I need an efficient way of calculating the minimum edit distance between two unordered collections of symbols. Like in the Levenshtein distance, which only works for sequences, I require insertions, deletions, and substitutions with different per-symbol costs. I'm also interested in recovering the edit script.
Since what I'm trying to accomplish is very similar to calculating string edit distance, I figured it might be called unordered string edit distance or maybe just set edit distance. However, Google doesn't turn up anything with those search terms, so I'm interested to learn if the problem is known by another name?
To clarify, the problem would be solved by
def unordered_edit_distance(target, source):
return min(edit_distance(target, source_perm)
for source_perm in permuations(source))
So for instance, the unordered_edit_distance('abc', 'cba') would be 0, whereas edit_distance('abc', 'cba') is 2. Unfortunately, the number of permutations grows large very quickly and is not practical even for moderately sized inputs.
EDIT Make it clearer that operations are associated with different costs.
Sort them (not necessary), then remove items which are same (and in equal numbers!) in both sets.
Then if the sets are equal in size, you need that numer of substitutions; if one is greater, then you also need some insertions or deletions. Anyway you need the number of operations equal the size of the greater set remaining after the first phase.
Although your observation is kind of correct, but you are actually make a simple problem more complex.
Since source can be any permutation of the original source, you first need check the difference in character level.
Have two map each map count the number of individual characters in your target and source string:
for example:
a: 2
c: 1
d: 100
Now compare two map, if you missing any character of course you need to insert it, and if you have extra character you delete it. Thats it.
Let's ignore substitutions for a moment.
Now it becomes a fairly trivial problem of determining the elements only in the first set (which would count as deletions) and those only in the second set (which would count as insertions). This can easily be done by either:
Sorting the sets and iterating through both at the same time, or
Inserting each element from the first set into a hash table, then removing each element from the second set from the hash table, with each element not found being an insertion and each element remaining in the hash table after we're done being a deletion
Now, to include substitutions, all that remains is finding the optimal pairing of inserted elements to deleted elements. This is actually the stable marriage problem:
The stable marriage problem (SMP) is the problem of finding a stable matching between two sets of elements given a set of preferences for each element. A matching is a mapping from the elements of one set to the elements of the other set. A matching is stable whenever it is not the case that both:
Some given element A of the first matched set prefers some given element B of the second matched set over the element to which A is already matched, and
B also prefers A over the element to which B is already matched
Which can be solved with the Gale-Shapley algorithm:
The Gale–Shapley algorithm involves a number of "rounds" (or "iterations"). In the first round, first a) each unengaged man proposes to the woman he prefers most, and then b) each woman replies "maybe" to her suitor she most prefers and "no" to all other suitors. She is then provisionally "engaged" to the suitor she most prefers so far, and that suitor is likewise provisionally engaged to her. In each subsequent round, first a) each unengaged man proposes to the most-preferred woman to whom he has not yet proposed (regardless of whether the woman is already engaged), and then b) each woman replies "maybe" to her suitor she most prefers (whether her existing provisional partner or someone else) and rejects the rest (again, perhaps including her current provisional partner). The provisional nature of engagements preserves the right of an already-engaged woman to "trade up" (and, in the process, to "jilt" her until-then partner).
We just need to get the cost correct. To pair an insertion and deletion, making it a substitution, we'll lose both the cost of the insertion and the deletion, and gain the cost of the substitution, so the net cost of the pairing would be substitutionCost - insertionCost - deletionCost.
Now the above algorithm guarantees that all insertion or deletions gets paired - we don't necessarily want this, but there's an easy fix - just create a bunch of "stay-as-is" elements (on both the insertion and deletion side) - any insertion or deletion paired with a "stay-as-is" element would have a cost of 0 and would result in it remaining an insertion or deletion and nothing would happen for two "stay-as-is" elements ending up paired.
Key observation: you are only concerned with how many 'a's, 'b's, ..., 'z's or other alphabet characters are in your strings, since you can reorder all the characters in each string.
