Given an input, print all exponents where the base and power sum to the input
up vote
18
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So this is my first challenge on this site.
What this challenge is is that you input a number, and your code spits out numbers which can be calculated as an exponent. The sum of the exponent and base in the number must equal the input.
The numbers must start from $1^{n-1}$ to $(n)^0$.
Example
Given input 5, the program will print:
1
8
9
4
1
$1^4$ is 1 and $1+4=5$
$2^3$ is 8 and $2+3=5$
$3^2$ is 9 and $3+2=5$
$4^1$ is 4 and $4+1=5$
$5^0$ is 1 and $5+0=5$
Input and Output
Input will be in the form of an integer.
Output will be a list of numbers, delimited by either commas or new lines.
This is code-golf, so shortest code wins.
code-golf math arithmetic
New contributor
|
show 5 more comments
up vote
18
down vote
favorite
So this is my first challenge on this site.
What this challenge is is that you input a number, and your code spits out numbers which can be calculated as an exponent. The sum of the exponent and base in the number must equal the input.
The numbers must start from $1^{n-1}$ to $(n)^0$.
Example
Given input 5, the program will print:
1
8
9
4
1
$1^4$ is 1 and $1+4=5$
$2^3$ is 8 and $2+3=5$
$3^2$ is 9 and $3+2=5$
$4^1$ is 4 and $4+1=5$
$5^0$ is 1 and $5+0=5$
Input and Output
Input will be in the form of an integer.
Output will be a list of numbers, delimited by either commas or new lines.
This is code-golf, so shortest code wins.
code-golf math arithmetic
New contributor
5
the comma/newline detail should be omitted, it is normal practice around here to let output of lists be in any convenient format, including as a list/array object being returned by a function
– Sparr
2 days ago
3
Is the input always greater than 0 or do we have to deal with 0 and negatives?
– Veskah
2 days ago
Inputs will always be positive
– Embodiment of Ignorance
2 days ago
6
Two equally short answers doesn't matter. If you feel like accepting an answer, choose the earliest posted one. However I strongly recommend waiting at least a few days, and would suggest never accepting an answer (to encourage more submissions).
– Οurous
2 days ago
1
Shouldn't the title be "Given an integer, print all the powers obtained with a base and an exponent that sum to the input"?
– Nicola Sap
2 days ago
|
show 5 more comments
up vote
18
down vote
favorite
up vote
18
down vote
favorite
So this is my first challenge on this site.
What this challenge is is that you input a number, and your code spits out numbers which can be calculated as an exponent. The sum of the exponent and base in the number must equal the input.
The numbers must start from $1^{n-1}$ to $(n)^0$.
Example
Given input 5, the program will print:
1
8
9
4
1
$1^4$ is 1 and $1+4=5$
$2^3$ is 8 and $2+3=5$
$3^2$ is 9 and $3+2=5$
$4^1$ is 4 and $4+1=5$
$5^0$ is 1 and $5+0=5$
Input and Output
Input will be in the form of an integer.
Output will be a list of numbers, delimited by either commas or new lines.
This is code-golf, so shortest code wins.
code-golf math arithmetic
New contributor
So this is my first challenge on this site.
What this challenge is is that you input a number, and your code spits out numbers which can be calculated as an exponent. The sum of the exponent and base in the number must equal the input.
The numbers must start from $1^{n-1}$ to $(n)^0$.
Example
Given input 5, the program will print:
1
8
9
4
1
$1^4$ is 1 and $1+4=5$
$2^3$ is 8 and $2+3=5$
$3^2$ is 9 and $3+2=5$
$4^1$ is 4 and $4+1=5$
$5^0$ is 1 and $5+0=5$
Input and Output
Input will be in the form of an integer.
Output will be a list of numbers, delimited by either commas or new lines.
This is code-golf, so shortest code wins.
code-golf math arithmetic
code-golf math arithmetic
New contributor
New contributor
edited 2 days ago
Οurous
5,99311032
5,99311032
New contributor
asked 2 days ago
Embodiment of Ignorance
1507
1507
New contributor
New contributor
5
the comma/newline detail should be omitted, it is normal practice around here to let output of lists be in any convenient format, including as a list/array object being returned by a function
– Sparr
2 days ago
3
Is the input always greater than 0 or do we have to deal with 0 and negatives?
– Veskah
2 days ago
Inputs will always be positive
– Embodiment of Ignorance
2 days ago
6
Two equally short answers doesn't matter. If you feel like accepting an answer, choose the earliest posted one. However I strongly recommend waiting at least a few days, and would suggest never accepting an answer (to encourage more submissions).
– Οurous
2 days ago
1
Shouldn't the title be "Given an integer, print all the powers obtained with a base and an exponent that sum to the input"?
– Nicola Sap
2 days ago
|
show 5 more comments
5
the comma/newline detail should be omitted, it is normal practice around here to let output of lists be in any convenient format, including as a list/array object being returned by a function
– Sparr
2 days ago
3
Is the input always greater than 0 or do we have to deal with 0 and negatives?
– Veskah
2 days ago
Inputs will always be positive
– Embodiment of Ignorance
2 days ago
6
Two equally short answers doesn't matter. If you feel like accepting an answer, choose the earliest posted one. However I strongly recommend waiting at least a few days, and would suggest never accepting an answer (to encourage more submissions).
– Οurous
2 days ago
1
Shouldn't the title be "Given an integer, print all the powers obtained with a base and an exponent that sum to the input"?
– Nicola Sap
2 days ago
5
5
the comma/newline detail should be omitted, it is normal practice around here to let output of lists be in any convenient format, including as a list/array object being returned by a function
– Sparr
2 days ago
the comma/newline detail should be omitted, it is normal practice around here to let output of lists be in any convenient format, including as a list/array object being returned by a function
– Sparr
2 days ago
3
3
Is the input always greater than 0 or do we have to deal with 0 and negatives?
– Veskah
2 days ago
Is the input always greater than 0 or do we have to deal with 0 and negatives?
– Veskah
2 days ago
Inputs will always be positive
– Embodiment of Ignorance
2 days ago
Inputs will always be positive
– Embodiment of Ignorance
2 days ago
6
6
Two equally short answers doesn't matter. If you feel like accepting an answer, choose the earliest posted one. However I strongly recommend waiting at least a few days, and would suggest never accepting an answer (to encourage more submissions).
– Οurous
2 days ago
Two equally short answers doesn't matter. If you feel like accepting an answer, choose the earliest posted one. However I strongly recommend waiting at least a few days, and would suggest never accepting an answer (to encourage more submissions).
– Οurous
2 days ago
1
1
Shouldn't the title be "Given an integer, print all the powers obtained with a base and an exponent that sum to the input"?
– Nicola Sap
2 days ago
Shouldn't the title be "Given an integer, print all the powers obtained with a base and an exponent that sum to the input"?
– Nicola Sap
2 days ago
|
show 5 more comments
34 Answers
34
active
oldest
votes
1 2
next
up vote
5
down vote
APL (Dyalog Unicode), 8 5 bytes
⍳*⊢-⍳
Try it online!
Anonymous prefix tacit function. TIO tests for the range [1..10].
Thanks @lirtosiast for 3 bytes.
How:
⍳*⊢-⍳ ⍝ Tacit function
⍳ ⍝ Range. ⍳n generates the vector [1..n].
⊢- ⍝ Subtracted from the argument. The vector is now [n-1,n-2,...,0]
⍳* ⍝ Exponentiate using the range [1..n] as base. The result is the vector
⍝ [1^(n-1), 2^(n-2), 3^(n-3),...]
2
⍳*⊢-⍳
is 5 bytes, using⎕IO←1
.
– lirtosiast
2 days ago
@lirtosiast took me a while to figure out why does that work, but I got it. Thanks.
– J. Sallé
yesterday
add a comment |
up vote
4
down vote
Japt, 5 bytes
õ_p´U
Try it
õ :Range [1,input]
_ :Map
p : Raise to the power of
´U : Input decremented
add a comment |
up vote
4
down vote
Perl 6, 19 bytes
{^$_+1 Z**[R,] ^$_}
Try it online!
