Number of rotations
up vote
7
down vote
favorite
Task
Write a function or a program to find the number of rotations required by a wheel to travel a given distance, given its radius.
Rules
Input can be 2 positive rational numbers and can be taken in any convenient format.
Both inputs are of same unit.
There must not be any digits 0-9 in your code.
The output will be an integer (in case of float, round to infinity)
This is code-golf so shortest code wins
Examples
distance radius output
10 1 2
50 2 4
52.22 4 3
3.4 0.08 7
12.5663 0.9999 3
code-golf restricted-source
add a comment |
up vote
7
down vote
favorite
Task
Write a function or a program to find the number of rotations required by a wheel to travel a given distance, given its radius.
Rules
Input can be 2 positive rational numbers and can be taken in any convenient format.
Both inputs are of same unit.
There must not be any digits 0-9 in your code.
The output will be an integer (in case of float, round to infinity)
This is code-golf so shortest code wins
Examples
distance radius output
10 1 2
50 2 4
52.22 4 3
3.4 0.08 7
12.5663 0.9999 3
code-golf restricted-source
4
You probably should add that digits are also forbidden in compiler options (or anywhere else): if you limit this constraint to code only, with gcc we can do something like-DP=3.14
in compiler flags, that would defineP
as an approximation of pi, which is probably not what you intended
– Annyo
Nov 21 at 16:50
add a comment |
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Task
Write a function or a program to find the number of rotations required by a wheel to travel a given distance, given its radius.
Rules
Input can be 2 positive rational numbers and can be taken in any convenient format.
Both inputs are of same unit.
There must not be any digits 0-9 in your code.
The output will be an integer (in case of float, round to infinity)
This is code-golf so shortest code wins
Examples
distance radius output
10 1 2
50 2 4
52.22 4 3
3.4 0.08 7
12.5663 0.9999 3
code-golf restricted-source
Task
Write a function or a program to find the number of rotations required by a wheel to travel a given distance, given its radius.
Rules
Input can be 2 positive rational numbers and can be taken in any convenient format.
Both inputs are of same unit.
There must not be any digits 0-9 in your code.
The output will be an integer (in case of float, round to infinity)
This is code-golf so shortest code wins
Examples
distance radius output
10 1 2
50 2 4
52.22 4 3
3.4 0.08 7
12.5663 0.9999 3
code-golf restricted-source
code-golf restricted-source
edited Nov 22 at 12:54
Dennis♦
184k32295732
184k32295732
asked Nov 21 at 12:42
Vedant Kandoi
65618
65618
4
You probably should add that digits are also forbidden in compiler options (or anywhere else): if you limit this constraint to code only, with gcc we can do something like-DP=3.14
in compiler flags, that would defineP
as an approximation of pi, which is probably not what you intended
– Annyo
Nov 21 at 16:50
add a comment |
4
You probably should add that digits are also forbidden in compiler options (or anywhere else): if you limit this constraint to code only, with gcc we can do something like-DP=3.14
in compiler flags, that would defineP
as an approximation of pi, which is probably not what you intended
– Annyo
Nov 21 at 16:50
4
4
You probably should add that digits are also forbidden in compiler options (or anywhere else): if you limit this constraint to code only, with gcc we can do something like
-DP=3.14
in compiler flags, that would define P
as an approximation of pi, which is probably not what you intended– Annyo
Nov 21 at 16:50
You probably should add that digits are also forbidden in compiler options (or anywhere else): if you limit this constraint to code only, with gcc we can do something like
-DP=3.14
in compiler flags, that would define P
as an approximation of pi, which is probably not what you intended– Annyo
Nov 21 at 16:50
add a comment |
31 Answers
31
active
oldest
votes
1 2
next
up vote
5
down vote
MathGolf, 5 4 bytes
τ/╠ü
Try it online!
Explanation
τ Push tau (2*pi)
/ Divide the first argument (total distance) by tau
╠ Reverse divide (computes (distance/tau)/radius)
ü Ceiling
add a comment |
up vote
4
down vote
APL+WIN, 9 bytes
Prompts for radius followed by distance:
⌈⎕÷○r+r←⎕
Try it online! Courtesy of Dyalog Classic
Explanation:
○r+r←⎕ prompt for radius and double it and multiply by pie
⌈⎕÷ prompt for distance, divide by result above and take ceiling
⌈⎕÷○+⍨⎕
works for 7 bytes.
– J. Sallé
Nov 22 at 13:04
@J.Sallé Thanks but unfortunately my ancient APL+WIN interpreter does not have the ⍨ operator
– Graham
Nov 22 at 15:51
add a comment |
up vote
4
down vote
Java 8, 32 30 bytes
a->b->-~(int)(a/b/Math.PI/'')
Contains unprintable u0002
between the single quotes.
Port of @jOKing's Perl 6 answer.
Try it online.
Is that the digit '1' in your code? I think that might not be allowed.
– ouflak
Nov 21 at 14:18
4
@ouflak Looks like it can be fixed like this.
– Erik the Outgolfer
Nov 21 at 14:22
@ouflak Woops, that was a pretty stupid mistake.. Using the unprintable so I don't use the digit2
, and then just use digit1
... Luckily Erik is indeed right that a simple negative unary has the same effect as+1
(often used to get rid of parenthesis since the negative and unary have higher precedence than most other operators).
– Kevin Cruijssen
Nov 21 at 18:21
add a comment |
up vote
4
down vote
Perl 6, 15 12 bytes
-3 bytes tjanks to nwellnhof reminding me about tau
*/*/τ+|$+!$
Try it online!
Anonymous Whatever lambda that uses the formula (a/b/tau).floor+1
. Tau is two times pi. The two anonymous variables $
are coerced to the number 0
, which is used to floor the number +|0
(bitwise or 0) and add one +!$
(plus not zero).
There must not be any digits 0-9 in your code.
– Titus
Nov 21 at 13:35
@Titus I can't believe I forgot that. Thanks, fixed!
– Jo King
Nov 21 at 13:37
Are digits in exponents also allowed?
– ouflak
Nov 21 at 14:21
add a comment |
up vote
3
down vote
Python 2, 47 45 44 43 bytes
lambda l,r:l/(r+r)//math.pi+l/l
import math
Try it online!
- -2 bytes, thanks to flawr
- -1 byte, thanks to Jonathan Allan
Since inputs have been guaranteed to be both (strictly) positive and rational we never hit the edge-case of requiring an exact number of rotations, so I think we can dol/(r+r)//pi+l/l
and save a byte.
– Jonathan Allan
Nov 21 at 13:50
@JonathanAllan Thanks :)
– TFeld
Nov 21 at 14:16
add a comment |
up vote
2
down vote
05AB1E, 6 bytes
·/žq/î
Port of @flawr's Python 2 comment.
Takes the input in the order radius
,distance
.
Try it online or verify all test cases.
Explanation:
· # Double the first (implicit) input
/ # Divide the second (implicit) input by it
žq/ # Divide it by PI
î # Ceil it (and output implicitly)
add a comment |
up vote
2
down vote
C, 46 bytes
f(float a,float b){return ceil(a/(b+b)/M_PI);}
I'm new to PPCG, so I'm not sure wether I have to count other parts in the byte count, such as the
include <math.h>
needed for the ceil function, which will rise the count to 64 bytes
New contributor
Welcome to PPCG! This is a nice first answer. Yes, you do need to count#include
and the like towards your byte total. A link to an online test suite is always appreciated, here's one you are free to incorporate into your post: tio.run/…
– O.O.Balance
Nov 21 at 14:53
@O.O.Balance Digits are not allowed in the code for this challenge ;)
– Annyo
Nov 21 at 15:25
@Annyo I knew I was forgetting something :(
– O.O.Balance
Nov 21 at 15:26
add a comment |
up vote
2
down vote
Catholicon, 8 bytes
ċ//ĊǓĊ`Ė
Explanation:
/ĊǓĊ divide the first input by the doubled second input
/ `Ė divide that by pi
ċ ceil
New version (pi builtin made one byte, division parameters swapped), 5 bytes
ċ/π/Ǔ
add a comment |
up vote
2
down vote
Stax, 5 bytes
Vt*/e
Run and debug it
Vt* multiply by tau (2pi)
/ divide
e ceiling
add a comment |
up vote
2
down vote
MathGolf, 6 5 bytes
∞/π/ü
Semi-port of @flawr's Python 2 comment.
Takes the input in the order radius distance
.
-1 byte because ceil
builtin has just been added, replacing the floor+1
.
Try it online.
Explanation:
∞ # Double the first (implicit) input
/ # Divide the second (implicit) input by it
π/ # Divide it by PI
ü # Ceil (and output implicitly)
add a comment |
up vote
2
down vote
C (gcc), 45 47 45 bytes
f(d,r,R)float d,r;{R=ceil(d/r/'G'/'n'*'q');}
A reasonable approximation of pi is 355/113. Since circumference C = 2 * r * PI, we can instead of pi use tau, which is then of course ~710/113. 710 happens to have the convenient factors 2 * 5 * 71, which is compactly expressed as 'G' * 'n'
. We add one (r/r
) to force rounding to infinity.
