セットは、一緒に追加されたときに2つの（必ずしも別個ではない）要素がセット自体の一部でない場合、合計がありません。

たとえば、{1, 5, 7}すべてのメンバーが奇数であり、合計2つの奇数が常に偶数であるため、合計は無料です。一方、セットのメンバーの{2, 4, 9, 13}いずれか2 + 2 = 4または合計として、合計はありません4 + 9 = 13。

入力としてセットを受け取り、セットが和のない場合はTruthy値を出力し、そうでない場合はFalsyを出力するプログラムまたは関数を作成します。

例：

Sum-free:
{}
{4}
{1, 5, 7}
{16, 1, 4, 9}

Not sum-free:
{0}
{1, 4, 5, 7}
{3, 0}
{16, 1, 4, 8}

Pyth- 8 5バイト

3バイト節約してくれた@FryAmTheEggmanに感謝します。

!@sM*

テストスイート。

!             Logical not. This makes the empty intersection true and vice versa.
 @    Q       Setwise intersection with input (implictly).
  sM          Map sum to all the pairs.
   *QQ        Get all pairs by doing cartesian product with input*input (implicit).

Python 2、41バイト

lambda s:s==s-{a+b for a in s for b in s}

s Pythonセットである必要があります。

楽しい事実：sum-free私の名前のアナグラムです。

ゼリー、5 バイト

ṗ3ḅ-P

オンラインでお試しください！

使い方

ṗ3ḅ-P  Main link. Argument: A (array)

ṗ3     Take the third Cartesian power of A, i.e., generate all triplets that
       consist of elements of A.
  ḅ-   Convert each triplet from base -1 to integer.
       This maps [a, b, c] to a - b + c = (a + c) - b.
       If (a + c) belong to A, this will yield 0 for some b.
    P  Take the product of all resulting integers. 

JavaScript、86 42 41バイト

n=>!n.some(m=>n.some(o=>n.includes(m+o)))

CᴏɴᴏʀO'B off、括弧/中括弧から大量のバイトを節約してくれてありがとう。また、関数が本来あるべき反対のブール値を返していたことを指摘してくれたニールにも感謝します。

再定義してバイト数を削減しようとしましたn.someが、残念ながらプロトタイプ関数であるため機能しません。Array.prototype.mapJSではより良い解決策があるかもしれませんが、いくつかの機能は本当に楽しいです。

.includes.indexOfのようなものを使用して1を追加するよりも短い方法があるのではないかと思っています（数値が含まれている場合は、真の値が得られます）。

テスト：

> (n=>!n.some(m=>n.some(o=>n.includes(m+o))))([1,5,7]);
true
> (n=>!n.some(m=>n.some(o=>n.includes(m+o))))([1,5,7,12]);
false

MATL、5バイト

t&+m~

これは、すべてのエントリが真である場合は真実であり、1そうでない場合は偽である配列を出力します。MATLでさまざまな真偽値を示すデモを次に示します。

オンラインで試す

説明

        % Implicitly grab input
t       % Duplicate
&+      % Compute sum of each element with every other element (2D Matrix)
m       % Check which members of the input are present in this matrix of sums
~       % Negate the result to yield a truthy value for sum-free sets
        % Implicitly display truthy/falsey value

Mathematica、23バイト

{}==#⋂Tr/@#~Tuples~2&

Haskell, 32, 30 bytes

簡単なソリューション：

f x=and[a+b/=c|a<-x,b<-x,c<-x]

Two bytes saved by @Lynn

Julia, 18 bytes

!x=x∩(x'.+x)==[]

Try it online!

J, 18 10 8 bytes

8 bytes saved thanks to miles, and 2 thanks to FrownyFrog!

-:]-.+/~

Matches the original list with the set difference of tabulated sums. This is equivalent to:

(-: (] -. +/~)) y

for input y. This translates to:

y -: (] -. +/~) y
y -: (y -. +/~ y)

+/~ returns a table of sums using y. For y =: 16 1 4 9, this gives:

   +/~ 16 1 4 9
32 17 20 25
17  2  5 10
20  5  8 13
25 10 13 18

Then, we use -., which produces a list consisting of all elements in y not in this table. If the list is sum-free, this will produce the same list. Then, -: checks for equality of lists, which produces the desired output.

Old, 18 bytes

[:-.[:>./^:_+/~e.]

+/~ creates a table of the values of the set added to itself, and e. checks if those members are in the original set. The rest of that is negating the maximal element.