Anonymous code block that takes a number and returns a list. Zip exponents the range 1 to input
and the range input-1 to 0
add a comment |
up vote
4
down vote
Aheui (esotope), 193 164 bytes (56 chars)
방빠싹받분샥퍼붇바파쟈뿌차샦히망맣여
타빠바푸투반또분뽀뿌서썪삯타삯받반타
석차샦져쌲볼어타토싻삭빠쏛ㅇ또섞썪뻐
Try it online!
Try it on AVIS(Korean); just copy and paste code above, press start button, input a number, see how it moves. To see output, press the >_ icon on left side.
It's not golfed much, but I give it a shot.
Is it possible to chose a character set, so that each character is stored in 2 bytes?
– tsh
yesterday
@tsh According to Aheui specification, an Aheui code consists of only UTF-8 characters.
– cobaltp
yesterday
add a comment |
up vote
3
down vote
Jelly, 5 bytes
R*ḶU$
Try it online!
R [1,...,n]
* to the power of
ḶU$ [0,...,n-1] reversed
add a comment |
up vote
3
down vote
Pyth, 5 bytes
_m^-Q
Try it online!
Optimally encoded this would be 4.106 bytes.
_ reverse of the following list:
m map the following lambda d:
^ (N-d)**d
-Qd
d
Q over [0,...,N-1]
add a comment |
up vote
3
down vote
Haskell, 23 bytes
f i=[x^(i-x)|x<-[1..i]]
Try it online!
Alternative version, also 23 bytes:
f i=(^)<*>(i-)<$>[1..i]
add a comment |
up vote
3
down vote
J, 10 bytes
(>:^|.)@i.
Try it online!
If we really need to separate the numbers by a newline:
J, 13 bytes
,.@(>:^|.)@i.
Try it online!
add a comment |
up vote
3
down vote
PHP, 32 bytes
while($argn)echo++$i**--$argn,_;
Run as pipe with -nR
or try it online.
add a comment |
up vote
3
down vote
Octave, 18 bytes
@(n)(t=1:n).^(n-t)
Try it online!
Thanks Luis Mendo, using internal variable saves 3 bytes.
add a comment |
up vote
2
down vote
Wolfram Language (Mathematica), 24 20 18 bytes
(x=Range@#)^(#-x)&
Try it online!
-4 thanks @lirtosiast.
add a comment |
up vote
2
down vote
MathGolf, 6 bytes
rx╒m#
Try it online!
I have implemented reverse subtraction, multiplication and division, but it looks like a reverse power operator could come in handy?
– maxb
2 days ago
add a comment |
up vote
2
down vote
Python 2, 40 bytes
lambda n:[i**(n-i)for i in range(1,n+1)] #Outputs a list
Try it online!
Python 2, 41 bytes
n,i=input(),0
exec"print(n-i)**i;i+=1;"*n #Prints in reversed order
Try it online!
add a comment |
up vote
2
down vote
Ruby, 27 bytes
->n{(1..n).map{|r|r**n-=1}}
Try it online!
add a comment |
up vote
2
down vote
Retina, 35 bytes
.+
*
_
$$.($.'*$($.>`$*)_¶
%~`^
.+¶
Try it online! Explanation:
.+
*
Convert the input to unary.
_
Match each position. This then sets several replacement variables. $`
becomes the left of the match; $>`
modifies this to be the left and match; $.>`
modifies this to take the length, i.e. the current index. $'
meanwhile is the right of the match, so $.'
is the length i.e. the current exponent.
$$.($.'*$($.>`$*)_¶
Create a string $.(
plus $.'
repetitions of $.>`*
plus _
. For an example, for an index of 2 in an original input of 5, $.'
is 3 and $.>`
is 2 so the resulting string is $.(2*2*2*_
. This conveniently is a Retina replacement expression that caluclates 2³. Each string is output on its own line.
%~`^
.+¶
For each line generated by the previous stage, prefix a line .+
to it, turning it into a replacement stage, and evaluate that stage, thereby calculating the expression.
add a comment |
up vote
2
down vote
QBasic, 3533 bytes
Thank you @Neil for 2 bytes!
INPUT a
FOR b=1TO a
?b^(a-b)
NEXT
Slightly expanded version on REPL.IT because the interpreter in't entirely up-to-spec.
Output
QBasic (qb.js)
Copyright (c) 2010 Steve Hanov
5
1
8
9
4
1
Save 2 bytes by outputting the list in the correct order! (b^(a-b)
forb=1..a
)
– Neil
2 days ago
@Neil Thanks, I've worked it in!
– steenbergh
2 days ago
add a comment |
up vote
2
down vote
F# (.NET Core), 42 bytes
let f x=Seq.map(fun y->pown y (x-y))[1..x]
Try it online!
add a comment |
up vote
2
down vote
JavaScript (Node.js), 33 32 bytes
n=>(g=i=>--n?++i**n+[,g(i)]:1)``
Try it online!
-3 bytes with credits to @Shaggy, and -1 byte by @l4m2!
JavaScript (Node.js), 36 bytes
f=(n,i=1)=>n--?[i++**n,...f(n,i)]:
Try it online!
JavaScript (Node.js), 37 bytes
n=>[...Array(n)].map(x=>++i**--n,i=0)
Try it online!
33 bytes
– Shaggy
2 days ago
32
– l4m2
2 days ago
add a comment |
up vote
2
down vote
C# (Visual C# Interactive Compiler), 46 bytes
x=>new int[x].Select((_,i)=>Math.Pow(i+1,--x))
Try it online!
add a comment |
up vote
2
down vote
MATL, 5 bytes
:Gy-^
Try it online!
Explanation
Consider input 5
as an example.
: % Implicit input. Range
% STACK: [1 2 3 4 5]
G % Push input again
% STACK: [1 2 3 4 5], 5
y % Duplicate from below
% STACK: [1 2 3 4 5], 5, [1 2 3 4 5]
- % Subtract, element-wise
% STACK: [1 2 3 4 5], [4 3 2 1 0]
^ % Power, element-wise. Implicit display
% STACK: [1 8 9 4 1]
add a comment |
up vote
2
down vote
Java, 59 Bytes
for(int i=1;a+1>i;i++)System.out.println(Math.pow(i,a-i));
Welcome to PPCG. It looks like this requires "input" be assigned to the predefined variablea
, which we don't allow.
– Shaggy
yesterday
Hello, here's a fix for you:n->{for(int i=0;i++<n;)System.out.println(Math.pow(i,n-i));}
60 bytes (code and test cases in the link)
– Olivier Grégoire
14 hours ago
add a comment |
up vote
2
down vote
Jelly, 4 bytes
*ạ¥€
Try it online!
add a comment |
up vote
1
down vote
Clean, 37 bytes
import StdEnv
$n=[i^(n-i)\i<-[1..n]]
Try it online!
Defines $ :: Int -> [Int]
taking an integer and returning the list of results.
$ n // function $ of n
= [i ^ (n-i) // i to the power of n minus i
\ i <- [1..n] // for each i in 1 to n
]
add a comment |
up vote
1
down vote
R, 34 bytes
x=1:scan();cat(x^rev(x-1),sep=',')
Try it online!
Is the default "sep" not a space? Would that not work?
– stuart stevenson
yesterday
1
@stuartstevenson "Output will be a list of numbers, delimited by either commas or new lines."
– Giuseppe
yesterday
add a comment |
up vote
1
down vote
05AB1E, 5 bytes
LD<Rm
Port of @lirtosiast's Jelly answer.
Try it online.
Explanation:
L # List in the range [1, (implicit) input integer]
# i.e. 5 → [1,2,3,4,5]
D< # Duplicate this list, and subtract 1 to make the range [0, input)
# i.e. [1,2,3,4,5] → [0,1,2,3,4]
R # Reverse it to make the range (input, 0]
# i.e. [0,1,2,3,4] → [4,3,2,1,0]
m # Take the power of the numbers in the lists (at the same indices)
# (and output implicitly)
# i.e. [1,2,3,4,5] and [4,3,2,1,0] → [1,8,9,4,1]
add a comment |
up vote
1
down vote
Lua, 43 41 bytes
-2 bytes thanks to @Shaggy
s=io.read()for i=1,s do print(i^(s-i))end
Try it online!
1
I don't think you need the+0
; seems to work without it.