Edit: My trick was too clever for its own good: it of course made it fail if the distance was a multiple of the circumference.
Try it online!
add a comment |
up vote
2
down vote
Julia 1.0, 20 bytes
f(d,r)=cld(d/π,r+r)
Try it online!
add a comment |
up vote
2
down vote
R, 39 32 bytes
-7 bytes Thanks to Giuseppe
function(d,r)ceiling(d/(r+r)/pi)
Try it online!
I feel like this could definitely be golfed, but I am a bit lazy right now to do anything about it
add a comment |
up vote
1
down vote
PHP, 47 bytes
<?=ceil($argv[++$i]/M_PI/(($b=end($argv))+$b));
Try it online.
add a comment |
up vote
1
down vote
Jelly, 6 bytes
÷÷ØPHĊ
Try it online!
add a comment |
up vote
1
down vote
Ruby, 29 bytes
->l,r{(l/Math::PI/r+=r).ceil}
Try it online!
add a comment |
up vote
1
down vote
J, 10 9 bytes
>.@%o.@+:
Try it online!
add a comment |
up vote
1
down vote
Japt, 7 bytes
/MT/V c
Try it here
add a comment |
up vote
1
down vote
JavaScript (Babel Node), 25 bytes
-2 bytes using @flawr comment =D. -1 from @Kevin. -7 from @Shaggy
a=>b=>-~(a/(b+b)/Math.PI)
Try it online!
Justa=>b=>Math.ceil(a/(b+b)/Math.PI)
is 32 bytes. :)
– Kevin Cruijssen
Nov 21 at 13:06
25 bytes
– Shaggy
Nov 21 at 17:09
add a comment |
up vote
1
down vote
min, 16 bytes
/ tau / ceil int
Takes the distance and radius put on the stack in that order. Then divides by tau, rounds, and makes int.
add a comment |
up vote
1
down vote
Dart, 47 46 bytes
import'dart:math';f(a,b)=>(a/(b+b)/pi).ceil();
Try it online!
- -1 byte thanks to @Shaggy
46 bytes
– Shaggy
Nov 22 at 12:30
add a comment |
up vote
1
down vote
JavaScript (Babel Node), 23 bytes
s=>r=>-~(s/2/r/Math.PI)
Try it online!
2
There must not be any digits 0-9 in your code.
– Dennis♦
Nov 22 at 12:55
add a comment |
up vote
1
down vote
Haskell, 25 bytes
f d r=ceiling(d/(r+r)/pi)
New contributor
add a comment |
up vote
1
down vote
Lua, 61 58 57 49 bytes
function(s,r)return math.ceil(s/(r+r)/math.pi)end
Try it online!
Thanks to KirillL. -8 bytes.
I don't know much Lua (so maybe it's still too long), but it appears to be shorter as a function: 49 bytes
– Kirill L.
Nov 22 at 11:18
@KirillL., I'm still learning the rules here. The OP's challenge is pretty open on the input. So my question is, would we have to count your program call() against the byte count? If not, your's definitely shaves off a nice chunk.
– ouflak
Nov 22 at 11:43
A quite common style of submission here is an anonymous function (so that we don't have to count the name, unless it is recursive), which outputs by its return value. The footer section with function calls and actual printing to console is then basically used for visualizing the results and doesn't count towards your score. BTW, you may add more of the OP's test examples to the footer, so that they can be conveniently viewed all at once. Note that in some cases a full program may actually turn out to be golfier!
– Kirill L.
Nov 22 at 11:58
add a comment |
up vote
1
down vote
Common Lisp, 36 bytes
(lambda(a b)(ceiling(/ a(+ b b)pi)))
Try it online!
add a comment |
up vote
1
down vote
Tcl, 50 bytes
proc N d r {expr ceil($d/(($r+$r)*acos(-$r/$r)))}
Try it online!
Tcl, 53 bytes
proc N d r {expr ceil($d/(($r+$r)*acos(-[incr i])))}
Try it online!
Lack of a pi constant or function makes me lose the golf competition!
Do I need to remove the .0 at end of each output? It would make me consume more bytes!
– sergiol
Nov 22 at 18:55
1
[incr i]
is quite clever but I think you can use$d/$d
or$r/$r
instead.
– david
Nov 22 at 20:17
Saved some bytes thanks to @david's idea!
– sergiol
Nov 22 at 22:56
add a comment |
up vote
1
down vote
PowerShell, 53 52 bytes
-1 byte thanks to @mazzy
param($d,$r);($a=[math])::ceiling($d/($r+$r)/$a::pi)
Takes input from two commandline parameters, distance -d
and radius -r
.
?param($d,$r);($a=[math])::ceiling($d/($r+$r)/$a::pi)
– mazzy
2 days ago
add a comment |
up vote
0
down vote
Mathematica, 8 bytes
#/(2π r)&
So that
r =22.9;
#/(2π r)& @ 32
gives
(* 0.2224 *)
2
There must not be any digits 0-9 in your code. Also, could you please show how to call your function, maybe in an online interpreter like tio.run/#mathematica? I'm not sure how to supply ther
.
– Dennis♦
Nov 22 at 13:03
add a comment |
up vote
0
down vote
Clojure, 50 bytes
(fn[a b](int(Math/ceil(/ a Math/PI(count" ")b))))
An anonymous function that accepts two integers a
and b
as arguments: the distance and the wheel's radius, respectively.
Try it online!
(count " ")
evaluates to 2
, so this function implements $lceil dfrac a{2pi b} rceil$.
add a comment |
up vote
0
down vote
TI-Basic (83 series), 12 bytes
-int(-Tmax⁻¹min(e^(ΔList(ln(Ans
Takes input as a list of radius and distance in Ans
: for example, {0.9999:12.5663:prgmX
.
e^(ΔList(ln(Ans
will take the ratio of those distances, and min(
turns this into a number. Then we divide by Tmax
, which is a graphing parameter that's equal to 2π by default. Finally, -int(-
takes the ceiling.
add a comment |
1 2
next
31 Answers
31
active
oldest
votes
31 Answers
31
active
oldest
votes
active
oldest
votes
active
oldest
votes
1 2
next
up vote
5
down vote
MathGolf, 5 4 bytes
τ/╠ü
Try it online!
Explanation
τ Push tau (2*pi)
/ Divide the first argument (total distance) by tau
╠ Reverse divide (computes (distance/tau)/radius)
ü Ceiling
add a comment |
up vote
5
down vote
MathGolf, 5 4 bytes
τ/╠ü
Try it online!
Explanation
τ Push tau (2*pi)
/ Divide the first argument (total distance) by tau
╠ Reverse divide (computes (distance/tau)/radius)
ü Ceiling
add a comment |
up vote
5
down vote
up vote
5
down vote
MathGolf, 5 4 bytes
τ/╠ü
Try it online!
Explanation
τ Push tau (2*pi)
/ Divide the first argument (total distance) by tau
╠ Reverse divide (computes (distance/tau)/radius)
ü Ceiling
MathGolf, 5 4 bytes
τ/╠ü
Try it online!
Explanation
τ Push tau (2*pi)
/ Divide the first argument (total distance) by tau
╠ Reverse divide (computes (distance/tau)/radius)
ü Ceiling
edited Nov 21 at 14:59
answered Nov 21 at 14:41
maxb
2,1081923
2,1081923
add a comment |
add a comment |
up vote
4
down vote
APL+WIN, 9 bytes
Prompts for radius followed by distance:
⌈⎕÷○r+r←⎕
Try it online! Courtesy of Dyalog Classic
Explanation:
○r+r←⎕ prompt for radius and double it and multiply by pie
⌈⎕÷ prompt for distance, divide by result above and take ceiling
⌈⎕÷○+⍨⎕
works for 7 bytes.
– J. Sallé
Nov 22 at 13:04
@J.Sallé Thanks but unfortunately my ancient APL+WIN interpreter does not have the ⍨ operator
– Graham
Nov 22 at 15:51
add a comment |
up vote
4
down vote
APL+WIN, 9 bytes
Prompts for radius followed by distance:
⌈⎕÷○r+r←⎕
Try it online! Courtesy of Dyalog Classic
Explanation:
○r+r←⎕ prompt for radius and double it and multiply by pie
⌈⎕÷ prompt for distance, divide by result above and take ceiling
⌈⎕÷○+⍨⎕
works for 7 bytes.
