I need to convert this function using the functional programming paradigm but I don't know how, I can use reducer or map creating an array but I don't know how to implement it, i can't use divide operator, loop or recursion;
function divide(dividend, divisor) {
var result = 0;
while (dividend >= divisor) {
dividend -= divisor;
result++;
}
return result;
}
console.log(divide(100, 2));The way to do it declaratively is with a recursive function....
const divide = (t, b, depth = 0) => t < b ? depth : divide(t-b, b, depth+1);
console.log(`150 / 3 = ${divide(150, 3)}`);
console.log(`24 / 3 = ${divide(24, 3)}`);
console.log(`4 / 3 = ${divide(4, 3)}`);I'm a bit puzzled by the requirements. My understanding is that loops or recursion aren't prohibited in functional programming. Assuming this is an exercise (it has to be) then here's another way to look at it:
To solve a / b you can count how many b you can fit in a. So for example:
10 / 2 -> [2, 2, 2, 2, 2] -> 5
or:
+2 +2 +2 +2 (map)
10 / 2 -> [2, 4, 6, 8, 10] -> 5
^ ^^
(x) (pred)
So we can unfold the divisor into a list of sums of itself:
const unfold = (pred, map, x) => {
const ys = [];
for (let y = x; pred(y); y = map(y)) ys.push(y);
return ys;
}
unfold(x => x <= 10, x => x + 2, 2);
//=> [2, 4, 6, 8, 10]
Now we can implement divide with unfold and return the length of the list:
const divide = (a, b) =>
unfold(x => x <= a, x => x + b, b)
.length;
divide(10, 2);
//=> 5
My Final solution is this:
const adition = (a, b) => a + b;
const subtraction = (a, b) => a - b;
const multiplication = (a, b) => {
return b >= 0 ? [...Array(b)].reduce((acc) => adition(acc, a), 0) : [...Array(a)].reduce((acc) => adition(acc, b), 0);
};
const division = (a, b) => {
return a === 0 || b === 0 ? 'Error' : b > 1 ? [...Array(a).keys()].reduce((acc, num) => multiplication(num, b) <= a ? adition(acc, 1) : acc, -1) : a;
};