want to sort: data[] = [5, 2, 4, 6, 1, 3]
Here are the steps involved in the sorting process:
- Step 1: Initialize the tapes and divide the data among them. We create three empty tapes and divide the data among them like this: tapes = [[], [], []]
for i, x in enumerate(data):
tapes[i % 3].append(x)
# tapes is now [[5, 4, 1], [2, 6, 3], []]
- Step 2: Sort the data on each tape using an internal sorting technique. We can use the built-in sort function to sort the data on each:
for tape in tapes:
tape.sort()
# tapes is now [[1, 4, 5], [2, 3, 6], []]
- Step 3: Repeatedly make passes over the tapes, merging pairs of tapes at a time. On each pass, we divide the tapes into pairs and merge the data from each pair into a temporary output tape using the merge function:
while len(tapes) > 1:
new_tapes = []
for i in range(0, len(tapes), 2):
tape1 = tapes[i]
tape2 = tapes[i+1] if i+1 < len(tapes) else None
new_tapes.append(merge(tape1, tape2))
tapes = new_tapes
# After the first pass, tapes is [[1, 2, 3, 4, 5, 6]]
- Step 4: Copy the data from the temporary output tapes back to the original tapes. Since we only have one tape at this point, we can skip this step.
- Step 5: Repeat the process until all of the data is on a single tape. Since we only have one tape at this point, the sorting process is complete.
The final result is a sorted version of the original data:
sorted_data = tapes[0]
# sorted_data is [1, 2, 3, 4, 5, 6]
The steps involved in implementing the balanced merge sorting algorithm are:
- Initialize the tapes and divide the data among them.
- Sort the data on each tape using an internal sorting technique.
- Repeatedly make passes over the tapes, merging pairs of tapes at a time.
- Copy the data from the temporary output tapes back to the original tapes.
- Repeat the process until all of the data is on a single tape.
Here is the code that demonstrates this approach:
// C++ code for the above approach #include <bits/stdc++.h> using namespace std;
vector< int > merge(vector< int > tape1, vector< int > tape2) {
if (tape2.size() == 0) {
return tape1;
}
// Merge the data from tape1 and tape2
// into a temporary output tape
vector< int > output_tape;
int i = 0, j = 0;
// Merge all the tapes left
while (i < tape1.size() && j < tape2.size()) {
if (tape1[i] < tape2[j]) {
output_tape.push_back(tape1[i]);
i++;
} else {
output_tape.push_back(tape2[j]);
j++;
}
}
output_tape.insert(output_tape.end(), tape1.begin() + i, tape1.end());
output_tape.insert(output_tape.end(), tape2.begin() + j, tape2.end());
return output_tape;
} // Function for Balanced merge sort vector< int > balanced_merge_sort(vector< int > data, int num_tapes) {
// Initialize the tapes
vector<vector< int >> tapes(num_tapes);
// Divide the data among the tapes
for ( int i = 0; i < data.size(); i++) {
tapes[i % num_tapes].push_back(data[i]);
}
// Sort each tape
for ( int i = 0; i < num_tapes; i++) {
sort(tapes[i].begin(), tapes[i].end());
}
// Repeatedly make passes over the tapes,
// merging pairs of tapes at a time
while (tapes.size() > 1) {
vector<vector< int >> new_tapes;
for ( int i = 0; i < tapes.size(); i += 2) {
vector< int > tape1 = tapes[i];
vector< int > tape2 = (i + 1 < tapes.size()) ? tapes[i + 1] : vector< int >();
new_tapes.push_back(merge(tape1, tape2));
}
tapes = new_tapes;
}
// Return the final merged tape
return tapes[0];
} // Driver code int main()
{ // Input
vector< int > data = {5, 2, 4, 6, 1, 3};
// Function call
vector< int > sorted_data = balanced_merge_sort(data, 3);
for ( int i = 0; i < sorted_data.size(); i++) {
cout << sorted_data[i] << " " ;
}
cout << endl;
return 0;
} // This code is contributed by lokeshpotta20. |
