How would I go about justifying this algorithm is O(log n)?
public static long exponentiation(long x, int n){
if(n == 0){
return 1;
}
else if (n % 2 == 0){
x = exponentiation(x, n / 2);
return x * x;
}
else{
return x * exponentiation(x, n-1);
}
}
Each recursive call to method exponentiation is a multiplication step. Hence you need to count the number of recursive calls. There are several ways to achieve this. I chose to add another parameter to the method.
public static long exponentiation(long x, int n, int count) {
if (n == 0) {
System.out.println("steps = " + count);
return 1;
}
else if (n % 2 == 0) {
x = exponentiation(x, n / 2, count + 1);
return x * x;
}
else {
return x * exponentiation(x, n - 1, count + 1);
}
}
Here is the initial call to method exponentiation
exponentiation(2, 63, 0);
When I run the above code, the following is printed
steps = 11
You can use a static counter as well (without changing the prototype of the function):
public static long counter = 0;
public static long exponentiation(long x, int n){
if(n == 0){
return 1;
}
else if (n % 2 == 0){
x = exponentiation(x, n / 2);
counter++;
return x * x;
}
else{
counter++;
return x * exponentiation(x, n-1);
}
}
However, you need to reset the counter before calling the function each time, i.e., set counter = 0.
Theoretical Analysis
Note that you need to the counter to prove that it is in O(log(n)). To prove the complexity, just you need to find the complexity term by looking at the flow of the code. Suppose T(n) is the number of multiplications for computing x^n. So, based on the written code, T(n) = T(n/2) + 1, if n is even, and T(n) = T(n-1) + 1, if n is odd. Now, at least in one of two consecutive recursions, input n is even. Therefore, at most 2 log(n) is required to reach to n = 0. Because, for each even input, the next input will be halved. So, we can conclude that the algorithm is in O(log(n)).
I need to write a recursive method using Java called power that takes a double x and an integer n and that returns x^n. Here is what I have so far.
public static double power(double x, int n) {
if (n == 0)
return 1;
if (n == 1)
return x;
else
return x * (power(x, n-1));
}
This code works as expected. However, I am trying to go the extra mile and perform the following optional exercise:
"Optional challenge: you can make this method more efficient, when n is even, using x^n = (x^(n/2))^2."
I am not sure how to implement that last formula when n is even. I do not think I can use recursion for that. I have tried to implement the following, but it also does not work because I cannot take a double to the power of an int.
if (n%2 == 0)
return (x^(n/2))^2;
Can somebody point me in the right direction? I feel like I am missing something obvious. All help appreciated.
It's exactly the same principle as for x^n == x*(x^(n-1)): Insert your recursive function for x^(n/2) and (...)^2, but make sure you don't enter an infinite recursion for n == 2 (as 2 is even, too):
if (n % 2 == 0 && n > 2)
return power(power(x, n / 2), 2);
}
Alternatively, you could just use an intermediate variable:
if (n % 2 == 0) {
double s = power(x, n / 2);
return s * s;
}
I'd probably just handle 2 as a special case, too -- and avoid the "and"-condition and extra variable:
public static double power(double x, int n) {
if (n == 0) return 1;
if (n == 1) return x;
if (n == 2) return x * x;
if (n % 2 == 0) return power(power(x, n / 2), 2);
return x * (power(x, n - 1));
}
P.S. I think this should work, too :)
public static double power(double x, int n) {
if (n == 0) return 1;
if (n == 1) return x;
if (n == 2) return x * x;
return power(x, n % 2) * power(power(x, n / 2), 2);
}
When n is even, the formula is exactly what you wrote: divide n by two, call power recursively, and square the result.
When n is odd, the formula is a little more complex: subtract 1 from n, make a recursive call for n/2, square the result, and multiply by x.
if (n%2 == 0)
return (x^(n/2))^2;
else
return x*(x^(n/2))^2;
n/2 truncates the result, so subtraction of 1 is not done explicitly. Here is an implementation in Java:
public static double power(double x, int n) {
if (n == 0) return 1;
if (n == 1) return x;
double pHalf = power(x, n/2);
if (n%2 == 0) {
return pHalf*pHalf;
} else {
return x*pHalf*pHalf;
}
}
Demo.
