mburke05

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mburke05

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mburke05

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mburke05

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mburke05

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mburke05

import unittest

test_string = ["spare", "hello", "pears", "world", "reaps"]

def find_anagrams(strings, word):
    """Takes a list of strings as an argument and returns a list of strings that are anagrams of
    the provided word like so:
    find_anagrams(["spare", "hello", "pears", "world", "reaps"], "parse") = ["spare", "pears", "reaps"]
    """
    return [string for string in strings if sorted(string) == sorted(word)]

mburke05

import unittest

NUMB = str(731671765313306249192251196744265747423553491949349698352031277450632623957831801698480186947885184385861560789112949495459501737958331952853208805511125406987471585238630507156932909632952274430435576689664895044524452316173185640309871112172238311362229893423380308135336276614282806444486645238749303589072962904915604407723907138105158593079608667017242712188399879790879227492190169972088809377665727333001053367881220235421809751254540594752243525849077116705560136048395864467063244157221553975369781797784617406495514929086256932197846862248283972241375657056057490261407972968652414535100474821663704844031998900088952434506585412275886668811642717147992444292823086346567481391912316282458617866458359124566529476545682848912883142607690042242190226710556263211111093705442175069416589604080719840385096245544436298123098787992724428490918884580156166097919133875499200524063689912560717606058861164671094050775410022569831552000559357297257163626956188267042825248360082325753042075296

mburke05

from math import floor, sqrt

def find_next_prime(n):
     """ We use is_prime passed through a while loop to check if a given value (starting at n+1) is_prime
     and exit if we've found our number. I think this too can be accomplished potentially using itertools to 
     greater effect and would love to hear criticism. I know that while True conditionals are generally frowned upon
     in python."""
    i = n+1
    while True:
        if is_prime(i):

mburke05

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mburke05

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