New answers tagged integer
0
Julia, 7 bytes
ndigits
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0
Julia, 7 bytes
ndigits
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1
PowerShell Core, 20 bytes
$args|% *m '-'|% Le*
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0
Arn -s, 2 bytes
.|
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Take absolute value, -s takes size. Implicit casting.
3
Haskell, 34 bytes
signum.minimum.(zipWith(-)<*>tail)
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Outputs 1,0,-1 for strictly monotonic, non-increasing, and otherwise, respectively.
Takes the sign of the minimum of the differences between adjacent pairs.
(zipWith(-)<*>tail applied to l computes l - tail l elementwise).
This is competitive with the existing Haskell answers ...
0
Python 3, 230 bytes
def f(n):
s="1"
while 2>1:
if int(s)%n==0:
return s
if "0" in s:
for j in range(0,len(s)):
if s[j]=="0":
q=j
t=s[:q]
for k in range(q,len(s)):
t=t+str(1-int(s[k]))
s=t
else:
s=str(10**len(s))
0
Whispers v2, 39 bytes
> Input
>> |1|
>> "2"
>> #3
>> Output 4
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Boring answer. Get input, absolute value, convert to string, return length.
2
V (vim), 20 bytes
:s/-
C<C-r>=strlen(<C-r>")
<esc>
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Remove minuses(if any), cut the line, and get its length.
Not sure if I need the <esc> at the end.
0
R, 38 bytes
f=function(x)`if`(x^2<100,1,1+f(x/10))
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Unusually for R, a recursive function seems to be shorter here than the previous answer using built-in functions (although this is mostly gained simply because we can skip abs(x) by using x^2<100 as a condition instead)...
2
Japt, 12 bytes
Based off Etheryte's solution, posted with permission.
_øB ªU´}f¤ÔÄ
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3
Husk, 9 7 bytes
ḟoΛεd∫∞
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-2 bytes from Leo.
1
PowerShell, 41 bytes
for(;++$n%"$args"-or$n-match'[2-9]'){}
$n
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PowerShell, 41 bytes
for(;++$n%"$args"-or$n-replace'0|1'){}
$n
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1
PowerShell, 74 bytes
param($x)for(){$n=iex ([convert]::tostring(++$i,2));if(!($n%$x)){$n;exit}}
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0
Husk, 9 bytes
#ṗuṁS:Lgp
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Slight improvement over Mr. XCoder's answer.
0
APL(Dyalog Extended), 13 bytes SBCS
+/×/¨0j1×⊆⍨⊤⎕
Try it on APLgolf!
A tradfn submission which takes a decimal number.
1
Ruby, 112 ... 102 bytes
->m{1until m==[:min,:max].map{|x|m=m.transpose|[];m-=m.combination(2).map{|a,b|a.zip(b).map &x}}[1];m}
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2
jq (74 bytes)
Since shortness is the goal ...
def f:def i:.,i|([0]+.,[1]+.);nth(.-1;[1]|i|select(index([1,1])|not))+[1];
Example:
65|f #=> [0,1,0,0,1,0,0,0,1,1]
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1
Charcoal, 40 bytes
Ｆ⊗Ｌ⁺θ§θ⁰≔Ｅ§θ⁰±ＥΦθ⬤θ⎇⁼μνπ⊙μ‹ρ§ξς§μλθＩθ
Try it online! Link is to verbose version of code. Outputs using Charcoal's default format (each cell on its own line with rows double-spaced from each other). Explanation:
Ｆ⊗Ｌ⁺θ§θ⁰
Loop a large enough number of times (twice the number of rows and columns; probably overkill, so 4 bytes could be ...
2
Husk, 14 12 bytes
Edit: -2 bytes thanks to Leo
↔:1!fo¬V&mḋN
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Based on Etheryte's approach: upvote that!
5
J, 43 41 40 39 bytes
f=.(#~1=1#.]*/ .>:|:)@~.
f&.:-&.|:@f^:_
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tldr how
Create a function f to filter out dominated rows.
Apply plain f, followed by f "under" a negative transpose, until we reach a fixed point.
4
Jelly, 22 21 bytes
Qµ>Ạ€TLʋ€’ỊTịNZ
Ç⁺$ƬṪ
Try it online! Or see the test-suite.
How?
Ç⁺$ƬṪ - Link: matrix
Ƭ - collect up inputs to this until no change:
$ - last two links as a monad:
Ç - call Link 1 (see below)
⁺ - repeat that
Ṫ - tail
Qµ>Ạ€TLʋ€’ỊTịNZ - Link 1: matrix
Q - de-duplicate (since one of any ...
3
Haskell, 106 101 100 bytes
-5 bytes thanks to kops.
until=<<((==)<*>)$g.g
g m=transpose.map(map(0-))$nub m\\(m>>=tail.mapM(\i->[i..9]))
import Data.List
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The relevant function is until=<<((==)<*>)$g.g, which takes as input a matrix in the format described in the statement (not transposed and without flipped ...
