Mixing
Find if a list is an ABC-triple
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Three positive integers A, B, C are ABC-triple if they are coprime,
with A < B and satisfying the relation : A + B = C
Examples :
1, 8, 9is an ABC-triple since they are coprime, 1 < 8 and 1 + 8 = 9
6, 8, 14is not because they are not coprime
7, 5, 12is not because 7 > 5
You can see this Frits Beukers 2005 presentation for more details about ABC-triples.
Input/Output
Three integers, decimal written. May be separated values or
list. Output had to be a truthy/falsy value whether the three
integers are an ABC-triple.
Note: It is important to respect integers order in the list, for example: 1, 8, 9 is not considered as the same list as 9, 1, 8 or any other combination. So first is an ABC-triple and second is not.
Thus A is the first element of the list, B the second and C the third.
Test cases
Each of the following lists should output a truthy value
[1, 8, 9]
[2, 3, 5]
[2, 6436341, 6436343]
[4, 121, 125]
[121, 48234375, 48234496]
Each of the following lists should output a falsey value
[1, 1, 2]
[1, 2, 5]
[1, 9, 8]
[4, 12872682, 12872686]
[6, 8, 14]
[7, 5, 12]
code-golf sequence decision-problem number-theory
asked Jan 3 at 11:53
daviddavid
199111
$endgroup$
|
show 4 more comments
$begingroup$
Three positive integers A, B, C are ABC-triple if they are coprime,
with A < B and satisfying the relation : A + B = C
Examples :
1, 8, 9is an ABC-triple since they are coprime, 1 < 8 and 1 + 8 = 9
6, 8, 14is not because they are not coprime
7, 5, 12is not because 7 > 5
You can see this Frits Beukers 2005 presentation for more details about ABC-triples.
Input/Output
Three integers, decimal written. May be separated values or
list. Output had to be a truthy/falsy value whether the three
integers are an ABC-triple.
Note: It is important to respect integers order in the list, for example: 1, 8, 9 is not considered as the same list as 9, 1, 8 or any other combination. So first is an ABC-triple and second is not.
Thus A is the first element of the list, B the second and C the third.
Test cases
Each of the following lists should output a truthy value
[1, 8, 9]
[2, 3, 5]
[2, 6436341, 6436343]
[4, 121, 125]
[121, 48234375, 48234496]
Each of the following lists should output a falsey value
[1, 1, 2]
[1, 2, 5]
[1, 9, 8]
[4, 12872682, 12872686]
[6, 8, 14]
[7, 5, 12]
code-golf sequence decision-problem number-theory
asked Jan 3 at 11:53
daviddavid
199111
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Does the output have to be only one of two values, or can we output different truthy/falsy values for different inputs?
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– Luis Mendo
Jan 3 at 12:04
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I think it should be consistent: your code have to output one kind of truthy/falsy values whatever the input. But the truthy/falsy couple can be what you want as far as it does the job: differentiate lists.
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– david
Jan 3 at 12:33
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If we take the input as list of three values, does the input have to be in the order[A,B,C], or are we also allowed to take the input in the order[C,B,A]or[C,A,B]?
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– Kevin Cruijssen
Jan 3 at 13:06
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You have to respect order since A < B is a criteria in the challenge.
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– david
Jan 3 at 13:08
1
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I don't think requiring a particular list order is compatible with allowing input to be taken as separate values, as separate values are inherently unordered and may be taken as a list.
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– Dennis♦
Jan 3 at 15:30
|
show 4 more comments
$begingroup$
Three positive integers A, B, C are ABC-triple if they are coprime,
with A < B and satisfying the relation : A + B = C
Examples :
1, 8, 9is an ABC-triple since they are coprime, 1 < 8 and 1 + 8 = 9
6, 8, 14is not because they are not coprime
7, 5, 12is not because 7 > 5
You can see this Frits Beukers 2005 presentation for more details about ABC-triples.
Input/Output
Three integers, decimal written. May be separated values or
list. Output had to be a truthy/falsy value whether the three
integers are an ABC-triple.
Note: It is important to respect integers order in the list, for example: 1, 8, 9 is not considered as the same list as 9, 1, 8 or any other combination. So first is an ABC-triple and second is not.
Thus A is the first element of the list, B the second and C the third.
Test cases
Each of the following lists should output a truthy value
[1, 8, 9]
[2, 3, 5]
[2, 6436341, 6436343]
[4, 121, 125]
[121, 48234375, 48234496]
Each of the following lists should output a falsey value
[1, 1, 2]
[1, 2, 5]
[1, 9, 8]
[4, 12872682, 12872686]
[6, 8, 14]
[7, 5, 12]
code-golf sequence decision-problem number-theory
asked Jan 3 at 11:53
daviddavid
199111
$endgroup$
Three positive integers A, B, C are ABC-triple if they are coprime,
with A < B and satisfying the relation : A + B = C
Examples :
1, 8, 9is an ABC-triple since they are coprime, 1 < 8 and 1 + 8 = 9
6, 8, 14is not because they are not coprime
7, 5, 12is not because 7 > 5
You can see this Frits Beukers 2005 presentation for more details about ABC-triples.
Input/Output
Three integers, decimal written. May be separated values or
list. Output had to be a truthy/falsy value whether the three
integers are an ABC-triple.
Note: It is important to respect integers order in the list, for example: 1, 8, 9 is not considered as the same list as 9, 1, 8 or any other combination. So first is an ABC-triple and second is not.
Thus A is the first element of the list, B the second and C the third.
Test cases
Each of the following lists should output a truthy value
[1, 8, 9]
[2, 3, 5]
[2, 6436341, 6436343]
[4, 121, 125]
[121, 48234375, 48234496]
Each of the following lists should output a falsey value
[1, 1, 2]
[1, 2, 5]
[1, 9, 8]
[4, 12872682, 12872686]
[6, 8, 14]
[7, 5, 12]
code-golf sequence decision-problem number-theory
code-golf sequence decision-problem number-theory
asked Jan 3 at 11:53
daviddavid
199111
asked Jan 3 at 11:53
daviddavid
199111
edited Jan 3 at 15:23
david
asked Jan 3 at 11:53
daviddavid
199111
asked Jan 3 at 11:53
daviddavid
199111
asked Jan 3 at 11:53
daviddavid
199111
199111
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Does the output have to be only one of two values, or can we output different truthy/falsy values for different inputs?
$endgroup$
– Luis Mendo
Jan 3 at 12:04
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I think it should be consistent: your code have to output one kind of truthy/falsy values whatever the input. But the truthy/falsy couple can be what you want as far as it does the job: differentiate lists.
$endgroup$
– david
Jan 3 at 12:33
$begingroup$
If we take the input as list of three values, does the input have to be in the order[A,B,C], or are we also allowed to take the input in the order[C,B,A]or[C,A,B]?
$endgroup$
– Kevin Cruijssen
Jan 3 at 13:06
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You have to respect order since A < B is a criteria in the challenge.
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– david
Jan 3 at 13:08
1
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I don't think requiring a particular list order is compatible with allowing input to be taken as separate values, as separate values are inherently unordered and may be taken as a list.
$endgroup$
– Dennis♦
Jan 3 at 15:30
|
show 4 more comments
$begingroup$
Does the output have to be only one of two values, or can we output different truthy/falsy values for different inputs?
$endgroup$
– Luis Mendo
Jan 3 at 12:04
$begingroup$
I think it should be consistent: your code have to output one kind of truthy/falsy values whatever the input. But the truthy/falsy couple can be what you want as far as it does the job: differentiate lists.
$endgroup$
– david
Jan 3 at 12:33
$begingroup$
If we take the input as list of three values, does the input have to be in the order[A,B,C], or are we also allowed to take the input in the order[C,B,A]or[C,A,B]?
$endgroup$
– Kevin Cruijssen
Jan 3 at 13:06
$begingroup$
You have to respect order since A < B is a criteria in the challenge.
$endgroup$
– david
Jan 3 at 13:08
1
$begingroup$
I don't think requiring a particular list order is compatible with allowing input to be taken as separate values, as separate values are inherently unordered and may be taken as a list.
$endgroup$
– Dennis♦
Jan 3 at 15:30
$begingroup$
Does the output have to be only one of two values, or can we output different truthy/falsy values for different inputs?
$endgroup$
– Luis Mendo
Jan 3 at 12:04
$begingroup$
Does the output have to be only one of two values, or can we output different truthy/falsy values for different inputs?
$endgroup$
– Luis Mendo
Jan 3 at 12:04
$begingroup$
I think it should be consistent: your code have to output one kind of truthy/falsy values whatever the input. But the truthy/falsy couple can be what you want as far as it does the job: differentiate lists.
$endgroup$
– david
Jan 3 at 12:33
$begingroup$
I think it should be consistent: your code have to output one kind of truthy/falsy values whatever the input. But the truthy/falsy couple can be what you want as far as it does the job: differentiate lists.
$endgroup$
– david
Jan 3 at 12:33
$begingroup$
If we take the input as list of three values, does the input have to be in the order
[A,B,C], or are we also allowed to take the input in the order [C,B,A] or [C,A,B]?$endgroup$
– Kevin Cruijssen
Jan 3 at 13:06
$begingroup$
If we take the input as list of three values, does the input have to be in the order
[A,B,C], or are we also allowed to take the input in the order [C,B,A] or [C,A,B]?$endgroup$
– Kevin Cruijssen
Jan 3 at 13:06
$begingroup$
You have to respect order since A < B is a criteria in the challenge.
$endgroup$
– david
Jan 3 at 13:08
$begingroup$
You have to respect order since A < B is a criteria in the challenge.
$endgroup$
– david
Jan 3 at 13:08
1
1
$begingroup$
I don't think requiring a particular list order is compatible with allowing input to be taken as separate values, as separate values are inherently unordered and may be taken as a list.
$endgroup$
– Dennis♦
Jan 3 at 15:30
$begingroup$
I don't think requiring a particular list order is compatible with allowing input to be taken as separate values, as separate values are inherently unordered and may be taken as a list.
$endgroup$
– Dennis♦
Jan 3 at 15:30
|
show 4 more comments
20 Answers
20
active
oldest
votes
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Jelly, 10 9 bytes
Ṫ=S×</=g/
Try it online!
How it works
Ṫ=S×</=g/ Main link. Argument: [a, b, c] (positive integers)
Ṫ Tail; pop and yield c.
S Take the sum of [a, b], yielding (a + b).
= Yield t := (c == a + b).
</ Reduce by less than, yielding (a < b).
× Multiply, yielding t(a < b).
g/ Reduce by GCD, yielding gcd(a, b).
