Given an input, print all exponents where the base and power sum to the input
So this is my first challenge on this site.
The challenge is to take in an input integer $n$, which will be positive, and print, in ascending order ($1$ to $n$, including n), the output of $i^{(n-i)}$ (where $i$ is the current integer).
Example
Given the 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 a positive 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
|
show 5 more comments
So this is my first challenge on this site.
The challenge is to take in an input integer $n$, which will be positive, and print, in ascending order ($1$ to $n$, including n), the output of $i^{(n-i)}$ (where $i$ is the current integer).
Example
Given the 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 a positive 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
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
Nov 28 '18 at 4:20
3
Is the input always greater than 0 or do we have to deal with 0 and negatives?
– Veskah
Nov 28 '18 at 4:21
Inputs will always be positive
– Embodiment of Ignorance
Nov 28 '18 at 5:01
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
Nov 28 '18 at 5:21
2
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
Nov 28 '18 at 14:12
|
show 5 more comments
So this is my first challenge on this site.
The challenge is to take in an input integer $n$, which will be positive, and print, in ascending order ($1$ to $n$, including n), the output of $i^{(n-i)}$ (where $i$ is the current integer).
Example
Given the 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 a positive 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
So this is my first challenge on this site.
The challenge is to take in an input integer $n$, which will be positive, and print, in ascending order ($1$ to $n$, including n), the output of $i^{(n-i)}$ (where $i$ is the current integer).
Example
Given the 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 a positive 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
edited Jan 1 at 22:26
Embodiment of Ignorance
asked Nov 28 '18 at 4:00
Embodiment of IgnoranceEmbodiment of Ignorance
601115
601115
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
Nov 28 '18 at 4:20
3
Is the input always greater than 0 or do we have to deal with 0 and negatives?
– Veskah
Nov 28 '18 at 4:21
Inputs will always be positive
– Embodiment of Ignorance
Nov 28 '18 at 5:01
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
Nov 28 '18 at 5:21
2
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
Nov 28 '18 at 14:12
|
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
Nov 28 '18 at 4:20
3
Is the input always greater than 0 or do we have to deal with 0 and negatives?
– Veskah
Nov 28 '18 at 4:21
Inputs will always be positive
– Embodiment of Ignorance
Nov 28 '18 at 5:01
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
Nov 28 '18 at 5:21
2
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
Nov 28 '18 at 14:12
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
Nov 28 '18 at 4:20
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
Nov 28 '18 at 4:20
3
3
Is the input always greater than 0 or do we have to deal with 0 and negatives?
– Veskah
Nov 28 '18 at 4:21
Is the input always greater than 0 or do we have to deal with 0 and negatives?
– Veskah
Nov 28 '18 at 4:21
Inputs will always be positive
– Embodiment of Ignorance
Nov 28 '18 at 5:01
Inputs will always be positive
– Embodiment of Ignorance
Nov 28 '18 at 5:01
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
Nov 28 '18 at 5:21
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
Nov 28 '18 at 5:21
2
2
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
Nov 28 '18 at 14:12
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
Nov 28 '18 at 14:12
|
show 5 more comments
38 Answers
38
active
oldest
votes
1 2
next
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
Nov 28 '18 at 18:43
@lirtosiast took me a while to figure out why does that work, but I got it. Thanks.
– J. Sallé
Nov 29 '18 at 12:20
add a comment |
Haskell, 23 bytes
f i=[x^(i-x)|x<-[1..i]]
Try it online!
Alternative version, also 23 bytes:
f i=(^)<*>(i-)<$>[1..i]
add a comment |
Japt, 5 bytes
õ_p´U
Try it
õ :Range [1,input]
_ :Map
p : Raise to the power of
´U : Input decremented
add a comment |
Perl 6, 19 bytes
{^$_+1 Z**[R,] ^$_}
Try it online!
Anonymous code block that takes a number and returns a list. Zip exponents the range 1 to input
and the range input-1 to 0
add a comment |
Aheui (esotope), 193 164 bytes (56 chars)
방빠싹받분샥퍼붇바파쟈뿌차샦히망맣여
타빠바푸투반또분뽀뿌서썪삯타삯받반타
석차샦져쌲볼어타토싻삭빠쏛ㅇ또섞썪뻐
Try it online!
Try it on AVIS(Korean); just copy and paste code above, press start button, input a number, see how it moves. To see output, press the >_ icon on left side.
