[Tutor] Rounding to n significant digits? (original) (raw)

Gregor Lingl glingl at aon.at
Sun Jul 4 13:43:57 EDT 2004

Alan Gauld schrieb:

Hi Gregor,

def engroundton(x,n): man,exp = roundton(x, n).split("e") # contains error-handling sh=int(exp)%3 return "%.*fe%+04i" % (n-1-sh,float(man)*10**sh,int(exp)-sh)

Eek! Are you sure that last line is Python and not Perl? Winner of this months obfuscated tutor code I think... Hi Alan!

What a nice reply! You are certainly right ... nevertheless I've been laughing about it for at least 5 minutes . (Although: it's only the 4th of July today, who knows what will follow ...)

Of course I'd be interested in a more elegant solution ... Unfortunately I cannot work on it right now, because it's only one hour to the beginning of the finale of then European football chamionship. I'm going to visit friends to watch it in television ...

Last week at the Linuxtag in Germany I participated (by accident) in a so called "coding contest". (Of course I didn't win!) We had to write a program to solve a certain problem with the sourcecode as short as possible (!? - strange task!).

Later after some correspondence about it we arrived at:

import sys for i in range(3): r='' for c in sys.argv[1]: n=int('2l7y202nn6lx6xh0ve',36)//8**(3int(c)+i)%8;r+=' |_ _ _| |__|'[2n:2*n+3]_ print r

You may guess what it does. Input has to be an arbitrary sequence of digits from 0 to 9, ( a so called integer) eg.: 1237583490 ... ... or try it out. The winning program had less than 140 bytes, as far as I know it was in Ruby.(Didn't see it)

Regards, Gregor

P.S.: Which code will win the obfuscated code prize now?

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