So, the problem boils down to the following: having s['a'] characters 'a', s['b'] characters 'b', ..., s['z'] characters 'z', transform them into t['a'] characters 'a', t['b'] characters 'b', ..., t['z'] characters 'z'. If your alphabet is short, s[] and t[] can be arrays; generally, they are mappings from the alphabet to integers, like map <char, int> in C++, dict in Python, etc.
Now, for each character c, you know s[c] and t[c]. If s[c] > t[c], you must remove s[c] - t[c] characters c from the first unordered string (s). If s[c] < t[c], you must add t[c] - s[c] characters c to the second unordered string (t).
Take X, the sum of s[c] - t[c] for all c such that s[c] > t[c], and you will get the number of characters you have to remove from s in total. Take Y, the sum of t[c] - s[c] for all c such that s[c] < t[c], and you will get the number of characters you have to remove from t in total.
Now, let Z = min (X, Y). We can have Z substitutions, and what's left is X - Z insertions and Y - Z deletions. Thus the total number of operations is Z + (X - Z) + (Y - Z), or X + Y - min (X, Y).
Related
Imagine there is a list of elements as follow:
1a, 2a, 3a, 4a, 5b, 6b, 7b, 8b
Now we need to randomize it such that not more than 2 "a"s or 2 "b"s get next to each other. For instance the following list is not allowed because of the 2nd, third and fourth elements:
3a, 7b, 8b, 5b, 2a, 1a, 5b, 4a
How can we write write an efficient code without generating many random sequences and many triad comparisons?
Create two bins, one for the a's and one for the b's. Pick from a random bin and record the bin. Pick a second number from a random bin. If the bin is not the same as before just record the bin. If the bin is the same as before then force the next pick to be from the other bin. Carry on forward, only forcing a bin when you have two picks in succession from the same bin.
I'm going to assume that:
There are only two kinds of element, a and b, and
There aren't "too many" of either kind (say, less than 30) or that you're willing to use a bignum package.
The basic idea is to (conceptually) first construct a valid sequence of as and bs, and then randomly assign the actual elements to the as and bs in the sequence. In practice, you could do both of these steps in parallel; every time you add an a to the sequence, you select a random a element from the set of such elements not yet assigned, and similarly with b elements.
The (slightly) complicated part is constructing the valid sequence without bias, and that's what I'm going to focus on.
As is often the case, the key is to be able to count the number of possible sequences, in a way which leads to an enumeration. We don't actually enumerate the possibilities -- that would take really a long time for even moderately long sequences -- but we do need to know for every prefix how to enumerate the sequences starting with that prefix.
Rather than produce the sequence element by element, we'll produce it in chunks of one or two elements of the same kind. Since we don't allow more than two consecutive elements of the same kind, the final sequence must be a series of alternating chunks. In effect, at every point except the very beginning, the choice is whether to select one or two of the "other" kind. At the beginning, we must select one or two of either kind, so we must first choose the starting kind, after which all the kinds are fixed; we merely need a sequence of 1's and 2's -- representing one element or two elements of the same kind -- with the kind alternating at each step. The sequence of 1s and 2s is constrained by the fact that we know how many elements there are of each kind, which corresponds to the sum of the numbers in the even and odd positions of the {1,2}-sequence.
Now, let's define f(m,n) as the count of sequences whose even and odd sums are m and n. (Using CS rather than maths rules, we'll assume that the first position is 0 (even) but it actually makes absolutely no difference.) Suppose that we have 6 as and 4 bs. There are then f(6,4) sequences which start with an a, and f(4,6) sequences which start with a b, so that the total count of valid sequences is f(6,4)+f(4,6).
Now, suppose we need to compute f(m,n). Assuming m is large enough, we have exactly two options: choose one of the m elements of the even kind or choose two of the m elements of the even kind. After that, we will swap even and odd because the next choice applies to the other kind.
