Art of Computer Programming Volume 4: Fascicle 3 has a ton of these that might fit your particular situation better than how I describe.

## Gray Codes

An issue that you will come across is of course memory and pretty quickly, you'll have problems by 20 elements in your set -- ^{20}C_{3} = 1140. And if you want to iterate over the set it's best to use a modified gray code algorithm so you aren't holding all of them in memory. These generate the next combination from the previous and avoid repetitions. There are many of these for different uses. Do we want to maximize the differences between successive combinations? minimize? et cetera.

Some of the original papers describing gray codes:

- Some Hamilton Paths and a Minimal Change Algorithm
- Adjacent Interchange Combination Generation Algorithm

Here are some other papers covering the topic:

- An Efficient Implementation of the Eades, Hickey, Read Adjacent Interchange Combination Generation Algorithm (PDF, with code in Pascal)
- Combination Generators
- Survey of Combinatorial Gray Codes (PostScript)
- An Algorithm for Gray Codes

## Chase's Twiddle (algorithm)

Phillip J Chase, `Algorithm 382: Combinations of M out of N Objects' (1970)

The algorithm in C...

## Index of Combinations in Lexicographical Order (Buckles Algorithm 515)

You can also reference a combination by its index (in lexicographical order). Realizing that the index should be some amount of change from right to left based on the index we can construct something that should recover a combination.

So, we have a set {1,2,3,4,5,6}... and we want three elements. Let's say {1,2,3} we can say that the difference between the elements is one and in order and minimal. {1,2,4} has one change and is lexicographically number 2. So the number of 'changes' in the last place accounts for one change in the lexicographical ordering. The second place, with one change {1,3,4} has one change but accounts for more change since it's in the second place (proportional to the number of elements in the original set).

The method I've described is a deconstruction, as it seems, from set to the index, we need to do the reverse – which is much trickier. This is how Buckles solves the problem. I wrote some C to compute them, with minor changes – I used the index of the sets rather than a number range to represent the set, so we are always working from 0...n.
Note:

- Since combinations are unordered, {1,3,2} = {1,2,3} --we order them to be lexicographical.
- This method has an implicit 0 to start the set for the first difference.

## Index of Combinations in Lexicographical Order (McCaffrey)

There is another way:, its concept is easier to grasp and program but it's without the optimizations of Buckles. Fortunately, it also does not produce duplicate combinations:

The set that maximizes , where .

For an example: `27 = C(6,4) + C(5,3) + C(2,2) + C(1,1)`

. So, the 27th lexicographical combination of four things is: {1,2,5,6}, those are the indexes of whatever set you want to look at. Example below (OCaml), requires `choose`

function, left to reader:

```
(* this will find the [x] combination of a [set] list when taking [k] elements *)
let combination_maccaffery set k x =
(* maximize function -- maximize a that is aCb *)
(* return largest c where c < i and choose(c,i) <= z *)
let rec maximize a b x =
if (choose a b ) <= x then a else maximize (a-1) b x
in
let rec iterate n x i = match i with
| 0 -> []
| i ->
let max = maximize n i x in
max :: iterate n (x - (choose max i)) (i-1)
in
if x < 0 then failwith "errors" else
let idxs = iterate (List.length set) x k in
List.map (List.nth set) (List.sort (-) idxs)
```

## A small and simple combinations iterator

The following two algorithms are provided for didactic purposes. They implement an iterator and (a more general) folder overall combinations.
They are as fast as possible, having the complexity O(^{n}C_{k}). The memory consumption is bound by `k`

.

We will start with the iterator, which will call a user provided function for each combination

```
let iter_combs n k f =
let rec iter v s j =
if j = k then f v
else for i = s to n - 1 do iter (i::v) (i+1) (j+1) done in
iter [] 0 0
```

A more general version will call the user provided function along with the state variable, starting from the initial state. Since we need to pass the state between different states we won't use the for-loop, but instead, use recursion,

```
let fold_combs n k f x =
let rec loop i s c x =
if i < n then
loop (i+1) s c @@
let c = i::c and s = s + 1 and i = i + 1 in
if s < k then loop i s c x else f c x
else x in
loop 0 0 [] x
```

I usually go with something like the implementation given in Josh Bloch's *fabulous* Effective Java. It's fast and creates a pretty good hash which is unlikely to cause collisions. Pick two different prime numbers, e.g. 17 and 23, and do:

```
public override int GetHashCode()
{
unchecked // Overflow is fine, just wrap
{
int hash = 17;
// Suitable nullity checks etc, of course :)
hash = hash * 23 + field1.GetHashCode();
hash = hash * 23 + field2.GetHashCode();
hash = hash * 23 + field3.GetHashCode();
return hash;
}
}
```

As noted in comments, you may find it's better to pick a large prime to multiply by instead. Apparently 486187739 is good... and although most examples I've seen with small numbers tend to use primes, there are at least similar algorithms where non-prime numbers are often used. In the not-quite-FNV example later, for example, I've used numbers which apparently work well - but the initial value isn't a prime. (The multiplication constant *is* prime though. I don't know quite how important that is.)

