Posts by Moshi
HTML, 25 bytes <style>*{background:#000} Not really sure if this is a legal solution, but technically you can open it in your browser...
Sclipting, (UTF-16) 26 bytes 갠긂갠밂乘감뒄뀢감雙갿및剩 Uses a formula I found here unsigned char b; // reverse this (8-bit) byte b = (b * 0x0202020202ULL & 0x010884422010ULL) % 1023;
Python, 89 bytes lambda l,g=lambda c:ord(c).bit_count():[i for i in l if min(map(g,l))<g(i)<max(map(g,l))] Attempt This Online!
Sclipting, (UTF-16) 22 bytes 놵멡뉖땯더굅낔뭕뇐虛移 Nothing complicated, just replaces the regex [^aeiouAEIOU] with the empty string.
Try it online doesn't support the latest JavaScript features such as optional chaining or null coalescence, which are very useful character-saving operators. Is there another site like TIO, but whi...
Sclipting, (UTF-16) 86 bytes 標갠 관上가①要❶❹剩걉눑감⓶重右갰雙⓷加⓶❸分終⓷棄終并❶訂乾⓶折❷同終長괐縮⒈棄丟 Explanation: Input n 標 갠 관上 for b from 2 to 16 가 s = 0 ① n' = n 要 while n' is not 0 ❶❹剩 x = n...
Python 3, 325 322 bytes -3 bytes thanks to steffan153! from base64 import* n=int.from_bytes(b85decode('8C^#Z9Uor=^}*nk`yLA}A<z=aTm_V+z!I{Y1+v7z89f>WmC>M4TN(sPp`fvySOeqafPpzk0!*=hK2{h9D...
JavaScript (Node.js), 38 32 bytes -6 bytes thanks to Shaggy! a=>a.map(v=>d[v]=-~d[v],d={})&&d Try it online! Basically just this but returning the result
Challenge It's a bootstrapping challenge this time! Write a full program that, once run, writes the source code of another program that in turn, once run, writes the source code for another progra...
JavaScript (Node.js), 31 bytes for(i='';;console.log(i+='*')); Try it online!
Python 3, 5 unique bytes, 28 total (()==())/((()==())+(()==())) (, ), =, /, + Uses (()==()) to get True with only 3 unique characters, as well as making the later grouping of (...+...) free. ...
Japt, 6 bytes -1 byte thanks to Shaggy! k ä¦ e Try it First time doing Japt so this is probably pretty bad. Just factorizes and checks that there are no duplicate factors.
JavaScript (Node.js), 160 bytes (r,d,n,l=r.length,s=d.length,p=(n,i)=>n?"01110110"[4*p(--n,i-1)+2*p(n,i)+1*p(n,i+1)]:i>=0&i<s?d[i]:r[(i%l+l)%l])=>[...Array(4*l+s+2*n)].map((_,i)=...
JavaScript (Node.js), 101 82 bytes -19 bytes thanks to Shaggy! x=>x.replace(/(..)\1{0,62}/g,(c,g)=>c>'c'||c[5]?(192+c.length/2).toString(16)+g:c) Try it online! Everything can be sol...
Python 3, 399 bytes def f(n): if n<1:return'zero' t=1000;m=t*t;r='r fif six seven eigh nine ';b=f'one two three four five six seven eight nine ten eleven twelve thir fou{r}twen thir fo{r}'....
Python 3, 191 bytes lambda p:[c for c in chain(*[C(p,r)for r in range(len(p)+1)])if all(all(i==(sum(p.values())/2>=sum(p[n]for n in d))for d in C(c,len(c)-i))for i in[0,1])] from itertools imp...
Sclipting, (UTF-16) 454 406 bytes Yes. 塊匱❸곴김分倘標⓶❹演긆륩닆롩닶밎併鈉不終⓷곴김剩❶갾밈分倘⓷標⓷❺演긇끨닷녳눖멤併鈉⓶不終갾밈剩❶뉀分倘⓷標⓷❺演긆둵닦끲뉖밄併鈉⓶不終뉀剩⓶글會⓶倘❷長是긆굮뉂밀⓶終긇끨늗긠뉦뭵댢걦늖눠댶땸긇꽥덦녮긆녩뉶될닦땮뉐❷굀瀰是❶銅긂건덶녮融壹坼❸겠分掘덇밉⓸겠剩倘껐⓶⓹逆⓸終終긆뭮뉒건덶묠덆둲뉖넠뉦뭵댢걦늗뉥긇꽩...
Use the unpacking * instead of list If you want to convert an iterator/generator into a list, use the * operator instead of using the list function, e.g. Instead of list(iterable) Use [*iter...
Sclipting, (UTF-16) 58 52 bytes 가匱❸增平갠下氫終要鈮貶⓶梴감⓶上❸乘殲終終并鈮掘增 Gives the nth ludic number, 0-indexed. 가 Start list of Ludics with 0 匱❹增平갠下氫終 Create a list of numbers from 2 to n+1 squared 要 W...
Python 3, 46 bytes lambda a,x:sum(c*x**i for i,c in enumerate(a)) Try it online!
Python 2, 58 54 bytes lambda l:''.join(filter(None,sum(map(None,'',*l),()))) Try it online! Amazingly, shorter than my Python 3 answer! Uses this method to zip unequal-length lists and this me...
Python 3, 72 69 bytes lambda l:''.join(sum(zip(*[[*i]+['']*max(map(len,l))for i in l]),())) Try it online! Uses this method to flat zip, with modifications to pad the shorter lists.
Sclipting, (UTF-16) 44 34 32 bytes 貶要❶갠剩❷隔❸增갰乘嗎終并長貶 Because comparing with 1 is expensive (requires copying and decrementing), we instead use a modified version of the Collatz sequence - namely...
