The modulo parity party

Question

You are given an array A of n strictly positive integers, with n ≥ 2.

Your task is to map each entry A_i to:

1 if A_j mod A_i is odd for each j such that 1 ≤ j ≤ n and j ≠ i
2 if A_j mod A_i is even for each j such that 1 ≤ j ≤ n and j ≠ i
0 otherwise (mixed parities)

Example

For A = [ 73, 50, 61 ], we have:

50 mod 73 = 50, 61 mod 73 = 61 → mixed
73 mod 50 = 23, 61 mod 50 = 11 → all odd
73 mod 61 = 12, 50 mod 61 = 50 → all even

Therefore, the expected output is [ 0, 1, 2 ].

Rules

You may use any three distinct values (of any type) instead of 0, 1 and 2 as long as they're consistent. Please specify your mapping if you're not using the one described in the challenge.
Should there be any doubt about that, zero is even.
This is code-golf, so the shortest answer in bytes wins!

Test cases

[ 1, 2 ] --> [ 2, 1 ]
[ 3, 4 ] --> [ 1, 1 ]
[ 1, 2, 3 ] --> [ 2, 1, 0 ]
[ 4, 4, 4 ] --> [ 2, 2, 2 ]
[ 73, 50, 61 ] --> [ 0, 1, 2 ]
[ 941, 459, 533 ] --> [ 1, 0, 0 ]
[ 817, 19, 928, 177 ] --> [ 1, 2, 1, 1 ]
[ 312, 463, 336, 729, 513 ] --> [ 0, 2, 0, 0, 0 ]
[ 53, 47, 33, 87, 81, 3, 17 ] --> [ 0, 0, 0, 1, 0, 2, 0 ]

Do the output values have to be integers or would [1], [0, 1], and [1, 1] work? — Dennis, Commented Feb 6, 2018 at 18:33
@Dennis Any consistent values are fine. So yes, that would work! — Arnauld, Commented Feb 6, 2018 at 18:40

Rod · Accepted Answer · 2018-02-06 18:26:29Z

9

Python 2, 68 67 66 bytes

^{-1 byte thanks to Mr. Xcoder}
^{-1 byte thanks to ovs}

x=input()
for j in x:k=sum(i%j%2for i in x);print(k<len(x)-1)+0**k

Try it online!

Returns 1,0,2 instead 0,1,2.

edited Feb 6, 2018 at 18:26

answered Feb 6, 2018 at 17:54

Rod

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\$\begingroup\$ replace (k<1) with 0**k for -1 byte. \$\endgroup\$
– ovs
Commented Feb 6, 2018 at 18:16

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Dennis · Accepted Answer · 2018-02-06 19:30:17Z

5

Jelly, 9 bytes

%þœ-€0Ḃ‘Ṭ

Returns [1, 1], [0, 1], [1] instead of 0, 1, 2.

Try it online!

How it works

%þœ-€0Ḃ‘Ṭ  Main link. Argument: A (array)

%þ           Build the modulus table.
  œ-€0       Remove one 0 from each list of moduli.
      Ḃ      Take the last bit of each.
       ‘     Increment, mapping 0 and 1 to 1 and 2.
        Ṭ    Untruth; map each array to an aray of 1's at the specified indices.
             This yields:
                 [1] if the array contains only 1's (all even).
                 [0, 1] if the array contains only 2's (all odd).
                 [1, 1] if the array contains 1's and 2's.

edited Feb 6, 2018 at 19:30

answered Feb 6, 2018 at 17:51

Dennis

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\$\begingroup\$ Could you replace ‘ṬUḄ with Q€Ḅ to save a byte? \$\endgroup\$
– Jonathan Allan
Commented Feb 6, 2018 at 18:25
\$\begingroup\$ Sadly, no. Q€ could return [0, 1] or [1, 0]. \$\endgroup\$
– Dennis
Commented Feb 6, 2018 at 18:30
\$\begingroup\$ Oh, right. I think [1], [1,1], and [0,1] are three distinct values so %þœ-€0Ḃ‘Ṭ should be acceptable for 9. EDIT - ah I see you asked this exact question :) \$\endgroup\$
– Jonathan Allan
Commented Feb 6, 2018 at 18:55
\$\begingroup\$ Another 9 byte alternative is ¹-Ƥ%"%2‘Ṭ \$\endgroup\$
– miles
Commented Feb 7, 2018 at 13:53

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Community · Accepted Answer · 2020-06-17 09:04:33Z

MATL, 12 bytes

!G\o~tAws1=-

This uses 0, -1, 1 instead of 0, 1, 2 respectively.

