What is a general algorithm to reorder a list of letters such that it minimizes the cost for a fixed list of words?

Algorithm input

To be optimized

  • a list of unique letters: [a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z]


  • a list of words: [cat, dog, apple]
    • this will be replaced with a list of my own words which uses only the letters in letters above (for example, there will not be the German ß in my words)

Algorithm output

  • a reordered list of letters such that the sum of all word cost is minimized for the input list of words
    • if there are multiple such lists, then any of them is acceptable


  • word cost: the cost of a word is the lowest integer $i$ such that the set of letters in word is a subsset of $\{letters_1, letters_2, ..., letters_{i - 1}, letters_i\}$


Text description

For example, if letters = [a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z] and my list of words were:


then the cost would be 20 (looking at a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t) + 15 (looking at a, b, c, d, e, f, g, h, i, j, k, l, m, n, o) + 16 (looking at a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p`, which totals to 51.

However, we can optimize the letters to:

letters = [p, l, e, c, a, t, d, o, g, b, f, h, i, j, k, m, n, q, r, s, u, v, w, x, y, z].

The cost would then be 6 (looking at p, l, e, c, a, t) + 9 (looking at p, l, e, c, a, t, d, o, g) + 5 (looking at p, l, e, c, a), which totals to 20.

The heuristic of sorting letters by the frequency in words appears to do well, but I cannot prove that it is optimal. This heuristic has been shown to be non-optimal by @greybeard with letters = ["g", "d", "o", "t", "c", "a", "p", "l", "e", ...] with cost of 18 using the heuristic "start with empty list. For each word, for each letter, add into the new list if it is not seen before" (this is not always optimal. Consider words = [a, b, bb] and letters = [a, b]).


Try it online! (using the "most frequent letter heuristic")

from collections import Counter

# TODO: optimize this function by returning a new list of letters such that cost is minimized
# Currently, it uses the heuristic of sorting letters by the frequency in words
def optimize_letters(letters, words):
    letter_freq = Counter()

    for word in words:
        letter_freq += Counter(word)

    optimized_ordered_letters = [letter for (letter, count) in letter_freq.most_common()]

    # add in letters that did appear in words at the back
    for letter in letters:
        if letter not in optimized_ordered_letters:

    return optimized_ordered_letters

words = ["cat", "dog", "apple"]

letters = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"]
# letters = ["p", "l", "e", "c", "a", "t", "d", "o", "g", "b", "f", "h", "i", "j", "k", "m", "n", "q", "r", "s", "u", "v", "w", "x", "y", "z"]
letters = optimize_letters(letters, words)

def word_cost(letters, word):
    word_letters = set(word)

    for i in range(len(letters)):
        if set(letters[0:i + 1]) >= word_letters:
            return i + 1

    raise Exception("letters does not contain all letters in word")

for word in words:
    cost = word_cost(letters, word)
    print(f"cost of '{word}' is {cost}")

total_cost = sum(word_cost(letters, word) for word in words)
print(f"total cost of the list of words {words} is {total_cost} for the list of letters {letters}")
cost of 'cat' is 4
cost of 'dog' is 7
cost of 'apple' is 9
total cost of the list of words ['cat', 'dog', 'apple'] is 20 for the list of letters ['a', 'p', 'c', 't', 'd', 'o', 'g', 'l', 'e', 'b', 'f', 'h', 'i', 'j', 'k', 'm', 'n', 'q', 'r', 's', 'u', 'v', 'w', 'x', 'y', 'z']

Through a brute force approach, the minimum cost for the example is 18.

  • $\begingroup$ @D.W. is it better now? do you have an example template of a general specification of the problem? $\endgroup$
    – wjwrpoyob
    Commented May 10 at 3:02
  • $\begingroup$ [one heuristic] appears to do well, but I cannot prove a valid approach, and a scientific twist to it. If you can't prove something, try proving its negation. Start "gdotcaple". $\endgroup$
    – greybeard
    Commented May 10 at 4:52
  • $\begingroup$ Looks great, thank you! $\endgroup$
    – D.W.
    Commented May 10 at 5:24
  • $\begingroup$ @greybeard I attempted to prove it is non-optimal with a counterexample, but I was not able to do that either. Thank you for your counterexample. How did you come up with it? $\endgroup$
    – wjwrpoyob
    Commented May 10 at 5:35
  • $\begingroup$ I attempted [something else without success] negative results contribute to science: Mention in your questions what you tried, successful or not! (I asked myself: What does make a difference? What doesn't?) $\endgroup$
    – greybeard
    Commented May 10 at 5:41

1 Answer 1


I don't know whether there is any efficient algorithm. One approach is to formulate this as an instance of integer linear programming and use an off-the-shelf ILP solver.

Introduce zero-or-one variables $x_{\ell,i}$, where $\ell$ is a letter and $i$ is an index, where $x_{\ell,i}$ means that letter $\ell$ appears in the $i$th position in the letters array. The requirement that letters be a permutation of all letters can be enforced by requiring $\sum_i x_{\ell,i}=1$ for all $\ell$ and $\sum_\ell x_{\ell,i}=1$ for all $i$.

Introduce integer variables $y_{\ell}$ to represent the position of letter $\ell$ in the letters array. Notice that $y_{\ell}=x_{\ell,1}+2x_{\ell,2}+3x_{\ell,3} + \cdots$.

Introduce integer variable $z_w$ to represent the cost of word $w$. We can ensure this by setting $y_\ell \le z_w$ for all letters $\ell$ in the word $w$.

Finally, the goal is to minimize $\sum_w z_w$, where the sum is taken over all words. Apply an ILP solver to this, and if it completes, you will have the optimal solution. If it takes too long, many ILP solvers allow you to terminate the search after a fixed timeout and have it output the best solution found so far.

  • $\begingroup$ I haven't used ILP solvers before, so it might take a while for me to write the ILP code and verify this answer. $\endgroup$
    – wjwrpoyob
    Commented May 10 at 6:08

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