Cubic Concatenation

Question

Challenge

Given three non-negative integers \$a, b\$ and \$c\$, decide if the sum of their cubes is equal to the concatenation of those numbers, aka: $$ a^{3}+b^{3}+c^{3} = a^\frown b ^\frown c $$

Test cases

Truthy

(1,5,3) // 1^3 + 5^3 + 3^3 = 153 (2,2,13) (4,0,7) (10,0,0) (10,0,1) (22,18,59) (98,28,27) (166,500,333) (828,538,472)

Falsy

(1,2,3) // 1^3 + 2^3 + 3^3 = 32 != 123 (4,5,6) (6,0,0) (166,500,334) (200,0,200)

You can assume there are no leading zeroes.

This is code-golf, so the shortest code wins.

@Jitse well, I didn't include any test cases with leading zeroes because it may cause issues with, for example, "001" being different from "1" and I just wanted people to focus on writing shortest code without getting into edge cases too much. — Blue
– Blue, Commented Aug 5, 2024 at 11:51

Neil · Accepted Answer · 2024-08-06 18:48:02Z

Charcoal, 9 bytes

⁼Ｉ⪫θωΣＸθ³

Try it online! Link is to verbose version of code. Takes input as an array and outputs a Charcoal boolean, i.e. - if the sum of cubes equals the concatenation, nothing if not. Explanation:

 θ Input array ⪫ Joined by ω Predefined variable empty string Ｉ Cast to integer ⁼ Equals θ Input array Ｘ Vectorised raised to power ³ Literal integer `3` Σ Take the sum Implicitly print

Jonah · Accepted Answer · 2024-08-07 00:38:50Z

2

J, 15 bytes

;(=1#.^&3)&:".>

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Takes input as list of boxed strings.

answered Aug 7, 2024 at 0:38

Jonah

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Glory2Ukraine · Accepted Answer · 2024-08-07 08:04:27Z

gnuplot, 33 bytes

f(a,b,c)=a**3+b**3+c**3=="".a.b.c

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A function that inputs 3 integer values and outputs 1 if their sum of cubes is equal to the concatenation of those numbers, otherwise the output is 0.

chunes · Accepted Answer · 2024-08-08 22:18:37Z

2

Lua, 38 bytes

x,y,z=...print(x^3+y^3+z^3==0|x..y..z)

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answered Aug 8, 2024 at 22:18

chunes

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tsh · Accepted Answer · 2024-08-09 09:33:10Z

2

JavaScript (Node.js), 31 bytes

(x,y,z)=>[x]+y+z-x**3-y**3-z**3

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A boring trivial solution is short...

answered Aug 9, 2024 at 9:33

tsh

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\$\begingroup\$ You could save 2 bytes by taking the arguments as strings. \$\endgroup\$

Shaggy
– Shaggy

2024-08-16 15:38:59 +00:00
Commented Aug 16, 2024 at 15:38

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Kevin Cruijssen · Accepted Answer · 2024-08-12 12:20:20Z

05AB1E, 6 bytes

3mOIJQ

Try it online or verify all test cases.

The 3m could alternatively be ** or n* for the same byte-count; or the order could be swapped to JI3mOQ as well:
Try it online.

Explanation:

3m # Take the cube of each value in the (implicit) input-triplet O # Sum those together I # Push the input-triplet again J # Join them together Q # Check if the two are the same # (after which the result is output implicitly)

bigyihsuan · Accepted Answer · 2024-08-12 13:49:50Z

Go, 108 bytes

import."fmt" func f(a,b,c int)bool{n:=0 Sscanf(Sprintf("%d%d%d",a,b,c),"%d",&n) return a*a*a+b*b*b+c*c*c==n}

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C K · Accepted Answer · 2024-08-16 17:25:05Z

2

AWK, 24 21 bytes

$1$2$3~$1^3+$2^3+$3^3

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Prints the digits if true, or returns nothing.

edited Aug 16, 2024 at 17:25

answered Aug 16, 2024 at 16:59

C K

1214 bronze badges

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chunes · Accepted Answer · 2024-08-05 18:21:47Z

Factor, ⁴⁶ 45 bytes

[ 4 dupn v* v. swap "%d%d%d"vsprintf dec> = ]

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Explanation

[ ! start function 4 dupn ! put 4 total copies of input list on stack v* ! multiply top two v. ! dot product w/ third swap ! put input on top "%d%d%d"vsprintf ! concat all three to string dec> ! to number = ! check equality ] ! end function

rst · Accepted Answer · 2024-08-06 12:03:45Z

1

MATLAB, 41 bytes

f=@(x)sum(x.^3)==str2num(sprintf('%d',x))

An @-function that accepts input as column vector.