That rather directly leads to the recursion
f(m, n) = f(n, m-1) + f(n, m-2)
which we might think of as a kind of two-dimensional fibonacci recursion. (Recall that fib(m) = fib(m-1) + fib(m-2); the difference here is the second argument, and the fact that the argument order flip-flops at each recursion.
As with Fibonacci numbers, computing the values naively without memoization leads to exponential blow-up of recursive calls, and a more efficient strategy is to compute the entire table starting from f(0,0) (which has the value 1, obviously); in essence, a dynamic programming approach. We could also just do the recursive computation with memoization, which is slightly less efficient but possibly easier to read.
For now, let's just assume that we've arranged for the computation of f(m,n) to be suitably fast, either because we've prebuilt the entire array of possibilities up to the largest values of m and n we will need, or because we're using a memoizing recursive solution so that we only need to do the slow computation once for any given m,n. Now let's construct the random sequence.
Suppose there are na a-elements and nb b-elements. Since we don't know whether the random sequence will start with an a or a b, we need to first make that decision. We know there are f(na,nb) valid sequences which start a and f(nb,na) valid sequences starting with a b, so we start by generating a random non-negative integer less than f(na,nb) + f(nb,na). If the random is less than f(na,nb) then we'll start with a-elements; otherwise we'll start with b elements.
Having made that decision, we'll proceed as follows. We know what the next element kind is and how many elements remain of each kind, so we only need to know whether to select one or two elements of the correct kind. To make that choice, we generate a non-negative random integer less than f(m, n); if it is less than f(n, m-1) then we select one element; otherwise we select two elements. Then we swap the element sets, fix the counts, and continue until m and n are both 0.
Given a dictionary as a hashtable. Find the minimum # of
deletions needed for a given word in order to make it match any word in the
dictionary.
Is there some clever trick to solve this problem in less than exponential complexity (trying all possible combinations)?
For starters, suppose that you have a single word w in the the hash table and that your word is x. You can delete letters from x to form w if and only if w is a subsequence of x, and in that case the number of letters you need to delete from x to form w is given by |x - w|. So certainly one option would be to just iterate over the hash table and, for each word, to see if x is a subsequence of that word, taking the best match you find across the table.
To analyze the runtime of this operation, let's suppose that there are n total words in your hash table and that their total length is L. Then the runtime of this operation is O(L), since you'll process each character across all the words at most once. The complexity of your initial approach is O(|x| · 2|x|) because there are 2|x| possible words you can make by deleting letters from x and you'll spend O(|x|) time processing each one. Depending on the size of your dictionary and the size of your word, one algorithm might be better than the other, but we can say that the runtime is O(min{L, |x|·2|x|) if you take the better of the two approaches.
You can build a trie and then see where your given word fits into it. The difference in the depth of your word and the closest existing parent is the number of deletions required.
I was asked a question
You are given a list of characters, a score associated with each character and a dictionary of valid words ( say normal English dictionary ). you have to form a word out of the character list such that the score is maximum and the word is valid.
I could think of a solution involving a trie made out of dictionary and backtracking with available characters, but could not formulate properly. Does anyone know the correct approach or come up with one?
First iterate over your letters and count how many times do you have each of the characters in the English alphabet. Store this in a static, say a char array of size 26 where first cell corresponds to a second to b and so on. Name this original array cnt. Now iterate over all words and for each word form a similar array of size 26. For each of the cells in this array check if you have at least as many occurrences in cnt. If that is the case, you can write the word otherwise you can't. If you can write the word you compute its score and maximize the score in a helper variable.
This approach will have linear complexity and this is also the best asymptotic complexity you can possibly have(after all the input you're given is of linear size).
Inspired by Programmer Person's answer (initially I thought that approach was O(n!) so I discarded it). It needs O(nr of words) setup and then O(2^(chars in query)) for each question. This is exponential, but in Scrabble you only have 7 letter tiles at a time; so you need to check only 128 possibilities!