= Check if t(a < b) == gcd(a, b).
answered Jan 3 at 13:21
Dennis♦Dennis
187k32297737
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add a comment |
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Haskell, 48 38 29 bytes
-10 bytes due to TFeld's gcd trick!
-7 bytes thanks to HPWiz for improving the co-primality test and spotting a superfluous space!
-2 bytes thanks to nimi for suggesting an infix-operator!
(a!b)c=a<b&&a+b==c&&gcd a b<2
Try it online!
Explanation
The first two conditions a < b and a + b == c are fairly obvious, the third one uses that $gcd(a,b) = gcd(a,c) = gcd(b,c)$:
Writing $gcd(a,c) = U cdot a + V cdot c$ using Bézout's identity and substituting $c = a + b$ gives:
$$
U cdot a + V cdot (a + b) = (U + V) cdot a + V cdot b
$$
Since the $gcd$ is the minimal positive solution to that identity it follows that $gcd(a,b) = gcd(a,c)$. The other case is symmetric.
answered Jan 3 at 12:27
BMOBMO
11.8k22188
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1
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Also, I believe you only need thatgcd a b==1. Sincegcd a bdividesa+b=c. i.egcd(gcd a b)c=gcd a b
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– H.PWiz
Jan 3 at 14:01
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@HPWiz: Ah yes,of course, thanks! Will edit later when not on mobile..
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– BMO
Jan 3 at 14:07
add a comment |
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Perl 6, 33 32 bytes
-1 byte thanks to nwellnhof
{(.sum/.[2]/2*[<] $_)==[gcd] $_}
Try it online!
Anonymous code block that takes a list of three numbers and returns True or False.
Explanation
{ } # Anonymous code block
[gcd] $_ # Is the gcd of all the numbers
( )== # Equal to
.sum # Whether the sum of numbes
/ # Is equal to
.[2]/2 # The last element doubled
*[<] $_ # And elements are in ascending order
answered Jan 3 at 12:29
Jo KingJo King
21.3k248110
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2
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32 bytes
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– nwellnhof
Jan 3 at 13:09
add a comment |
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Java 10, 65 64 bytes
(a,b,c)->{var r=a<b&a+b==c;for(;b>0;a=b,b=c)c=a%b;return r&a<2;}
-1 byte thank to @Shaggy.
Try it online.
Explanation:
(a,b,c)->{ // Method with three integer parameters and boolean return-type
var r= // Result-boolean, starting at:
a<b // Check if `a` is smaller than `b`
&a+b==c; // And if `a+b` is equal to `c`
for(;b>0 // Then loop as long as `b` is not 0 yet
; // After every iteration:
a=b, // Set `a` to the current `b`
b=c) // And set `b` to the temp value `c`
c=a%b; // Set the temp value `c` to `a` modulo-`b`
// (we no longer need `c` at this point)
return r // Return if the boolean-result is true
&a<2;} // And `a` is now smaller than 2
answered Jan 3 at 15:04
Kevin CruijssenKevin Cruijssen
36.4k555192
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a==1->a<2to save a byte.
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– Shaggy
Jan 3 at 15:09
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@Shaggy Thanks!
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– Kevin Cruijssen
Jan 3 at 16:23
add a comment |
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05AB1E, 12 11 10 bytes
Saved 1 byte thanks to Kevin Cruijssen
ÂÆ_*`‹*¿Θ
Try it online!
or as a Test Suite
Explanation
ÂÆ # reduce a reversed copy of the input by subtraction
_ # logically negate
* # multiply with input
` # push the values of the resulting list separately to stack
# remove the top (last) value
‹ # is a < b ?
* # multiply by the input list
¿ # calculate the gcd of the result
Θ # is it true ?
answered Jan 3 at 12:41
EmignaEmigna
45.6k432139
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Oops.. deleted my comment.. >.> So again: you can save a byte by using multiples instead of swaps with product:RÆ_*`‹*¿ΘTest Suite.
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– Kevin Cruijssen
Jan 3 at 14:30
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@KevinCruijssen: Thanks! Yeah, usually when you have that many swaps, you're doing something wrong :P
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– Emigna
Jan 3 at 16:45
add a comment |
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Excel, 33 bytes
=AND(A1+B1=C1,GCD(A1:C1)=1,A1<B1)
answered Jan 3 at 14:27
WernischWernisch
1,608317
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add a comment |
$begingroup$
Python 2, 69 67 63 62 55 bytes
lambda a,b,c:(c-b==a<b)/gcd(a,b)
from fractions import*
Try it online!
Python 3, 58 51 bytes
lambda a,b,c:(c-b==a<b)==gcd(a,b)
from math import*
Try it online!
-7 bytes, thanks to H.PWiz
answered Jan 3 at 12:28
TFeldTFeld
14.5k21240
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is thegcdingcdtrick valid? What ifais not coprime withc?
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– Jo King
Jan 3 at 12:31
2
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@jo-king If p divides a and c, it should divide c-a so b.
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– david
Jan 3 at 12:41
2
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@JoKing: It is in this case, but not in general (you can prove it via Bezout's identity).
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– BMO
Jan 3 at 12:42
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You can take it one step further and usegcd(a,b), sincegcd(a,b)dividesa+b
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– H.PWiz
Jan 3 at 14:04
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@H.PWiz Thanks :)
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– TFeld
Jan 3 at 14:37
|
show 1 more comment
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Japt, 16 14 13 11 bytes
<V¥yU «NÔr-
Try it
:Implicit input of integers U=A, V=B & W=C
<V :Is U less than V?
¥ :Test that for equality with
yU :The GCD of V & U
« :Logical AND with the negation of
N :The array of inputs
Ô :Reversed
r- :Reduced by subtraction
answered Jan 3 at 12:42
ShaggyShaggy
19.2k21666
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Here is another 11 byte solution, though on closer inspection it isn't much different from yours in its actual logic.
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– Kamil Drakari
Jan 3 at 15:13
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@KamilDrakari, had a variation on that at one stage, too. It could be 10 bytes if variables were auto-inserted when>follows©.
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– Shaggy
Jan 3 at 18:05
add a comment |
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JavaScript (ES6), 54 43 42 40 bytes
Thanks to @Shaggy for pointing out that we don't need to compute $gcd(a,c)$. Saved 11 bytes by rewriting the code accordingly.
Takes input as 3 separate integers. Returns $true$ for an ABC-triple, or either $0$ or $false$ otherwise.
f=(a,b,c)=>c&&a/b|a+b-c?0:b?f(b,a%b):a<2
Try it online!
answered Jan 3 at 12:18
ArnauldArnauld
73.4k689308
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1
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I don't think you need to testgcd(c,a).
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– Shaggy
Jan 3 at 15:13
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@Shaggy Thanks! I've rewritten the code entirely.
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– Arnauld
Jan 3 at 15:47
add a comment |
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bash, 61 bytes
factor $@|grep -vzP '( .+b).*n.*1b'&&(($1<$2&&$1+$2==$3))
Try it online!
Input as command line arguments,
output in the exit code
(also produces output on stdout as a side effect, but this can be ignored).
The second part (starting from &&(() is pretty standard,
but the interesting bit is the coprime test:
factor $@ # produces output of the form "6: 2 3n8: 2 2 2n14: 2 7n"
|grep - # regex search on the result
v # invert the match (return truthy for strings that don't match)
z # zero-terminated, allowing us to match newlines
P # perl (extended) regex
'( .+b)' # match one or more full factors
'.*n.*' # and somewhere on the next line...
'1b' # find the same full factors
answered Jan 3 at 15:13
Doorknob♦Doorknob
54.3k17113347
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add a comment |
$begingroup$
Wolfram Language 24 30 28 bytes
With 2 bytes saved by Doorknob,
#<#2&&CoprimeQ@##&&#+#2==#3&
answered Jan 3 at 14:24
DavidCDavidC
23.9k243102
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I think you should also be able to useCoprimeQ@##to save 2 bytes.
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– Doorknob♦
Jan 3 at 16:48
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@Doorknob, If the first and second numbers are coprime, are they necessarily coprime with their sum?
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– DavidC
Jan 3 at 17:27
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They are, but the original definition actually states that A, B, and C should be coprime. Most answers check only A and B just because it's usually shorter.
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– Doorknob♦
Jan 3 at 17:32
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 59 bytes
(a,b,c)=>Enumerable.Range(2,a).All(i=>a%i+b%i>0)&a<b&a+b==c
Try it online!
answered Jan 3 at 17:19
danadana
63145
$endgroup$
add a comment |
$begingroup$
J, 27 bytes
(+/=2*{:)*({.<1{])*1=+./ .*
Try it online!
Inspired by Jo King's Perl solution
answered Jan 3 at 13:07
Galen IvanovGalen Ivanov
6,49711032
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 90 bytes
n=>new int[(int)1e8].Where((_,b)=>n[0]%++b<1&n[1]%b<1).Count()<2&n[0]+n[1]==n[2]&n[0]<n[1]
Runs for numbers up to 1e8, takes about 35 seconds on my machine. Instead of calculating the gcd like others, the function just instantiate a huge array and filter the indexes that aren't divisors of a or b, and check how many elements are left. Next it check if element one plus element two equals element three. Lastly, it checks if the first element is less than the second.
Try it online!
answered Jan 3 at 16:54
Embodiment of IgnoranceEmbodiment of Ignorance
631115
$endgroup$
add a comment |
$begingroup$
C# (.NET Core), 68 bytes
Without Linq.
(a,b,c)=>{var t=a<b&a+b==c;while(b>0){c=b;b=a%b;a=c;}return t&a<2;};
Try it online!
answered Jan 3 at 17:26
DestroigoDestroigo
1815
$endgroup$
add a comment |
$begingroup$
Stax, 12 bytes
ü╡v╕7+Pü°╔|g
Run and debug it
answered Jan 3 at 17:37
wastlwastl
2,084425
$endgroup$
add a comment |
$begingroup$
Pari/GP, 30 bytes
Saved 2 bytes thanks to @Shaggy.
(a,b,c)->a<b==gcd(a,b)&&a+b==c
Try it online!
answered Jan 3 at 16:23
alephalphaalephalpha
21.2k32991
$endgroup$
1
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30 bytes(?)
$endgroup$
– Shaggy
Jan 3 at 20:13
add a comment |
$begingroup$
Befunge-98 (FBBI), 83 bytes
&:&:03p&:04pw>03g04g:v_1w03g04g+w1.@
00: 7j@.0[^j7 _^;>0.@;j7;>0.@;:%g00p
Try it online!
The input which is a triple of integers [A,B,C] is feeded into Befunge as space-separated integers C B A.
answered Jan 4 at 1:37
WisławWisław
1413
$endgroup$
add a comment |
$begingroup$
Clean, 43 bytes
import StdEnv
$a b c=a<b&&a+b==c&&gcd a b<2
Try it online!
