Prerequisite: Page Replacement Algorithms
In operating systems that use paging for memory management, page replacement algorithm are needed to decide which page needed to be replaced when new page comes in. Whenever a new page is referred and not present in memory, page fault occurs and Operating System replaces one of the existing pages with newly needed page. Different page replacement algorithms suggest different ways to decide which page to replace. The target for all algorithms is to reduce number of page faults.
In Least Recently Used (LRU) algorithm is a Greedy algorithm where the page to be replaced is least recently used. The idea is based on locality of reference, the least recently used page is not likely
Let say the page reference string 7 0 1 2 0 3 0 4 2 3 0 3 2 . Initially we have 4 page slots empty.
Initially all slots are empty, so when 7 0 1 2 are allocated to the empty slots —> 4 Page faults
0 is already there so —> 0 Page fault.
when 3 came it will take the place of 7 because it is least recently used —>1 Page fault
0 is already in memory so —> 0 Page fault.
4 will takes place of 1 —> 1 Page Fault
Now for the further page reference string —> 0 Page fault because they are already available in the memory.
Given memory capacity (as number of pages it can hold) and a string representing pages to be referred, write a function to find number of page faults.
Let capacity be the number of pages that
memory can hold. Let set be the current
set of pages in memory.
1- Start traversing the pages.
i) If set holds less pages than capacity.
a) Insert page into the set one by one until
the size of set reaches capacity or all
page requests are processed.
b) Simultaneously maintain the recent occurred
index of each page in a map called indexes.
c) Increment page fault
ii) Else
If current page is present in set, do nothing.
Else
a) Find the page in the set that was least
recently used. We find it using index array.
We basically need to replace the page with
minimum index.
b) Replace the found page with current page.
c) Increment page faults.
d) Update index of current page.
2. Return page faults.
Below is implementation of above steps.
C++
//C++ implementation of above algorithm #include<bits/stdc++.h> using namespace std; // Function to find page faults using indexes int pageFaults(int pages[], int n, int capacity) { // To represent set of current pages. We use // an unordered_set so that we quickly check // if a page is present in set or not unordered_set<int> s; // To store least recently used indexes // of pages. unordered_map<int, int> indexes; // Start from initial page int page_faults = 0; for (int i=0; i<n; i++) { // Check if the set can hold more pages if (s.size() < capacity) { // Insert it into set if not present // already which represents page fault if (s.find(pages[i])==s.end()) { s.insert(pages[i]); // increment page fault page_faults++; } // Store the recently used index of // each page indexes[pages[i]] = i; } // If the set is full then need to perform lru // i.e. remove the least recently used page // and insert the current page else { // Check if current page is not already // present in the set if (s.find(pages[i]) == s.end()) { // Find the least recently used pages // that is present in the set int lru = INT_MAX, val; for (auto it=s.begin(); it!=s.end(); it++) { if (indexes[*it] < lru) { lru = indexes[*it]; val = *it; } } // Remove the indexes page s.erase(val); // insert the current page s.insert(pages[i]); // Increment page faults page_faults++; } // Update the current page index indexes[pages[i]] = i; } } return page_faults; } // Driver code int main() { int pages[] = {7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2}; int n = sizeof(pages)/sizeof(pages[0]); int capacity = 4; cout << pageFaults(pages, n, capacity); return 0; } |
Java
// Java implementation of above algorithm import java.util.HashMap; import java.util.HashSet; import java.util.Iterator; class Test { // Method to find page faults using indexes static int pageFaults(int pages[], int n, int capacity) { // To represent set of current pages. We use // an unordered_set so that we quickly check // if a page is present in set or not HashSet<Integer> s = new HashSet<>(capacity); // To store least recently used indexes // of pages. HashMap<Integer, Integer> indexes = new HashMap<>(); // Start from initial page int page_faults = 0; for (int i=0; i<n; i++) { // Check if the set can hold more pages if (s.size() < capacity) { // Insert it into set if not present // already which represents page fault if (!s.contains(pages[i])) { s.add(pages[i]); // increment page fault page_faults++; } // Store the recently used index of // each page indexes.put(pages[i], i); } // If the set is full then need to perform lru // i.e. remove the least recently used page // and insert the current page else { // Check if current page is not already // present in the set if (!s.contains(pages[i])) { // Find the least recently used pages // that is present in the set int lru = Integer.MAX_VALUE, val=Integer.MIN_VALUE; Iterator<Integer> itr = s.iterator(); while (itr.hasNext()) { int temp = itr.next(); if (indexes.get(temp) < lru) { lru = indexes.get(temp); val = temp; } } // Remove the indexes page s.remove(val); //remove lru from hashmap indexes.remove(val); // insert the current page s.add(pages[i]); // Increment page faults page_faults++; } // Update the current page index indexes.put(pages[i], i); } } return page_faults; } // Driver method public static void main(String args[]) { int pages[] = {7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2}; int capacity = 4; System.out.println(pageFaults(pages, pages.length, capacity)); } } // This code is contributed by Gaurav Miglani |