It's not golfed much, but I give it a shot.
Is it possible to chose a character set, so that each character is stored in 2 bytes?
– tsh
Nov 29 '18 at 2:52
@tsh According to Aheui specification, an Aheui code consists of only UTF-8 characters.
– cobaltp
Nov 29 '18 at 5:07
add a comment |
Jelly, 5 bytes
R*ḶU$
Try it online!
R [1,...,n]
* to the power of
ḶU$ [0,...,n-1] reversed
add a comment |
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 |
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 |
PHP, 32 bytes
while($argn)echo++$i**--$argn,_;
Run as pipe with -nR
or try it online.
add a comment |
Octave, 18 bytes
@(n)(t=1:n).^(n-t)
Try it online!
Thanks Luis Mendo, using internal variable saves 3 bytes.
add a comment |
Jelly, 4 bytes
*ạ¥€
Try it online!
add a comment |
Wolfram Language (Mathematica), 24 20 18 bytes
(x=Range@#)^(#-x)&
Try it online!
-4 thanks @lirtosiast.
add a comment |
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
Nov 28 '18 at 7:33
add a comment |
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 |
Ruby, 27 bytes
->n{(1..n).map{|r|r**n-=1}}
Try it online!
add a comment |
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 |
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
Nov 28 '18 at 12:14
@Neil Thanks, I've worked it in!
– steenbergh
Nov 28 '18 at 13:26
add a comment |
F# (.NET Core), 42 bytes
let f x=Seq.map(fun y->pown y (x-y))[1..x]
Try it online!
add a comment |
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
Nov 28 '18 at 8:03
32
– l4m2
Nov 28 '18 at 14:25
add a comment |
C# (Visual C# Interactive Compiler), 46 bytes
x=>new int[x].Select((_,i)=>Math.Pow(i+1,--x))
Try it online!
add a comment |
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 |
Java, 59 Bytes
for(int i=1;a+1>i;i++)System.out.println(Math.pow(i,a-i));
1
Welcome to PPCG. It looks like this requires "input" be assigned to the predefined variablea
, which we don't allow.
– Shaggy
Nov 29 '18 at 21:01
2
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
Nov 30 '18 at 9:20
add a comment |
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 |
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
Nov 29 '18 at 15:51
1
@stuartstevenson "Output will be a list of numbers, delimited by either commas or new lines."
– Giuseppe
Nov 29 '18 at 15:53
add a comment |
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 |
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
Nov 28 '18 at 10:06
add a comment |
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 |
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 |
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 |
Perl 5 -n
, 21 bytes
say++$**--$_ while$_
Try it online!
add a comment |
1 2
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38 Answers
38
active
oldest
votes
38 Answers
38
active
oldest
votes
active
oldest
votes
active
oldest
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1 2
next
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
Nov 28 '18 at 18:43
@lirtosiast took me a while to figure out why does that work, but I got it. Thanks.
– J. Sallé
Nov 29 '18 at 12:20
add a comment |
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
Nov 28 '18 at 18:43
@lirtosiast took me a while to figure out why does that work, but I got it. Thanks.
– J. Sallé
Nov 29 '18 at 12:20
add a comment |
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 Nov 29 '18 at 12:26
answered Nov 28 '18 at 14:16
J. SalléJ. Sallé
1,903322
1,903322
2
⍳*⊢-⍳
is 5 bytes, using⎕IO←1
.
– lirtosiast
Nov 28 '18 at 18:43
@lirtosiast took me a while to figure out why does that work, but I got it. Thanks.
– J. Sallé
Nov 29 '18 at 12:20
add a comment |
2
⍳*⊢-⍳
is 5 bytes, using⎕IO←1
.
– lirtosiast
Nov 28 '18 at 18:43
@lirtosiast took me a while to figure out why does that work, but I got it. Thanks.
– J. Sallé
Nov 29 '18 at 12:20
2
2
⍳*⊢-⍳
is 5 bytes, using ⎕IO←1
.– lirtosiast
Nov 28 '18 at 18:43
⍳*⊢-⍳
is 5 bytes, using ⎕IO←1
.– lirtosiast
Nov 28 '18 at 18:43
@lirtosiast took me a while to figure out why does that work, but I got it. Thanks.
– J. Sallé
Nov 29 '18 at 12:20
@lirtosiast took me a while to figure out why does that work, but I got it. Thanks.