Similar to basically everything else because the direct test is the same.
answered Jan 3 at 17:52
ΟurousΟurous
6,55211033
$endgroup$
add a comment |
$begingroup$
Retina 0.8.2, 40 bytes
d+
$*
A`^(11+)+,1+,
^(1+),(1+1),12$
Try it online! Link includes test cases. Explanation:
d+
$*
Convert to unary.
A`^(11+)+,1+,
Check that A and B have no common factor.
^(1+),(1+1),12$
Check that A < B and A + B = C.
answered Jan 3 at 22:13
NeilNeil
79.8k744177
$endgroup$
add a comment |
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20 Answers
20
active
oldest
votes
20 Answers
20
active
oldest
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$begingroup$
Jelly, 10 9 bytes
Ṫ=S×</=g/
Try it online!
How it works
Ṫ=S×</=g/ Main link. Argument: [a, b, c] (positive integers)
Ṫ Tail; pop and yield c.
S Take the sum of [a, b], yielding (a + b).
= Yield t := (c == a + b).
</ Reduce by less than, yielding (a < b).
× Multiply, yielding t(a < b).
g/ Reduce by GCD, yielding gcd(a, b).
= Check if t(a < b) == gcd(a, b).
answered Jan 3 at 13:21
Dennis♦Dennis
187k32297737
$endgroup$
add a comment |
$begingroup$
Jelly, 10 9 bytes
Ṫ=S×</=g/
Try it online!
How it works
Ṫ=S×</=g/ Main link. Argument: [a, b, c] (positive integers)
Ṫ Tail; pop and yield c.
S Take the sum of [a, b], yielding (a + b).
= Yield t := (c == a + b).
</ Reduce by less than, yielding (a < b).
× Multiply, yielding t(a < b).
g/ Reduce by GCD, yielding gcd(a, b).
= Check if t(a < b) == gcd(a, b).
answered Jan 3 at 13:21
Dennis♦Dennis
187k32297737
$endgroup$
add a comment |
$begingroup$
Jelly, 10 9 bytes
Ṫ=S×</=g/
Try it online!
How it works
Ṫ=S×</=g/ Main link. Argument: [a, b, c] (positive integers)
Ṫ Tail; pop and yield c.
S Take the sum of [a, b], yielding (a + b).
= Yield t := (c == a + b).
</ Reduce by less than, yielding (a < b).
× Multiply, yielding t(a < b).
g/ Reduce by GCD, yielding gcd(a, b).
= Check if t(a < b) == gcd(a, b).
answered Jan 3 at 13:21
Dennis♦Dennis
187k32297737
$endgroup$
Jelly, 10 9 bytes
Ṫ=S×</=g/
Try it online!
How it works
Ṫ=S×</=g/ Main link. Argument: [a, b, c] (positive integers)
Ṫ Tail; pop and yield c.
S Take the sum of [a, b], yielding (a + b).
= Yield t := (c == a + b).
</ Reduce by less than, yielding (a < b).
× Multiply, yielding t(a < b).
g/ Reduce by GCD, yielding gcd(a, b).
= Check if t(a < b) == gcd(a, b).
answered Jan 3 at 13:21
Dennis♦Dennis
187k32297737
edited Jan 3 at 16:34
answered Jan 3 at 13:21
Dennis♦Dennis
187k32297737
answered Jan 3 at 13:21
Dennis♦Dennis
187k32297737
answered Jan 3 at 13:21
Dennis♦Dennis
187k32297737
187k32297737
add a comment |
add a comment |
$begingroup$
Haskell, 48 38 29 bytes
-10 bytes due to TFeld's gcd trick!
-7 bytes thanks to HPWiz for improving the co-primality test and spotting a superfluous space!
-2 bytes thanks to nimi for suggesting an infix-operator!
(a!b)c=a<b&&a+b==c&&gcd a b<2
Try it online!
Explanation
The first two conditions a < b and a + b == c are fairly obvious, the third one uses that $gcd(a,b) = gcd(a,c) = gcd(b,c)$:
Writing $gcd(a,c) = U cdot a + V cdot c$ using Bézout's identity and substituting $c = a + b$ gives:
$$
U cdot a + V cdot (a + b) = (U + V) cdot a + V cdot b
$$
Since the $gcd$ is the minimal positive solution to that identity it follows that $gcd(a,b) = gcd(a,c)$. The other case is symmetric.
answered Jan 3 at 12:27
BMOBMO
11.8k22188
$endgroup$
1
$begingroup$
Also, I believe you only need thatgcd a b==1. Sincegcd a bdividesa+b=c. i.egcd(gcd a b)c=gcd a b
$endgroup$
– H.PWiz
Jan 3 at 14:01
$begingroup$
@HPWiz: Ah yes,of course, thanks! Will edit later when not on mobile..
$endgroup$
– BMO
Jan 3 at 14:07
add a comment |
$begingroup$
Haskell, 48 38 29 bytes
-10 bytes due to TFeld's gcd trick!
-7 bytes thanks to HPWiz for improving the co-primality test and spotting a superfluous space!
-2 bytes thanks to nimi for suggesting an infix-operator!
(a!b)c=a<b&&a+b==c&&gcd a b<2
Try it online!
Explanation
The first two conditions a < b and a + b == c are fairly obvious, the third one uses that $gcd(a,b) = gcd(a,c) = gcd(b,c)$:
Writing $gcd(a,c) = U cdot a + V cdot c$ using Bézout's identity and substituting $c = a + b$ gives:
$$
U cdot a + V cdot (a + b) = (U + V) cdot a + V cdot b
$$
Since the $gcd$ is the minimal positive solution to that identity it follows that $gcd(a,b) = gcd(a,c)$. The other case is symmetric.
answered Jan 3 at 12:27
BMOBMO
11.8k22188
$endgroup$
1
$begingroup$
Also, I believe you only need thatgcd a b==1. Sincegcd a bdividesa+b=c. i.egcd(gcd a b)c=gcd a b
$endgroup$
– H.PWiz
Jan 3 at 14:01
$begingroup$
@HPWiz: Ah yes,of course, thanks! Will edit later when not on mobile..
$endgroup$
– BMO
Jan 3 at 14:07
add a comment |
$begingroup$
Haskell, 48 38 29 bytes
-10 bytes due to TFeld's gcd trick!
-7 bytes thanks to HPWiz for improving the co-primality test and spotting a superfluous space!
-2 bytes thanks to nimi for suggesting an infix-operator!
(a!b)c=a<b&&a+b==c&&gcd a b<2
Try it online!
Explanation
The first two conditions a < b and a + b == c are fairly obvious, the third one uses that $gcd(a,b) = gcd(a,c) = gcd(b,c)$:
Writing $gcd(a,c) = U cdot a + V cdot c$ using Bézout's identity and substituting $c = a + b$ gives:
$$
U cdot a + V cdot (a + b) = (U + V) cdot a + V cdot b
$$
Since the $gcd$ is the minimal positive solution to that identity it follows that $gcd(a,b) = gcd(a,c)$. The other case is symmetric.
answered Jan 3 at 12:27
BMOBMO
11.8k22188
$endgroup$
Haskell, 48 38 29 bytes
-10 bytes due to TFeld's gcd trick!
-7 bytes thanks to HPWiz for improving the co-primality test and spotting a superfluous space!
-2 bytes thanks to nimi for suggesting an infix-operator!
(a!b)c=a<b&&a+b==c&&gcd a b<2
Try it online!
Explanation
The first two conditions a < b and a + b == c are fairly obvious, the third one uses that $gcd(a,b) = gcd(a,c) = gcd(b,c)$:
Writing $gcd(a,c) = U cdot a + V cdot c$ using Bézout's identity and substituting $c = a + b$ gives:
$$
U cdot a + V cdot (a + b) = (U + V) cdot a + V cdot b
$$
Since the $gcd$ is the minimal positive solution to that identity it follows that $gcd(a,b) = gcd(a,c)$. The other case is symmetric.
answered Jan 3 at 12:27
BMOBMO
11.8k22188
edited Jan 3 at 17:30
answered Jan 3 at 12:27
BMOBMO
11.8k22188
answered Jan 3 at 12:27
BMOBMO
11.8k22188
answered Jan 3 at 12:27
BMOBMO
11.8k22188
11.8k22188
1
$begingroup$
Also, I believe you only need thatgcd a b==1. Sincegcd a bdividesa+b=c. i.egcd(gcd a b)c=gcd a b
$endgroup$
– H.PWiz
Jan 3 at 14:01
$begingroup$
@HPWiz: Ah yes,of course, thanks! Will edit later when not on mobile..
$endgroup$
– BMO
Jan 3 at 14:07
add a comment |
1
$begingroup$
Also, I believe you only need thatgcd a b==1. Sincegcd a bdividesa+b=c. i.egcd(gcd a b)c=gcd a b
$endgroup$
– H.PWiz
Jan 3 at 14:01
$begingroup$
@HPWiz: Ah yes,of course, thanks! Will edit later when not on mobile..
$endgroup$
– BMO
Jan 3 at 14:07
1
1
$begingroup$
Also, I believe you only need that
gcd a b==1. Since gcd a b divides a+b=c. i.e gcd(gcd a b)c=gcd a b$endgroup$
– H.PWiz
Jan 3 at 14:01
$begingroup$
Also, I believe you only need that
gcd a b==1. Since gcd a b divides a+b=c. i.e gcd(gcd a b)c=gcd a b$endgroup$
– H.PWiz
Jan 3 at 14:01
$begingroup$
@HPWiz: Ah yes,of course, thanks! Will edit later when not on mobile..
$endgroup$
– BMO
Jan 3 at 14:07
$begingroup$
@HPWiz: Ah yes,of course, thanks! Will edit later when not on mobile..
$endgroup$
– BMO
Jan 3 at 14:07
add a comment |
$begingroup$
Perl 6, 33 32 bytes
-1 byte thanks to nwellnhof
{(.sum/.[2]/2*[<] $_)==[gcd] $_}
Try it online!
Anonymous code block that takes a list of three numbers and returns True or False.
Explanation
{ } # Anonymous code block
[gcd] $_ # Is the gcd of all the numbers
( )== # Equal to
.sum # Whether the sum of numbes
/ # Is equal to
.[2]/2 # The last element doubled
*[<] $_ # And elements are in ascending order
answered Jan 3 at 12:29
Jo KingJo King
21.3k248110
$endgroup$
2
$begingroup$
32 bytes
$endgroup$
– nwellnhof
Jan 3 at 13:09
add a comment |
$begingroup$
Perl 6, 33 32 bytes
-1 byte thanks to nwellnhof
{(.sum/.[2]/2*[<] $_)==[gcd] $_}
Try it online!