Hint: The ^ operation won't perform exponentiation in Java, but the function you wrote, power will.
Also, don't forget squaring a number is the same as just multiplying it by itself. No function call needed.
Making a small change to your function, it will reduce the number of recursive calls made:
public static double power(double x, int n) {
if (n == 0) {
return 1;
}
if (n == 1) {
return x;
}
if (n % 2 == 0) {
double temp = power(x, n / 2);
return temp * temp;
} else {
return x * (power(x, n - 1));
}
}
Since
x^(2n) = (x^n)^2
you can add this rule to your method, either using the power function you wrote, as Stefan Haustein suggested, or using the regular multiplication operator, since it seems you are allowed to do that.
Note that there is no need for both the base cases n=1 and n=0, one of them suffices (prefferably use the base case n=0, since otherwise your method would not be defined for n=0).
public static double power(double x, int n) {
if (n == 0)
return 1;
else if (n % 2 == 0)
double val = power(x, n/2);
return val * val;
else
return x * (power(x, n-1));
}
There is no need to check that n>2 in any of the cases.
This just reminds me more optimisation could be done
and this following code.
class Solution:
# #param x, a float
# #param n, a integer
# #return a float
def pow(self, x, n):
if n<0:
return 1.0/self.pow(x,-n)
elif n==0:
return 1.0
elif n==1:
return x
else:
m = n & (-n)
if( m==n ):
r1 = self.pow(x,n>>1)
return r1*r1
else:
return self.pow(x,m)*self.pow(x,n-m)
what is more intermediate result could be memorised and avoid redundant computation.
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I didn't find it, yet. Did I miss something?
I know a factorial method is a common example program for beginners. But wouldn't it be useful to have a standard implementation for this one to reuse?
I could use such a method with standard types (Eg. int, long...) and with BigInteger / BigDecimal, too.
Apache Commons Math has a few factorial methods in the MathUtils class.
public class UsefulMethods {
public static long factorial(int number) {
long result = 1;
for (int factor = 2; factor <= number; factor++) {
result *= factor;
}
return result;
}
}
Big Numbers version by HoldOffHunger:
public static BigInteger factorial(BigInteger number) {
BigInteger result = BigInteger.valueOf(1);
for (long factor = 2; factor <= number.longValue(); factor++) {
result = result.multiply(BigInteger.valueOf(factor));
}
return result;
}
I don't think it would be useful to have a library function for factorial. There is a good deal of research into efficient factorial implementations. Here is a handful of implementations.
Bare naked factorials are rarely needed in practice. Most often you will need one of the following:
1) divide one factorial by another, or
2) approximated floating-point answer.
In both cases, you'd be better with simple custom solutions.
In case (1), say, if x = 90! / 85!, then you'll calculate the result just as x = 86 * 87 * 88 * 89 * 90, without a need to hold 90! in memory :)
In case (2), google for "Stirling's approximation".
Use Guava's BigIntegerMath as follows:
BigInteger factorial = BigIntegerMath.factorial(n);
(Similar functionality for int and long is available in IntMath and LongMath respectively.)
Although factorials make a nice exercise for the beginning programmer, they're not very useful in most cases, and everyone knows how to write a factorial function, so they're typically not in the average library.
i believe this would be the fastest way, by a lookup table:
private static final long[] FACTORIAL_TABLE = initFactorialTable();
private static long[] initFactorialTable() {
final long[] factorialTable = new long[21];
factorialTable[0] = 1;
for (int i=1; i<factorialTable.length; i++)
factorialTable[i] = factorialTable[i-1] * i;
return factorialTable;
}
/**
* Actually, even for {#code long}, it works only until 20 inclusively.
