New Order #2: Turn My WayIncrementing Gray CodesHow many consecutive descending numbers in my number?Give me the Gray Code list of bit width nNew Order #1: How does this feel?Give me the Gray Code list of bit width nPrint the intersection of sequencesHarmonious “Convergence”Sylvester's sequenceIncrementing Gray CodesThe dragon Curve sequenceAlternating bit smearingJumping NumbersMake me a metasequenceNew Order #1: How does this feel?
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New Order #2: Turn My Way
Incrementing Gray CodesHow many consecutive descending numbers in my number?Give me the Gray Code list of bit width nNew Order #1: How does this feel?Give me the Gray Code list of bit width nPrint the intersection of sequencesHarmonious “Convergence”Sylvester's sequenceIncrementing Gray CodesThe dragon Curve sequenceAlternating bit smearingJumping NumbersMake me a metasequenceNew Order #1: How does this feel?
$begingroup$
Introduction (may be ignored)
Putting all positive numbers in its regular order (1, 2, 3, ...) is a bit boring, isn't it? So here is a series of challenges around permutations (reshuffelings) of all positive numbers. This is the second challenge in this series. The first challenge can be found here.
In this challenge, we use Gray codes to rearrage the natural numbers. A Gray code, or "reflected binary code" is a binary encoding in such a way that two successive values differ in only one bit. A practical application of this encoding is to use it in rotary encoders, hence my reference to "Turn My Way".
Note that this encoding leaves some degree of freedom. For example, following binary 1100, there are four possible following codes: 1101, 1110, 1000 and 0100. This is why I will define $a(n)$ as the smallest, not previously used value that differs only one character in binary encoding. This sequence corresponds with A163252.
Since this is a "pure sequence" challenge, the task is to output $a(n)$ for a given $n$ as input, where $a(n)$ is A163252.
Task
Given an integer input $n$, output $a(n)$ in integer format (not in binary format).
$a(n)$ is defined as the least positive integer not occurring earlier in the sequence such that $a(n-1)$ and $a(n)$ differ in only one bit when written in binary.
Note: 1-based indexing is assumed here; you may use 0-based indexing, so $a(0) = 1; a(1) = 3$, etc. Please mention this in your answer if you choose to use this.
Test cases
Input | Output
--------------
1 | 1
5 | 4
20 | 18
50 | 48
123 | 121
1234 | 1333
3000 | 3030
9999 | 9997
Rules
- Input and output are integers (your program should at least support input and output in the range of 1 up to 32767)
- Invalid input (0, floats, strings, negative values, etc.) may lead to unpredicted output, errors or (un)defined behaviour. In A163252, $a(0)$ is defined as 0. For this challenge, we will ignore this.
- Default I/O rules apply.
Default loopholes are forbidden.- This is code-golf, so the shortest answers in bytes wins
Final note
See the following related (but not equal) PP&CG questions:
- Finding the next Gray code (input and output in binary)
- Generate the all Gray codes of length n
code-golf sequence
asked Mar 19 at 16:18
agtoeveragtoever
1,163421
$endgroup$
add a comment |
$begingroup$
Introduction (may be ignored)
Putting all positive numbers in its regular order (1, 2, 3, ...) is a bit boring, isn't it? So here is a series of challenges around permutations (reshuffelings) of all positive numbers. This is the second challenge in this series. The first challenge can be found here.
In this challenge, we use Gray codes to rearrage the natural numbers. A Gray code, or "reflected binary code" is a binary encoding in such a way that two successive values differ in only one bit. A practical application of this encoding is to use it in rotary encoders, hence my reference to "Turn My Way".
Note that this encoding leaves some degree of freedom. For example, following binary 1100, there are four possible following codes: 1101, 1110, 1000 and 0100. This is why I will define $a(n)$ as the smallest, not previously used value that differs only one character in binary encoding. This sequence corresponds with A163252.
Since this is a "pure sequence" challenge, the task is to output $a(n)$ for a given $n$ as input, where $a(n)$ is A163252.
Task
Given an integer input $n$, output $a(n)$ in integer format (not in binary format).
$a(n)$ is defined as the least positive integer not occurring earlier in the sequence such that $a(n-1)$ and $a(n)$ differ in only one bit when written in binary.
Note: 1-based indexing is assumed here; you may use 0-based indexing, so $a(0) = 1; a(1) = 3$, etc. Please mention this in your answer if you choose to use this.
Test cases
Input | Output
--------------
1 | 1
5 | 4
20 | 18
50 | 48
123 | 121
1234 | 1333
3000 | 3030
9999 | 9997
Rules
- Input and output are integers (your program should at least support input and output in the range of 1 up to 32767)
- Invalid input (0, floats, strings, negative values, etc.) may lead to unpredicted output, errors or (un)defined behaviour. In A163252, $a(0)$ is defined as 0. For this challenge, we will ignore this.
- Default I/O rules apply.
Default loopholes are forbidden.- This is code-golf, so the shortest answers in bytes wins
Final note
See the following related (but not equal) PP&CG questions:
- Finding the next Gray code (input and output in binary)
- Generate the all Gray codes of length n
code-golf sequence
asked Mar 19 at 16:18
agtoeveragtoever
1,163421
$endgroup$
add a comment |
$begingroup$
Introduction (may be ignored)
Putting all positive numbers in its regular order (1, 2, 3, ...) is a bit boring, isn't it? So here is a series of challenges around permutations (reshuffelings) of all positive numbers. This is the second challenge in this series. The first challenge can be found here.
In this challenge, we use Gray codes to rearrage the natural numbers. A Gray code, or "reflected binary code" is a binary encoding in such a way that two successive values differ in only one bit. A practical application of this encoding is to use it in rotary encoders, hence my reference to "Turn My Way".
Note that this encoding leaves some degree of freedom. For example, following binary 1100, there are four possible following codes: 1101, 1110, 1000 and 0100. This is why I will define $a(n)$ as the smallest, not previously used value that differs only one character in binary encoding. This sequence corresponds with A163252.
Since this is a "pure sequence" challenge, the task is to output $a(n)$ for a given $n$ as input, where $a(n)$ is A163252.
Task
Given an integer input $n$, output $a(n)$ in integer format (not in binary format).
$a(n)$ is defined as the least positive integer not occurring earlier in the sequence such that $a(n-1)$ and $a(n)$ differ in only one bit when written in binary.
Note: 1-based indexing is assumed here; you may use 0-based indexing, so $a(0) = 1; a(1) = 3$, etc. Please mention this in your answer if you choose to use this.
Test cases
Input | Output
--------------
1 | 1
5 | 4
20 | 18
50 | 48
123 | 121
1234 | 1333
3000 | 3030
9999 | 9997
Rules
- Input and output are integers (your program should at least support input and output in the range of 1 up to 32767)
- Invalid input (0, floats, strings, negative values, etc.) may lead to unpredicted output, errors or (un)defined behaviour. In A163252, $a(0)$ is defined as 0. For this challenge, we will ignore this.
- Default I/O rules apply.
Default loopholes are forbidden.- This is code-golf, so the shortest answers in bytes wins
Final note
See the following related (but not equal) PP&CG questions:
- Finding the next Gray code (input and output in binary)
- Generate the all Gray codes of length n
code-golf sequence
asked Mar 19 at 16:18
agtoeveragtoever
1,163421
$endgroup$
Introduction (may be ignored)
Putting all positive numbers in its regular order (1, 2, 3, ...) is a bit boring, isn't it? So here is a series of challenges around permutations (reshuffelings) of all positive numbers. This is the second challenge in this series. The first challenge can be found here.
In this challenge, we use Gray codes to rearrage the natural numbers. A Gray code, or "reflected binary code" is a binary encoding in such a way that two successive values differ in only one bit. A practical application of this encoding is to use it in rotary encoders, hence my reference to "Turn My Way".
