## § Ranking and Sorting

We we want to sort an arrray xs and write the results into an array ys. In both cases, the invariant to be satisfied is that ys is xs in ascending order. I'll be considering two broad ways to implement such a thing:
• 1. (RANK) Index straight into ys, reindexed into xs. We name the reindexingarray as rs (rs for ranks).
for(int i = 0; i < N; ++i) {
ys[rs[i]] = xs[i]
}

• 2. (SELECT) Reindex into ys, index straight into xs. We name the reindexing array ss (ss for selects).
for(int i = 0; i < N; ++i) {
ys[i] = xs[ss[i]]
}


#### § Defnition of rank

In the case of (RANK), the specification is that the rank of an element e is the number of elements that are:
1. less than e.
2. equal to e but occur before e.
This ensures that rank is a permutation: that is, every element e is given a unique index.
int rank(const int *xs, int i) {
int cnt = 0;
for (int j = 0; j < N; ++j) {
if (xs[j] < xs[i] || (xs[j] == xs[i] && j < i)) cnt += 1;
}
return cnt;
}

for(int i = 0; i < N; ++i) {
rs[i] = rank(xs, i);
}


#### § Rank: Alternative 1

An alternative way to look at our definition of rank is that we are sorting the tuples (xs[i], i) using lex ordering. So if two indeces i, j have the same value, then we sort on the index.

#### § Rank: Alternative 2

We could have also defined rank as:
int rank(int *xs, int i) {
int cnt = 0;
for (int j = 0; j < i; ++j) {
if (xs[j] <= xs[i]) cnt += 1
}
return cnt;
}

I declined from doing so because I wanted to show the lex-ordering interpretation of rank will be be useful later.

#### § Definition of select

For (SELECT), we'll need to think a little harder. Let's try to rewrite our way to glory:
• 1. From definition of rank:
ys[rs[i]] = xs[i]

• 2. move i -> ss[i] where ss[i] is guaranteed to be a permutation,so we will write each index eventually:
ys[rs[ss[i]]] = xs[ss[i]]

• 3. Stipulate that rs[ss[i]] = i, since we want a ys[i]:
ys[i] = xs[ss[i]]

This gives us necessary and sufficient conditions on how to find an ss: ss must be a permutation that is an inverse permutation to rs.

#### § How to invert a permutation?

How does one invert a permutation? We have a permutation rs[i] that maps i to rs[i]. We want to find a new permutation ss[i] such that
rs[ss[i]] = i

Equivalently:
// ss[i] is an index 'k'...
ss[i] = k
// 'k' is the location of 'i' in rs.
rs[k] = i

So if we first tag each location of rs with its index, and then sort with the values of rs being the keys, then ss[i] will be the values. In pictures:
rs[i]          [3 4 0 1 2]
rs_tagged[i]   [(3, 0) (4, 1) (0, 2) (1, 3) (2, 4)]
rs_sorted[i]   [(0, 2) (1, 3) (2, 4) (3, 0) (4, 1)]
ss[i]          [    2      3      4      0      1 ]
rs[ss[i]]      [ rs[2]  rs[3]  rs[4]  rs[0]  rs[1]]
rs[ss[i]]      [    0      1      2      3      4 ]


#### § Rank and Select are inverses

It's nice, there are a couple of miracles that lined up here:
• Rank starts with r and select starts with s so we get nice naming.
• Rank and select are true inverse permutations of each other.

#### § Generalized rank and select are adjoint

We had to "fix" our definition of rank to avoid equal elements in the array. Hence, we had the rule that we also sort by indexes if the elements are equal. However, if we now decide to ignore this rule, we will then recreate the classical adjunction between rank and select.

#### § References

• Richard Bird: Pearls of functional algorithm design