– J. Sallé
Nov 22 at 13:04
@J.Sallé Thanks but unfortunately my ancient APL+WIN interpreter does not have the ⍨ operator
– Graham
Nov 22 at 15:51
add a comment |
up vote
4
down vote
up vote
4
down vote
APL+WIN, 9 bytes
Prompts for radius followed by distance:
⌈⎕÷○r+r←⎕
Try it online! Courtesy of Dyalog Classic
Explanation:
○r+r←⎕ prompt for radius and double it and multiply by pie
⌈⎕÷ prompt for distance, divide by result above and take ceiling
APL+WIN, 9 bytes
Prompts for radius followed by distance:
⌈⎕÷○r+r←⎕
Try it online! Courtesy of Dyalog Classic
Explanation:
○r+r←⎕ prompt for radius and double it and multiply by pie
⌈⎕÷ prompt for distance, divide by result above and take ceiling
edited Nov 21 at 14:10
answered Nov 21 at 13:59
Graham
2,12668
2,12668
⌈⎕÷○+⍨⎕
works for 7 bytes.
– J. Sallé
Nov 22 at 13:04
@J.Sallé Thanks but unfortunately my ancient APL+WIN interpreter does not have the ⍨ operator
– Graham
Nov 22 at 15:51
add a comment |
⌈⎕÷○+⍨⎕
works for 7 bytes.
– J. Sallé
Nov 22 at 13:04
@J.Sallé Thanks but unfortunately my ancient APL+WIN interpreter does not have the ⍨ operator
– Graham
Nov 22 at 15:51
⌈⎕÷○+⍨⎕
works for 7 bytes.– J. Sallé
Nov 22 at 13:04
⌈⎕÷○+⍨⎕
works for 7 bytes.– J. Sallé
Nov 22 at 13:04
@J.Sallé Thanks but unfortunately my ancient APL+WIN interpreter does not have the ⍨ operator
– Graham
Nov 22 at 15:51
@J.Sallé Thanks but unfortunately my ancient APL+WIN interpreter does not have the ⍨ operator
– Graham
Nov 22 at 15:51
add a comment |
up vote
4
down vote
Java 8, 32 30 bytes
a->b->-~(int)(a/b/Math.PI/'')
Contains unprintable u0002
between the single quotes.
Port of @jOKing's Perl 6 answer.
Try it online.
Is that the digit '1' in your code? I think that might not be allowed.
– ouflak
Nov 21 at 14:18
4
@ouflak Looks like it can be fixed like this.
– Erik the Outgolfer
Nov 21 at 14:22
@ouflak Woops, that was a pretty stupid mistake.. Using the unprintable so I don't use the digit2
, and then just use digit1
... Luckily Erik is indeed right that a simple negative unary has the same effect as+1
(often used to get rid of parenthesis since the negative and unary have higher precedence than most other operators).
– Kevin Cruijssen
Nov 21 at 18:21
add a comment |
up vote
4
down vote
Java 8, 32 30 bytes
a->b->-~(int)(a/b/Math.PI/'')
Contains unprintable u0002
between the single quotes.
Port of @jOKing's Perl 6 answer.
Try it online.
Is that the digit '1' in your code? I think that might not be allowed.
– ouflak
Nov 21 at 14:18
4
@ouflak Looks like it can be fixed like this.
– Erik the Outgolfer
Nov 21 at 14:22
@ouflak Woops, that was a pretty stupid mistake.. Using the unprintable so I don't use the digit2
, and then just use digit1
... Luckily Erik is indeed right that a simple negative unary has the same effect as+1
(often used to get rid of parenthesis since the negative and unary have higher precedence than most other operators).
– Kevin Cruijssen
Nov 21 at 18:21
add a comment |
up vote
4
down vote
up vote
4
down vote
Java 8, 32 30 bytes
a->b->-~(int)(a/b/Math.PI/'')
Contains unprintable u0002
between the single quotes.
Port of @jOKing's Perl 6 answer.
Try it online.
Java 8, 32 30 bytes
a->b->-~(int)(a/b/Math.PI/'')
Contains unprintable u0002
between the single quotes.
Port of @jOKing's Perl 6 answer.
Try it online.
edited Nov 21 at 18:19
answered Nov 21 at 13:05
Kevin Cruijssen
34.4k554182
34.4k554182
Is that the digit '1' in your code? I think that might not be allowed.
– ouflak
Nov 21 at 14:18
4
@ouflak Looks like it can be fixed like this.
– Erik the Outgolfer
Nov 21 at 14:22
@ouflak Woops, that was a pretty stupid mistake.. Using the unprintable so I don't use the digit2
, and then just use digit1
... Luckily Erik is indeed right that a simple negative unary has the same effect as+1
(often used to get rid of parenthesis since the negative and unary have higher precedence than most other operators).
– Kevin Cruijssen
Nov 21 at 18:21
add a comment |
Is that the digit '1' in your code? I think that might not be allowed.
– ouflak
Nov 21 at 14:18
4
@ouflak Looks like it can be fixed like this.
– Erik the Outgolfer
Nov 21 at 14:22
@ouflak Woops, that was a pretty stupid mistake.. Using the unprintable so I don't use the digit2
, and then just use digit1
... Luckily Erik is indeed right that a simple negative unary has the same effect as+1
(often used to get rid of parenthesis since the negative and unary have higher precedence than most other operators).
– Kevin Cruijssen
Nov 21 at 18:21
Is that the digit '1' in your code? I think that might not be allowed.
– ouflak
Nov 21 at 14:18
Is that the digit '1' in your code? I think that might not be allowed.
– ouflak
Nov 21 at 14:18
4
4
@ouflak Looks like it can be fixed like this.
– Erik the Outgolfer
Nov 21 at 14:22
@ouflak Looks like it can be fixed like this.
– Erik the Outgolfer
Nov 21 at 14:22
@ouflak Woops, that was a pretty stupid mistake.. Using the unprintable so I don't use the digit
2
, and then just use digit 1
... Luckily Erik is indeed right that a simple negative unary has the same effect as +1
(often used to get rid of parenthesis since the negative and unary have higher precedence than most other operators).– Kevin Cruijssen
Nov 21 at 18:21
@ouflak Woops, that was a pretty stupid mistake.. Using the unprintable so I don't use the digit
2
, and then just use digit 1
... Luckily Erik is indeed right that a simple negative unary has the same effect as +1
(often used to get rid of parenthesis since the negative and unary have higher precedence than most other operators).– Kevin Cruijssen
Nov 21 at 18:21
add a comment |
up vote
4
down vote
Perl 6, 15 12 bytes
-3 bytes tjanks to nwellnhof reminding me about tau
*/*/τ+|$+!$
Try it online!
Anonymous Whatever lambda that uses the formula (a/b/tau).floor+1
. Tau is two times pi. The two anonymous variables $
are coerced to the number 0
, which is used to floor the number +|0
(bitwise or 0) and add one +!$
(plus not zero).
There must not be any digits 0-9 in your code.
– Titus
Nov 21 at 13:35
@Titus I can't believe I forgot that. Thanks, fixed!
– Jo King
Nov 21 at 13:37
Are digits in exponents also allowed?
– ouflak
Nov 21 at 14:21
add a comment |
up vote
4
down vote
Perl 6, 15 12 bytes
-3 bytes tjanks to nwellnhof reminding me about tau
*/*/τ+|$+!$
Try it online!
Anonymous Whatever lambda that uses the formula (a/b/tau).floor+1
. Tau is two times pi. The two anonymous variables $
are coerced to the number 0
, which is used to floor the number +|0
(bitwise or 0) and add one +!$
(plus not zero).
There must not be any digits 0-9 in your code.
– Titus
Nov 21 at 13:35
@Titus I can't believe I forgot that. Thanks, fixed!
– Jo King
Nov 21 at 13:37
Are digits in exponents also allowed?
– ouflak
Nov 21 at 14:21
add a comment |
up vote
4
down vote
up vote
4
down vote
Perl 6, 15 12 bytes
-3 bytes tjanks to nwellnhof reminding me about tau
*/*/τ+|$+!$
Try it online!
Anonymous Whatever lambda that uses the formula (a/b/tau).floor+1
. Tau is two times pi. The two anonymous variables $
are coerced to the number 0
, which is used to floor the number +|0
(bitwise or 0) and add one +!$
(plus not zero).
Perl 6, 15 12 bytes
-3 bytes tjanks to nwellnhof reminding me about tau
*/*/τ+|$+!$
Try it online!
Anonymous Whatever lambda that uses the formula (a/b/tau).floor+1
. Tau is two times pi. The two anonymous variables $
are coerced to the number 0
, which is used to floor the number +|0
(bitwise or 0) and add one +!$
(plus not zero).
edited Nov 21 at 20:17
answered Nov 21 at 13:31
Jo King
19.4k245102
19.4k245102
There must not be any digits 0-9 in your code.
– Titus
Nov 21 at 13:35
@Titus I can't believe I forgot that. Thanks, fixed!
– Jo King
Nov 21 at 13:37
Are digits in exponents also allowed?
– ouflak
Nov 21 at 14:21
add a comment |
There must not be any digits 0-9 in your code.
– Titus
Nov 21 at 13:35
@Titus I can't believe I forgot that. Thanks, fixed!
– Jo King
Nov 21 at 13:37
Are digits in exponents also allowed?