Note that this encoding leaves some degree of freedom. For example, following binary 1100, there are four possible following codes: 1101, 1110, 1000 and 0100. This is why I will define $a(n)$ as the smallest, not previously used value that differs only one character in binary encoding. This sequence corresponds with A163252.
Since this is a "pure sequence" challenge, the task is to output $a(n)$ for a given $n$ as input, where $a(n)$ is A163252.
Task
Given an integer input $n$, output $a(n)$ in integer format (not in binary format).
$a(n)$ is defined as the least positive integer not occurring earlier in the sequence such that $a(n-1)$ and $a(n)$ differ in only one bit when written in binary.
Note: 1-based indexing is assumed here; you may use 0-based indexing, so $a(0) = 1; a(1) = 3$, etc. Please mention this in your answer if you choose to use this.
Test cases
Input | Output
--------------
1 | 1
5 | 4
20 | 18
50 | 48
123 | 121
1234 | 1333
3000 | 3030
9999 | 9997
Rules
- Input and output are integers (your program should at least support input and output in the range of 1 up to 32767)
- Invalid input (0, floats, strings, negative values, etc.) may lead to unpredicted output, errors or (un)defined behaviour. In A163252, $a(0)$ is defined as 0. For this challenge, we will ignore this.
- Default I/O rules apply.
Default loopholes are forbidden.- This is code-golf, so the shortest answers in bytes wins
Final note
See the following related (but not equal) PP&CG questions:
- Finding the next Gray code (input and output in binary)
- Generate the all Gray codes of length n
code-golf sequence
code-golf sequence
asked Mar 19 at 16:18
agtoeveragtoever
1,163421
asked Mar 19 at 16:18
agtoeveragtoever
1,163421
asked Mar 19 at 16:18
agtoeveragtoever
1,163421
asked Mar 19 at 16:18
agtoeveragtoever
1,163421
asked Mar 19 at 16:18
agtoeveragtoever
1,163421
1,163421
add a comment |
add a comment |
11 Answers
11
active
oldest
votes
$begingroup$
JavaScript (ES6), 65 bytes
1-indexed.
n=>for(o=p=[k=1];o[k]
Try it online!
Commented
n => // n = index of requested term
for( // for loop:
o = // o = storage object for the terms of the sequence
p = // p = last term found in the sequence
[k = 1]; // k = current term
o[k] // end
answered Mar 19 at 17:35
ArnauldArnauld
79.5k796330
$endgroup$
$begingroup$
On TIO, I get a stack overflow for n > ~1024. Any suggestions on how tot deal with that in Abu other environment? Rule: "your program should at least support input and output in theorie range of 1 up tot 32767"
$endgroup$
– agtoever
Mar 19 at 20:46
1
$begingroup$
@agtoever I've updated it to a non-recursive version.
$endgroup$
– Arnauld
Mar 19 at 21:16
add a comment |
$begingroup$
Jelly, 26 20 bytes
ṀBLŻ2*^1ị$ḟ⁸Ṃ;
0Ç⁸¡Ḣ
Try it online!
A full program that takes n as the single argument. Works for all test cases. Also note that, although not required, it handles n=0.
Explanation
Helper link: find next term and prepend
Ṁ | maximum of list so far
B | convert to binary
L | number of binary digits
Ż | 0..above number
2* | 2 to the power of each of the above
^ | exclusive or with...
1ị$ | ... the most recent term in the list so far
ḟ⁸ | filter out anything used already
Ṃ | find the minimum
; | prepend to existing list
Main link
0 | start with zero
Ç | call the above link
⁸¡ | and repeat n times
Ḣ | take the last term added
answered Mar 19 at 20:24
Nick KennedyNick Kennedy
91147
$endgroup$
add a comment |
$begingroup$
Java (JDK), 142 138 124 123 132 130 98 bytes
- increased to account for import, saved a byte thanks to @kevin-cruijssen
- switched collection to int array thanks to @olivier-grégoire
n->n.bitCount(j^k)>1;);return k;
Try it online!
answered Mar 20 at 7:00
Daniel WiddisDaniel Widdis
1595
$endgroup$
1
$begingroup$
I'm afraid imports has to be included in the byte-count. You can however golf theimport java.util.*;+Set s=new HashSet();tovar s=new java.util.HashSet();. In addition, the rest can be golfed to:Integer i=0,j,k=0;for(;i++<n;s.add(k=j))for(j=0;s.contains(++j)|i.bitCount(j^k)>1;);return k;. Nice answer nonetheless, so +1 from me. :)
$endgroup$
– Kevin Cruijssen
Mar 20 at 9:02
1
$begingroup$
Saved 2 more bytes usingStackrather thanHashSet. A lot slower but works!
$endgroup$
– Daniel Widdis
Mar 20 at 10:00
1
$begingroup$
Ah, of course, smart. And no matter how slow, if we can save a byte it's worth it for code-golf challenges. ;p I once had an answer that went from complexity $O(n)$ to $O(n^n)$ by saving a byte, haha xD
$endgroup$
– Kevin Cruijssen
Mar 20 at 10:01
2
$begingroup$
You can still golf it to 126 bytes with the second golf I suggested in my first comment. :)
$endgroup$
– Kevin Cruijssen
Mar 20 at 10:27
2
$begingroup$
98 bytes.
$endgroup$
– Olivier Grégoire
Mar 20 at 10:50
|
show 11 more comments
$begingroup$
Wolfram Language (Mathematica), 74 bytes
Last@Nest[#~Join~Min[BitXor[Last@#,2^Range[0,20]]~Complement~#]&,0,#]&
Try it online!
answered Mar 19 at 21:13
J42161217J42161217
13.5k21252
$endgroup$
add a comment |
$begingroup$
APL (Dyalog Extended), 46 bytes
⍵⌷2∘(~⍺∊⍵)∧1=≢⍸≠⌿↑⌽∘⊤¨⍺,⊃⌽⍵:⍵,⍺⋄⍵∇⍨⍺+1⍣⍵⊢1
Try it online!
answered Mar 19 at 22:25
voidhawkvoidhawk
1,36125
$endgroup$
add a comment |
$begingroup$
Stax, 19 bytes
±↔Φd┌û╓°╡ñ|Q5┌¿Éúb≈
Run and debug it
It's very slow, but not nearly as slow as the 17 byte solution I was working on. I think there's still some golfing potential. I will add an explanation after I get rid of the bloat.
answered Mar 19 at 23:11
recursiverecursive
5,6391322
$endgroup$
$begingroup$
You're 1 byte behind the shortest 05AB1E answer. Do you plan on optimizing this further? Otherwise I'll accept Kevin's answer...
$endgroup$
– agtoever
6 hours ago
add a comment |
$begingroup$
Charcoal, 65 bytes
≔⁰θＦＮ«⊞υθ≔¹ηＷ¬‹θ⊗η≦⊗ηＷ∧›η¹∨¬＆θη№υ⁻θη≧÷²ηＷ№υ⁻｜θη＆θη≦⊗η≔⁻｜θη＆θηθ»Ｉθ
Try it online! Link is to verbose version of code. Explanation:
≔⁰θ
Initialise the result to 0.
ＦＮ«
Loop n times.
⊞υθ
Save the previous result so that we don't use it again.
≔¹ηＷ¬‹θ⊗η≦⊗η
Find the highest bit in the previous result.
Ｗ∧›η¹∨¬＆θη№υ⁻θη≧÷²η
While that bit is greater than 1, if the bit is set in the previous result, try subtracting that bit to see if the result is an unseen result. This ensures that the potential results are tried in ascending order of value.
Ｗ№υ⁻｜θη＆θη≦⊗η
Now try XORing that bit with the previous result, doubling the bit until an unseen result is found. This handles the cases when a bit needs to be set, again in ascending order of value, but also the case when the least significant bit needs to be toggled, which the previous loop doesn't bother to test (because it's golfier to test for that here). If the previous loop found an unseen result then this loop never runs; if it didn't then this loop will uselessly retest those results.
