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Given a number n (0 <= n <= 2642245), check if n and n3 have the same set of digits, and output a truthy or falsey value accordingly.
For example, let's check the number 100.
1003 is 1000000.
The set of digits in 100 is {0, 1}.
The set of digits in 1000000 is {0, 1}.
Therefore, 100 should give a truthy value.
0 -> True
1 -> True
10 -> True
107624 -> True
251894 -> True
251895 -> False
102343 -> False
Remember, this is code-golf, so the code with the fewest bytes wins.
Takes input on the command line. Makes n into a character class and tests the string n ^ 3 against it.
a=arg[1]print(not("%u"):format(a^3):find("[^"..a.."]"))
[ dup 3 ^ [ 10 >base unique ] bi@ = ]
(passing) Tests:
: func ( x -- ? ) dup 3 ^ [ number>string unique ] bi@ = ;
{ t } [ 0 func ] unit-test
{ t } [ 1 func ] unit-test
{ t } [ 10 func ] unit-test
{ t } [ 107624 func ] unit-test
{ t } [ 251894 func ] unit-test
{ f } [ 251895 func ] unit-test
{ f } [ 102343 func ] unit-test
using System.Linq;i=>(i*i*i+"").All((i+"").Contains)&&(i+"").All((i*i*i+"").Contains);
Because it is a lambda, in practical use, you will need to assign it to type Func<decimal, bool>. This should, according to MSDN, be useful until you go beyond 28 digits on the i*i*i calculation. If you use the BigInteger package, you can theoretically go to however many digits you have memory for. Just substitute BigInteger for decimal. The function itself does not change.
nUnit Test:
using System;
using System.Linq;
using NUnit.Framework;
namespace CodeGolf
{
[TestFixture]
public class CodeGolfTestFixture
{
[TestCase(0, true)]
[TestCase(1, true)]
[TestCase(10, true)]
[TestCase(107624, true)]
[TestCase(251894, true)]
[TestCase(251895, false)]
[TestCase(102343, false)]
[TestCase(106239, false)]
[TestCase(2103869, true)]
public void Test(int inp, bool expectedResult)
{
Func<decimal, bool> subject = i =>
(i*i*i + "").All((i + "").Contains) &&
(i + "").All((i*i*i + "").Contains);
var observedResult = subject(inp);
Assert.AreEqual(expectedResult, observedResult);
}
}
}
using System.Linq. I'd like to discuss its requirement. For example: codegolf.stackexchange.com/a/84789/62301
\$\endgroup\$
[ dup 3 ^ [ present ] bi@ set= ]
dup Make a copy of the input.
Stack (e.g.): 100 100
3 ^ Cube.
Stack: 100 1000000
[ present ] bi@ Convert them both to strings.
Stack: "100" "1000000"
set= Are they equal when viewed as sets?
Stack: t
3e⊍
An empty set is truthy and a set with items is falsey. If that ain't an acceptable format for output:
3e⊍¬
3e⊍
3e # input ^ 3
⊍ # symmetric set difference between the above and input
# ¬ will negate that result, making it either `0` or `1`
It was difficult for me to find a function to convert to a string while searching the online tutorial. I just guessed ` and was right. I suppose this site is better for searching.
q.{`^Q3.{`
Explanation
Q evaluated input (implicit Q at the end as well)
^ 3 cubed
` ` str() - it's actually repr, but doesn't have an 'L' at the end of longs
.{ .{ convert each string to set()
q check if equal
implicit print
vz is Q and you can remove the last z
\$\endgroup\$
@set/an=%1,c=n*n*n,e=1
@for /l %%a in (0,1,9)do @call:l %%a
@exit/b%e%
:l
@set f=1
@call set s=%%n:%1=%%
@if "%s%"=="%n%" set/af=1-f
@call set s=%%c:%1=%%
@if "%s%"=="%c%" set/af=1-f
@set/ae*=f
Note: Only works up to 1290 due to the limited range of Batch's data type. Takes input on the command line and returns an error level of 1 if the cube uses the same set of digits, 0 otherwise. Works by looping through each digit, checking to see whether an even number of the two strings contains the digit.