First observation is that the order of characters in query or word doesn't matter, so you want to process your list into a set of bag of chars. A way to do that is to 'sort' the word so "bac", "cab" become "abc".
Now you take your query, and iterate all possible answers. All variants of keep/discard for each letter. It's easier to see in binary form: 1111 to keep all, 1110 to discard the last letter...
Then check if each possibility exists in your dictionary (hash map for simplicity), then return the one with the maximum score.
import nltk
from string import ascii_lowercase
from itertools import product
scores = {c:s for s, c in enumerate(ascii_lowercase)}
sanitize = lambda w: "".join(c for c in w.lower() if c in scores)
anagram = lambda w: "".join(sorted(w))
anagrams = {anagram(sanitize(w)):w for w in nltk.corpus.words.words()}
while True:
query = input("What do you have?")
if not query: break
# make it look like our preprocessed word list
query = anagram(sanitize(query))
results = {}
# all variants for our query
for mask in product((True, False), repeat=len(query)):
# get the variant given the mask
masked = "".join(c for i, c in enumerate(query) if mask[i])
# check if it's valid
if masked in anagrams:
# score it, also getting the word back would be nice
results[sum(scores[c] for c in masked)] = anagrams[masked]
print(*max(results.items()))
Build a lookup trie of just the sorted-anagram of each word of the dictionary. This is a one time cost.
By sorted anagram I mean: if the word is eat you represent it as aet. It the word is tea, you represent it as aet, bubble is represent as bbbelu etc
Since this is scrabble, assuming you have 8 tiles (say you want to use one from the board), you will need to maximum check 2^8 possibilities.
For any subset of the tiles from the set of 8, you sort the tiles, and lookup in the anagram trie.
There are at most 2^8 such subsets, and this could potentially be optimized (in case of repeating tiles) by doing a more clever subset generation.
If this is a more general problem, where 2^{number of tiles} could be much higher than the total number of anagram-words in the dictionary, it might be better to use frequency counts as in Ivaylo's answer, and the lookups potentially can be optimized using multi-dimensional range queries. (In this case 26 dimensions!)
Sorry, this might not help you as-is (I presume you are trying to do some exercise and have constraints), but I hope this will help the future readers who don't have those constraints.
If the number of dictionary entries is relatively small (up to a few million) you can use brute force: For each word, create a 32 bit mask. Preprocess the data: Set one bit if the letter a/b/c/.../z is used. For the six most common English characters etaoin set another bit if the letter is used twice.
Create a similar bitmap for the letters that you have. Then scan the dictionary for words where all bits that are needed for the word are set in the bitmap for the available letters. You have reduced the problem to words where you have all needed characters once, and the six most common characters twice if the are needed twice. You'll still have to check if a word can be formed in case you have a word like "bubble" and the first test only tells you that you have letters b,u,l,e but not necessarily 3 b's.
By also sorting the list of words by point values before doing the check, the first hit is the best one. This has another advantage: You can count the points that you have, and don't bother checking words with more points. For example, bubble has 12 points. If you have only 11 points, then there is no need to check this word at all (have a small table with the indexes of the first word with any given number of points).
To improve anagrams: In the table, only store different bitmasks with equal number of points (so we would have entries for bubble and blue because they have different point values, but not for team and mate). Then store all the possible words, possibly more than one, for each bit mask and check them all. This should reduce the number of bit masks to check.
Here is a brute force approach in python, using an english dictionary containing 58,109 words. This approach is actually quite fast timing at about .3 seconds on each run.
from random import shuffle
from string import ascii_lowercase
import time
def getValue(word):
return sum(map( lambda x: key[x], word))
if __name__ == '__main__':
v = range(26)
shuffle(v)
key = dict(zip(list(ascii_lowercase), v))
with open("/Users/james_gaddis/PycharmProjects/Unpack Sentance/hard/words.txt", 'r') as f:
wordDict = f.read().splitlines()
f.close()
valued = map(lambda x: (getValue(x), x), wordDict)
print max(valued)
Here is the dictionary I used, with one hyphenated entry removed for convenience.