>> f([828;538;472]) ans = logical 1 >> f([200;0;200]) ans = logical 0

answered Aug 6, 2024 at 12:03

rst

111 bronze badge

\$\begingroup\$ Welcome to the site and nice first answer! You may skip f= from your bytecount. Also for the demonstration purposes and to have the bytes counted you can use TIO website. MATLAB isn't available there, but you try GNU Octave . See this example. \$\endgroup\$

Glory2Ukraine
– Glory2Ukraine

2024-08-07 08:34:00 +00:00
Commented Aug 7, 2024 at 8:34

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Glory2Ukraine · Accepted Answer · 2024-08-06 18:31:26Z

Pyth, 9 bytes

qsjkQsm**

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First time golfing in Pyth, so this might be not optimal.

Explanation:

sm** calculates the sum of the cubes;

jkQ joins the list elements into a string and s converts the result to a number;

q: are the values equal?

Neil · Accepted Answer · 2024-08-06 18:46:13Z

Retina, 54 bytes

,(\d+),(\d+) $1$2,$.(***_$2*$2*$2*_$`*$`*$`* ^(.+),\1$

Try it online! Link includes test cases. Explanation:

,(\d+),(\d+)

Match the first number in $`, the second number in $1 and the third number in $2.

$1$2,$.(***_$2*$2*$2*_$`*$`*$`*

Concatenate the input numbers, then compute the sum of the cube of $1 (implicitly because it's the first number in the match), the cube of $2 and the cube of $`. (Retina 1 will do this calculation using arbitrary-precision integers as it knows it will be converting the result back to decimal.)

^(.+),\1$

Check that the two results are the same.

Donat · Accepted Answer · 2024-08-06 20:02:47Z

Bash, 33 bytes

[ $1$2$3 = $[$1**3+$2**3+$3**3] ]

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The script returns a shell exit code: 1 for True, 0 for False.

David Conrad · Accepted Answer · 2024-08-08 17:22:30Z

Java (JDK), 120 bytes

t->t.stream().mapToInt(x->x*x*x).sum()==Integer.parseInt(t.stream().map(Object::toString).collect(Collectors.joining()))

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I'm not sure if the import of java.util.List needs to be counted. The predicate takes a List and some old advice says that all the types for lambdas need to be included, with any imports, but in JShell it is imported by default and in JDK 23 (which is currently only a preview edition) you also get it for free in a Java program that is all in one file (you can even skip declaring a class or having a full public-static-void-main), but TIO doesn't support JShell or recent versions of Java.

att · Accepted Answer · 2024-08-16 13:10:34Z

1

Wolfram Language (Mathematica), 31 30 bytes

boring wins

s[#.#^2]==s/@#<>""& s=ToString

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edited Aug 16, 2024 at 13:10

answered Aug 16, 2024 at 13:04

att

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Julian · Accepted Answer · 2024-09-02 01:43:40Z

PowerShell Core, 45 bytes

-join$args-(($args|%{"$_*"*3+1})-join'+'|iex)

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Takes 3 integers as parameters
Returns \$0\$ for true, otherwise false=

NikoNyrh · Accepted Answer · 2024-09-03 14:56:03Z

Clojure, 56 bytes

#(=(seq(str(apply +(for[i %](* i i i)))))(mapcat str %))

TIO. This was surprisingly painless, thanks to str and mapcat. Takes the input as a list of numbers.

Glory2Ukraine · Accepted Answer · 2024-09-11 13:55:16Z

Husk, 8 bytes

=dṁd¹ṁ^3

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Takes a list of numbers and outputs 1 (truthy) "if the sum of their cubes is equal to the concatenation of those numbers", otherwise 0 (falsy).

Explanation:

 ṁ^3 # sum up the cubes ṁd¹ # split every number into its digits and flatten the list d # join digits back into a single number = # check for the equality

xrs · Accepted Answer · 2024-11-18 20:36:45Z

AWK, 24 bytes

$0=$1^3+$2^3+$3^3~$1$2$3

awk '$0=$1^3+$2^3+$3^3~$1$2$3' << "1 5 3"

Prints 1 if truthy, or nothing.