// Java code for the above approach import java.io.*;
import java.util.*;
class GFG {
public static List<Integer> merge(List<Integer> tape1,
List<Integer> tape2)
{
if (tape2.size() == 0 ) {
return tape1;
}
// Merge the data from tape1 and tape2
// into a temporary output tape
List<Integer> outputTape = new ArrayList<>();
int i = 0 , j = 0 ;
// Merge all the tapes left
while (i < tape1.size() && j < tape2.size()) {
if (tape1.get(i) < tape2.get(j)) {
outputTape.add(tape1.get(i));
i++;
}
else {
outputTape.add(tape2.get(j));
j++;
}
}
outputTape.addAll(tape1.subList(i, tape1.size()));
outputTape.addAll(tape2.subList(j, tape2.size()));
return outputTape;
}
// Function for Balanced merge sort
public static List<Integer>
balancedMergeSort(List<Integer> data, int numTapes)
{
// Initialize the tapes
List<List<Integer> > tapes = new ArrayList<>();
for ( int i = 0 ; i < numTapes; i++)
tapes.add( new ArrayList<>());
// Divide the data among the tapes
for ( int i = 0 ; i < data.size(); i++) {
tapes.get(i % numTapes).add(data.get(i));
}
// Sort each tape
for ( int i = 0 ; i < numTapes; i++) {
tapes.get(i).sort(Integer::compareTo);
}
// Repeatedly make passes over the tapes,
// merging pairs of tapes at a time
while (tapes.size() > 1 ) {
List<List<Integer> > newTapes
= new ArrayList<>();
for ( int i = 0 ; i < tapes.size(); i += 2 ) {
List<Integer> tape1 = tapes.get(i);
List<Integer> tape2
= (i + 1 < tapes.size())
? tapes.get(i + 1 )
: new ArrayList<>();
newTapes.add(merge(tape1, tape2));
}
tapes = newTapes;
}
// Return the final merged tape
return tapes.get( 0 );
}
public static void main(String[] args)
{
// Input
List<Integer> data = List.of( 5 , 2 , 4 , 6 , 1 , 3 );
// Function call
List<Integer> sortedData
= balancedMergeSort(data, 3 );
for ( int i = 0 ; i < sortedData.size(); i++) {
System.out.print(sortedData.get(i) + " " );
}
System.out.println();
}
} // This code is contributed by lokesh. |
# Python code for the above approach # Function for Balanced merge sort def balanced_merge_sort(data, num_tapes):
# Initialize the tapes
tapes = [[] for _ in range (num_tapes)]
# Divide the data among the tapes
for i, x in enumerate (data):
tapes[i % num_tapes].append(x)
# Sort each tape
for tape in tapes:
tape.sort()
# Repeatedly make passes over the tapes,
# merging pairs of tapes at a time
while len (tapes) > 1 :
new_tapes = []
for i in range ( 0 , len (tapes), 2 ):
tape1 = tapes[i]
tape2 = tapes[i + 1 ] if i + 1 < len (tapes) else None
new_tapes.append(merge(tape1, tape2))
tapes = new_tapes
# Return the final merged tape
return tapes[ 0 ]
def merge(tape1, tape2):
if tape2 is None :
return tape1
# Merge the data from tape1 and tape2
# into a temporary output tape
output_tape = []
i = j = 0
# Merge all the tapes left
while i < len (tape1) and j < len (tape2):
if tape1[i] < tape2[j]:
output_tape.append(tape1[i])
i + = 1
else :
output_tape.append(tape2[j])
j + = 1
output_tape.extend(tape1[i:])
output_tape.extend(tape2[j:])
return output_tape
# Driver code # Input data = [ 5 , 2 , 4 , 6 , 1 , 3 ]
# Function call sorted_data = balanced_merge_sort(data, 3 )
print (sorted_data)
|
// C# code for the above approach using System;
using System.Collections.Generic;
public class GFG {
public static List< int > merge(List< int > tape1,
List< int > tape2)
{
if (tape2.Count == 0) {
return tape1;
}
// Merge the data from tape1 and tape2
// into a temporary output tape
List< int > outputTape = new List< int >();
int i = 0, j = 0;
// Merge all the tapes left
while (i < tape1.Count && j < tape2.Count) {
if (tape1[i] < tape2[j]) {
outputTape.Add(tape1[i]);
i++;
}
else {
outputTape.Add(tape2[j]);
j++;
}
}
outputTape.AddRange(
tape1.GetRange(i, tape1.Count - i));
outputTape.AddRange(
