Given two non-zero integers M and N, the problem is to compute the result of the Ackermann function based on some particular equations.
Ackermann function is defined as:
Examples:
Input: M = 2, N = 2
Output: 7Input: M = 2, N = 7
Output: 6141004759
The approach for Ackermann function described in this article, takes a very huge amount of time to compute the value for even small values of (M, N) or in most cases doesn’t result in anything.
Dynamic Programming approach:
Here are the following Ackermann equations that would be used to come up with efficient solution.
A(m, n) = A(m-1, A(m, n-1)) —– (Eq 1)
A(0, n) = n+1 —– (Eq 2)
A(m, 0) = A(m-1, 1) —– (Eq 3)
Let’s assume the value of m = 2 and n = 2
A 2d DP table of size ( (m+1) x (n+1) ) is created for storing the result of each sub-problem.
Following are the steps demonstrated to fill up the table.
- Empty Table – Initial Step
Filled using A ( 0, n ) = n + 1 The very next method is to fill all the base values, by taking the help of equation-2.
In the next step the whole 1st row would be filled,
A ( 1, 0 ) = A ( 0, 1 ) —– (refer Eq (3)) Since A ( 0, 1 ) = 2 Therefore, A ( 1, 0 ) = 2 —–(Eq 4)
A ( 1, 1 ) = A ( 0, A ( 1, 0 ) ) —– refer Eq (1) = A ( 0, 2 ) —– refer Eq (4) = 3 —– refer Eq (2) So, A ( 1, 1 ) = 3 —–(Eq 5)
A ( 1, 2 ) = A ( 0, A ( 1, 1 ) ) —– refer equation (1) = A ( 0, 3 ) —– refer equation (5) = 4 —– refer equation (2) So, A ( 1, 2 ) = 4 —–(Eq 6)
- Fill the table using equations and stored values
Let’s just fill the first column of the last row i.e (2, 0) in the same manner as above, because for the other two columns there are some unsolved values.
A ( 2, 0 ) = A ( 1, 1 ) —– refer equation (3) A ( 1, 1 ) = 3 So, A ( 2, 0 ) = 3 —– (Eq. 7)
Solving for A ( 2, 1 ) and A ( 2, 2 ).
For simplicity the process for solving above functions is divided into two steps,
- In the first one, the problem is identified.
A ( 2, 1 ) = A ( 1, A ( 2, 0 ) ) —– refer equation (1) A ( 2, 0 ) = 3 A ( 2, 1 ) = A ( 1, 3 ) So to compute this value, again use equation (1) A ( 1, 3 ) = A ( 0, A ( 1, 2 ) ) A ( 1, 2 ) = 4 A ( 1, 3 ) = A ( 0, 4 ) —– refer equation(2) = 5 Therefore A ( 2, 1 ) = A ( 1, 3 ) = 5 — (Eq 7)
- In the next one, methodology is described in detail and a generic formula is obtained to be logical while being used in program
Let’s solve A ( 2, 2 ) , with a theory upfront
A ( 2, 2 ) = A ( 1, A ( 2, 1 ) ) —– refer equation (1) A ( 2, 1) = 5 A ( 2, 2 ) = A ( 1, 5 )
To compute A(1, 5) in a generic manner, observe how it reduces itself!