– Shaggy
2 days ago
add a comment |
up vote
1
down vote
R, 22 bytes
n=scan();(1:n)^(n:1-1)
Fairly self-explanatory; note that the :
operator is higher precendence than the -
operator so that n:1-1
is shorter than (n-1):0
If we are allowed to start at 0, then we can lose two bytes by using (0:n)^(n:0)
avoiding the need for a -1.
add a comment |
up vote
1
down vote
Charcoal, 9 bytes
I⮌ENX⁻θιι
Try it online! Link is to verbose version of code. Explanation:
N Input as a number
E Map over implicit range
ι Current value
⁻ Subtracted from
θ First input
X Raised to power
ι Current value
⮌ Reverse list
I Cast to string
Implicitly print on separate lines
add a comment |
up vote
1
down vote
C# (Visual C# Interactive Compiler), 55 bytes
v=>Enumerable.Range(0,v--).Select(i=>Math.Pow(i+1,v--))
Try it online!
add a comment |
up vote
1
down vote
Perl 5 -n
, 21 bytes
say++$**--$_ while$_
Try it online!
add a comment |
1 2
next
34 Answers
34
active
oldest
votes
34 Answers
34
active
oldest
votes
active
oldest
votes
active
oldest
votes
1 2
next
up vote
5
down vote
APL (Dyalog Unicode), 8 5 bytes
⍳*⊢-⍳
Try it online!
Anonymous prefix tacit function. TIO tests for the range [1..10].
Thanks @lirtosiast for 3 bytes.
How:
⍳*⊢-⍳ ⍝ Tacit function
⍳ ⍝ Range. ⍳n generates the vector [1..n].
⊢- ⍝ Subtracted from the argument. The vector is now [n-1,n-2,...,0]
⍳* ⍝ Exponentiate using the range [1..n] as base. The result is the vector
⍝ [1^(n-1), 2^(n-2), 3^(n-3),...]
2
⍳*⊢-⍳
is 5 bytes, using⎕IO←1
.
– lirtosiast
2 days ago
@lirtosiast took me a while to figure out why does that work, but I got it. Thanks.
– J. Sallé
yesterday
add a comment |
up vote
5
down vote
APL (Dyalog Unicode), 8 5 bytes
⍳*⊢-⍳
Try it online!
Anonymous prefix tacit function. TIO tests for the range [1..10].
Thanks @lirtosiast for 3 bytes.
How:
⍳*⊢-⍳ ⍝ Tacit function
⍳ ⍝ Range. ⍳n generates the vector [1..n].
⊢- ⍝ Subtracted from the argument. The vector is now [n-1,n-2,...,0]
⍳* ⍝ Exponentiate using the range [1..n] as base. The result is the vector
⍝ [1^(n-1), 2^(n-2), 3^(n-3),...]
2
⍳*⊢-⍳
is 5 bytes, using⎕IO←1
.
– lirtosiast
2 days ago
@lirtosiast took me a while to figure out why does that work, but I got it. Thanks.
– J. Sallé
yesterday
add a comment |
up vote
5
down vote
up vote
5
down vote
APL (Dyalog Unicode), 8 5 bytes
⍳*⊢-⍳
Try it online!
Anonymous prefix tacit function. TIO tests for the range [1..10].
Thanks @lirtosiast for 3 bytes.
How:
⍳*⊢-⍳ ⍝ Tacit function
⍳ ⍝ Range. ⍳n generates the vector [1..n].
⊢- ⍝ Subtracted from the argument. The vector is now [n-1,n-2,...,0]
⍳* ⍝ Exponentiate using the range [1..n] as base. The result is the vector
⍝ [1^(n-1), 2^(n-2), 3^(n-3),...]
APL (Dyalog Unicode), 8 5 bytes
⍳*⊢-⍳
Try it online!
Anonymous prefix tacit function. TIO tests for the range [1..10].
Thanks @lirtosiast for 3 bytes.
How:
⍳*⊢-⍳ ⍝ Tacit function
⍳ ⍝ Range. ⍳n generates the vector [1..n].
⊢- ⍝ Subtracted from the argument. The vector is now [n-1,n-2,...,0]
⍳* ⍝ Exponentiate using the range [1..n] as base. The result is the vector
⍝ [1^(n-1), 2^(n-2), 3^(n-3),...]
edited yesterday
answered 2 days ago
J. Sallé
1,853322
1,853322
2
⍳*⊢-⍳
is 5 bytes, using⎕IO←1
.
– lirtosiast
2 days ago
@lirtosiast took me a while to figure out why does that work, but I got it. Thanks.
– J. Sallé
yesterday
add a comment |
2
⍳*⊢-⍳
is 5 bytes, using⎕IO←1
.
– lirtosiast
2 days ago
@lirtosiast took me a while to figure out why does that work, but I got it. Thanks.
– J. Sallé
yesterday
2
2
⍳*⊢-⍳
is 5 bytes, using ⎕IO←1
.– lirtosiast
2 days ago
⍳*⊢-⍳
is 5 bytes, using ⎕IO←1
.– lirtosiast
2 days ago
@lirtosiast took me a while to figure out why does that work, but I got it. Thanks.
– J. Sallé
yesterday
@lirtosiast took me a while to figure out why does that work, but I got it. Thanks.
– J. Sallé
yesterday
add a comment |
up vote
4
down vote
Japt, 5 bytes
õ_p´U
Try it
õ :Range [1,input]
_ :Map
p : Raise to the power of
´U : Input decremented
add a comment |
up vote
4
down vote
Japt, 5 bytes
õ_p´U
Try it
õ :Range [1,input]
_ :Map
p : Raise to the power of
´U : Input decremented
add a comment |
up vote
4
down vote
up vote
4
down vote
Japt, 5 bytes
õ_p´U
Try it
õ :Range [1,input]
_ :Map
p : Raise to the power of
´U : Input decremented
Japt, 5 bytes
õ_p´U
Try it
õ :Range [1,input]
_ :Map
p : Raise to the power of
´U : Input decremented
edited 2 days ago
answered 2 days ago
Shaggy
18.4k21663
18.4k21663
add a comment |
add a comment |
up vote
4
down vote
Perl 6, 19 bytes
{^$_+1 Z**[R,] ^$_}
Try it online!
Anonymous code block that takes a number and returns a list. Zip exponents the range 1 to input
and the range input-1 to 0
add a comment |
up vote
4
down vote
Perl 6, 19 bytes
{^$_+1 Z**[R,] ^$_}
Try it online!
Anonymous code block that takes a number and returns a list. Zip exponents the range 1 to input
and the range input-1 to 0
add a comment |
up vote
4
down vote
up vote
4
down vote
Perl 6, 19 bytes
{^$_+1 Z**[R,] ^$_}
Try it online!
Anonymous code block that takes a number and returns a list. Zip exponents the range 1 to input
and the range input-1 to 0
Perl 6, 19 bytes
{^$_+1 Z**[R,] ^$_}
Try it online!
Anonymous code block that takes a number and returns a list. Zip exponents the range 1 to input
and the range input-1 to 0
edited 2 days ago
answered 2 days ago
Jo King
19.7k245105
19.7k245105
add a comment |
add a comment |
up vote
4
down vote
Aheui (esotope), 193 164 bytes (56 chars)
방빠싹받분샥퍼붇바파쟈뿌차샦히망맣여
타빠바푸투반또분뽀뿌서썪삯타삯받반타
석차샦져쌲볼어타토싻삭빠쏛ㅇ또섞썪뻐
Try it online!
Try it on AVIS(Korean); just copy and paste code above, press start button, input a number, see how it moves. To see output, press the >_ icon on left side.
It's not golfed much, but I give it a shot.
Is it possible to chose a character set, so that each character is stored in 2 bytes?
– tsh
yesterday
@tsh According to Aheui specification, an Aheui code consists of only UTF-8 characters.
– cobaltp
yesterday
add a comment |
up vote
4
down vote
Aheui (esotope), 193 164 bytes (56 chars)
방빠싹받분샥퍼붇바파쟈뿌차샦히망맣여
타빠바푸투반또분뽀뿌서썪삯타삯받반타
석차샦져쌲볼어타토싻삭빠쏛ㅇ또섞썪뻐
Try it online!
Try it on AVIS(Korean); just copy and paste code above, press start button, input a number, see how it moves. To see output, press the >_ icon on left side.
It's not golfed much, but I give it a shot.
