New answers tagged binary
3
Red, 79 bytes
func[x][p: 0 x: replace/all x"111111""1111110"forall x[prin p: 49 - x/1 xor p]]
Try it online!
Both input and output are strings of \$1\$s and \$0\$s. Output is done by printing.
Despite the unwieldy string replacement, other ways to account for the extra zero bit tend to be even longer.
1
Japt -R, 40 36 bytes
This'll need another pass once I'm properly caffeinated!
5Æ17î-³i+ÃíUs'#iS)ùG ®û3 i|ÃòG)cf ú|
Try it
5Æ17î-³i+ÃíUs'#iS)ùG ®û3 i|ÃòG)cf ú| :Implicit input of integer U
5Æ :Map the range [0,5)
17î : Repeat the following to length 17
-³ ...
0
05AB1E, 12 bytes
∞ʒLDb1ö*y¢2@
Outputs the infinite sequence.
Try it online.
Explanation:
∞ # Push an infinite positive list: [1,2,3,...]
ʒ # Filter each value `y` by:
L # Pop and create a list in the range [1,`y`]
D # Duplicate this list
b # Convert each inner value to a binary-String
1ö # ...
1
Vyxal j, 47 bytes
b:L16-(0p)4ẇƛƛ‛ #$i`| % `$%;ṅ\|+;\+:3-4*pw5ẋf$Y
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A big mess
1
Python 3 (63 bytes)
def a(n):m=len(bin(n))-3;r=2**m;return n and a(n%r)+m*r/2+n%r+1
Here is my attempt 2, since dingledooper noted that my previous attempt was not polylog time complexity.
Also, the golfiness can probably be improved by using walrus operator in python3.8
I used the formula found from "a curious PARI program on the OEIS page":
a(n) ...
3
Haskell, 139 bytes
([]%)
a%(x:y)|x=='_'=['a'..]!!foldl1((+).(2*))(take 5a):drop 5a%y|x>'@',x<'['=[x..]!!32:(a++[1])%y|x>'`',x<'{'=x:(a++[0])%y|1<2=x:a%y
a%e=e
Try it online!
2
Haskell, 85 bytes
map("JK"!!).(0%)
j%(1:1:1:1:1:1:y)=j:j:j:j:j:j:j%(0:y)
j%(i:y)|k<-i^j*j^i=k:k%y
j%e=e
Try it online!
3
Red, 79 bytes
func[x][p: s: 0 collect[foreach i x[s: any[all[s < 6 keep p = i s + 1]0]p: i]]]
Try it online!
Takes input as a block of \$1\$s and \$0\$s, outputs a block of booleans.
Test suite on TIO uses two additional transformation functions that allow reading inputs from JK-strings as in task specification and convert output to numbers for nicer ...
1
C (gcc), 84 80 78 bytes
-6 thanks to @ceilingcat
x;c;a(char*p){for(c=0,x=1;*p;c>5&&a(L"10"+x))putchar(x^=*p<49),c=*p++/49*-~c;}
Try it online! Takes input as the characters 0 and 1, outputs with bytes 0x01 for J and 0x00 for K.
For output as actual J and K,
Try it online!
Decoder
3
<>^v, 161 bytes
1±n"0"o"1"i0q.h"J"j"K"kJtv
v v=ICc®NHn)N v=N-?1¿H<
>Q)7=v_qv >`! ~
vcOn(Nq0< v >Ktv T
v< <>TJ=^Jtv ^
>C O=^ > ^
Explanation
First line
1±n"0"o"1"i0q.h"J"j"K"kJtv
1 ...
2
K (ngn/k), 21 bytes
0=\.'(t,$0)/(t:6#$1)\
Try it online!
Takes input as a string of "1"s and "0"s. Returns a list of 0s (for J) and 1s (for K).
do a string replace of "111111" with "1111110"
(t:6#$1) store "111111" in variable t
(t,$0) append a "0" to the end of t, generating "1111110"
...
4
C (gcc), 79 76 bytes
-3 thanks to @ceilingcat
x;c;a(char*p){x=74;for(c=1;c*=*p==x,x=*p++;c=c>5?x=*p++,1:c+1)putchar(!!c);}
Try it online! Takes input as J and K, outputs with bytes 0x00 and 0x01 for 0 and 1.
For output as numeric 0 and 1,
Try it online! for 78 bytes.
