Program for Least Recently Used (LRU) Page Replacement algorithm

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++ 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 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 

# 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 = 4 print(pageFaults(pages, n, capacity)) # This code is contributed by ishankhandelwals. 

// 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 

<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 :

1. Using a deque data structure, the program implements the page replacement algorithm.
2. A predetermined number of pages are kept in memory by the algorithm, and they are replaced as new pages are requested.
3. Using an integer array to stimulate page requests, the code keeps track the number of page faults that occur throughout the simulation.
4. The deque data structure, which is built using STL in C++, is used to maintain the pages in memory.
5. 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++ 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 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 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# 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 

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.