A ( 1, 5 ) = A ( 0, A ( 1, 4 ) ) A ( 1, 4 ) = A( 0, A ( 1, 3 ) ) A ( 1, 3 ) = A ( 0, A ( 1, 2 ) ) A ( 1, 2 ) = 4 Returning back from the function we get, A ( 1, 3 ) = A ( 0, 4 ) = 5 —– refer equation (2) A ( 1, 4 ) = A ( 0, A (1, 3 ) ) = A ( 0, 5 ) = 6 —–Since A ( 1, 3 ) = 5 A ( 1, 5 ) = A ( 0, A ( 1, 4 ) ) = A ( 0, 6 ) = 7 So, A ( 2, 2 ) = 7 ——- (Eq 9)
- Final Table with result of each sub-problem
Below is the implementation of the above approach:
C++
#include <iostream>#include <string>#include <vector>using namespace std;// Bottom Up Approach long long int Ackermann(int m, int n){ // creating 2D LIST vector<vector<long long int>> cache(m + 1, vector<long long int>(n + 1, 0)); for (int rows = 0 ; rows < m + 1 ; rows++){ for (int cols = 0 ; cols < n + 1 ; cols++){ // base case A ( 0, n ) = n + 1 if (rows == 0){ cache[rows][cols] = cols + 1; } // base case A ( m, 0 ) = // A ( m-1, 1) [Computed already] else if (cols == 0) { cache[rows][cols] = cache[rows-1][1]; } else { // if rows and cols > 0 // then applying A ( m, n ) = // A ( m-1, A ( m, n-1 ) ) long long int r = rows - 1; long long int c = cache[rows][cols-1]; long long int ans; // applying equation (2) // here A ( 0, n ) = n + 1 if (r == 0){ ans = c + 1; }else if (c <= n){ // using stored value in cache ans = cache[rows-1][cache[rows][cols-1]]; }else{ // Using the Derived Formula // to compute mystery values in O(1) time ans = (c-n)*(r) + cache[r][n]; } cache[rows][cols] = ans; } } } return cache[m][n];}// Driver codeint main(){ // very small values int m = 2; int n = 2; // a bit higher value int m1 = 5; int n1 = 7; cout << "Ackermann value for m = " << m << " and n = " << n << " is -> " << Ackermann(m, n) << endl; cout << "Ackermann value for m = " << m1 << " and n = " << n1 << " is -> " << Ackermann(m1, n1) << endl; return 0;}// This code is contributed by subhamgoyal2014. |
Java
// Java program to implement the approachimport java.util.*;class GFG { // Bottom Up Approach static long Ackermann(int m, int n) { // creating 2D LIST ArrayList<ArrayList<Long> > cache = new ArrayList<ArrayList<Long> >(); for (int i = 0; i <= m; i++) { cache.add(new ArrayList<Long>()); for (int j = 0; j <= n; j++) { ArrayList<Long> l1 = cache.get(i); l1.add(0L); cache.set(i, l1); } } for (int rows = 0; rows < m + 1; rows++) { for (int cols = 0; cols < n + 1; cols++) { // base case A ( 0, n ) = n + 1 if (rows == 0) { ArrayList<Long> l1 = cache.get(rows); l1.set(cols, cols + 1L); cache.set(rows, l1); } // base case A ( m, 0 ) = // A ( m-1, 1) [Computed already] else if (cols == 0) { ArrayList<Long> l1 = cache.get(rows); l1.set(cols, cache.get(rows - 1).get(1)); cache.set(rows, l1); } else { // if rows and cols > 0 // then applying A ( m, n ) = // A ( m-1, A ( m, n-1 ) ) long r = rows - 1; long c = cache.get(rows).get(cols - 1); long ans; // applying equation (2) // here A ( 0, n ) = n + 1 if (r == 0) { ans = c + 1; } else if (c <= n) { // using stored value in cache ans = cache.get(rows - 1).get( Math.toIntExact( cache.get(rows).get( (int)cols - 1))); } else { // Using the Derived Formula // to compute mystery values in O(1) // time ans = (c - n) * (r) + cache.get((int)r).get(n); } ArrayList<Long> l1 = cache.get(rows); l1.set(cols, ans); cache.set(rows, l1); } } } return cache.get(m).get(n); } // Driver code public static void main(String[] args) { // very small values int m = 2; int n = 2; // a bit higher value int m1 = 5; int n1 = 7; System.out.println("Ackermann value for m = " + m + " and n = " + n + " is -> " + Ackermann(m, n)); System.out.println("Ackermann value for m = " + m1 + " and n = " + n1 + " is -> " + Ackermann(m1, n1)); }}// This code is contributed by phasing17. |
Python3