Python3
# Python implementation of above algorithm def pageFaults(pages, n, capacity): # To represent set of current pages. We use # an unordered_set so that we quickly check # if a page is present in set or not s = set() # To store least recently used indexes # of pages. indexes = {} # Start from initial page page_faults = 0 for i in range(n): # Check if the set can hold more pages if len(s) < capacity: # Insert it into set if not present # already which represents page fault if pages[i] not in s: s.add(pages[i]) # increment page fault page_faults += 1 # Store the recently used index of # each page indexes[pages[i]] = i # If the set is full then need to perform lru # i.e. remove the least recently used page # and insert the current page else: # Check if current page is not already # present in the set if pages[i] not in s: # Find the least recently used pages # that is present in the set lru = float('inf') for page in s: if indexes[page] < lru: lru = indexes[page] val = page # Remove the indexes page s.remove(val) # insert the current page s.add(pages[i]) # increment page fault page_faults += 1 # Update the current page index indexes[pages[i]] = i return page_faults # Driver code pages = [7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2] n = len(pages) capacity = 4print(pageFaults(pages, n, capacity)) # This code is contributed by ishankhandelwals. |
C#
// C# implementation of above algorithm using System; using System.Collections.Generic; class GFG { // Method to find page faults // using indexes static int pageFaults(int []pages, int n, int capacity) { // To represent set of current pages. // We use an unordered_set so that // we quickly check if a page is // present in set or not HashSet<int> s = new HashSet<int>(capacity); // To store least recently used indexes // of pages. Dictionary<int, int> indexes = new Dictionary<int, int>(); // Start from initial page int page_faults = 0; for (int i = 0; i < n; i++) { // Check if the set can hold more pages if (s.Count < capacity) { // Insert it into set if not present // already which represents page fault if (!s.Contains(pages[i])) { s.Add(pages[i]); // increment page fault page_faults++; } // Store the recently used index of // each page if(indexes.ContainsKey(pages[i])) indexes[pages[i]] = i; else indexes.Add(pages[i], i); } // If the set is full then need to // perform lru i.e. remove the least // recently used page and insert // the current page else { // Check if current page is not // already present in the set if (!s.Contains(pages[i])) { // Find the least recently used pages // that is present in the set int lru = int.MaxValue, val = int.MinValue; foreach (int itr in s) { int temp = itr; if (indexes[temp] < lru) { lru = indexes[temp]; val = temp; } } // Remove the indexes page s.Remove(val); //remove lru from hashmap indexes.Remove(val); // insert the current page s.Add(pages[i]); // Increment page faults page_faults++; } // Update the current page index if(indexes.ContainsKey(pages[i])) indexes[pages[i]] = i; else indexes.Add(pages[i], i); } } return page_faults; } // Driver Code public static void Main(String []args) { int []pages = {7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2}; int capacity = 4; Console.WriteLine(pageFaults(pages, pages.Length, capacity)); } } // This code is contributed by 29AjayKumar |
Javascript
<script> // JavaScript implementation of above algorithm // Method to find page faults using indexes function pageFaults(pages,n,capacity) { // To represent set of current pages. We use // an unordered_set so that we quickly check // if a page is present in set or not let s = new Set(); // To store least recently used indexes // of pages. let indexes = new Map(); // Start from initial page let page_faults = 0; for (let i=0; i<n; i++) { // Check if the set can hold more pages if (s.size < capacity) { // Insert it into set if not present // already which represents page fault if (!s.has(pages[i])) { s.add(pages[i]); // increment page fault page_faults++; } // Store the recently used index of // each page indexes.set(pages[i], i); } // If the set is full then need to perform lru // i.e. remove the least recently used page // and insert the current page else { // Check if current page is not already // present in the set if (!s.has(pages[i])) { // Find the least recently used pages // that is present in the set let lru = Number.MAX_VALUE, val=Number.MIN_VALUE; for(let itr of s.values()) { let temp = itr; if (indexes.get(temp) < lru) { lru = indexes.get(temp); val = temp; } } // Remove the indexes page s.delete(val); //remove lru from hashmap indexes.delete(val); // insert the current page s.add(pages[i]); // Increment page faults page_faults++; } // Update the current page index indexes.set(pages[i], i); } } return page_faults; } // Driver method let pages=[7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2]; let capacity = 4; document.write(pageFaults(pages, pages.length, capacity)); // This code is contributed by rag2127 </script> |
Output:
6
Complexity Analysis :
- Time Complexity : average time complexity of set and map operations is O(1) and the worst-case time complexity is O(n) but O(n) is the dominant term.
