how can we find the nth 3 word combination from a word corpus of 3000 words - combinations

Question

I have a word corpus of say 3000 words such as [hello, who, this ..].
I want to find the nth 3 word combination from this corpus.I am fine with any order as long as the algorithm gives consistent output.
What would be the time complexity of the algorithm.
I have seen this answer but was looking for something simple.

Answer 1

(Note that I will be using 1-based indexes and ranks throughout this answer.)
To generate all combinations of 3 elements from a list of n elements, we'd take all elements from 1 to n-2 as the first element, then for each of these we'd take all elements after the first element up to n-1 as the second element, then for each of these we'd take all elements after the second element up to n as the third element. This gives us a fixed order, and a direct relation between the rank and a specific combination.
If we take element i as the first element, there are (n-i choose 2) possibilities for the second and third element, and thus (n-i choose 2) combinations with i as the first element. If we then take element j as the second element, there are (n-j choose 1) = n-j possibilities for the third element, and thus n-j combinations with i and j as the first two elements.
Linear search in tables of binomial coefficients
With tables of these binomial coefficients, we can quickly find a specific combination, given its rank. Let's look at a simplified example with a list of 10 elements; these are the number of combinations with element i as the first element:
i
1 C(9,2) = 36
2 C(8,2) = 28
3 C(7,2) = 21
4 C(6,2) = 15
5 C(5,2) = 10
6 C(4,2) = 6
7 C(3,2) = 3
8 C(2,2) = 1
---
120 = C(10,3)
And these are the number of combinations with element j as the second element:
j
2 C(8,1) = 8
3 C(7,1) = 7
4 C(6,1) = 6
5 C(5,1) = 5
6 C(4,1) = 4
7 C(3,1) = 3
8 C(2,1) = 2
9 C(1,1) = 1
So if we're looking for the combination with e.g. rank 96, we look at the number of combinations for each choice of first element i, until we find which group of combinations the combination ranked 96 is in:
i
1 36 96 > 36 96 - 36 = 60
2 28 60 > 28 60 - 28 = 32
3 21 32 > 21 32 - 21 = 11
4 15 11 <= 15
So we know that the first element i is 4, and that within the 15 combinations with i=4, we're looking for the eleventh combination. Now we look at the number of combinations for each choice of second element j, starting after 4:
j
5 5 11 > 5 11 - 5 = 6
6 4 6 > 4 6 - 4 = 2
7 3 2 <= 3
So we know that the second element j is 7, and that the third element is the second combination with j=7, which is k=9. So the combination with rank 96 contains the elements 4, 7 and 9.
Binary search in tables of running total of binomial coefficients
Instead of creating a table of the binomial coefficients and then performing a linear search, it is of course more efficient to create a table of the running total of the binomial coefficient, and then perform a binary search on it. This will improve the time complexity from O(N) to O(logN); in the case of N=3000, the two look-ups can be done in log2(3000) = 12 steps.
So we'd store:
i
1 36
2 64
3 85
4 100
5 110
6 116
7 119
8 120
and:
j
2 8
3 15
4 21
5 26
6 30
7 33
8 35
9 36
Note that when finding j in the second table, you have to subtract the sum corresponding with i from the sums. Let's walk through the example of rank 96 and combination [4,7,9] again; we find the first value that is greater than or equal to the rank:
3 85 96 > 85
4 100 96 <= 100
So we know that i=4; we then subtract the previous sum next to i-1, to get:
96 - 85 = 11
Now we look at the table for j, but we start after j=4, and subtract the sum corresponding to 4, which is 21, from the sums. then again, we find the first value that is greater than or equal to the rank we're looking for (which is now 11):
