New order #4: World
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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 fourth challenge in this series (links to the first, second and third challenge).
In this challenge, we will explore not one permutation of the natural numbers, but an entire world of permutations!
In 2000, Clark Kimberling posed a problem in the 26th issue of Crux Mathematicorum, a scientific journal of mathematics published by the Canadian Mathematical Society. The problem was:
$text{Sequence }a = begin{cases}
a_1 = 1\
a_n = lfloor frac{a_{n-1}}{2} rfloortext{ if }lfloor frac{a_{n-1}}{2} rfloor notin {0, a_1, ... , a_{n-1}}\
a_n = 3 a_{n-1}text{ otherwise}
end{cases}$
Does every positive integer occur exactly once in this sequence?
In 2004, Mateusz Kwasnicki provided positive proof in the same journal and in 2008, he published a more formal and (compared to the original question) a more general proof. He formulated the sequence with parameters $p$ and $q$:
$begin{cases}
a_1 = 1\
a_n = lfloor frac{a_{n-1}}{q} rfloortext{ if }lfloor frac{a_{n-1}}{q} rfloor notin {0, a_1, ... , a_{n-1}}\
a_n = p a_{n-1}text{ otherwise}
end{cases}$
He proved that for any $p, q>1$ such that $log_p(q)$ is irrational, the sequence is a permutation of the natural numbers. Since there are an infinite number of $p$ and $q$ values for which this is true, this is truly an entire world of permutations of the natural numbers. We will stick with the original $(p, q)=(3, 2)$, and for these paramters, the sequence can be found as A050000 in the OEIS. Its first 20 elements are:
1, 3, 9, 4, 2, 6, 18, 54, 27, 13, 39, 19, 57, 28, 14, 7, 21, 10, 5, 15
Since this is a "pure sequence" challenge, the task is to output $a(n)$ for a given $n$ as input, where $a(n)$ is A050000.
Task
Given an integer input $n$, output $a(n)$ in integer format, where:
$begin{cases}
a(1) = 1\
a(n) = lfloor frac{a(n-1)}{2} rfloortext{ if }lfloor frac{a(n-1)}{2} rfloor notin {0, a_1, ... , a(n-1)}\
a(n) = 3 a(n-1)text{ otherwise}
end{cases}$
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 | 2
20 | 15
50 | 165
78 | 207
123 | 94
1234 | 3537
3000 | 2245
9999 | 4065
29890 | 149853
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.
- Default I/O rules apply.
Default loopholes are forbidden.- This is code-golf, so the shortest answers in bytes wins
code-golf sequence
asked yesterday
agtoeveragtoever
1,322424
$endgroup$
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 fourth challenge in this series (links to the first, second and third challenge).
In this challenge, we will explore not one permutation of the natural numbers, but an entire world of permutations!
In 2000, Clark Kimberling posed a problem in the 26th issue of Crux Mathematicorum, a scientific journal of mathematics published by the Canadian Mathematical Society. The problem was:
$text{Sequence }a = begin{cases}
a_1 = 1\
a_n = lfloor frac{a_{n-1}}{2} rfloortext{ if }lfloor frac{a_{n-1}}{2} rfloor notin {0, a_1, ... , a_{n-1}}\
a_n = 3 a_{n-1}text{ otherwise}
end{cases}$
Does every positive integer occur exactly once in this sequence?
In 2004, Mateusz Kwasnicki provided positive proof in the same journal and in 2008, he published a more formal and (compared to the original question) a more general proof. He formulated the sequence with parameters $p$ and $q$:
$begin{cases}
a_1 = 1\
a_n = lfloor frac{a_{n-1}}{q} rfloortext{ if }lfloor frac{a_{n-1}}{q} rfloor notin {0, a_1, ... , a_{n-1}}\
a_n = p a_{n-1}text{ otherwise}
end{cases}$
He proved that for any $p, q>1$ such that $log_p(q)$ is irrational, the sequence is a permutation of the natural numbers. Since there are an infinite number of $p$ and $q$ values for which this is true, this is truly an entire world of permutations of the natural numbers. We will stick with the original $(p, q)=(3, 2)$, and for these paramters, the sequence can be found as A050000 in the OEIS. Its first 20 elements are:
1, 3, 9, 4, 2, 6, 18, 54, 27, 13, 39, 19, 57, 28, 14, 7, 21, 10, 5, 15
Since this is a "pure sequence" challenge, the task is to output $a(n)$ for a given $n$ as input, where $a(n)$ is A050000.
Task
Given an integer input $n$, output $a(n)$ in integer format, where:
$begin{cases}
a(1) = 1\
a(n) = lfloor frac{a(n-1)}{2} rfloortext{ if }lfloor frac{a(n-1)}{2} rfloor notin {0, a_1, ... , a(n-1)}\
a(n) = 3 a(n-1)text{ otherwise}
end{cases}$
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 | 2
20 | 15
50 | 165
78 | 207
123 | 94
1234 | 3537
3000 | 2245
9999 | 4065
29890 | 149853
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.
- Default I/O rules apply.
Default loopholes are forbidden.- This is code-golf, so the shortest answers in bytes wins
code-golf sequence
asked yesterday
agtoeveragtoever
1,322424
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Sharp! My mistake. Corrected it. Sorry if this caused any trouble or confusion.
$endgroup$
– agtoever
yesterday
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 fourth challenge in this series (links to the first, second and third challenge).
In this challenge, we will explore not one permutation of the natural numbers, but an entire world of permutations!
In 2000, Clark Kimberling posed a problem in the 26th issue of Crux Mathematicorum, a scientific journal of mathematics published by the Canadian Mathematical Society. The problem was:
$text{Sequence }a = begin{cases}
a_1 = 1\
a_n = lfloor frac{a_{n-1}}{2} rfloortext{ if }lfloor frac{a_{n-1}}{2} rfloor notin {0, a_1, ... , a_{n-1}}\
a_n = 3 a_{n-1}text{ otherwise}
end{cases}$
Does every positive integer occur exactly once in this sequence?
In 2004, Mateusz Kwasnicki provided positive proof in the same journal and in 2008, he published a more formal and (compared to the original question) a more general proof. He formulated the sequence with parameters $p$ and $q$:
$begin{cases}
a_1 = 1\
a_n = lfloor frac{a_{n-1}}{q} rfloortext{ if }lfloor frac{a_{n-1}}{q} rfloor notin {0, a_1, ... , a_{n-1}}\
a_n = p a_{n-1}text{ otherwise}
end{cases}$
He proved that for any $p, q>1$ such that $log_p(q)$ is irrational, the sequence is a permutation of the natural numbers. Since there are an infinite number of $p$ and $q$ values for which this is true, this is truly an entire world of permutations of the natural numbers. We will stick with the original $(p, q)=(3, 2)$, and for these paramters, the sequence can be found as A050000 in the OEIS. Its first 20 elements are:
1, 3, 9, 4, 2, 6, 18, 54, 27, 13, 39, 19, 57, 28, 14, 7, 21, 10, 5, 15
Since this is a "pure sequence" challenge, the task is to output $a(n)$ for a given $n$ as input, where $a(n)$ is A050000.
Task
Given an integer input $n$, output $a(n)$ in integer format, where:
$begin{cases}
a(1) = 1\
a(n) = lfloor frac{a(n-1)}{2} rfloortext{ if }lfloor frac{a(n-1)}{2} rfloor notin {0, a_1, ... , a(n-1)}\
a(n) = 3 a(n-1)text{ otherwise}
end{cases}$
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 | 2
20 | 15
50 | 165
78 | 207
123 | 94
1234 | 3537
3000 | 2245
9999 | 4065
29890 | 149853
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.
- Default I/O rules apply.
Default loopholes are forbidden.- This is code-golf, so the shortest answers in bytes wins
code-golf sequence
asked yesterday
agtoeveragtoever
1,322424
$endgroup$
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 fourth challenge in this series (links to the first, second and third challenge).
In this challenge, we will explore not one permutation of the natural numbers, but an entire world of permutations!
In 2000, Clark Kimberling posed a problem in the 26th issue of Crux Mathematicorum, a scientific journal of mathematics published by the Canadian Mathematical Society. The problem was:
$text{Sequence }a = begin{cases}
a_1 = 1\
a_n = lfloor frac{a_{n-1}}{2} rfloortext{ if }lfloor frac{a_{n-1}}{2} rfloor notin {0, a_1, ... , a_{n-1}}\
a_n = 3 a_{n-1}text{ otherwise}
end{cases}$
Does every positive integer occur exactly once in this sequence?
