Bit floating sequence

Question 1

A bit floats from the LSB to the MSB moving one position each time until it floats to the top of the container:

Once one bit floats to the top, another bit begins its journey and it stops when it meets other bit:

1001
1010
1100

This happens until the container is filled with bits:

1101
1110
1111

Challenge

Given an integer number, output the "Bit floating sequence" for a container of that number of bits.

Each term of the sequence can be separated by any separator of your choice.
Edit: Sequence must be shown as decimal integer numbers, starting by the first therm: 0.
The container size sould be greater than zero and up to the number of bits of the bigest integer suported by the language of your choice. You can assume that the input always match this requirement.

Examples

Only the numeric sequence is required, binary representation is shown as example:

For 1: 0 1
```
0 -> 0
1 -> 1
```

For 3: 0 1 2 4 5 6 7

000 -> 0
001 -> 1
010 -> 2
100 -> 4
101 -> 5
110 -> 6
111 -> 7

For 4: 0 1 2 4 8 9 10 12 13 14 15

For 8: 0 1 2 4 8 16 32 64 128 129 130 132 136 144 160 192 193 194 196 200 208 224 225 226 228 232 240 241 242 244 248 249 250 252 253 254 255

00000000 -> 0
00000001 -> 1
00000010 -> 2
00000100 -> 4
00001000 -> 8
...
...
...
11111000 -> 248
11111001 -> 249
11111010 -> 250
11111100 -> 252
11111101 -> 253
11111110 -> 254
11111111 -> 255

Question 2

May we output the sequence in any order (i.e. reversed), or do they have to be sorted from lowest to highest?

Question 3

May we output as floats? E.g. [0.0, 1.0]

Question 4

May we output using the binary representation?

Question 5

May we output the zero-indexed sequence? i.e. 0 -> [0, 1]

Question 6

05AB1E, 10 bytes

LRL ̃Íoî.\ï

Try it online!

L # range [1..input]
 R # reversed
 L # convert each to a range: [[1..input], [1..input-1], ..., [1]]
 ̃ # flatten
 Í # subtract 2 from each
 o # 2**each
 î # round up (returns a float)
 ï # convert to integer
 .\ # undelta

Question 7

I think there is a meta-post somewhere allowing floats with .0 by default for integers, but not sure. I personally usually put the ï in the footer to pretty-print and don't include it in the byte-count.

Question 8

Python 2, 45 bytes

y=n=2**input()
while y:print n-y;y=y&y-1or~-y

Try it online!

It turns out shorter to generate 2**n minus each term in the sequence for input n. If we look at their binary expansion, below for n=5, we see a nice pattern of triangles of 1's in the binary expansions.

Each number is obtained from the previous one by removing the rightmost one in the binary expansion, except if that would make the number 0, we subtract 1 instead, creating a new block of 1's that starts a new smaller triangle. This is implemented as y=y&y-1or~-y, where y&y-1 is a bit trick to remove the rightmost 1, and or~-y gives y-1 instead if that value was 0.

Python 2, 49 bytes

def f(n,x=0):1%n;print x;f(n-x%2,x+(x%2**n or 1))

Try it online!

A function that prints, terminating with error. The more nice program below turned out longer.

51 bytes

n=input()
x=0
while n:n-=x%2;print x;x+=x%2**n or 1

Try it online!

Question 9

Jelly, (削除) 11 (削除ここまで) 10 bytes

RUḶ’F2*ĊÄŻ

Port of @Grimy's 05AB1E answer, so make sure to upvote him!
-1 byte thanks to @Grimy.

Try it online.

Explanation:

R # Create a list in the range [1, (implicit) argument]
 U # Reverse it to [argument, 1]
 Ḷ # Create an inner list in the range [0, N) for each value N in this list
 ’ # Decrease each by 1
 F # Flatten the list of lists
 2* # Take 2 to the power each
 Ċ # Ceil
 Ä # Undelta (cumulative sum) the list
 Ż # And add a leading 0
 # (after which the result is output implicitly)

Question 10

R_2 -> Ḷ’ for -1. Ḷ is the only sensible range, I really wish 05AB1E had a single-byter for it.

Question 11

@Grimy Ah, how did I miss that one. I searched for range and must have skipped passed it somehow.. >.> Thanks!