DesmosEnthusiast · Accepted Answer · 2025-02-02 20:15:59Z

Desmos, 66 bytes

d(n)=10^{floor(log(n+0^n))+1} f(a,b,c)=a^3+b^3+c^3-c-d(c)(b+d(b)a)

This solution evolved very close to Aiden Chow's solution, but it does improve it by one byte. 0 represents truthy outputs, any any other integer represents falsey.

Try it on Desmos!

ATaco · Accepted Answer · 2025-02-04 04:52:14Z

Jalapeño, 4 bytes

J=Σₓ³

Explained

J=Σₓ³ J # Join (implicit input) together = # Is equal to Σₓ # The sum of (implicit input) mapped by ³ # Cubed

Hex-Dump of Bytecode

 0 1 2 3 4 5 6 7 8 9 A B C D E F 0000: 14 59 ea 62

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Rosario · Accepted Answer · 2025-02-05 16:44:22Z

APL(NARS), 15 chars

{(∊⍕¨⍵)≡⍕+/⍵*3}

test:

 {(∊⍕¨⍵)≡⍕+/⍵*3}10 0 0 1 ~ {(∊⍕¨⍵)≡⍕+/⍵*3}10 0 1 1 ~ {(∊⍕¨⍵)≡⍕+/⍵*3}¨(1 5 3)(2 2 13)(4 0 7)(828 538 472) ┌4───────┐ │ 1 1 1 1│ └~───────┘ {(∊⍕¨⍵)≡⍕+/⍵*3}¨(1 2 3)(4 5 6)(6 0 0)(166 500 334) ┌4───────┐ │ 0 0 0 0│ └~───────┘

ayaan098 · Accepted Answer · 2025-02-05 21:41:50Z

C (gcc), 174 bytes

char c[99];j;k; #define F(x)atoi(x)*atoi(x)*atoi(x) main(int a,char**v){strcpy(c,v[1]);while(k++-2)strcpy(c+(j+=strlen(v[k])),v[k+1]);exit(atoi(c)!=F(v[1])+F(v[2])+F(v[3]));}

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\$\begingroup\$ 108 bytes \$\endgroup\$

ceilingcat
– ceilingcat

2025-02-07 09:01:45 +00:00
Commented Feb 7 at 9:01 — ceilingcat
– ceilingcat, Commented Feb 7 at 9:01

Stack Exchange Network

Cubic Concatenation

Challenge

Test cases

53 Answers 53

Charcoal, 9 bytes

J, 15 bytes

gnuplot, 33 bytes

Lua, 38 bytes

JavaScript (Node.js), 31 bytes

05AB1E, 6 bytes

Go, 108 bytes

AWK, 24 21 bytes

Factor, ⁴⁶ 45 bytes

Explanation

MATLAB, 41 bytes

Pyth, 9 bytes

Explanation:

Retina, 54 bytes

Bash, 33 bytes

Java (JDK), 120 bytes

Wolfram Language (Mathematica), 31 30 bytes

PowerShell Core, 45 bytes

Clojure, 56 bytes

Husk, 8 bytes

AWK, 24 bytes

Desmos, 66 bytes

Jalapeño, 4 bytes

Explained

Hex-Dump of Bytecode

APL(NARS), 15 chars

C (gcc), 174 bytes

Hot Network Questions

Cubic Concatenation

Challenge

Test cases

53 Answers 53

Charcoal, 9 bytes

J, 15 bytes

gnuplot, 33 bytes

Lua, 38 bytes

JavaScript (Node.js), 31 bytes

05AB1E, 6 bytes

Go, 108 bytes

AWK, 24 21 bytes

Factor, 46 45 bytes

Explanation

MATLAB, 41 bytes

Pyth, 9 bytes

Explanation:

Retina, 54 bytes

Bash, 33 bytes

Java (JDK), 120 bytes

Wolfram Language (Mathematica), 31 30 bytes

PowerShell Core, 45 bytes

Clojure, 56 bytes

Husk, 8 bytes

AWK, 24 bytes

Desmos, 66 bytes

Jalapeño, 4 bytes

Explained

Hex-Dump of Bytecode

APL(NARS), 15 chars

C (gcc), 174 bytes

Related

Hot Network Questions

Factor, ⁴⁶ 45 bytes