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Activity for Arpad Horvath‭

Type On... Excerpt Status Date
Edit Post #294525 Post edited:
using >>= (like a bind operator)
12 months ago
Edit Post #294525 Post edited:
I need to revert += solution as it gives reserved solution
12 months ago
Edit Post #294525 Post edited:
courtesy
about 1 year ago
Edit Post #294525 Post edited:
-2 bytes
about 1 year ago
Edit Post #290206 Post edited:
Some fixes
about 1 year ago
Edit Post #290206 Post edited:
Make the first line a header
about 1 year ago
Edit Post #290206 Post edited:
Make the first line a header
about 1 year ago
Edit Post #294525 Post edited:
string concatenation instead of list
about 1 year ago
Edit Post #294525 Post edited:
about 1 year ago
Edit Post #294525 Initial revision about 1 year ago
Answer A: Convert to Hexadecimal
[Python 3.8 (pre-release)], 70 bytes -1 bytes using default arg & -1 bytes using >>= (thanks trichoplax) def h(n,a=''): while n>0:a="0123456789ABCDEF"[n%16]+a;n>>=4 return a Try it online!
(more)
about 1 year ago
Comment Post #293336 I am not familiar with TIO. (And also with APL, I've used Q/KDB+ earlier, and they are somewhat similar.) If I remember correctly assigning the function to a name is not part of the code so I've added the assignment (f←) in the footer. If that counts too, than it is 2 characters longer.
(more)
over 1 year ago
Edit Post #293336 Post edited:
over 1 year ago
Edit Post #293336 Initial revision over 1 year ago
Answer A: Expand a greyscale/colour hex code
[APL (Dyalog Classic)], 31 bytes '#',{4=≢⍵:⍵[2 2 3 3 4 4]⋄6⍴1↓⍵} Try it online!
(more)
over 1 year ago
Comment Post #292945 with lambda function it is little bit shorter Instead of def f(n,d):return (19 chars with the space at the end) You can write: lambda n, d: (11 chars) As far as I know you don't need to assign it to a name, so it is a solution.
(more)
over 1 year ago
Edit Post #293328 Post edited:
Testing
over 1 year ago
Edit Post #293328 Initial revision over 1 year ago
Answer A: Expand a greyscale/colour hex code
Q/KDB+, 39 bytes {"#",?[4=count x;x 1 1 2 2 3 3;6#1x]} Testing the function: q)cc:{"#",?[4=count x;x 1 1 2 2 3 3;6#1x]} q)cc "#ABCDEF" "#ABCDEF" q)cc "#A" "#AAAAAA" q)cc "#AB" "#ABABAB" q)cc "#ABC" "#AABBCC" q) is the prompt, other lin...
(more)
over 1 year ago
Edit Post #291390 Initial revision over 2 years ago
Answer A: Round trip stones
[Python], 104 bytes ```python f=lambda n:"1.0000000.7272730.7285710.7368420.7437890.7491640.7533440.7566570.7593360.761542"[8n-8:8n] ``` Try it online! Not too clever. I am curious whether it can be beaten with a Python code that really calculates the probability and prints the re...
(more)
over 2 years ago
Edit Post #291320 Post edited:
version
over 2 years ago
Edit Post #291320 Post edited:
just python
over 2 years ago
Edit Post #291320 Post edited:
over 2 years ago
Comment Post #291320 There are 8 kind of characters, " IXLCDM". 2 characters are 6 bits that can be easily mapped to printable characters (after 32 code point). This string is much shorter then the "DD DM IC ..." string itself. Even if I add the extraction it is still somewhat shorter. I just regenerate the " IXLC...
(more)
over 2 years ago
Edit Post #291324 Initial revision over 2 years ago
Answer A: The 50 substrings that validate any string of Roman numerals
[Haskell], 271 bytes import Data.Char import Data.List n=" IVXLCDM" r x=maximum[if(s`isInfixOf`x)then s else"A"|s(n!!div x 8):[(n!!(x-8(div x 8)))])$map(\x->ord x-32)"&PW!H.!@/$HF$@G\"H6\"@7\"03#P?%N%O%U%]%^%&N&O!!+!1!=!9!<!:!;$=$<\"\"+#M#N#L#O#K#C#=#<%MH))\'X;;"] Try...
(more)
over 2 years ago
Edit Post #291320 Post edited:
even shorter 220
over 2 years ago
Comment Post #291320 Sorry for making that much work for you. My bad. I don't know exactly what did I do then but does't work for me now. But fortunately the current one is just 1 bytes longer. It works if I use square brackets inside te max function, but then it is longer then the current one.
(more)
over 2 years ago
Edit Post #291320 Post edited:
Fixing the code
over 2 years ago
Edit Post #291320 Post edited:
3.9
over 2 years ago
Edit Post #291320 Post edited:
over 2 years ago
Edit Post #291320 Post edited:
shorter
over 2 years ago
Edit Post #291320 Post edited:
over 2 years ago
Edit Post #291320 Initial revision over 2 years ago
Answer A: The 50 substrings that validate any string of Roman numerals
[Python], 220 bytes Works with Python 3.8 or newer. n=" IVXLCDM" x=("".join([n[(d:=ord(c)-32)//8]+n[d-8(d//8)]for c in'&PW!H.!@/$HF$@G"H6"@7"03#P?%N%O%U%]%^%&N&O!!+!1!=!9! r=lambda c:max(s if s in c else"A"for s in'DD DM IC ID IL IM LC LD LL LM VC VD VL VM VV VX XD XM CC...
(more)
over 2 years ago
Edit Post #291299 Post edited:
small fix
over 2 years ago
Edit Post #291299 Post edited:
remove test
over 2 years ago
Edit Post #291299 Post edited:
remove duplicate
over 2 years ago
Edit Post #291299 Post edited:
was a long variable name
over 2 years ago
Edit Post #291299 Post edited:
try it online after the code
over 2 years ago
Edit Post #291299 Post edited:
you can try it online
over 2 years ago
Edit Post #291299 Post edited:
Doesn't matter whether the assignments are in one line or several lines, the number of bytes are the same, so I have here the more readable version with new lines.
over 2 years ago
Edit Post #291299 Post edited:
584
over 2 years ago
Edit Post #291299 Post edited:
shorter version
over 2 years ago
Comment Post #291299 Thank you, At the end I have realized, that using 0,1 and 2 instead of the variables is even shorter.
(more)
over 2 years ago
Edit Post #291299 Post edited:
remove set
over 2 years ago
Edit Post #291299 Post edited:
over 2 years ago
Edit Post #291299 Post edited:
reordering long version
over 2 years ago
Edit Post #291299 Post edited:
test cases
over 2 years ago