Created
May 4, 2011 20:51
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Google Nexus S Challenge Puzzle #6
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| """ | |
| My answer was: "1 vowel; 3-5 digits; letters in ascending order, but sequence | |
| can restart after first 2 if rest of letters are unique. #googlenexus #pattern" | |
| Though it may seem a bit iffy, this is the most specific pattern I could find | |
| that would validate all strings marked as "yes" and invalidate all strings | |
| marked as "no". After 5 hours, I'm not prepared to explore this puzzle any | |
| further. | |
| The following code verifies that my solution holds true. | |
| """ | |
| from string import ascii_lowercase, digits | |
| yes = """ | |
| adgjl24680 | |
| ABBBB12345 | |
| adgjl24680 | |
| asd12nx123 | |
| ASDFG24680 | |
| acdfg13579 | |
| acdfg35719 | |
| bcdef13579 | |
| 13579acdfg | |
| ghijk57913 | |
| defgh88129 | |
| ABDFG24680 | |
| 13bcd579ef | |
| a1b3c5d7f9 | |
| abxyz24688 | |
| abcdz13579 | |
| rstuv12345 | |
| mnopq24688 | |
| 13acd579fg | |
| a1b2b3b4b5 | |
| abbb1b3579 | |
| BBBBE54321 | |
| aglmn24688 | |
| ablmn24688 | |
| abbbbbb123 | |
| acccccc120 | |
| effffff120 | |
| affffff135 | |
| abcdgh2345 | |
| ABBBBB1234 | |
| abbbbb1223 | |
| """ | |
| no = """ | |
| b873ahfj11 | |
| QWANG16235 | |
| AHTPG60342 | |
| omgwtf1234 | |
| e11111cccc | |
| asd2ds1234 | |
| a2c3n5g8g9 | |
| shryo167ly | |
| BBBBA54321 | |
| oblmn24688 | |
| eblmn24688 | |
| abbbbbbb20 | |
| 29dknkfpqa | |
| ab12345678 | |
| ABC0123456 | |
| abbb112358 | |
| ablm246888 | |
| samp012345 | |
| BBBA123456 | |
| a1p357975p | |
| a111111111 | |
| googl63987 | |
| n3xu5n3xu5 | |
| 12345abcde | |
| a2c4e6g8i0 | |
| Abcde13579 | |
| acdhi13579 | |
| cdfgh02468 | |
| GHJMK18576 | |
| jklmn39751 | |
| CCCCC12345 | |
| a1b2c3d4e5 | |
| ZY8X7W6V5S | |
| v1llainrom | |
| hdjs63yh5d | |
| qsxt293xpm | |
| googlegogo | |
| 57d3hjs351 | |
| sbnn295pwd | |
| gnannwxyzh | |
| gmflcokvn4 | |
| AAAAAAAAAA | |
| nexuss235i | |
| a1b2c3d4e5 | |
| dfg2jklrd5 | |
| ayawanapat | |
| bbbbbbbbbb | |
| n3xvsr0cks | |
| 1abcdefgha | |
| fdmmm619vb | |
| sdbbn769wm | |
| cddbtn538p | |
| abcde11379 | |
| cgwmt70vv3 | |
| eghkljmnbh | |
| gtttn234tg | |
| ABC0123456 | |
| abbb112358 | |
| dffffff120 | |
| bcccccc120 | |
| almnu76498 | |
| b359gjmpwz | |
| hw30lrll0d | |
| 2468824688 | |
| c1c2c3c4c5 | |
| icecream71 | |
| msnvm90231 | |
| w4ntn3x0ss | |
| 1BBBBC2345 | |
| """ | |
| yes = yes.lower().strip().split('\n') | |
| no = no.lower().strip().split('\n') | |
| def evaluate(s): | |
| # Check if only one vowel | |
| if len([c for c in s if c in 'aeiou']) != 1: | |
| return False | |
| # Check if between 3 and 5 digits | |
| if len([c for c in s if c in digits]) not in range(3,6): | |
| return False | |
| # Extract the character codes of the Latin letters in the string | |
| letters = [ord(c) for c in s if c in ascii_lowercase] | |
| # Get the differences in position in the alphabet for each sequential | |
| # pair of letters | |
| diffs = [] | |
| for i in range(len(letters) - 1): | |
| diffs.append(letters[i+1] - letters[i]) | |
| # First check if all letters sequential | |
| if all(diff >= 0 for diff in diffs): | |
| return True | |
| # If not, check if the first 2 are sequential, and no duplicates occur | |
| # in the subsequent letters | |
| else: | |
| return (diffs[0] >= 0 and diffs[1] < 0 and all(diff > 0 for diff in diffs[2:])) | |
| assert all(map(evaluate, yes)), 'FAIL: All "yes" strings must evaluate to True!' | |
| assert not any(map(evaluate, no)), 'FAIL: All "no" strings must evaluate to False!' | |
| print 'SUCCESS: Evaluate function is correct for the data set!' |
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