In 2004, Mateusz Kwasnicki provided positive proof in the same journal and in 2008, he published a more formal and (compared to the original question) a more general proof. He formulated the sequence with parameters $p$ and $q$:
$begin{cases}
a_1 = 1\
a_n = lfloor frac{a_{n-1}}{q} rfloortext{ if }lfloor frac{a_{n-1}}{q} rfloor notin {0, a_1, ... , a_{n-1}}\
a_n = p a_{n-1}text{ otherwise}
end{cases}$
He proved that for any $p, q>1$ such that $log_p(q)$ is irrational, the sequence is a permutation of the natural numbers. Since there are an infinite number of $p$ and $q$ values for which this is true, this is truly an entire world of permutations of the natural numbers. We will stick with the original $(p, q)=(3, 2)$, and for these paramters, the sequence can be found as A050000 in the OEIS. Its first 20 elements are:
1, 3, 9, 4, 2, 6, 18, 54, 27, 13, 39, 19, 57, 28, 14, 7, 21, 10, 5, 15
Since this is a "pure sequence" challenge, the task is to output $a(n)$ for a given $n$ as input, where $a(n)$ is A050000.
Task
Given an integer input $n$, output $a(n)$ in integer format, where:
$begin{cases}
a(1) = 1\
a(n) = lfloor frac{a(n-1)}{2} rfloortext{ if }lfloor frac{a(n-1)}{2} rfloor notin {0, a_1, ... , a(n-1)}\
a(n) = 3 a(n-1)text{ otherwise}
end{cases}$
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 | 2
20 | 15
50 | 165
78 | 207
123 | 94
1234 | 3537
3000 | 2245
9999 | 4065
29890 | 149853
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.
- Default I/O rules apply.
Default loopholes are forbidden.- This is code-golf, so the shortest answers in bytes wins
code-golf sequence
code-golf sequence
asked yesterday
agtoeveragtoever
1,322424
asked yesterday
agtoeveragtoever
1,322424
edited yesterday
agtoever
asked yesterday
agtoeveragtoever
1,322424
asked yesterday
agtoeveragtoever
1,322424
asked yesterday
agtoeveragtoever
1,322424
1,322424
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Sharp! My mistake. Corrected it. Sorry if this caused any trouble or confusion.
$endgroup$
– agtoever
yesterday
add a comment |
$begingroup$
Sharp! My mistake. Corrected it. Sorry if this caused any trouble or confusion.
$endgroup$
– agtoever
yesterday
$begingroup$
Sharp! My mistake. Corrected it. Sorry if this caused any trouble or confusion.
$endgroup$
– agtoever
yesterday
$begingroup$
Sharp! My mistake. Corrected it. Sorry if this caused any trouble or confusion.
$endgroup$
– agtoever
yesterday
add a comment |
18 Answers
18
active
oldest
votes
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Jelly, 15 bytes
µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ
A full program accepting the integer, n (1-based), from STDIN which prints the result.
Try it online!
How?
µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ - Main Link: no arguments (implicit left argument = 0)
µ µ¡ - repeat this monadic chain STDIN times (starting with x=0)
- e.g. x = ... 0 [1,0] [9,3,1,0]
×3 - multiply by 3 0 [3,0] [27,9,3,0]
H - halve 0 [1.5,0] [4.5,1.5,0.5,0]
ż - zip together [0,0] [[3,1.5],[0,0]] [[27,4.5],[9,1.5],[3,0.5],[0,0]]
Ḟ - floor [0,0] [[3,1],[0,0]] [[27,4],[9,1],[3,0],[0,0]]
Ḣ - head 0 [3,1] [27,4]
ḟ - filter discard if in x [3] [27,4]
ȯ1 - logical OR with 1 1 [3] [27,4]
Ṫ - tail 1 3 4
; - concatenate with x [1,0] [3,1,0] [4,9,3,1,0]
Ḣ - head 1 3 4
- implicit print
answered yesterday
Jonathan AllanJonathan Allan
54k536174
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JavaScript (ES6), 55 51 50 bytes
Saved 1 byte thanks to @EmbodimentofIgnorance
Saved 1 byte thanks to @tsh
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")
Try it online!
answered yesterday
ArnauldArnauld
80.6k797334
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55 bytes
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– Embodiment of Ignorance
yesterday
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@EmbodimentofIgnorance I usually avoid that trick, as the eval'ed code is much slower. But the difference is barely noticeable for that one, so I guess that's fine.
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– Arnauld
yesterday
1
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But this is code-golf, we don't care about speed, as long as it gets the job done
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– Embodiment of Ignorance
yesterday
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n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")
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– tsh
yesterday
add a comment |
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Perl 6, 51 bytes
{(1,3,{first *∉@_,@_[*-1]+>1,3*@_[*-1]}...*)[$_]}
Try it online!
Returns the 0-indexed element in the sequence. You can change this to 1-indexed by changing the starting elements to 0,1 instead of 1,3
Explanation:
{ } # Anonymous code block
( ...*)[$_] # Index into the infinite sequence
1,3 # That starts with 1,3
,{ } # And each element is
first # The first of
@_[*-1]+>1 # The previous element bitshifted one
,3*@_[*-1] # Triple the previous element
*∉@_, # That hasn't appeared in the sequence
answered yesterday
Jo KingJo King
26.6k364132
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J, 47 40 bytes
[:{:0 1(],<.@-:@{:@](e.{[,3*{:@])])^:[~]
Try it online!
ungolfed
[: {: 0 1 (] , <.@-:@{:@] (e. { [ , 3 * {:@]) ])^:[~ ]
Direct translation of the definition into J. It builds bottom up by using ^: to iterate from the starting value the required number of times.
answered yesterday
JonahJonah
2,6611017
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add a comment |
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05AB1E, 16 15 bytes
Saved 1 byte thanks to Kevin Cruijssen.
0-indexed.
¾ˆ$FDˆx3*‚;ï¯Kн
Try it online!
Explanation
Using n=1 as example
¾ˆ # initialize global array as [0]
$ # initialize stack with 1, input
F # input times do:
Dˆ # duplicate current item (initially 1) and add one copy to global array
# STACK: 1, GLOBAL_ARRAY: [0, 1]
x # push Top_of_stack*2
# STACK: 1, 2, GLOBAL_ARRAY: [0, 1]
3* # multiply by 3
# STACK: 1, 6, GLOBAL_ARRAY: [0, 1]
‚;ï # pair and integer divide both by 2
# STACK: [0, 3], GLOBAL_ARRAY: [0, 1]
¯K # remove any numbers already in the global array
# STACK: [3], GLOBAL_ARRAY: [0, 1]
н # and take the head
# STACK: 3
answered 22 hours ago
EmignaEmigna
47.6k433145
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15 bytes
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– Kevin Cruijssen
16 hours ago
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@KevinCruijssen: Thanks! I thought of using the global array, but assumed it would be the same length as a list on the stack and never tried it :/
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– Emigna
16 hours ago
add a comment |
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Jelly, 21 bytes
Ø.;0ị×3$:2$:2eɗ?Ɗ$⁸¡Ṫ
Try it online!
A monadic link that takes zero-indexed $n$ as the argument and returns $a(n)$.
answered yesterday
Nick KennedyNick Kennedy
1,36649
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Java 10, 120 99 bytes
n->{var L=" 1 0 ";int r=1,t;for(;n-->0;L+=r+" ")if(L.contains(" "+(r=(t=r)/2)+" "))r=t*3;return r;}
Try it online.
Explanation:
n->{ // Method with integer as both parameter and return-type
var L=" 1 0 "; // Create a String that acts as 'List', starting at [1,0]
int r=1, // Result-integer, starting at 1
t; // Temp-integer, uninitialized
for(;n-->0; // Loop the input amount of times:
L+=r+" ")) // After every iteration: add the result to the 'List'
t=r // Create a copy of the result in `t`
r=(...)/2 // Then integer-divide the result by 2
if(L.contains(" "+(...)+" ")) // If the 'List' contains this result//2:
r=t*3; // Set the result to `t` multiplied by 3 instead
return r;} // Return the result
answered 16 hours ago
Kevin CruijssenKevin Cruijssen
42.4k570217
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Japt, 15 14 bytes
1-indexed.
@[X*3Xz]kZ Ì}g
Try it
@[X*3Xz]kZ Ì}g :Implicit input of integer U
g :Starting with the array [0,1] do the following U times, pushing the result to the array each time
@ : Pass the last element X in the array Z through the following function
[ : Build an array containing
X*3 : X multiplied by 3
Xz : X floor divided by 2
] : Close array
kZ : Remove all elements contained in Z
Ì : Get the last element
} : End function
:Implicit output of the last element in the array
answered 14 hours ago
ShaggyShaggy
18.9k21768
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Haskell, 67 65 bytes
(h[1,0]!!)
h l@(a:o)|elem(div a 2)o=a:h(3*a:l)|1>0=a:h(div a 2:l)
Try it online!
Uses 0-based indexing.
EDIT: saved 2 bytes by using elem instead of notElem and switching conditions
answered 22 hours ago
user1472751user1472751
1,25126
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Python 2, 66 bytes
l=lambda n,p=1,s=[0]:p*(n<len(s))or l(n,3*p*(p/2in s)or p/2,[p]+s)
Try it online!