≔⁻｜θη＆θηθ
Update the result by actually XORing the bit with it.
»Ｉθ
Output the final result at the end of the loop.
answered Mar 19 at 23:51
NeilNeil
82k745178
$endgroup$
add a comment |
$begingroup$
05AB1E, 21 20 18 bytes
ÎFˆ∞.Δ¯θy^bSO¯yå_*
Pretty inefficient, so the larger the input, the longer it takes to get the result. Does work for input 0 as well, though.
Try it online or verify the first $n$ terms.
Explanation:
Î # Push 0 and the input
F # Loop the input amount of times:
ˆ # Pop the current number and add it to the global_array
∞.Δ # Inner loop starting at 1 to find the first number which is truthy for:
¯θy^ # XOR the last number of the global_array with the loop-number `y`
b # Convert it to binary
SO # Sum it's binary digits
¯yå_ # Check if the loop-number `y` is NOT in the global_array yet
* # Multiply both (only if this is 1 (truthy), the inner loop will stop)
# (after the loops, output the top of the stack implicitly)
answered Mar 20 at 9:26
Kevin CruijssenKevin Cruijssen
41.3k567213
$endgroup$
add a comment |
$begingroup$
Python 2, 81 bytes
1-based indexing
l=[0];p=0
exec"n=0nwhile(p^n)&(p^n)-1or n in l:n+=1np=n;l+=p,;"*input()
print p
Try it online!
Python 2, 79 bytes
This takes a lot of time (9999 wasn't finished after running locally for 7 minutes)
l=0;p=0;n=input()
exec'p=min(p^2**k for k in range(n)-l);l|=p;'*n
print p
Try it online!
answered Mar 19 at 20:02
ovsovs
19.3k21160
$endgroup$
1
$begingroup$
Maximum input 32767 isn't supported (the default recursion depth isn't system-dependent).
$endgroup$
– Erik the Outgolfer
Mar 19 at 21:37
$begingroup$
Even the given test case 9999 isn't supported. :)
$endgroup$
– Daniel Widdis
Mar 20 at 8:32
$begingroup$
@EriktheOutgolfer Changed it to an iterative approach, probably still doesn't finish in time on TIO, but runs locally just fine.
$endgroup$
– ovs
Mar 20 at 10:23
$begingroup$
@ovs Oh, timeouts alone don't matter.
$endgroup$
– Erik the Outgolfer
Mar 20 at 12:06
add a comment |
$begingroup$
Haskell, 101 bytes
import Data.Bits
(u!n)0=n
(u!n)m|q<-minimum[x|r<-[0..62],x<-[xor(2^r)n],notElem x u]=(n:u)!q$m-1
[]!0
Try it online!
It seems a shame to incur an import just for xor, but I haven't found a good work-around yet. I also wonder if there's a better way to express the loop.
answered Mar 20 at 21:11
dfeuerdfeuer
894910
$endgroup$
add a comment |
$begingroup$
R, 90 bytes
function(n)A=1
while(sum(A
Try it online!
answered 2 days ago
GiuseppeGiuseppe
17k31152
$endgroup$
add a comment |
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11 Answers
11
active
oldest
votes
11 Answers
11
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
JavaScript (ES6), 65 bytes
1-indexed.
n=>for(o=p=[k=1];o[k]
Try it online!
Commented
n => // n = index of requested term
for( // for loop:
o = // o = storage object for the terms of the sequence
p = // p = last term found in the sequence
[k = 1]; // k = current term
o[k] // end
answered Mar 19 at 17:35
ArnauldArnauld
79.5k796330
$endgroup$
$begingroup$
On TIO, I get a stack overflow for n > ~1024. Any suggestions on how tot deal with that in Abu other environment? Rule: "your program should at least support input and output in theorie range of 1 up tot 32767"
$endgroup$
– agtoever
Mar 19 at 20:46
1
$begingroup$
@agtoever I've updated it to a non-recursive version.
$endgroup$
– Arnauld
Mar 19 at 21:16
add a comment |
$begingroup$
JavaScript (ES6), 65 bytes
1-indexed.
n=>for(o=p=[k=1];o[k]
Try it online!
Commented
n => // n = index of requested term
for( // for loop:
o = // o = storage object for the terms of the sequence
p = // p = last term found in the sequence
[k = 1]; // k = current term
o[k] // end
answered Mar 19 at 17:35
ArnauldArnauld
79.5k796330
$endgroup$
$begingroup$
On TIO, I get a stack overflow for n > ~1024. Any suggestions on how tot deal with that in Abu other environment? Rule: "your program should at least support input and output in theorie range of 1 up tot 32767"
$endgroup$
– agtoever
Mar 19 at 20:46
1
$begingroup$
@agtoever I've updated it to a non-recursive version.
$endgroup$
– Arnauld
Mar 19 at 21:16
add a comment |
$begingroup$
JavaScript (ES6), 65 bytes
1-indexed.
n=>for(o=p=[k=1];o[k]
Try it online!
Commented
n => // n = index of requested term
for( // for loop:
o = // o = storage object for the terms of the sequence
p = // p = last term found in the sequence
[k = 1]; // k = current term
o[k] // end
answered Mar 19 at 17:35
ArnauldArnauld
79.5k796330
$endgroup$
JavaScript (ES6), 65 bytes
1-indexed.
n=>for(o=p=[k=1];o[k]
Try it online!
Commented
n => // n = index of requested term
for( // for loop:
o = // o = storage object for the terms of the sequence
p = // p = last term found in the sequence
[k = 1]; // k = current term
o[k] // end
answered Mar 19 at 17:35
ArnauldArnauld
79.5k796330
edited Mar 19 at 21:37
answered Mar 19 at 17:35
ArnauldArnauld
79.5k796330
answered Mar 19 at 17:35
ArnauldArnauld
79.5k796330
answered Mar 19 at 17:35
ArnauldArnauld
79.5k796330
79.5k796330
$begingroup$
On TIO, I get a stack overflow for n > ~1024. Any suggestions on how tot deal with that in Abu other environment? Rule: "your program should at least support input and output in theorie range of 1 up tot 32767"
$endgroup$
– agtoever
Mar 19 at 20:46
1
$begingroup$
@agtoever I've updated it to a non-recursive version.
$endgroup$
– Arnauld
Mar 19 at 21:16
add a comment |
$begingroup$
On TIO, I get a stack overflow for n > ~1024. Any suggestions on how tot deal with that in Abu other environment? Rule: "your program should at least support input and output in theorie range of 1 up tot 32767"
$endgroup$
– agtoever
Mar 19 at 20:46
1
$begingroup$
@agtoever I've updated it to a non-recursive version.
$endgroup$
– Arnauld
Mar 19 at 21:16
$begingroup$
On TIO, I get a stack overflow for n > ~1024. Any suggestions on how tot deal with that in Abu other environment? Rule: "your program should at least support input and output in theorie range of 1 up tot 32767"
$endgroup$
– agtoever
Mar 19 at 20:46
$begingroup$
On TIO, I get a stack overflow for n > ~1024. Any suggestions on how tot deal with that in Abu other environment? Rule: "your program should at least support input and output in theorie range of 1 up tot 32767"
$endgroup$
– agtoever
Mar 19 at 20:46
1
1
$begingroup$
@agtoever I've updated it to a non-recursive version.
$endgroup$
– Arnauld
Mar 19 at 21:16
$begingroup$
@agtoever I've updated it to a non-recursive version.
$endgroup$
– Arnauld
Mar 19 at 21:16
add a comment |
$begingroup$
Jelly, 26 20 bytes
ṀBLŻ2*^1ị$ḟ⁸Ṃ;
0Ç⁸¡Ḣ
Try it online!
