Reputation: 1365

Recursive Fibonacci memoization

I need some help with a program I'm writing for my Programming II class at universtiy. The question asks that one calculates the Fibonacci sequence using recursion. One must store the calculated Fibonacci numbers in an array to stop unnecessary repeated calculations and to cut down to the calculation time.

I managed to get the program working without the array and memorization, now I'm trying to implement that and I'm stuck. I'm not sure how to structure it. I've Googled and skimmed through some books but haven't found much to help me solve how to implement a solution.

import javax.swing.JOptionPane;
public class question2
{
static int count = 0;
static int [] dictionary;

public static void main(String[] args)
{

int answer;
int num = Integer.parseInt(javax.swing.JOptionPane.showInputDialog("Enter n:"));

javax.swing.JOptionPane.showMessageDialog(null, 
        "About to calculate fibonacci(" + num + ")");

//giving the array "n" elements
dictionary= new int [num];

if (dictionary.length>=0)
dictionary[0]= 0;

if (dictionary.length>=1)
dictionary[0]= 0;
dictionary[1]= 1;


//method call
answer = fibonacci(num);

//output
JOptionPane.showMessageDialog(null,"Fibonacci("+num+") is "+answer+" (took "+count+" calls)");
}



  static int fibonacci(int n)
  {
count++;

// Only defined for n >= 0
if (n < 0) {
  System.out.println("ERROR: fibonacci sequence not defined for negative numbers.");
  System.exit(1);
}

// Base cases: f(0) is 0, f(1) is 1
// Other cases: f(n) = f(n-1) + f(n-2)/
if (n == 0) 
{
  return dictionary[0];
}

else if (n == 1) 
{
  return dictionary[1];
}

else
return dictionary[n] = fibonacci(n-1) + fibonacci(n-2);
  
 

}

}

The above is incorrect, the end of my fib method is the main problem. I've no idea how to get it to add the numbers recursively to the correctly parts of the array.

Upvotes: 14

Answers (15)

kumar shivam

Reputation: 69

using System; using System.Collections.Generic;

namespace Fibonacci { public class FibonacciSeries {

        static void Main(string[] args)
    {
        int n;
        Dictionary<int, long> dict = new Dictionary<int, long>();
        Console.WriteLine("ENTER NUMBER::");
        n = Convert.ToInt32(Console.ReadLine());
        for (int j = 0; j <= n; j++)
        {
            Console.WriteLine(Fib(j, dict));
        }
    }

   

    public static long Fib(int n, Dictionary<int, long> dict)
    {
        if (n <= 1)
            return n;
        if (dict.ContainsKey(n))
            return dict[n]; 
       
        var value = Fib(n - 1,dict) + Fib(n - 2,dict);
        dict[n] = value;
        return value;
    }

}

}

Upvotes: 0

multitudes

Reputation: 3495

In Swift 5.3

This is a very quick one using memoisation. First I initialise my cache dictionary.

var cache = [Int:Int]()

Then I create my Fibonacci number generator. Since it is a recursive function, every call to the function would theoretically compute the whole Fibonacci sequence again up to the requested number. This is why we use the cache, to speed up the recursive function:

func fibonacci(_ number: Int) -> Int {
    // if the value is in the dictionary I just return it
    if let value = cache[number] { return value }

    // Otherwise I calculate it recursively. 
    // Every recursion will check the cache again, 
    // this is why memoisation is faster!
    let newValue = number < 2 ? number : fibonacci(number - 1) + fibonacci(number - 2)
    cache[number] = newValue
    return newValue
}

I can save my sequence in an array like this:

var numbers = Array(0..<10).map(fibonacci) //[0, 1, 1, 2, 3, 5, 8, 13, 21, 34]

Or use the function in a loop.

Upvotes: 0

LearningJava

Reputation: 1

public class FiboSeries {

    // first two terms of Fibonacci
    int x1 = 0;
    int x2 = 1;
    long xn; // nth number in Fibo series
    long[] array; // an array for implementing memoization

    // print the Nth number of Fibonacci - logic is f(n) = f(n-1) + f(n-2)

    long fibo(int n) {

        // initialize the array having n elements if it does not exist already
        if (array == null) {

            array = new long[n + 1];

        }

        // Fetch the memoized value from the array instead of recursion
        // for instance, fibo(3) will be calculated just once and stored inside this
        // array for next call
        if (array[n] != 0)

        {
            xn = array[n];
            return xn;
        }

        // value of fibo(1)
        if (n == 1) {
            xn = x1;

        }

        // value of fibo(2)
        if (n == 2) {
            xn = x2;

        }

        // value of Fibo(n) using non linear recursion
        if (n > 2) {

            xn = fibo(n - 1) + fibo(n - 2);
        }

        // before returning the value - store it at nth position of an array
        // However, before saving the value into array, check if the position is already 
        //full or not

        if (array[n] == 0) {

            array[n] = xn;
        }

        return xn;