– ouflak
Nov 21 at 14:21
There must not be any digits 0-9 in your code.
– Titus
Nov 21 at 13:35
There must not be any digits 0-9 in your code.
– Titus
Nov 21 at 13:35
@Titus I can't believe I forgot that. Thanks, fixed!
– Jo King
Nov 21 at 13:37
@Titus I can't believe I forgot that. Thanks, fixed!
– Jo King
Nov 21 at 13:37
Are digits in exponents also allowed?
– ouflak
Nov 21 at 14:21
Are digits in exponents also allowed?
– ouflak
Nov 21 at 14:21
add a comment |
up vote
3
down vote
Python 2, 47 45 44 43 bytes
lambda l,r:l/(r+r)//math.pi+l/l
import math
Try it online!
- -2 bytes, thanks to flawr
- -1 byte, thanks to Jonathan Allan
Since inputs have been guaranteed to be both (strictly) positive and rational we never hit the edge-case of requiring an exact number of rotations, so I think we can dol/(r+r)//pi+l/l
and save a byte.
– Jonathan Allan
Nov 21 at 13:50
@JonathanAllan Thanks :)
– TFeld
Nov 21 at 14:16
add a comment |
up vote
3
down vote
Python 2, 47 45 44 43 bytes
lambda l,r:l/(r+r)//math.pi+l/l
import math
Try it online!
- -2 bytes, thanks to flawr
- -1 byte, thanks to Jonathan Allan
Since inputs have been guaranteed to be both (strictly) positive and rational we never hit the edge-case of requiring an exact number of rotations, so I think we can dol/(r+r)//pi+l/l
and save a byte.
– Jonathan Allan
Nov 21 at 13:50
@JonathanAllan Thanks :)
– TFeld
Nov 21 at 14:16
add a comment |
up vote
3
down vote
up vote
3
down vote
Python 2, 47 45 44 43 bytes
lambda l,r:l/(r+r)//math.pi+l/l
import math
Try it online!
- -2 bytes, thanks to flawr
- -1 byte, thanks to Jonathan Allan
Python 2, 47 45 44 43 bytes
lambda l,r:l/(r+r)//math.pi+l/l
import math
Try it online!
- -2 bytes, thanks to flawr
- -1 byte, thanks to Jonathan Allan
edited Nov 21 at 14:23
answered Nov 21 at 12:51
TFeld
13.7k21139
13.7k21139
Since inputs have been guaranteed to be both (strictly) positive and rational we never hit the edge-case of requiring an exact number of rotations, so I think we can dol/(r+r)//pi+l/l
and save a byte.
– Jonathan Allan
Nov 21 at 13:50
@JonathanAllan Thanks :)
– TFeld
Nov 21 at 14:16
add a comment |
Since inputs have been guaranteed to be both (strictly) positive and rational we never hit the edge-case of requiring an exact number of rotations, so I think we can dol/(r+r)//pi+l/l
and save a byte.
– Jonathan Allan
Nov 21 at 13:50
@JonathanAllan Thanks :)
– TFeld
Nov 21 at 14:16
Since inputs have been guaranteed to be both (strictly) positive and rational we never hit the edge-case of requiring an exact number of rotations, so I think we can do
l/(r+r)//pi+l/l
and save a byte.– Jonathan Allan
Nov 21 at 13:50
Since inputs have been guaranteed to be both (strictly) positive and rational we never hit the edge-case of requiring an exact number of rotations, so I think we can do
l/(r+r)//pi+l/l
and save a byte.– Jonathan Allan
Nov 21 at 13:50
@JonathanAllan Thanks :)
– TFeld
Nov 21 at 14:16
@JonathanAllan Thanks :)
– TFeld
Nov 21 at 14:16
add a comment |
up vote
2
down vote
05AB1E, 6 bytes
·/žq/î
Port of @flawr's Python 2 comment.
Takes the input in the order radius
,distance
.
Try it online or verify all test cases.
Explanation:
· # Double the first (implicit) input
/ # Divide the second (implicit) input by it
žq/ # Divide it by PI
î # Ceil it (and output implicitly)
add a comment |
up vote
2
down vote
05AB1E, 6 bytes
·/žq/î
Port of @flawr's Python 2 comment.
Takes the input in the order radius
,distance
.
Try it online or verify all test cases.
Explanation:
· # Double the first (implicit) input
/ # Divide the second (implicit) input by it
žq/ # Divide it by PI
î # Ceil it (and output implicitly)
add a comment |
up vote
2
down vote
up vote
2
down vote
05AB1E, 6 bytes
·/žq/î
Port of @flawr's Python 2 comment.
Takes the input in the order radius
,distance
.
Try it online or verify all test cases.
Explanation:
· # Double the first (implicit) input
/ # Divide the second (implicit) input by it
žq/ # Divide it by PI
î # Ceil it (and output implicitly)
05AB1E, 6 bytes
·/žq/î
Port of @flawr's Python 2 comment.
Takes the input in the order radius
,distance
.
Try it online or verify all test cases.
Explanation:
· # Double the first (implicit) input
/ # Divide the second (implicit) input by it
žq/ # Divide it by PI
î # Ceil it (and output implicitly)
answered Nov 21 at 13:11
Kevin Cruijssen
34.4k554182
34.4k554182
add a comment |
add a comment |
up vote
2
down vote
C, 46 bytes
f(float a,float b){return ceil(a/(b+b)/M_PI);}
I'm new to PPCG, so I'm not sure wether I have to count other parts in the byte count, such as the
include <math.h>
needed for the ceil function, which will rise the count to 64 bytes
New contributor
Welcome to PPCG! This is a nice first answer. Yes, you do need to count#include
and the like towards your byte total. A link to an online test suite is always appreciated, here's one you are free to incorporate into your post: tio.run/…
– O.O.Balance
Nov 21 at 14:53
@O.O.Balance Digits are not allowed in the code for this challenge ;)
– Annyo
Nov 21 at 15:25
@Annyo I knew I was forgetting something :(
– O.O.Balance
Nov 21 at 15:26
add a comment |
up vote
2
down vote
C, 46 bytes
f(float a,float b){return ceil(a/(b+b)/M_PI);}
I'm new to PPCG, so I'm not sure wether I have to count other parts in the byte count, such as the
include <math.h>
needed for the ceil function, which will rise the count to 64 bytes
New contributor
Welcome to PPCG! This is a nice first answer. Yes, you do need to count#include
and the like towards your byte total. A link to an online test suite is always appreciated, here's one you are free to incorporate into your post: tio.run/…
– O.O.Balance
Nov 21 at 14:53
@O.O.Balance Digits are not allowed in the code for this challenge ;)
– Annyo
Nov 21 at 15:25
@Annyo I knew I was forgetting something :(
– O.O.Balance
Nov 21 at 15:26
add a comment |
up vote
2
down vote
up vote
2
down vote
C, 46 bytes
f(float a,float b){return ceil(a/(b+b)/M_PI);}
I'm new to PPCG, so I'm not sure wether I have to count other parts in the byte count, such as the
include <math.h>
needed for the ceil function, which will rise the count to 64 bytes
New contributor
C, 46 bytes
f(float a,float b){return ceil(a/(b+b)/M_PI);}
I'm new to PPCG, so I'm not sure wether I have to count other parts in the byte count, such as the
include <math.h>
needed for the ceil function, which will rise the count to 64 bytes
New contributor
New contributor
answered Nov 21 at 14:30
bznein
1212
1212
New contributor
New contributor
Welcome to PPCG! This is a nice first answer. Yes, you do need to count#include
and the like towards your byte total. A link to an online test suite is always appreciated, here's one you are free to incorporate into your post: tio.run/…
– O.O.Balance
Nov 21 at 14:53
@O.O.Balance Digits are not allowed in the code for this challenge ;)
– Annyo
Nov 21 at 15:25
@Annyo I knew I was forgetting something :(
– O.O.Balance
Nov 21 at 15:26
add a comment |
Welcome to PPCG! This is a nice first answer. Yes, you do need to count#include
and the like towards your byte total. A link to an online test suite is always appreciated, here's one you are free to incorporate into your post: tio.run/…
– O.O.Balance
Nov 21 at 14:53
@O.O.Balance Digits are not allowed in the code for this challenge ;)
– Annyo
Nov 21 at 15:25
@Annyo I knew I was forgetting something :(
– O.O.Balance
Nov 21 at 15:26
Welcome to PPCG! This is a nice first answer. Yes, you do need to count
#include
and the like towards your byte total. A link to an online test suite is always appreciated, here's one you are free to incorporate into your post: tio.run/…– O.O.Balance
Nov 21 at 14:53
Welcome to PPCG! This is a nice first answer. Yes, you do need to count
#include
and the like towards your byte total. A link to an online test suite is always appreciated, here's one you are free to incorporate into your post: tio.run/…– O.O.Balance
Nov 21 at 14:53
@O.O.Balance Digits are not allowed in the code for this challenge ;)
– Annyo
Nov 21 at 15:25
@O.O.Balance Digits are not allowed in the code for this challenge ;)
– Annyo
Nov 21 at 15:25
@Annyo I knew I was forgetting something :(
– O.O.Balance
Nov 21 at 15:26
@Annyo I knew I was forgetting something :(
– O.O.Balance
Nov 21 at 15:26
add a comment |
up vote
2
down vote
Catholicon, 8 bytes
ċ//ĊǓĊ`Ė
Explanation:
/ĊǓĊ divide the first input by the doubled second input
/ `Ė divide that by pi
ċ ceil
New version (pi builtin made one byte, division parameters swapped), 5 bytes
ċ/π/Ǔ
add a comment |
up vote
2
down vote
Catholicon, 8 bytes
ċ//ĊǓĊ`Ė
Explanation:
/ĊǓĊ divide the first input by the doubled second input
/ `Ė divide that by pi
ċ ceil
New version (pi builtin made one byte, division parameters swapped), 5 bytes
ċ/π/Ǔ
add a comment |
up vote
2
down vote
up vote
2
down vote
Catholicon, 8 bytes
ċ//ĊǓĊ`Ė
Explanation:
/ĊǓĊ divide the first input by the doubled second input
/ `Ė divide that by pi
ċ ceil
New version (pi builtin made one byte, division parameters swapped), 5 bytes
ċ/π/Ǔ
Catholicon, 8 bytes
ċ//ĊǓĊ`Ė
Explanation:
/ĊǓĊ divide the first input by the doubled second input
/ `Ė divide that by pi
ċ ceil
New version (pi builtin made one byte, division parameters swapped), 5 bytes
ċ/π/Ǔ
edited Nov 21 at 14:32
answered Nov 21 at 14:23
Okx
12.4k27100
12.4k27100
add a comment |
add a comment |
up vote
2
down vote
Stax, 5 bytes
Vt*/e
Run and debug it
Vt* multiply by tau (2pi)
/ divide
e ceiling
add a comment |
up vote
2
down vote
Stax, 5 bytes
Vt*/e
Run and debug it
Vt* multiply by tau (2pi)
/ divide
e ceiling
add a comment |
up vote
2
down vote
up vote
2
down vote
Stax, 5 bytes
Vt*/e
Run and debug it
Vt* multiply by tau (2pi)
/ divide
e ceiling
Stax, 5 bytes
Vt*/e
Run and debug it
Vt* multiply by tau (2pi)
/ divide
e ceiling
answered Nov 21 at 17:29
recursive
4,9941221
4,9941221
add a comment |
add a comment |
up vote
2
down vote
MathGolf, 6 5 bytes
∞/π/ü
Semi-port of @flawr's Python 2 comment.