Is it possible to chose a character set, so that each character is stored in 2 bytes?
– tsh
yesterday
@tsh According to Aheui specification, an Aheui code consists of only UTF-8 characters.
– cobaltp
yesterday
add a comment |
up vote
4
down vote
up vote
4
down vote
Aheui (esotope), 193 164 bytes (56 chars)
방빠싹받분샥퍼붇바파쟈뿌차샦히망맣여
타빠바푸투반또분뽀뿌서썪삯타삯받반타
석차샦져쌲볼어타토싻삭빠쏛ㅇ또섞썪뻐
Try it online!
Try it on AVIS(Korean); just copy and paste code above, press start button, input a number, see how it moves. To see output, press the >_ icon on left side.
It's not golfed much, but I give it a shot.
Aheui (esotope), 193 164 bytes (56 chars)
방빠싹받분샥퍼붇바파쟈뿌차샦히망맣여
타빠바푸투반또분뽀뿌서썪삯타삯받반타
석차샦져쌲볼어타토싻삭빠쏛ㅇ또섞썪뻐
Try it online!
Try it on AVIS(Korean); just copy and paste code above, press start button, input a number, see how it moves. To see output, press the >_ icon on left side.
It's not golfed much, but I give it a shot.
edited yesterday
answered 2 days ago
cobaltp
3417
3417
Is it possible to chose a character set, so that each character is stored in 2 bytes?
– tsh
yesterday
@tsh According to Aheui specification, an Aheui code consists of only UTF-8 characters.
– cobaltp
yesterday
add a comment |
Is it possible to chose a character set, so that each character is stored in 2 bytes?
– tsh
yesterday
@tsh According to Aheui specification, an Aheui code consists of only UTF-8 characters.
– cobaltp
yesterday
Is it possible to chose a character set, so that each character is stored in 2 bytes?
– tsh
yesterday
Is it possible to chose a character set, so that each character is stored in 2 bytes?
– tsh
yesterday
@tsh According to Aheui specification, an Aheui code consists of only UTF-8 characters.
– cobaltp
yesterday
@tsh According to Aheui specification, an Aheui code consists of only UTF-8 characters.
– cobaltp
yesterday
add a comment |
up vote
3
down vote
Jelly, 5 bytes
R*ḶU$
Try it online!
R [1,...,n]
* to the power of
ḶU$ [0,...,n-1] reversed
add a comment |
up vote
3
down vote
Jelly, 5 bytes
R*ḶU$
Try it online!
R [1,...,n]
* to the power of
ḶU$ [0,...,n-1] reversed
add a comment |
up vote
3
down vote
up vote
3
down vote
Jelly, 5 bytes
R*ḶU$
Try it online!
R [1,...,n]
* to the power of
ḶU$ [0,...,n-1] reversed
Jelly, 5 bytes
R*ḶU$
Try it online!
R [1,...,n]
* to the power of
ḶU$ [0,...,n-1] reversed
edited 2 days ago
answered 2 days ago
lirtosiast
15.6k436105
15.6k436105
add a comment |
add a comment |
up vote
3
down vote
Pyth, 5 bytes
_m^-Q
Try it online!
Optimally encoded this would be 4.106 bytes.
_ reverse of the following list:
m map the following lambda d:
^ (N-d)**d
-Qd
d
Q over [0,...,N-1]
add a comment |
up vote
3
down vote
Pyth, 5 bytes
_m^-Q
Try it online!
Optimally encoded this would be 4.106 bytes.
_ reverse of the following list:
m map the following lambda d:
^ (N-d)**d
-Qd
d
Q over [0,...,N-1]
add a comment |
up vote
3
down vote
up vote
3
down vote
Pyth, 5 bytes
_m^-Q
Try it online!
Optimally encoded this would be 4.106 bytes.
_ reverse of the following list:
m map the following lambda d:
^ (N-d)**d
-Qd
d
Q over [0,...,N-1]
Pyth, 5 bytes
_m^-Q
Try it online!
Optimally encoded this would be 4.106 bytes.
_ reverse of the following list:
m map the following lambda d:
^ (N-d)**d
-Qd
d
Q over [0,...,N-1]
edited 2 days ago
answered 2 days ago
lirtosiast
15.6k436105
15.6k436105
add a comment |
add a comment |
up vote
3
down vote
Haskell, 23 bytes
f i=[x^(i-x)|x<-[1..i]]
Try it online!
Alternative version, also 23 bytes:
f i=(^)<*>(i-)<$>[1..i]
add a comment |
up vote
3
down vote
Haskell, 23 bytes
f i=[x^(i-x)|x<-[1..i]]
Try it online!
Alternative version, also 23 bytes:
f i=(^)<*>(i-)<$>[1..i]
add a comment |
up vote
3
down vote
up vote
3
down vote
Haskell, 23 bytes
f i=[x^(i-x)|x<-[1..i]]
Try it online!
Alternative version, also 23 bytes:
f i=(^)<*>(i-)<$>[1..i]
Haskell, 23 bytes
f i=[x^(i-x)|x<-[1..i]]
Try it online!
Alternative version, also 23 bytes:
f i=(^)<*>(i-)<$>[1..i]
answered 2 days ago
nimi
30.9k31985
30.9k31985
add a comment |
add a comment |
up vote
3
down vote
J, 10 bytes
(>:^|.)@i.
Try it online!
If we really need to separate the numbers by a newline:
J, 13 bytes
,.@(>:^|.)@i.
Try it online!
add a comment |
up vote
3
down vote
J, 10 bytes
(>:^|.)@i.
Try it online!
If we really need to separate the numbers by a newline:
J, 13 bytes
,.@(>:^|.)@i.
Try it online!
add a comment |
up vote
3
down vote
up vote
3
down vote
J, 10 bytes
(>:^|.)@i.
Try it online!
If we really need to separate the numbers by a newline:
J, 13 bytes
,.@(>:^|.)@i.
Try it online!
J, 10 bytes
(>:^|.)@i.
Try it online!
If we really need to separate the numbers by a newline:
J, 13 bytes
,.@(>:^|.)@i.
Try it online!
edited 2 days ago
answered 2 days ago
Galen Ivanov
5,99711032
5,99711032
add a comment |
add a comment |
up vote
3
down vote
PHP, 32 bytes
while($argn)echo++$i**--$argn,_;
Run as pipe with -nR
or try it online.
add a comment |
up vote
3
down vote
PHP, 32 bytes
while($argn)echo++$i**--$argn,_;
Run as pipe with -nR
or try it online.
add a comment |
up vote
3
down vote
up vote
3
down vote
PHP, 32 bytes
while($argn)echo++$i**--$argn,_;
Run as pipe with -nR
or try it online.
PHP, 32 bytes
while($argn)echo++$i**--$argn,_;
Run as pipe with -nR
or try it online.
answered 2 days ago
Titus
12.9k11237
12.9k11237
add a comment |
add a comment |
up vote
3
down vote
Octave, 18 bytes
@(n)(t=1:n).^(n-t)
Try it online!
Thanks Luis Mendo, using internal variable saves 3 bytes.
add a comment |
up vote
3
down vote
Octave, 18 bytes
@(n)(t=1:n).^(n-t)
Try it online!
Thanks Luis Mendo, using internal variable saves 3 bytes.
add a comment |
up vote
3
down vote
up vote
3
down vote
Octave, 18 bytes
@(n)(t=1:n).^(n-t)
Try it online!
Thanks Luis Mendo, using internal variable saves 3 bytes.
Octave, 18 bytes
@(n)(t=1:n).^(n-t)
Try it online!
Thanks Luis Mendo, using internal variable saves 3 bytes.
edited yesterday
answered 2 days ago
tsh
8,08011346
8,08011346
add a comment |
add a comment |
up vote
2
down vote
Wolfram Language (Mathematica), 24 20 18 bytes
(x=Range@#)^(#-x)&
Try it online!
-4 thanks @lirtosiast.
add a comment |
up vote
2
down vote
Wolfram Language (Mathematica), 24 20 18 bytes
(x=Range@#)^(#-x)&
Try it online!
-4 thanks @lirtosiast.
add a comment |
up vote
2
down vote
up vote
2
down vote
Wolfram Language (Mathematica), 24 20 18 bytes
(x=Range@#)^(#-x)&
Try it online!
-4 thanks @lirtosiast.