2
Python 3, 15 bytes (Courtesy of @Makonede)
print(*input())
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Python 3, 24 bytes (~0.03s)
print(' '.join(input()))
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Should be compact and fast
PowerShell, 23 bytes
("$args"|% t*y)-join' '
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This dosen't deserve seperate answer
1
Stax, 16 bytes
è}╘⌐ƒw┴L≤⌂IÑ═eαü
Run and debug it
3
Factor, 52 bytes
{ 1/2 } [ dup stack. dup 1 v+n append 2 v/n t ] loop
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Prints each fraction on its own line forever. Keeps track of the array of all terms with equal denominator.
{ 1/2 } ! Push the initial array
[ ... t ] loop ! Loop forever:
dup stack. ! Print all fractions separated by a newline
dup 1 v+n ! Copy and ...
1
Mini-Flak (BrainHack), 104 bytes
([{}]((()))){({}(())({{}[()](({}({}))){(()[{}]){(({}{}[()]))}}{}}{}(())){[({})]({}[()](()[()]))}{}{})}{}
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Starts at 11 and iteratively computes the encoding of the next number n-1 times.
# Create 11 while computing 1-n.
([{}]((())))
# Repeat
{
# Add 1 to counter (going up to zero) and allow loop to start
(...
0
R, 21 bytes, no builtin
x=scan();x/abs(x^!!x)
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Takis input from STDIN. Avoids the builtin sign. This comes out 4 bytes shorter than JAD's solution with if. We cannot use directly x/abs(x) since we would divide by 0 when x=0.
After coercion to integer, !!x will be equal to 1 for all input except 0 (when it is equal to 0). Since 0^0=1, this ...
0
Python 3.9, 20 bytes
import math
math.lcm
Now Python 3.9 has a builtin in math module to calculate LCM of two or more integers.
4
Python 3.8 (pre-release), 51 bytes
Another based on Etheryte's Japt answer
f=lambda n,k=0:-n*f'1{k-1:b}'or f(n-(k&k//2<1),k+1)
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2
Red, 135 124 bytes
func[n][i: k: 0 until[i: i + 1 unless find do
b:[enbase/base to#{01}i 2]"11"[k: k + 1]n = k]reverse
rejoin["1"find do b"1"]]
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Uses @Etheryte's method
As an excercise - the step by step method:
Red, 153 bytes
func[n][c: charset 1 until[k: 0 a: 0 b: 1 until[k: k + 1
t: a a: b b: t + b b > n]c/(...
2
JavaScript (Node.js), 43 bytes
n=>(g=p=>n<p||2*g(p+q,q=p)+(n>(n%=p)))(q=1)
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It returns an integer whose binary form is reversed Fibonacci Encoding.
// I have a code for base 2 convention
n=>(g=p=>n<p 2*g( )+(n>(n%=p)))( )
// I have a code for Fibonacci sequence
n=>(g=p=> g(p+q,q=p) )(q=1)
...
8
C (gcc), 51 50 bytes
k;f(n){for(k=0;n-=!(++k&k/2););n=log2(k);k+=2<<n;}
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Uses the formula from Etheryte's Japt answer.
Returns the Fibonacci code word for input \$n\$ in reversed binary.
Explanation
k;f(n){ // function taking an integer n > 0
for(k=0; ;); // loop over values of ...
1
Charcoal, 38 bytes
ＮθＦ²⊞υ⊕ιＷ‹⌈υθ⊞υΣ…⮌υ²←1ＦΦ⮌υκ«←Ｉ¬›ιθ≧﹪ιθ
Try it online! Link is to verbose version of code. Explanation: Outputs digits in reverse order, but this is the least of Charcoal's problems because it can simply print them backwards.
Ｎθ
Input n.
Ｆ²⊞υ⊕ι
Start with 1 and 2 from the Fibonacci sequence.
Ｗ‹⌈υθ⊞υΣ…⮌υ²
Keep pushing terms until one ...
2
05AB1E, 14 bytes
∞ʒb11å>}s<èb1ì
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∞ʒb11å>}s<èb1ì # full program
<è # push the...
s # implicit input...
<è # -th...
∞ # natural number...
ʒ } # which...
> # does not...
å # contain...
11 # literal...
b # ...
1
Retina 0.8.2, 63 bytes
.+
$*
(\G1|(?>\2?)(\1))*1
$#1$*01¶
+`^(.*).(.*¶)(\1.)¶
$3$2
¶
1
Try it online! Link includes test cases. Explanation:
.+
$*
Convert to unary.
(\G1|(?>\2?)(\1))*1
Use a tweaked version of @MartinEnder's answer to Am I a Fibonacci Number? to match Fibonacci numbers that sum to the input.
$#1$*01¶
For each match, generate the ...
3
Scala, 69 bytes
Based on Etheryte's amazing answer
"1"+Stream.from(0).map(_.toBinaryString).filter(!_.contains("11"))(_)
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Outputs a String (reversed)
Original answer, 123 bytes
n=>(f.takeWhile(n>=_):\(n,1::Nil)){case(x,n->c)=>if(x>n)(n,0::c)else(n-x,1::c)}._2
def f:Stream[Int]=1#::f.scanLeft(2)(_+_)
Try it ...
