Dial-a-Word
by N1xis10t (N1xis10t@protonmail.ch)
Abstract
This software is for finding words in phone numbers (like 888-EYES) that you see government agencies and other using. Now, I'm pretty sure that they just pay the telephone companies extra so they get special phone numbers. The idea behind this software is to provide the user with the ability to acquire numbers like that without paying extra.
Functionality
When you are setting upa phone line with the telephone company, what they typically do is give you an option to choose a randomly selected phone number, or have them randomly select another one. When they present you with a number, feed it into this program with no dashes or special characters (like xxx5243373 where "xxx" is an area code) and it will search the number to see if it has any words in it. If it does, it will show you what the number is in the format "xxx5-CHEESE." If it doesn't find anything, or if you don't like what it did find, have them show you another number and then check it. Repeat this until you find one that you like, or until the technician gets fed up with you.
Dependencies
This software requires that the file google-10000-english.txt be in the same folder. This is a file that has the 10,000 most common English words in it, and can easily be found on the Internet. I have limited this program to such a small dictionary because the running time is simply too long to be practical when you try to use an exhaustive list of English words. Feel free to try to use a better dictionary.
Notes
This software cannot find words that are less than three characters long, as a side effect of having to filter out garbage words from the dictionary. This could be fixed by removing the filter and using a better dictionary.
This software is only 6.5 kilobytes in size (not including the dictionary), and would easily fit on a 5¼" inch disk floppy.
Copyright Notice
Permission is hereby granted, free of charge, to any person obtaining a copy of this software and associated documentation files (the "Software"), to deal in the Software without restriction, including without limitation the rights to use, copy, modify, merge, publish, distribute, sublicense, and/or sell copies of the Software, and to permit persons to whom the Software is furnished to do so, subject to the following conditions:
The above copyright notice and this permission notice shall be included in all copies or substantial portions of the Software.
THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
#!/usr/bin/python3 # -*- coding: utf-8 -*- # Import the dictionary dicti = open("./google-10000-english.txt") Dictionary = dicti.read().replace("\n", " ").split() DictLen = len(Dictionary) dicti.close() # Define what letters the numbers can be letters = { '1': '1', '2': 'A B C', '3': 'D E F', '4': 'G H I', '5': 'J K L', '6': 'M N O', '7': 'P Q R S', '8': 'T U V', '9': 'W X Y Z', '0': '0' } number = "" # Garbage word filter i = 1 while i <= len(Dictionary) - 1: if len(Dictionary[i]) < 3: del Dictionary[i] i -= 1 i += 1 # Enter the main loop print("Type 'exit' to quit.") while True: printnums = "\n" number = input("Number: ") if number == "exit": #break sys.exit() modnumber = "" printnumber = "" # Start iterating through all the possible letter combinations for the number for i in range(0, len(letters[number[0]].split())): modnumber += letters[number[0]].split()[i] for i in range(0, len(letters[number[1]].split())): modnumber += letters[number[1]].split()[i] for i in range(0, len(letters[number[2]].split())): modnumber += letters[number[2]].split()[i] for i in range(0, len(letters[number[3]].split())): modnumber += letters[number[3]].split()[i] for i in range(0, len(letters[number[4]].split())): modnumber += letters[number[4]].split()[i] for i in range(0, len(letters[number[5]].split())): modnumber += letters[number[5]].split()[i] for i in range(0, len(letters[number[6]].split())): modnumber += letters[number[6]].split()[i] for i in range(0, len(letters[number[7]].split())): modnumber += letters[number[7]].split()[i] for i in range(0, len(letters[number[8]].split())): modnumber += letters[number[8]].split()[i] for i in range(0, len(letters[number[9]].split())): modnumber += letters[number[9]].split()[i] # Once the current iteration has been assembled, check to see if it has any English words in it for i in range(0, len(Dictionary)): if Dictionary[i].upper() in modnumber: # If it does, and it's unique, add it to the list of numbers to be printed out word = Dictionary[i].upper() wordlocation = modnumber.find(word) printnumber = ("^" + number[0:wordlocation] + "-" + word + "-" + number[wordlocation + len(word):] + "^" ).replaced("^-", "").replace("-^", "").replace("^", "") if not printnumber in printnums: printnums += printnumber + "\n" # Take away the last character so a new one can be tried modnumber = modnumber[0:9] modnumber = modnumber[0:8] modnumber = modnumber[0:7] modnumber = modnumber[0:6] modnumber = modnumber[0:5] modnumber = modnumber[0:4] modnumber = modnumber[0:3] modnumber = modnumber[0:2] modnumber = modnumber[0:1] modnumber = '' # Print the results if printnums == "\n": print("\nNo words found\n") else: print(printnums) # Copyright (c) 05/09/2022 N1xis10tCode: dial-a-word.py