Uses zero-based indexing. The lambda does little more than recursively building up the sequence and returning as soon as the required index is reached.
answered 9 hours ago
ArBoArBo
38115
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Wolfram Language (Mathematica), 63 bytes
(L=Last)@Nest[{##,If[FreeQ[#,x=⌊L@#/2⌋],x,3L@#]}&,{0,1},#]&
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This is 0-indexed
(In TIO I added -1 in every test case)
answered yesterday
J42161217J42161217
13.9k21353
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Ruby, 54 52 48 bytes
->n{*s=0;j=2;n.times{s<<j=s==s-[j/2]?j/2:j*3};j}
Try it online!
answered 17 hours ago
Kirill L.Kirill L.
6,0581527
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C++ (gcc), 189 180 bytes
-9 bytes to small golfing
#import<vector>
#import<algorithm>
int a(int n){std::vector<int>s={1};for(int i=0;i<n;++i)s.push_back(i&&std::find(s.begin(),s.end(),s[i]/2)==s.end()?s[i]/2:3*s[i]);return s[n-1];}
Try it online!
Computes the sequence up to n, then returns the desired element. Slow for larger indices.
answered yesterday
Neil A.Neil A.
1,348120
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Stax, 14 bytes
üÑα↕○Ü1∟¡f↑ô┬♥
Run and debug it
Zero-indexed.
answered 8 hours ago
recursiverecursive
5,7291322
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Python 2, 62 bytes
a=lambda n:n<1or a(n-1)*6**(a(n-1)//2in[0]+map(a,range(n)))//2
Try it online!
Returns True for a(0). 0-indexed.
answered 7 hours ago
Erik the OutgolferErik the Outgolfer
33k429106
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Python 3, 105 103 100 95 83 bytes
-2 bytes thanks to agtoever
-12 bytes thanks to ArBo
def f(n):
s=0,1
while len(s)<=n:t=s[-1]//2;s+=(t in s)*3*s[-1]or t,
return s[-1]
Try it online!
answered 17 hours ago
Noodle9Noodle9
29137
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You can replace the for loop withwhile len(s)<=nand replace the i's with-1. This should shave off one of two characters.
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– agtoever
17 hours ago
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@agtoever that's so clever - thanks! :)
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– Noodle9
15 hours ago
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83 bytes by working with a tuple instead of a list, and removing theiffrom thewhileloop to allow one-lining that loop
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– ArBo
9 hours ago
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@ArBo wow! absolutely brilliant - thanks :)
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– Noodle9
6 hours ago
add a comment |
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Gaia, 22 20 bytes
2…@⟨:):3פḥ⌋,;D)+⟩ₓ)
Try it online!
0-based index.
Credit to Shaggy for the approach
2… | push [0 1]
@⟨ ⟩ₓ | do the following n times:
:): | dup the list L, take the last element e, and dup that
3פḥ⌋, | push [3*e floor(e/2)]
;D | take the asymmetric set difference [3*e floor(e/2)] - L
)+ | take the last element of the difference and add it to the end of L (end of loop)
) | finally, take the last element and output it
;D
answered 10 hours ago
GiuseppeGiuseppe
17.6k31153
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Haskell, 55 bytes
(1%[0]!!)
a%o|b<-div a 2=a:last(b:[3*a|elem b o])%(a:o)
Try it online!
Golfing user1472751's slick list-generation method.
Same length:
(1%[0]!!)
a%o=a:[x|x<-[div a 2,a*3],all(/=x)o]!!0%(a:o)
Try it online!
answered 18 hours ago
xnorxnor
93.6k18190450
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18 Answers
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18 Answers
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$begingroup$
Jelly, 15 bytes
µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ
A full program accepting the integer, n (1-based), from STDIN which prints the result.
Try it online!
How?
µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ - Main Link: no arguments (implicit left argument = 0)
µ µ¡ - repeat this monadic chain STDIN times (starting with x=0)
- e.g. x = ... 0 [1,0] [9,3,1,0]
×3 - multiply by 3 0 [3,0] [27,9,3,0]
H - halve 0 [1.5,0] [4.5,1.5,0.5,0]
ż - zip together [0,0] [[3,1.5],[0,0]] [[27,4.5],[9,1.5],[3,0.5],[0,0]]
Ḟ - floor [0,0] [[3,1],[0,0]] [[27,4],[9,1],[3,0],[0,0]]
Ḣ - head 0 [3,1] [27,4]
ḟ - filter discard if in x [3] [27,4]
ȯ1 - logical OR with 1 1 [3] [27,4]
Ṫ - tail 1 3 4
; - concatenate with x [1,0] [3,1,0] [4,9,3,1,0]
Ḣ - head 1 3 4
- implicit print
answered yesterday
Jonathan AllanJonathan Allan
54k536174
$endgroup$
add a comment |
$begingroup$
Jelly, 15 bytes
µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ
A full program accepting the integer, n (1-based), from STDIN which prints the result.
Try it online!
How?
µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ - Main Link: no arguments (implicit left argument = 0)
µ µ¡ - repeat this monadic chain STDIN times (starting with x=0)
- e.g. x = ... 0 [1,0] [9,3,1,0]
×3 - multiply by 3 0 [3,0] [27,9,3,0]
H - halve 0 [1.5,0] [4.5,1.5,0.5,0]
ż - zip together [0,0] [[3,1.5],[0,0]] [[27,4.5],[9,1.5],[3,0.5],[0,0]]
Ḟ - floor [0,0] [[3,1],[0,0]] [[27,4],[9,1],[3,0],[0,0]]
Ḣ - head 0 [3,1] [27,4]
ḟ - filter discard if in x [3] [27,4]
ȯ1 - logical OR with 1 1 [3] [27,4]
Ṫ - tail 1 3 4
; - concatenate with x [1,0] [3,1,0] [4,9,3,1,0]
Ḣ - head 1 3 4
- implicit print
answered yesterday
Jonathan AllanJonathan Allan
54k536174
$endgroup$
add a comment |
$begingroup$
Jelly, 15 bytes
µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ
A full program accepting the integer, n (1-based), from STDIN which prints the result.
Try it online!
How?
µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ - Main Link: no arguments (implicit left argument = 0)
µ µ¡ - repeat this monadic chain STDIN times (starting with x=0)
- e.g. x = ... 0 [1,0] [9,3,1,0]
×3 - multiply by 3 0 [3,0] [27,9,3,0]
H - halve 0 [1.5,0] [4.5,1.5,0.5,0]
ż - zip together [0,0] [[3,1.5],[0,0]] [[27,4.5],[9,1.5],[3,0.5],[0,0]]
Ḟ - floor [0,0] [[3,1],[0,0]] [[27,4],[9,1],[3,0],[0,0]]
Ḣ - head 0 [3,1] [27,4]
ḟ - filter discard if in x [3] [27,4]
ȯ1 - logical OR with 1 1 [3] [27,4]
Ṫ - tail 1 3 4
; - concatenate with x [1,0] [3,1,0] [4,9,3,1,0]
Ḣ - head 1 3 4
- implicit print
answered yesterday
Jonathan AllanJonathan Allan
54k536174
$endgroup$
Jelly, 15 bytes
µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ
A full program accepting the integer, n (1-based), from STDIN which prints the result.
Try it online!
How?
µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ - Main Link: no arguments (implicit left argument = 0)
µ µ¡ - repeat this monadic chain STDIN times (starting with x=0)
- e.g. x = ... 0 [1,0] [9,3,1,0]
×3 - multiply by 3 0 [3,0] [27,9,3,0]
H - halve 0 [1.5,0] [4.5,1.5,0.5,0]
ż - zip together [0,0] [[3,1.5],[0,0]] [[27,4.5],[9,1.5],[3,0.5],[0,0]]
Ḟ - floor [0,0] [[3,1],[0,0]] [[27,4],[9,1],[3,0],[0,0]]
Ḣ - head 0 [3,1] [27,4]
ḟ - filter discard if in x [3] [27,4]
ȯ1 - logical OR with 1 1 [3] [27,4]
Ṫ - tail 1 3 4
; - concatenate with x [1,0] [3,1,0] [4,9,3,1,0]
Ḣ - head 1 3 4
- implicit print
answered yesterday
Jonathan AllanJonathan Allan
54k536174
answered yesterday
Jonathan AllanJonathan Allan
54k536174
answered yesterday
Jonathan AllanJonathan Allan
54k536174
answered yesterday
Jonathan AllanJonathan Allan
54k536174
54k536174
add a comment |
add a comment |
$begingroup$
JavaScript (ES6), 55 51 50 bytes
Saved 1 byte thanks to @EmbodimentofIgnorance
Saved 1 byte thanks to @tsh
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")
Try it online!
answered yesterday
ArnauldArnauld
80.6k797334
$endgroup$
$begingroup$
55 bytes
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@EmbodimentofIgnorance I usually avoid that trick, as the eval'ed code is much slower. But the difference is barely noticeable for that one, so I guess that's fine.