Wolfram Language (Mathematica), 24 20 18 bytes
(x=Range@#)^(#-x)&
Try it online!
-4 thanks @lirtosiast.
edited 2 days ago
answered 2 days ago
Shieru Asakoto
2,380314
2,380314
add a comment |
add a comment |
up vote
2
down vote
MathGolf, 6 bytes
rx╒m#
Try it online!
I have implemented reverse subtraction, multiplication and division, but it looks like a reverse power operator could come in handy?
– maxb
2 days ago
add a comment |
up vote
2
down vote
MathGolf, 6 bytes
rx╒m#
Try it online!
I have implemented reverse subtraction, multiplication and division, but it looks like a reverse power operator could come in handy?
– maxb
2 days ago
add a comment |
up vote
2
down vote
up vote
2
down vote
MathGolf, 6 bytes
rx╒m#
Try it online!
MathGolf, 6 bytes
rx╒m#
Try it online!
answered 2 days ago
Jo King
19.7k245105
19.7k245105
I have implemented reverse subtraction, multiplication and division, but it looks like a reverse power operator could come in handy?
– maxb
2 days ago
add a comment |
I have implemented reverse subtraction, multiplication and division, but it looks like a reverse power operator could come in handy?
– maxb
2 days ago
I have implemented reverse subtraction, multiplication and division, but it looks like a reverse power operator could come in handy?
– maxb
2 days ago
I have implemented reverse subtraction, multiplication and division, but it looks like a reverse power operator could come in handy?
– maxb
2 days ago
add a comment |
up vote
2
down vote
Python 2, 40 bytes
lambda n:[i**(n-i)for i in range(1,n+1)] #Outputs a list
Try it online!
Python 2, 41 bytes
n,i=input(),0
exec"print(n-i)**i;i+=1;"*n #Prints in reversed order
Try it online!
add a comment |
up vote
2
down vote
Python 2, 40 bytes
lambda n:[i**(n-i)for i in range(1,n+1)] #Outputs a list
Try it online!
Python 2, 41 bytes
n,i=input(),0
exec"print(n-i)**i;i+=1;"*n #Prints in reversed order
Try it online!
add a comment |
up vote
2
down vote
up vote
2
down vote
Python 2, 40 bytes
lambda n:[i**(n-i)for i in range(1,n+1)] #Outputs a list
Try it online!
Python 2, 41 bytes
n,i=input(),0
exec"print(n-i)**i;i+=1;"*n #Prints in reversed order
Try it online!
Python 2, 40 bytes
lambda n:[i**(n-i)for i in range(1,n+1)] #Outputs a list
Try it online!
Python 2, 41 bytes
n,i=input(),0
exec"print(n-i)**i;i+=1;"*n #Prints in reversed order
Try it online!
answered 2 days ago
Vedant Kandoi
76021
76021
add a comment |
add a comment |
up vote
2
down vote
Ruby, 27 bytes
->n{(1..n).map{|r|r**n-=1}}
Try it online!
add a comment |
up vote
2
down vote
Ruby, 27 bytes
->n{(1..n).map{|r|r**n-=1}}
Try it online!
add a comment |
up vote
2
down vote
up vote
2
down vote
Ruby, 27 bytes
->n{(1..n).map{|r|r**n-=1}}
Try it online!
Ruby, 27 bytes
->n{(1..n).map{|r|r**n-=1}}
Try it online!
answered 2 days ago
G B
7,5861328
7,5861328
add a comment |
add a comment |
up vote
2
down vote
Retina, 35 bytes
.+
*
_
$$.($.'*$($.>`$*)_¶
%~`^
.+¶
Try it online! Explanation:
.+
*
Convert the input to unary.
_
Match each position. This then sets several replacement variables. $`
becomes the left of the match; $>`
modifies this to be the left and match; $.>`
modifies this to take the length, i.e. the current index. $'
meanwhile is the right of the match, so $.'
is the length i.e. the current exponent.
$$.($.'*$($.>`$*)_¶
Create a string $.(
plus $.'
repetitions of $.>`*
plus _
. For an example, for an index of 2 in an original input of 5, $.'
is 3 and $.>`
is 2 so the resulting string is $.(2*2*2*_
. This conveniently is a Retina replacement expression that caluclates 2³. Each string is output on its own line.
%~`^
.+¶
For each line generated by the previous stage, prefix a line .+
to it, turning it into a replacement stage, and evaluate that stage, thereby calculating the expression.
add a comment |
up vote
2
down vote
Retina, 35 bytes
.+
*
_
$$.($.'*$($.>`$*)_¶
%~`^
.+¶
Try it online! Explanation:
.+
*
Convert the input to unary.
_
Match each position. This then sets several replacement variables. $`
becomes the left of the match; $>`
modifies this to be the left and match; $.>`
modifies this to take the length, i.e. the current index. $'
meanwhile is the right of the match, so $.'
is the length i.e. the current exponent.
$$.($.'*$($.>`$*)_¶
Create a string $.(
plus $.'
repetitions of $.>`*
plus _
. For an example, for an index of 2 in an original input of 5, $.'
is 3 and $.>`
is 2 so the resulting string is $.(2*2*2*_
. This conveniently is a Retina replacement expression that caluclates 2³. Each string is output on its own line.
%~`^
.+¶
For each line generated by the previous stage, prefix a line .+
to it, turning it into a replacement stage, and evaluate that stage, thereby calculating the expression.
add a comment |
up vote
2
down vote
up vote
2
down vote
Retina, 35 bytes
.+
*
_
$$.($.'*$($.>`$*)_¶
%~`^
.+¶
Try it online! Explanation:
.+
*
Convert the input to unary.
_
Match each position. This then sets several replacement variables. $`
becomes the left of the match; $>`
modifies this to be the left and match; $.>`
modifies this to take the length, i.e. the current index. $'
meanwhile is the right of the match, so $.'
is the length i.e. the current exponent.
$$.($.'*$($.>`$*)_¶
Create a string $.(
plus $.'
repetitions of $.>`*
plus _
. For an example, for an index of 2 in an original input of 5, $.'
is 3 and $.>`
is 2 so the resulting string is $.(2*2*2*_
. This conveniently is a Retina replacement expression that caluclates 2³. Each string is output on its own line.
%~`^
.+¶
For each line generated by the previous stage, prefix a line .+
to it, turning it into a replacement stage, and evaluate that stage, thereby calculating the expression.
Retina, 35 bytes
.+
*
_
$$.($.'*$($.>`$*)_¶
%~`^
.+¶
Try it online! Explanation:
.+
*
Convert the input to unary.
_
Match each position. This then sets several replacement variables. $`
becomes the left of the match; $>`
modifies this to be the left and match; $.>`
modifies this to take the length, i.e. the current index. $'
meanwhile is the right of the match, so $.'
is the length i.e. the current exponent.
$$.($.'*$($.>`$*)_¶
Create a string $.(
plus $.'
repetitions of $.>`*
plus _
. For an example, for an index of 2 in an original input of 5, $.'
is 3 and $.>`
is 2 so the resulting string is $.(2*2*2*_
. This conveniently is a Retina replacement expression that caluclates 2³. Each string is output on its own line.
%~`^
.+¶
For each line generated by the previous stage, prefix a line .+
to it, turning it into a replacement stage, and evaluate that stage, thereby calculating the expression.
answered 2 days ago
Neil
78.4k744175
78.4k744175
add a comment |
add a comment |
up vote
2
down vote
QBasic, 3533 bytes
Thank you @Neil for 2 bytes!
INPUT a
FOR b=1TO a
?b^(a-b)
NEXT
Slightly expanded version on REPL.IT because the interpreter in't entirely up-to-spec.
Output
QBasic (qb.js)
Copyright (c) 2010 Steve Hanov
5
1
8
9
4
1
Save 2 bytes by outputting the list in the correct order! (b^(a-b)
forb=1..a
)
– Neil
2 days ago
@Neil Thanks, I've worked it in!
– steenbergh
2 days ago
add a comment |
up vote
2
down vote
QBasic, 3533 bytes
Thank you @Neil for 2 bytes!
INPUT a
FOR b=1TO a
?b^(a-b)
NEXT
Slightly expanded version on REPL.IT because the interpreter in't entirely up-to-spec.