This is better than the common practice of `XOR`

ing hashcodes for two main reasons. Suppose we have a type with two `int`

fields:

```
XorHash(x, x) == XorHash(y, y) == 0 for all x, y
XorHash(x, y) == XorHash(y, x) for all x, y
```

By the way, the earlier algorithm is the one currently used by the C# compiler for anonymous types.

This page gives quite a few options. I think for most cases the above is "good enough" and it's incredibly easy to remember and get right. The FNV alternative is similarly simple, but uses different constants and `XOR`

instead of `ADD`

as a combining operation. It looks *something* like the code below, but the normal FNV algorithm operates on individual bytes, so this would require modifying to perform one iteration per byte, instead of per 32-bit hash value. FNV is also designed for variable lengths of data, whereas the way we're using it here is always for the same number of field values. Comments on this answer suggest that the code here doesn't actually work as well (in the sample case tested) as the addition approach above.

```
// Note: Not quite FNV!
public override int GetHashCode()
{
unchecked // Overflow is fine, just wrap
{
int hash = (int) 2166136261;
// Suitable nullity checks etc, of course :)
hash = (hash * 16777619) ^ field1.GetHashCode();
hash = (hash * 16777619) ^ field2.GetHashCode();
hash = (hash * 16777619) ^ field3.GetHashCode();
return hash;
}
}
```

Note that one thing to be aware of is that ideally you should prevent your equality-sensitive (and thus hashcode-sensitive) state from changing after adding it to a collection that depends on the hash code.

As per the documentation:

You can override GetHashCode for immutable reference types. In general, for mutable reference types, you should override GetHashCode only if:

- You can compute the hash code from fields that are not mutable; or
- You can ensure that the hash code of a mutable object does not change while the object is contained in a collection that relies on its hash code.

The link to the FNV article is broken but here is a copy in the Internet Archive: Eternally Confuzzled - The Art of Hashing

## Best Solution

There are several other answers already, but I'd like to show you the approach I took to solve it: First, let's check out how Stack Overflow handles normal cases and edge cases. Each of my pages displays 10 results, so to find out what it does for 1 page, find a tag that has less than 11 entries: usability works today. We can see nothing is displayed, which makes sense.

How about 2 pages? Find a tag that has between 11 and 20 entries (emacs works today). We see: "

12 Next" or "Prev 12", depending on which page we're on.3 pages? "

12 3 ... 3 Next", "Prev 123 Next", and "Prev 1 ... 23". Interestingly, we can see that Stack Overflow itself doesn't handle this edge case very well: it should display "12 ... 3 Next"4 pages? "

12 3 ... 4 Next", "Prev 123 ... 4 Next", "Prev 1 ... 234 Next" and "Prev 1 ... 34"Finally let's look at the general case, N pages: "

12 3 ... N Next", "Prev 123 ... N Next", "Prev 1 ... 234 ... N Next", "Prev 1 ... 345 ... N Next", etc.Let's generalize based on what we've seen: The algorithm seems to have these traits in common:

Let's ignore the edge case of a single page and make a good first attempt at the algorithm: (As has been mentioned, the code to actually print out the links would be more complicated. Imagine each place we place a page number, Prev or Next as a function call that will return the correct URL.)

This function works ok, but it doesn't take into account whether we're near the first or last page. Looking at the above examples, we only want to display the ... if the current page is two or more away.

As you can see, we have some duplication here. We can go ahead and clean that up for readibility:

There are only two problems left. First, we don't print out correctly for one page, and secondly, we'll print out "1" twice if we're on the first or last page. Let's clean those both up in one go:

Actually, I lied: We have one remaining issue. When you have at least 4 pages and are on the first or last page, you get an extra page in your display. Instead of "

12 ... 10 Next" you get "12 3 ... 10 Next". To match what's going on at Stack Overflow exactly, you'll have to check for this situation:I hope this helps!