- Space Complexity : O(capacity) which is a constant and depends on the size of the input array and the size of the memory buffer.
Another approach: (Without using HashMap)
Following are the steps to solve this problem :
- Using a deque data structure, the program implements the page replacement algorithm.
- A predetermined number of pages are kept in memory by the algorithm, and they are replaced as new pages are requested.
- Using an integer array to stimulate page requests, the code keeps track the number of page faults that occur throughout the simulation.
- The deque data structure, which is built using STL in C++, is used to maintain the pages in memory.
- The total number of page faults that occurred throughout the simulation is given as output by the code.
Below is the implementation of the above approach :
C++
// C++ program for page replacement algorithms #include <iostream> #include<bits/stdc++.h> using namespace std; int main() { int capacity = 4; int arr[] = {7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2}; deque<int> q(capacity); int count=0; int page_faults=0; deque<int>::iterator itr; q.clear(); for(int i:arr) { // Insert it into set if not present // already which represents page fault itr = find(q.begin(),q.end(),i); if(!(itr != q.end())) { ++page_faults; // Check if the set can hold equal pages if(q.size() == capacity) { q.erase(q.begin()); q.push_back(i); } else{ q.push_back(i); } } else { // Remove the indexes page q.erase(itr); // insert the current page q.push_back(i); } } cout<<page_faults; } // This code is contributed by Akshit Saxena |
Java
// Java program for page replacement algorithms import java.util.ArrayList; public class LRU { // Driver method public static void main(String[] args) { int capacity = 4; int arr[] = {7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2}; // To represent set of current pages.We use // an Arraylist ArrayList<Integer> s=new ArrayList<>(capacity); int count=0; int page_faults=0; for(int i:arr) { // Insert it into set if not present // already which represents page fault if(!s.contains(i)) { // Check if the set can hold equal pages if(s.size()==capacity) { s.remove(0); s.add(capacity-1,i); } else s.add(count,i); // Increment page faults page_faults++; ++count; } else { // Remove the indexes page s.remove((Object)i); // insert the current page s.add(s.size(),i); } } System.out.println(page_faults); } } |
Python3
# Python3 program for page replacement algorithm # Driver code capacity = 4 processList = [ 7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2] # List of current pages in Main Memory s = [] pageFaults = 0# pageHits = 0 for i in processList: # If i is not present in currentPages list if i not in s: # Check if the list can hold equal pages if(len(s) == capacity): s.remove(s[0]) s.append(i) else: s.append(i) # Increment Page faults pageFaults +=1 # If page is already there in # currentPages i.e in Main else: # Remove previous index of current page s.remove(i) # Now append it, at last index s.append(i) print("{}".format(pageFaults)) # This code is contributed by mahi_07 |
C#
// C# program for page replacement algorithms using System; using System.Collections.Generic; class LRU { // Driver method public static void Main(String[] args) { int capacity = 4; int []arr = {7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2}; // To represent set of current pages. // We use an Arraylist List<int> s = new List<int>(capacity); int count = 0; int page_faults = 0; foreach(int i in arr) { // Insert it into set if not present // already which represents page fault if(!s.Contains(i)) { // Check if the set can hold equal pages if(s.Count == capacity) { s.RemoveAt(0); s.Insert(capacity - 1, i); } else s.Insert(count, i); // Increment page faults page_faults++; ++count; } else { // Remove the indexes page s.Remove(i); // insert the current page s.Insert(s.Count, i); } } Console.WriteLine(page_faults); } } // This code is contributed by Rajput-Ji |
Javascript
let capacity = 4; let arr = [7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2]; let q = []; let count = 0; let page_faults = 0; let itr; q.length = 0; for (let i of arr) { // Insert it into set if not present // already which represents page fault itr = q.indexOf(i); if (itr == -1) { page_faults++; // Check if the set can hold equal pages if (q.length == capacity) { q.shift(); q.push(i); } else { q.push(i); } } else { // Remove the indexes page q.splice(itr, 1); // insert the current page q.push(i); } } console.log(page_faults); // This code is contributed by ishankhandelwals. |
Output:
6
Complexity Analysis :
- Time Complexity : O(n), as it performs a constant amount of work for each page request.
- Space Complexity : O(n+4), where n is the size of the input array and 4 is the size of the memory buffer.
Note : We can also find the number of page hits. Just have to maintain a separate count.
If the current page is already in the memory then that must be count as Page-hit.
We will discuss other Page-replacement Algorithms in further sets.
This article is contributed by Sahil Chhabra. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