Takes the input in the order radius distance
.
-1 byte because ceil
builtin has just been added, replacing the floor+1
.
Try it online.
Explanation:
∞ # Double the first (implicit) input
/ # Divide the second (implicit) input by it
π/ # Divide it by PI
ü # Ceil (and output implicitly)
add a comment |
up vote
2
down vote
MathGolf, 6 5 bytes
∞/π/ü
Semi-port of @flawr's Python 2 comment.
Takes the input in the order radius distance
.
-1 byte because ceil
builtin has just been added, replacing the floor+1
.
Try it online.
Explanation:
∞ # Double the first (implicit) input
/ # Divide the second (implicit) input by it
π/ # Divide it by PI
ü # Ceil (and output implicitly)
add a comment |
up vote
2
down vote
up vote
2
down vote
MathGolf, 6 5 bytes
∞/π/ü
Semi-port of @flawr's Python 2 comment.
Takes the input in the order radius distance
.
-1 byte because ceil
builtin has just been added, replacing the floor+1
.
Try it online.
Explanation:
∞ # Double the first (implicit) input
/ # Divide the second (implicit) input by it
π/ # Divide it by PI
ü # Ceil (and output implicitly)
MathGolf, 6 5 bytes
∞/π/ü
Semi-port of @flawr's Python 2 comment.
Takes the input in the order radius distance
.
-1 byte because ceil
builtin has just been added, replacing the floor+1
.
Try it online.
Explanation:
∞ # Double the first (implicit) input
/ # Divide the second (implicit) input by it
π/ # Divide it by PI
ü # Ceil (and output implicitly)
edited Nov 21 at 18:14
answered Nov 21 at 13:25
Kevin Cruijssen
34.4k554182
34.4k554182
add a comment |
add a comment |
up vote
2
down vote
C (gcc), 45 47 45 bytes
f(d,r,R)float d,r;{R=ceil(d/r/'G'/'n'*'q');}
A reasonable approximation of pi is 355/113. Since circumference C = 2 * r * PI, we can instead of pi use tau, which is then of course ~710/113. 710 happens to have the convenient factors 2 * 5 * 71, which is compactly expressed as 'G' * 'n'
. We add one (r/r
) to force rounding to infinity.
Edit: My trick was too clever for its own good: it of course made it fail if the distance was a multiple of the circumference.
Try it online!
add a comment |
up vote
2
down vote
C (gcc), 45 47 45 bytes
f(d,r,R)float d,r;{R=ceil(d/r/'G'/'n'*'q');}
A reasonable approximation of pi is 355/113. Since circumference C = 2 * r * PI, we can instead of pi use tau, which is then of course ~710/113. 710 happens to have the convenient factors 2 * 5 * 71, which is compactly expressed as 'G' * 'n'
. We add one (r/r
) to force rounding to infinity.
Edit: My trick was too clever for its own good: it of course made it fail if the distance was a multiple of the circumference.
Try it online!
add a comment |
up vote
2
down vote
up vote
2
down vote
C (gcc), 45 47 45 bytes
f(d,r,R)float d,r;{R=ceil(d/r/'G'/'n'*'q');}
A reasonable approximation of pi is 355/113. Since circumference C = 2 * r * PI, we can instead of pi use tau, which is then of course ~710/113. 710 happens to have the convenient factors 2 * 5 * 71, which is compactly expressed as 'G' * 'n'
. We add one (r/r
) to force rounding to infinity.
Edit: My trick was too clever for its own good: it of course made it fail if the distance was a multiple of the circumference.
Try it online!
C (gcc), 45 47 45 bytes
f(d,r,R)float d,r;{R=ceil(d/r/'G'/'n'*'q');}
A reasonable approximation of pi is 355/113. Since circumference C = 2 * r * PI, we can instead of pi use tau, which is then of course ~710/113. 710 happens to have the convenient factors 2 * 5 * 71, which is compactly expressed as 'G' * 'n'
. We add one (r/r
) to force rounding to infinity.
Edit: My trick was too clever for its own good: it of course made it fail if the distance was a multiple of the circumference.
Try it online!
edited Nov 21 at 18:24
answered Nov 21 at 17:57
gastropner
1,8701410
1,8701410
add a comment |
add a comment |
up vote
2
down vote
Julia 1.0, 20 bytes
f(d,r)=cld(d/π,r+r)
Try it online!
add a comment |
up vote
2
down vote
Julia 1.0, 20 bytes
f(d,r)=cld(d/π,r+r)
Try it online!
add a comment |
up vote
2
down vote
up vote
2
down vote
Julia 1.0, 20 bytes
f(d,r)=cld(d/π,r+r)
Try it online!
Julia 1.0, 20 bytes
f(d,r)=cld(d/π,r+r)
Try it online!
edited Nov 21 at 21:11
answered Nov 21 at 18:45
gggg
1,22656
1,22656
add a comment |
add a comment |
up vote
2
down vote
R, 39 32 bytes
-7 bytes Thanks to Giuseppe
function(d,r)ceiling(d/(r+r)/pi)
Try it online!
I feel like this could definitely be golfed, but I am a bit lazy right now to do anything about it
add a comment |
up vote
2
down vote
R, 39 32 bytes
-7 bytes Thanks to Giuseppe
function(d,r)ceiling(d/(r+r)/pi)
Try it online!
I feel like this could definitely be golfed, but I am a bit lazy right now to do anything about it
add a comment |
up vote
2
down vote
up vote
2
down vote
R, 39 32 bytes
-7 bytes Thanks to Giuseppe
function(d,r)ceiling(d/(r+r)/pi)
Try it online!
I feel like this could definitely be golfed, but I am a bit lazy right now to do anything about it
R, 39 32 bytes
-7 bytes Thanks to Giuseppe
function(d,r)ceiling(d/(r+r)/pi)
Try it online!
I feel like this could definitely be golfed, but I am a bit lazy right now to do anything about it
edited Nov 21 at 21:45
answered Nov 21 at 20:23
Sumner18
3215
3215
add a comment |
add a comment |
up vote
1
down vote
PHP, 47 bytes
<?=ceil($argv[++$i]/M_PI/(($b=end($argv))+$b));
Try it online.
add a comment |
up vote
1
down vote
PHP, 47 bytes
<?=ceil($argv[++$i]/M_PI/(($b=end($argv))+$b));
Try it online.
add a comment |
up vote
1
down vote
up vote
1
down vote
PHP, 47 bytes
<?=ceil($argv[++$i]/M_PI/(($b=end($argv))+$b));
Try it online.