$endgroup$
– Arnauld
yesterday
1
$begingroup$
But this is code-golf, we don't care about speed, as long as it gets the job done
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")
$endgroup$
– tsh
yesterday
add a comment |
$begingroup$
JavaScript (ES6), 55 51 50 bytes
Saved 1 byte thanks to @EmbodimentofIgnorance
Saved 1 byte thanks to @tsh
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")
Try it online!
answered yesterday
ArnauldArnauld
80.6k797334
$endgroup$
$begingroup$
55 bytes
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@EmbodimentofIgnorance I usually avoid that trick, as the eval'ed code is much slower. But the difference is barely noticeable for that one, so I guess that's fine.
$endgroup$
– Arnauld
yesterday
1
$begingroup$
But this is code-golf, we don't care about speed, as long as it gets the job done
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")
$endgroup$
– tsh
yesterday
add a comment |
$begingroup$
JavaScript (ES6), 55 51 50 bytes
Saved 1 byte thanks to @EmbodimentofIgnorance
Saved 1 byte thanks to @tsh
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")
Try it online!
answered yesterday
ArnauldArnauld
80.6k797334
$endgroup$
JavaScript (ES6), 55 51 50 bytes
Saved 1 byte thanks to @EmbodimentofIgnorance
Saved 1 byte thanks to @tsh
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")
Try it online!
answered yesterday
ArnauldArnauld
80.6k797334
edited 20 hours ago
answered yesterday
ArnauldArnauld
80.6k797334
answered yesterday
ArnauldArnauld
80.6k797334
answered yesterday
ArnauldArnauld
80.6k797334
80.6k797334
$begingroup$
55 bytes
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@EmbodimentofIgnorance I usually avoid that trick, as the eval'ed code is much slower. But the difference is barely noticeable for that one, so I guess that's fine.
$endgroup$
– Arnauld
yesterday
1
$begingroup$
But this is code-golf, we don't care about speed, as long as it gets the job done
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")
$endgroup$
– tsh
yesterday
add a comment |
$begingroup$
55 bytes
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@EmbodimentofIgnorance I usually avoid that trick, as the eval'ed code is much slower. But the difference is barely noticeable for that one, so I guess that's fine.
$endgroup$
– Arnauld
yesterday
1
$begingroup$
But this is code-golf, we don't care about speed, as long as it gets the job done
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")
$endgroup$
– tsh
yesterday
$begingroup$
55 bytes
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
55 bytes
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@EmbodimentofIgnorance I usually avoid that trick, as the eval'ed code is much slower. But the difference is barely noticeable for that one, so I guess that's fine.
$endgroup$
– Arnauld
yesterday
$begingroup$
@EmbodimentofIgnorance I usually avoid that trick, as the eval'ed code is much slower. But the difference is barely noticeable for that one, so I guess that's fine.
$endgroup$
– Arnauld
yesterday
1
1
$begingroup$
But this is code-golf, we don't care about speed, as long as it gets the job done
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
But this is code-golf, we don't care about speed, as long as it gets the job done
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")$endgroup$
– tsh
yesterday
$begingroup$
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")$endgroup$
– tsh
yesterday
add a comment |
$begingroup$
Perl 6, 51 bytes
{(1,3,{first *∉@_,@_[*-1]+>1,3*@_[*-1]}...*)[$_]}
Try it online!
Returns the 0-indexed element in the sequence. You can change this to 1-indexed by changing the starting elements to 0,1 instead of 1,3
Explanation:
{ } # Anonymous code block
( ...*)[$_] # Index into the infinite sequence
1,3 # That starts with 1,3
,{ } # And each element is
first # The first of
@_[*-1]+>1 # The previous element bitshifted one
,3*@_[*-1] # Triple the previous element
*∉@_, # That hasn't appeared in the sequence
answered yesterday
Jo KingJo King
26.6k364132
$endgroup$
add a comment |
$begingroup$
Perl 6, 51 bytes
{(1,3,{first *∉@_,@_[*-1]+>1,3*@_[*-1]}...*)[$_]}
Try it online!
Returns the 0-indexed element in the sequence. You can change this to 1-indexed by changing the starting elements to 0,1 instead of 1,3
Explanation:
{ } # Anonymous code block
( ...*)[$_] # Index into the infinite sequence
1,3 # That starts with 1,3
,{ } # And each element is
first # The first of
@_[*-1]+>1 # The previous element bitshifted one
,3*@_[*-1] # Triple the previous element
*∉@_, # That hasn't appeared in the sequence
answered yesterday
Jo KingJo King
26.6k364132
$endgroup$
add a comment |
$begingroup$
Perl 6, 51 bytes
{(1,3,{first *∉@_,@_[*-1]+>1,3*@_[*-1]}...*)[$_]}
Try it online!
Returns the 0-indexed element in the sequence. You can change this to 1-indexed by changing the starting elements to 0,1 instead of 1,3
Explanation:
{ } # Anonymous code block
( ...*)[$_] # Index into the infinite sequence
1,3 # That starts with 1,3
,{ } # And each element is
first # The first of
@_[*-1]+>1 # The previous element bitshifted one
,3*@_[*-1] # Triple the previous element
*∉@_, # That hasn't appeared in the sequence
answered yesterday
Jo KingJo King
26.6k364132
$endgroup$
Perl 6, 51 bytes
{(1,3,{first *∉@_,@_[*-1]+>1,3*@_[*-1]}...*)[$_]}
Try it online!
Returns the 0-indexed element in the sequence. You can change this to 1-indexed by changing the starting elements to 0,1 instead of 1,3
Explanation:
{ } # Anonymous code block
( ...*)[$_] # Index into the infinite sequence
1,3 # That starts with 1,3
,{ } # And each element is
first # The first of
@_[*-1]+>1 # The previous element bitshifted one
,3*@_[*-1] # Triple the previous element
*∉@_, # That hasn't appeared in the sequence
answered yesterday
Jo KingJo King
26.6k364132
answered yesterday
Jo KingJo King
26.6k364132
answered yesterday
Jo KingJo King
26.6k364132
answered yesterday
Jo KingJo King
26.6k364132
26.6k364132
add a comment |
add a comment |
$begingroup$
J, 47 40 bytes
[:{:0 1(],<.@-:@{:@](e.{[,3*{:@])])^:[~]
Try it online!
ungolfed
[: {: 0 1 (] , <.@-:@{:@] (e. { [ , 3 * {:@]) ])^:[~ ]
Direct translation of the definition into J. It builds bottom up by using ^: to iterate from the starting value the required number of times.
answered yesterday
JonahJonah
2,6611017
$endgroup$
add a comment |
$begingroup$
J, 47 40 bytes
[:{:0 1(],<.@-:@{:@](e.{[,3*{:@])])^:[~]
Try it online!
ungolfed
[: {: 0 1 (] , <.@-:@{:@] (e. { [ , 3 * {:@]) ])^:[~ ]
Direct translation of the definition into J. It builds bottom up by using ^: to iterate from the starting value the required number of times.
answered yesterday
JonahJonah
2,6611017
$endgroup$
add a comment |
$begingroup$
J, 47 40 bytes
[:{:0 1(],<.@-:@{:@](e.{[,3*{:@])])^:[~]
Try it online!
ungolfed
[: {: 0 1 (] , <.@-:@{:@] (e. { [ , 3 * {:@]) ])^:[~ ]
Direct translation of the definition into J. It builds bottom up by using ^: to iterate from the starting value the required number of times.
answered yesterday
JonahJonah
2,6611017
$endgroup$
J, 47 40 bytes
[:{:0 1(],<.@-:@{:@](e.{[,3*{:@])])^:[~]
Try it online!
ungolfed
[: {: 0 1 (] , <.@-:@{:@] (e. { [ , 3 * {:@]) ])^:[~ ]
Direct translation of the definition into J. It builds bottom up by using ^: to iterate from the starting value the required number of times.
answered yesterday
JonahJonah
2,6611017
edited yesterday
answered yesterday
JonahJonah
2,6611017
answered yesterday
JonahJonah
2,6611017
answered yesterday
JonahJonah
2,6611017
2,6611017
add a comment |
add a comment |
$begingroup$
05AB1E, 16 15 bytes
Saved 1 byte thanks to Kevin Cruijssen.
0-indexed.
¾ˆ$FDˆx3*‚;ï¯Kн
Try it online!