@=(f\ss[sort I;unq;str])#0f ^#0 3
Usage:
(@=(f\ss[sort I;unq;str])#0f ^#0 3)500
@
= #. check if the following are equal
(f\ #. store the following to f:
ss[sort I;unq;str] #. composition of to string, unique, sort
)
#0 #. apply f to the arg
f ^ #0 3 #. apply f to arg cubed
Golfed
n=>(x=(a,b)=>!RegExp(`[^${a}]`).test(b))(m=n*n*n,n)&x(n,m)
Test
N=n=>(x=(a,b)=>!RegExp(`[^${a}]`).test(b))(m=n*n*n,n)&x(n,m)
N(0)
1 (true)
N(1)
1 (true)
N(10)
1 (true)
N(107624)
1 (true)
N(251894)
1 (true)
N(251895)
0 (false)
N(102343)
0 (false)
N(106239)
0 (false)
def n(x:BigInt)=x.toString.toSet==(x*x*x).toString.toSet
=s?
\$\endgroup\$
Inspired by @Yodle, but I can't put a comment on his answer because I lack the reputation, I just parsed the string in a separate variable, this saves 5 characters:
var n=long.Parse(i);return new HashSet<char>(i).SetEquals(new HashSet<char>(n*n*n+""));
I know it's not the best answer, but it is my first one:
l:Fi3#s$L|FL|$=
Explanation:
l Read input
:F Put it in the var F
i Convert to int
3# Raise to the 3rd power
s Back to a string
$ Sort
L| Remove dupes
FL| Remove dupes from the original input
$ Sort
= Check if they are equal
3g.sh:
c=`bc<<<$1^3`;[ ${c//[$1]}${1//[$c]} ];echo $?
Usage from file:
$ mksh 3g.sh 100
1
$ mksh 3g.sh 11
0
$ mksh 3g.sh 251894
1
Or from command line displaying return value outside (38 bytes):
$ mksh -c 'c=`bc<<<$0^3`;[ ${c//[$0]}${0//[$c]} ]' 2103869;echo $?
1
$ mksh -c 'c=`bc<<<$0^3`;[ ${c//[$0]}${0//[$c]} ]' 102343;echo $?
0
$ mksh -c 'c=`bc<<<$0^3`;[ ${c//[$0]}${0//[$c]} ]' 106239;echo $?
0
My question: is my 2nd oneliner a valid solution? The result should be printed on terminal or is it enough in an invisible return value? ( I'm new to codegolf. )
1 is true, and 0 is false. meta.codegolf.stackexchange.com/questions/2190/…
\$\endgroup\$
n=input()
print([i for i in str(int(n)**3)if i not in n]==[])
Or in python 2 (60 Bytes):
n=input()
print[i for i in str(n**3) if i not in str(n)]==[]
lambda n:[i for i in str(n**3) if i not in str(n)]==[]
\$\endgroup\$
Nov 17, 2016 at 15:10
lambda x:set(`x`)==set(str(x**3))
Can't use backticks (repr) on large numbers because it includes the 'L'.
This takes input from top of stack, and leaves a result on top of stack. The result is the count of symmetric difference between the digit sets, so zero indicates true, and other numbers indicate false.
d[O~1r:ad0<f]dsfx+3^[O~1r:bd0<f]dsfxsaO[1-ddd;ar;b-d*la+sad0<m]dsmxla
Expanded version as commented full program with I/O to standard streams:
#!/usr/bin/dc
# read input
?