Question 12

Perl 5 (`-n`), (削除) 41 (削除ここまで) 40 bytes

-1 byte thanls to Xcali

map{/01.*1/||say oct}glob"0b"."{0,1}"x$_

TIO

"{0,1}"x$_ : the string "{0,1}" repeated n times
"0b". : concatenate to "0b"
glob : glob expansion (cartesian product)
map{...} : for each element
/01.*1/|| : to skip when 01 followed by something then 1
say oct : to convert to decimal and say

Question 13

40

Question 14

JavaScript (ES6), 43 bytes

When in doubt, use xnor's method.

n=>(g=x=>x?[n-x,...g(x&--x||x)]:[])(n=1<<n)

Try it online!

JavaScript (ES6), (削除) 59 57 55 (削除ここまで) 52 bytes

f=(n,x=0)=>x>>n?[]:[x,...f(n,x+=x+(x&=-x)>>n|!x||x)]

Try it online!

How?

We define \$p(x)\$ as the highest power of \2ドル\$ dividing \$x\$, with \$p(0)=0\$ by convention.

This function can be implemented with a simple bitwise AND of \$x\$ and \$-x\$ to isolate the lowest bit set to \1ドル\$ in \$x\$. For instance:

$$p(52)=52 \operatorname{AND}-52=4$$

Using \$p\$, the sequence \$a_n\$ can be defined as \$a_n(0)=0\$ and:

$$a_n(k+1)=\cases{ a_n(k)+p(a_n(k)), & \text{if $p(a_n(k))\neq0$ and $a_n(k)+p(a_n(k))<2^n$}\\ a_n(k)+1, & \text{otherwise} }$$

Commented

f = ( // f is a recursive function taking:
 n, // n = input
 x = 0 // x = current term of the sequence
) => //
 x >> n ? // if x is greater than or equal to 2**n:
 [] // stop recursion
 : // else:
 [ // update the sequence:
 x, // append the current term to the sequence
 ...f( // do a recursive call:
 n, // pass n unchanged
 x += // update x:
 x + (x &= -x) // given x' = lowest bit of x set to 1:
 >> n // if x + x' is greater than or equal to 2**n
 | !x // or x' is equal to 0: add 1 to x
 || x // otherwise, add x' to x
 ) // end of recursive call
 ] // end of sequence update

Question 15

Python 2, (削除) 95 (削除ここまで) 76 bytes

n=input()
a=0
print 0
while n:
 for j in range(n):print a+2**j
 n-=1;a+=2**n

Try it online!

Question 16

Perl 6, 43 bytes

{0 x$_,{say :2($_);S/(0)1|0$/10ドル/}...1 x$_}

Try it online!

Anonymous code block that takes a number and outputs the sequence separated by newlines. This works by starting with 0 repeated n times then replacing either 01 with 10 or the last 0 with a 1 until the number is just ones.

Or 40 bytes, using Nahuel Fouilleul's approach

{grep /010*1/|{say :2($_)},[X~] ^2 xx$_}

Try it online!

Question 17

"then replacing either 01 with 10 or the last 0 with a 1 until the number is just ones" That's a genius move!

Question 18

Python 2, 60 bytes

f=lambda i,n=0,b=1:[n][i:]or[n]+f(i-1/b,n^b+b/2,b>>i or 2*b)

Try it online!

Python 3, 76 bytes

f=lambda n:[0][n:]or[0]+[2**i for i in range(n-1)]+[x|1<<n-1for x in f(n-1)]

Try it online!

Question 19

Python 2, 67 bytes

n=0
i=2**input()-1
while n<=i:print n;d=n&(~-n^i)or 1;n+=n+d>i or d

Try it online!

Question 20

Python 3, 62 bytes

def f(n,c=0):
 while c<2**n:yield c;r=c&-c;c+=c+r>>n or r or 1

Try it online!

The idea is more or less the same as @Arnauld's solution.

Another 65-byte solution:

lambda n:g(2**n-1)
g=lambda c:[0][c:]or g(c-((c&-c)//2 or 1))+[c]

Try it online!

Question 21

Jelly, 12 bytes

=þ‘ṚÄUo€ƊẎQḄ

Try it online!

Question 22

05AB1E, (削除) 13 (削除ここまで) 12 bytes

Tsãʒ1ÛSO2‹}C{

-1 byte thanks to @Grimy (also take a look at his shorter approach here).