Explanation
Using n=1 as example
¾ˆ # initialize global array as [0]
$ # initialize stack with 1, input
F # input times do:
Dˆ # duplicate current item (initially 1) and add one copy to global array
# STACK: 1, GLOBAL_ARRAY: [0, 1]
x # push Top_of_stack*2
# STACK: 1, 2, GLOBAL_ARRAY: [0, 1]
3* # multiply by 3
# STACK: 1, 6, GLOBAL_ARRAY: [0, 1]
‚;ï # pair and integer divide both by 2
# STACK: [0, 3], GLOBAL_ARRAY: [0, 1]
¯K # remove any numbers already in the global array
# STACK: [3], GLOBAL_ARRAY: [0, 1]
н # and take the head
# STACK: 3
answered 22 hours ago
EmignaEmigna
47.6k433145
$endgroup$
$begingroup$
15 bytes
$endgroup$
– Kevin Cruijssen
16 hours ago
$begingroup$
@KevinCruijssen: Thanks! I thought of using the global array, but assumed it would be the same length as a list on the stack and never tried it :/
$endgroup$
– Emigna
16 hours ago
add a comment |
$begingroup$
05AB1E, 16 15 bytes
Saved 1 byte thanks to Kevin Cruijssen.
0-indexed.
¾ˆ$FDˆx3*‚;ï¯Kн
Try it online!
Explanation
Using n=1 as example
¾ˆ # initialize global array as [0]
$ # initialize stack with 1, input
F # input times do:
Dˆ # duplicate current item (initially 1) and add one copy to global array
# STACK: 1, GLOBAL_ARRAY: [0, 1]
x # push Top_of_stack*2
# STACK: 1, 2, GLOBAL_ARRAY: [0, 1]
3* # multiply by 3
# STACK: 1, 6, GLOBAL_ARRAY: [0, 1]
‚;ï # pair and integer divide both by 2
# STACK: [0, 3], GLOBAL_ARRAY: [0, 1]
¯K # remove any numbers already in the global array
# STACK: [3], GLOBAL_ARRAY: [0, 1]
н # and take the head
# STACK: 3
answered 22 hours ago
EmignaEmigna
47.6k433145
$endgroup$
$begingroup$
15 bytes
$endgroup$
– Kevin Cruijssen
16 hours ago
$begingroup$
@KevinCruijssen: Thanks! I thought of using the global array, but assumed it would be the same length as a list on the stack and never tried it :/
$endgroup$
– Emigna
16 hours ago
add a comment |
$begingroup$
05AB1E, 16 15 bytes
Saved 1 byte thanks to Kevin Cruijssen.
0-indexed.
¾ˆ$FDˆx3*‚;ï¯Kн
Try it online!
Explanation
Using n=1 as example
¾ˆ # initialize global array as [0]
$ # initialize stack with 1, input
F # input times do:
Dˆ # duplicate current item (initially 1) and add one copy to global array
# STACK: 1, GLOBAL_ARRAY: [0, 1]
x # push Top_of_stack*2
# STACK: 1, 2, GLOBAL_ARRAY: [0, 1]
3* # multiply by 3
# STACK: 1, 6, GLOBAL_ARRAY: [0, 1]
‚;ï # pair and integer divide both by 2
# STACK: [0, 3], GLOBAL_ARRAY: [0, 1]
¯K # remove any numbers already in the global array
# STACK: [3], GLOBAL_ARRAY: [0, 1]
н # and take the head
# STACK: 3
answered 22 hours ago
EmignaEmigna
47.6k433145
$endgroup$
05AB1E, 16 15 bytes
Saved 1 byte thanks to Kevin Cruijssen.
0-indexed.
¾ˆ$FDˆx3*‚;ï¯Kн
Try it online!
Explanation
Using n=1 as example
¾ˆ # initialize global array as [0]
$ # initialize stack with 1, input
F # input times do:
Dˆ # duplicate current item (initially 1) and add one copy to global array
# STACK: 1, GLOBAL_ARRAY: [0, 1]
x # push Top_of_stack*2
# STACK: 1, 2, GLOBAL_ARRAY: [0, 1]
3* # multiply by 3
# STACK: 1, 6, GLOBAL_ARRAY: [0, 1]
‚;ï # pair and integer divide both by 2
# STACK: [0, 3], GLOBAL_ARRAY: [0, 1]
¯K # remove any numbers already in the global array
# STACK: [3], GLOBAL_ARRAY: [0, 1]
н # and take the head
# STACK: 3
answered 22 hours ago
EmignaEmigna
47.6k433145
edited 16 hours ago
answered 22 hours ago
EmignaEmigna
47.6k433145
answered 22 hours ago
EmignaEmigna
47.6k433145
answered 22 hours ago
EmignaEmigna
47.6k433145
47.6k433145
$begingroup$
15 bytes
$endgroup$
– Kevin Cruijssen
16 hours ago
$begingroup$
@KevinCruijssen: Thanks! I thought of using the global array, but assumed it would be the same length as a list on the stack and never tried it :/
$endgroup$
– Emigna
16 hours ago
add a comment |
$begingroup$
15 bytes
$endgroup$
– Kevin Cruijssen
16 hours ago
$begingroup$
@KevinCruijssen: Thanks! I thought of using the global array, but assumed it would be the same length as a list on the stack and never tried it :/
$endgroup$
– Emigna
16 hours ago
$begingroup$
15 bytes
$endgroup$
– Kevin Cruijssen
16 hours ago
$begingroup$
15 bytes
$endgroup$
– Kevin Cruijssen
16 hours ago
$begingroup$
@KevinCruijssen: Thanks! I thought of using the global array, but assumed it would be the same length as a list on the stack and never tried it :/
$endgroup$
– Emigna
16 hours ago
$begingroup$
@KevinCruijssen: Thanks! I thought of using the global array, but assumed it would be the same length as a list on the stack and never tried it :/
$endgroup$
– Emigna
16 hours ago
add a comment |
$begingroup$
Jelly, 21 bytes
Ø.;0ị×3$:2$:2eɗ?Ɗ$⁸¡Ṫ
Try it online!
A monadic link that takes zero-indexed $n$ as the argument and returns $a(n)$.
answered yesterday
Nick KennedyNick Kennedy
1,36649
$endgroup$
add a comment |
$begingroup$
Jelly, 21 bytes
Ø.;0ị×3$:2$:2eɗ?Ɗ$⁸¡Ṫ
Try it online!
A monadic link that takes zero-indexed $n$ as the argument and returns $a(n)$.
answered yesterday
Nick KennedyNick Kennedy
1,36649
$endgroup$
add a comment |
$begingroup$
Jelly, 21 bytes
Ø.;0ị×3$:2$:2eɗ?Ɗ$⁸¡Ṫ
Try it online!
A monadic link that takes zero-indexed $n$ as the argument and returns $a(n)$.
answered yesterday
Nick KennedyNick Kennedy
1,36649
$endgroup$
Jelly, 21 bytes
Ø.;0ị×3$:2$:2eɗ?Ɗ$⁸¡Ṫ
Try it online!
A monadic link that takes zero-indexed $n$ as the argument and returns $a(n)$.
answered yesterday
Nick KennedyNick Kennedy
1,36649
answered yesterday
Nick KennedyNick Kennedy
1,36649
answered yesterday
Nick KennedyNick Kennedy
1,36649
answered yesterday
Nick KennedyNick Kennedy
1,36649
1,36649
add a comment |
add a comment |
$begingroup$
Java 10, 120 99 bytes
n->{var L=" 1 0 ";int r=1,t;for(;n-->0;L+=r+" ")if(L.contains(" "+(r=(t=r)/2)+" "))r=t*3;return r;}
Try it online.
Explanation:
n->{ // Method with integer as both parameter and return-type
var L=" 1 0 "; // Create a String that acts as 'List', starting at [1,0]
int r=1, // Result-integer, starting at 1
t; // Temp-integer, uninitialized
for(;n-->0; // Loop the input amount of times:
L+=r+" ")) // After every iteration: add the result to the 'List'
t=r // Create a copy of the result in `t`
r=(...)/2 // Then integer-divide the result by 2
if(L.contains(" "+(...)+" ")) // If the 'List' contains this result//2:
r=t*3; // Set the result to `t` multiplied by 3 instead
return r;} // Return the result
answered 16 hours ago
Kevin CruijssenKevin Cruijssen
42.4k570217
$endgroup$
add a comment |
$begingroup$
Java 10, 120 99 bytes
n->{var L=" 1 0 ";int r=1,t;for(;n-->0;L+=r+" ")if(L.contains(" "+(r=(t=r)/2)+" "))r=t*3;return r;}
Try it online.
Explanation:
n->{ // Method with integer as both parameter and return-type
var L=" 1 0 "; // Create a String that acts as 'List', starting at [1,0]
int r=1, // Result-integer, starting at 1
t; // Temp-integer, uninitialized
for(;n-->0; // Loop the input amount of times:
L+=r+" ")) // After every iteration: add the result to the 'List'
t=r // Create a copy of the result in `t`
r=(...)/2 // Then integer-divide the result by 2
if(L.contains(" "+(...)+" ")) // If the 'List' contains this result//2:
r=t*3; // Set the result to `t` multiplied by 3 instead
return r;} // Return the result
answered 16 hours ago
Kevin CruijssenKevin Cruijssen
42.4k570217
$endgroup$
add a comment |
$begingroup$
Java 10, 120 99 bytes
n->{var L=" 1 0 ";int r=1,t;for(;n-->0;L+=r+" ")if(L.contains(" "+(r=(t=r)/2)+" "))r=t*3;return r;}
Try it online.