Anonymous code block that takes a list of three numbers and returns True or False.
Explanation
{ } # Anonymous code block
[gcd] $_ # Is the gcd of all the numbers
( )== # Equal to
.sum # Whether the sum of numbes
/ # Is equal to
.[2]/2 # The last element doubled
*[<] $_ # And elements are in ascending order
answered Jan 3 at 12:29
Jo KingJo King
21.3k248110
$endgroup$
2
$begingroup$
32 bytes
$endgroup$
– nwellnhof
Jan 3 at 13:09
add a comment |
$begingroup$
Perl 6, 33 32 bytes
-1 byte thanks to nwellnhof
{(.sum/.[2]/2*[<] $_)==[gcd] $_}
Try it online!
Anonymous code block that takes a list of three numbers and returns True or False.
Explanation
{ } # Anonymous code block
[gcd] $_ # Is the gcd of all the numbers
( )== # Equal to
.sum # Whether the sum of numbes
/ # Is equal to
.[2]/2 # The last element doubled
*[<] $_ # And elements are in ascending order
answered Jan 3 at 12:29
Jo KingJo King
21.3k248110
$endgroup$
Perl 6, 33 32 bytes
-1 byte thanks to nwellnhof
{(.sum/.[2]/2*[<] $_)==[gcd] $_}
Try it online!
Anonymous code block that takes a list of three numbers and returns True or False.
Explanation
{ } # Anonymous code block
[gcd] $_ # Is the gcd of all the numbers
( )== # Equal to
.sum # Whether the sum of numbes
/ # Is equal to
.[2]/2 # The last element doubled
*[<] $_ # And elements are in ascending order
answered Jan 3 at 12:29
Jo KingJo King
21.3k248110
edited Jan 3 at 21:14
answered Jan 3 at 12:29
Jo KingJo King
21.3k248110
answered Jan 3 at 12:29
Jo KingJo King
21.3k248110
answered Jan 3 at 12:29
Jo KingJo King
21.3k248110
21.3k248110
2
$begingroup$
32 bytes
$endgroup$
– nwellnhof
Jan 3 at 13:09
add a comment |
2
$begingroup$
32 bytes
$endgroup$
– nwellnhof
Jan 3 at 13:09
2
2
$begingroup$
32 bytes
$endgroup$
– nwellnhof
Jan 3 at 13:09
$begingroup$
32 bytes
$endgroup$
– nwellnhof
Jan 3 at 13:09
add a comment |
$begingroup$
Java 10, 65 64 bytes
(a,b,c)->{var r=a<b&a+b==c;for(;b>0;a=b,b=c)c=a%b;return r&a<2;}
-1 byte thank to @Shaggy.
Try it online.
Explanation:
(a,b,c)->{ // Method with three integer parameters and boolean return-type
var r= // Result-boolean, starting at:
a<b // Check if `a` is smaller than `b`
&a+b==c; // And if `a+b` is equal to `c`
for(;b>0 // Then loop as long as `b` is not 0 yet
; // After every iteration:
a=b, // Set `a` to the current `b`
b=c) // And set `b` to the temp value `c`
c=a%b; // Set the temp value `c` to `a` modulo-`b`
// (we no longer need `c` at this point)
return r // Return if the boolean-result is true
&a<2;} // And `a` is now smaller than 2
answered Jan 3 at 15:04
Kevin CruijssenKevin Cruijssen
36.4k555192
$endgroup$
$begingroup$
a==1->a<2to save a byte.
$endgroup$
– Shaggy
Jan 3 at 15:09
$begingroup$
@Shaggy Thanks!
$endgroup$
– Kevin Cruijssen
Jan 3 at 16:23
add a comment |
$begingroup$
Java 10, 65 64 bytes
(a,b,c)->{var r=a<b&a+b==c;for(;b>0;a=b,b=c)c=a%b;return r&a<2;}
-1 byte thank to @Shaggy.
Try it online.
Explanation:
(a,b,c)->{ // Method with three integer parameters and boolean return-type
var r= // Result-boolean, starting at:
a<b // Check if `a` is smaller than `b`
&a+b==c; // And if `a+b` is equal to `c`
for(;b>0 // Then loop as long as `b` is not 0 yet
; // After every iteration:
a=b, // Set `a` to the current `b`
b=c) // And set `b` to the temp value `c`
c=a%b; // Set the temp value `c` to `a` modulo-`b`
// (we no longer need `c` at this point)
return r // Return if the boolean-result is true
&a<2;} // And `a` is now smaller than 2
answered Jan 3 at 15:04
Kevin CruijssenKevin Cruijssen
36.4k555192
$endgroup$
$begingroup$
a==1->a<2to save a byte.
$endgroup$
– Shaggy
Jan 3 at 15:09
$begingroup$
@Shaggy Thanks!
$endgroup$
– Kevin Cruijssen
Jan 3 at 16:23
add a comment |
$begingroup$
Java 10, 65 64 bytes
(a,b,c)->{var r=a<b&a+b==c;for(;b>0;a=b,b=c)c=a%b;return r&a<2;}
-1 byte thank to @Shaggy.
Try it online.
Explanation:
(a,b,c)->{ // Method with three integer parameters and boolean return-type
var r= // Result-boolean, starting at:
a<b // Check if `a` is smaller than `b`
&a+b==c; // And if `a+b` is equal to `c`
for(;b>0 // Then loop as long as `b` is not 0 yet
; // After every iteration:
a=b, // Set `a` to the current `b`
b=c) // And set `b` to the temp value `c`
c=a%b; // Set the temp value `c` to `a` modulo-`b`
// (we no longer need `c` at this point)
return r // Return if the boolean-result is true
&a<2;} // And `a` is now smaller than 2
answered Jan 3 at 15:04
Kevin CruijssenKevin Cruijssen
36.4k555192
$endgroup$
Java 10, 65 64 bytes
(a,b,c)->{var r=a<b&a+b==c;for(;b>0;a=b,b=c)c=a%b;return r&a<2;}
-1 byte thank to @Shaggy.
Try it online.
Explanation:
(a,b,c)->{ // Method with three integer parameters and boolean return-type
var r= // Result-boolean, starting at:
a<b // Check if `a` is smaller than `b`
&a+b==c; // And if `a+b` is equal to `c`
for(;b>0 // Then loop as long as `b` is not 0 yet
; // After every iteration:
a=b, // Set `a` to the current `b`
b=c) // And set `b` to the temp value `c`
c=a%b; // Set the temp value `c` to `a` modulo-`b`
// (we no longer need `c` at this point)
return r // Return if the boolean-result is true
&a<2;} // And `a` is now smaller than 2
answered Jan 3 at 15:04
Kevin CruijssenKevin Cruijssen
36.4k555192
edited Jan 3 at 16:23
answered Jan 3 at 15:04
Kevin CruijssenKevin Cruijssen
36.4k555192
answered Jan 3 at 15:04
Kevin CruijssenKevin Cruijssen
36.4k555192
answered Jan 3 at 15:04
Kevin CruijssenKevin Cruijssen
36.4k555192
36.4k555192
$begingroup$
a==1->a<2to save a byte.
$endgroup$
– Shaggy
Jan 3 at 15:09
$begingroup$
@Shaggy Thanks!
$endgroup$
– Kevin Cruijssen
Jan 3 at 16:23
add a comment |
$begingroup$
a==1->a<2to save a byte.
$endgroup$
– Shaggy
Jan 3 at 15:09
$begingroup$
@Shaggy Thanks!
$endgroup$
– Kevin Cruijssen
Jan 3 at 16:23
$begingroup$
a==1 -> a<2 to save a byte.$endgroup$
– Shaggy
Jan 3 at 15:09
$begingroup$
a==1 -> a<2 to save a byte.$endgroup$
– Shaggy
Jan 3 at 15:09
$begingroup$
@Shaggy Thanks!
$endgroup$
– Kevin Cruijssen
Jan 3 at 16:23
$begingroup$
@Shaggy Thanks!
$endgroup$
– Kevin Cruijssen
Jan 3 at 16:23
add a comment |
$begingroup$
05AB1E, 12 11 10 bytes
Saved 1 byte thanks to Kevin Cruijssen
ÂÆ_*`‹*¿Θ
Try it online!
or as a Test Suite
Explanation
ÂÆ # reduce a reversed copy of the input by subtraction
_ # logically negate
* # multiply with input
` # push the values of the resulting list separately to stack
# remove the top (last) value
‹ # is a < b ?
* # multiply by the input list
¿ # calculate the gcd of the result
Θ # is it true ?
answered Jan 3 at 12:41
EmignaEmigna
45.6k432139
$endgroup$
$begingroup$
Oops.. deleted my comment.. >.> So again: you can save a byte by using multiples instead of swaps with product:RÆ_*`‹*¿ΘTest Suite.
$endgroup$
– Kevin Cruijssen
Jan 3 at 14:30
$begingroup$
@KevinCruijssen: Thanks! Yeah, usually when you have that many swaps, you're doing something wrong :P
$endgroup$
– Emigna
Jan 3 at 16:45
add a comment |
$begingroup$
05AB1E, 12 11 10 bytes
Saved 1 byte thanks to Kevin Cruijssen
ÂÆ_*`‹*¿Θ
Try it online!
or as a Test Suite
Explanation
ÂÆ # reduce a reversed copy of the input by subtraction
_ # logically negate
* # multiply with input
` # push the values of the resulting list separately to stack
# remove the top (last) value
‹ # is a < b ?
* # multiply by the input list
¿ # calculate the gcd of the result
Θ # is it true ?
answered Jan 3 at 12:41
EmignaEmigna
45.6k432139
$endgroup$
$begingroup$
Oops.. deleted my comment.. >.> So again: you can save a byte by using multiples instead of swaps with product:RÆ_*`‹*¿ΘTest Suite.
$endgroup$
– Kevin Cruijssen
Jan 3 at 14:30
$begingroup$
@KevinCruijssen: Thanks! Yeah, usually when you have that many swaps, you're doing something wrong :P
$endgroup$
– Emigna
Jan 3 at 16:45
add a comment |
$begingroup$
05AB1E, 12 11 10 bytes
Saved 1 byte thanks to Kevin Cruijssen
ÂÆ_*`‹*¿Θ
Try it online!
or as a Test Suite
Explanation
ÂÆ # reduce a reversed copy of the input by subtraction
_ # logically negate
* # multiply with input
` # push the values of the resulting list separately to stack
# remove the top (last) value
‹ # is a < b ?