18
Japt, 13 bytes
@!X¤øB}iU
¤ÔÄ
Explanation:
@!X¤øB}iU
@ }iU // Get the U-th (input) number
! // which doesn't have
øB // "11" in its
X¤ // binary representation
¤ÔÄ
¤ // Take that number's binary representation,
Ô // reverse it
Ä // and add one to the end.
Try it here.
3
Stax, 16 bytes
âî}g♥▼«zΩ╥;QbB]╔
Run and debug it
Based on Etheryte's observation, which shaved a off a ton of bytes. Check out their answer!
Explanation
Z{|B.11#!}{ge|B'1s+
Z push 0 under the input(counter)
ge generator: find the first n elements, return the last
{ empty block: get the next element by ...
6
Haskell, 76 bytes
1#2
(u#v)x|x<v=["11"|x==u]|z<-u+v=map('0':)(v#z$x)++map("10"++)(z#(z+v)$x-u)
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Outputs a string of 0's and 1's wrapped in a singleton list (e.g. ["001011"]).
How?
(u#v)x: we want this function to return a list of all the possible Fibonacci encodings for x using only Fibonacci numbers ...
2
Haskell (GHC), 118 104 bytes
x@(_:v)=1:scanl(+)1x
f n=snd(foldr(\v(x,l)->if v<=x then(x-v,1:l)else(x,0:l))(n,[])$fst$span(<=n)v)++[1]
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Original:
Outputs a list of reversed bits.
import Data.List
x@(_:v)=1:scanl(+)1x
f n=1:snd(mapAccumL(\a b->if b<=a then(a-b,1)else(a,0))n$reverse$fst$span(<=n)v)
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2
Wolfram Language (Mathematica), 127 bytes
(f@n_:=(k=1;While[(F=Fibonacci)[++k]<=n];k-1);s=Table[0,f[t=#]];(s[[#]]=1)&/@(f/@NestWhileList[#-F@f@#&,t,#>0&]);RotateLeft@s)&
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5
JavaScript (ES6), 46 44 bytes
Based on @Etheryte's insight.
Returns a string of bits in reverse order.
f=(n,k)=>k&k/2||n--?f(n,-~k):1+k.toString(2)
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Commented
f = ( // f is a recursive function taking:
n, // n = input
k // k = counter, initially undefined (coerced
) => // ...
6
Jelly, 21 12 bytes
Using Etheryte's observation from their Japt answer is much terser and more efficient to boot (it may be possible in even less, let's see...)
1BSƝỊẠƲ#ṪB1;
A monadic Link accepting a positive integer that yields a list of 1s and 0s (in reversed order as allowed in the rules).
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A faster 12 byter:
1Ḥ^:3ƲƑ#ṪB1;
Try it online!...
0
Python 3, 172 bytes
def z(n):
if n==0:return [0]
y=[2,1]
while y[0]<n:y[0:0]=[sum(y[:2])]
d=[]
for f in y:
if f<=n:d,n=d+[1],n-f
else:d+=[0]
return ''.join(map(str,d[::-1]))[:-1]+'1'
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2
Factor, 68 66 bytes
Saved 2 bytes thanks to @chunes
[ 1 2 [ 2dup / . [ 2 + ] dip 2dup > [ 2 * 1 swap ] when t ] loop ]
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Outputs the sequence infinitely.
[
1 2 ! First numerator and denominator
[ ! Loop body
2dup / ! Divide to get next element of sequence
. !...
1
Factor + math.unicode, 30 29 24 bytes
Saved 1 byte thanks to @Bubbler
Saved 5 bytes thanks to @chunes!
[ 1 rot neg <range> Π ]
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Takes k (which -th factorial) and then n. First we push an n onto the stack, then rot makes the stack look like n 1 k. neg negates k, which will act as the step for a range from n to 1 (made using <range>...
2
BRASCA, 8 bytes
,'/-1$%n
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Explanation
, - Reverse the stack to take the first character from input
'/- - Subtract it by 47
1$% - 1 % that value
n - And output it
0
x86_64 machine code - 26 24 bytes
This is callable function, input in r8 output in rax.
-3 bytes using cbw instruction is 2 bytes instead of movzx rax, al is 4 bytes.
0000000000400080 <output_the_sign>:
400080: 48 31 c0 xor rax,rax
400083: 4d 85 c0 test r8,r8
400086: 74 11 je 400099 &...
2
BRASCA, 15 bytes
Ci,:::SSL{+,SS=
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Explanation
Ci - Set implicit output to output numbers, and turn "0-9" into 0-9
, - Reverse stack
:::SSL{+ - First digit of input, concatenated to itself twice, + 12
,SS= - Concatenate the digits of the original, then check if both numbers are ...
3
Jelly, 11 bytes
Obligatory Jelly answer.
ḃØ⁵ịØJ⁾“»jV
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How?
ḃØ⁵ịØJ⁾“»jV - Link: integer, n
ḃ - convert (n) to bijective base:
Ø⁵ - 250
ị - index into:
ØJ - Jelly's code-page
⁾“» - list of characters = "“»"
j - join -> "“...»"
V - evaluate as Jelly code
Top 50 recent answers are included
