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Activity for dino‭

Type On... Excerpt Status Date
Edit Post #287444 Initial revision almost 2 years ago
Answer A: Encode with ROT13.5
[Python 3], 145 bytes lambda s:[dict({chr(i+c):chr((i+13)%26+c)for i in range(26)for c in(65,97)},{chr(i+48):chr((i+5)%10+48)for i in range(10)}).get(c,c)for c in s] Try it online! Generates and concatonates 2 dicts: `{chr(i+c):chr((i+13)%26+c)for i in range(26)for c in(65,97)...
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almost 2 years ago
Edit Post #287442 Initial revision almost 2 years ago
Answer A: Lowercase, but not just the letters
[Python 3], 35 bytes lambda x:[chr(ord(i)|32)for i in x] Try it online! Performs a list comprehension on the input string; for each character it: transforms it into the ASCII character code integer representation using `ord()` performs a `binary or` with `32` (or `0b100000`...
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almost 2 years ago
Edit Post #286801 Post edited:
almost 2 years ago
Edit Post #286801 Post edited:
almost 2 years ago
Edit Post #287352 Post edited:
almost 2 years ago
Edit Post #287352 Initial revision almost 2 years ago
Answer A: Sort letters by height
[Python 3], 43 bytes lambda x:sorted(x,key='tibdfghklpqyj'.find) Try it online! Sorts based on index in a sorted string. Inputs as a string and outputs as a list. -14 bytes by replacing `.index()` with `.find()`
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almost 2 years ago
Edit Post #286801 Post edited:
about 2 years ago
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about 2 years ago
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over 2 years ago
Edit Post #286801 Post edited:
over 2 years ago
Edit Post #286801 Initial revision over 2 years ago
Answer A: Cumulative Counts
[Python 3.8 (pre-release)], 69 bytes def f(x,y={},z=[]): for i in x:y[i]=y.get(i,0)+1;z+=[y[i]] return z Try it online! Bonus: theoretical answer if python allowed named assignment with subscript (54 bytes) ```python lambda x,y={}:[y[i]:=y[i]+1if i in y else 1for ...
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over 2 years ago
Edit Post #286431 Initial revision over 2 years ago
Answer A: Word Count Tool
[Python 3.8], 73 bytes lambda x:[len(x.split()),len(x.replace('\n','')),len(''.join(x.split()))] Try it online!
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over 2 years ago
Edit Post #286073 Initial revision over 2 years ago
Answer A: Solve Goldbach's Conjecture
[Python 3.8], 112 bytes def f(x):y=[i for i in range(2,x)if not[j for j in range(2,i)if i%j<1]];return(q,x-q)for q in y if x-q in y Try it online!
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over 2 years ago
Edit Post #285271 Post edited:
almost 3 years ago
Comment Post #285508 My hope with this solution was that I could extract the duplicate " bogey" to make it shorter, but I couldn't get it shorter than 110 bytes with the bogey extracted: https://tio.run/##TY@xjsIwDED3foWVKYFwulJAJ@56EhVIjAxspUMKKRep10SxGRDi20uKBIoXP1vPlu2u9Ge7rG/yQ9@q//qkwElcsq1tNZgObKfZiOPPVFjPdspDYc/6C...
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almost 3 years ago
Edit Post #285508 Initial revision almost 3 years ago
Answer A: Golf golf challenge
[Python 3], 108 bytes lambda p,s:s<2and"Hole in one"or"Par:Bogey:Double bogey:Triple bogey:Albatross:Eagle:Birdie".split(':')[s-p] Try it online! Very similar to my other solution, but handles the "Hole in one" case independently and uses wrap-around indexing for the other cases ...
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almost 3 years ago
Edit Post #285503 Initial revision almost 3 years ago
Answer A: Golf golf challenge
[Python 3], 108 bytes lambda p,s:"Hole in one:Albatross:Eagle:Birdie:Par:Bogey:Double bogey:Triple bogey".split(':')[s-1and s-p+4] Try it online!
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almost 3 years ago
Comment Post #285479 Very nice, thank you!
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almost 3 years ago
Edit Post #285479 Post edited:
almost 3 years ago
Edit Post #285479 Post edited:
Implement suggested changes from Moshi
almost 3 years ago
Edit Post #285479 Post edited:
almost 3 years ago
Comment Post #285479 Excellent suggestions, thank you!
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almost 3 years ago
Edit Post #285479 Post edited:
Implement suggested changes from celtschk
almost 3 years ago
Edit Post #285479 Post edited:
almost 3 years ago
Edit Post #285479 Post edited:
almost 3 years ago
Edit Post #285479 Post edited:
almost 3 years ago
Comment Post #285479 That's... a very fair point
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almost 3 years ago
Edit Post #285479 Post edited:
almost 3 years ago
Edit Post #285479 Post edited:
almost 3 years ago
Edit Post #285479 Post edited:
almost 3 years ago
Edit Post #285479 Post edited:
almost 3 years ago
Comment Post #285479 I /could/ remove some bytes by changing `int(sqrt(n)+1)` to `n`, but some part of me really doesn't like sacrificing time complexity like that…
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almost 3 years ago
Edit Post #285479 Initial revision almost 3 years ago
Answer A: Determine whether an integer is square-free
Python3, 39 bytes ```python lambda n:all(n%i2for i in range(2,n)) ``` Try it online! Makes a list comprehension from the numbers `2` through `n` of the remainder of the square, and then checks whether the list contains `0` -8 bytes thanks to celtschk‭ -5 bytes thanks to Moshi
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almost 3 years ago
Edit Post #285271 Post edited:
almost 3 years ago
Edit Post #285271 Post edited:
almost 3 years ago
Edit Post #285271 Post edited:
almost 3 years ago
Edit Post #285271 Post edited:
almost 3 years ago
Edit Post #285271 Post edited:
almost 3 years ago
Edit Post #285271 Post edited:
almost 3 years ago
Edit Post #285271 Post edited:
almost 3 years ago
Edit Post #285271 Post edited:
almost 3 years ago