Try it online or verify all test cases.

Explanation:

T # Push 10
 sã # Swap to get the (implicit) input, and get the cartesian product with "10"
 ʒ # Filter it by:
 1Û # Remove leading 1s
 SO # Get the sum of the remaining digits
 ! # Check that the sum is either 0 or 1 by taking the factorial
 # (NOTE: Only 1 is truthy in 05AB1E)
 }C # After the filter: convert all remaining strings from binary to integer
 { # And sort (reverse) them
 # (after which the result is output implicitly)

Question 23

Alternate 13: oL<ʒbIj1Û1¢2‹. Doesn't look like I can get it lower.

Question 24

@Grimy I just had oL<ʒbIj1ÛSO2‹ and was trying to see where my error was. :) But I'm glad to see you aren't able to find a shorter version for one of my answers for a change. ;p (inb4 you find a shorter one after all xD)

Question 25

@Grimy I have the feeling SO2‹ can be 3 bytes somehow perhaps, but I'm not seeing it and also not entirely sure.. There are some alternatives, like SO1~ or SÆ>d, but I'm unable to find a 3-byter.

Question 26

10 with a completely different approach

Question 27

Your feeling was right, I just found a 3-byter: SO!. Pretty sure I have some old answers using 2‹ that could benefit from this as well.

Question 28

Retina, 26 bytes

.+
*0
L$w`.(.*)
$.`*1$'11ドル

Try it online! Outputs in binary. If that's not acceptable, then for 39 bytes:

.+
*0
L$w`.(.*)
$.`*1$'11ドル
+`10
011
%`1

Try it online! Explanation:

.+
*0

Convert the input into a string of n zeros.

L$w`.(.*)

Match all possible non-empty substrings.

$.`*1$'11ドル

For each substring, output: the prefix with 0s changed to 1s; the suffix; the match with the initial 0 changed to 1.

+`10
011
%`1

Convert from binary to decimal.

Question 29

Brachylog, 27 bytes

1;0|⟦5;2z^(mLtT&-1↰+m↙T,L,0

Try it online!

Outputs out of order and with duplicates. If that's not okay, tack do onto the end.

Question 30

Charcoal, 19 bytes

I⮌E⊕θEι+−X2IθX2ιX2λ

Try it online! Link is to verbose version of code. Explanation:

 θ Input
 ⊕ Incremented
 E Map over implicit range
 ι Outer index
 E Map over implicit range
 Iθ Input cast to integer
 ι Outer index
 λ Inner index
 X2 X2 X2 Power of 2
 +− Subtract and add
 ⮌ Reverse outer list
I Cast to string
 Implicitly print

Question 31

Perl 5, 40 bytes

map{say$-;$-+=2**$_}0,0..$_-2;$_--&&redo

Try it online!

Question 32

Retina, 24 bytes

.+
*0
/0/+<0`(0)1|0$
11ドル

Outputs in binary. Input should have a trailing newline.

Attempt at explanation:

.+ #match the entire input
*0 #replace it with that many zeroes
/0/+<0`(0)1|0$ #while the string has a 0, substitute the first match and output
11ドル #if 01 is present in the string, replace it with 10, else replace the last character with $

I tried to avoid the 3 bytes long /0/ regex option by rearranging the options, but couldn't.

Try it online!

Question 33

I don't think outputting in binary is allowed. There's a comment asking if it is allowed, but it's better to assume that you can't until the asker replies

Question 34

C (clang), 73 bytes

o,j,y;f(x){for(o=j=0;printf("%d ",o),x;o+=y+!y,y+=y+!y)j=!j?y=0,--x:--j;}

Try it online!

for(o=j=0;printf("%d ",o),x; o+=y+!y, y+=y+!y) 
// adds 1, 1+1=>2 , 2+2=> 4 .... sequence
 j=!j?y=0,--x:--j; 
// uses ternary instead of nested loop to decrement 'x' when 'j' go to 0

Question 35

k4, (削除) 28 (削除ここまで) 24 bytes

0,+\"j"2ドル xexp,/-1+|,\!:

@Grimy's approach ported to k4

edit: -4 thanks to ngn!