Encoder
1
Python 3, 64 bytes
f=lambda n:n and(2*f(n//2)+n//2+1if n%2else f(n/2)+f(n/2-1)+n/2)
Try it online!
from the recurrence relation: a(0) = 0, a(2n) = a(n)+a(n-1)+n, a(2n+1) = 2a(n)+n+1
2
Charcoal, 24 bytes
≔⪫⪪Ｓ×1⁶⁺×1⁶0θ⭆θ§JK№…θ⊕κ0
Try it online! Link is to verbose version of code. Explanation:
≔⪫⪪Ｓ×1⁶⁺×1⁶0θ
Perform bit-stuffing.
⭆θ§JK№…θ⊕κ0
For each prefix of the string count the number of 0s and output K if it is odd and J if it is even.
5
Husk, 11 bytes
G=Jḋ126xḋ63
Try it online!
Input as array of 0s and 1s, output as array of 1s (for state J) and 0s (for state K): TIO header converts to Js and Ks.
x # split input on
ḋ63 # binary digits of 63 = [1,1,1,1,1,1]
J # then join back together using
ḋ126 # binary digits of 126 = [1,1,1,1,1,1,0]
G= # and ...
4
Jelly, 12 bytes
7Ṭ¬©Ṗœṣ@j®⁼\
A monadic Link that accepts a list of integers (from \$[0,1]\$) and yields a list of integers (from \$[0,1]\$) where \$0\$ represents \$K\$.
Try it online! Or see the test-suite.
How?
7Ṭ¬©Ṗœṣ@j®⁼\ - Link: list of integers in [0..1], A
7 - seven
Ṭ - untruth -> [0,0,0,0,0,0,1]
¬ - logical NOT -&...
3
Vyxal, 16 bytes
\16*:0+V1$(nɽ߬…
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\16* # Six ones
:0+ # Duplicate and append a 0
V # Replace in input
1$ # Starting with a one
( # Iterating through the string
߬ # Flip if...
nɽ # It's a 1
… # Print without popping
Returns 1 for J ...
3
R, 59 bytes
function(x)cumsum(49-utf8ToInt(gsub('(1{6})','\\10',x)))%%2
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Regex-based bit-stuffing stolen from inspired by pajonk's R answer to the paired usb-decoding challenge.
The state-switching is achieved here by determining whether the cumulative sum is odd or even.
3
convey, 31 bytes
Outputs KJ as 01.
v|6
0"("=0
+*6!<}
""~,="
{ 5 1<
Try it online!
In the top left is a counter that gets incremented by the input 1s + and reset whenever it reaches 6 |6 or the input has a 0 *. If it reaches 6, it also injects a 0 into the stream (the ("=0 6!< part). Because the input stream would be ...
4
JavaScript (ES6), 53 bytes
Expects a binary string. Returns an array of 0's and 1's.
s=>[...s.replace(/1{6}/g,'$&0')].map(q=c=>q^=c^!!++q)
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Commented
This is a rather straightforward implementation in two steps.
s => // s = input string
// STEP 1 - Bit stuffing
//
[...s.replace( // replace ...
3
Jelly, 14 bytes
“?~‘Bœṣjƭƒ^¬¥\
Try it online!
-2 bytes thanks to Nick Kennedy
“?~‘Bœṣjƭƒ^¬¥\ Main Link; accepts a list of [0, 1] on the left
“?~‘ 63, 126
B To binary ([1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 0])
ƒ Reduce starting with the input:
ƭ Tie; cycle
œṣ The first time, split on sublist ...
2
Retina 0.8.2, 33 32 bytes
1{6}
$&0
{+rT`d`JK`K.
1T`d`KJ`\d
Try it online! Link includes test cases. Explanation:
1{6}
$&0
Perform bit-stuffing.
{`
Make as many replacements as possible.
+rT`d`JK`K.
Change 0 after a K to J and 1 after a K to K. Change starting with the character after the last K (and working back, not that it matters), and keep ...
1
Python 3, 81 bytes
def f(A):
c=0
while A>1:
a=len(bin(A))-3;A=A%(2**a);c+=2**a*a/2+A
return c
Try it online!
For exponentials of 2 -1 every bit is flipped 50% of the time so the total sum of bits is n*(# of bits)/2 this gets done recursively. For every further recursion we also need to add up the first bits that get progessively ignored.
2
05AB1E, 13 bytes
0ìø€ËJƵPbD¨.:
Try it online! Uses 0 for J and 1 for K.
0ìø€ËJƵPbD¨.: # full program
.: # replace all instances of...
ƵP # 126...
b # in binary...
.: # in...
J # joined...
Ë # are all digits of...
€ # each element in...
# implicit input...
...