– J. Sallé
Nov 29 '18 at 12:20
add a comment |
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 |
Haskell, 23 bytes
f i=[x^(i-x)|x<-[1..i]]
Try it online!
Alternative version, also 23 bytes:
f i=(^)<*>(i-)<$>[1..i]
add a comment |
Haskell, 23 bytes
f i=[x^(i-x)|x<-[1..i]]
Try it online!
Alternative version, also 23 bytes:
f i=(^)<*>(i-)<$>[1..i]
Haskell, 23 bytes
f i=[x^(i-x)|x<-[1..i]]
Try it online!
Alternative version, also 23 bytes:
f i=(^)<*>(i-)<$>[1..i]
answered Nov 28 '18 at 5:00
niminimi
31.5k32185
31.5k32185
add a comment |
add a comment |
Japt, 5 bytes
õ_p´U
Try it
õ :Range [1,input]
_ :Map
p : Raise to the power of
´U : Input decremented
add a comment |
Japt, 5 bytes
õ_p´U
Try it
õ :Range [1,input]
_ :Map
p : Raise to the power of
´U : Input decremented
add a comment |
Japt, 5 bytes
õ_p´U
Try it
õ :Range [1,input]
_ :Map
p : Raise to the power of
´U : Input decremented
Japt, 5 bytes
õ_p´U
Try it
õ :Range [1,input]
_ :Map
p : Raise to the power of
´U : Input decremented
edited Nov 28 '18 at 10:02
answered Nov 28 '18 at 7:59
ShaggyShaggy
19.2k21666
19.2k21666
add a comment |
add a comment |
Perl 6, 19 bytes
{^$_+1 Z**[R,] ^$_}
Try it online!
Anonymous code block that takes a number and returns a list. Zip exponents the range 1 to input
and the range input-1 to 0
add a comment |
Perl 6, 19 bytes
{^$_+1 Z**[R,] ^$_}
Try it online!
Anonymous code block that takes a number and returns a list. Zip exponents the range 1 to input
and the range input-1 to 0
add a comment |
Perl 6, 19 bytes
{^$_+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 Nov 28 '18 at 11:10
answered Nov 28 '18 at 4:34
Jo KingJo King
21.2k248110
21.2k248110
add a comment |
add a comment |
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
Nov 29 '18 at 2:52
@tsh According to Aheui specification, an Aheui code consists of only UTF-8 characters.
– cobaltp
Nov 29 '18 at 5:07
add a comment |
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
Nov 29 '18 at 2:52
@tsh According to Aheui specification, an Aheui code consists of only UTF-8 characters.
– cobaltp
Nov 29 '18 at 5:07
add a comment |
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 Nov 29 '18 at 6:22
answered Nov 28 '18 at 6:50
cobaltpcobaltp
35618
35618
Is it possible to chose a character set, so that each character is stored in 2 bytes?
– tsh
Nov 29 '18 at 2:52
@tsh According to Aheui specification, an Aheui code consists of only UTF-8 characters.
– cobaltp
Nov 29 '18 at 5:07
add a comment |
Is it possible to chose a character set, so that each character is stored in 2 bytes?
– tsh
Nov 29 '18 at 2:52
@tsh According to Aheui specification, an Aheui code consists of only UTF-8 characters.
– cobaltp
Nov 29 '18 at 5:07
Is it possible to chose a character set, so that each character is stored in 2 bytes?
– tsh
Nov 29 '18 at 2:52
Is it possible to chose a character set, so that each character is stored in 2 bytes?
– tsh
Nov 29 '18 at 2:52
@tsh According to Aheui specification, an Aheui code consists of only UTF-8 characters.
– cobaltp
Nov 29 '18 at 5:07
@tsh According to Aheui specification, an Aheui code consists of only UTF-8 characters.
– cobaltp
Nov 29 '18 at 5:07
add a comment |
Jelly, 5 bytes
R*ḶU$
Try it online!
R [1,...,n]
* to the power of
ḶU$ [0,...,n-1] reversed
add a comment |
Jelly, 5 bytes
R*ḶU$
Try it online!
