We have a number of items which the end user will be able to organize into a desired order. The set of items is unordered, but each item contains a sort key which can be modified.

We're looking for an algorithm that would allow generating a new sort key for an item that is added or moved to be either the first item, last item, or between any two items. We're hoping to only have to modify the sort key of the item being moved.

An example algorithm would be to have each sort key be a floating point number, and when placing an item between two items, set the sort key to be their average. Placing an item first or last would take the outermost value +- 1.

The problem here is that floating point precision could cause the sorting to fail. Using two integers to represent a fractional number could similarly have the numbers become so large that they couldn't be accurately represented in regular numerical types (e.g. when transferring as JSON). We wouldn't want to use BigInts.

Is there a suitable algorithm for this that would work, for example, using strings, which wouldn't be affected by these shortcomings?

We're not looking to support huge numbers of moves, but the algorithm described above could fail on a double-precision floating points number after about 50 moves.

  • Strings are an obvious choice, because you can just keep adding characters to the end of them to bifurcate. That said, I feel like there's a better way to approach this. May 26, 2016 at 1:40
  • From the top of my head I don't see how to make it work using strings either without modifying other items' keys.
    – Sampo
    May 26, 2016 at 1:44
  • 3
    The problem you are describing is called the Order Maintenance Problem May 26, 2016 at 2:39
  • 1
    Why are you concerned with not modifying the other items in the list?
    – GER
    May 26, 2016 at 16:46
  • 1
    How you make it work with strings is like this: A, B, C -- A, AA, B, C -- A, AA, AB, B, C -- A, AA, AAA, AAB, AAC, AB, AC, B, C. Of course, you would probably want to space your letters out more so that the strings don't grow so quickly, but it can be done. May 26, 2016 at 17:04

3 Answers 3


As a summary of all comments and answers:

TL;DR - Using double-precision floating point numbers with the initially proposed algorithm should be sufficient for most practical (at least manually-ordered) needs. Maintaining a separate ordered list of the elements should be considered as well. Other sort key solutions are somewhat cumbersome.

The two problematic operations are inserting items at the beginning / end over and over again, and repeatedly inserting or moving items to the same spot (e.g. with three elements repeatedly moving the third element between the first two, or repeatedly adding new elements as the second element).

From a theoretical point of view (i.e. allowing infinite reordering), the only solution I can think of is using two unlimited size integers as a fractional a/b. This allows infinite precision for the mid-inserts, but the numbers can become increasingly large.

Strings may be able to support a large number of updates (though I'm still having trouble figuring out the algorithm for both operations), but not infinite, because you can not add infinitely many at the first position (at least using regular string sort comparison).

Integers would require choosing an initial spacing for the sort keys, which limits how many mid-inserts you can perform. If you initially space sort keys 1024 apart, you can only perform 10 worst-case mid-inserts before you have adjacent numbers. Choosing a larger initial spacing limits how many first / last inserts you can perform. Using a 64-bit integer, you are limited to ~63 operations either way, which you need to split up between mid-inserts and first/last inserts a priori.

Using floating point values removes the need for selecting spacing a priori. The algorithm is simple:

  1. The first element inserted has a sort key 0.0
  2. An element inserted (or moved) first or last has the sort key of the first element - 1.0 or last element + 1.0, respectively.
  3. An element inserted (or moved) between two element has a sort key equal to the average of the two.

Using a double-precision float allows 52 worst-case mid-inserts and effectively infinite (about 1e15) first/last inserts. In practice when moving items around the algorithm should self-correct itself, as every time you move an item first or last it expands the range that can be used.

Double-precision floats also have the benefit that they are supported by all platforms and easily stored and transported by practically all transport formats and libraries. This was what we ended up using.


I wrote up a solution in TypeScript based on @Sampo's summary. The code can be found below.

A couple of insights gained along the way.

  • Only insertion in the middle between two existing sort keys need to generate a new sort key, swapping (i.e. rearranging) does not cause splits (i.e. new midpoints). If you move two items and you only touched one of them you are losing information about what two elements changed position in the list. Even if it was a requirement to begin with, note sure it's a good idea

  • Every 1074:th midpoint split we need to normalize the floating point range. We detect this by simply checking if the new midpoint satisfy the invariant

    a.sortKey < m && m < b.sortKey

  • Scaling doesn't matter, as the sort keys are normalized, normalization still happens every 1074 midpoint split. The situation wouldn't improve if we distributed the numbers more widely to begin with.

  • Sort key normalization is incredibly rare. You will amortize this cost to the point where the normalization is not noticable. Though, I would be careful with this approach if you have more than 1000s of elements.

export interface HasSortKey {
  sortKey: number;

function normalizeList<T extends HasSortKey>(list: Array<T>) {
  const normalized = new Array<T>(list.length);
  for (let i = 0; i < list.length; i++) {
    normalized[i] = { ...list[i], sortKey: i };
  return normalized;

function insertItem<T extends HasSortKey>(
  list: Array<T>,
  index: number,
  item: Partial<T>
): Array<T> {
  if (list.length === 0) {
    list.push({ ...item, sortKey: 0 } as T);
  } else {
    // list is non-empty

    if (index === 0) {
      list.splice(0, 0, { ...item, sortKey: list[0].sortKey - 1 } as T);
    } else if (index < list.length) {
      // midpoint, index is non-zero and less than length

      const a = list[index - 1];
      const b = list[index];

      const m = (a.sortKey + b.sortKey) / 2;

      if (!(a.sortKey < m && m < b.sortKey)) {
        return insertItem(normalizeList(list), index, item);

      list.splice(index, 0, { ...item, sortKey: m } as T);
    } else if (index === list.length) {
      list.push({ ...item, sortKey: list[list.length - 1].sortKey + 1 } as T);
  return list;

export function main() {
  const normalized: Array<number> = [];

  let list: Array<{ n: number } & HasSortKey> = [];

  list = insertItem(list, 0, { n: 0 });

  for (let n = 1; n < 10 * 1000; n++) {
    const list2 = insertItem(list, 1, { n });
    if (list2 !== list) {
    list = list2;

  let m = normalized[0];

    normalized.slice(1).map(n => {
      const k = n - m;
      m = n;
      return k;

Been there, done that, may have to do that again. Use a string as the sort key, then you can always find a key that is between two given keys. If the strings get too long for your taste, you have to modify multiple or all sort keys.

  • 2
    You can't always find a key that is before another string key though.
    – Sampo
    Oct 14, 2019 at 7:13

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