tape2.GetRange(j, tape2.Count - j));
return outputTape;
}
// Function for Balanced merge sort
public static List< int >
balancedMergeSort(List< int > data, int numTapes)
{
// Initialize the tapes
List<List< int > > tapes = new List<List< int > >();
for ( int i = 0; i < numTapes; i++)
tapes.Add( new List< int >());
// Divide the data among the tapes
for ( int i = 0; i < data.Count; i++) {
tapes[i % numTapes].Add(data[i]);
}
// Sort each tape
for ( int i = 0; i < numTapes; i++) {
tapes[i].Sort();
}
// Repeatedly make passes over the tapes,
// merging pairs of tapes at a time
while (tapes.Count > 1) {
List<List< int > > newTapes
= new List<List< int > >();
for ( int i = 0; i < tapes.Count; i += 2) {
List< int > tape1 = tapes[i];
List< int > tape2 = (i + 1 < tapes.Count)
? tapes[i + 1]
: new List< int >();
newTapes.Add(merge(tape1, tape2));
}
tapes = newTapes;
}
// Return the final merged tape
return tapes[0];
}
static public void Main()
{
// Input
List< int > data = new List< int >{ 5, 2, 4, 6, 1, 3 };
// Function call
List< int > sortedData = balancedMergeSort(data, 3);
for ( int i = 0; i < sortedData.Count; i++) {
Console.Write(sortedData[i] + " " );
}
Console.WriteLine();
}
} // This code is contributed by lokeshmvs21. |
// JavaScript code for the above approach // Function to merge two sorted tapes function merge(tape1, tape2) {
// If one of the tapes is empty,
// return the other tape
if (tape2.length == 0) {
return tape1;
}
// Merge the data from tape1 and tape2
// into a temporary output tape
let output_tape = [];
let i = 0, j = 0;
// Merge all the tapes left
while (i < tape1.length && j < tape2.length) {
if (tape1[i] < tape2[j]) {
output_tape.push(tape1[i]);
i++;
} else {
output_tape.push(tape2[j]);
j++;
}
}
output_tape.push(...tape1.slice(i));
output_tape.push(...tape2.slice(j));
return output_tape;
} // Function for Balanced merge sort function balanced_merge_sort(data, num_tapes) {
// Initialize the tapes
let tapes = new Array(num_tapes);
for (let i=0; i<num_tapes; i++) {
tapes[i] = [];
}
// Divide the data among the tapes
for (let i = 0; i < data.length; i++) {
tapes[i % num_tapes].push(data[i]);
}
// Sort each tape
for (let i = 0; i < num_tapes; i++) {
tapes[i].sort( function (a, b) {
return a - b;
});
}
// Repeatedly make passes over the tapes,
// merging pairs of tapes at a time
while (tapes.length > 1) {
let new_tapes = [];
for (let i = 0; i < tapes.length; i += 2) {
let tape1 = tapes[i];
let tape2 = (i + 1 < tapes.length) ? tapes[i + 1] : [];
new_tapes.push(merge(tape1, tape2));
}
tapes = new_tapes;
}
// Return the final merged tape
return tapes[0];
} // Driver code let data = [5, 2, 4, 6, 1, 3]; // Input
let sorted_data = balanced_merge_sort(data, 3); console.log(sorted_data); // Output
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[1, 2, 3, 4, 5, 6]
Time Complexity: O(n log n)
Auxiliary Space: O(n)
Advantages of Balanced Merge:
- Can take advantage of parallelism to speed up the sorting process
- The time complexity of O(n log n), which is generally considered to be efficient for large amounts of data
- Can handle large amounts of data
Disadvantages of Balanced Merge:
- Requires additional space to store the data on the tapes and the temporary output tapes during the merge process
- May not be the most efficient choice for smaller datasets
Benefits of using this over other sorting techniques:
- Good choice for sorting large amounts of data when multiple tapes (or other forms of sequential storage) are available
- Can be used to take advantage of parallelism to speed up the sorting process
- Maybe a good choice when you need to sort data in an external storage device (such as a tape drive or hard drive)