*/
public static long factorial(final int n) {
if ((n < 0) || (n > 20))
throw new OutOfRangeException("n", 0, 20);
return FACTORIAL_TABLE[n];
}
For the native type long (8 bytes), it can only hold up to 20!
20! = 2432902008176640000(10) = 0x 21C3 677C 82B4 0000
Obviously, 21! will cause overflow.
Therefore, for native type long, only a maximum of 20! is allowed, meaningful, and correct.
Because factorial grows so quickly, stack overflow is not an issue if you use recursion. In fact, the value of 20! is the largest one can represent in a Java long. So the following method will either calculate factorial(n) or throw an IllegalArgumentException if n is too big.
public long factorial(int n) {
if (n > 20) throw new IllegalArgumentException(n + " is out of range");
return (1 > n) ? 1 : n * factorial(n - 1);
}
Another (cooler) way to do the same stuff is to use Java 8's stream library like this:
public long factorial(int n) {
if (n > 20) throw new IllegalArgumentException(n + " is out of range");
return LongStream.rangeClosed(1, n).reduce(1, (a, b) -> a * b);
}
Read more on Factorials using Java 8's streams
Apache Commons Math package has a factorial method, I think you could use that.
Short answer is: use recursion.
You can create one method and call that method right inside the same method recursively:
public class factorial {
public static void main(String[] args) {
System.out.println(calc(10));
}
public static long calc(long n) {
if (n <= 1)
return 1;
else
return n * calc(n - 1);
}
}
Try this
public static BigInteger factorial(int value){
if(value < 0){
throw new IllegalArgumentException("Value must be positive");
}
BigInteger result = BigInteger.ONE;
for (int i = 2; i <= value; i++) {
result = result.multiply(BigInteger.valueOf(i));
}
return result;
}
You can use recursion.
public static int factorial(int n){
if (n == 0)
return 1;
else
return(n * factorial(n-1));
}
and then after you create the method(function) above:
System.out.println(factorial(number of your choice));
//direct example
System.out.println(factorial(3));
I found an amazing trick to find factorials in just half the actual multiplications.
Please be patient as this is a little bit of a long post.
For Even Numbers:
To halve the multiplication with even numbers, you will end up with n/2 factors. The first factor will be the number you are taking the factorial of, then the next will be that number plus that number minus two. The next number will be the previous number plus the lasted added number minus two. You are done when the last number you added was two (i.e. 2). That probably didn't make much sense, so let me give you an example.
8! = 8 * (8 + 6 = 14) * (14 + 4 = 18) * (18 + 2 = 20)
8! = 8 * 14 * 18 * 20 which is **40320**
Note that I started with 8, then the first number I added was 6, then 4, then 2, each number added being two less then the number added before it. This method is equivalent to multiplying the least numbers with the greatest numbers, just with less multiplication, like so:
8! = 1 * 2 * 3 * 4 * 5 * 6 * 7 *
8! = (1 * 8) * (2 * 7) * (3 * 6) * (4 * 5)
8! = 8 * 14 * 18 * 20
Simple isn't it :)
Now For Odd Numbers: If the number is odd, the adding is the same, as in you subtract two each time, but you stop at three. The number of factors however changes. If you divide the number by two, you will end up with some number ending in .5. The reason is that if we multiply the ends together, that we are left with the middle number. Basically, this can all be solved by solving for a number of factors equal to the number divided by two, rounded up. This probably didn't make much sense either to minds without a mathematical background, so let me do an example:
9! = 9 * (9 + 7 = 16) * (16 + 5 = 21) * (21 + 3 = 24) * (roundUp(9/2) = 5)
9! = 9 * 16 * 21 * 24 * 5 = **362880**
Note: If you don't like this method, you could also just take the factorial of the even number before the odd (eight in this case) and multiply it by the odd number (i.e. 9! = 8! * 9).