* # multiply by the input list
¿ # calculate the gcd of the result
Θ # is it true ?
answered Jan 3 at 12:41
EmignaEmigna
45.6k432139
$endgroup$
05AB1E, 12 11 10 bytes
Saved 1 byte thanks to Kevin Cruijssen
ÂÆ_*`‹*¿Θ
Try it online!
or as a Test Suite
Explanation
ÂÆ # reduce a reversed copy of the input by subtraction
_ # logically negate
* # multiply with input
` # push the values of the resulting list separately to stack
# remove the top (last) value
‹ # is a < b ?
* # multiply by the input list
¿ # calculate the gcd of the result
Θ # is it true ?
answered Jan 3 at 12:41
EmignaEmigna
45.6k432139
edited Jan 3 at 16:44
answered Jan 3 at 12:41
EmignaEmigna
45.6k432139
answered Jan 3 at 12:41
EmignaEmigna
45.6k432139
answered Jan 3 at 12:41
EmignaEmigna
45.6k432139
45.6k432139
$begingroup$
Oops.. deleted my comment.. >.> So again: you can save a byte by using multiples instead of swaps with product:RÆ_*`‹*¿ΘTest Suite.
$endgroup$
– Kevin Cruijssen
Jan 3 at 14:30
$begingroup$
@KevinCruijssen: Thanks! Yeah, usually when you have that many swaps, you're doing something wrong :P
$endgroup$
– Emigna
Jan 3 at 16:45
add a comment |
$begingroup$
Oops.. deleted my comment.. >.> So again: you can save a byte by using multiples instead of swaps with product:RÆ_*`‹*¿ΘTest Suite.
$endgroup$
– Kevin Cruijssen
Jan 3 at 14:30
$begingroup$
@KevinCruijssen: Thanks! Yeah, usually when you have that many swaps, you're doing something wrong :P
$endgroup$
– Emigna
Jan 3 at 16:45
$begingroup$
Oops.. deleted my comment.. >.> So again: you can save a byte by using multiples instead of swaps with product:
RÆ_*`‹*¿Θ Test Suite.$endgroup$
– Kevin Cruijssen
Jan 3 at 14:30
$begingroup$
Oops.. deleted my comment.. >.> So again: you can save a byte by using multiples instead of swaps with product:
RÆ_*`‹*¿Θ Test Suite.$endgroup$
– Kevin Cruijssen
Jan 3 at 14:30
$begingroup$
@KevinCruijssen: Thanks! Yeah, usually when you have that many swaps, you're doing something wrong :P
$endgroup$
– Emigna
Jan 3 at 16:45
$begingroup$
@KevinCruijssen: Thanks! Yeah, usually when you have that many swaps, you're doing something wrong :P
$endgroup$
– Emigna
Jan 3 at 16:45
add a comment |
$begingroup$
Excel, 33 bytes
=AND(A1+B1=C1,GCD(A1:C1)=1,A1<B1)
answered Jan 3 at 14:27
WernischWernisch
1,608317
$endgroup$
add a comment |
$begingroup$
Excel, 33 bytes
=AND(A1+B1=C1,GCD(A1:C1)=1,A1<B1)
answered Jan 3 at 14:27
WernischWernisch
1,608317
$endgroup$
add a comment |
$begingroup$
Excel, 33 bytes
=AND(A1+B1=C1,GCD(A1:C1)=1,A1<B1)
answered Jan 3 at 14:27
WernischWernisch
1,608317
$endgroup$
Excel, 33 bytes
=AND(A1+B1=C1,GCD(A1:C1)=1,A1<B1)
answered Jan 3 at 14:27
WernischWernisch
1,608317
answered Jan 3 at 14:27
WernischWernisch
1,608317
answered Jan 3 at 14:27
WernischWernisch
1,608317
answered Jan 3 at 14:27
WernischWernisch
1,608317
1,608317
add a comment |
add a comment |
$begingroup$
Python 2, 69 67 63 62 55 bytes
lambda a,b,c:(c-b==a<b)/gcd(a,b)
from fractions import*
Try it online!
Python 3, 58 51 bytes
lambda a,b,c:(c-b==a<b)==gcd(a,b)
from math import*
Try it online!
-7 bytes, thanks to H.PWiz
answered Jan 3 at 12:28
TFeldTFeld
14.5k21240
$endgroup$
$begingroup$
is thegcdingcdtrick valid? What ifais not coprime withc?
$endgroup$
– Jo King
Jan 3 at 12:31
2
$begingroup$
@jo-king If p divides a and c, it should divide c-a so b.
$endgroup$
– david
Jan 3 at 12:41
2
$begingroup$
@JoKing: It is in this case, but not in general (you can prove it via Bezout's identity).
$endgroup$
– BMO
Jan 3 at 12:42
$begingroup$
You can take it one step further and usegcd(a,b), sincegcd(a,b)dividesa+b
$endgroup$
– H.PWiz
Jan 3 at 14:04
$begingroup$
@H.PWiz Thanks :)
$endgroup$
– TFeld
Jan 3 at 14:37
|
show 1 more comment
$begingroup$
Python 2, 69 67 63 62 55 bytes
lambda a,b,c:(c-b==a<b)/gcd(a,b)
from fractions import*
Try it online!
Python 3, 58 51 bytes
lambda a,b,c:(c-b==a<b)==gcd(a,b)
from math import*
Try it online!
-7 bytes, thanks to H.PWiz
answered Jan 3 at 12:28
TFeldTFeld
14.5k21240
$endgroup$
$begingroup$
is thegcdingcdtrick valid? What ifais not coprime withc?
$endgroup$
– Jo King
Jan 3 at 12:31
2
$begingroup$
@jo-king If p divides a and c, it should divide c-a so b.
$endgroup$
– david
Jan 3 at 12:41
2
$begingroup$
@JoKing: It is in this case, but not in general (you can prove it via Bezout's identity).
$endgroup$
– BMO
Jan 3 at 12:42
$begingroup$
You can take it one step further and usegcd(a,b), sincegcd(a,b)dividesa+b
$endgroup$
– H.PWiz
Jan 3 at 14:04
$begingroup$
@H.PWiz Thanks :)
$endgroup$
– TFeld
Jan 3 at 14:37
|
show 1 more comment
$begingroup$
Python 2, 69 67 63 62 55 bytes
lambda a,b,c:(c-b==a<b)/gcd(a,b)
from fractions import*
Try it online!
Python 3, 58 51 bytes
lambda a,b,c:(c-b==a<b)==gcd(a,b)
from math import*
Try it online!
-7 bytes, thanks to H.PWiz
answered Jan 3 at 12:28
TFeldTFeld
14.5k21240
$endgroup$
Python 2, 69 67 63 62 55 bytes
lambda a,b,c:(c-b==a<b)/gcd(a,b)
from fractions import*
Try it online!
Python 3, 58 51 bytes
lambda a,b,c:(c-b==a<b)==gcd(a,b)
from math import*
Try it online!
-7 bytes, thanks to H.PWiz
answered Jan 3 at 12:28
TFeldTFeld
14.5k21240
edited Jan 3 at 14:37
answered Jan 3 at 12:28
TFeldTFeld
14.5k21240
answered Jan 3 at 12:28
TFeldTFeld
14.5k21240
answered Jan 3 at 12:28
TFeldTFeld
14.5k21240
14.5k21240
$begingroup$
is thegcdingcdtrick valid? What ifais not coprime withc?
$endgroup$
– Jo King
Jan 3 at 12:31
2
$begingroup$
@jo-king If p divides a and c, it should divide c-a so b.
$endgroup$
– david
Jan 3 at 12:41
2
$begingroup$
@JoKing: It is in this case, but not in general (you can prove it via Bezout's identity).
$endgroup$
– BMO
Jan 3 at 12:42
$begingroup$
You can take it one step further and usegcd(a,b), sincegcd(a,b)dividesa+b
$endgroup$
– H.PWiz
Jan 3 at 14:04
$begingroup$
@H.PWiz Thanks :)
$endgroup$
– TFeld
Jan 3 at 14:37
|
show 1 more comment
$begingroup$
is thegcdingcdtrick valid? What ifais not coprime withc?
$endgroup$
– Jo King
Jan 3 at 12:31
2
$begingroup$
@jo-king If p divides a and c, it should divide c-a so b.
$endgroup$
– david
Jan 3 at 12:41
2
$begingroup$
@JoKing: It is in this case, but not in general (you can prove it via Bezout's identity).
$endgroup$
– BMO
Jan 3 at 12:42
$begingroup$
You can take it one step further and usegcd(a,b), sincegcd(a,b)dividesa+b
$endgroup$
– H.PWiz
Jan 3 at 14:04
$begingroup$
@H.PWiz Thanks :)
$endgroup$
– TFeld
Jan 3 at 14:37
$begingroup$
is the
gcd in gcd trick valid? What if a is not coprime with c?$endgroup$
– Jo King
Jan 3 at 12:31
$begingroup$
is the
gcd in gcd trick valid? What if a is not coprime with c?$endgroup$
– Jo King
Jan 3 at 12:31
2
2
$begingroup$
@jo-king If p divides a and c, it should divide c-a so b.
$endgroup$
– david
Jan 3 at 12:41
$begingroup$
@jo-king If p divides a and c, it should divide c-a so b.
$endgroup$
– david
Jan 3 at 12:41
2
2
$begingroup$
@JoKing: It is in this case, but not in general (you can prove it via Bezout's identity).
$endgroup$
– BMO
Jan 3 at 12:42
$begingroup$
@JoKing: It is in this case, but not in general (you can prove it via Bezout's identity).
$endgroup$
– BMO
Jan 3 at 12:42
$begingroup$
You can take it one step further and use
gcd(a,b), since gcd(a,b) divides a+b$endgroup$
– H.PWiz
Jan 3 at 14:04
$begingroup$
You can take it one step further and use
gcd(a,b), since gcd(a,b) divides a+b$endgroup$
– H.PWiz
Jan 3 at 14:04
$begingroup$
@H.PWiz Thanks :)
$endgroup$
– TFeld
Jan 3 at 14:37
$begingroup$
@H.PWiz Thanks :)
$endgroup$
– TFeld
Jan 3 at 14:37
|
show 1 more comment
$begingroup$
Japt, 16 14 13 11 bytes
<V¥yU «NÔr-
Try it
:Implicit input of integers U=A, V=B & W=C
<V :Is U less than V?
¥ :Test that for equality with
yU :The GCD of V & U
« :Logical AND with the negation of
N :The array of inputs
Ô :Reversed
r- :Reduced by subtraction
answered Jan 3 at 12:42
ShaggyShaggy
19.2k21666
$endgroup$
$begingroup$
Here is another 11 byte solution, though on closer inspection it isn't much different from yours in its actual logic.