Try it online! Or verify all test cases.

Explanation

!    % Implicit input: row vector. Transpose into a column
G    % Push input again
\    % Modulus, element-wise with broadcast. Gives a square matrix
o    % Parity: gives 1 for odd, 0 for even
~    % Logical negate: 0 for odd, 1 for even
t    % Duplicate
A    % All: gives 1 for columns that contain only 1
w    % Swap
s    % Sum of each column
1    % Push 1
=    % Is equal? Gives 1 if the column sum was 1, 0 otherwise
-    % Subtract, element-wise. Implicit display

David Bevan · Accepted Answer · 2018-02-08 09:22:32Z

3

Mathematica, 57 49 48 bytes

(s=#;And@@#.Or@@#&@OddQ@Rest@Sort[s~Mod~#]&)/@#&

This returns:

False.True for 0 (mixed)
True.True for 1 (all odd)
False.False for 2 (all even)

Try it online!

Here's a slightly longer alternative (49 bytes):

Sign[(s=#;Tr@Mod[s~Mod~#,2]&)/@#/.Tr[1^#]-1->-1]&

This one returns:

1 for 0 (mixed)
-1 for 1 (all odd)
0 for 2 (all even)

Try it online!

edited Feb 8, 2018 at 9:22

answered Feb 7, 2018 at 17:27

David Bevan

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Jonathan Frech · Accepted Answer · 2020-05-24 22:52:20Z

3

C (gcc), 118 114 97 92 91 86 bytes

Thanks to Peter Cordes for bugfixing.
Saved ~~four~~ twenty-one bytes thanks to Peter Cordes; suggesting to use a different output value mapping; [0 1 2] ~ [3 2 1].
Saved five bytes; using yet another mapping; [0 1 2] ~ [  ].
Saved a byte; golfed for(i=0;i<n;i++,putchar... to for(i=~0;++i<n;putchar....
Saved five bytes thanks to ceilingcat.

i,j,r;f(A,n)int*A;{for(i=~0;++i<n;putchar(r))for(r=0,j=n;j--;)r|=j-i?A[j]%A[i]%2+1:0;}

Try it online!

edited May 24, 2020 at 22:52

answered Feb 6, 2018 at 20:22

Jonathan Frech

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\$\begingroup\$ Your test functions on TIO don't pass enough args, and this undefined behaviour leads to a division by zero (SIGFPE) from the last test case. f(I,7) overwrites the first element of I[] (A[] in f()) with one of the args that you're using as locals. f() assumes that arg was passed on the stack by the caller, but the caller didn't know that, and what's actually on the stack above the return address is A[0]. (i.e. this UB caused t and A[0] to have the same address). Anyway, this is only UB in your test function on TIO. \$\endgroup\$
– Peter Cordes
Commented Feb 7, 2018 at 4:02
\$\begingroup\$ And BTW, I couldn't repro the crash locally so I had to add execlp("/usr/bin/objdump", "objdump", "-drwC", "-Mintel", argv[0], 0); to main to get the asm from TIO's gcc 7.2.1, which didn't exactly match my Arch Linux gcc 7.2.1. After turning that disassembly back into asm source for the calling function, I could repro it locally inside gdb and confirm exactly what was happening. \$\endgroup\$
– Peter Cordes
Commented Feb 7, 2018 at 4:03
\$\begingroup\$ You might save bytes using a different mapping, like 1 for even, 2 for odd, 3 for mixed, so you can o|=1<<(A[j]%A[i]%2) without needing any fancy decoding for o. \$\endgroup\$
– Peter Cordes
Commented Feb 7, 2018 at 4:05
\$\begingroup\$ @PeterCordes Thank you for noting, even though I still do not fully understand why the first array entry gets overwritten. I now chose to use global variables instead of local ones, removing the undefined behaviour. \$\endgroup\$
– Jonathan Frech
Commented Feb 7, 2018 at 12:06
\$\begingroup\$ @PeterCordes I also took your golfing suggestion and managed to save four bytes. However, I do not know if this really was what you were suggesting, as you wrote o|=1<<... instead of something along the lines of o|=1<<(t=.... \$\endgroup\$
– Jonathan Frech
Commented Feb 7, 2018 at 12:08