R [1,...,n]
* to the power of
ḶU$ [0,...,n-1] reversed
add a comment |
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 Nov 28 '18 at 4:22
answered Nov 28 '18 at 4:16
lirtosiastlirtosiast
15.8k436107
15.8k436107
add a comment |
add a comment |
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 |
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 |
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 Nov 28 '18 at 4:48
answered Nov 28 '18 at 4:43
lirtosiastlirtosiast
15.8k436107
15.8k436107
add a comment |
add a comment |
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 |
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 |
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 Nov 28 '18 at 9:24
answered Nov 28 '18 at 9:11
Galen IvanovGalen Ivanov
6,44711032
6,44711032
add a comment |
add a comment |
PHP, 32 bytes
while($argn)echo++$i**--$argn,_;
Run as pipe with -nR
or try it online.
add a comment |
PHP, 32 bytes
while($argn)echo++$i**--$argn,_;
Run as pipe with -nR
or try it online.
add a comment |
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 Nov 28 '18 at 11:43
TitusTitus
13k11238
13k11238
add a comment |
add a comment |
Octave, 18 bytes
@(n)(t=1:n).^(n-t)
Try it online!
Thanks Luis Mendo, using internal variable saves 3 bytes.
add a comment |
Octave, 18 bytes
@(n)(t=1:n).^(n-t)
Try it online!
Thanks Luis Mendo, using internal variable saves 3 bytes.
add a comment |
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 Nov 29 '18 at 2:37
answered Nov 28 '18 at 5:32
tshtsh
8,49511547
8,49511547
add a comment |
add a comment |
Jelly, 4 bytes
*ạ¥€
Try it online!
add a comment |
Jelly, 4 bytes
*ạ¥€
Try it online!
add a comment |
Jelly, 4 bytes
*ạ¥€
Try it online!
Jelly, 4 bytes
*ạ¥€
Try it online!
answered Nov 29 '18 at 22:22
Erik the OutgolferErik the Outgolfer
31.4k429103
31.4k429103
add a comment |
add a comment |
Wolfram Language (Mathematica), 24 20 18 bytes
(x=Range@#)^(#-x)&
Try it online!
-4 thanks @lirtosiast.
add a comment |
Wolfram Language (Mathematica), 24 20 18 bytes
(x=Range@#)^(#-x)&
Try it online!
-4 thanks @lirtosiast.
add a comment |
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 Nov 28 '18 at 4:34
answered Nov 28 '18 at 4:31
Shieru AsakotoShieru Asakoto
2,380315
2,380315
add a comment |
add a comment |
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
Nov 28 '18 at 7:33
add a comment |
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
Nov 28 '18 at 7:33
add a comment |
MathGolf, 6 bytes
rx╒m#
Try it online!
MathGolf, 6 bytes
rx╒m#
Try it online!
answered Nov 28 '18 at 5:00
Jo KingJo King
21.2k248110
21.2k248110
I have implemented reverse subtraction, multiplication and division, but it looks like a reverse power operator could come in handy?
– maxb
Nov 28 '18 at 7:33
add a comment |
I have implemented reverse subtraction, multiplication and division, but it looks like a reverse power operator could come in handy?
– maxb
Nov 28 '18 at 7:33
I have implemented reverse subtraction, multiplication and division, but it looks like a reverse power operator could come in handy?
– maxb
Nov 28 '18 at 7:33
I have implemented reverse subtraction, multiplication and division, but it looks like a reverse power operator could come in handy?
– maxb
Nov 28 '18 at 7:33
add a comment |
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 |
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 |
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 Nov 28 '18 at 5:58
Vedant KandoiVedant Kandoi
1,073226
1,073226
add a comment |
add a comment |
Ruby, 27 bytes
->n{(1..n).map{|r|r**n-=1}}
Try it online!
add a comment |
Ruby, 27 bytes
->n{(1..n).map{|r|r**n-=1}}
Try it online!
add a comment |
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 Nov 28 '18 at 7:06
G BG B
7,6961328
7,6961328
add a comment |
add a comment |
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 |
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 |
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 Nov 28 '18 at 12:06
NeilNeil
79.6k744177
79.6k744177
add a comment |
add a comment |
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
Nov 28 '18 at 12:14
@Neil Thanks, I've worked it in!
– steenbergh
Nov 28 '18 at 13:26
add a comment |
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
Nov 28 '18 at 12:14
@Neil Thanks, I've worked it in!
