# Alt Facts, Semi Facts

This problem wasn’t overly complicated, it just took awhile to assemble all the parts.

The factorial of a positive number n is the product of all numbers from 1 to n.

5! = 5 x 4 x 3 x 2 x 1 = 120

The semifactorial (also known as the double factorial) of a positive number n is the product of all numbers from 1 to n that have the same parity (i.e. odd or even) as n.

12!! = 12 x 10 x 8 x 6 x 4 x 2 = 46,080

7!! = 7 x 5 x 3 x 1 = 105

The alternating factorial of a positive number n is the sum of the consecutive factorials from n to 1, where every other factorial is multiplied by -1.

Alternating factorial of 1:

af(1) = 1!

Alternating factorial of 2:

af(2) = 2! + (-1)x(1!) = 2! - 1! = 2 -1 = 1

Alternating factorial of 3:

af(3) = 3! - 2! + 1! = 6 - 2 + 1 = 5

Create a function that takes a number n and returns the difference between the alternating factorial and semifactorial of n. Examples

altSemi(1) ➞ 0

altSemi(2) ➞ -1

altSemi(3)➞ 2

The factorial of a positive number n is the product of all numbers from 1 to n.

5! = 5 x 4 x 3 x 2 x 1 = 120

The semifactorial (also known as the double factorial) of a positive number n is the product of all numbers from 1 to n that have the same parity (i.e. odd or even) as n.

12!! = 12 x 10 x 8 x 6 x 4 x 2 = 46,080

7!! = 7 x 5 x 3 x 1 = 105

The alternating factorial of a positive number n is the sum of the consecutive factorials from n to 1, where every other factorial is multiplied by -1.

Alternating factorial of 1:

af(1) = 1!

Alternating factorial of 2:

af(2) = 2! + (-1)x(1!) = 2! - 1! = 2 -1 = 1

Alternating factorial of 3:

af(3) = 3! - 2! + 1! = 6 - 2 + 1 = 5

Create a function that takes a number n and returns the difference between the alternating factorial and semifactorial of n. Examples

altSemi(1) ➞ 0

altSemi(2) ➞ -1

altSemi(3)➞ 2

## My solution

```
function altSemi(n) {
function semifactorial(num) {
const check = num
let numbers = []
for (let i=0; i<num; i-1) {
numbers.push(num--)
}
return check % 2 === 0 ?
numbers.filter(x=> x % 2 === 0).reduce((a, c) => a * c) :
numbers.filter(x=> x % 2 !== 0).reduce((a, c) => a * c)
}
function alterFactorial(num) {
let numbers = []
for (let i=0; i<num; i-1) {
numbers.push(num--)
}
function factorial(n) {
let lnums = []
for (let i=0; i<n; i-1) {
lnums.push(n--)
}
return lnums.reduce((a,c)=> a * c)
}
let factorials = numbers.map(x=> factorial(x))
return factorials.map((e,i)=> i % 2 !== 0 ? e * -1 : e).reduce((a,c)=> a + c)
}
return alterFactorial(n) - semifactorial(n)
}
altSemi(1) // 0
altSemi(2) // -1
altSemi(3) // 2
```

## Other solutions

Both of these solutions are much more concise than mine (as usual). Note how the first has three separate functions on each line.

```
function altSemi(n) {
const fctril=n=>n==1?1:n*fctril(n-1)
const semiF=n=>n<=1?1:n*semiF(n-2)
const altF=n=>n==1?1:fctril(n)-altF(n-1)
return altF(n)-semiF(n)
}
//Adamqwerty
```

```
const factorial = n => n === 1 ? 1 : n * factorial(n-1);
const sFactorial = n => n === 0 || n === 1 ? 1 : n * sFactorial(n-2);
function aFactorial(n) {
let result = 0, multiplier = 1;
while(n !== 0) {
result += factorial(n) * multiplier;
multiplier *= -1;
--n;
}
return result;
}
const altSemi = n => aFactorial(n) - sFactorial(n);
//el rookie
```