PHP, 47 bytes
<?=ceil($argv[++$i]/M_PI/(($b=end($argv))+$b));
Try it online.
edited Nov 21 at 13:55
answered Nov 21 at 13:47
Titus
12.9k11237
12.9k11237
add a comment |
add a comment |
up vote
1
down vote
Jelly, 6 bytes
÷÷ØPHĊ
Try it online!
add a comment |
up vote
1
down vote
Jelly, 6 bytes
÷÷ØPHĊ
Try it online!
add a comment |
up vote
1
down vote
up vote
1
down vote
Jelly, 6 bytes
÷÷ØPHĊ
Try it online!
Jelly, 6 bytes
÷÷ØPHĊ
Try it online!
answered Nov 21 at 14:02
Erik the Outgolfer
30.7k429102
30.7k429102
add a comment |
add a comment |
up vote
1
down vote
Ruby, 29 bytes
->l,r{(l/Math::PI/r+=r).ceil}
Try it online!
add a comment |
up vote
1
down vote
Ruby, 29 bytes
->l,r{(l/Math::PI/r+=r).ceil}
Try it online!
add a comment |
up vote
1
down vote
up vote
1
down vote
Ruby, 29 bytes
->l,r{(l/Math::PI/r+=r).ceil}
Try it online!
Ruby, 29 bytes
->l,r{(l/Math::PI/r+=r).ceil}
Try it online!
answered Nov 21 at 14:09
G B
7,5561328
7,5561328
add a comment |
add a comment |
up vote
1
down vote
J, 10 9 bytes
>.@%o.@+:
Try it online!
add a comment |
up vote
1
down vote
J, 10 9 bytes
>.@%o.@+:
Try it online!
add a comment |
up vote
1
down vote
up vote
1
down vote
J, 10 9 bytes
>.@%o.@+:
Try it online!
J, 10 9 bytes
>.@%o.@+:
Try it online!
edited Nov 21 at 15:11
answered Nov 21 at 14:59
Galen Ivanov
5,95711032
5,95711032
add a comment |
add a comment |
up vote
1
down vote
Japt, 7 bytes
/MT/V c
Try it here
add a comment |
up vote
1
down vote
Japt, 7 bytes
/MT/V c
Try it here
add a comment |
up vote
1
down vote
up vote
1
down vote
Japt, 7 bytes
/MT/V c
Try it here
Japt, 7 bytes
/MT/V c
Try it here
answered Nov 21 at 17:06
Shaggy
18.2k21663
18.2k21663
add a comment |
add a comment |
up vote
1
down vote
JavaScript (Babel Node), 25 bytes
-2 bytes using @flawr comment =D. -1 from @Kevin. -7 from @Shaggy
a=>b=>-~(a/(b+b)/Math.PI)
Try it online!
Justa=>b=>Math.ceil(a/(b+b)/Math.PI)
is 32 bytes. :)
– Kevin Cruijssen
Nov 21 at 13:06
25 bytes
– Shaggy
Nov 21 at 17:09
add a comment |
up vote
1
down vote
JavaScript (Babel Node), 25 bytes
-2 bytes using @flawr comment =D. -1 from @Kevin. -7 from @Shaggy
a=>b=>-~(a/(b+b)/Math.PI)
Try it online!
Justa=>b=>Math.ceil(a/(b+b)/Math.PI)
is 32 bytes. :)
– Kevin Cruijssen
Nov 21 at 13:06
25 bytes
– Shaggy
Nov 21 at 17:09
add a comment |
up vote
1
down vote
up vote
1
down vote
JavaScript (Babel Node), 25 bytes
-2 bytes using @flawr comment =D. -1 from @Kevin. -7 from @Shaggy
a=>b=>-~(a/(b+b)/Math.PI)
Try it online!
JavaScript (Babel Node), 25 bytes
-2 bytes using @flawr comment =D. -1 from @Kevin. -7 from @Shaggy
a=>b=>-~(a/(b+b)/Math.PI)
Try it online!
edited Nov 21 at 18:01
answered Nov 21 at 12:52
Luis felipe De jesus Munoz
4,01421253
4,01421253
Justa=>b=>Math.ceil(a/(b+b)/Math.PI)
is 32 bytes. :)
– Kevin Cruijssen
Nov 21 at 13:06
25 bytes
– Shaggy
Nov 21 at 17:09
add a comment |
Justa=>b=>Math.ceil(a/(b+b)/Math.PI)
is 32 bytes. :)
– Kevin Cruijssen
Nov 21 at 13:06
25 bytes
– Shaggy
Nov 21 at 17:09
Just
a=>b=>Math.ceil(a/(b+b)/Math.PI)
is 32 bytes. :)– Kevin Cruijssen
Nov 21 at 13:06
Just
a=>b=>Math.ceil(a/(b+b)/Math.PI)
is 32 bytes. :)– Kevin Cruijssen
Nov 21 at 13:06
25 bytes
– Shaggy
Nov 21 at 17:09
25 bytes
– Shaggy
Nov 21 at 17:09
add a comment |
up vote
1
down vote
min, 16 bytes
/ tau / ceil int
Takes the distance and radius put on the stack in that order. Then divides by tau, rounds, and makes int.
add a comment |
up vote
1
down vote
min, 16 bytes
/ tau / ceil int
Takes the distance and radius put on the stack in that order. Then divides by tau, rounds, and makes int.
add a comment |
up vote
1
down vote
up vote
1
down vote
min, 16 bytes
/ tau / ceil int
Takes the distance and radius put on the stack in that order. Then divides by tau, rounds, and makes int.
min, 16 bytes
/ tau / ceil int
Takes the distance and radius put on the stack in that order. Then divides by tau, rounds, and makes int.
answered Nov 22 at 11:53
Panda0nEarth
1013
1013
add a comment |
add a comment |
up vote
1
down vote
Dart, 47 46 bytes
import'dart:math';f(a,b)=>(a/(b+b)/pi).ceil();
Try it online!
- -1 byte thanks to @Shaggy
46 bytes
– Shaggy
Nov 22 at 12:30
add a comment |
up vote
1
down vote
Dart, 47 46 bytes
import'dart:math';f(a,b)=>(a/(b+b)/pi).ceil();
Try it online!
- -1 byte thanks to @Shaggy
46 bytes
– Shaggy
Nov 22 at 12:30
add a comment |
up vote
1
down vote
up vote
1
down vote
Dart, 47 46 bytes
import'dart:math';f(a,b)=>(a/(b+b)/pi).ceil();
Try it online!
- -1 byte thanks to @Shaggy
Dart, 47 46 bytes
import'dart:math';f(a,b)=>(a/(b+b)/pi).ceil();
Try it online!
- -1 byte thanks to @Shaggy
edited Nov 22 at 12:42
answered Nov 22 at 7:31
Elcan
30115
30115
46 bytes
– Shaggy
Nov 22 at 12:30
add a comment |
46 bytes
– Shaggy
Nov 22 at 12:30
46 bytes
– Shaggy
Nov 22 at 12:30
46 bytes
– Shaggy
Nov 22 at 12:30
add a comment |
up vote
1
down vote
JavaScript (Babel Node), 23 bytes
s=>r=>-~(s/2/r/Math.PI)
Try it online!
2
There must not be any digits 0-9 in your code.
– Dennis♦
Nov 22 at 12:55
add a comment |
up vote
1
down vote
JavaScript (Babel Node), 23 bytes
s=>r=>-~(s/2/r/Math.PI)
Try it online!
2
There must not be any digits 0-9 in your code.
– Dennis♦
Nov 22 at 12:55
add a comment |
up vote
1
down vote
up vote
1
down vote
JavaScript (Babel Node), 23 bytes
s=>r=>-~(s/2/r/Math.PI)
Try it online!
JavaScript (Babel Node), 23 bytes
s=>r=>-~(s/2/r/Math.PI)
Try it online!
answered Nov 22 at 12:43
Igor Sowinski
211
211
2
There must not be any digits 0-9 in your code.
– Dennis♦
Nov 22 at 12:55
add a comment |
2
There must not be any digits 0-9 in your code.
– Dennis♦
Nov 22 at 12:55
2
2
There must not be any digits 0-9 in your code.
– Dennis♦
Nov 22 at 12:55
There must not be any digits 0-9 in your code.
– Dennis♦
Nov 22 at 12:55
add a comment |
up vote
1
down vote
Haskell, 25 bytes
f d r=ceiling(d/(r+r)/pi)
New contributor
add a comment |
up vote
1
down vote
Haskell, 25 bytes
f d r=ceiling(d/(r+r)/pi)
New contributor
add a comment |
up vote
1
down vote
up vote
1
down vote
Haskell, 25 bytes
f d r=ceiling(d/(r+r)/pi)
New contributor
Haskell, 25 bytes
f d r=ceiling(d/(r+r)/pi)
New contributor
edited Nov 22 at 12:59
New contributor
answered Nov 22 at 12:32
memo
1113
1113
New contributor
New contributor
add a comment |
add a comment |
up vote
1
down vote
Lua, 61 58 57 49 bytes
function(s,r)return math.ceil(s/(r+r)/math.pi)end
Try it online!