6 30 - 21 = 9 11 > 9
7 33 - 21 = 12 11 <= 12
So we know that j=7; we subtract the previous sum corresponding to j-1, to get:
11 - 9 = 2
So we know that the second element j is 7, and that the third element is the second combination with j=7, which is k=9. So the combination with rank 96 contains the elements 4, 7 and 9.
Hard-coding the look-up tables
It is of course unnecessary to generate these look-up tables again every time we want to perform a look-up. We only need to generate them once, and then hard-code them into the rank-to-combination algorithm; this should take only 2998 * 64-bit + 2998 * 32-bit = 35kB of space, and make the algorithm incredibly fast.
Inverse algorithm
The inverse algorithm, to find the rank given a combination of elements [i,j,k] then means:
Finding the index of the elements in the list; if the list is sorted (e.g. words sorted alphabetically) this can be done with a binary search in O(logN).
Find the sum in the table for i that corresponds with i-1.
Add to that the sum in the table for j that corresponds with j-1, minus the sum that corresponds with i.
Add to that k-j.
Let's look again at the same example with the combination of elements [4,7,9]:
i=4 -> table_i[3] = 85
j=7 -> table_j[6] - table_j[4] = 30 - 21 = 9
k=9 -> k-j = 2
rank = 85 + 9 + 2 = 96
Look-up tables for N=3000
This snippet generates the look-up table with the running total of the binomial coefficients for i = 1 to 2998:
function C(n, k) { // binomial coefficient (Pascal's triangle)
if (k < 0 || k > n) return 0;
if (k > n - k) k = n - k;
if (! C.t) C.t = [[1]];
while (C.t.length <= n) {
C.t.push([1]);
var l = C.t.length - 1;
for (var i = 1; i < l / 2; i++)
C.t[l].push(C.t[l - 1][i - 1] + C.t[l - 1][i]);
if (l % 2 == 0)
C.t[l].push(2 * C.t[l - 1][(l - 2) / 2]);
}
return C.t[n][k];
}
for (var total = 0, x = 2999; x > 1; x--) {
total += C(x, 2);
document.write(total + ", ");
}
This snippet generates the look-up table with the running total of the binomial coefficients for j = 2 to 2999:
for (var total = 0, x = 2998; x > 0; x--) {
total += x;
document.write(total + ", ");
}
Code example
Here's a quick code example, unfortunately without the full hardcoded look-up tables, because of the size restriction on answers on SO. Run the snippets above and paste the results into the arrays iTable and jTable (after the leading zeros) to get the faster version with hard-coded look-up tables.
function combinationToRank(i, j, k) {
return iTable[i - 1] + jTable[j - 1] - jTable[i] + k - j;
}
function rankToCombination(rank) {
var i = binarySearch(iTable, rank, 1);
rank -= iTable[i - 1];
rank += jTable[i];
var j = binarySearch(jTable, rank, i + 1);
rank -= jTable[j - 1];
var k = j + rank;
return [i, j, k];
function binarySearch(array, value, first) {
var last = array.length - 1;
while (first < last - 1) {
var middle = Math.floor((last + first) / 2);
if (value > array[middle]) first = middle;
else last = middle;
}
return (value <= array[first]) ? first : last;
}
}
var iTable = [0]; // append look-up table values here
var jTable = [0, 0]; // and here
// remove this part when using hard-coded look-up tables
function C(n,k){if(k<0||k>n)return 0;if(k>n-k)k=n-k;if(!C.t)C.t=[[1]];while(C.t.length<=n){C.t.push([1]);var l=C.t.length-1;for(var i=1;i<l/2;i++)C.t[l].push(C.t[l-1][i-1]+C.t[l-1][i]);if(l%2==0)C.t[l].push(2*C.t[l-1][(l-2)/2])}return C.t[n][k]}
for (var iTotal = 0, jTotal = 0, x = 2999; x > 1; x--) {
iTable.push(iTotal += C(x, 2));
jTable.push(jTotal += x - 1);
}
document.write(combinationToRank(500, 1500, 2500) + "<br>");
document.write(rankToCombination(1893333750) + "<br>");