Question 36

!:'1+|!: -> |,\!:

Question 37

you can remove the space after xexp

Question 38

@ngn, agh |,\!: seems so obvious now that i see it!

Grimmy 15.8k1 gold badge36 silver badges64 bronze badges · Answer 1 · 2019-09-06 13:08:31Z

7

\$\begingroup\$

05AB1E, 10 bytes

LRL ̃Íoî.\ï

Try it online!

L # range [1..input]
 R # reversed
 L # convert each to a range: [[1..input], [1..input-1], ..., [1]]
 ̃ # flatten
 Í # subtract 2 from each
 o # 2**each
 î # round up (returns a float)
 ï # convert to integer
 .\ # undelta

Share

Improve this answer

answered Sep 6, 2019 at 13:08

Grimmy's user avatar

Grimmy

15.8k1 gold badge36 silver badges64 bronze badges

\$\endgroup\$

1

2

\$\begingroup\$ I think there is a meta-post somewhere allowing floats with .0 by default for integers, but not sure. I personally usually put the ï in the footer to pretty-print and don't include it in the byte-count. \$\endgroup\$

Kevin Cruijssen
– Kevin Cruijssen

2019年09月06日 13:38:23 +00:00
Commented Sep 6, 2019 at 13:38

Add a comment |

xnor 150k26 gold badges287 silver badges676 bronze badges · Answer 2 · 2019-09-07 00:07:07Z

Python 2, 45 bytes

y=n=2**input()
while y:print n-y;y=y&y-1or~-y

Try it online!

It turns out shorter to generate 2**n minus each term in the sequence for input n. If we look at their binary expansion, below for n=5, we see a nice pattern of triangles of 1's in the binary expansions.

Each number is obtained from the previous one by removing the rightmost one in the binary expansion, except if that would make the number 0, we subtract 1 instead, creating a new block of 1's that starts a new smaller triangle. This is implemented as y=y&y-1or~-y, where y&y-1 is a bit trick to remove the rightmost 1, and or~-y gives y-1 instead if that value was 0.

Python 2, 49 bytes

def f(n,x=0):1%n;print x;f(n-x%2,x+(x%2**n or 1))

Try it online!

A function that prints, terminating with error. The more nice program below turned out longer.

51 bytes

n=input()
x=0
while n:n-=x%2;print x;x+=x%2**n or 1

Try it online!

Kevin Cruijssen 136k14 gold badges154 silver badges394 bronze badges · Answer 3 · 2019-09-06 13:35:33Z

Jelly, (削除) 11 (削除ここまで) 10 bytes

RUḶ’F2*ĊÄŻ

Port of @Grimy's 05AB1E answer, so make sure to upvote him!
-1 byte thanks to @Grimy.

Try it online.

Explanation:

R # Create a list in the range [1, (implicit) argument]
 U # Reverse it to [argument, 1]
 Ḷ # Create an inner list in the range [0, N) for each value N in this list
 ’ # Decrease each by 1
 F # Flatten the list of lists
 2* # Take 2 to the power each
 Ċ # Ceil
 Ä # Undelta (cumulative sum) the list
 Ż # And add a leading 0
 # (after which the result is output implicitly)

R_2 -> Ḷ’ for -1. Ḷ is the only sensible range, I really wish 05AB1E had a single-byter for it.
@Grimy Ah, how did I miss that one. I searched for range and must have skipped passed it somehow.. >.> Thanks!

Nahuel Fouilleul 8,6571 gold badge11 silver badges19 bronze badges · Answer 4 · 2019-09-06 10:33:56Z

Perl 5 (`-n`), (削除) 41 (削除ここまで) 40 bytes

-1 byte thanls to Xcali

map{/01.*1/||say oct}glob"0b"."{0,1}"x$_

TIO

"{0,1}"x$_ : the string "{0,1}" repeated n times
"0b". : concatenate to "0b"
glob : glob expansion (cartesian product)
map{...} : for each element
/01.*1/|| : to skip when 01 followed by something then 1
say oct : to convert to decimal and say

\$\begingroup\$ 40 \$\endgroup\$

Xcali
– Xcali

2019年09月06日 14:37:02 +00:00
Commented Sep 6, 2019 at 14:37

Arnauld 206k21 gold badges187 silver badges670 bronze badges · Answer 5 · 2019-09-06 11:25:41Z

JavaScript (ES6), 43 bytes

When in doubt, use xnor's method.

n=>(g=x=>x?[n-x,...g(x&--x||x)]:[])(n=1<<n)

Try it online!