– steenbergh
Nov 28 '18 at 13:26
add a comment |
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 Nov 28 '18 at 13:26
answered Nov 28 '18 at 8:49
steenberghsteenbergh
6,81411739
6,81411739
Save 2 bytes by outputting the list in the correct order! (b^(a-b)
forb=1..a
)
– Neil
Nov 28 '18 at 12:14
@Neil Thanks, I've worked it in!
– steenbergh
Nov 28 '18 at 13:26
add a comment |
Save 2 bytes by outputting the list in the correct order! (b^(a-b)
forb=1..a
)
– Neil
Nov 28 '18 at 12:14
@Neil Thanks, I've worked it in!
– steenbergh
Nov 28 '18 at 13:26
Save 2 bytes by outputting the list in the correct order! (
b^(a-b)
for b=1..a
)– Neil
Nov 28 '18 at 12:14
Save 2 bytes by outputting the list in the correct order! (
b^(a-b)
for b=1..a
)– Neil
Nov 28 '18 at 12:14
@Neil Thanks, I've worked it in!
– steenbergh
Nov 28 '18 at 13:26
@Neil Thanks, I've worked it in!
– steenbergh
Nov 28 '18 at 13:26
add a comment |
F# (.NET Core), 42 bytes
let f x=Seq.map(fun y->pown y (x-y))[1..x]
Try it online!
add a comment |
F# (.NET Core), 42 bytes
let f x=Seq.map(fun y->pown y (x-y))[1..x]
Try it online!
add a comment |
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 Nov 28 '18 at 15:18
danadana
58135
58135
add a comment |
add a comment |
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
Nov 28 '18 at 8:03
32
– l4m2
Nov 28 '18 at 14:25
add a comment |
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
Nov 28 '18 at 8:03
32
– l4m2
Nov 28 '18 at 14:25
add a comment |
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 Nov 28 '18 at 15:53
answered Nov 28 '18 at 4:15
Shieru AsakotoShieru Asakoto
2,380315
2,380315
33 bytes
– Shaggy
Nov 28 '18 at 8:03
32
– l4m2
Nov 28 '18 at 14:25
add a comment |
33 bytes
– Shaggy
Nov 28 '18 at 8:03
32
– l4m2
Nov 28 '18 at 14:25
33 bytes
– Shaggy
Nov 28 '18 at 8:03
33 bytes
– Shaggy
Nov 28 '18 at 8:03
32
– l4m2
Nov 28 '18 at 14:25
32
– l4m2
Nov 28 '18 at 14:25
add a comment |
C# (Visual C# Interactive Compiler), 46 bytes
x=>new int[x].Select((_,i)=>Math.Pow(i+1,--x))
Try it online!
add a comment |
C# (Visual C# Interactive Compiler), 46 bytes
x=>new int[x].Select((_,i)=>Math.Pow(i+1,--x))
Try it online!
add a comment |
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 Nov 28 '18 at 16:03
danadana
58135
58135
add a comment |
add a comment |
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 |
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 |
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 Nov 28 '18 at 18:49
Luis MendoLuis Mendo
74k886291
74k886291
add a comment |
add a comment |
Java, 59 Bytes
for(int i=1;a+1>i;i++)System.out.println(Math.pow(i,a-i));
1
Welcome to PPCG. It looks like this requires "input" be assigned to the predefined variablea
, which we don't allow.
– Shaggy
Nov 29 '18 at 21:01
2
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
Nov 30 '18 at 9:20
add a comment |
Java, 59 Bytes
for(int i=1;a+1>i;i++)System.out.println(Math.pow(i,a-i));
1
Welcome to PPCG. It looks like this requires "input" be assigned to the predefined variablea
, which we don't allow.
– Shaggy
Nov 29 '18 at 21:01
2
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
Nov 30 '18 at 9:20
add a comment |
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 Nov 29 '18 at 21:00
Shaggy
19.2k21666
19.2k21666
answered Nov 28 '18 at 15:57
isaaceisaace
1914
1914
1
Welcome to PPCG. It looks like this requires "input" be assigned to the predefined variablea
, which we don't allow.
– Shaggy
Nov 29 '18 at 21:01
2
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
Nov 30 '18 at 9:20
add a comment |
1
Welcome to PPCG. It looks like this requires "input" be assigned to the predefined variablea
, which we don't allow.