$endgroup$
– Kamil Drakari
Jan 3 at 15:13
$begingroup$
@KamilDrakari, had a variation on that at one stage, too. It could be 10 bytes if variables were auto-inserted when>follows©.
$endgroup$
– Shaggy
Jan 3 at 18:05
add a comment |
$begingroup$
Japt, 16 14 13 11 bytes
<V¥yU «NÔr-
Try it
:Implicit input of integers U=A, V=B & W=C
<V :Is U less than V?
¥ :Test that for equality with
yU :The GCD of V & U
« :Logical AND with the negation of
N :The array of inputs
Ô :Reversed
r- :Reduced by subtraction
answered Jan 3 at 12:42
ShaggyShaggy
19.2k21666
$endgroup$
$begingroup$
Here is another 11 byte solution, though on closer inspection it isn't much different from yours in its actual logic.
$endgroup$
– Kamil Drakari
Jan 3 at 15:13
$begingroup$
@KamilDrakari, had a variation on that at one stage, too. It could be 10 bytes if variables were auto-inserted when>follows©.
$endgroup$
– Shaggy
Jan 3 at 18:05
add a comment |
$begingroup$
Japt, 16 14 13 11 bytes
<V¥yU «NÔr-
Try it
:Implicit input of integers U=A, V=B & W=C
<V :Is U less than V?
¥ :Test that for equality with
yU :The GCD of V & U
« :Logical AND with the negation of
N :The array of inputs
Ô :Reversed
r- :Reduced by subtraction
answered Jan 3 at 12:42
ShaggyShaggy
19.2k21666
$endgroup$
Japt, 16 14 13 11 bytes
<V¥yU «NÔr-
Try it
:Implicit input of integers U=A, V=B & W=C
<V :Is U less than V?
¥ :Test that for equality with
yU :The GCD of V & U
« :Logical AND with the negation of
N :The array of inputs
Ô :Reversed
r- :Reduced by subtraction
answered Jan 3 at 12:42
ShaggyShaggy
19.2k21666
edited Jan 3 at 14:55
answered Jan 3 at 12:42
ShaggyShaggy
19.2k21666
answered Jan 3 at 12:42
ShaggyShaggy
19.2k21666
answered Jan 3 at 12:42
ShaggyShaggy
19.2k21666
19.2k21666
$begingroup$
Here is another 11 byte solution, though on closer inspection it isn't much different from yours in its actual logic.
$endgroup$
– Kamil Drakari
Jan 3 at 15:13
$begingroup$
@KamilDrakari, had a variation on that at one stage, too. It could be 10 bytes if variables were auto-inserted when>follows©.
$endgroup$
– Shaggy
Jan 3 at 18:05
add a comment |
$begingroup$
Here is another 11 byte solution, though on closer inspection it isn't much different from yours in its actual logic.
$endgroup$
– Kamil Drakari
Jan 3 at 15:13
$begingroup$
@KamilDrakari, had a variation on that at one stage, too. It could be 10 bytes if variables were auto-inserted when>follows©.
$endgroup$
– Shaggy
Jan 3 at 18:05
$begingroup$
Here is another 11 byte solution, though on closer inspection it isn't much different from yours in its actual logic.
$endgroup$
– Kamil Drakari
Jan 3 at 15:13
$begingroup$
Here is another 11 byte solution, though on closer inspection it isn't much different from yours in its actual logic.
$endgroup$
– Kamil Drakari
Jan 3 at 15:13
$begingroup$
@KamilDrakari, had a variation on that at one stage, too. It could be 10 bytes if variables were auto-inserted when
> follows ©.$endgroup$
– Shaggy
Jan 3 at 18:05
$begingroup$
@KamilDrakari, had a variation on that at one stage, too. It could be 10 bytes if variables were auto-inserted when
> follows ©.$endgroup$
– Shaggy
Jan 3 at 18:05
add a comment |
$begingroup$
JavaScript (ES6), 54 43 42 40 bytes
Thanks to @Shaggy for pointing out that we don't need to compute $gcd(a,c)$. Saved 11 bytes by rewriting the code accordingly.
Takes input as 3 separate integers. Returns $true$ for an ABC-triple, or either $0$ or $false$ otherwise.
f=(a,b,c)=>c&&a/b|a+b-c?0:b?f(b,a%b):a<2
Try it online!
answered Jan 3 at 12:18
ArnauldArnauld
73.4k689308
$endgroup$
1
$begingroup$
I don't think you need to testgcd(c,a).
$endgroup$
– Shaggy
Jan 3 at 15:13
$begingroup$
@Shaggy Thanks! I've rewritten the code entirely.
$endgroup$
– Arnauld
Jan 3 at 15:47
add a comment |
$begingroup$
JavaScript (ES6), 54 43 42 40 bytes
Thanks to @Shaggy for pointing out that we don't need to compute $gcd(a,c)$. Saved 11 bytes by rewriting the code accordingly.
Takes input as 3 separate integers. Returns $true$ for an ABC-triple, or either $0$ or $false$ otherwise.
f=(a,b,c)=>c&&a/b|a+b-c?0:b?f(b,a%b):a<2
Try it online!
answered Jan 3 at 12:18
ArnauldArnauld
73.4k689308
$endgroup$
1
$begingroup$
I don't think you need to testgcd(c,a).
$endgroup$
– Shaggy
Jan 3 at 15:13
$begingroup$
@Shaggy Thanks! I've rewritten the code entirely.
$endgroup$
– Arnauld
Jan 3 at 15:47
add a comment |
$begingroup$
JavaScript (ES6), 54 43 42 40 bytes
Thanks to @Shaggy for pointing out that we don't need to compute $gcd(a,c)$. Saved 11 bytes by rewriting the code accordingly.
Takes input as 3 separate integers. Returns $true$ for an ABC-triple, or either $0$ or $false$ otherwise.
f=(a,b,c)=>c&&a/b|a+b-c?0:b?f(b,a%b):a<2
Try it online!
answered Jan 3 at 12:18
ArnauldArnauld
73.4k689308
$endgroup$
JavaScript (ES6), 54 43 42 40 bytes
Thanks to @Shaggy for pointing out that we don't need to compute $gcd(a,c)$. Saved 11 bytes by rewriting the code accordingly.
Takes input as 3 separate integers. Returns $true$ for an ABC-triple, or either $0$ or $false$ otherwise.
f=(a,b,c)=>c&&a/b|a+b-c?0:b?f(b,a%b):a<2
Try it online!
answered Jan 3 at 12:18
ArnauldArnauld
73.4k689308
edited Jan 3 at 16:05
answered Jan 3 at 12:18
ArnauldArnauld
73.4k689308
answered Jan 3 at 12:18
ArnauldArnauld
73.4k689308
answered Jan 3 at 12:18
ArnauldArnauld
73.4k689308
73.4k689308
1
$begingroup$
I don't think you need to testgcd(c,a).
$endgroup$
– Shaggy
Jan 3 at 15:13
$begingroup$
@Shaggy Thanks! I've rewritten the code entirely.
$endgroup$
– Arnauld
Jan 3 at 15:47
add a comment |
1
$begingroup$
I don't think you need to testgcd(c,a).
$endgroup$
– Shaggy
Jan 3 at 15:13
$begingroup$
@Shaggy Thanks! I've rewritten the code entirely.
$endgroup$
– Arnauld
Jan 3 at 15:47
1
1
$begingroup$
I don't think you need to test
gcd(c,a).$endgroup$
– Shaggy
Jan 3 at 15:13
$begingroup$
I don't think you need to test
gcd(c,a).$endgroup$
– Shaggy
Jan 3 at 15:13
$begingroup$
@Shaggy Thanks! I've rewritten the code entirely.
$endgroup$
– Arnauld
Jan 3 at 15:47
$begingroup$
@Shaggy Thanks! I've rewritten the code entirely.
$endgroup$
– Arnauld
Jan 3 at 15:47
add a comment |
$begingroup$
bash, 61 bytes
factor $@|grep -vzP '( .+b).*n.*1b'&&(($1<$2&&$1+$2==$3))
Try it online!
Input as command line arguments,
output in the exit code
(also produces output on stdout as a side effect, but this can be ignored).
The second part (starting from &&(() is pretty standard,
but the interesting bit is the coprime test:
factor $@ # produces output of the form "6: 2 3n8: 2 2 2n14: 2 7n"
|grep - # regex search on the result
v # invert the match (return truthy for strings that don't match)
z # zero-terminated, allowing us to match newlines
P # perl (extended) regex
'( .+b)' # match one or more full factors
'.*n.*' # and somewhere on the next line...
'1b' # find the same full factors
answered Jan 3 at 15:13
Doorknob♦Doorknob
54.3k17113347
$endgroup$
add a comment |
$begingroup$
bash, 61 bytes
factor $@|grep -vzP '( .+b).*n.*1b'&&(($1<$2&&$1+$2==$3))
Try it online!
Input as command line arguments,
output in the exit code
(also produces output on stdout as a side effect, but this can be ignored).
The second part (starting from &&(() is pretty standard,
but the interesting bit is the coprime test:
factor $@ # produces output of the form "6: 2 3n8: 2 2 2n14: 2 7n"
|grep - # regex search on the result
v # invert the match (return truthy for strings that don't match)
z # zero-terminated, allowing us to match newlines
P # perl (extended) regex
'( .+b)' # match one or more full factors
'.*n.*' # and somewhere on the next line...
'1b' # find the same full factors
answered Jan 3 at 15:13
Doorknob♦Doorknob
54.3k17113347
$endgroup$
add a comment |
$begingroup$
bash, 61 bytes
factor $@|grep -vzP '( .+b).*n.*1b'&&(($1<$2&&$1+$2==$3))
Try it online!
Input as command line arguments,
output in the exit code
(also produces output on stdout as a side effect, but this can be ignored).
The second part (starting from &&(() is pretty standard,
but the interesting bit is the coprime test:
factor $@ # produces output of the form "6: 2 3n8: 2 2 2n14: 2 7n"
|grep - # regex search on the result
v # invert the match (return truthy for strings that don't match)
z # zero-terminated, allowing us to match newlines
P # perl (extended) regex
'( .+b)' # match one or more full factors
'.*n.*' # and somewhere on the next line...
'1b' # find the same full factors
answered Jan 3 at 15:13
Doorknob♦Doorknob
54.3k17113347
$endgroup$
bash, 61 bytes
factor $@|grep -vzP '( .+b).*n.*1b'&&(($1<$2&&$1+$2==$3))
Try it online!
Input as command line arguments,
output in the exit code
(also produces output on stdout as a side effect, but this can be ignored).