Output
QBasic (qb.js)
Copyright (c) 2010 Steve Hanov
5
1
8
9
4
1
Save 2 bytes by outputting the list in the correct order! (b^(a-b)
forb=1..a
)
– Neil
2 days ago
@Neil Thanks, I've worked it in!
– steenbergh
2 days ago
add a comment |
up vote
2
down vote
up vote
2
down vote
QBasic, 3533 bytes
Thank you @Neil for 2 bytes!
INPUT a
FOR b=1TO a
?b^(a-b)
NEXT
Slightly expanded version on REPL.IT because the interpreter in't entirely up-to-spec.
Output
QBasic (qb.js)
Copyright (c) 2010 Steve Hanov
5
1
8
9
4
1
QBasic, 3533 bytes
Thank you @Neil for 2 bytes!
INPUT a
FOR b=1TO a
?b^(a-b)
NEXT
Slightly expanded version on REPL.IT because the interpreter in't entirely up-to-spec.
Output
QBasic (qb.js)
Copyright (c) 2010 Steve Hanov
5
1
8
9
4
1
edited 2 days ago
answered 2 days ago
steenbergh
6,79411739
6,79411739
Save 2 bytes by outputting the list in the correct order! (b^(a-b)
forb=1..a
)
– Neil
2 days ago
@Neil Thanks, I've worked it in!
– steenbergh
2 days ago
add a comment |
Save 2 bytes by outputting the list in the correct order! (b^(a-b)
forb=1..a
)
– Neil
2 days ago
@Neil Thanks, I've worked it in!
– steenbergh
2 days ago
Save 2 bytes by outputting the list in the correct order! (
b^(a-b)
for b=1..a
)– Neil
2 days ago
Save 2 bytes by outputting the list in the correct order! (
b^(a-b)
for b=1..a
)– Neil
2 days ago
@Neil Thanks, I've worked it in!
– steenbergh
2 days ago
@Neil Thanks, I've worked it in!
– steenbergh
2 days ago
add a comment |
up vote
2
down vote
F# (.NET Core), 42 bytes
let f x=Seq.map(fun y->pown y (x-y))[1..x]
Try it online!
add a comment |
up vote
2
down vote
F# (.NET Core), 42 bytes
let f x=Seq.map(fun y->pown y (x-y))[1..x]
Try it online!
add a comment |
up vote
2
down vote
up vote
2
down vote
F# (.NET Core), 42 bytes
let f x=Seq.map(fun y->pown y (x-y))[1..x]
Try it online!
F# (.NET Core), 42 bytes
let f x=Seq.map(fun y->pown y (x-y))[1..x]
Try it online!
answered 2 days ago
dana
25114
25114
add a comment |
add a comment |
up vote
2
down vote
JavaScript (Node.js), 33 32 bytes
n=>(g=i=>--n?++i**n+[,g(i)]:1)``
Try it online!
-3 bytes with credits to @Shaggy, and -1 byte by @l4m2!
JavaScript (Node.js), 36 bytes
f=(n,i=1)=>n--?[i++**n,...f(n,i)]:
Try it online!
JavaScript (Node.js), 37 bytes
n=>[...Array(n)].map(x=>++i**--n,i=0)
Try it online!
33 bytes
– Shaggy
2 days ago
32
– l4m2
2 days ago
add a comment |
up vote
2
down vote
JavaScript (Node.js), 33 32 bytes
n=>(g=i=>--n?++i**n+[,g(i)]:1)``
Try it online!
-3 bytes with credits to @Shaggy, and -1 byte by @l4m2!
JavaScript (Node.js), 36 bytes
f=(n,i=1)=>n--?[i++**n,...f(n,i)]:
Try it online!
JavaScript (Node.js), 37 bytes
n=>[...Array(n)].map(x=>++i**--n,i=0)
Try it online!
33 bytes
– Shaggy
2 days ago
32
– l4m2
2 days ago
add a comment |
up vote
2
down vote
up vote
2
down vote
JavaScript (Node.js), 33 32 bytes
n=>(g=i=>--n?++i**n+[,g(i)]:1)``
Try it online!
-3 bytes with credits to @Shaggy, and -1 byte by @l4m2!
JavaScript (Node.js), 36 bytes
f=(n,i=1)=>n--?[i++**n,...f(n,i)]:
Try it online!
JavaScript (Node.js), 37 bytes
n=>[...Array(n)].map(x=>++i**--n,i=0)
Try it online!
JavaScript (Node.js), 33 32 bytes
n=>(g=i=>--n?++i**n+[,g(i)]:1)``
Try it online!
-3 bytes with credits to @Shaggy, and -1 byte by @l4m2!
JavaScript (Node.js), 36 bytes
f=(n,i=1)=>n--?[i++**n,...f(n,i)]:
Try it online!
JavaScript (Node.js), 37 bytes
n=>[...Array(n)].map(x=>++i**--n,i=0)
Try it online!
edited 2 days ago
answered 2 days ago
Shieru Asakoto
2,380314
2,380314
33 bytes
– Shaggy
2 days ago
32
– l4m2
2 days ago
add a comment |
33 bytes
– Shaggy
2 days ago
32
– l4m2
2 days ago
33 bytes
– Shaggy
2 days ago
33 bytes
– Shaggy
2 days ago
32
– l4m2
2 days ago
32
– l4m2
2 days ago
add a comment |
up vote
2
down vote
C# (Visual C# Interactive Compiler), 46 bytes
x=>new int[x].Select((_,i)=>Math.Pow(i+1,--x))
Try it online!
add a comment |
up vote
2
down vote
C# (Visual C# Interactive Compiler), 46 bytes
x=>new int[x].Select((_,i)=>Math.Pow(i+1,--x))
Try it online!
add a comment |
up vote
2
down vote
up vote
2
down vote
C# (Visual C# Interactive Compiler), 46 bytes
x=>new int[x].Select((_,i)=>Math.Pow(i+1,--x))
Try it online!
C# (Visual C# Interactive Compiler), 46 bytes
x=>new int[x].Select((_,i)=>Math.Pow(i+1,--x))
Try it online!
answered 2 days ago
dana
25114
25114
add a comment |
add a comment |
up vote
2
down vote
MATL, 5 bytes
:Gy-^
Try it online!
Explanation
Consider input 5
as an example.
: % Implicit input. Range
% STACK: [1 2 3 4 5]
G % Push input again
% STACK: [1 2 3 4 5], 5
y % Duplicate from below
% STACK: [1 2 3 4 5], 5, [1 2 3 4 5]
- % Subtract, element-wise
% STACK: [1 2 3 4 5], [4 3 2 1 0]
^ % Power, element-wise. Implicit display
% STACK: [1 8 9 4 1]
add a comment |
up vote
2
down vote
MATL, 5 bytes
:Gy-^
Try it online!
Explanation
Consider input 5
as an example.
: % Implicit input. Range
% STACK: [1 2 3 4 5]
G % Push input again
% STACK: [1 2 3 4 5], 5
y % Duplicate from below
% STACK: [1 2 3 4 5], 5, [1 2 3 4 5]
- % Subtract, element-wise
% STACK: [1 2 3 4 5], [4 3 2 1 0]
^ % Power, element-wise. Implicit display
% STACK: [1 8 9 4 1]
add a comment |
up vote
2
down vote
up vote
2
down vote
MATL, 5 bytes
:Gy-^
Try it online!
Explanation
Consider input 5
as an example.
: % Implicit input. Range
% STACK: [1 2 3 4 5]
G % Push input again
% STACK: [1 2 3 4 5], 5
y % Duplicate from below
% STACK: [1 2 3 4 5], 5, [1 2 3 4 5]
- % Subtract, element-wise
% STACK: [1 2 3 4 5], [4 3 2 1 0]
^ % Power, element-wise. Implicit display
% STACK: [1 8 9 4 1]
MATL, 5 bytes
:Gy-^
Try it online!
Explanation
Consider input 5
as an example.