JavaScript (ES6), (削除) 59 57 55 (削除ここまで) 52 bytes

f=(n,x=0)=>x>>n?[]:[x,...f(n,x+=x+(x&=-x)>>n|!x||x)]

Try it online!

How?

We define \$p(x)\$ as the highest power of \2ドル\$ dividing \$x\$, with \$p(0)=0\$ by convention.

This function can be implemented with a simple bitwise AND of \$x\$ and \$-x\$ to isolate the lowest bit set to \1ドル\$ in \$x\$. For instance:

$$p(52)=52 \operatorname{AND}-52=4$$

Using \$p\$, the sequence \$a_n\$ can be defined as \$a_n(0)=0\$ and:

$$a_n(k+1)=\cases{ a_n(k)+p(a_n(k)), & \text{if $p(a_n(k))\neq0$ and $a_n(k)+p(a_n(k))<2^n$}\\ a_n(k)+1, & \text{otherwise} }$$

Commented

f = ( // f is a recursive function taking:
 n, // n = input
 x = 0 // x = current term of the sequence
) => //
 x >> n ? // if x is greater than or equal to 2**n:
 [] // stop recursion
 : // else:
 [ // update the sequence:
 x, // append the current term to the sequence
 ...f( // do a recursive call:
 n, // pass n unchanged
 x += // update x:
 x + (x &= -x) // given x' = lowest bit of x set to 1:
 >> n // if x + x' is greater than or equal to 2**n
 | !x // or x' is equal to 0: add 1 to x
 || x // otherwise, add x' to x
 ) // end of recursive call
 ] // end of sequence update

TFeld 19.9k3 gold badges21 silver badges63 bronze badges · Answer 6 · 2019-09-06 10:03:40Z

3

\$\begingroup\$

Python 2, (削除) 95 (削除ここまで) 76 bytes

n=input()
a=0
print 0
while n:
 for j in range(n):print a+2**j
 n-=1;a+=2**n

Try it online!

Share

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edited Sep 6, 2019 at 10:46

answered Sep 6, 2019 at 10:03

TFeld's user avatar

TFeld

19.9k3 gold badges21 silver badges63 bronze badges

\$\endgroup\$

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Jo King 48.1k6 gold badges130 silver badges187 bronze badges · Answer 7 · 2019-09-06 10:44:34Z

Perl 6, 43 bytes

{0 x$_,{say :2($_);S/(0)1|0$/10ドル/}...1 x$_}

Try it online!

Anonymous code block that takes a number and outputs the sequence separated by newlines. This works by starting with 0 repeated n times then replacing either 01 with 10 or the last 0 with a 1 until the number is just ones.

Or 40 bytes, using Nahuel Fouilleul's approach

{grep /010*1/|{say :2($_)},[X~] ^2 xx$_}

Try it online!

"then replacing either 01 with 10 or the last 0 with a 1 until the number is just ones" That's a genius move!

ovs 61.2k3 gold badges49 silver badges164 bronze badges · Answer 8 · 2019-09-06 11:11:52Z

Python 2, 60 bytes

f=lambda i,n=0,b=1:[n][i:]or[n]+f(i-1/b,n^b+b/2,b>>i or 2*b)

Try it online!

Python 3, 76 bytes

f=lambda n:[0][n:]or[0]+[2**i for i in range(n-1)]+[x|1<<n-1for x in f(n-1)]

Try it online!

negative seven 2,2111 gold badge11 silver badges19 bronze badges · Answer 9 · 2019-09-06 10:38:14Z

3

\$\begingroup\$

Python 2, 67 bytes

n=0
i=2**input()-1
while n<=i:print n;d=n&(~-n^i)or 1;n+=n+d>i or d

Try it online!

Share

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edited Sep 6, 2019 at 14:28

answered Sep 6, 2019 at 10:38

negative seven's user avatar

negative seven

2,2111 gold badge11 silver badges19 bronze badges

\$\endgroup\$

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Joel 1,7617 silver badges12 bronze badges · Answer 10 · 2019-09-06 21:21:09Z

Python 3, 62 bytes

def f(n,c=0):
 while c<2**n:yield c;r=c&-c;c+=c+r>>n or r or 1

Try it online!

