9
Jelly, 20 bytes
+Ø.÷€AĊc2ḋIɓN;æl¥ƒ-Ḋ
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Port of my own APL answer. Trailing ɓ chain trick is indeed strong :P
How it works
+Ø.÷€AĊc2ḋIɓN;æl¥ƒ-Ḋ Dyadic link; left=range, right=divisors
xxxxxxxxxxxɓyyyyyyyy x(left, y(right, left))
N;æl¥ƒ-Ḋ y: subset LCMs, negative for even-sized ones
N Negate the divisors
...
8
APL (Dyalog Extended), 31 bytes
-/+/(⎕+⍳2)(⊢×2!⌈⍤|⍤÷)\∊(⊢,∧)\-⎕
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Full program which takes the list of divisors, and then the (start,end) pair. Uses a modified trick to generate all nonempty subsequences. Also abuses the definition of LCM which is LCM(a,b) = a * b / GCD(a,b) and GCD is always positive, so LCM follows the sign of a * b.
How it ...
6
Haskell, 131 bytes
f=fromIntegral
u=unwords.map show
a%b|a>b=(a-b)%b|1<2=a
(r#c)s="P2":u[c,r,99]:[u[round$f j%(f i**s)/f r**s*99|j<-[1..c]]|i<-[1..r]]
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(r#c)s returns the lines of a PGM file.
After writing it like this, I learned I could probably just return a matrix of float values but I don't think that's very ...
6
J, 51 bytes
load'viewmat'
1 :'[:viewmat(|~^&u)"0~/&(1+i.)%u^~['
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A J adverb, which uses the library function viewmat to do all the heavy lifting -- we merely need to construct the matrix values.
Assuming the adverb has been assigned to f, called like:
500 (0.8 f) 800
500 800 0.8
200 250 0.3
5
Python 3, 229 bytes
lambda d,s,e:sum(-(-1)**len(k)*(t(e//l(*k))-t(~-s//l(*k)))*l(*k)for k in p(d)[1:])
t=lambda x:x*-~x//2
p=lambda k:k and[j+a for j in[[],k[:1]]for a in p(k[1:])]or[[]]
l=lambda a,*b:b and a*l(*b)//math.gcd(a,l(*b))or a
import math
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-5 bytes thanks to ovs
There's probably a less direct approach. I haven't properly golfed in ...
5
Vyxal s, 32 bytes
$ṗḢƛ:λ₌*ġḭ;R:¹‹÷›"$ḭ:›*2ḭ¯unLe**
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5
Jelly, 7 bytes
O%32Ṅ€S
A full program that accepts a string of three letters and prints their alphabetic indices followed by their sum on four lines.
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How?
O%32Ṅ€S - Main Link: list of characters, C
O - cast to ordinals A:65 ... Z:90 / a:97 ... z:122
%32 - modulo 32 A:1 ... Z:26 / a:1 ... z:26
Ṅ€ - print each followed ...
4
05AB1E, 25 bytes
æ¦ε.¿DIāÉ-s÷D>*2÷Æ®ygmP}O
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æ¦ε } # map over each non-empty subset of the divisors
.¿D # take the LCM of the subset
IāÉ- # push the range with the lower bound decreased
s÷ # integer divide modified range by LCM
D>*2÷...
4
Python 3, 116 114118 bytes
-2 thanks to some basic pointers from hyper-neutrino
+4 to correct an off-by-one error, thanks Tipping Octopus
from matplotlib.pylab import*
def M(R,C,s):
imshow([[j%i**s/R**s for j in range(1,C+1)]for i in range(1,R+1)]);show()
Fairly straightforward, my first code golf attempt so I may be missing something easily golfable. ...
4
JavaScript + HTML, 156 bytes
(R,C,s)=>{for((w=x=>document.write(x))`<table cellspacing=0>`,i=0;i++<R;)for(w`<tr>`,j=0;j<C;)w(`<td bgcolor=#${((++j%i**s*16/R**s|0)*273).toString(16)}>`)}
-7 bytes by Shaggy
; (function run() {
f=
(R,C,s)=>{for((w=x=>document.write(x))`<table cellspacing=0>`,i=0;i++<R;)for(w`&...
4
Wolfram Language (Mathematica), 39 bytes
sImage@Array[Mod[#2,#^s]&,{##}]/#^s&
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Input [s][R, C].
4
K (oK) + iKe, 61 bytes
{w::y;h::x;p::pow[;z];,(;gray;+_255*((p 1+!h)!\:/:1+!w)%p w)}
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Shortened heavily and made to work with the help of JohnE and coltim at the k tree.
A function which takes input as R, C, s.
4
R, 66 bytes
function(R,C,s)cat('P2',C,R,99,(99*outer(1:C,(1:R)^s,`%%`))%/%R^s)
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I kinda think that pajonk's answer is close to the shortest possible using R's built-in graphics... so here's a completely different approach, which actually turns-out to be 4 bytes shorter...
Outputs the contents of a greyscale PGM file. At least on my laptop ...
4
Python 3, 90 73 bytes
from pylab import*;ion()
def M(R,C,s):imshow(-~r_[:C]%c_[1:R+1]**s/R**s)
Heavily based on Danica's answer, I would have commented but I have 0 reputation.
