Created
March 25, 2014 23:40
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Basic hash creation methods for a given dictionary.
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| #Hash function which converts hash_me("leep", dictionary) => 13427273 | |
| def hash_me(text, dictionary) | |
| h = 7 | |
| dictionary = "acdegilmnoprstuw" | |
| text.size.times do |i| | |
| h = (h * 37) + dictionary.index(text[i]).to_i | |
| end | |
| h | |
| end | |
| def find_position(number, dictionary) | |
| dictionary.split("").each_with_index do |l, i| | |
| div, mod = (number - i).divmod(37) | |
| if mod == 0 | |
| return div, l | |
| break | |
| end | |
| end | |
| ["", 0] | |
| end | |
| #Unhash function which converts unhash_me(13427273, dictionary) => "leep | |
| def unhash_me(number, dictionary) | |
| h = 7 | |
| text = "" | |
| while number > h | |
| number, letter = find_position(number, dictionary) | |
| text << letter | |
| end | |
| text.reverse | |
| end | |
| #Usage: set the dictionary and the number to be decoded | |
| dictionary = "acdegilmnoprstuw" | |
| number = 910897038977002 | |
| #Call the unhash fuction | |
| text = unhash_me(number, dictionary) | |
| #Print the victory mark if everything went ok. | |
| puts "Sucess!!" if number == hash_me(text, dictionary) |
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I Like your solution, and I've translated to Java as a fork (here https://gist.github.com/MuhammadHewedy/11380342)