: % Implicit input. Range
% STACK: [1 2 3 4 5]
G % Push input again
% STACK: [1 2 3 4 5], 5
y % Duplicate from below
% STACK: [1 2 3 4 5], 5, [1 2 3 4 5]
- % Subtract, element-wise
% STACK: [1 2 3 4 5], [4 3 2 1 0]
^ % Power, element-wise. Implicit display
% STACK: [1 8 9 4 1]
answered 2 days ago
Luis Mendo
73.8k885290
73.8k885290
add a comment |
add a comment |
up vote
2
down vote
Java, 59 Bytes
for(int i=1;a+1>i;i++)System.out.println(Math.pow(i,a-i));
Welcome to PPCG. It looks like this requires "input" be assigned to the predefined variablea
, which we don't allow.
– Shaggy
yesterday
Hello, here's a fix for you:n->{for(int i=0;i++<n;)System.out.println(Math.pow(i,n-i));}
60 bytes (code and test cases in the link)
– Olivier Grégoire
14 hours ago
add a comment |
up vote
2
down vote
Java, 59 Bytes
for(int i=1;a+1>i;i++)System.out.println(Math.pow(i,a-i));
Welcome to PPCG. It looks like this requires "input" be assigned to the predefined variablea
, which we don't allow.
– Shaggy
yesterday
Hello, here's a fix for you:n->{for(int i=0;i++<n;)System.out.println(Math.pow(i,n-i));}
60 bytes (code and test cases in the link)
– Olivier Grégoire
14 hours ago
add a comment |
up vote
2
down vote
up vote
2
down vote
Java, 59 Bytes
for(int i=1;a+1>i;i++)System.out.println(Math.pow(i,a-i));
Java, 59 Bytes
for(int i=1;a+1>i;i++)System.out.println(Math.pow(i,a-i));
edited yesterday
Shaggy
18.4k21663
18.4k21663
answered 2 days ago
isaace
1614
1614
Welcome to PPCG. It looks like this requires "input" be assigned to the predefined variablea
, which we don't allow.
– Shaggy
yesterday
Hello, here's a fix for you:n->{for(int i=0;i++<n;)System.out.println(Math.pow(i,n-i));}
60 bytes (code and test cases in the link)
– Olivier Grégoire
14 hours ago
add a comment |
Welcome to PPCG. It looks like this requires "input" be assigned to the predefined variablea
, which we don't allow.
– Shaggy
yesterday
Hello, here's a fix for you:n->{for(int i=0;i++<n;)System.out.println(Math.pow(i,n-i));}
60 bytes (code and test cases in the link)
– Olivier Grégoire
14 hours ago
Welcome to PPCG. It looks like this requires "input" be assigned to the predefined variable
a
, which we don't allow.– Shaggy
yesterday
Welcome to PPCG. It looks like this requires "input" be assigned to the predefined variable
a
, which we don't allow.– Shaggy
yesterday
Hello, here's a fix for you:
n->{for(int i=0;i++<n;)System.out.println(Math.pow(i,n-i));}
60 bytes (code and test cases in the link)– Olivier Grégoire
14 hours ago
Hello, here's a fix for you:
n->{for(int i=0;i++<n;)System.out.println(Math.pow(i,n-i));}
60 bytes (code and test cases in the link)– Olivier Grégoire
14 hours ago
add a comment |
up vote
2
down vote
Jelly, 4 bytes
*ạ¥€
Try it online!
add a comment |
up vote
2
down vote
Jelly, 4 bytes
*ạ¥€
Try it online!
add a comment |
up vote
2
down vote
up vote
2
down vote
Jelly, 4 bytes
*ạ¥€
Try it online!
Jelly, 4 bytes
*ạ¥€
Try it online!
answered yesterday
Erik the Outgolfer
30.8k429102
30.8k429102
add a comment |
add a comment |
up vote
1
down vote
Clean, 37 bytes
import StdEnv
$n=[i^(n-i)\i<-[1..n]]
Try it online!
Defines $ :: Int -> [Int]
taking an integer and returning the list of results.
$ n // function $ of n
= [i ^ (n-i) // i to the power of n minus i
\ i <- [1..n] // for each i in 1 to n
]
add a comment |
up vote
1
down vote
Clean, 37 bytes
import StdEnv
$n=[i^(n-i)\i<-[1..n]]
Try it online!
Defines $ :: Int -> [Int]
taking an integer and returning the list of results.
$ n // function $ of n
= [i ^ (n-i) // i to the power of n minus i
\ i <- [1..n] // for each i in 1 to n
]
add a comment |
up vote
1
down vote
up vote
1
down vote
Clean, 37 bytes
import StdEnv
$n=[i^(n-i)\i<-[1..n]]
Try it online!
Defines $ :: Int -> [Int]
taking an integer and returning the list of results.
$ n // function $ of n
= [i ^ (n-i) // i to the power of n minus i
\ i <- [1..n] // for each i in 1 to n
]
Clean, 37 bytes
import StdEnv
$n=[i^(n-i)\i<-[1..n]]
Try it online!
Defines $ :: Int -> [Int]
taking an integer and returning the list of results.
$ n // function $ of n
= [i ^ (n-i) // i to the power of n minus i
\ i <- [1..n] // for each i in 1 to n
]
edited 2 days ago
answered 2 days ago
Οurous
5,99311032
5,99311032
add a comment |
add a comment |
up vote
1
down vote
R, 34 bytes
x=1:scan();cat(x^rev(x-1),sep=',')
Try it online!
Is the default "sep" not a space? Would that not work?
– stuart stevenson
yesterday
1
@stuartstevenson "Output will be a list of numbers, delimited by either commas or new lines."
– Giuseppe
yesterday
add a comment |
up vote
1
down vote
R, 34 bytes
x=1:scan();cat(x^rev(x-1),sep=',')
Try it online!
Is the default "sep" not a space? Would that not work?
– stuart stevenson
yesterday
1
@stuartstevenson "Output will be a list of numbers, delimited by either commas or new lines."
– Giuseppe
yesterday
add a comment |
up vote
1
down vote
up vote
1
down vote
R, 34 bytes
x=1:scan();cat(x^rev(x-1),sep=',')
Try it online!
R, 34 bytes
x=1:scan();cat(x^rev(x-1),sep=',')
Try it online!
answered 2 days ago
Giuseppe
16.1k31052
16.1k31052
Is the default "sep" not a space? Would that not work?
– stuart stevenson
yesterday
1
@stuartstevenson "Output will be a list of numbers, delimited by either commas or new lines."
– Giuseppe
yesterday
add a comment |
Is the default "sep" not a space? Would that not work?
– stuart stevenson
yesterday
1
@stuartstevenson "Output will be a list of numbers, delimited by either commas or new lines."
– Giuseppe
yesterday
Is the default "sep" not a space? Would that not work?
– stuart stevenson
yesterday
Is the default "sep" not a space? Would that not work?
– stuart stevenson
yesterday
1
1
@stuartstevenson "Output will be a list of numbers, delimited by either commas or new lines."
– Giuseppe
yesterday
@stuartstevenson "Output will be a list of numbers, delimited by either commas or new lines."
– Giuseppe
yesterday
add a comment |
up vote
1
down vote
05AB1E, 5 bytes
LD<Rm
Port of @lirtosiast's Jelly answer.
Try it online.
Explanation:
L # List in the range [1, (implicit) input integer]
# i.e. 5 → [1,2,3,4,5]
D< # Duplicate this list, and subtract 1 to make the range [0, input)
# i.e. [1,2,3,4,5] → [0,1,2,3,4]
R # Reverse it to make the range (input, 0]
# i.e. [0,1,2,3,4] → [4,3,2,1,0]
m # Take the power of the numbers in the lists (at the same indices)
# (and output implicitly)
# i.e. [1,2,3,4,5] and [4,3,2,1,0] → [1,8,9,4,1]
add a comment |
up vote
1
down vote
05AB1E, 5 bytes
LD<Rm
Port of @lirtosiast's Jelly answer.
Try it online.
Explanation:
L # List in the range [1, (implicit) input integer]
# i.e. 5 → [1,2,3,4,5]
D< # Duplicate this list, and subtract 1 to make the range [0, input)
# i.e. [1,2,3,4,5] → [0,1,2,3,4]
R # Reverse it to make the range (input, 0]
# i.e. [0,1,2,3,4] → [4,3,2,1,0]
m # Take the power of the numbers in the lists (at the same indices)
# (and output implicitly)
# i.e. [1,2,3,4,5] and [4,3,2,1,0] → [1,8,9,4,1]
add a comment |
up vote
1
down vote
up vote
1
down vote
05AB1E, 5 bytes
LD<Rm
Port of @lirtosiast's Jelly answer.
