Posts by celtschk
If you have any tips for golfing in Python, add them as answers to this post.
Given a non-negative integer up to $999\\,999\\,999$, write it in English. The input number can be in any form other than English (though you'll typically want to use the native integer type of yo...
Unlambda, 1 solution Since Unlambda code that does something always contains the backtick character, there cannot be more than one solution. ```.2.5.6i
C (gcc), 232 bytes f(n,w,p){printf("%i bottle%s of beer%s%s",n,"s"+(n<2)," on the wall"+w,p);}main(i){for(i=99;i;){f(i,0,", ");f(i,12,".\n");printf(i>1?"Take one down and pass it around, "...
Given two strings, I define their product as follows: If any of the two strings is empty, the product is the empty string. If the second string consists of a single character, the result ...
C (gcc), 114 bytes f(int n,int*p,int*i,int**o){if(n){int*m=i,k;for(;*m!=*p;++m);k=m-i;f(k,p+1,i,o);f(n-k-1,p+k+1,m+1,o);*(*o)++=*p;}} Try it online! Arguments: n is the length of the arra...
On my challenge Looping Counter there are several entries like this which contains a function which needs to be called with the right parameters to give the correct output. In this particular case,...
Python, 89 Bytes import math as m def f(*n):return m.gcd(*n)==1and all(m.gcd(a,b)!=1for a in n for b in n) Note: The outdated Python interpreter at tio.run won't run this code (nor the ungolfe...
Your task is to implements the product of polynomials modulo 2 $(\mathbb F_2[x]$) in integer representation. Background The field $\mathbb F_2$ represents the integers modulo 2, or equivalently, ...
C (gcc), 53 bytes This uses the shortcut behaviour of logical or (||) to only multiply if it is a divisor; the loop end condition then makes sure it's a proper divisor. i=1;p=1;f(n){for(;i<n...
Python 3, 35 bytes lambda m:list(zip(*m[::-1]))[::-1] Try it online! The format is a list of tuples. The content of the tuples could be any type; in my tests I used single-digit strings bec...
C (gcc), 117 bytes b,r;f(char*s){char*p=s;for(;*p;p++)*p-=47+7*(*p>57),b=b<*p?*p:b;if(b<2)return p-s;for(;*s;s++)r*=b,r+=*s-1;return r;} Try it online!
Python 3, 74 bytes def f(a): d={x:0 for x in a};r=[] for x in a:d[x]+=1;r+=[d[x]] return r Try it online!
Python 3, 59 bytes lambda n,m:sum(choices(range(m),k=n))+n from random import* Try it online!
Lundin just suggested in a comment under another question that I tag the challenge Looping Counter as kolmogorov-complexity. Now I'm not sure if it actually qualifies for that tag, for the followi...
SOS, 155 Bytes !+!-!!+!-!!!!+!!-!!+!-!+!-!+!!-!+!!-!!!+!!-!+!!-!!!+!!-!+!!!!-!!+!-!+!!-!!!!+!-!!!!!!+!-!+!-!+!!!-!+!!-!+!!!!-!+!!!-!!+!-!!+!!-!+!!-!!!+!!-!!+!-!!+!-!!!!+! Explanation: Each ! o...
Given a string s of text and a line length l, your task is to word-wrap that text to that line length. Concretely, the given string consists of words separated by single spaces, with no leading or...
C (gcc), 103 bytes Using a different approach than my previous solution, therefore posting as new answer as suggested here. i;main(){while(i++<100){char s[]="FizzBuzz",*t=s+4*!!(i%3);if(i%5)...
Python 3, 75 bytes lambda a:[[i,k]for i in r(l(a))for k in r(l(a[0]))if a[i][k]] r=range;l=len Try it online!
Unlambda, 40 bytes `````````````.H.e.l.l.o.,. .W.o.r.l.d.!i Try it online!
Python 3, 91 90 bytes Saved one byte thanks to Mark Giraffe in the comments lambda l:"".join(["".join(x)for x in zip_longest(*l,fillvalue='')]) from itertools import* Try it online!
C (gcc), 40 bytes i;f(n){for(i=1;n>0;n-=i,i+=2);return!n;} Try it online!
Python 3, 64 62 bytes for i in range(1,101):print("Fizz"*(i%3<1)+"Buzz"*(i%5<1)or i) Try it online! Saved two bytes thanks to Moshi in the comments.
C (gcc), 87 bytes P1: i;main(){scanf("%d",&i);printf("i;main(){scanf(\"%%d\",&i);printf(\"%%d\",%d+i);}",i);} Try it online! Generated P2 for input 10: i;main(){scanf("%d",&i...
C (gcc), 50 bytes f(int(*o)(i,j),int n){return n-1?o(f(o,n-1),n):1;} Try it online!
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