Thanks to KirillL. -8 bytes.
I don't know much Lua (so maybe it's still too long), but it appears to be shorter as a function: 49 bytes
– Kirill L.
Nov 22 at 11:18
@KirillL., I'm still learning the rules here. The OP's challenge is pretty open on the input. So my question is, would we have to count your program call() against the byte count? If not, your's definitely shaves off a nice chunk.
– ouflak
Nov 22 at 11:43
A quite common style of submission here is an anonymous function (so that we don't have to count the name, unless it is recursive), which outputs by its return value. The footer section with function calls and actual printing to console is then basically used for visualizing the results and doesn't count towards your score. BTW, you may add more of the OP's test examples to the footer, so that they can be conveniently viewed all at once. Note that in some cases a full program may actually turn out to be golfier!
– Kirill L.
Nov 22 at 11:58
add a comment |
up vote
1
down vote
Lua, 61 58 57 49 bytes
function(s,r)return math.ceil(s/(r+r)/math.pi)end
Try it online!
Thanks to KirillL. -8 bytes.
I don't know much Lua (so maybe it's still too long), but it appears to be shorter as a function: 49 bytes
– Kirill L.
Nov 22 at 11:18
@KirillL., I'm still learning the rules here. The OP's challenge is pretty open on the input. So my question is, would we have to count your program call() against the byte count? If not, your's definitely shaves off a nice chunk.
– ouflak
Nov 22 at 11:43
A quite common style of submission here is an anonymous function (so that we don't have to count the name, unless it is recursive), which outputs by its return value. The footer section with function calls and actual printing to console is then basically used for visualizing the results and doesn't count towards your score. BTW, you may add more of the OP's test examples to the footer, so that they can be conveniently viewed all at once. Note that in some cases a full program may actually turn out to be golfier!
– Kirill L.
Nov 22 at 11:58
add a comment |
up vote
1
down vote
up vote
1
down vote
Lua, 61 58 57 49 bytes
function(s,r)return math.ceil(s/(r+r)/math.pi)end
Try it online!
Thanks to KirillL. -8 bytes.
Lua, 61 58 57 49 bytes
function(s,r)return math.ceil(s/(r+r)/math.pi)end
Try it online!
Thanks to KirillL. -8 bytes.
edited Nov 22 at 13:12
answered Nov 21 at 14:49
ouflak
171310
171310
I don't know much Lua (so maybe it's still too long), but it appears to be shorter as a function: 49 bytes
– Kirill L.
Nov 22 at 11:18
@KirillL., I'm still learning the rules here. The OP's challenge is pretty open on the input. So my question is, would we have to count your program call() against the byte count? If not, your's definitely shaves off a nice chunk.
– ouflak
Nov 22 at 11:43
A quite common style of submission here is an anonymous function (so that we don't have to count the name, unless it is recursive), which outputs by its return value. The footer section with function calls and actual printing to console is then basically used for visualizing the results and doesn't count towards your score. BTW, you may add more of the OP's test examples to the footer, so that they can be conveniently viewed all at once. Note that in some cases a full program may actually turn out to be golfier!
– Kirill L.
Nov 22 at 11:58
add a comment |
I don't know much Lua (so maybe it's still too long), but it appears to be shorter as a function: 49 bytes
– Kirill L.
Nov 22 at 11:18
@KirillL., I'm still learning the rules here. The OP's challenge is pretty open on the input. So my question is, would we have to count your program call() against the byte count? If not, your's definitely shaves off a nice chunk.
– ouflak
Nov 22 at 11:43
A quite common style of submission here is an anonymous function (so that we don't have to count the name, unless it is recursive), which outputs by its return value. The footer section with function calls and actual printing to console is then basically used for visualizing the results and doesn't count towards your score. BTW, you may add more of the OP's test examples to the footer, so that they can be conveniently viewed all at once. Note that in some cases a full program may actually turn out to be golfier!
– Kirill L.
Nov 22 at 11:58
I don't know much Lua (so maybe it's still too long), but it appears to be shorter as a function: 49 bytes
– Kirill L.
Nov 22 at 11:18
I don't know much Lua (so maybe it's still too long), but it appears to be shorter as a function: 49 bytes
– Kirill L.
Nov 22 at 11:18
@KirillL., I'm still learning the rules here. The OP's challenge is pretty open on the input. So my question is, would we have to count your program call() against the byte count? If not, your's definitely shaves off a nice chunk.
– ouflak
Nov 22 at 11:43
@KirillL., I'm still learning the rules here. The OP's challenge is pretty open on the input. So my question is, would we have to count your program call() against the byte count? If not, your's definitely shaves off a nice chunk.
– ouflak
Nov 22 at 11:43
A quite common style of submission here is an anonymous function (so that we don't have to count the name, unless it is recursive), which outputs by its return value. The footer section with function calls and actual printing to console is then basically used for visualizing the results and doesn't count towards your score. BTW, you may add more of the OP's test examples to the footer, so that they can be conveniently viewed all at once. Note that in some cases a full program may actually turn out to be golfier!
– Kirill L.
Nov 22 at 11:58
A quite common style of submission here is an anonymous function (so that we don't have to count the name, unless it is recursive), which outputs by its return value. The footer section with function calls and actual printing to console is then basically used for visualizing the results and doesn't count towards your score. BTW, you may add more of the OP's test examples to the footer, so that they can be conveniently viewed all at once. Note that in some cases a full program may actually turn out to be golfier!
– Kirill L.
Nov 22 at 11:58
add a comment |
up vote
1
down vote
Common Lisp, 36 bytes
(lambda(a b)(ceiling(/ a(+ b b)pi)))
Try it online!
add a comment |
up vote
1
down vote
Common Lisp, 36 bytes
(lambda(a b)(ceiling(/ a(+ b b)pi)))
Try it online!
add a comment |
up vote
1
down vote
up vote
1
down vote
Common Lisp, 36 bytes
(lambda(a b)(ceiling(/ a(+ b b)pi)))
Try it online!
Common Lisp, 36 bytes
(lambda(a b)(ceiling(/ a(+ b b)pi)))
Try it online!
edited Nov 22 at 15:14
answered Nov 22 at 15:07
Renzo
1,600516
1,600516
add a comment |
add a comment |
up vote
1
down vote
Tcl, 50 bytes
proc N d r {expr ceil($d/(($r+$r)*acos(-$r/$r)))}
Try it online!
Tcl, 53 bytes
proc N d r {expr ceil($d/(($r+$r)*acos(-[incr i])))}
Try it online!
Lack of a pi constant or function makes me lose the golf competition!
Do I need to remove the .0 at end of each output? It would make me consume more bytes!
– sergiol
Nov 22 at 18:55
1
[incr i]
is quite clever but I think you can use$d/$d
or$r/$r
instead.
– david
Nov 22 at 20:17
Saved some bytes thanks to @david's idea!
– sergiol
Nov 22 at 22:56
add a comment |
up vote
1
down vote
Tcl, 50 bytes
proc N d r {expr ceil($d/(($r+$r)*acos(-$r/$r)))}
Try it online!
Tcl, 53 bytes
proc N d r {expr ceil($d/(($r+$r)*acos(-[incr i])))}
Try it online!
Lack of a pi constant or function makes me lose the golf competition!
Do I need to remove the .0 at end of each output? It would make me consume more bytes!
– sergiol
Nov 22 at 18:55
1
[incr i]
is quite clever but I think you can use$d/$d
or$r/$r
instead.
– david
Nov 22 at 20:17
Saved some bytes thanks to @david's idea!
– sergiol
Nov 22 at 22:56
add a comment |
up vote
1
down vote
up vote
1
down vote
Tcl, 50 bytes
proc N d r {expr ceil($d/(($r+$r)*acos(-$r/$r)))}
Try it online!
Tcl, 53 bytes
proc N d r {expr ceil($d/(($r+$r)*acos(-[incr i])))}
Try it online!
Lack of a pi constant or function makes me lose the golf competition!
Tcl, 50 bytes
proc N d r {expr ceil($d/(($r+$r)*acos(-$r/$r)))}
Try it online!
Tcl, 53 bytes
proc N d r {expr ceil($d/(($r+$r)*acos(-[incr i])))}
Try it online!
Lack of a pi constant or function makes me lose the golf competition!
edited Nov 22 at 22:56
answered Nov 22 at 18:55
sergiol
2,2921825
2,2921825
Do I need to remove the .0 at end of each output? It would make me consume more bytes!
– sergiol
Nov 22 at 18:55
1
[incr i]
is quite clever but I think you can use$d/$d
or$r/$r
instead.