# Python code for the above approach# Bottom Up Approachdef Ackermann(m, n): # creating 2D LIST cache = [[0 for i in range(n + 1)] for j in range(m + 1)] for rows in range(m + 1): for cols in range(n + 1): # base case A ( 0, n ) = n + 1 if rows == 0: cache[rows][cols] = cols + 1 # base case A ( m, 0 ) = # A ( m-1, 1) [Computed already] elif cols == 0: cache[rows][cols] = cache[rows-1][1] else: # if rows and cols > 0 # then applying A ( m, n ) = # A ( m-1, A ( m, n-1 ) ) r = rows - 1 c = cache[rows][cols-1] # applying equation (2) # here A ( 0, n ) = n + 1 if r == 0: ans = c + 1 elif c <= n: # using stored value in cache ans = cache[rows-1][cache[rows][cols-1]] else: # Using the Derived Formula # to compute mystery values in O(1) time ans = (c-n)*(r) + cache[r][n] cache[rows][cols] = ans return cache[m][n]# very small valuesm = 2n = 2# a bit higher valuem1 = 5n1 = 7print("Ackermann value for m = ", m, " and n = ", n, "is -> ", Ackermann(m, n))print("Ackermann value for m = ", m1, " and n = ", n1, "is -> ", Ackermann(m1, n1)) |
C#
// C# program to implement the approachusing System;using System.Collections.Generic;class GFG { // Bottom Up Approach static long Ackermann(int m, int n) { // creating 2D LIST List<List<long> > cache = new List<List<long> >(); for (int i = 0; i <= m; i++) { cache.Add(new List<long>()); for (int j = 0; j <= n; j++) cache[i].Add(0); } for (int rows = 0; rows < m + 1; rows++) { for (int cols = 0; cols < n + 1; cols++) { // base case A ( 0, n ) = n + 1 if (rows == 0) { cache[rows][cols] = cols + 1; } // base case A ( m, 0 ) = // A ( m-1, 1) [Computed already] else if (cols == 0) { cache[rows][cols] = cache[rows - 1][1]; } else { // if rows and cols > 0 // then applying A ( m, n ) = // A ( m-1, A ( m, n-1 ) ) long r = rows - 1; long c = cache[rows][cols - 1]; long ans; // applying equation (2) // here A ( 0, n ) = n + 1 if (r == 0) { ans = c + 1; } else if (c <= n) { // using stored value in cache ans = cache[rows - 1][( int)cache[rows][cols - 1]]; } else { // Using the Derived Formula // to compute mystery values in O(1) // time ans = (c - n) * (r) + cache[(int)r][n]; } cache[rows][cols] = ans; } } } return cache[m][n]; } // Driver code public static void Main(string[] args) { // very small values int m = 2; int n = 2; // a bit higher value int m1 = 5; int n1 = 7; Console.WriteLine("Ackermann value for m = " + m + " and n = " + n + " is -> " + Ackermann(m, n)); Console.WriteLine("Ackermann value for m = " + m1 + " and n = " + n1 + " is -> " + Ackermann(m1, n1)); }}// This code is contributed by phasing17. |
Javascript
// JS program to implement the approach// Bottom Up Approach function Ackermann(m, n){ // creating 2D LIST let cache = new Array(m + 1); for (var i = 0; i <= m; i++) cache[i] = new Array(n + 1).fill(0); for (var rows = 0 ; rows < m + 1 ; rows++){ for (var cols = 0 ; cols < n + 1 ; cols++){ // base case A ( 0, n ) = n + 1 if (rows == 0){ cache[rows][cols] = cols + 1; } // base case A ( m, 0 ) = // A ( m-1, 1) [Computed already] else if (cols == 0) { cache[rows][cols] = cache[rows-1][1]; } else { // if rows and cols > 0 // then applying A ( m, n ) = // A ( m-1, A ( m, n-1 ) ) let r = rows - 1; let c = cache[rows][cols-1]; let ans; // applying equation (2) // here A ( 0, n ) = n + 1 if (r == 0){ ans = c + 1; }else if (c <= n){ // using stored value in cache ans = cache[rows-1][cache[rows][cols-1]]; }else{ // Using the Derived Formula // to compute mystery values in O(1) time ans = (c-n)*(r) + cache[r][n]; } cache[rows][cols] = ans; } } } return cache[m][n];}// Driver code// very small values let m = 2; let n = 2;// a bit higher value let m1 = 5; let n1 = 7;console.log("Ackermann value for m = " + m + " and n = " + n + " is -> " + Ackermann(m, n));console.log("Ackermann value for m = " + m1 + " and n = " + n1 + " is -> " + Ackermann(m1, n1)); // This code is contributed by phasing17. |
Ackermann value for m = 2 and n = 2 is -> 7 Ackermann value for m = 5 and n = 7 is -> 6141004759
Time complexity: O( M * N )
Auxiliary Space: O( M * N )
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