The second part (starting from &&(() is pretty standard,
but the interesting bit is the coprime test:
factor $@ # produces output of the form "6: 2 3n8: 2 2 2n14: 2 7n"
|grep - # regex search on the result
v # invert the match (return truthy for strings that don't match)
z # zero-terminated, allowing us to match newlines
P # perl (extended) regex
'( .+b)' # match one or more full factors
'.*n.*' # and somewhere on the next line...
'1b' # find the same full factors
answered Jan 3 at 15:13
Doorknob♦Doorknob
54.3k17113347
answered Jan 3 at 15:13
Doorknob♦Doorknob
54.3k17113347
answered Jan 3 at 15:13
Doorknob♦Doorknob
54.3k17113347
answered Jan 3 at 15:13
Doorknob♦Doorknob
54.3k17113347
54.3k17113347
add a comment |
add a comment |
$begingroup$
Wolfram Language 24 30 28 bytes
With 2 bytes saved by Doorknob,
#<#2&&CoprimeQ@##&&#+#2==#3&
answered Jan 3 at 14:24
DavidCDavidC
23.9k243102
$endgroup$
$begingroup$
I think you should also be able to useCoprimeQ@##to save 2 bytes.
$endgroup$
– Doorknob♦
Jan 3 at 16:48
$begingroup$
@Doorknob, If the first and second numbers are coprime, are they necessarily coprime with their sum?
$endgroup$
– DavidC
Jan 3 at 17:27
$begingroup$
They are, but the original definition actually states that A, B, and C should be coprime. Most answers check only A and B just because it's usually shorter.
$endgroup$
– Doorknob♦
Jan 3 at 17:32
add a comment |
$begingroup$
Wolfram Language 24 30 28 bytes
With 2 bytes saved by Doorknob,
#<#2&&CoprimeQ@##&&#+#2==#3&
answered Jan 3 at 14:24
DavidCDavidC
23.9k243102
$endgroup$
$begingroup$
I think you should also be able to useCoprimeQ@##to save 2 bytes.
$endgroup$
– Doorknob♦
Jan 3 at 16:48
$begingroup$
@Doorknob, If the first and second numbers are coprime, are they necessarily coprime with their sum?
$endgroup$
– DavidC
Jan 3 at 17:27
$begingroup$
They are, but the original definition actually states that A, B, and C should be coprime. Most answers check only A and B just because it's usually shorter.
$endgroup$
– Doorknob♦
Jan 3 at 17:32
add a comment |
$begingroup$
Wolfram Language 24 30 28 bytes
With 2 bytes saved by Doorknob,
#<#2&&CoprimeQ@##&&#+#2==#3&
answered Jan 3 at 14:24
DavidCDavidC
23.9k243102
$endgroup$
Wolfram Language 24 30 28 bytes
With 2 bytes saved by Doorknob,
#<#2&&CoprimeQ@##&&#+#2==#3&
answered Jan 3 at 14:24
DavidCDavidC
23.9k243102
edited Jan 3 at 17:46
answered Jan 3 at 14:24
DavidCDavidC
23.9k243102
answered Jan 3 at 14:24
DavidCDavidC
23.9k243102
answered Jan 3 at 14:24
DavidCDavidC
23.9k243102
23.9k243102
$begingroup$
I think you should also be able to useCoprimeQ@##to save 2 bytes.
$endgroup$
– Doorknob♦
Jan 3 at 16:48
$begingroup$
@Doorknob, If the first and second numbers are coprime, are they necessarily coprime with their sum?
$endgroup$
– DavidC
Jan 3 at 17:27
$begingroup$
They are, but the original definition actually states that A, B, and C should be coprime. Most answers check only A and B just because it's usually shorter.
$endgroup$
– Doorknob♦
Jan 3 at 17:32
add a comment |
$begingroup$
I think you should also be able to useCoprimeQ@##to save 2 bytes.
$endgroup$
– Doorknob♦
Jan 3 at 16:48
$begingroup$
@Doorknob, If the first and second numbers are coprime, are they necessarily coprime with their sum?
$endgroup$
– DavidC
Jan 3 at 17:27
$begingroup$
They are, but the original definition actually states that A, B, and C should be coprime. Most answers check only A and B just because it's usually shorter.
$endgroup$
– Doorknob♦
Jan 3 at 17:32
$begingroup$
I think you should also be able to use
CoprimeQ@## to save 2 bytes.$endgroup$
– Doorknob♦
Jan 3 at 16:48
$begingroup$
I think you should also be able to use
CoprimeQ@## to save 2 bytes.$endgroup$
– Doorknob♦
Jan 3 at 16:48
$begingroup$
@Doorknob, If the first and second numbers are coprime, are they necessarily coprime with their sum?
$endgroup$
– DavidC
Jan 3 at 17:27
$begingroup$
@Doorknob, If the first and second numbers are coprime, are they necessarily coprime with their sum?
$endgroup$
– DavidC
Jan 3 at 17:27
$begingroup$
They are, but the original definition actually states that A, B, and C should be coprime. Most answers check only A and B just because it's usually shorter.
$endgroup$
– Doorknob♦
Jan 3 at 17:32
$begingroup$
They are, but the original definition actually states that A, B, and C should be coprime. Most answers check only A and B just because it's usually shorter.
$endgroup$
– Doorknob♦
Jan 3 at 17:32
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 59 bytes
(a,b,c)=>Enumerable.Range(2,a).All(i=>a%i+b%i>0)&a<b&a+b==c
Try it online!
answered Jan 3 at 17:19
danadana
63145
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 59 bytes
(a,b,c)=>Enumerable.Range(2,a).All(i=>a%i+b%i>0)&a<b&a+b==c
Try it online!
answered Jan 3 at 17:19
danadana
63145
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 59 bytes
(a,b,c)=>Enumerable.Range(2,a).All(i=>a%i+b%i>0)&a<b&a+b==c
Try it online!
answered Jan 3 at 17:19
danadana
63145
$endgroup$
C# (Visual C# Interactive Compiler), 59 bytes
(a,b,c)=>Enumerable.Range(2,a).All(i=>a%i+b%i>0)&a<b&a+b==c
Try it online!
answered Jan 3 at 17:19
danadana
63145
edited Jan 3 at 18:29
answered Jan 3 at 17:19
danadana
63145
answered Jan 3 at 17:19
danadana
63145
answered Jan 3 at 17:19
danadana
63145
63145
add a comment |
add a comment |
$begingroup$
J, 27 bytes
(+/=2*{:)*({.<1{])*1=+./ .*
Try it online!
Inspired by Jo King's Perl solution
answered Jan 3 at 13:07
Galen IvanovGalen Ivanov
6,49711032
$endgroup$
add a comment |
$begingroup$
J, 27 bytes
(+/=2*{:)*({.<1{])*1=+./ .*
Try it online!
Inspired by Jo King's Perl solution
answered Jan 3 at 13:07
Galen IvanovGalen Ivanov
6,49711032
$endgroup$
add a comment |
$begingroup$
J, 27 bytes
(+/=2*{:)*({.<1{])*1=+./ .*
Try it online!
Inspired by Jo King's Perl solution
answered Jan 3 at 13:07
Galen IvanovGalen Ivanov
6,49711032
$endgroup$
J, 27 bytes
(+/=2*{:)*({.<1{])*1=+./ .*
Try it online!
Inspired by Jo King's Perl solution
answered Jan 3 at 13:07
Galen IvanovGalen Ivanov
6,49711032
edited Jan 3 at 13:14
answered Jan 3 at 13:07
Galen IvanovGalen Ivanov
6,49711032
answered Jan 3 at 13:07
Galen IvanovGalen Ivanov
6,49711032
answered Jan 3 at 13:07
Galen IvanovGalen Ivanov
6,49711032
6,49711032
add a comment |
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 90 bytes
n=>new int[(int)1e8].Where((_,b)=>n[0]%++b<1&n[1]%b<1).Count()<2&n[0]+n[1]==n[2]&n[0]<n[1]
Runs for numbers up to 1e8, takes about 35 seconds on my machine. Instead of calculating the gcd like others, the function just instantiate a huge array and filter the indexes that aren't divisors of a or b, and check how many elements are left. Next it check if element one plus element two equals element three. Lastly, it checks if the first element is less than the second.
Try it online!
answered Jan 3 at 16:54
Embodiment of IgnoranceEmbodiment of Ignorance
631115
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 90 bytes
n=>new int[(int)1e8].Where((_,b)=>n[0]%++b<1&n[1]%b<1).Count()<2&n[0]+n[1]==n[2]&n[0]<n[1]
Runs for numbers up to 1e8, takes about 35 seconds on my machine. Instead of calculating the gcd like others, the function just instantiate a huge array and filter the indexes that aren't divisors of a or b, and check how many elements are left. Next it check if element one plus element two equals element three. Lastly, it checks if the first element is less than the second.
Try it online!
answered Jan 3 at 16:54
Embodiment of IgnoranceEmbodiment of Ignorance
631115
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 90 bytes
n=>new int[(int)1e8].Where((_,b)=>n[0]%++b<1&n[1]%b<1).Count()<2&n[0]+n[1]==n[2]&n[0]<n[1]
Runs for numbers up to 1e8, takes about 35 seconds on my machine. Instead of calculating the gcd like others, the function just instantiate a huge array and filter the indexes that aren't divisors of a or b, and check how many elements are left. Next it check if element one plus element two equals element three. Lastly, it checks if the first element is less than the second.
Try it online!
answered Jan 3 at 16:54
Embodiment of IgnoranceEmbodiment of Ignorance
631115
$endgroup$
C# (Visual C# Interactive Compiler), 90 bytes
n=>new int[(int)1e8].Where((_,b)=>n[0]%++b<1&n[1]%b<1).Count()<2&n[0]+n[1]==n[2]&n[0]<n[1]
Runs for numbers up to 1e8, takes about 35 seconds on my machine. Instead of calculating the gcd like others, the function just instantiate a huge array and filter the indexes that aren't divisors of a or b, and check how many elements are left. Next it check if element one plus element two equals element three. Lastly, it checks if the first element is less than the second.
Try it online!
answered Jan 3 at 16:54
Embodiment of IgnoranceEmbodiment of Ignorance
631115
answered Jan 3 at 16:54
Embodiment of IgnoranceEmbodiment of Ignorance
631115
answered Jan 3 at 16:54
Embodiment of IgnoranceEmbodiment of Ignorance
631115
answered Jan 3 at 16:54
Embodiment of IgnoranceEmbodiment of Ignorance
631115
631115
add a comment |
add a comment |
$begingroup$
C# (.NET Core), 68 bytes
Without Linq.