Retina, 45 44 bytes

\d+
<$&$*1>
$
$`$`
M`(<1*)>.*<(1*>).*\1\2
^0

Input is a decimal list of comma-separated numbers. Output is 0 (falsy) or 1 (truthy).

Try it online! (The first line enables a linefeed-separated test suite.)

Explanation

Stage 1: Substitution

\d+
<$&$*1>

This converts all elements of the input to unary and wraps them in <...>. The purpose of the angle brackets is to distinguish a list containing only 0 from an empty list (since the unary representation of 0 is empty itself).

Stage 2: Substitution

$
$`$`

We repeat the string 3 times by append it twice at the end.

Stage 3: Match

M`(<1*)>.*<(1*>).*\1\2

We now try to find three numbers in the result such that the first two add up to the third. Those matches are counted (this doesn't actually count all such tuples, because matches cannot overlap, but if such a tuple exists it will be found). Hence, we get 0 for sum-free sets and something positive otherwise.

Stage 4: Match

^0

Since the previous stage gave the opposite of what we want, we negate the result by counting the matches of ^0 which is 1 for input 0 and 0 for everything else.

Octave, 29 21 25 bytes

@(s)~[ismember(s,s+s') 0]

Thanks to Suever! It returns an array. I added 0 at the end to make [] become sum-free. To verify truthy and falsey in Octave, you can do this:

> f=@(s)~[ismember(s,s+s') 0]

> if f([]) "sum-free" else "not sum-free" end
ans = sum-free

> if f([0]) "sum-free" else "not sum-free" end
ans = not sum-free

> if f([4]) "sum-free" else "not sum-free" end
ans = sum-free

> if f([1 3]) "sum-free" else "not sum-free" end
ans = sum-free

> if f([2 4]) "sum-free" else "not sum-free" end
ans = not sum-free

An alternative that returns 0 or 1 is:

@(s)~numel(intersect(s+s',s))

Clojure, 47 37 bytes

#(=(for[a % b % :when(%(+ a b))]a)[])

quite plain solution. uses list comprehension to find all elements which sum is equal to another element.

38 bytes variant:

#(every? nil?(for[a % b %](%(+ a b))))

Perl 6,  24 21 20  19 bytes

{not any (@_ X+@_)X==@_}
{so all (@_ X+@_)X!==@_}
{not @_ (&)(@_ X+@_)}
{not @_∩(@_ X+@_)}

{!(@_∩(@_ X+@_))}

Input is any Positional value like a List.
( a Set is an Associative so you would have to call .keys on it. )

Test:

#! /usr/bin/env perl6
use v6.c;
use Test;

my @sum-free = (
  (),
  (4,),
  (1, 5, 7),
  (16, 1, 4, 9),
);

my @not-sum-free = (
  (0,),
  (1, 4, 5, 7),
  (3, 0),
  (16, 1, 4, 8),
);

my @tests = ( |(@sum-free X=> True), |(@not-sum-free X=> False) );

plan +@tests;

# store the lambda in lexical namespace for clarity
my &sum-free-set = {!(@_∩(@_ X+@_))}

for @tests -> $_ ( :key(@list), :value($expected) ) {
  is sum-free-set(@list), $expected, .gist
}

1..8
ok 1 - () => True
ok 2 - (4) => True
ok 3 - (1 5 7) => True
ok 4 - (16 1 4 9) => True
ok 5 - (0) => False
ok 6 - (1 4 5 7) => False
ok 7 - (3 0) => False
ok 8 - (16 1 4 8) => False

Mathematica 63 62 42 bytes

This shorter version benefitted from A Simmons' submission. No element needs to be removed from the list before IntegerPartitions is applied.

If an element cannot be partitioned into two integers (each from the list), then IntegerPartitions[#,{2},#]=={} holds. And checks whether this holds for every element in the list. If so, the list is sum-free.

And@@(IntegerPartitions[#,{2},#]=={}&/@#)&

Examples

 And@@(IntegerPartitions[#,{2},#]=={}&/@ #)&@{2, 4, 9, 13}

False

 And@@(IntegerPartitions[#,{2},#]=={}&/@ #)&@{1, 5, 7}

True

There is a 2, but no odd numbers that differ by 2.