Can we assume that the dictionary is fixed and the score are fixed and that only the letters available will change (as in scrabble) ? Otherwise, I think there is no better than looking up each word of the dictionnary as previously suggested.
So let's assume that we are in this setting. Pick an order < that respects the costs of letters. For instance Q > Z > J > X > K > .. > A >E >I .. > U.
Replace your dictionary D with a dictionary D' made of the anagrams of the words of D with letters ordered by the previous order (so the word buzz is mapped to zzbu, for instance), and also removing duplicates and words of length > 8 if you have at most 8 letters in your game.
Then construct a trie with the words of D' where the children nodes are ordered by the value of their letters (so the first child of the root would be Q, the second Z, .., the last child one U). On each node of the trie, also store the maximal value of a word going through this node.
Given a set of available characters, you can explore the trie in a depth first manner, going from left to right, and keeping in memory the current best value found. Only explore branches whose node's value is larger than you current best value. This way, you will explore only a few branches after the first ones (for instance, if you have a Z in your game, exploring any branch that start with a one point letter as A is discarded, because it will score at most 8x1 which is less than the value of Z). I bet that you will explore only a very few branches each time.
An autogram is a sentence which describes the characters it contains, usually enumerating each letter of the alphabet, but possibly also the punctuation it contains. Here is the example given in the wiki page.
This sentence employs two a’s, two c’s, two d’s, twenty-eight e’s, five f’s, three g’s, eight h’s, eleven i’s, three l’s, two m’s, thirteen n’s, nine o’s, two p’s, five r’s, twenty-five s’s, twenty-three t’s, six v’s, ten w’s, two x’s, five y’s, and one z.
Coming up with one is hard, because you don't know how many letters it contains until you finish the sentence. Which is what prompts me to ask: is it possible to write an algorithm which could create an autogram? For example, a given parameter would be the start of the sentence as an input e.g. "This sentence employs", and assuming that it uses the same format as the above "x a's, ... y z's".
I'm not asking for you to actually write an algorithm, although by all means I'd love to see if you know one to exist or want to try and write one; rather I'm curious as to whether the problem is computable in the first place.
You are asking two different questions.
"is it possible to write an algorithm which could create an autogram?"
There are algorithms to find autograms. As far as I know, they use randomization, which means that such an algorithm might find a solution for a given start text, but if it doesn't find one, then this doesn't mean that there isn't one. This takes us to the second question.
"I'm curious as to whether the problem is computable in the first place."
Computable would mean that there is an algorithm which for a given start text either outputs a solution, or states that there isn't one. The above-mentioned algorithms can't do that, and an exhaustive search is not workable. Therefore I'd say that this problem is not computable. However, this is rather of academic interest. In practice, the randomized algorithms work well enough.
Let's assume for the moment that all counts are less than or equal to some maximum M, with M < 100. As mentioned in the OP's link, this means that we only need to decide counts for the 16 letters that appear in these number words, as counts for the other 10 letters are already determined by the specified prefix text and can't change.
One property that I think is worth exploiting is the fact that, if we take some (possibly incorrect) solution and rearrange the number-words in it, then the total letter counts don't change. IOW, if we ignore the letters spent "naming themselves" (e.g. the c in two c's) then the total letter counts only depend on the multiset of number-words that are actually present in the sentence. What that means is that instead of having to consider all possible ways of assigning one of M number-words to each of the 16 letters, we can enumerate just the (much smaller) set of all multisets of number-words of size 16 or less, having elements taken from the ground set of number-words of size M, and for each multiset, look to see whether we can fit the 16 letters to its elements in a way that uses each multiset element exactly once.
Note that a multiset of numbers can be uniquely represented as a nondecreasing list of numbers, and this makes them easy to enumerate.
