other

A Set object, or set-like object.

true if this set has no elements in common with the other set, and false otherwise.

Two sets are disjoint if they have no elements in common. In mathematical notation:

$$A\text{ is disjoint from }B \Leftrightarrow A\cap B = \empty$$

And using Venn diagram:

isDisjointFrom() accepts set-like objects as the other parameter. It requires this to be an actual Set instance, because it directly retrieves the underlying data stored in this without invoking any user code. Then, its behavior depends on the sizes of this and other:

• If there are more elements in this than other.size, then it iterates over other by calling its keys() method, and if any element in other is present in this, it returns false (and closes the keys() iterator by calling its return() method). Otherwise, it returns true.
• Otherwise, it iterates over the elements in this, and returns false if any element e in this causes other.has(e) to return a truthy value. Otherwise, it returns true.

Because of this implementation, the efficiency of isDisjointFrom() mostly depends on the size of the smaller set between this and other (assuming sets can be accessed in sublinear time).

The set of perfect squares (<20) is disjoint from the set of prime numbers (<20), because a perfect square is by definition decomposable into the product of two integers, while 1 is also not considered a prime number:

const primes = new Set([2, 3, 5, 7, 11, 13, 17, 19]);
const squares = new Set([1, 4, 9, 16]);
console.log(primes.isDisjointFrom(squares)); // true

The set of perfect squares (<20) is not disjoint from the set of composite numbers (<20), because all non-1 perfect squares are by definition composite numbers:

const composites = new Set([4, 6, 8, 9, 10, 12, 14, 15, 16, 18]);
const squares = new Set([1, 4, 9, 16]);
console.log(composites.isDisjointFrom(squares)); // false

• Polyfill of Set.prototype.isDisjointFrom in core-js
• Set.prototype.difference()
• Set.prototype.intersection()
• Set.prototype.isSubsetOf()
• Set.prototype.isSupersetOf()
• Set.prototype.symmetricDifference()
• Set.prototype.union()