    }

    public static void main(String[] args) {

        FiboSeries f = new FiboSeries();

        int n = 50;
        long number = f.fibo(n);

        System.out.println(number);

    }


}

Upvotes: -1

Manish Singh

Reputation: 310

Might be too old but here is my solution for swift

class Recursion {
    func fibonacci(_ input: Int) {
        var dictioner: [Int: Int] = [:]
        dictioner[0] = 0
        dictioner[1] = 1
       print(fibonacciCal(input, dictioner: &dictioner))
    }

    func fibonacciCal(_ input: Int, dictioner: inout [Int: Int]) -> Int {
        if let va = dictioner[input]{
            return va
        } else {
            let firstPart = fibonacciCal(input-1, dictioner: &dictioner)

            let secondPart = fibonacciCal(input-2, dictioner: &dictioner)

            if dictioner[input] == nil {
                dictioner[input] = firstPart+secondPart
            }

            return firstPart+secondPart
        }
    }
}

// 0,1,1,2,3,5,8
class TestRecursion {
    func testRecursion () {
        let t = Recursion()
        t.fibonacci(3)
    }
}

Upvotes: -1

Naren

Reputation: 3

import java.util.HashMap;
import java.util.Map;

public class FibonacciSequence {

    public static int fibonacci(int n, Map<Integer, Integer> memo) {
        if (n < 2) {
            return n;
        }
        if (!memo.containsKey(n)) {
            memo.put(n, fibonacci(n - 1, memo) + fibonacci(n - 2, memo));
        }
        return memo.get(n);
    }

    public static int fibonacci(int n, int[] memo) {
        if (n < 2) {
            return n;
        }
        if (memo[n - 1] != 0) {
            return memo[n - 1];
        }
        return memo[n - 1] = fibonacci(n - 1, memo) + fibonacci(n - 2, memo);
    }

    public static void main(String[] s) {
        int n = 10;

        System.out.println("f(n) = " + fibonacci(n, new HashMap<Integer, Integer>()));
        System.out.println("f(n) = " + fibonacci(n, new int[n]));
    }
}

Upvotes: 0

Lin Endian

Reputation: 103

Here is my implementation.

private static int F(int N, int[] A) {
    if ((N == 0) || (N == 1)) return N;
    if (A[N] != 0) return A[N];

    if ((A[N - 1] != 0) && (A[N - 2] != 0)) {
        A[N] = A[N - 1] + A[N - 2];
        return A[N];
    }

    if (A[N-2] != 0) {
        A[N] = A[N - 2] + F(N - 1, A);
        return A[N];
    }
    if (A[N-1] != 0) {
        A[N] = A[N - 1] + F(N - 2, A);
        return A[N];
    }
    A[N] = F(N-1, A) + F(N-2, A);
    return A[N];
}

Upvotes: -2

MAHDI AKBARIZARKESH

Reputation: 23

#include <stdio.h>
long int A[100]={1,1};
long int fib(int n){
    if (A[n])
    {
        return A[n];
    }
    else
    {
         return A[n]=fib(n-1)+fib(n-2);
    }
}
int main(){
    printf("%ld",fib(30));
}

Upvotes: -2

AlexMelw

Reputation: 2624

Here is a fully-fledged class that leverages the memoization concept:

import java.util.HashMap;
import java.util.Map;

public class Fibonacci {

    public static Fibonacci getInstance() {
        return new Fibonacci();
    }

    public int fib(int n) {
        HashMap<Integer, Integer> memoizedMap = new HashMap<>();

        memoizedMap.put(0, 0);
        memoizedMap.put(1, 1);

        return fib(n, memoizedMap);
    }

    private int fib(int n, Map<Integer, Integer> map) {
        if (map.containsKey(n))
            return map.get(n);

        int fibFromN = fib(n - 1, map) + fib(n - 2, map);

        // MEMOIZE the computed value
        map.put(n, fibFromN);

        return fibFromN;
    }
}

Notice that

memoizedMap.put(0, 0);
memoizedMap.put(1, 1);

are used to eliminate the necessity of the following check

if (n == 0) return 0;
if (n == 1) return 1;

at each recursive function call.

Upvotes: 2

user3403187

Reputation: 129

public static int fib(int n, Map<Integer,Integer> map){

    if(n ==0){
        return 0;
    }

    if(n ==1){
        return 1;
    }

    if(map.containsKey(n)){
        return map.get(n);
    }

    Integer fibForN = fib(n-1,map) + fib(n-2,map);
    map.put(n, fibForN);

    return fibForN; 

}

Similar to most solutions above but using a Map instead.