What does it mean for a letter to "fit" a multiset? Suppose we have a multiset W of number-words; this determines total letter counts for each of the 16 letters (for each letter, just sum the counts of that letter across all the number-words in W; also add a count of 1 for the letter "S" for each number-word besides "one", to account for the pluralisation). Call these letter counts f["A"] for the frequency of "A", etc. Pretend we have a function etoi() that operates like C's atoi(), but returns the numeric value of a number-word. (This is just conceptual; of course in practice we would always generate the number-word from the integer value (which we would keep around), and never the other way around.) Then a letter x fits a particular number-word w in W if and only if f[x] + 1 = etoi(w), since writing the letter x itself into the sentence will increase its frequency by 1, thereby making the two sides of the equation equal.
This does not yet address the fact that if more than one letter fits a number-word, only one of them can be assigned it. But it turns out that it is easy to determine whether a given multiset W of number-words, represented as a nondecreasing list of integers, simultaneously fits any set of letters:
Calculate the total letter frequencies f[] that W implies.
Sort these frequencies.
Skip past any zero-frequency letters. Suppose there were k of these.
For each remaining letter, check whether its frequency is equal to one less than the numeric value of the number-word in the corresponding position. I.e. check that f[k] + 1 == etoi(W[0]), f[k+1] + 1 == etoi(W[1]), etc.
If and only if all these frequencies agree, we have a winner!
The above approach is naive in that it assumes that we choose words to put in the multiset from a size M ground set. For M > 20 there is a lot of structure in this set that can be exploited, at the cost of slightly complicating the algorithm. In particular, instead of enumerating straight multisets of this ground set of all allowed numbers, it would be much better to enumerate multisets of {"one", "two", ..., "nineteen", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"}, and then allow the "fit detection" step to combine the number-words for multiples of 10 with the single-digit number-words.
Suppose there is given two String:
String s1= "MARTHA"
String s2= "MARHTA"
here we exchange positions of T and H. I am interested to write code which counts how many changes are necessary to transform from one String to another String.
There are several edit distance algorithms, the given Wikipeida link has links to a few.
Assuming that the distance counts only swaps, here is an idea based on permutations, that runs in linear time.
The first step of the algorithm is ensuring that the two strings are really equivalent in their character contents. This can be done in linear time using a hash table (or a fixed array that covers all the alphabet). If they are not, then s2 can't be considered a permutation of s1, and the "swap count" is irrelevant.
The second step counts the minimum number of swaps required to transform s2 to s1. This can be done by inspecting the permutation p that corresponds to the transformation from s1 to s2. For example, if s1="abcde" and s2="badce", then p=(2,1,4,3,5), meaning that position 1 contains element #2, position 2 contains element #1, etc. This permutation can be broke up into permutation cycles in linear time. The cycles in the example are (2,1) (4,3) and (5). The minimum swap count is the total count of the swaps required per cycle. A cycle of length k requires k-1 swaps in order to "fix it". Therefore, The number of swaps is N-C, where N is the string length and C is the number of cycles. In our example, the result is 2 (swap 1,2 and then 3,4).
Now, there are two problems here, and I think I'm too tired to solve them right now :)
1) My solution assumes that no character is repeated, which is not always the case. Some adjustment is needed to calculate the swap count correctly.
2) My formula #MinSwaps=N-C needs a proof... I didn't find it in the web.
Your problem is not so easy, since before counting the swaps you need to ensure that every swap reduces the "distance" (in equality) between these two strings. Then actually you look for the count but you should look for the smallest count (or at least I suppose), otherwise there exists infinite ways to swap a string to obtain another one.
You should first check which charaters are already in place, then for every character that is not look if there is a couple that can be swapped so that the next distance between strings is reduced. Then iterate over until you finish the process.
If you don't want to effectively do it but just count the number of swaps use a bit array in which you have 1 for every well-placed character and 0 otherwise. You will finish when every bit is 1.