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Fibonacci optimized using binet's formula #3728

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Merged
merged 6 commits into from
Oct 30, 2022
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Fibonacci optimized using binet's formula #3728

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de502bc
Fibonacci optimized using binet's formula
Aayush2111 Oct 28, 2022
bcd539c
Merge branch 'TheAlgorithms:master' into patch-1
Aayush2111 Oct 29, 2022
1810d4a
Update Fibonacci.java
Aayush2111 Oct 29, 2022
62de576
Update Fibonacci.java
Aayush2111 Oct 29, 2022
1f9c2bb
updated changes
Aayush2111 Oct 29, 2022
49a7061
Merge branch 'master' into patch-1
Oct 30, 2022
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19 changes: 18 additions & 1 deletion src/main/java/com/thealgorithms/dynamicprogramming/Fibonacci.java
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Original file line number Diff line number Diff line change
Expand Up @@ -19,6 +19,7 @@ public static void main(String[] args) {
System.out.println(fibMemo(n));
System.out.println(fibBotUp(n));
System.out.println(fibOptimized(n));
System.out.println(fibBinet(n));
sc.close();
}

Expand Down Expand Up @@ -90,5 +91,21 @@ public static int fibOptimized(int n) {
res = next;
}
return res;
}
}

/**
* We have only defined the nth Fibonacci number in terms of the two before it. Now, we will look at Binet's formula to calculate the nth Fibonacci number in constant time.
* The Fibonacci terms maintain a ratio called golden ratio denoted by Φ, the Greek character pronounced ‘phi'.
* First, let's look at how the golden ratio is calculated: Φ = ( 1 + √5 )/2 = 1.6180339887...
* Now, let's look at Binet's formula: Sn = Φⁿ–(– Φ⁻ⁿ)/√5
* We first calculate the squareRootof5 and phi and store them in variables. Later, we apply Binet's formula to get the required term.
* Time Complexity will be O(1)
*/

public static int fibBinet(int n) {
double squareRootOf5 = Math.sqrt(5);
double phi = (1 + squareRootOf5)/2;
int nthTerm = (int) ((Math.pow(phi, n) - Math.pow(-phi, -n))/squareRootOf5);
return nthTerm;
}
}