 And@@(IntegerPartitions[#,{2},#]=={}&/@#)&@{2, 3, 7, 11, 17, 23, 29, 37, 41, 47, 53, 59, 67, 71}

True

Ruby, 36 bytes

Constructs a cartesian product of the set against itself and finds the sum of all elements, then checks for intersection with the original set. Input is arrays, but in Ruby they have enough set operations to make it work out nicely anyways.

-1 byte over my original solution (used & instead of - and compared with []) because of inspiration from @feersum

Try it here!

->s{s-s.product(s).map{|x,y|x+y}==s}

Python, 40 bytes

lambda s:s^{a+b for a in s for b in s}>s

^ = symmetric difference, new set with elements in either sets but not both

> True if the left set is a superset of the right set.

Brachylog, 13 bytes

'(p:?+L:?x'L)

Explanation

'(          )  True if what's in the parentheses is impossible, false otherwise
  p            Get a permutation of Input
   :?+L        L is the list of element-wise sums of the permutation with Input
       :?x'L   There is at least one element of Input in L

R, 39 36 bytes

w<-function(s)!any(outer(s,s,'+')%in%s)

Call as w(s), where s is the set (actually vector) of values. Here is the output for some test cases:

> w(numeric(0)) # The empty set
[1] TRUE
> w(0)
[1] FALSE
> w(1)
[1] TRUE
> w(c(1, 5, 7))
[1] TRUE
> w(c(2, 4, 9, 13))
[1] FALSE

Where c() is the concatenation function that takes a bunch of values and makes it a vector.

EDIT: Making it an anonymous function to save 3 bytes, thanks to @MickyT.

function(s)!any(outer(s,s,'+')%in%s)

Racket, 58 bytes

(λ(l)(andmap(λ(m)(andmap(λ(n)(not(memq(+ n m)l)))l))l))

Explanation:

(λ(l)(andmap(λ(m)(andmap(λ(n)(not(memq(+ n m)l)))l))l))
(λ(l)                                                 ) # Define a lambda function that takes in one parameter
     (andmap(λ(m)                                  )l)  # If for all m in l
                 (andmap(λ(n)                   )l)     # If for all n in l
                             (not(memq(+ n m)l))        # n + m is not in l

05AB1E, 9 5 bytes

Saved 4 bytes thanks to Magic Octopus Urn

ãOå_P

Try it online!

Explanation

ã       # cartesian product
 O      # sum
  å     # check each if it exists in input
   _    # logical negation
    P   # product

APL, 8 bytes

⊢≡⊢~∘.+⍨

Explanation:

⊢         argument
 ≡        equals
  ⊢       argument
   ~      without 
    ∘.+⍨  sums of its elements

Test:

      ( ⊢≡⊢~∘.+⍨ ) ¨ (1 5 7)(2 4 9 13)
1 0

Haskell, 30 bytes

f s=and[x+y/=z|x<-s,y<-s,z<-s]

I think there exists a shorter solution that's more interesting, but I haven't found it.

These are 33 and 34 bytes:

f s=and$((/=)<$>s<*>)$(+)<$>s<*>s
f s|q<-((-)<$>s<*>)=all(/=0)$q$q$s

Actually, 7 bytes

;;∙♂Σ∩Y

Try it online!

;;∙♂Σ∩Y              Stack: [1,5,7]
;         duplicate         [1,5,7] [1,5,7]
 ;        duplicate         [1,5,7] [1,5,7] [1,5,7]
  ∙       cartesian product [1,5,7] [[1,1],[1,5],[1,7],[5,1],[5,5],[5,7],[7,1],[7,5],[7,7]]
   ♂Σ     sum each          [1,5,7] [2,6,8,6,10,12,8,12,14]
     ∩    intersect         []
      Y   negate            1

TSQL, 47 bytes

CREATE table T(a int)
INSERT T values(1),(5),(7),(12)

SELECT min(iif(a.a+b.a<>T.a,1,0))FROM T a,T b,T

Note: This will only run once, then the table needs to be deleted or dropped to run again. The fiddle editor doesn't allow creation of tables. Therefore the fiddle included in my answer uses 2 extra bytes to compensate for this - the fiddle version doesn't require cleanup.

Fiddle

Perl, 46 bytes

45 bytes code + 1 byte command line (-p)

$_="$_ $_ $_"!~/(\b\d++.*)((?1))(??{$1+$2})/

Uses a single regex match with Perl's support for 'code expressions' inside the regex to allow for evaluation within a match.