Explanation:
n->{ // Method with integer as both parameter and return-type
var L=" 1 0 "; // Create a String that acts as 'List', starting at [1,0]
int r=1, // Result-integer, starting at 1
t; // Temp-integer, uninitialized
for(;n-->0; // Loop the input amount of times:
L+=r+" ")) // After every iteration: add the result to the 'List'
t=r // Create a copy of the result in `t`
r=(...)/2 // Then integer-divide the result by 2
if(L.contains(" "+(...)+" ")) // If the 'List' contains this result//2:
r=t*3; // Set the result to `t` multiplied by 3 instead
return r;} // Return the result
answered 16 hours ago
Kevin CruijssenKevin Cruijssen
42.4k570217
$endgroup$
Java 10, 120 99 bytes
n->{var L=" 1 0 ";int r=1,t;for(;n-->0;L+=r+" ")if(L.contains(" "+(r=(t=r)/2)+" "))r=t*3;return r;}
Try it online.
Explanation:
n->{ // Method with integer as both parameter and return-type
var L=" 1 0 "; // Create a String that acts as 'List', starting at [1,0]
int r=1, // Result-integer, starting at 1
t; // Temp-integer, uninitialized
for(;n-->0; // Loop the input amount of times:
L+=r+" ")) // After every iteration: add the result to the 'List'
t=r // Create a copy of the result in `t`
r=(...)/2 // Then integer-divide the result by 2
if(L.contains(" "+(...)+" ")) // If the 'List' contains this result//2:
r=t*3; // Set the result to `t` multiplied by 3 instead
return r;} // Return the result
answered 16 hours ago
Kevin CruijssenKevin Cruijssen
42.4k570217
edited 16 hours ago
answered 16 hours ago
Kevin CruijssenKevin Cruijssen
42.4k570217
answered 16 hours ago
Kevin CruijssenKevin Cruijssen
42.4k570217
answered 16 hours ago
Kevin CruijssenKevin Cruijssen
42.4k570217
42.4k570217
add a comment |
add a comment |
$begingroup$
Japt, 15 14 bytes
1-indexed.
@[X*3Xz]kZ Ì}g
Try it
@[X*3Xz]kZ Ì}g :Implicit input of integer U
g :Starting with the array [0,1] do the following U times, pushing the result to the array each time
@ : Pass the last element X in the array Z through the following function
[ : Build an array containing
X*3 : X multiplied by 3
Xz : X floor divided by 2
] : Close array
kZ : Remove all elements contained in Z
Ì : Get the last element
} : End function
:Implicit output of the last element in the array
answered 14 hours ago
ShaggyShaggy
18.9k21768
$endgroup$
add a comment |
$begingroup$
Japt, 15 14 bytes
1-indexed.
@[X*3Xz]kZ Ì}g
Try it
@[X*3Xz]kZ Ì}g :Implicit input of integer U
g :Starting with the array [0,1] do the following U times, pushing the result to the array each time
@ : Pass the last element X in the array Z through the following function
[ : Build an array containing
X*3 : X multiplied by 3
Xz : X floor divided by 2
] : Close array
kZ : Remove all elements contained in Z
Ì : Get the last element
} : End function
:Implicit output of the last element in the array
answered 14 hours ago
ShaggyShaggy
18.9k21768
$endgroup$
add a comment |
$begingroup$
Japt, 15 14 bytes
1-indexed.
@[X*3Xz]kZ Ì}g
Try it
@[X*3Xz]kZ Ì}g :Implicit input of integer U
g :Starting with the array [0,1] do the following U times, pushing the result to the array each time
@ : Pass the last element X in the array Z through the following function
[ : Build an array containing
X*3 : X multiplied by 3
Xz : X floor divided by 2
] : Close array
kZ : Remove all elements contained in Z
Ì : Get the last element
} : End function
:Implicit output of the last element in the array
answered 14 hours ago
ShaggyShaggy
18.9k21768
$endgroup$
Japt, 15 14 bytes
1-indexed.
@[X*3Xz]kZ Ì}g
Try it
@[X*3Xz]kZ Ì}g :Implicit input of integer U
g :Starting with the array [0,1] do the following U times, pushing the result to the array each time
@ : Pass the last element X in the array Z through the following function
[ : Build an array containing
X*3 : X multiplied by 3
Xz : X floor divided by 2
] : Close array
kZ : Remove all elements contained in Z
Ì : Get the last element
} : End function
:Implicit output of the last element in the array
answered 14 hours ago
ShaggyShaggy
18.9k21768
edited 13 hours ago
answered 14 hours ago
ShaggyShaggy
18.9k21768
answered 14 hours ago
ShaggyShaggy
18.9k21768
answered 14 hours ago
ShaggyShaggy
18.9k21768
18.9k21768
add a comment |
add a comment |
$begingroup$
Haskell, 67 65 bytes
(h[1,0]!!)
h l@(a:o)|elem(div a 2)o=a:h(3*a:l)|1>0=a:h(div a 2:l)
Try it online!
Uses 0-based indexing.
EDIT: saved 2 bytes by using elem instead of notElem and switching conditions
answered 22 hours ago
user1472751user1472751
1,25126
$endgroup$
add a comment |
$begingroup$
Haskell, 67 65 bytes
(h[1,0]!!)
h l@(a:o)|elem(div a 2)o=a:h(3*a:l)|1>0=a:h(div a 2:l)
Try it online!
Uses 0-based indexing.
EDIT: saved 2 bytes by using elem instead of notElem and switching conditions
answered 22 hours ago
user1472751user1472751
1,25126
$endgroup$
add a comment |
$begingroup$
Haskell, 67 65 bytes
(h[1,0]!!)
h l@(a:o)|elem(div a 2)o=a:h(3*a:l)|1>0=a:h(div a 2:l)
Try it online!
Uses 0-based indexing.
EDIT: saved 2 bytes by using elem instead of notElem and switching conditions
answered 22 hours ago
user1472751user1472751
1,25126
$endgroup$
Haskell, 67 65 bytes
(h[1,0]!!)
h l@(a:o)|elem(div a 2)o=a:h(3*a:l)|1>0=a:h(div a 2:l)
Try it online!
Uses 0-based indexing.
EDIT: saved 2 bytes by using elem instead of notElem and switching conditions
answered 22 hours ago
user1472751user1472751
1,25126
edited 13 hours ago
answered 22 hours ago
user1472751user1472751
1,25126
answered 22 hours ago
user1472751user1472751
1,25126
answered 22 hours ago
user1472751user1472751
1,25126
1,25126
add a comment |
add a comment |
$begingroup$
Python 2, 66 bytes
l=lambda n,p=1,s=[0]:p*(n<len(s))or l(n,3*p*(p/2in s)or p/2,[p]+s)
Try it online!
Uses zero-based indexing. The lambda does little more than recursively building up the sequence and returning as soon as the required index is reached.
answered 9 hours ago
ArBoArBo
38115
$endgroup$
add a comment |
$begingroup$
Python 2, 66 bytes
l=lambda n,p=1,s=[0]:p*(n<len(s))or l(n,3*p*(p/2in s)or p/2,[p]+s)
Try it online!
Uses zero-based indexing. The lambda does little more than recursively building up the sequence and returning as soon as the required index is reached.
answered 9 hours ago
ArBoArBo
38115
$endgroup$
add a comment |
$begingroup$
Python 2, 66 bytes
l=lambda n,p=1,s=[0]:p*(n<len(s))or l(n,3*p*(p/2in s)or p/2,[p]+s)
Try it online!
Uses zero-based indexing. The lambda does little more than recursively building up the sequence and returning as soon as the required index is reached.
answered 9 hours ago
ArBoArBo
38115
$endgroup$
Python 2, 66 bytes
l=lambda n,p=1,s=[0]:p*(n<len(s))or l(n,3*p*(p/2in s)or p/2,[p]+s)
Try it online!
Uses zero-based indexing. The lambda does little more than recursively building up the sequence and returning as soon as the required index is reached.
answered 9 hours ago
ArBoArBo
38115
answered 9 hours ago
ArBoArBo
38115
answered 9 hours ago
ArBoArBo
38115
answered 9 hours ago
ArBoArBo
38115
38115
add a comment |
add a comment |
$begingroup$
Wolfram Language (Mathematica), 63 bytes
(L=Last)@Nest[{##,If[FreeQ[#,x=⌊L@#/2⌋],x,3L@#]}&,{0,1},#]&
Try it online!