Upvotes: 13

Jack Ca

Reputation: 93

Here is my implementation of recursive fibonacci memoization. Using BigInteger and ArrayList allows to calculate 100th or even larger term. I tried 1000th terms, and result is returned in a matter of milliseconds, here is the code:

    private static List<BigInteger> dict = new ArrayList<BigInteger>();
    public static void printFebonachiRecursion (int num){
    if (num==1){
        printFebonachiRecursion(num-1);
        System.out.printf("Term %d: %d%n",num,1);
        dict.add(BigInteger.ONE);
    }
    else if (num==0){
        System.out.printf("Term %d: %d%n",num,0);
        dict.add(BigInteger.ZERO);
    }
    else {
    printFebonachiRecursion(num-1);
    dict.add(dict.get(num-2).add(dict.get(num-1)));
    System.out.printf("Term %d: %d%n",num,dict.get(num));
    }
}

Output example

printFebonachiRecursion(100);

Term 0: 0
Term 1: 1
Term 2: 1
Term 3: 2
...
Term 98: 135301852344706746049
Term 99: 218922995834555169026
Term 100: 354224848179261915075

Upvotes: 4

Frank

Reputation: 41

This is another way to approach memoization for recursive fibonacci() method using a static array of values -

public static long fibArray[]=new long[50];\\Keep it as large as you need

public static long fibonacci(long n){
long fibValue=0;
if(n==0 ){
    return 0;
}else if(n==1){
    return 1;
}else if(fibArray[(int)n]!=0){
    return fibArray[(int)n];    
}
else{
    fibValue=fibonacci(n-1)+fibonacci(n-2);
    fibArray[(int) n]=fibValue;
    return fibValue;
}
}

Note that this method uses a global(class level) static array fibArray[]. To have a look at the whole code with explanation you can also see the following - http://www.javabrahman.com/gen-java-programs/recursive-fibonacci-in-java-with-memoization/

Upvotes: 1

Rohit

Reputation: 6613

Program to print first n fibonacci numbers using Memoization.

int[] dictionary; 
// Get Fibonacci with Memoization
public int getFibWithMem(int n) {
    if (dictionary == null) {
        dictionary = new int[n];
    }

    if (dictionary[n - 1] == 0) {
        if (n <= 2) {
            dictionary[n - 1] = n - 1;
        } else {
            dictionary[n - 1] = getFibWithMem(n - 1) + getFibWithMem(n - 2);
        }
    }

    return dictionary[n - 1];
}

public void printFibonacci()
{
    for (int curr : dictionary) {
        System.out.print("F[" + i++ + "]:" + curr + ", ");
    }
}

Upvotes: 6

aksinghdce

Reputation: 193

int F(int Num){
int i =0;
int* A = NULL;
if(Num > 0)
{
 A = (int*) malloc(Num * sizeof(int));
}
else
 return Num;

for(;i<Num;i++)
 A[i] = -1;

return F_M(Num, &A);


}

int F_M(int Num, int** Ap){
int Num1 = 0;
int Num2 = 0;

if((*Ap)[Num - 1] < 0)
{
  Num1 = F_M(Num - 1, Ap);
  (*Ap)[Num -1] = Num1;
  printf("Num1:%d\n",Num1);
}
else
  Num1 = (*Ap)[Num - 1];

if((*Ap)[Num - 2] < 0)
{
  Num2 = F_M(Num - 2, Ap);
  (*Ap)[Num -2] = Num2;
  printf("Num2:%d\n",Num2);
}
else
  Num2 = (*Ap)[Num - 2];

if(0 == Num || 1 == Num)
{
 (*Ap)[Num] = Num;
 return Num;
}
else{
//  return ((*Ap)[Num - 2] > 0?(*Ap)[Num - 2] = F_M(Num -2, Ap): (*Ap)[Num - 2]  ) +     ((*Ap)[Num - 1] > 0?(*Ap)[Num - 1] = F_M(Num -1, Ap): (*Ap)[Num - 1]  );
  return (Num1 + Num2);
}

}

int main(int argc, char** argv){
int Num = 0;
if(argc>1){
sscanf(argv[1], "%d", &Num);
}

printf("F(%d) = %d", Num, F(Num));

return 0;

}

Upvotes: 1

Joachim Sauer

Reputation: 308001

You need to distinguish between already calculated number and not calculated numbers in the dictionary, which you currently don't: you always recalculate the numbers.

if (n == 0) 
{
  // special case because fib(0) is 0
  return dictionary[0];
}
else 
{
  int f = dictionary[n];
  if (f == 0) {
    // number wasn't calculated yet.
    f = fibonacci(n-1) + fibonacci(n-2);
    dictionary[n] = f;
  }
  return f;
}

Upvotes: 25

aioobe

Reputation: 420951

I believe you forget to actually look up stuff in your dictionary.

Change

else
    return dictionary[n] = fibonacci(n-1) + fibonacci(n-2);

else {
    if (dictionary[n] > 0)
        return dictionary[n];

    return dictionary[n] = fibonacci(n - 1) + fibonacci(n - 2);
}

and it works just fine (tested it myself :)

Upvotes: 4

Recursive Fibonacci memoization

Answers (15)

In Swift 5.3

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