(a,b,c)=>{var t=a<b&a+b==c;while(b>0){c=b;b=a%b;a=c;}return t&a<2;};
Try it online!
answered Jan 3 at 17:26
DestroigoDestroigo
1815
$endgroup$
add a comment |
$begingroup$
C# (.NET Core), 68 bytes
Without Linq.
(a,b,c)=>{var t=a<b&a+b==c;while(b>0){c=b;b=a%b;a=c;}return t&a<2;};
Try it online!
answered Jan 3 at 17:26
DestroigoDestroigo
1815
$endgroup$
add a comment |
$begingroup$
C# (.NET Core), 68 bytes
Without Linq.
(a,b,c)=>{var t=a<b&a+b==c;while(b>0){c=b;b=a%b;a=c;}return t&a<2;};
Try it online!
answered Jan 3 at 17:26
DestroigoDestroigo
1815
$endgroup$
C# (.NET Core), 68 bytes
Without Linq.
(a,b,c)=>{var t=a<b&a+b==c;while(b>0){c=b;b=a%b;a=c;}return t&a<2;};
Try it online!
answered Jan 3 at 17:26
DestroigoDestroigo
1815
answered Jan 3 at 17:26
DestroigoDestroigo
1815
answered Jan 3 at 17:26
DestroigoDestroigo
1815
answered Jan 3 at 17:26
DestroigoDestroigo
1815
1815
add a comment |
add a comment |
$begingroup$
Stax, 12 bytes
ü╡v╕7+Pü°╔|g
Run and debug it
answered Jan 3 at 17:37
wastlwastl
2,084425
$endgroup$
add a comment |
$begingroup$
Stax, 12 bytes
ü╡v╕7+Pü°╔|g
Run and debug it
answered Jan 3 at 17:37
wastlwastl
2,084425
$endgroup$
add a comment |
$begingroup$
Stax, 12 bytes
ü╡v╕7+Pü°╔|g
Run and debug it
answered Jan 3 at 17:37
wastlwastl
2,084425
$endgroup$
Stax, 12 bytes
ü╡v╕7+Pü°╔|g
Run and debug it
answered Jan 3 at 17:37
wastlwastl
2,084425
answered Jan 3 at 17:37
wastlwastl
2,084425
answered Jan 3 at 17:37
wastlwastl
2,084425
answered Jan 3 at 17:37
wastlwastl
2,084425
2,084425
add a comment |
add a comment |
$begingroup$
Pari/GP, 30 bytes
Saved 2 bytes thanks to @Shaggy.
(a,b,c)->a<b==gcd(a,b)&&a+b==c
Try it online!
answered Jan 3 at 16:23
alephalphaalephalpha
21.2k32991
$endgroup$
1
$begingroup$
30 bytes(?)
$endgroup$
– Shaggy
Jan 3 at 20:13
add a comment |
$begingroup$
Pari/GP, 30 bytes
Saved 2 bytes thanks to @Shaggy.
(a,b,c)->a<b==gcd(a,b)&&a+b==c
Try it online!
answered Jan 3 at 16:23
alephalphaalephalpha
21.2k32991
$endgroup$
1
$begingroup$
30 bytes(?)
$endgroup$
– Shaggy
Jan 3 at 20:13
add a comment |
$begingroup$
Pari/GP, 30 bytes
Saved 2 bytes thanks to @Shaggy.
(a,b,c)->a<b==gcd(a,b)&&a+b==c
Try it online!
answered Jan 3 at 16:23
alephalphaalephalpha
21.2k32991
$endgroup$
Pari/GP, 30 bytes
Saved 2 bytes thanks to @Shaggy.
(a,b,c)->a<b==gcd(a,b)&&a+b==c
Try it online!
answered Jan 3 at 16:23
alephalphaalephalpha
21.2k32991
edited Jan 3 at 23:20
answered Jan 3 at 16:23
alephalphaalephalpha
21.2k32991
answered Jan 3 at 16:23
alephalphaalephalpha
21.2k32991
answered Jan 3 at 16:23
alephalphaalephalpha
21.2k32991
21.2k32991
1
$begingroup$
30 bytes(?)
$endgroup$
– Shaggy
Jan 3 at 20:13
add a comment |
1
$begingroup$
30 bytes(?)
$endgroup$
– Shaggy
Jan 3 at 20:13
1
1
$begingroup$
30 bytes(?)
$endgroup$
– Shaggy
Jan 3 at 20:13
$begingroup$
30 bytes(?)
$endgroup$
– Shaggy
Jan 3 at 20:13
add a comment |
$begingroup$
Befunge-98 (FBBI), 83 bytes
&:&:03p&:04pw>03g04g:v_1w03g04g+w1.@
00: 7j@.0[^j7 _^;>0.@;j7;>0.@;:%g00p
Try it online!
The input which is a triple of integers [A,B,C] is feeded into Befunge as space-separated integers C B A.
answered Jan 4 at 1:37
WisławWisław
1413
$endgroup$
add a comment |
$begingroup$
Befunge-98 (FBBI), 83 bytes
&:&:03p&:04pw>03g04g:v_1w03g04g+w1.@
00: 7j@.0[^j7 _^;>0.@;j7;>0.@;:%g00p
Try it online!
The input which is a triple of integers [A,B,C] is feeded into Befunge as space-separated integers C B A.
answered Jan 4 at 1:37
WisławWisław
1413
$endgroup$
add a comment |
$begingroup$
Befunge-98 (FBBI), 83 bytes
&:&:03p&:04pw>03g04g:v_1w03g04g+w1.@
00: 7j@.0[^j7 _^;>0.@;j7;>0.@;:%g00p
Try it online!
The input which is a triple of integers [A,B,C] is feeded into Befunge as space-separated integers C B A.
answered Jan 4 at 1:37
WisławWisław
1413
$endgroup$
Befunge-98 (FBBI), 83 bytes
&:&:03p&:04pw>03g04g:v_1w03g04g+w1.@
00: 7j@.0[^j7 _^;>0.@;j7;>0.@;:%g00p
Try it online!
The input which is a triple of integers [A,B,C] is feeded into Befunge as space-separated integers C B A.
answered Jan 4 at 1:37
WisławWisław
1413
answered Jan 4 at 1:37
WisławWisław
1413
answered Jan 4 at 1:37
WisławWisław
1413
answered Jan 4 at 1:37
WisławWisław
1413
1413
add a comment |
add a comment |
$begingroup$
Clean, 43 bytes
import StdEnv
$a b c=a<b&&a+b==c&&gcd a b<2
Try it online!
Similar to basically everything else because the direct test is the same.
answered Jan 3 at 17:52
ΟurousΟurous
6,55211033
$endgroup$
add a comment |
$begingroup$
Clean, 43 bytes
import StdEnv
$a b c=a<b&&a+b==c&&gcd a b<2
Try it online!
Similar to basically everything else because the direct test is the same.
answered Jan 3 at 17:52
ΟurousΟurous
6,55211033
$endgroup$
add a comment |
$begingroup$
Clean, 43 bytes
import StdEnv
$a b c=a<b&&a+b==c&&gcd a b<2
Try it online!
Similar to basically everything else because the direct test is the same.
answered Jan 3 at 17:52
ΟurousΟurous
6,55211033
$endgroup$
Clean, 43 bytes
import StdEnv
$a b c=a<b&&a+b==c&&gcd a b<2
Try it online!
Similar to basically everything else because the direct test is the same.
answered Jan 3 at 17:52
ΟurousΟurous
6,55211033
answered Jan 3 at 17:52
ΟurousΟurous
6,55211033
answered Jan 3 at 17:52
ΟurousΟurous
6,55211033
answered Jan 3 at 17:52
ΟurousΟurous
6,55211033
6,55211033
add a comment |
add a comment |
$begingroup$
Retina 0.8.2, 40 bytes
d+
$*
A`^(11+)+,1+,
^(1+),(1+1),12$
Try it online! Link includes test cases. Explanation:
d+
$*
Convert to unary.
A`^(11+)+,1+,
Check that A and B have no common factor.
^(1+),(1+1),12$
Check that A < B and A + B = C.
answered Jan 3 at 22:13
NeilNeil
79.8k744177
$endgroup$
add a comment |
$begingroup$
Retina 0.8.2, 40 bytes
d+
$*
A`^(11+)+,1+,
^(1+),(1+1),12$
Try it online! Link includes test cases. Explanation:
d+
$*
Convert to unary.
A`^(11+)+,1+,
Check that A and B have no common factor.
^(1+),(1+1),12$
Check that A < B and A + B = C.
answered Jan 3 at 22:13
NeilNeil
79.8k744177
$endgroup$
add a comment |
$begingroup$
Retina 0.8.2, 40 bytes
d+
$*
A`^(11+)+,1+,
^(1+),(1+1),12$
Try it online! Link includes test cases. Explanation:
d+
$*
Convert to unary.
A`^(11+)+,1+,
Check that A and B have no common factor.
^(1+),(1+1),12$
Check that A < B and A + B = C.
answered Jan 3 at 22:13
NeilNeil
79.8k744177
$endgroup$
Retina 0.8.2, 40 bytes
d+
$*
A`^(11+)+,1+,
^(1+),(1+1),12$
Try it online! Link includes test cases. Explanation:
d+
$*
Convert to unary.
A`^(11+)+,1+,
Check that A and B have no common factor.
^(1+),(1+1),12$
Check that A < B and A + B = C.
answered Jan 3 at 22:13
NeilNeil
79.8k744177
answered Jan 3 at 22:13
NeilNeil
79.8k744177
answered Jan 3 at 22:13
NeilNeil
79.8k744177
answered Jan 3 at 22:13
NeilNeil
79.8k744177
79.8k744177
add a comment |
add a comment |
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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$begingroup$
Does the output have to be only one of two values, or can we output different truthy/falsy values for different inputs?
$endgroup$
– Luis Mendo
Jan 3 at 12:04
$begingroup$
I think it should be consistent: your code have to output one kind of truthy/falsy values whatever the input. But the truthy/falsy couple can be what you want as far as it does the job: differentiate lists.
$endgroup$
– david
Jan 3 at 12:33
$begingroup$
If we take the input as list of three values, does the input have to be in the order
[A,B,C], or are we also allowed to take the input in the order[C,B,A]or[C,A,B]?$endgroup$
– Kevin Cruijssen
Jan 3 at 13:06
$begingroup$
You have to respect order since A < B is a criteria in the challenge.
$endgroup$
– david
Jan 3 at 13:08
1
$begingroup$
I don't think requiring a particular list order is compatible with allowing input to be taken as separate values, as separate values are inherently unordered and may be taken as a list.
$endgroup$
– Dennis♦
Jan 3 at 15:30