A full program that takes n as the single argument. Works for all test cases. Also note that, although not required, it handles n=0.
Explanation
Helper link: find next term and prepend
Ṁ | maximum of list so far
B | convert to binary
L | number of binary digits
Ż | 0..above number
2* | 2 to the power of each of the above
^ | exclusive or with...
1ị$ | ... the most recent term in the list so far
ḟ⁸ | filter out anything used already
Ṃ | find the minimum
; | prepend to existing list
Main link
0 | start with zero
Ç | call the above link
⁸¡ | and repeat n times
Ḣ | take the last term added
answered Mar 19 at 20:24
Nick KennedyNick Kennedy
91147
$endgroup$
add a comment |
$begingroup$
Jelly, 26 20 bytes
ṀBLŻ2*^1ị$ḟ⁸Ṃ;
0Ç⁸¡Ḣ
Try it online!
A full program that takes n as the single argument. Works for all test cases. Also note that, although not required, it handles n=0.
Explanation
Helper link: find next term and prepend
Ṁ | maximum of list so far
B | convert to binary
L | number of binary digits
Ż | 0..above number
2* | 2 to the power of each of the above
^ | exclusive or with...
1ị$ | ... the most recent term in the list so far
ḟ⁸ | filter out anything used already
Ṃ | find the minimum
; | prepend to existing list
Main link
0 | start with zero
Ç | call the above link
⁸¡ | and repeat n times
Ḣ | take the last term added
answered Mar 19 at 20:24
Nick KennedyNick Kennedy
91147
$endgroup$
add a comment |
$begingroup$
Jelly, 26 20 bytes
ṀBLŻ2*^1ị$ḟ⁸Ṃ;
0Ç⁸¡Ḣ
Try it online!
A full program that takes n as the single argument. Works for all test cases. Also note that, although not required, it handles n=0.
Explanation
Helper link: find next term and prepend
Ṁ | maximum of list so far
B | convert to binary
L | number of binary digits
Ż | 0..above number
2* | 2 to the power of each of the above
^ | exclusive or with...
1ị$ | ... the most recent term in the list so far
ḟ⁸ | filter out anything used already
Ṃ | find the minimum
; | prepend to existing list
Main link
0 | start with zero
Ç | call the above link
⁸¡ | and repeat n times
Ḣ | take the last term added
answered Mar 19 at 20:24
Nick KennedyNick Kennedy
91147
$endgroup$
Jelly, 26 20 bytes
ṀBLŻ2*^1ị$ḟ⁸Ṃ;
0Ç⁸¡Ḣ
Try it online!
A full program that takes n as the single argument. Works for all test cases. Also note that, although not required, it handles n=0.
Explanation
Helper link: find next term and prepend
Ṁ | maximum of list so far
B | convert to binary
L | number of binary digits
Ż | 0..above number
2* | 2 to the power of each of the above
^ | exclusive or with...
1ị$ | ... the most recent term in the list so far
ḟ⁸ | filter out anything used already
Ṃ | find the minimum
; | prepend to existing list
Main link
0 | start with zero
Ç | call the above link
⁸¡ | and repeat n times
Ḣ | take the last term added
answered Mar 19 at 20:24
Nick KennedyNick Kennedy
91147
edited Mar 19 at 23:13
answered Mar 19 at 20:24
Nick KennedyNick Kennedy
91147
answered Mar 19 at 20:24
Nick KennedyNick Kennedy
91147
answered Mar 19 at 20:24
Nick KennedyNick Kennedy
91147
91147
add a comment |
add a comment |
$begingroup$
Java (JDK), 142 138 124 123 132 130 98 bytes
- increased to account for import, saved a byte thanks to @kevin-cruijssen
- switched collection to int array thanks to @olivier-grégoire
n->n.bitCount(j^k)>1;);return k;
Try it online!
answered Mar 20 at 7:00
Daniel WiddisDaniel Widdis
1595
$endgroup$
1
$begingroup$
I'm afraid imports has to be included in the byte-count. You can however golf theimport java.util.*;+Set s=new HashSet();tovar s=new java.util.HashSet();. In addition, the rest can be golfed to:Integer i=0,j,k=0;for(;i++<n;s.add(k=j))for(j=0;s.contains(++j)|i.bitCount(j^k)>1;);return k;. Nice answer nonetheless, so +1 from me. :)
$endgroup$
– Kevin Cruijssen
Mar 20 at 9:02
1
$begingroup$
Saved 2 more bytes usingStackrather thanHashSet. A lot slower but works!
$endgroup$
– Daniel Widdis
Mar 20 at 10:00
1
$begingroup$
Ah, of course, smart. And no matter how slow, if we can save a byte it's worth it for code-golf challenges. ;p I once had an answer that went from complexity $O(n)$ to $O(n^n)$ by saving a byte, haha xD
$endgroup$
– Kevin Cruijssen
Mar 20 at 10:01
2
$begingroup$
You can still golf it to 126 bytes with the second golf I suggested in my first comment. :)
$endgroup$
– Kevin Cruijssen
Mar 20 at 10:27
2
$begingroup$
98 bytes.
$endgroup$
– Olivier Grégoire
Mar 20 at 10:50
|
show 11 more comments
$begingroup$
Java (JDK), 142 138 124 123 132 130 98 bytes
- increased to account for import, saved a byte thanks to @kevin-cruijssen
- switched collection to int array thanks to @olivier-grégoire
n->n.bitCount(j^k)>1;);return k;
Try it online!
answered Mar 20 at 7:00
Daniel WiddisDaniel Widdis
1595
$endgroup$
1
$begingroup$
I'm afraid imports has to be included in the byte-count. You can however golf theimport java.util.*;+Set s=new HashSet();tovar s=new java.util.HashSet();. In addition, the rest can be golfed to:Integer i=0,j,k=0;for(;i++<n;s.add(k=j))for(j=0;s.contains(++j)|i.bitCount(j^k)>1;);return k;. Nice answer nonetheless, so +1 from me. :)
$endgroup$
– Kevin Cruijssen
Mar 20 at 9:02
1
$begingroup$
Saved 2 more bytes usingStackrather thanHashSet. A lot slower but works!
$endgroup$
– Daniel Widdis
Mar 20 at 10:00
1
$begingroup$
Ah, of course, smart. And no matter how slow, if we can save a byte it's worth it for code-golf challenges. ;p I once had an answer that went from complexity $O(n)$ to $O(n^n)$ by saving a byte, haha xD
$endgroup$
– Kevin Cruijssen
Mar 20 at 10:01
2
$begingroup$
You can still golf it to 126 bytes with the second golf I suggested in my first comment. :)
$endgroup$
– Kevin Cruijssen
Mar 20 at 10:27
2
$begingroup$
98 bytes.
$endgroup$
– Olivier Grégoire
Mar 20 at 10:50
|
show 11 more comments
$begingroup$
Java (JDK), 142 138 124 123 132 130 98 bytes
- increased to account for import, saved a byte thanks to @kevin-cruijssen
- switched collection to int array thanks to @olivier-grégoire
n->n.bitCount(j^k)>1;);return k;
Try it online!
answered Mar 20 at 7:00
Daniel WiddisDaniel Widdis
1595
$endgroup$
Java (JDK), 142 138 124 123 132 130 98 bytes
- increased to account for import, saved a byte thanks to @kevin-cruijssen
- switched collection to int array thanks to @olivier-grégoire
n->n.bitCount(j^k)>1;);return k;
Try it online!
answered Mar 20 at 7:00
Daniel WiddisDaniel Widdis
1595
edited Mar 20 at 19:54
answered Mar 20 at 7:00
Daniel WiddisDaniel Widdis
1595
answered Mar 20 at 7:00
Daniel WiddisDaniel Widdis
1595
answered Mar 20 at 7:00
Daniel WiddisDaniel Widdis
1595
1595
1
$begingroup$
I'm afraid imports has to be included in the byte-count. You can however golf theimport java.util.*;+Set s=new HashSet();tovar s=new java.util.HashSet();. In addition, the rest can be golfed to:Integer i=0,j,k=0;for(;i++<n;s.add(k=j))for(j=0;s.contains(++j)|i.bitCount(j^k)>1;);return k;. Nice answer nonetheless, so +1 from me. :)
$endgroup$
– Kevin Cruijssen
Mar 20 at 9:02
1
$begingroup$
Saved 2 more bytes usingStackrather thanHashSet. A lot slower but works!