Shorter import statement
ion() to show instead of ;show()
arange (from numpy.arange) array approach for faster performance and fewer bytes
remove def function indent
from pylab ...
4
Factor + combinators.extras, 42 35 bytes
0 [ read1 32 mod dup . + ] thrice .
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-7 bytes thanks to @MarcMush
0 Push 0, our sum, to the data stack.
[ ... ] thrice Call a quotation three times.
read1 Read one code point from standard input.
32 mod Modulo 32.
dup . Output on its own line non-destructively.
+ Add it to the sum.
. Print the sum.
3
Vyxal, 37 bytes
$ṗḢƛ:λ₌*ġḭ;R:¹÷$‹$"$ḭ:²+½$*÷-$Lu$e*;∑
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This is extremely messy. In case it wasn't clear, I am terrible with stack-based languages.
3
J, 72 69 bytes
1#.(_1*_1^1#.g)*(_1 0+[)(]*[:-~/(2!>:)@<.@%/)]*./@#~g=.1}.2#:@i.@^#@]
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-3 thanks to nC2 trick from Bubbler's answer
There's some more golf's I believe I can port from Bubbler's APL answer. Will try tomorrow.
3
Jelly, 30 29 bytes
-1 thanks to caird coinheringaahing!
’1¦:בHIƲḢ×ʋⱮⱮœcⱮJæl/€€Ɗ}§Ṛḅ-
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How?
’1¦:בHIƲḢ×ʋⱮⱮœcⱮJæl/€€Ɗ}§Ṛḅ- - Link: [start, stop]; divisors
1¦ - apply to index 1 (of [start, stop]):
’ - decrement -> [start-1, stop]
} - using right (divisors):...
3
Vyxal, 5 bytes
-*∑+c
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Inputs: attacks, growths, hits, heads
Explanation
- subtract: growths - attacks
* multiply: that * hits
∑ sum
+ add: that + heads
c contains: is it in attacks?
3
Python, 359 (184 without whitespace) 224 212 bytes, ~$$H_{ψ(ψ_Ι(0))[7]}(8)$$
*S,n=0,2,9
while S:
i=r=d=0
for a in S[::-1]:
i-=1
if d<S[-1]-a:
r=r or i;d+=S[-1]-a-d>1
if[k+d for k in S[i:]]<S[r:]:break
p=i+1
S.pop();n*=n
if r:S=S[:p]+[d*i+k for i in range(n)for k in S[p:]]
I'm new to codegolf so I'm not sure if all the whitespace ...
3
R, 84 70 bytes
-14 bytes thanks to @Dominic
function(R,C,s)image(outer(1:C,(R:1)^s,`%%`)/R^s,c=rainbow(64),as=R/C)
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Try it on rdrr.io with graphical output
3
APL (Dyalog Extended), 10 bytes
Full program. Prompts via stdin and prints to stdout
+⌿⎕←⍪⎕A⍳⌈⍞
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⍞ prompt via stdin
⌈ uppercase
⎕A⍳ indices in uppercase Alphabet
⍪ make into column
⎕← send to stdout
+⌿ sum (and implicitly print to stdout)
3
Elixir, 70 bytes
&(IO.puts Enum.reduce &1,0,fn b,c->IO.puts b=b-(b>91&&96||64)
c+b
end)
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3
Perl 5 + -pF, 27 bytes
-6 bytes thanks to @Sisyphus!
$\+=$;*say$;=31&ord for@F}{
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2
K (ngn/k), 16 bytes
{|/x*c=&/c:#'$x}
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$x convert the input to a list of strings
c:#' count the number of characters/digits in each input, storing in c
c=&/c generate boolean mask indicating which indices contain the minimal number of digits
x* "mask-out" inputs that do not have the minimal number of digits (i.e., convert ...
2
Clojure, 90 bytes
#(for[i[0 1]j((split-at(/(count %)2%2)(partition-all %2%))i)k(([seq reverse]i)(sort j))]k)
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2
javascript 77 bytes
let i=0;while(true){console.log(i);if(i.toString().includes('00'))break;i++;}
2
Vyxal, 21 7 bytes
⌈E∑ƒ\/j
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No rational numbers here...
-14 thanks to lyxal.
Or, if I can take it in a very specific form:
Vyxal, 4 bytes
*+$"
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Input as denominator,number,numerator.
2
Perl 5 (ppencode-compatible), 64 bytes
You didn't clarify that I must separate each with exactly one character, so here it is mine.
print length uc xor s qq q xor print while length ne ord qw q eq
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Explained
# print(length) did not work for zero as $_ is not defined at then
print length uc xor
s qq q xor
# delimiter
print
while
...
2
Python 3, 146 bytes
lambda d,s,e:(e*e+e-s*s+s+g(d,e)-g(d,s-1))//2
g=lambda d,n,m=1:g(d[1:],n,m)-g(d[1:],n,m*d[0]//math.gcd(m,d[0]))if d else~n//m*(n//m)*m
import math
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Some tricks explained: ~n//m is a shorter way to write -(n//m+1), and e*e+e-s*s+s is twice the sum of integers in the interval.
Python 3.9, 138 bytes
lambda d,s,e:(e*e+e-s*s+...
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