Now let's implement it in Java:
public static int getFactorial(int num)
{
int factorial=1;
int diffrennceFromActualNum=0;
int previousSum=num;
if(num==0) //Returning 1 as factorial if number is 0
return 1;
if(num%2==0)// Checking if Number is odd or even
{
while(num-diffrennceFromActualNum>=2)
{
if(!isFirst)
{
previousSum=previousSum+(num-diffrennceFromActualNum);
}
isFirst=false;
factorial*=previousSum;
diffrennceFromActualNum+=2;
}
}
else // In Odd Case (Number * getFactorial(Number-1))
{
factorial=num*getFactorial(num-1);
}
return factorial;
}
isFirst is a boolean variable declared as static; it is used for the 1st case where we do not want to change the previous sum.
Try with even as well as for odd numbers.
The only business use for a factorial that I can think of is the Erlang B and Erlang C formulas, and not everyone works in a call center or for the phone company. A feature's usefulness for business seems to often dictate what shows up in a language - look at all the data handling, XML, and web functions in the major languages.
It is easy to keep a factorial snippet or library function for something like this around.
A very simple method to calculate factorials:
private double FACT(double n) {
double num = n;
double total = 1;
if(num != 0 | num != 1){
total = num;
}else if(num == 1 | num == 0){
total = 1;
}
double num2;
while(num > 1){
num2 = num - 1;
total = total * num2;
num = num - 1;
}
return total;
}
I have used double because they can hold massive numbers, but you can use any other type like int, long, float, etc.
P.S. This might not be the best solution but I am new to coding and it took me ages to find a simple code that could calculate factorials so I had to write the method myself but I am putting this on here so it helps other people like me.
You can use recursion version as well.
static int myFactorial(int i) {
if(i == 1)
return;
else
System.out.prinln(i * (myFactorial(--i)));
}
Recursion is usually less efficient because of having to push and pop recursions, so iteration is quicker. On the other hand, recursive versions use fewer or no local variables which is advantage.
We need to implement iteratively. If we implement recursively, it will causes StackOverflow if input becomes very big (i.e. 2 billions). And we need to use unbound size number such as BigInteger to avoid an arithmatic overflow when a factorial number becomes bigger than maximum number of a given type (i.e. 2 billion for int). You can use int for maximum 14 of factorial and long for maximum 20
of factorial before the overflow.
public BigInteger getFactorialIteratively(BigInteger input) {
if (input.compareTo(BigInteger.ZERO) <= 0) {
throw new IllegalArgumentException("zero or negatives are not allowed");
}
BigInteger result = BigInteger.ONE;
for (BigInteger i = BigInteger.ONE; i.compareTo(input) <= 0; i = i.add(BigInteger.ONE)) {
result = result.multiply(i);
}
return result;
}
If you can't use BigInteger, add an error checking.
public long getFactorialIteratively(long input) {
if (input <= 0) {
throw new IllegalArgumentException("zero or negatives are not allowed");
} else if (input == 1) {
return 1;
}
long prev = 1;
long result = 0;
for (long i = 2; i <= input; i++) {
result = prev * i;
if (result / prev != i) { // check if result holds the definition of factorial
// arithmatic overflow, error out
throw new RuntimeException("value "+i+" is too big to calculate a factorial, prev:"+prev+", current:"+result);
}
prev = result;
}
return result;
}
Factorial is highly increasing discrete function.So I think using BigInteger is better than using int.
I have implemented following code for calculation of factorial of non-negative integers.I have used recursion in place of using a loop.
public BigInteger factorial(BigInteger x){
if(x.compareTo(new BigInteger("1"))==0||x.compareTo(new BigInteger("0"))==0)
return new BigInteger("1");
else return x.multiply(factorial(x.subtract(new BigInteger("1"))));
}
Here the range of big integer is
-2^Integer.MAX_VALUE (exclusive) to +2^Integer.MAX_VALUE,
where Integer.MAX_VALUE=2^31.
However the range of the factorial method given above can be extended up to twice by using unsigned BigInteger.