$endgroup$
– Daniel Widdis
Mar 20 at 10:00
1
$begingroup$
Ah, of course, smart. And no matter how slow, if we can save a byte it's worth it for code-golf challenges. ;p I once had an answer that went from complexity $O(n)$ to $O(n^n)$ by saving a byte, haha xD
$endgroup$
– Kevin Cruijssen
Mar 20 at 10:01
2
$begingroup$
You can still golf it to 126 bytes with the second golf I suggested in my first comment. :)
$endgroup$
– Kevin Cruijssen
Mar 20 at 10:27
2
$begingroup$
98 bytes.
$endgroup$
– Olivier Grégoire
Mar 20 at 10:50
|
show 11 more comments
1
$begingroup$
I'm afraid imports has to be included in the byte-count. You can however golf theimport java.util.*;+Set s=new HashSet();tovar s=new java.util.HashSet();. In addition, the rest can be golfed to:Integer i=0,j,k=0;for(;i++<n;s.add(k=j))for(j=0;s.contains(++j)|i.bitCount(j^k)>1;);return k;. Nice answer nonetheless, so +1 from me. :)
$endgroup$
– Kevin Cruijssen
Mar 20 at 9:02
1
$begingroup$
Saved 2 more bytes usingStackrather thanHashSet. A lot slower but works!
$endgroup$
– Daniel Widdis
Mar 20 at 10:00
1
$begingroup$
Ah, of course, smart. And no matter how slow, if we can save a byte it's worth it for code-golf challenges. ;p I once had an answer that went from complexity $O(n)$ to $O(n^n)$ by saving a byte, haha xD
$endgroup$
– Kevin Cruijssen
Mar 20 at 10:01
2
$begingroup$
You can still golf it to 126 bytes with the second golf I suggested in my first comment. :)
$endgroup$
– Kevin Cruijssen
Mar 20 at 10:27
2
$begingroup$
98 bytes.
$endgroup$
– Olivier Grégoire
Mar 20 at 10:50
1
1
$begingroup$
I'm afraid imports has to be included in the byte-count. You can however golf the
import java.util.*;+Set s=new HashSet(); to var s=new java.util.HashSet();. In addition, the rest can be golfed to: Integer i=0,j,k=0;for(;i++<n;s.add(k=j))for(j=0;s.contains(++j)|i.bitCount(j^k)>1;);return k;. Nice answer nonetheless, so +1 from me. :)$endgroup$
– Kevin Cruijssen
Mar 20 at 9:02
$begingroup$
I'm afraid imports has to be included in the byte-count. You can however golf the
import java.util.*;+Set s=new HashSet(); to var s=new java.util.HashSet();. In addition, the rest can be golfed to: Integer i=0,j,k=0;for(;i++<n;s.add(k=j))for(j=0;s.contains(++j)|i.bitCount(j^k)>1;);return k;. Nice answer nonetheless, so +1 from me. :)$endgroup$
– Kevin Cruijssen
Mar 20 at 9:02
1
1
$begingroup$
Saved 2 more bytes using
Stack rather than HashSet. A lot slower but works!$endgroup$
– Daniel Widdis
Mar 20 at 10:00
$begingroup$
Saved 2 more bytes using
Stack rather than HashSet. A lot slower but works!$endgroup$
– Daniel Widdis
Mar 20 at 10:00
1
1
$begingroup$
Ah, of course, smart. And no matter how slow, if we can save a byte it's worth it for code-golf challenges. ;p I once had an answer that went from complexity $O(n)$ to $O(n^n)$ by saving a byte, haha xD
$endgroup$
– Kevin Cruijssen
Mar 20 at 10:01
$begingroup$
Ah, of course, smart. And no matter how slow, if we can save a byte it's worth it for code-golf challenges. ;p I once had an answer that went from complexity $O(n)$ to $O(n^n)$ by saving a byte, haha xD
$endgroup$
– Kevin Cruijssen
Mar 20 at 10:01
2
2
$begingroup$
You can still golf it to 126 bytes with the second golf I suggested in my first comment. :)
$endgroup$
– Kevin Cruijssen
Mar 20 at 10:27
$begingroup$
You can still golf it to 126 bytes with the second golf I suggested in my first comment. :)
$endgroup$
– Kevin Cruijssen
Mar 20 at 10:27
2
2
$begingroup$
98 bytes.
$endgroup$
– Olivier Grégoire
Mar 20 at 10:50
$begingroup$
98 bytes.
$endgroup$
– Olivier Grégoire
Mar 20 at 10:50
|
show 11 more comments
$begingroup$
Wolfram Language (Mathematica), 74 bytes
Last@Nest[#~Join~Min[BitXor[Last@#,2^Range[0,20]]~Complement~#]&,0,#]&
Try it online!
answered Mar 19 at 21:13
J42161217J42161217
13.5k21252
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 74 bytes
Last@Nest[#~Join~Min[BitXor[Last@#,2^Range[0,20]]~Complement~#]&,0,#]&
Try it online!
answered Mar 19 at 21:13
J42161217J42161217
13.5k21252
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 74 bytes
Last@Nest[#~Join~Min[BitXor[Last@#,2^Range[0,20]]~Complement~#]&,0,#]&
Try it online!
answered Mar 19 at 21:13
J42161217J42161217
13.5k21252
$endgroup$
Wolfram Language (Mathematica), 74 bytes
Last@Nest[#~Join~Min[BitXor[Last@#,2^Range[0,20]]~Complement~#]&,0,#]&
Try it online!
answered Mar 19 at 21:13
J42161217J42161217
13.5k21252
answered Mar 19 at 21:13
J42161217J42161217
13.5k21252
answered Mar 19 at 21:13
J42161217J42161217
13.5k21252
answered Mar 19 at 21:13
J42161217J42161217
13.5k21252
13.5k21252
add a comment |
add a comment |
$begingroup$
APL (Dyalog Extended), 46 bytes
⍵⌷2∘(~⍺∊⍵)∧1=≢⍸≠⌿↑⌽∘⊤¨⍺,⊃⌽⍵:⍵,⍺⋄⍵∇⍨⍺+1⍣⍵⊢1
Try it online!
answered Mar 19 at 22:25
voidhawkvoidhawk
1,36125
$endgroup$
add a comment |
$begingroup$
APL (Dyalog Extended), 46 bytes
⍵⌷2∘(~⍺∊⍵)∧1=≢⍸≠⌿↑⌽∘⊤¨⍺,⊃⌽⍵:⍵,⍺⋄⍵∇⍨⍺+1⍣⍵⊢1
Try it online!
answered Mar 19 at 22:25
voidhawkvoidhawk
1,36125
$endgroup$
add a comment |
$begingroup$
APL (Dyalog Extended), 46 bytes
⍵⌷2∘(~⍺∊⍵)∧1=≢⍸≠⌿↑⌽∘⊤¨⍺,⊃⌽⍵:⍵,⍺⋄⍵∇⍨⍺+1⍣⍵⊢1
Try it online!
answered Mar 19 at 22:25
voidhawkvoidhawk
1,36125
$endgroup$
APL (Dyalog Extended), 46 bytes
⍵⌷2∘(~⍺∊⍵)∧1=≢⍸≠⌿↑⌽∘⊤¨⍺,⊃⌽⍵:⍵,⍺⋄⍵∇⍨⍺+1⍣⍵⊢1
Try it online!
answered Mar 19 at 22:25
voidhawkvoidhawk
1,36125
answered Mar 19 at 22:25
voidhawkvoidhawk
1,36125
answered Mar 19 at 22:25
voidhawkvoidhawk
1,36125
answered Mar 19 at 22:25
voidhawkvoidhawk
1,36125
1,36125
add a comment |
add a comment |
$begingroup$
Stax, 19 bytes
±↔Φd┌û╓°╡ñ|Q5┌¿Éúb≈
Run and debug it
It's very slow, but not nearly as slow as the 17 byte solution I was working on. I think there's still some golfing potential. I will add an explanation after I get rid of the bloat.
answered Mar 19 at 23:11
recursiverecursive
5,6391322
$endgroup$
$begingroup$
You're 1 byte behind the shortest 05AB1E answer. Do you plan on optimizing this further? Otherwise I'll accept Kevin's answer...