# store 1 in a[i] for each digit i
d[O~
1r:a
d0<f]dsfx
# cube the original number
+3^
# record its digits in b[]
[O~
1r:b
d0<f]dsfx
# 0 left on stack used to initialize accumulator
sa
# for i in 9..0, add (b[i]-a[i])^2
# accumulate in register 'a'
O[1-d
dd;ar;b-d*
la+sa
d0<m]dsmx
# load result from accumulator
la
# print output
p
I hoped I could re-use the first function to store to both a[] and b[] but I couldn't find an easy way to do it. Arrays can't be pushed or popped, and it was too hard to add an indirection to the function.
Here's the test cases from the question, plus those suggested in comments:
0 -> 0
1 -> 0
10 -> 0
107624 -> 0
251894 -> 0
251895 -> 4
102343 -> 6
106239 -> 1
2103869 -> 0
And here's the first 50 terms of A029795, of 536 that I identified with this program by testing the numbers up to ten million:
0
1
10
100
1000
10000
100000
107624
109573
132485
138624
159406
165640
192574
205738
215806
251894
281536
318725
419375
427863
568314
642510
713960
953867
954086
963218
965760
1000000
1008529
1023479
1023674
1026258
1028537
1028565
1028756
1032284
1035743
1037689
1039725
1045573
1046783
1062851
1062854
1063279
1063724
1066254
1072399
1073824
1076240
filter f{"$("$_"[0..99]|sort -u)"}(($a=[long]$args[0])|f)-eq($a*$a*$a|f)
Pretty much the same solution idea as TimmyD had, just a lot shorter (and arrived at independently). I also think there aren't that many more obvious PowerShell solutions to this.
-Unique parameter on the Sort-Object. Otherwise, nice tricks.
\$\endgroup\$
Nov 19, 2016 at 18:54
(f=Sign@*DigitCount)@#==f[#^3]&
because Sign<Union and DigitCount<IntegerDigits
Đ³ąỤ⇹ąỤ\
Outputs an empty list if true
Đ³ąỤ⇹ąỤ\
Implicit input
Đ Duplicate input
³ Cube
ą Make array of digits
Ụ Filter so you have no duplicates
⇹ swap
ąỤ make array of digits and filter duplicates
\ set difference
d.&^₃dp
-2 bytes thanks to Fatalize for fixing the interaction between zero and p.
The predicate succeeds if the input has the same set of digits as its cube, and fails otherwise. If run as a standalone program, success prints true. and failure prints false.
d The input with duplicates removed
. is the output variable
&^₃ and the input cubed
d with duplicates removed
p is a permutation of the output variable.
0 and p has been fixed in the latest commit. The 7 bytes d.&^₃dp should now work properly (once the commit is pulled on TIO).
\$\endgroup\$
ⁿαÉ▒▀z=
-2 bytes thanks to KevinCruijssen. By changing the map to a for loop, we save one byte from not having to do a map (m), and another byte from not having to dump the array onto the stack (~).
ⁿ pop a : push(a*a*a)
α wrap last two elements in array
É start block of length 3
▒ split to list of chars/digits
▀ unique elements of string/list
z reverse sort(array)
= pop(a, b), push(a==b)
.~3?^!
. # Copy the input "102343" "102343"
~ # Parse it to an integer "102343" 102343
3? # Cube it "102343" 1071949756947607
^ # Get the different digits "237956"
! # Is it an empty string? 0
V3esuFsux#
Like most other answers here, it just generates sets for both numbers and checks for equality:
\ Implicit: Input on stack 1
V \ Copy into stack 2
3e \ Cube (stack 1)
su \ Split into digits and create sorted set
Fsu \ ^ Do the same to the second stack (not cubed)
x# \ Check cross-stack equality and output (0 or 1)
Hopefully can take another char off.
@I.{`^Q3`
! Negation, tells if set is empty
- Setwise subtraction
`^Q3 String repr of input^3
` String repr of implicit input
2103869 -> True. This (or a larger one) is necessary to test a language with alongdatatype. \$\endgroup\$