– Shaggy
Nov 29 '18 at 21:01
2
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
Nov 30 '18 at 9:20
1
1
Welcome to PPCG. It looks like this requires "input" be assigned to the predefined variable
a
, which we don't allow.– Shaggy
Nov 29 '18 at 21:01
Welcome to PPCG. It looks like this requires "input" be assigned to the predefined variable
a
, which we don't allow.– Shaggy
Nov 29 '18 at 21:01
2
2
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
Nov 30 '18 at 9:20
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
Nov 30 '18 at 9:20
add a comment |
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 |
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 |
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 Nov 28 '18 at 4:26
answered Nov 28 '18 at 4:12
ΟurousΟurous
6,54211033
6,54211033
add a comment |
add a comment |
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
Nov 29 '18 at 15:51
1
@stuartstevenson "Output will be a list of numbers, delimited by either commas or new lines."
– Giuseppe
Nov 29 '18 at 15:53
add a comment |
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
Nov 29 '18 at 15:51
1
@stuartstevenson "Output will be a list of numbers, delimited by either commas or new lines."
– Giuseppe
Nov 29 '18 at 15:53
add a comment |
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 Nov 28 '18 at 4:42
GiuseppeGiuseppe
16.6k31052
16.6k31052
Is the default "sep" not a space? Would that not work?
– stuart stevenson
Nov 29 '18 at 15:51
1
@stuartstevenson "Output will be a list of numbers, delimited by either commas or new lines."
– Giuseppe
Nov 29 '18 at 15:53
add a comment |
Is the default "sep" not a space? Would that not work?
– stuart stevenson
Nov 29 '18 at 15:51
1
@stuartstevenson "Output will be a list of numbers, delimited by either commas or new lines."
– Giuseppe
Nov 29 '18 at 15:53
Is the default "sep" not a space? Would that not work?
– stuart stevenson
Nov 29 '18 at 15:51
Is the default "sep" not a space? Would that not work?
– stuart stevenson
Nov 29 '18 at 15:51
1
1
@stuartstevenson "Output will be a list of numbers, delimited by either commas or new lines."
– Giuseppe
Nov 29 '18 at 15:53
@stuartstevenson "Output will be a list of numbers, delimited by either commas or new lines."
– Giuseppe
Nov 29 '18 at 15:53
add a comment |
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 |
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 |
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 Nov 28 '18 at 10:42
Kevin CruijssenKevin Cruijssen
36.1k554190
36.1k554190
add a comment |
add a comment |
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
Nov 28 '18 at 10:06
add a comment |
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
Nov 28 '18 at 10:06
add a comment |
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 Nov 28 '18 at 11:01
answered Nov 28 '18 at 10:01
ouflakouflak
1931311
1931311
1
I don't think you need the+0
; seems to work without it.
– Shaggy
Nov 28 '18 at 10:06
add a comment |
1
I don't think you need the+0
; seems to work without it.
– Shaggy
Nov 28 '18 at 10:06
1
1
I don't think you need the
+0
; seems to work without it.– Shaggy
Nov 28 '18 at 10:06
I don't think you need the
+0
; seems to work without it.– Shaggy
Nov 28 '18 at 10:06
add a comment |
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 |
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 |
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 Nov 28 '18 at 11:02
JDLJDL
1,285410
1,285410
add a comment |
add a comment |
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 |
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 |
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 Nov 28 '18 at 11:49
NeilNeil
79.6k744177
79.6k744177
add a comment |
add a comment |
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 |
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 |
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 Nov 28 '18 at 11:59
auhmaanauhmaan
86637
86637
add a comment |
add a comment |
Perl 5 -n
, 21 bytes
say++$**--$_ while$_
Try it online!
add a comment |
Perl 5 -n
, 21 bytes
say++$**--$_ while$_
Try it online!
add a comment |
Perl 5 -n
, 21 bytes
say++$**--$_ while$_
Try it online!
Perl 5 -n
, 21 bytes
say++$**--$_ while$_
Try it online!
answered Nov 28 '18 at 18:04
XcaliXcali
5,218520
5,218520
add a comment |
add a comment |
1 2
next
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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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
Nov 28 '18 at 4:20
3
Is the input always greater than 0 or do we have to deal with 0 and negatives?
– Veskah
Nov 28 '18 at 4:21
Inputs will always be positive
– Embodiment of Ignorance
Nov 28 '18 at 5:01
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
Nov 28 '18 at 5:21
2
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
Nov 28 '18 at 14:12