– david
Nov 22 at 20:17
Saved some bytes thanks to @david's idea!
– sergiol
Nov 22 at 22:56
add a comment |
Do I need to remove the .0 at end of each output? It would make me consume more bytes!
– sergiol
Nov 22 at 18:55
1
[incr i]
is quite clever but I think you can use$d/$d
or$r/$r
instead.
– david
Nov 22 at 20:17
Saved some bytes thanks to @david's idea!
– sergiol
Nov 22 at 22:56
Do I need to remove the .0 at end of each output? It would make me consume more bytes!
– sergiol
Nov 22 at 18:55
Do I need to remove the .0 at end of each output? It would make me consume more bytes!
– sergiol
Nov 22 at 18:55
1
1
[incr i]
is quite clever but I think you can use $d/$d
or $r/$r
instead.– david
Nov 22 at 20:17
[incr i]
is quite clever but I think you can use $d/$d
or $r/$r
instead.– david
Nov 22 at 20:17
Saved some bytes thanks to @david's idea!
– sergiol
Nov 22 at 22:56
Saved some bytes thanks to @david's idea!
– sergiol
Nov 22 at 22:56
add a comment |
up vote
1
down vote
PowerShell, 53 52 bytes
-1 byte thanks to @mazzy
param($d,$r);($a=[math])::ceiling($d/($r+$r)/$a::pi)
Takes input from two commandline parameters, distance -d
and radius -r
.
?param($d,$r);($a=[math])::ceiling($d/($r+$r)/$a::pi)
– mazzy
2 days ago
add a comment |
up vote
1
down vote
PowerShell, 53 52 bytes
-1 byte thanks to @mazzy
param($d,$r);($a=[math])::ceiling($d/($r+$r)/$a::pi)
Takes input from two commandline parameters, distance -d
and radius -r
.
?param($d,$r);($a=[math])::ceiling($d/($r+$r)/$a::pi)
– mazzy
2 days ago
add a comment |
up vote
1
down vote
up vote
1
down vote
PowerShell, 53 52 bytes
-1 byte thanks to @mazzy
param($d,$r);($a=[math])::ceiling($d/($r+$r)/$a::pi)
Takes input from two commandline parameters, distance -d
and radius -r
.
PowerShell, 53 52 bytes
-1 byte thanks to @mazzy
param($d,$r);($a=[math])::ceiling($d/($r+$r)/$a::pi)
Takes input from two commandline parameters, distance -d
and radius -r
.
edited 2 days ago
answered Nov 24 at 2:05
Gabriel Mills
915
915
?param($d,$r);($a=[math])::ceiling($d/($r+$r)/$a::pi)
– mazzy
2 days ago
add a comment |
?param($d,$r);($a=[math])::ceiling($d/($r+$r)/$a::pi)
– mazzy
2 days ago
?
param($d,$r);($a=[math])::ceiling($d/($r+$r)/$a::pi)
– mazzy
2 days ago
?
param($d,$r);($a=[math])::ceiling($d/($r+$r)/$a::pi)
– mazzy
2 days ago
add a comment |
up vote
0
down vote
Mathematica, 8 bytes
#/(2π r)&
So that
r =22.9;
#/(2π r)& @ 32
gives
(* 0.2224 *)
2
There must not be any digits 0-9 in your code. Also, could you please show how to call your function, maybe in an online interpreter like tio.run/#mathematica? I'm not sure how to supply ther
.
– Dennis♦
Nov 22 at 13:03
add a comment |
up vote
0
down vote
Mathematica, 8 bytes
#/(2π r)&
So that
r =22.9;
#/(2π r)& @ 32
gives
(* 0.2224 *)
2
There must not be any digits 0-9 in your code. Also, could you please show how to call your function, maybe in an online interpreter like tio.run/#mathematica? I'm not sure how to supply ther
.
– Dennis♦
Nov 22 at 13:03
add a comment |
up vote
0
down vote
up vote
0
down vote
Mathematica, 8 bytes
#/(2π r)&
So that
r =22.9;
#/(2π r)& @ 32
gives
(* 0.2224 *)
Mathematica, 8 bytes
#/(2π r)&
So that
r =22.9;
#/(2π r)& @ 32
gives
(* 0.2224 *)
edited Nov 22 at 19:08
answered Nov 22 at 7:59
David G. Stork
1857
1857
2
There must not be any digits 0-9 in your code. Also, could you please show how to call your function, maybe in an online interpreter like tio.run/#mathematica? I'm not sure how to supply ther
.
– Dennis♦
Nov 22 at 13:03
add a comment |
2
There must not be any digits 0-9 in your code. Also, could you please show how to call your function, maybe in an online interpreter like tio.run/#mathematica? I'm not sure how to supply ther
.
– Dennis♦
Nov 22 at 13:03
2
2
There must not be any digits 0-9 in your code. Also, could you please show how to call your function, maybe in an online interpreter like tio.run/#mathematica? I'm not sure how to supply the
r
.– Dennis♦
Nov 22 at 13:03
There must not be any digits 0-9 in your code. Also, could you please show how to call your function, maybe in an online interpreter like tio.run/#mathematica? I'm not sure how to supply the
r
.– Dennis♦
Nov 22 at 13:03
add a comment |
up vote
0
down vote
Clojure, 50 bytes
(fn[a b](int(Math/ceil(/ a Math/PI(count" ")b))))
An anonymous function that accepts two integers a
and b
as arguments: the distance and the wheel's radius, respectively.
Try it online!
(count " ")
evaluates to 2
, so this function implements $lceil dfrac a{2pi b} rceil$.
add a comment |
up vote
0
down vote
Clojure, 50 bytes
(fn[a b](int(Math/ceil(/ a Math/PI(count" ")b))))
An anonymous function that accepts two integers a
and b
as arguments: the distance and the wheel's radius, respectively.
Try it online!
(count " ")
evaluates to 2
, so this function implements $lceil dfrac a{2pi b} rceil$.
add a comment |
up vote
0
down vote
up vote
0
down vote
Clojure, 50 bytes
(fn[a b](int(Math/ceil(/ a Math/PI(count" ")b))))
An anonymous function that accepts two integers a
and b
as arguments: the distance and the wheel's radius, respectively.
Try it online!
(count " ")
evaluates to 2
, so this function implements $lceil dfrac a{2pi b} rceil$.
Clojure, 50 bytes
(fn[a b](int(Math/ceil(/ a Math/PI(count" ")b))))
An anonymous function that accepts two integers a
and b
as arguments: the distance and the wheel's radius, respectively.
Try it online!
(count " ")
evaluates to 2
, so this function implements $lceil dfrac a{2pi b} rceil$.
answered Nov 23 at 2:02
TheGreatGeek
614
614
add a comment |
add a comment |
up vote
0
down vote
TI-Basic (83 series), 12 bytes
-int(-Tmax⁻¹min(e^(ΔList(ln(Ans
Takes input as a list of radius and distance in Ans
: for example, {0.9999:12.5663:prgmX
.
e^(ΔList(ln(Ans
will take the ratio of those distances, and min(
turns this into a number. Then we divide by Tmax
, which is a graphing parameter that's equal to 2π by default. Finally, -int(-
takes the ceiling.
add a comment |
up vote
0
down vote
TI-Basic (83 series), 12 bytes
-int(-Tmax⁻¹min(e^(ΔList(ln(Ans
Takes input as a list of radius and distance in Ans
: for example, {0.9999:12.5663:prgmX
.
e^(ΔList(ln(Ans
will take the ratio of those distances, and min(
turns this into a number. Then we divide by Tmax
, which is a graphing parameter that's equal to 2π by default. Finally, -int(-
takes the ceiling.
add a comment |
up vote
0
down vote
up vote
0
down vote
TI-Basic (83 series), 12 bytes
-int(-Tmax⁻¹min(e^(ΔList(ln(Ans
Takes input as a list of radius and distance in Ans
: for example, {0.9999:12.5663:prgmX
.
e^(ΔList(ln(Ans
will take the ratio of those distances, and min(
turns this into a number. Then we divide by Tmax
, which is a graphing parameter that's equal to 2π by default. Finally, -int(-
takes the ceiling.
TI-Basic (83 series), 12 bytes
-int(-Tmax⁻¹min(e^(ΔList(ln(Ans
Takes input as a list of radius and distance in Ans
: for example, {0.9999:12.5663:prgmX
.
e^(ΔList(ln(Ans
will take the ratio of those distances, and min(
turns this into a number. Then we divide by Tmax
, which is a graphing parameter that's equal to 2π by default. Finally, -int(-
takes the ceiling.
answered yesterday
Misha Lavrov
4,080424
4,080424
add a comment |
add a comment |
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4
You probably should add that digits are also forbidden in compiler options (or anywhere else): if you limit this constraint to code only, with gcc we can do something like
-DP=3.14
in compiler flags, that would defineP
as an approximation of pi, which is probably not what you intended– Annyo
Nov 21 at 16:50