The idea is more or less the same as @Arnauld's solution.

Another 65-byte solution:

lambda n:g(2**n-1)
g=lambda c:[0][c:]or g(c-((c&-c)//2 or 1))+[c]

Try it online!

Erik the Outgolfer 40.8k5 gold badges46 silver badges125 bronze badges · Answer 11 · 2019-09-06 12:27:17Z

2

\$\begingroup\$

Jelly, 12 bytes

=þ‘ṚÄUo€ƊẎQḄ

Try it online!

Share

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answered Sep 6, 2019 at 12:27

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Erik the Outgolfer

40.8k5 gold badges46 silver badges125 bronze badges

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Kevin Cruijssen 136k14 gold badges154 silver badges394 bronze badges · Answer 12 · 2019-09-06 11:59:21Z

2

\$\begingroup\$

05AB1E, (削除) 13 (削除ここまで) 12 bytes

Tsãʒ1ÛSO2‹}C{

-1 byte thanks to @Grimy (also take a look at his shorter approach here).

Try it online or verify all test cases.

Explanation:

T # Push 10
 sã # Swap to get the (implicit) input, and get the cartesian product with "10"
 ʒ # Filter it by:
 1Û # Remove leading 1s
 SO # Get the sum of the remaining digits
 ! # Check that the sum is either 0 or 1 by taking the factorial
 # (NOTE: Only 1 is truthy in 05AB1E)
 }C # After the filter: convert all remaining strings from binary to integer
 { # And sort (reverse) them
 # (after which the result is output implicitly)

Share

Improve this answer

edited Jun 17, 2020 at 9:04

Community's user avatar

Community Bot

1

answered Sep 6, 2019 at 11:59

Kevin Cruijssen's user avatar

Kevin Cruijssen

136k14 gold badges154 silver badges394 bronze badges

\$\endgroup\$

7

\$\begingroup\$ Alternate 13: oL<ʒbIj1Û1¢2‹. Doesn't look like I can get it lower. \$\endgroup\$

Grimmy
– Grimmy

2019年09月06日 13:03:55 +00:00
Commented Sep 6, 2019 at 13:03
1

\$\begingroup\$ @Grimy I just had oL<ʒbIj1ÛSO2‹ and was trying to see where my error was. :) But I'm glad to see you aren't able to find a shorter version for one of my answers for a change. ;p (inb4 you find a shorter one after all xD) \$\endgroup\$

Kevin Cruijssen
– Kevin Cruijssen

2019年09月06日 13:06:41 +00:00
Commented Sep 6, 2019 at 13:06
1

\$\begingroup\$ @Grimy I have the feeling SO2‹ can be 3 bytes somehow perhaps, but I'm not seeing it and also not entirely sure.. There are some alternatives, like SO1~ or SÆ>d, but I'm unable to find a 3-byter. \$\endgroup\$

Kevin Cruijssen
– Kevin Cruijssen

2019年09月06日 13:13:25 +00:00
Commented Sep 6, 2019 at 13:13
1

\$\begingroup\$ 10 with a completely different approach \$\endgroup\$

Grimmy
– Grimmy

2019年09月06日 13:13:52 +00:00
Commented Sep 6, 2019 at 13:13
1

\$\begingroup\$ Your feeling was right, I just found a 3-byter: SO!. Pretty sure I have some old answers using 2‹ that could benefit from this as well. \$\endgroup\$

Grimmy
– Grimmy

2019年09月06日 13:45:15 +00:00
Commented Sep 6, 2019 at 13:45

| Show 2 more comments

Neil 184k12 gold badges76 silver badges290 bronze badges · Answer 13 · 2019-09-06 22:21:02Z

Retina, 26 bytes

.+
*0
L$w`.(.*)
$.`*1$'11ドル

Try it online! Outputs in binary. If that's not acceptable, then for 39 bytes:

.+
*0
L$w`.(.*)
$.`*1$'11ドル
+`10
011
%`1

Try it online! Explanation:

.+
*0

Convert the input into a string of n zeros.

L$w`.(.*)

Match all possible non-empty substrings.

