Mathematical notation for Hash Tables

A matrix is nice for modelling arrays, eg

const A = [1, 2, 3, 4]
const a1 = A[0]
const a2 = A[1]
const a3 = A[2]
const a4 = A[3]


can be modelled with $$A = \begin{bmatrix} 1\\ 2\\ 3\\ 4 \end{bmatrix}$$

$$a1 = A_1\\ a2 = A_2\\ a3 = A_3\\ a4 = A_4$$

I am curious if there is some similar construct to this for modelling hash maps.

const A = {
"one": 1,
"two": 2,
"three": 3,
"four": 4
}
const A_one = A["one"]
const A_two = A["two"]
const A_three = A["three"]
const A_four = A["four"]


in a such a way that

$$a1 = A_{"one"} \\ a2 = A_{"two"} \\ a3 = A_{"three"}\\ a4 = A_{"four"}$$

I was considering augmenting a matrix, but it's not standard as far as I know.

$$\left[ \begin{array}{c|c} "one"&1\\ "two"&2\\ "three"&3\\ "four"&4 \end{array} \right]$$

I know that hash tables are not matrices; as matrices are ordered, hash tables are unordered, and matrices have a bunch of properties/operations that do not map directly 1:1 to hash tables.

I know that one way a hash tables can be constructed is given a bunch of key, value pairs a input

eg.

$$\left[ \begin{array}{c@{}c@{}} \left[ \begin{array}{cc} "one"\\ 1 \end{array} \right] \\ \left[ \begin{array}{cc} "two"\\ 2 \end{array} \right] \\ \left[ \begin{array}{cc} "three"\\ 3 \end{array} \right] \\ \left[ \begin{array}{cc} "four"\\ 4 \end{array} \right] \end{array} \right]$$

But an associator function to make this a proper map is still needed.

Surely there must be some existing way of representing hash tables/"spare" arrays without inventing my own notation for them, as maps are more natural in mathematics than matrices, as a matrix can be represented as an ordered map, but a maps can't be represented by matrices because the matrix construct cannot "skip" rows, and cannot have noninteger indexing.

Note:

It seems that unordered map can be represented by discontinuous functions, and ordered maps can be represented as a product of a function and an ordered key vector; eg

const f = x =>
x == "one" ? 1 :
x == "two" ? 2 :
x == "three" ? 3 :
x == "four" ? 4
: undefined
;

const keys = ["one", "two", "three", "four"];
const m = keys.map(x => [x, f(x)]);

// [ [ 'one', 1 ], [ 'two', 2 ], [ 'three', 3 ], [ 'four', 4 ] ]
console.log(m);


The above is effectively the same as

const f = {
"one": 1,
"two": 2,
"three": 3,
"four": 4
};
const m = Object.keys(f).map(x => [x, f[x]]);

// [ [ 'one', 1 ], [ 'two', 2 ], [ 'three', 3 ], [ 'four', 4 ] ]
console.log(m);


and they can be composed naturally to add entries

'use strict';

const UnorderedMap = (f, domain) => {
let res;
res = x => res[x];
domain.forEach(x => res[x] = f(x));
res['domain'] = domain;
res.composeWith = (f2, domain2) => {
var domainSet = new Set();

return UnorderedMap((x => {
return res(x) !== undefined ? res(x) : f2(x);
}), domainSet);
};

return res;
};

let myMap = UnorderedMap((x => x === "zero" ? 0 : undefined), new Set(["zero"]));
myMap = myMap.composeWith((x => x === "one" ? 1 : undefined), new Set(["one"]));
myMap = myMap.composeWith((x => x === "two" ? 2 : undefined), new Set(["two"]));
myMap = myMap.composeWith((x => x === "three" ? 3 : undefined), new Set(["three"]));

console.log(myMap("zero")); // 0
console.log(myMap["zero"]); // 0
console.log(myMap("one")); // 1
console.log(myMap["one"]); // 1
console.log(myMap("two")); // 2
console.log(myMap["two"]); // 2
console.log(myMap("three")); // 3
console.log(myMap["three"]); // 3
console.log(myMap["domain"]); // Set { 'zero', 'one', 'two', 'three' }

• What about a set of pairs? As in $\{(\text{"one"}, 1),\ldots\}$, which is a binary relation. – Fabio Somenzi Dec 30 '16 at 19:57
• it still only represents the set of entries(coproduct of products), but does not actually contsruct an indexer I can use. eg if I say let A = {("one",1),…} I have done existential quantification for A to the set, but a set has no indexer, so I can't do A_one, I must create a function f that performs the lookup, and it can no longer be named A_one since A is already quantified as a set construct, and set does not support indexing, another construct is needed to behave both like such a set AND have a lookup function. – Dmitry Dec 30 '16 at 20:08
• @Dmitry How do you think of a map or function mathematically? How do you think of a matrix mathematically? Can you give formal definitions? – k.stm Dec 30 '16 at 20:09
• basically, once i assign A to a Set, I am forbidden from indexing it since Sets do not support indexing, as sets are unordered. Matrices do allow indexing but only via integers. To act like hashes, a new construct that is neither a Set nor a Matrix, but acts like a Set with that does support indexing. A Matrix is essentially an Ordered Map as long as its keys are always ascending and have no gaps, otherwise they become a Sparse Array, which again has no representation as a Matrix nor a Set. – Dmitry Dec 30 '16 at 20:10
• Your proposal is basically a coproduct(union) of multiple objects in product of string and integer sets, which can be a basis for an unordered map if an associator for the constructor of such an object is defined(that assigns to each index indicates by the first value of the product the 2nd value of the right of the product). If it was an array of pairs, it would be ok for a basis for an ordered map, which does preserve order of entries, given a similar associator for the actual constructor of such objects. – Dmitry Dec 30 '16 at 20:19

If $V$ is a set of values and $K$ is a set of keys, then a hash is just a surjective map $h : V \to K$. If $k \in K$ is a key, then the subset $h^{-1} (k)$ is made up of all the values associated with the key $k$.

Notice that if $h$ is injective, then (since it is assumed surjective) it is bijective, therefore invertible, which defeats the practical purpose of a hash. Therefore, a hash function must be "as non-injective as practically possible". Notice, though, that a delicate balance must be attained, because if a hash function is "too non-injective", then it produces too many collisions (if $h(k) = h(k')$ the it is said that $k$ and $k'$ collide under $h$).

In computer science one usually takes $V$ and $K$ to be finite, and the cardinal of $K$ to be significantly smaller than the one of $V$. One also introduces some topology on $V$ and $K$ (such as the order topology) and requires $h$ to be "very discontinuous", in order to make approximation of zeros difficult (i.e. finding approximate solutions to the equation $f(v) = k$), i.e. making the hash function difficult to break.

• So this seems to agree with the idea that unordered maps are the same as discontinuous functions(since hashes in the programming sense never exhaustively map the domain; they could(eg map from byte domain is fairly cheap) but such hashes tend to be redefined in terms of functions), and ordered maps are just a product of an unordered map and an ordering vector with an implicitly deduced map operation which maps the unordered map based on its' ordering vector. – Dmitry Dec 30 '16 at 21:27