1
Jelly, 14 bytes
ŻI¬“~?‘B¤œṣjƭƒ
Try it online!
A monadic link taking the argument as a vector of 0 (=J) and 1 (=K) and returning a vector of 0 and 1.
2
K (ngn/k), 23 bytes
*|+(126=2/)_7'(&6),0=':
Try it online!
Takes J as 0, and K as 1. Returns a list of 0s and 1s.
0=': run an equals-each-prior, seeded with 0 on the (implicit) input, handling the non-return-to-zero part
7'(&6), prepend 6 0s to the input, then slice into length-7 sliding windows. the first 6 slices will contain some dummy 0s
(126=2/...
7
Husk, 11 bytes
σḋ126ḋ63Ẋ=Θ
Try it online!
Takes an array of bits(0=J, 1=K), outputs an array of bits(formatted as string for convenience)
Explanation
σḋ126ḋ63Ẋ=Θ
Θ Prepend 0
Ẋ= map overlapping pairs by equality
σḋ126ḋ63 replace binary digits of 126 with binary digits of 63 (stolen from Luis Mendo)
2
Stax, 16 bytes
Å3≈àE∩◄σ←π<δô╝ìà
Run and debug it
takes in a string/array of bytes 0x00 for J and 0x01 for K.
Outputs a string, same as the given testcases.
Sadly, this 14 byte program doesn't work since stax auto-replaces null bytes with spaces.
4
Python 3, 106 104 bytes
def f(A):
A+=1;a=0
for i in range(32):
x=2<<i;d,r=A//x,A%x;a+=d*x/2
if 2*r>x:a+=r-x/2
return a
Try It Online!!
This is designed for 32-bit integers, here I am using float to save some bytes from "//" operator, I hope you won't mind \$x.0\$ instead of \$x\$, further bytes can be reduced if I allow ...
2
Charcoal, 19 bytes
⪫⪪⭆θ⁼ι§⁺Jθκ⁺×1⁶0×1⁶
Try it online! Link is to verbose version of code. Explanation:
θ Input string
⭆ Map over characters and join
J Literal string `J`
⁺ Concatenated with
θ Input string
§ Indexed by
κ Current index
...
3
Python 2, 77 bytes
lambda s:''.join((`+(x==y)`for x,y in zip('J'+s,s))).replace('1'*6+'0','1'*6)
Try it online!
-2 bytes thanks to @ovs
11
convey, 31 25 bytes
Takes JK as 01.
>="*>0
",v+<"(6
{0>">>!`}
Try it online!
Compares = the string with itself with a J prepended ,0. The middle loop just adds the comparisons to a counter (starting with 0) that gets reset on a 0. Only take ! elements where the previous counter was less than 6 (6.
1
Retina 0.8.2, 31 bytes
(.)(?<=(\1.|^J)?)
$#2
1{6}0
6$*
Try it online! Link includes test cases. Explanation:
(.)(?<=(\1.|^J)?)
For each letter, look behind for a possible duplicate or a leading J.
$#2
If it was then replace it with a 1 otherwise replace it with a 0.
1{6}0
6$*
Remove the 0 that appears after a run of six 1s.
2
Charcoal, 28 bytes
ＮθＩΣＥ⮌↨θ²⁺×÷θＸ²⊕κＸ²κ×ι⊕﹪θＸ²κ
Try it online! Explanation: For a given bit position k, the sum of the bits at position k from 1 to n depends on whether the kth bit of n is 0 or 1:
If the kth bit of n is 0, then the bits at position k are of the form 0...1...0...1...0...1...0..0. Each run of 2ᵏ⁺¹ integers contributes 2ᵏ bits.
If the kth bit ...
3
PHP -F, 96 bytes
for($s=$r=$argn;$s[++$i]!='';)$r[$i]=+!abs($s[$i]-$s[$i-1]);echo str_replace(1111110,111111,$r);
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Well, I.. tried.. to make it short.. hum!
takes a string of 0 (J) and 1 (K) as input
3
Vyxal, 16 bytes
1p-Ṫv¬ṅ\16*:0+$V
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1p # Prepend a 1
- # Differences - truthy if unchanged, falsy if changed
Ṫ # Get rid of last
v¬ # Vectorised not
ṅ # Joined
\16* # Six ones
: # Duplicate
0+ # Append a 0 to the first
$...
1
Jelly, 14 bytes
Ż⁼Ɲ0ȯ1בʋ7Ƒ?ɼƇ
Try it online!
Takes J and K as 0 and 1.