$.`*1$'11ドル

For each substring, output: the prefix with 0s changed to 1s; the suffix; the match with the initial 0 changed to 1.

+`10
011
%`1

Convert from binary to decimal.

Unrelated String 23.9k3 gold badges37 silver badges61 bronze badges · Answer 14 · 2019-09-08 05:11:10Z

2

\$\begingroup\$

Brachylog, 27 bytes

1;0|⟦5;2z^(mLtT&-1↰+m↙T,L,0

Try it online!

Outputs out of order and with duplicates. If that's not okay, tack do onto the end.

Share

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answered Sep 8, 2019 at 5:11

Unrelated String's user avatar

Unrelated String

23.9k3 gold badges37 silver badges61 bronze badges

\$\endgroup\$

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Neil 184k12 gold badges76 silver badges290 bronze badges · Answer 15 · 2019-09-06 12:34:26Z

Charcoal, 19 bytes

I⮌E⊕θEι+−X2IθX2ιX2λ

Try it online! Link is to verbose version of code. Explanation:

 θ Input
 ⊕ Incremented
 E Map over implicit range
 ι Outer index
 E Map over implicit range
 Iθ Input cast to integer
 ι Outer index
 λ Inner index
 X2 X2 X2 Power of 2
 +− Subtract and add
 ⮌ Reverse outer list
I Cast to string
 Implicitly print

Grimmy 15.8k1 gold badge36 silver badges64 bronze badges · Answer 16 · 2019-09-06 13:36:51Z

1

\$\begingroup\$

Perl 5, 40 bytes

map{say$-;$-+=2**$_}0,0..$_-2;$_--&&redo

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answered Sep 6, 2019 at 13:36

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Grimmy

15.8k1 gold badge36 silver badges64 bronze badges

\$\endgroup\$

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the default. 8,4651 gold badge37 silver badges63 bronze badges · Answer 17 · 2019-09-07 06:10:03Z

Retina, 24 bytes

.+
*0
/0/+<0`(0)1|0$
11ドル

Outputs in binary. Input should have a trailing newline.

Attempt at explanation:

.+ #match the entire input
*0 #replace it with that many zeroes
/0/+<0`(0)1|0$ #while the string has a 0, substitute the first match and output
11ドル #if 01 is present in the string, replace it with 10, else replace the last character with $

I tried to avoid the 3 bytes long /0/ regex option by rearranging the options, but couldn't.

Try it online!

I don't think outputting in binary is allowed. There's a comment asking if it is allowed, but it's better to assume that you can't until the asker replies

AZTECCO 11k1 gold badge17 silver badges60 bronze badges · Answer 18 · 2019-09-13 14:26:17Z

C (clang), 73 bytes

o,j,y;f(x){for(o=j=0;printf("%d ",o),x;o+=y+!y,y+=y+!y)j=!j?y=0,--x:--j;}

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for(o=j=0;printf("%d ",o),x; o+=y+!y, y+=y+!y) 
// adds 1, 1+1=>2 , 2+2=> 4 .... sequence
 j=!j?y=0,--x:--j; 
// uses ternary instead of nested loop to decrement 'x' when 'j' go to 0

scrawl 1,3597 silver badges14 bronze badges · Answer 19 · 2019-09-13 15:11:38Z

1

\$\begingroup\$

k4, (削除) 28 (削除ここまで) 24 bytes

0,+\"j"2ドル xexp,/-1+|,\!:

@Grimy's approach ported to k4

edit: -4 thanks to ngn!

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edited Sep 17, 2019 at 10:33

answered Sep 13, 2019 at 15:11

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scrawl

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\$\endgroup\$

3

1

\$\begingroup\$ !:'1+|!: -> |,\!: \$\endgroup\$

ngn
– ngn

2019年09月16日 16:19:41 +00:00
Commented Sep 16, 2019 at 16:19
\$\begingroup\$ you can remove the space after xexp \$\endgroup\$

ngn
– ngn

2019年09月16日 16:21:49 +00:00
Commented Sep 16, 2019 at 16:21
\$\begingroup\$ @ngn, agh |,\!: seems so obvious now that i see it! \$\endgroup\$

scrawl
– scrawl

2019年09月17日 10:34:57 +00:00
Commented Sep 17, 2019 at 10:34

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Bit floating sequence

Challenge

Examples

19 Answers 19

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