$endgroup$
– agtoever
6 hours ago
add a comment |
$begingroup$
Stax, 19 bytes
±↔Φd┌û╓°╡ñ|Q5┌¿Éúb≈
Run and debug it
It's very slow, but not nearly as slow as the 17 byte solution I was working on. I think there's still some golfing potential. I will add an explanation after I get rid of the bloat.
answered Mar 19 at 23:11
recursiverecursive
5,6391322
$endgroup$
$begingroup$
You're 1 byte behind the shortest 05AB1E answer. Do you plan on optimizing this further? Otherwise I'll accept Kevin's answer...
$endgroup$
– agtoever
6 hours ago
add a comment |
$begingroup$
Stax, 19 bytes
±↔Φd┌û╓°╡ñ|Q5┌¿Éúb≈
Run and debug it
It's very slow, but not nearly as slow as the 17 byte solution I was working on. I think there's still some golfing potential. I will add an explanation after I get rid of the bloat.
answered Mar 19 at 23:11
recursiverecursive
5,6391322
$endgroup$
Stax, 19 bytes
±↔Φd┌û╓°╡ñ|Q5┌¿Éúb≈
Run and debug it
It's very slow, but not nearly as slow as the 17 byte solution I was working on. I think there's still some golfing potential. I will add an explanation after I get rid of the bloat.
answered Mar 19 at 23:11
recursiverecursive
5,6391322
edited Mar 19 at 23:30
answered Mar 19 at 23:11
recursiverecursive
5,6391322
answered Mar 19 at 23:11
recursiverecursive
5,6391322
answered Mar 19 at 23:11
recursiverecursive
5,6391322
5,6391322
$begingroup$
You're 1 byte behind the shortest 05AB1E answer. Do you plan on optimizing this further? Otherwise I'll accept Kevin's answer...
$endgroup$
– agtoever
6 hours ago
add a comment |
$begingroup$
You're 1 byte behind the shortest 05AB1E answer. Do you plan on optimizing this further? Otherwise I'll accept Kevin's answer...
$endgroup$
– agtoever
6 hours ago
$begingroup$
You're 1 byte behind the shortest 05AB1E answer. Do you plan on optimizing this further? Otherwise I'll accept Kevin's answer...
$endgroup$
– agtoever
6 hours ago
$begingroup$
You're 1 byte behind the shortest 05AB1E answer. Do you plan on optimizing this further? Otherwise I'll accept Kevin's answer...
$endgroup$
– agtoever
6 hours ago
add a comment |
$begingroup$
Charcoal, 65 bytes
≔⁰θＦＮ«⊞υθ≔¹ηＷ¬‹θ⊗η≦⊗ηＷ∧›η¹∨¬＆θη№υ⁻θη≧÷²ηＷ№υ⁻｜θη＆θη≦⊗η≔⁻｜θη＆θηθ»Ｉθ
Try it online! Link is to verbose version of code. Explanation:
≔⁰θ
Initialise the result to 0.
ＦＮ«
Loop n times.
⊞υθ
Save the previous result so that we don't use it again.
≔¹ηＷ¬‹θ⊗η≦⊗η
Find the highest bit in the previous result.
Ｗ∧›η¹∨¬＆θη№υ⁻θη≧÷²η
While that bit is greater than 1, if the bit is set in the previous result, try subtracting that bit to see if the result is an unseen result. This ensures that the potential results are tried in ascending order of value.
Ｗ№υ⁻｜θη＆θη≦⊗η
Now try XORing that bit with the previous result, doubling the bit until an unseen result is found. This handles the cases when a bit needs to be set, again in ascending order of value, but also the case when the least significant bit needs to be toggled, which the previous loop doesn't bother to test (because it's golfier to test for that here). If the previous loop found an unseen result then this loop never runs; if it didn't then this loop will uselessly retest those results.
≔⁻｜θη＆θηθ
Update the result by actually XORing the bit with it.
»Ｉθ
Output the final result at the end of the loop.
answered Mar 19 at 23:51
NeilNeil
82k745178
$endgroup$
add a comment |
$begingroup$
Charcoal, 65 bytes
≔⁰θＦＮ«⊞υθ≔¹ηＷ¬‹θ⊗η≦⊗ηＷ∧›η¹∨¬＆θη№υ⁻θη≧÷²ηＷ№υ⁻｜θη＆θη≦⊗η≔⁻｜θη＆θηθ»Ｉθ
Try it online! Link is to verbose version of code. Explanation:
≔⁰θ
Initialise the result to 0.
ＦＮ«
Loop n times.
⊞υθ
Save the previous result so that we don't use it again.
≔¹ηＷ¬‹θ⊗η≦⊗η
Find the highest bit in the previous result.
Ｗ∧›η¹∨¬＆θη№υ⁻θη≧÷²η
While that bit is greater than 1, if the bit is set in the previous result, try subtracting that bit to see if the result is an unseen result. This ensures that the potential results are tried in ascending order of value.
Ｗ№υ⁻｜θη＆θη≦⊗η
Now try XORing that bit with the previous result, doubling the bit until an unseen result is found. This handles the cases when a bit needs to be set, again in ascending order of value, but also the case when the least significant bit needs to be toggled, which the previous loop doesn't bother to test (because it's golfier to test for that here). If the previous loop found an unseen result then this loop never runs; if it didn't then this loop will uselessly retest those results.
≔⁻｜θη＆θηθ
Update the result by actually XORing the bit with it.
»Ｉθ
Output the final result at the end of the loop.
answered Mar 19 at 23:51
NeilNeil
82k745178
$endgroup$
add a comment |
$begingroup$
Charcoal, 65 bytes
≔⁰θＦＮ«⊞υθ≔¹ηＷ¬‹θ⊗η≦⊗ηＷ∧›η¹∨¬＆θη№υ⁻θη≧÷²ηＷ№υ⁻｜θη＆θη≦⊗η≔⁻｜θη＆θηθ»Ｉθ
Try it online! Link is to verbose version of code. Explanation:
≔⁰θ
Initialise the result to 0.
ＦＮ«
Loop n times.
⊞υθ
Save the previous result so that we don't use it again.
≔¹ηＷ¬‹θ⊗η≦⊗η
Find the highest bit in the previous result.
Ｗ∧›η¹∨¬＆θη№υ⁻θη≧÷²η
While that bit is greater than 1, if the bit is set in the previous result, try subtracting that bit to see if the result is an unseen result. This ensures that the potential results are tried in ascending order of value.
Ｗ№υ⁻｜θη＆θη≦⊗η
Now try XORing that bit with the previous result, doubling the bit until an unseen result is found. This handles the cases when a bit needs to be set, again in ascending order of value, but also the case when the least significant bit needs to be toggled, which the previous loop doesn't bother to test (because it's golfier to test for that here). If the previous loop found an unseen result then this loop never runs; if it didn't then this loop will uselessly retest those results.