This is 0-indexed
(In TIO I added -1 in every test case)
answered yesterday
J42161217J42161217
13.9k21353
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 63 bytes
(L=Last)@Nest[{##,If[FreeQ[#,x=⌊L@#/2⌋],x,3L@#]}&,{0,1},#]&
Try it online!
This is 0-indexed
(In TIO I added -1 in every test case)
answered yesterday
J42161217J42161217
13.9k21353
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 63 bytes
(L=Last)@Nest[{##,If[FreeQ[#,x=⌊L@#/2⌋],x,3L@#]}&,{0,1},#]&
Try it online!
This is 0-indexed
(In TIO I added -1 in every test case)
answered yesterday
J42161217J42161217
13.9k21353
$endgroup$
Wolfram Language (Mathematica), 63 bytes
(L=Last)@Nest[{##,If[FreeQ[#,x=⌊L@#/2⌋],x,3L@#]}&,{0,1},#]&
Try it online!
This is 0-indexed
(In TIO I added -1 in every test case)
answered yesterday
J42161217J42161217
13.9k21353
edited yesterday
answered yesterday
J42161217J42161217
13.9k21353
answered yesterday
J42161217J42161217
13.9k21353
answered yesterday
J42161217J42161217
13.9k21353
13.9k21353
add a comment |
add a comment |
$begingroup$
Ruby, 54 52 48 bytes
->n{*s=0;j=2;n.times{s<<j=s==s-[j/2]?j/2:j*3};j}
Try it online!
answered 17 hours ago
Kirill L.Kirill L.
6,0581527
$endgroup$
add a comment |
$begingroup$
Ruby, 54 52 48 bytes
->n{*s=0;j=2;n.times{s<<j=s==s-[j/2]?j/2:j*3};j}
Try it online!
answered 17 hours ago
Kirill L.Kirill L.
6,0581527
$endgroup$
add a comment |
$begingroup$
Ruby, 54 52 48 bytes
->n{*s=0;j=2;n.times{s<<j=s==s-[j/2]?j/2:j*3};j}
Try it online!
answered 17 hours ago
Kirill L.Kirill L.
6,0581527
$endgroup$
Ruby, 54 52 48 bytes
->n{*s=0;j=2;n.times{s<<j=s==s-[j/2]?j/2:j*3};j}
Try it online!
answered 17 hours ago
Kirill L.Kirill L.
6,0581527
edited 16 hours ago
answered 17 hours ago
Kirill L.Kirill L.
6,0581527
answered 17 hours ago
Kirill L.Kirill L.
6,0581527
answered 17 hours ago
Kirill L.Kirill L.
6,0581527
6,0581527
add a comment |
add a comment |
$begingroup$
C++ (gcc), 189 180 bytes
-9 bytes to small golfing
#import<vector>
#import<algorithm>
int a(int n){std::vector<int>s={1};for(int i=0;i<n;++i)s.push_back(i&&std::find(s.begin(),s.end(),s[i]/2)==s.end()?s[i]/2:3*s[i]);return s[n-1];}
Try it online!
Computes the sequence up to n, then returns the desired element. Slow for larger indices.
answered yesterday
Neil A.Neil A.
1,348120
$endgroup$
add a comment |
$begingroup$
C++ (gcc), 189 180 bytes
-9 bytes to small golfing
#import<vector>
#import<algorithm>
int a(int n){std::vector<int>s={1};for(int i=0;i<n;++i)s.push_back(i&&std::find(s.begin(),s.end(),s[i]/2)==s.end()?s[i]/2:3*s[i]);return s[n-1];}
Try it online!
Computes the sequence up to n, then returns the desired element. Slow for larger indices.
answered yesterday
Neil A.Neil A.
1,348120
$endgroup$
add a comment |
$begingroup$
C++ (gcc), 189 180 bytes
-9 bytes to small golfing
#import<vector>
#import<algorithm>
int a(int n){std::vector<int>s={1};for(int i=0;i<n;++i)s.push_back(i&&std::find(s.begin(),s.end(),s[i]/2)==s.end()?s[i]/2:3*s[i]);return s[n-1];}
Try it online!
Computes the sequence up to n, then returns the desired element. Slow for larger indices.
answered yesterday
Neil A.Neil A.
1,348120
$endgroup$
C++ (gcc), 189 180 bytes
-9 bytes to small golfing
#import<vector>
#import<algorithm>
int a(int n){std::vector<int>s={1};for(int i=0;i<n;++i)s.push_back(i&&std::find(s.begin(),s.end(),s[i]/2)==s.end()?s[i]/2:3*s[i]);return s[n-1];}
Try it online!
Computes the sequence up to n, then returns the desired element. Slow for larger indices.
answered yesterday
Neil A.Neil A.
1,348120
edited 9 hours ago
answered yesterday
Neil A.Neil A.
1,348120
answered yesterday
Neil A.Neil A.
1,348120
answered yesterday
Neil A.Neil A.
1,348120
1,348120
add a comment |
add a comment |
$begingroup$
Stax, 14 bytes
üÑα↕○Ü1∟¡f↑ô┬♥
Run and debug it
Zero-indexed.
answered 8 hours ago
recursiverecursive
5,7291322
$endgroup$
add a comment |
$begingroup$
Stax, 14 bytes
üÑα↕○Ü1∟¡f↑ô┬♥
Run and debug it
Zero-indexed.
answered 8 hours ago
recursiverecursive
5,7291322
$endgroup$
add a comment |
$begingroup$
Stax, 14 bytes
üÑα↕○Ü1∟¡f↑ô┬♥
Run and debug it
Zero-indexed.
answered 8 hours ago
recursiverecursive
5,7291322
$endgroup$
Stax, 14 bytes
üÑα↕○Ü1∟¡f↑ô┬♥
Run and debug it
Zero-indexed.
answered 8 hours ago
recursiverecursive
5,7291322
answered 8 hours ago
recursiverecursive
5,7291322
answered 8 hours ago
recursiverecursive
5,7291322
answered 8 hours ago
recursiverecursive
5,7291322
5,7291322
add a comment |
add a comment |
$begingroup$
Python 2, 62 bytes
a=lambda n:n<1or a(n-1)*6**(a(n-1)//2in[0]+map(a,range(n)))//2
Try it online!
Returns True for a(0). 0-indexed.
answered 7 hours ago
Erik the OutgolferErik the Outgolfer
33k429106
$endgroup$
add a comment |
$begingroup$
Python 2, 62 bytes
a=lambda n:n<1or a(n-1)*6**(a(n-1)//2in[0]+map(a,range(n)))//2
Try it online!
Returns True for a(0). 0-indexed.
answered 7 hours ago
Erik the OutgolferErik the Outgolfer
33k429106
$endgroup$
add a comment |
$begingroup$
Python 2, 62 bytes
a=lambda n:n<1or a(n-1)*6**(a(n-1)//2in[0]+map(a,range(n)))//2
Try it online!
Returns True for a(0). 0-indexed.
answered 7 hours ago
Erik the OutgolferErik the Outgolfer
33k429106
$endgroup$
Python 2, 62 bytes
a=lambda n:n<1or a(n-1)*6**(a(n-1)//2in[0]+map(a,range(n)))//2
Try it online!
Returns True for a(0). 0-indexed.
answered 7 hours ago
Erik the OutgolferErik the Outgolfer
33k429106
answered 7 hours ago
Erik the OutgolferErik the Outgolfer
33k429106
answered 7 hours ago
Erik the OutgolferErik the Outgolfer
33k429106
answered 7 hours ago
Erik the OutgolferErik the Outgolfer
33k429106
33k429106
add a comment |
add a comment |
$begingroup$
Python 3, 105 103 100 95 83 bytes
-2 bytes thanks to agtoever
-12 bytes thanks to ArBo
def f(n):
s=0,1
while len(s)<=n:t=s[-1]//2;s+=(t in s)*3*s[-1]or t,
return s[-1]
Try it online!
answered 17 hours ago
Noodle9Noodle9
29137
$endgroup$
$begingroup$
You can replace the for loop withwhile len(s)<=nand replace the i's with-1. This should shave off one of two characters.
$endgroup$
– agtoever
17 hours ago
$begingroup$
@agtoever that's so clever - thanks! :)
$endgroup$
– Noodle9
15 hours ago
$begingroup$
83 bytes by working with a tuple instead of a list, and removing theiffrom thewhileloop to allow one-lining that loop
$endgroup$
– ArBo
9 hours ago
$begingroup$
@ArBo wow! absolutely brilliant - thanks :)
$endgroup$
– Noodle9
6 hours ago
add a comment |
$begingroup$
Python 3, 105 103 100 95 83 bytes
-2 bytes thanks to agtoever
-12 bytes thanks to ArBo
def f(n):
s=0,1
while len(s)<=n:t=s[-1]//2;s+=(t in s)*3*s[-1]or t,
return s[-1]
Try it online!
answered 17 hours ago
Noodle9Noodle9
29137
$endgroup$
$begingroup$
You can replace the for loop withwhile len(s)<=nand replace the i's with-1. This should shave off one of two characters.