To get around the requirement that the input is unsorted, we repeat the input string three times. This guarantees that the result is after the two operands, and allows the same digit to be matched again (e.g. in the case of input 2 4).

Usage example:

echo "3 5 6 8" | perl -p entry.pl

Factor, 47 bytes

[ dup dup 2array [ Σ ] product-map ∩ { } = ]

• ∩ { } = is equivalent to but shorter than intersects?.
• Σ is shorter than but equivalent to sum.

Thanks, math.unicode!

testing code:

USING: arrays kernel math.unicode sequences sequences.product ;
IN: sum-free

: sum-free? ( seq -- ? )
  dup dup 2array [ Σ ] product-map ∩ { } = ;

USING: tools.test sum-free ;
IN: sum-free.tests

{ t } [ { 5 7 9 } sum-free? ] unit-test
{ f } [ { 2 4 9 13 } sum-free? ] unit-test
{ t } [ { } sum-free? ] unit-test
{ f } [ { 0 } sum-free? ] unit-test
{ t } [ { 1 } sum-free? ] unit-test
{ f } [ { 0 1 } sum-free? ] unit-test

I'm only confident the first two are correct. It's unclear from the question what the rest should be, so I think it's fine for now.

PHP, 73 bytes

+8 to turn the snippet into a program, -8 on obsolete variables thanks to insertusernamehere

<?foreach($argv as$p)foreach($argv as$q)if(in_array($p+$q,$argv))die;echo 1;

prints 1 for true, empty output for false
usage: php <filename> <value1> <value2> ...

qualified function for testing (94 86): returns 1 or nothing

function f($a){foreach($a as$p)foreach($a as$q)if(in_array($p+$q,$a))return;return 1;}

tests

function out($a){if(!is_array($a))return$a;$r=[];foreach($a as$v)$r[]=out($v);return'['.join(',',$r).']';}
function cmp($a,$b){if(is_numeric($a)&&is_numeric($b))return 1e-2<abs($a-$b);if(is_array($a)&&is_array($b)&&count($a)==count($b)){foreach($a as $v){$w = array_shift($b);if(cmp($v,$w))return true;}return false;}return strcmp($a,$b);}
function test($x,$e,$y){static $h='<table border=1><tr><th>input</th><th>output</th><th>expected</th><th>ok?</th></tr>';echo"$h<tr><td>",out($x),'</td><td>',out($y),'</td><td>',out($e),'</td><td>',cmp($e,$y)?'N':'Y',"</td></tr>";$h='';}
$samples = [
    [], 1,
    [0], false,
    [1], 1,
    [0,1], false,
    [2, 4, 9, 13], false,
    [1,5,7], 1
];
while($samples)
{
    $a=array_shift($samples);
    $e=array_shift($samples);
    test($a,$e,f($a));
}

Java, 67 bytes

s->!s.stream().anyMatch(p->s.stream().anyMatch(q->s.contains(p+q)))

Input is a Set<Integer>. Tests:

import java.util.Arrays;
import java.util.Set;
import java.util.function.Function;
import java.util.stream.Collectors;

public class SumFree {
    public static void main(String[] args) {
        sumFree(s->!s.stream()
            .anyMatch(p->s.stream()
                .anyMatch(q->s.contains(p+q)))); // formatted to avoid wrapping
    }

    public static void sumFree(Function<Set<Integer>, Boolean> func) {
        test(func);
        test(func, 4);
        test(func, 1, 5, 7);
        test(func, 16, 1, 4, 9);
        test(func, 1, 4, 5, 7);
        test(func, 0);
        test(func, 3, 0);
        test(func, 16, 1, 4, 8);
    }

    public static void test(Function<Set<Integer>, Boolean> func, Integer... vals) {
        Set<Integer> set = Arrays.stream(vals).collect(Collectors.toSet());
        System.out.format("%b %s%n", func.apply(set), set);
    }
}

Output:

true []
true [4]
true [1, 5, 7]
true [16, 1, 4, 9]
false [0]
false [1, 4, 5, 7]
false [0, 3]
false [16, 1, 4, 8]

Clojure, 34 bytes

#(not-any? %(for[i % j %](+ i j)))

I wrote this before noticing the earlier Clojure solution. Anyway, this one is more compact as it uses the input set as a pred function for not-any?.

Prolog (SWI), 66 56 49 bytes

A/B:-member(A,B).
-L:- \+ (E/L,F/L,G is E+F,G/L).

Try it online!