We have a single line to calculate it:
Long factorialNumber = LongStream.rangeClosed(2, N).reduce(1, Math::multiplyExact);
A fairly simple method
for ( int i = 1; i < n ; i++ )
{
answer = answer * i;
}
/**
import java liberary class
*/
import java.util.Scanner;
/* class to find factorial of a number
*/
public class factorial
{
public static void main(String[] args)
{
// scanner method for read keayboard values
Scanner factor= new Scanner(System.in);
int n;
double total = 1;
double sum= 1;
System.out.println("\nPlease enter an integer: ");
n = factor.nextInt();
// evaluvate the integer is greater than zero and calculate factorial
if(n==0)
{
System.out.println(" Factorial of 0 is 1");
}
else if (n>0)
{
System.out.println("\nThe factorial of " + n + " is " );
System.out.print(n);
for(int i=1;i<n;i++)
{
do // do while loop for display each integer in the factorial
{
System.out.print("*"+(n-i) );
}
while ( n == 1);
total = total * i;
}
// calculate factorial
sum= total * n;
// display sum of factorial
System.out.println("\n\nThe "+ n +" Factorial is : "+" "+ sum);
}
// display invalid entry, if enter a value less than zero
else
{
System.out.println("\nInvalid entry!!");
}System.exit(0);
}
}
public static int fact(int i){
if(i==0)
return 0;
if(i>1){
i = i * fact(--i);
}
return i;
}
public int factorial(int num) {
if (num == 1) return 1;
return num * factorial(num - 1);
}
while loop (for small numbers)
public class factorial {
public static void main(String[] args) {
int counter=1, sum=1;
while (counter<=10) {
sum=sum*counter;
counter++;
}
System.out.println("Factorial of 10 is " +sum);
}
}
I got this from EDX use it! its called recursion
public static int factorial(int n) {
if (n == 1) {
return 1;
} else {
return n * factorial(n-1);
}
}
with recursion:
public static int factorial(int n)
{
if(n == 1)
{
return 1;
}
return n * factorial(n-1);
}
with while loop:
public static int factorial1(int n)
{
int fact=1;
while(n>=1)
{
fact=fact*n;
n--;
}
return fact;
}
using recursion is the simplest method. if we want to find the factorial of
N, we have to consider the two cases where N = 1 and N>1 since in factorial
we keep multiplying N,N-1, N-2,,,,, until 1. if we go to N= 0 we will get 0
for the answer. in order to stop the factorial reaching zero, the following
recursive method is used. Inside the factorial function,while N>1, the return
value is multiplied with another initiation of the factorial function. this
will keep the code recursively calling the factorial() until it reaches the
N= 1. for the N=1 case, it returns N(=1) itself and all the previously built
up result of multiplied return N s gets multiplied with N=1. Thus gives the
factorial result.
static int factorial(int N) {
if(N > 1) {
return n * factorial(N - 1);
}
// Base Case N = 1
else {
return N;
}
public static long factorial(int number) {
if (number < 0) {
throw new ArithmeticException(number + " is negative");
}
long fact = 1;
for (int i = 1; i <= number; ++i) {
fact *= i;
}
return fact;
}
using recursion.
public static long factorial(int number) {
if (number < 0) {
throw new ArithmeticException(number + " is negative");
}
return number == 0 || number == 1 ? 1 : number * factorial(number - 1);
}
source
Using Java 9+, you can use this solution. This uses BigInteger, ideal for holding large numbers.
...
import java.math.BigInteger;
import java.util.stream.Stream;
...
String getFactorial(int n) {
return Stream.iterate(BigInteger.ONE, i -> i.add(BigInteger.ONE)).parallel()
.limit(n).reduce(BigInteger.ONE, BigInteger::multiply).toString();
}
USING DYNAMIC PROGRAMMING IS EFFICIENT
if you want to use it to calculate again and again (like caching)
Java code:
int fact[]=new int[n+1]; //n is the required number you want to find factorial for.
int factorial(int num)
{
if(num==0){
fact[num]=1;
return fact[num];
}
else
fact[num]=(num)*factorial(num-1);
return fact[num];
}