First creative solution for House Password by melpon - python coding challenges

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First solution in Creative category for House Password by melpon

f=lambda d,x:any(ord(t)&96==x for t in d)
checkio=lambda d:f(d,32)&f(d,64)&f(d,96)&(len(d)>9)

Feb. 14, 2014

Comments:

jollyca on April 23, 2014, 7:28 p.m.
I do not speak enough Python to understand this. What does it mean?

waynew on Jan. 23, 2015, 3:35 a.m.
This is a pretty clever use of bitwise manipulation, but I'm not sure that I could really explain it. To grok it, though, just break it out: what does `bin(ord('a'))` look like? `bin(96)`? & is the bitwise and.

Doppler_shan on March 18, 2016, 1:40 p.m.
I am also confused with this answer...

Snake2k on April 26, 2014, 2:37 a.m.
Hmmmm great!

jcwillian on May 24, 2014, 10:30 p.m.
good

qler89 on June 2, 2014, 8:24 a.m.
Do not like it

oduvan on June 8, 2014, 9:26 a.m.
why?

qler89 on June 13, 2014, 10:58 a.m.
Too complex for me :D

evgeniy.potseluev on May 14, 2018, 8:09 p.m.
unreadable

renanvicente on July 9, 2014, 5:11 a.m.
I have no ideia how it works :)

juanMarin on July 13, 2014, 10:37 a.m.
Nice, it took me a while to get it since I had no idea about ord

jackyl on July 15, 2014, 5:40 a.m.
I sort of get why it works, but the "&96==x" part confuses me. Can anyone help?

LorenzoPanichi on July 21, 2014, 7:24 a.m.
Nice Idea, but it looks bugged to me. It returns true with !!!!@@@{{{ (33, 64, 123). Am I wrong?

yzh07137 on Aug. 2, 2014, 4:29 a.m.
Really creative! But what if I need a more complicated match, will your code still do?

sndnyang on Sept. 6, 2014, 12:53 a.m.
I think this can be easily extend to any ASCII char. and I know little about unicode so i'm not sure.

bart.fiten on Nov. 16, 2014, 7:32 p.m.
I should spend some more time studying lambdas 'cause I have no clue how this works

EvanRavenelle on March 4, 2015, 6:23 a.m.
Seems counter-intuitive if one of Python's draws is it's readability. Maybe it's just that I can't read it (yet).

tommycarstensen on April 23, 2015, 9:08 p.m.
Cool :)

przemyslaw.daniel on July 24, 2017, 4:26 p.m.
This one is clever and can be simplified

dangyi1115 on Aug. 9, 2017, 3:52 a.m.
how it works? you check whether the password contain some element which ord(element) between 32-64, 64-96 ,96+ but how it satisfies the requirement ?

simon_wang on March 17, 2018, 12:13 a.m.
creative really, but it's a little bit hard for freshman of python.

cobbwho on March 30, 2018, 3:10 p.m.
Awesome.He cleverly use 96's high grade code(0110) and bitwise operation of '0,1' to judge if 't' is letter or number.ps:0&*=0, 1|？=1

nrsaty on April 25, 2019, 11:46 p.m.
This is not precious. Word like 'aaaabbcc[[[:::' return True in this algorythm. ':' is next character of '9', and '[' is next character of 'Z'. Obiously, this word not contain upper latin and digit. And the solution required 'The password contains only ASCII latin letters or digits.', so this answer is not fit, I think.

oduvan on April 26, 2019, 3:44 a.m.
hi, you right ‘aaaabbcc[[[:::’ returns True, but precondition of this mission says that [ and : will not be used for testing.

abrashitov on June 11, 2019, 11:04 a.m.
Test

educostadev on Sept. 6, 2019, 6:07 p.m.
<h1>This is Black Magic programming</h1>

Gugugu on Nov. 9, 2019, 8:36 p.m.
it's a pretty cool way to solve this problem. translate everything to unicode, then simplelize the phrase of compreration. thumb up. its should be the most creative one , but not so pythonic

Hey_Jack on Nov. 27, 2019, 7:45 p.m.
I would like to learn this way, but not doing this way, even couple days later, I may forget why I am doing this. :(

dalex68 on March 19, 2020, 5:25 a.m.
Cool!

Strings Theory

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