$endgroup$
– agtoever
17 hours ago
$begingroup$
@agtoever that's so clever - thanks! :)
$endgroup$
– Noodle9
15 hours ago
$begingroup$
83 bytes by working with a tuple instead of a list, and removing theiffrom thewhileloop to allow one-lining that loop
$endgroup$
– ArBo
9 hours ago
$begingroup$
@ArBo wow! absolutely brilliant - thanks :)
$endgroup$
– Noodle9
6 hours ago
add a comment |
$begingroup$
Python 3, 105 103 100 95 83 bytes
-2 bytes thanks to agtoever
-12 bytes thanks to ArBo
def f(n):
s=0,1
while len(s)<=n:t=s[-1]//2;s+=(t in s)*3*s[-1]or t,
return s[-1]
Try it online!
answered 17 hours ago
Noodle9Noodle9
29137
$endgroup$
Python 3, 105 103 100 95 83 bytes
-2 bytes thanks to agtoever
-12 bytes thanks to ArBo
def f(n):
s=0,1
while len(s)<=n:t=s[-1]//2;s+=(t in s)*3*s[-1]or t,
return s[-1]
Try it online!
answered 17 hours ago
Noodle9Noodle9
29137
edited 6 hours ago
answered 17 hours ago
Noodle9Noodle9
29137
answered 17 hours ago
Noodle9Noodle9
29137
answered 17 hours ago
Noodle9Noodle9
29137
29137
$begingroup$
You can replace the for loop withwhile len(s)<=nand replace the i's with-1. This should shave off one of two characters.
$endgroup$
– agtoever
17 hours ago
$begingroup$
@agtoever that's so clever - thanks! :)
$endgroup$
– Noodle9
15 hours ago
$begingroup$
83 bytes by working with a tuple instead of a list, and removing theiffrom thewhileloop to allow one-lining that loop
$endgroup$
– ArBo
9 hours ago
$begingroup$
@ArBo wow! absolutely brilliant - thanks :)
$endgroup$
– Noodle9
6 hours ago
add a comment |
$begingroup$
You can replace the for loop withwhile len(s)<=nand replace the i's with-1. This should shave off one of two characters.
$endgroup$
– agtoever
17 hours ago
$begingroup$
@agtoever that's so clever - thanks! :)
$endgroup$
– Noodle9
15 hours ago
$begingroup$
83 bytes by working with a tuple instead of a list, and removing theiffrom thewhileloop to allow one-lining that loop
$endgroup$
– ArBo
9 hours ago
$begingroup$
@ArBo wow! absolutely brilliant - thanks :)
$endgroup$
– Noodle9
6 hours ago
$begingroup$
You can replace the for loop with
while len(s)<=n and replace the i's with -1. This should shave off one of two characters.$endgroup$
– agtoever
17 hours ago
$begingroup$
You can replace the for loop with
while len(s)<=n and replace the i's with -1. This should shave off one of two characters.$endgroup$
– agtoever
17 hours ago
$begingroup$
@agtoever that's so clever - thanks! :)
$endgroup$
– Noodle9
15 hours ago
$begingroup$
@agtoever that's so clever - thanks! :)
$endgroup$
– Noodle9
15 hours ago
$begingroup$
83 bytes by working with a tuple instead of a list, and removing the
if from the while loop to allow one-lining that loop$endgroup$
– ArBo
9 hours ago
$begingroup$
83 bytes by working with a tuple instead of a list, and removing the
if from the while loop to allow one-lining that loop$endgroup$
– ArBo
9 hours ago
$begingroup$
@ArBo wow! absolutely brilliant - thanks :)
$endgroup$
– Noodle9
6 hours ago
$begingroup$
@ArBo wow! absolutely brilliant - thanks :)
$endgroup$
– Noodle9
6 hours ago
add a comment |
$begingroup$
Gaia, 22 20 bytes
2…@⟨:):3פḥ⌋,;D)+⟩ₓ)
Try it online!
0-based index.
Credit to Shaggy for the approach
2… | push [0 1]
@⟨ ⟩ₓ | do the following n times:
:): | dup the list L, take the last element e, and dup that
3פḥ⌋, | push [3*e floor(e/2)]
;D | take the asymmetric set difference [3*e floor(e/2)] - L
)+ | take the last element of the difference and add it to the end of L (end of loop)
) | finally, take the last element and output it
;D
answered 10 hours ago
GiuseppeGiuseppe
17.6k31153
$endgroup$
add a comment |
$begingroup$
Gaia, 22 20 bytes
2…@⟨:):3פḥ⌋,;D)+⟩ₓ)
Try it online!
0-based index.
Credit to Shaggy for the approach
2… | push [0 1]
@⟨ ⟩ₓ | do the following n times:
:): | dup the list L, take the last element e, and dup that
3פḥ⌋, | push [3*e floor(e/2)]
;D | take the asymmetric set difference [3*e floor(e/2)] - L
)+ | take the last element of the difference and add it to the end of L (end of loop)
) | finally, take the last element and output it
;D
answered 10 hours ago
GiuseppeGiuseppe
17.6k31153
$endgroup$
add a comment |
$begingroup$
Gaia, 22 20 bytes
2…@⟨:):3פḥ⌋,;D)+⟩ₓ)
Try it online!
0-based index.
Credit to Shaggy for the approach
2… | push [0 1]
@⟨ ⟩ₓ | do the following n times:
:): | dup the list L, take the last element e, and dup that
3פḥ⌋, | push [3*e floor(e/2)]
;D | take the asymmetric set difference [3*e floor(e/2)] - L
)+ | take the last element of the difference and add it to the end of L (end of loop)
) | finally, take the last element and output it
;D
answered 10 hours ago
GiuseppeGiuseppe
17.6k31153
$endgroup$
Gaia, 22 20 bytes
2…@⟨:):3פḥ⌋,;D)+⟩ₓ)
Try it online!
0-based index.
Credit to Shaggy for the approach
2… | push [0 1]
@⟨ ⟩ₓ | do the following n times:
:): | dup the list L, take the last element e, and dup that
3פḥ⌋, | push [3*e floor(e/2)]
;D | take the asymmetric set difference [3*e floor(e/2)] - L
)+ | take the last element of the difference and add it to the end of L (end of loop)
) | finally, take the last element and output it
;D
answered 10 hours ago
GiuseppeGiuseppe
17.6k31153
edited 4 hours ago
answered 10 hours ago
GiuseppeGiuseppe
17.6k31153
answered 10 hours ago
GiuseppeGiuseppe
17.6k31153
answered 10 hours ago
GiuseppeGiuseppe
17.6k31153
17.6k31153
add a comment |
add a comment |
$begingroup$
Haskell, 55 bytes
(1%[0]!!)
a%o|b<-div a 2=a:last(b:[3*a|elem b o])%(a:o)
Try it online!
Golfing user1472751's slick list-generation method.
Same length:
(1%[0]!!)
a%o=a:[x|x<-[div a 2,a*3],all(/=x)o]!!0%(a:o)
Try it online!
answered 18 hours ago
xnorxnor
93.6k18190450
$endgroup$
add a comment |
$begingroup$
Haskell, 55 bytes
(1%[0]!!)
a%o|b<-div a 2=a:last(b:[3*a|elem b o])%(a:o)
Try it online!
Golfing user1472751's slick list-generation method.
Same length:
(1%[0]!!)
a%o=a:[x|x<-[div a 2,a*3],all(/=x)o]!!0%(a:o)
Try it online!
answered 18 hours ago
xnorxnor
93.6k18190450
$endgroup$
add a comment |
$begingroup$
Haskell, 55 bytes
(1%[0]!!)
a%o|b<-div a 2=a:last(b:[3*a|elem b o])%(a:o)
Try it online!
Golfing user1472751's slick list-generation method.
Same length:
(1%[0]!!)
a%o=a:[x|x<-[div a 2,a*3],all(/=x)o]!!0%(a:o)
Try it online!
answered 18 hours ago
xnorxnor
93.6k18190450
$endgroup$
Haskell, 55 bytes
(1%[0]!!)
a%o|b<-div a 2=a:last(b:[3*a|elem b o])%(a:o)
Try it online!
Golfing user1472751's slick list-generation method.
Same length:
(1%[0]!!)
a%o=a:[x|x<-[div a 2,a*3],all(/=x)o]!!0%(a:o)
Try it online!
answered 18 hours ago
xnorxnor
93.6k18190450
edited 17 hours ago
answered 18 hours ago
xnorxnor
93.6k18190450
answered 18 hours ago
xnorxnor
93.6k18190450
answered 18 hours ago
xnorxnor
93.6k18190450
93.6k18190450
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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Sharp! My mistake. Corrected it. Sorry if this caused any trouble or confusion.
$endgroup$
– agtoever
yesterday