≔⁻｜θη＆θηθ
Update the result by actually XORing the bit with it.
»Ｉθ
Output the final result at the end of the loop.
answered Mar 19 at 23:51
NeilNeil
82k745178
$endgroup$
Charcoal, 65 bytes
≔⁰θＦＮ«⊞υθ≔¹ηＷ¬‹θ⊗η≦⊗ηＷ∧›η¹∨¬＆θη№υ⁻θη≧÷²ηＷ№υ⁻｜θη＆θη≦⊗η≔⁻｜θη＆θηθ»Ｉθ
Try it online! Link is to verbose version of code. Explanation:
≔⁰θ
Initialise the result to 0.
ＦＮ«
Loop n times.
⊞υθ
Save the previous result so that we don't use it again.
≔¹ηＷ¬‹θ⊗η≦⊗η
Find the highest bit in the previous result.
Ｗ∧›η¹∨¬＆θη№υ⁻θη≧÷²η
While that bit is greater than 1, if the bit is set in the previous result, try subtracting that bit to see if the result is an unseen result. This ensures that the potential results are tried in ascending order of value.
Ｗ№υ⁻｜θη＆θη≦⊗η
Now try XORing that bit with the previous result, doubling the bit until an unseen result is found. This handles the cases when a bit needs to be set, again in ascending order of value, but also the case when the least significant bit needs to be toggled, which the previous loop doesn't bother to test (because it's golfier to test for that here). If the previous loop found an unseen result then this loop never runs; if it didn't then this loop will uselessly retest those results.
≔⁻｜θη＆θηθ
Update the result by actually XORing the bit with it.
»Ｉθ
Output the final result at the end of the loop.
answered Mar 19 at 23:51
NeilNeil
82k745178
answered Mar 19 at 23:51
NeilNeil
82k745178
answered Mar 19 at 23:51
NeilNeil
82k745178
answered Mar 19 at 23:51
NeilNeil
82k745178
82k745178
add a comment |
add a comment |
$begingroup$
05AB1E, 21 20 18 bytes
ÎFˆ∞.Δ¯θy^bSO¯yå_*
Pretty inefficient, so the larger the input, the longer it takes to get the result. Does work for input 0 as well, though.
Try it online or verify the first $n$ terms.
Explanation:
Î # Push 0 and the input
F # Loop the input amount of times:
ˆ # Pop the current number and add it to the global_array
∞.Δ # Inner loop starting at 1 to find the first number which is truthy for:
¯θy^ # XOR the last number of the global_array with the loop-number `y`
b # Convert it to binary
SO # Sum it's binary digits
¯yå_ # Check if the loop-number `y` is NOT in the global_array yet
* # Multiply both (only if this is 1 (truthy), the inner loop will stop)
# (after the loops, output the top of the stack implicitly)
answered Mar 20 at 9:26
Kevin CruijssenKevin Cruijssen
41.3k567213
$endgroup$
add a comment |
$begingroup$
05AB1E, 21 20 18 bytes
ÎFˆ∞.Δ¯θy^bSO¯yå_*
Pretty inefficient, so the larger the input, the longer it takes to get the result. Does work for input 0 as well, though.
Try it online or verify the first $n$ terms.
Explanation:
Î # Push 0 and the input
F # Loop the input amount of times:
ˆ # Pop the current number and add it to the global_array
∞.Δ # Inner loop starting at 1 to find the first number which is truthy for:
¯θy^ # XOR the last number of the global_array with the loop-number `y`
b # Convert it to binary
SO # Sum it's binary digits
¯yå_ # Check if the loop-number `y` is NOT in the global_array yet
* # Multiply both (only if this is 1 (truthy), the inner loop will stop)
# (after the loops, output the top of the stack implicitly)
answered Mar 20 at 9:26
Kevin CruijssenKevin Cruijssen
41.3k567213
$endgroup$
add a comment |
$begingroup$
05AB1E, 21 20 18 bytes
ÎFˆ∞.Δ¯θy^bSO¯yå_*
Pretty inefficient, so the larger the input, the longer it takes to get the result. Does work for input 0 as well, though.
Try it online or verify the first $n$ terms.
Explanation:
Î # Push 0 and the input
F # Loop the input amount of times:
ˆ # Pop the current number and add it to the global_array
∞.Δ # Inner loop starting at 1 to find the first number which is truthy for:
¯θy^ # XOR the last number of the global_array with the loop-number `y`
b # Convert it to binary
SO # Sum it's binary digits
¯yå_ # Check if the loop-number `y` is NOT in the global_array yet
* # Multiply both (only if this is 1 (truthy), the inner loop will stop)
# (after the loops, output the top of the stack implicitly)
answered Mar 20 at 9:26
Kevin CruijssenKevin Cruijssen
41.3k567213
$endgroup$
05AB1E, 21 20 18 bytes
ÎFˆ∞.Δ¯θy^bSO¯yå_*
Pretty inefficient, so the larger the input, the longer it takes to get the result. Does work for input 0 as well, though.
Try it online or verify the first $n$ terms.
Explanation:
Î # Push 0 and the input
F # Loop the input amount of times:
ˆ # Pop the current number and add it to the global_array
∞.Δ # Inner loop starting at 1 to find the first number which is truthy for:
¯θy^ # XOR the last number of the global_array with the loop-number `y`
b # Convert it to binary
SO # Sum it's binary digits
¯yå_ # Check if the loop-number `y` is NOT in the global_array yet
* # Multiply both (only if this is 1 (truthy), the inner loop will stop)
# (after the loops, output the top of the stack implicitly)
answered Mar 20 at 9:26
Kevin CruijssenKevin Cruijssen
41.3k567213
edited Mar 20 at 10:30
answered Mar 20 at 9:26
Kevin CruijssenKevin Cruijssen
41.3k567213
answered Mar 20 at 9:26
Kevin CruijssenKevin Cruijssen
41.3k567213
answered Mar 20 at 9:26
Kevin CruijssenKevin Cruijssen
41.3k567213
41.3k567213
add a comment |
add a comment |
$begingroup$
Python 2, 81 bytes
1-based indexing
l=[0];p=0
exec"n=0nwhile(p^n)&(p^n)-1or n in l:n+=1np=n;l+=p,;"*input()
print p
Try it online!
Python 2, 79 bytes
This takes a lot of time (9999 wasn't finished after running locally for 7 minutes)
l=0;p=0;n=input()
exec'p=min(p^2**k for k in range(n)-l);l|=p;'*n
print p
Try it online!
answered Mar 19 at 20:02
ovsovs
19.3k21160
$endgroup$
1
$begingroup$
Maximum input 32767 isn't supported (the default recursion depth isn't system-dependent).
$endgroup$
– Erik the Outgolfer
Mar 19 at 21:37
$begingroup$
Even the given test case 9999 isn't supported. :)
$endgroup$
– Daniel Widdis
Mar 20 at 8:32
$begingroup$
@EriktheOutgolfer Changed it to an iterative approach, probably still doesn't finish in time on TIO, but runs locally just fine.
$endgroup$
– ovs
Mar 20 at 10:23
$begingroup$
@ovs Oh, timeouts alone don't matter.
$endgroup$
– Erik the Outgolfer
Mar 20 at 12:06
add a comment |
$begingroup$
Python 2, 81 bytes
1-based indexing
l=[0];p=0
exec"n=0nwhile(p^n)&(p^n)-1or n in l:n+=1np=n;l+=p,;"*input()
print p
Try it online!
Python 2, 79 bytes
This takes a lot of time (9999 wasn't finished after running locally for 7 minutes)
l=0;p=0;n=input()
exec'p=min(p^2**k for k in range(n)-l);l|=p;'*n
print p
Try it online!
answered Mar 19 at 20:02
ovsovs
19.3k21160
$endgroup$
1
$begingroup$
Maximum input 32767 isn't supported (the default recursion depth isn't system-dependent).
