8/12/2023 0 Comments Macports nmap![]() ![]() ![]() find all possible overlapping prefixes in a word using python.Overall you're still looking at exponential work in the worst case, but it shouldn't be too bad for shorter strings. These sorts of heuristics will probably work really well in practice. You could alternatively try to use the most constrained letters up first to decrease the branching factor. To do that, you could sort your word list from longest to shortest and try the words in that order. For example, you might want to start with larger words first to pull out as many letters as possible and keep the branching factor low. You might consider some other heuristics along the way. I'd imagine this recursion will dead-end a lot, and that you'll probably find all your answers without too much effort. You can use that to filter out words that can't be added in any more, and this essentially saves you the cost of having to check the whole dictionary each time. To do this, I'd use a backtracking recursion where at each point you maintain a histogram of the remaining letter counts. Once you have the unordered lists, you can generate the permutations from them pretty quickly. I'd begin by finding unordered lists of words that anagram to the target rather than all possible ordered lists of words, because there's many fewer of them to find. Once you have the set of words you can legally use in the anagram, from there you're left with the problem of finding which combinations of them can be used to form a complete anagram of the sentence. Your dictionary is likely going to be pretty colossal (150,000+, I'd suspect), so rescanning it after each decision point is going to be completely infeasible. I like your idea of filtering the word list down to just the words that could possibly be made with the input letters, and I like the idea of trying to string them together, but I think there are a few major optimizations you could put into place that would likely speed things up quite a bit.įor starters, rather than choosing a word and then rescanning the entire dictionary for what's left, I'd consider just doing a single filtering pass at the start to find all possible words that could be made with the letters that you have. django-pytest setup_method database issue.Validate and get the user using the jwt token inside a view or consumer.Connect Django to remote mysql server on a local network.Django ForeignKey issue related class not found.Dynamically create image thumbnails (using django).django m2m_changed with custom through model.'WSGIRequest' object has no attribute 'Post'.Find rows with a given difference between values in a column.How to pair rows with the same value in one column of a dataframe in R.How to select from dataframe when column is optional.How to make an array from for loop globally available in Julia?.Why can't statsmodels reproduce my R logistic regression results?.Convert five-year data to annual data and calculate new records in R. ![]()
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