6
MATL, 15 14 bytes
0ihd~126B63BZt
Input is an array with 0 for 'J' and 1 for 'K'.
Try it online! Or verify all test cases.
Explanation
0 % Push 0
i % Input: string
h % Concatenate
d % Consecutive differences
~ % Negate: converts nonzero to 0, and zero to 1. Gives a numeric vector (*)
126B % Push 126, convert to binary. Gives [...
4
R, 83 76 62 bytes
function(x)gsub('(1{6})0','\\1',Reduce(paste0,+!diff(c(0,x))))
Try it online!
Takes input as 0(J) and 1(K).
Outputs string of 1s and 0s.
Change of approach after looking at @Luis Mendo's answer.
function(x) # function taking x as input
gsub( # replace all occurences of
'(1{6})0', # 6 ones and a zero
'\\1', # with ...
4
JavaScript (ES6), 54 bytes
s=>s.replace(/./g,a=>s^(s=a>"J")?i=i>5?'':0:++i/i,i=0)
Try it online!
Take input as a string of "J" / "K". Output a string with "0" and "1".
And f=J=>J.replace(/./g,a=>J^(J=a>f)?i=i>5?'':0:++i/i,i=0) is 54 bytes too.
JavaScript (ES6), 44 bytes
s=>s....
3
J, 26 bytes
Takes in JK as 0 1.
t#~0=_6|.(6$1)E.t=:2=/\0,]
Try it online!
t#~0=_6|.(6$1)E.t=:2=/\0,]
0,] prepend a J
2=/\ pair-wise equal
t=: store as t
(6$1)E. bitmask of places where 6 ones starts
_6|. shifted by 6 bits
0= ...
1
JavaScript (Node.js), 61 bytes
n=>n.replace(/./g,(e,i)=>1-e^n[i-1]).replace(/1{6}0/g,111111)
Try it online!
I have tried to make sure it outputs numbers.
-10 bytes by pxeger
6
JavaScript (ES6), 51 bytes
f=(n,i=0,p=1<<i)=>n<p?0:f(n,i+1)+(n&p&&n%p+1+p*i/2)
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Let we denote
$$ n = \left(\overline{b_mb_{m-1}\dots b_2b_1b_0}\right)_2 $$
Then
$$ f\left(n\right)=\sum_{i=0}^{m} b_i \cdot \left(\left(n \bmod 2^i\right) + 1 + 2^{i-1}\cdot i\right) $$
This function loops from \$0\$ to \$m\$ with constant ...
13
JavaScript (ES6), 57 54 53 52 bytes
n=>(i=0,g=b=>b--&&(n>>b&1&&b+i+i++<<b)+g(b))(32)/2+i
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This is designed for 32-bit integers and always performs 32 iterations, no matter the input value.
The following version performs \$\lfloor log_2(n)\rfloor+1\$ iterations instead, but this is really pointless since ...
13
K (ngn/k), 45 32 bytes
{+/b*1+(q!'x)+-2!q*<q:2/=#b:2\x}
Try it online!
Golfing ideas thanks to @ngn and @coltim
The OEIS page contains a curious PARI program:
a(n)=if(n==0,0,m=logint(n,2);r=n%2^m;m*2^(m-1)+r+1+a(r));
which has a recursion depth of \$\mathcal{O}(\log n)\$ and each recursive call involves constant number of arithmetic operations (and ...
0
MMIX, 24 bytes (6 instrs)
00000000: 36ff0000 c8ffff00 3bffff01 c80000ff 6”¡¡Ṁ””¡;””¢Ṁ¡¡”
00000010: 73000001 f8010000 s¡¡¢ẏ¢¡¡
Disassembled
dragon NEGU $255,0,$0 // t = -n
AND $255,$255,$0 // t &= n (gets last 1 bit)
SLU $255,$255,1 // t <<= 1
AND $0,$0,$255 // n &= t
ZSZ $0,...
0
Vyxal, 8 bytes
bṘ:1ḟ›i¬
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Straightforward use of the formula
b # To bitlist
Ṙ # Reverse
: # Duplicate
1ḟ # First truthy value
› # Incremented
i # Indexed
¬ # Flipped
2
Python 3.8 (pre-release), 62 bytes
exec(a:="print(''.join(f'{i:08b}'for i in b'exec(a:=%r)'%a))")
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How it work :
exec(a:="print('exec(a:=%r)'%a)") is a quine
f'{i:08b}' converts i into its 8-digits binary representation
b'example_string' is a binary sting. When we iterate over it, it convert the characters by their ...
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