Try it online.
Explanation:
L # List in the range [1, (implicit) input integer]
# i.e. 5 → [1,2,3,4,5]
D< # Duplicate this list, and subtract 1 to make the range [0, input)
# i.e. [1,2,3,4,5] → [0,1,2,3,4]
R # Reverse it to make the range (input, 0]
# i.e. [0,1,2,3,4] → [4,3,2,1,0]
m # Take the power of the numbers in the lists (at the same indices)
# (and output implicitly)
# i.e. [1,2,3,4,5] and [4,3,2,1,0] → [1,8,9,4,1]
05AB1E, 5 bytes
LD<Rm
Port of @lirtosiast's Jelly answer.
Try it online.
Explanation:
L # List in the range [1, (implicit) input integer]
# i.e. 5 → [1,2,3,4,5]
D< # Duplicate this list, and subtract 1 to make the range [0, input)
# i.e. [1,2,3,4,5] → [0,1,2,3,4]
R # Reverse it to make the range (input, 0]
# i.e. [0,1,2,3,4] → [4,3,2,1,0]
m # Take the power of the numbers in the lists (at the same indices)
# (and output implicitly)
# i.e. [1,2,3,4,5] and [4,3,2,1,0] → [1,8,9,4,1]
answered 2 days ago
Kevin Cruijssen
34.6k554182
34.6k554182
add a comment |
add a comment |
up vote
1
down vote
Lua, 43 41 bytes
-2 bytes thanks to @Shaggy
s=io.read()for i=1,s do print(i^(s-i))end
Try it online!
1
I don't think you need the+0
; seems to work without it.
– Shaggy
2 days ago
add a comment |
up vote
1
down vote
Lua, 43 41 bytes
-2 bytes thanks to @Shaggy
s=io.read()for i=1,s do print(i^(s-i))end
Try it online!
1
I don't think you need the+0
; seems to work without it.
– Shaggy
2 days ago
add a comment |
up vote
1
down vote
up vote
1
down vote
Lua, 43 41 bytes
-2 bytes thanks to @Shaggy
s=io.read()for i=1,s do print(i^(s-i))end
Try it online!
Lua, 43 41 bytes
-2 bytes thanks to @Shaggy
s=io.read()for i=1,s do print(i^(s-i))end
Try it online!
edited 2 days ago
answered 2 days ago
ouflak
193311
193311
1
I don't think you need the+0
; seems to work without it.
– Shaggy
2 days ago
add a comment |
1
I don't think you need the+0
; seems to work without it.
– Shaggy
2 days ago
1
1
I don't think you need the
+0
; seems to work without it.– Shaggy
2 days ago
I don't think you need the
+0
; seems to work without it.– Shaggy
2 days ago
add a comment |
up vote
1
down vote
R, 22 bytes
n=scan();(1:n)^(n:1-1)
Fairly self-explanatory; note that the :
operator is higher precendence than the -
operator so that n:1-1
is shorter than (n-1):0
If we are allowed to start at 0, then we can lose two bytes by using (0:n)^(n:0)
avoiding the need for a -1.
add a comment |
up vote
1
down vote
R, 22 bytes
n=scan();(1:n)^(n:1-1)
Fairly self-explanatory; note that the :
operator is higher precendence than the -
operator so that n:1-1
is shorter than (n-1):0
If we are allowed to start at 0, then we can lose two bytes by using (0:n)^(n:0)
avoiding the need for a -1.
add a comment |
up vote
1
down vote
up vote
1
down vote
R, 22 bytes
n=scan();(1:n)^(n:1-1)
Fairly self-explanatory; note that the :
operator is higher precendence than the -
operator so that n:1-1
is shorter than (n-1):0
If we are allowed to start at 0, then we can lose two bytes by using (0:n)^(n:0)
avoiding the need for a -1.
R, 22 bytes
n=scan();(1:n)^(n:1-1)
Fairly self-explanatory; note that the :
operator is higher precendence than the -
operator so that n:1-1
is shorter than (n-1):0
If we are allowed to start at 0, then we can lose two bytes by using (0:n)^(n:0)
avoiding the need for a -1.
answered 2 days ago
JDL
1,275410
1,275410
add a comment |
add a comment |
up vote
1
down vote
Charcoal, 9 bytes
I⮌ENX⁻θιι
Try it online! Link is to verbose version of code. Explanation:
N Input as a number
E Map over implicit range
ι Current value
⁻ Subtracted from
θ First input
X Raised to power
ι Current value
⮌ Reverse list
I Cast to string
Implicitly print on separate lines
add a comment |
up vote
1
down vote
Charcoal, 9 bytes
I⮌ENX⁻θιι
Try it online! Link is to verbose version of code. Explanation:
N Input as a number
E Map over implicit range
ι Current value
⁻ Subtracted from
θ First input
X Raised to power
ι Current value
⮌ Reverse list
I Cast to string
Implicitly print on separate lines
add a comment |
up vote
1
down vote
up vote
1
down vote
Charcoal, 9 bytes
I⮌ENX⁻θιι
Try it online! Link is to verbose version of code. Explanation:
N Input as a number
E Map over implicit range
ι Current value
⁻ Subtracted from
θ First input
X Raised to power
ι Current value
⮌ Reverse list
I Cast to string
Implicitly print on separate lines
Charcoal, 9 bytes
I⮌ENX⁻θιι
Try it online! Link is to verbose version of code. Explanation:
N Input as a number
E Map over implicit range
ι Current value
⁻ Subtracted from
θ First input
X Raised to power
ι Current value
⮌ Reverse list
I Cast to string
Implicitly print on separate lines
answered 2 days ago
Neil
78.4k744175
78.4k744175
add a comment |
add a comment |
up vote
1
down vote
C# (Visual C# Interactive Compiler), 55 bytes
v=>Enumerable.Range(0,v--).Select(i=>Math.Pow(i+1,v--))
Try it online!
add a comment |
up vote
1
down vote
C# (Visual C# Interactive Compiler), 55 bytes
v=>Enumerable.Range(0,v--).Select(i=>Math.Pow(i+1,v--))
Try it online!
add a comment |
up vote
1
down vote
up vote
1
down vote
C# (Visual C# Interactive Compiler), 55 bytes
v=>Enumerable.Range(0,v--).Select(i=>Math.Pow(i+1,v--))
Try it online!
C# (Visual C# Interactive Compiler), 55 bytes
v=>Enumerable.Range(0,v--).Select(i=>Math.Pow(i+1,v--))
Try it online!
answered 2 days ago
auhmaan
86637
86637
add a comment |
add a comment |
up vote
1
down vote
Perl 5 -n
, 21 bytes
say++$**--$_ while$_
Try it online!
add a comment |
up vote
1
down vote
Perl 5 -n
, 21 bytes
say++$**--$_ while$_
Try it online!
add a comment |
up vote
1
down vote
up vote
1
down vote
Perl 5 -n
, 21 bytes
say++$**--$_ while$_
Try it online!
Perl 5 -n
, 21 bytes
say++$**--$_ while$_
Try it online!
answered 2 days ago
Xcali
5,030520
5,030520
add a comment |
add a comment |
1 2
next
Embodiment of Ignorance is a new contributor. Be nice, and check out our Code of Conduct.
Embodiment of Ignorance is a new contributor. Be nice, and check out our Code of Conduct.
Embodiment of Ignorance is a new contributor. Be nice, and check out our Code of Conduct.
Embodiment of Ignorance is a new contributor. Be nice, and check out our Code of Conduct.
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5
the comma/newline detail should be omitted, it is normal practice around here to let output of lists be in any convenient format, including as a list/array object being returned by a function
– Sparr
2 days ago
3
Is the input always greater than 0 or do we have to deal with 0 and negatives?
– Veskah
2 days ago
Inputs will always be positive
– Embodiment of Ignorance
2 days ago
6
Two equally short answers doesn't matter. If you feel like accepting an answer, choose the earliest posted one. However I strongly recommend waiting at least a few days, and would suggest never accepting an answer (to encourage more submissions).
– Οurous
2 days ago
1
Shouldn't the title be "Given an integer, print all the powers obtained with a base and an exponent that sum to the input"?
– Nicola Sap
2 days ago