$endgroup$
– Erik the Outgolfer
Mar 19 at 21:37
$begingroup$
Even the given test case 9999 isn't supported. :)
$endgroup$
– Daniel Widdis
Mar 20 at 8:32
$begingroup$
@EriktheOutgolfer Changed it to an iterative approach, probably still doesn't finish in time on TIO, but runs locally just fine.
$endgroup$
– ovs
Mar 20 at 10:23
$begingroup$
@ovs Oh, timeouts alone don't matter.
$endgroup$
– Erik the Outgolfer
Mar 20 at 12:06
add a comment |
$begingroup$
Python 2, 81 bytes
1-based indexing
l=[0];p=0
exec"n=0nwhile(p^n)&(p^n)-1or n in l:n+=1np=n;l+=p,;"*input()
print p
Try it online!
Python 2, 79 bytes
This takes a lot of time (9999 wasn't finished after running locally for 7 minutes)
l=0;p=0;n=input()
exec'p=min(p^2**k for k in range(n)-l);l|=p;'*n
print p
Try it online!
answered Mar 19 at 20:02
ovsovs
19.3k21160
$endgroup$
Python 2, 81 bytes
1-based indexing
l=[0];p=0
exec"n=0nwhile(p^n)&(p^n)-1or n in l:n+=1np=n;l+=p,;"*input()
print p
Try it online!
Python 2, 79 bytes
This takes a lot of time (9999 wasn't finished after running locally for 7 minutes)
l=0;p=0;n=input()
exec'p=min(p^2**k for k in range(n)-l);l|=p;'*n
print p
Try it online!
answered Mar 19 at 20:02
ovsovs
19.3k21160
edited Mar 20 at 10:37
answered Mar 19 at 20:02
ovsovs
19.3k21160
answered Mar 19 at 20:02
ovsovs
19.3k21160
answered Mar 19 at 20:02
ovsovs
19.3k21160
19.3k21160
1
$begingroup$
Maximum input 32767 isn't supported (the default recursion depth isn't system-dependent).
$endgroup$
– Erik the Outgolfer
Mar 19 at 21:37
$begingroup$
Even the given test case 9999 isn't supported. :)
$endgroup$
– Daniel Widdis
Mar 20 at 8:32
$begingroup$
@EriktheOutgolfer Changed it to an iterative approach, probably still doesn't finish in time on TIO, but runs locally just fine.
$endgroup$
– ovs
Mar 20 at 10:23
$begingroup$
@ovs Oh, timeouts alone don't matter.
$endgroup$
– Erik the Outgolfer
Mar 20 at 12:06
add a comment |
1
$begingroup$
Maximum input 32767 isn't supported (the default recursion depth isn't system-dependent).
$endgroup$
– Erik the Outgolfer
Mar 19 at 21:37
$begingroup$
Even the given test case 9999 isn't supported. :)
$endgroup$
– Daniel Widdis
Mar 20 at 8:32
$begingroup$
@EriktheOutgolfer Changed it to an iterative approach, probably still doesn't finish in time on TIO, but runs locally just fine.
$endgroup$
– ovs
Mar 20 at 10:23
$begingroup$
@ovs Oh, timeouts alone don't matter.
$endgroup$
– Erik the Outgolfer
Mar 20 at 12:06
1
1
$begingroup$
Maximum input 32767 isn't supported (the default recursion depth isn't system-dependent).
$endgroup$
– Erik the Outgolfer
Mar 19 at 21:37
$begingroup$
Maximum input 32767 isn't supported (the default recursion depth isn't system-dependent).
$endgroup$
– Erik the Outgolfer
Mar 19 at 21:37
$begingroup$
Even the given test case 9999 isn't supported. :)
$endgroup$
– Daniel Widdis
Mar 20 at 8:32
$begingroup$
Even the given test case 9999 isn't supported. :)
$endgroup$
– Daniel Widdis
Mar 20 at 8:32
$begingroup$
@EriktheOutgolfer Changed it to an iterative approach, probably still doesn't finish in time on TIO, but runs locally just fine.
$endgroup$
– ovs
Mar 20 at 10:23
$begingroup$
@EriktheOutgolfer Changed it to an iterative approach, probably still doesn't finish in time on TIO, but runs locally just fine.
$endgroup$
– ovs
Mar 20 at 10:23
$begingroup$
@ovs Oh, timeouts alone don't matter.
$endgroup$
– Erik the Outgolfer
Mar 20 at 12:06
$begingroup$
@ovs Oh, timeouts alone don't matter.
$endgroup$
– Erik the Outgolfer
Mar 20 at 12:06
add a comment |
$begingroup$
Haskell, 101 bytes
import Data.Bits
(u!n)0=n
(u!n)m|q<-minimum[x|r<-[0..62],x<-[xor(2^r)n],notElem x u]=(n:u)!q$m-1
[]!0
Try it online!
It seems a shame to incur an import just for xor, but I haven't found a good work-around yet. I also wonder if there's a better way to express the loop.
answered Mar 20 at 21:11
dfeuerdfeuer
894910
$endgroup$
add a comment |
$begingroup$
Haskell, 101 bytes
import Data.Bits
(u!n)0=n
(u!n)m|q<-minimum[x|r<-[0..62],x<-[xor(2^r)n],notElem x u]=(n:u)!q$m-1
[]!0
Try it online!
It seems a shame to incur an import just for xor, but I haven't found a good work-around yet. I also wonder if there's a better way to express the loop.
answered Mar 20 at 21:11
dfeuerdfeuer
894910
$endgroup$
add a comment |
$begingroup$
Haskell, 101 bytes
import Data.Bits
(u!n)0=n
(u!n)m|q<-minimum[x|r<-[0..62],x<-[xor(2^r)n],notElem x u]=(n:u)!q$m-1
[]!0
Try it online!
It seems a shame to incur an import just for xor, but I haven't found a good work-around yet. I also wonder if there's a better way to express the loop.
answered Mar 20 at 21:11
dfeuerdfeuer
894910
$endgroup$
Haskell, 101 bytes
import Data.Bits
(u!n)0=n
(u!n)m|q<-minimum[x|r<-[0..62],x<-[xor(2^r)n],notElem x u]=(n:u)!q$m-1
[]!0
Try it online!
It seems a shame to incur an import just for xor, but I haven't found a good work-around yet. I also wonder if there's a better way to express the loop.
answered Mar 20 at 21:11
dfeuerdfeuer
894910
edited Mar 20 at 21:35
answered Mar 20 at 21:11
dfeuerdfeuer
894910
answered Mar 20 at 21:11
dfeuerdfeuer
894910
answered Mar 20 at 21:11
dfeuerdfeuer
894910
894910
add a comment |
add a comment |
$begingroup$
R, 90 bytes
function(n)A=1
while(sum(A
Try it online!
answered 2 days ago
GiuseppeGiuseppe
17k31152
$endgroup$
add a comment |
$begingroup$
R, 90 bytes
function(n)A=1
while(sum(A
Try it online!
answered 2 days ago
GiuseppeGiuseppe
17k31152
$endgroup$
add a comment |
$begingroup$
R, 90 bytes
function(n)A=1
while(sum(A
Try it online!
answered 2 days ago
GiuseppeGiuseppe
17k31152
$endgroup$
R, 90 bytes
function(n)A=1
while(sum(A
Try it online!
answered 2 days ago
GiuseppeGiuseppe
17k31152
answered 2 days ago
GiuseppeGiuseppe
17k31152
answered 2 days ago
GiuseppeGiuseppe
17k31152
answered 2 days ago
GiuseppeGiuseppe
17k31152
17k31152
add a comment |
add a comment |
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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