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Asked: 2020-08-11 14:50:18 +0800 CST 2020-08-11 14:50:18 +0800 CST

How does a return to || b of a function in recursion in JavaScript?

  • 772

This is a recursive function to show on the screen the way to obtain X number through a mathematical calculation where only sumar 5 (+5)or can be done multiplicar por 3 (*3), based on the number 1:

Example:

If I want to get to 13 from 1, I first multiply by 3, then add 5, then add 5 again.

findSolution(13) === (((1 * 3) + 5) + 5)

Javascript code:

function findSolution(target) {
  function find(current, history) {
    if (current == target)
      return history;
    else if (current > target)
      return null;
    else
      return find(current + 5, "(" + history + " + 5)") ||
             find(current * 3, "(" + history + " * 3)");
  }
  return find(1, "1");
}

console.log(findSolution(13));
(((1 * 3) + 5) + 5)

Iterating only using +5:

6,1
11,1
16,1
null

Iterating multiplying by 3:

3,1
9,1
27,1
null

Therefore, I would like to know: How does the return find() || find()? work, since I don't know how the function does it to decide between +5 or *3, or if it automatically generates two calls to FIND each time it makes a return, and in turn two calls by option?, something similar to the following tree:

enter image description here

javascript

2 Answers

  1. Best Answer

    The operator ||in JavaScript evaluates the expression on the left and if it is not a falsy value (null in this case) it returns it, otherwise it returns the expression on the right.

    In this case this line:

    return find(current + 5, "(" + history + " + 5)") ||
         find(current * 3, "(" + history + " * 3)");

    It would be equivalent to:

    var left = find(current + 5, "(" + history + " + 5)");
return left ? left : find(current * 3, "(" + history + " * 3)");

    Or to this:

    var left = find(current + 5, "(" + history + " + 5)");
if (left) {
    return left;
} else {
    return find(current * 3, "(" + history + " * 3)");
}

    But without creating any intermediate variables or executing the expression on the left twice

    • 3

  2. To complement @Carlos Muñoz's answer:

    These would be the recursion calls:

    find(1, "1")
  find(6, "(1 + 5)")
    find(11, "((1 + 5) + 5)")
      find(16, "(((1 + 5) + 5) + 5)")
        muy grande
      find(33, "(((1 + 5) + 5) * 3)")
        muy grande
    find(18, "((1 + 5) * 3)")
      muy grande
  find(3, "(1 * 3)")
    find(8, "((1 * 3) + 5)")
      find(13, "(((1 * 3) + 5) + 5)")
        Encontrado!
    • 2