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Galen Ivanov · Accepted Answer · 2018-02-06 20:56:50Z

2

Red, 101 bytes

g: func[b][foreach n b[a: copy[]foreach m b[append a m % n % 2]sort a a: copy next a print unique a]]

Try it online!

Returns 1 0 for mixed, 1 for odd and 0 for even

g: func[b] [
    foreach n b [
        a: copy []
        foreach m b [
            append a m % n % 2
        ]
        sort a
        a: copy next a
        print unique a
    ]
]

answered Feb 6, 2018 at 20:56

Galen Ivanov

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Community · Accepted Answer · 2020-06-17 09:04:33Z

JavaScript (ES6), 46 bytes

a=>a.map(A=>a.map(B=>d+=B%A%2,d=0)|!a[d+1]-!d)

Returns -1 (even), 1 (odd), and 0 (mixed).

How it works:

The d accumulator will be:

Zero if all even moduli. (!a[d+1] == false, !d == 1, false - 1 == -1)
One less ^* than the length of the array if all odd moduli. (^* The accumulator includes an element modulused against itself, resulting in one even modulus.) (!a[d+1] == true, !d == 0, true - 0 == 1)
Two or more less than the length of the array if a mix. (!a[d+1] == false, !d == 0, false - 0 == 0)

Test cases:

let f=

a=>a.map(A=>a.map(B=>d+=B%A%2,d=0)|!a[d+1]-!d)

console.log(f([ 1, 2 ] ) + ''); // [ -1, 1 ]
console.log(f([ 3, 4 ] ) + ''); // [ 1, 1 ]
console.log(f([ 1, 2, 3 ] ) + ''); // [ -1, 1, 0 ]
console.log(f([ 4, 4, 4 ] ) + ''); // [ -1, -1, -1 ]
console.log(f([ 73, 50, 61 ] ) + ''); // [ 0, 1, -1 ]
console.log(f([ 941, 459, 533 ] ) + ''); // [ 1, 0, 0 ]
console.log(f([ 817, 19, 928, 177 ] ) + ''); // [ 1, -1, 1, 1 ]
console.log(f([ 312, 463, 336, 729, 513 ] ) + ''); // [ 0, -1, 0, 0, 0 ]
console.log(f([ 53, 47, 33, 87, 81, 3, 17 ] ) + ''); // [ 0, 0, 0, 1, 0, -1, 0 ]

Galen Ivanov · Accepted Answer · 2018-02-06 19:51:42Z

1

J, 27 20 bytes

[:<@~.@}:@\:"1~2||/~

Try it online!

Uses [1 0] [1] [0] instead of 0 1 2

Explanation:

|/~ - makes a table with remainders:

  |/~ 73 50 61 
 0 50 61
23  0 11
12 50  0

2| odd or even? :

   2||/~ 73 50 61 
0 0 1
1 0 1
0 0 0

<@~.@}:@\:"1 - sort down, drop last element (always a zero), keep the ùnique elements and box each row:

   <@~.@}:@\:"1~2||/~ 73 50 61 
┌───┬─┬─┐
│1 0│1│0│
└───┴─┴─┘

edited Feb 6, 2018 at 19:51

answered Feb 6, 2018 at 19:22

Galen Ivanov

21.2k3 gold badges24 silver badges61 bronze badges

1

\$\begingroup\$ 16 bytes with 2/:~@:|"1]|1]\.] returning a list of pairs. \$\endgroup\$
– miles
Commented Feb 7, 2018 at 13:56
\$\begingroup\$ @ miles Thank you! Is this output acceptable? \$\endgroup\$
– Galen Ivanov
Commented Feb 7, 2018 at 14:41
\$\begingroup\$ Actually no, I missed that part about distinct values. I'll get back to it in a bit. \$\endgroup\$
– miles
Commented Feb 7, 2018 at 14:55

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Reinstate Monica -- notmaynard · Accepted Answer · 2018-02-06 19:57:38Z

1

Ruby, 58 56 bytes

->r{r.map{m=r.shift;s=r.map{|e|e%m%2}.uniq.sort;r<<m;s}}

Returns [0, 1], [1], [0] instead of 0, 1, 2 (i.e. [0] for all even , [1] for all odd, and [0, 1] for mixed).

Try it online!

edited Feb 6, 2018 at 19:57

answered Feb 6, 2018 at 19:44

Reinstate Monica -- notmaynard

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Ton Hospel · Accepted Answer · 2018-02-06 20:43:25Z

1

Perl, 38 bytes

Includes +3 for -p

#!/usr/bin/perl -p
s/\d+/$@|=$_%$&%2+1for<$`$'>;$@/gee

Outputs 1 for all even, 2 for all odd, 3 for mixed

answered Feb 6, 2018 at 20:43

Ton Hospel

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Οurous · Accepted Answer · 2018-02-07 05:01:56Z

1

Clean, 95 65 63 bytes

import StdEnv

\l=[sum(removeDup[-1^(j rem i)\\j<-l|j<>i])\\i<-l]

Try it online!

As a lambda, taking [Int] and returning [Int], mapping to:

0: mixed
1: all even
-1: all odd

edited Feb 7, 2018 at 5:01

answered Feb 6, 2018 at 21:48

Οurous

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Xcali · Accepted Answer · 2018-02-07 19:34:27Z

1

Perl 5, 71 69 + 1 (`-a`) = 70 bytes

map{//;$j=$s='';$j++-$i&&($s+=$_%$'&1)for@F;say$s?0|$s==$#F:2;$i++}@F

Try it online!

edited Feb 7, 2018 at 19:34

answered Feb 6, 2018 at 20:16

Xcali

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alephalpha · Accepted Answer · 2018-02-13 17:19:03Z

1

Pari/GP, 42 bytes

a->[[!s=vecsum([x%y%2|x<-a]),s>#a-2]|y<-a]

Returns [0, 0], [0, 1], [1, 0] instead of 0, 1, 2.

Try it online!

edited Feb 13, 2018 at 17:19

answered Feb 13, 2018 at 16:55

alephalpha

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Community · Accepted Answer · 2020-06-17 09:04:33Z

Java 8, 91 89 bytes

a->{for(int z:a){int s=1;for(int y:a)s+=y%z%2;System.out.print(" "+(s<a.length)+(s<2));}}

Using truetrue instead of 2 for even
Using falsefalse instead of 1 for odd
Using truefalse instead of 0 for mixed

Explanation:

Try it online.

a->{                      // Method with integer-array parameter and no return-type
  for(int z:a){           //  Loop over the array
    int s=1;              //   Sum-integer, starting at 1
    for(int y:a)          //   Inner loop over the array again
      s+=y%z%2;           //    Increase the sum by `y` modulo-`z` modulo-2
    System.out.print(" "  //   Print a space
      +(s<a.length)       //    + "true" if the sum is smaller than the length of the array
                          //      (this means there is at least one even)
      +(s<2));}}          //    + "true" if the sum is still 1
                          //      (this means all are even)

NikoNyrh · Accepted Answer · 2018-02-14 01:56:31Z

0

Clojure, 82 bytes

#(for[R[(range(count %))]i R](set(for[j R :when(not= i j)](odd?(mod(% j)(% i))))))

A complete example with output conversion:

(def f #(for[R[(range(count %))]i R](set(for[j R :when(not= i j)](odd?(mod(% j)(% i)))))))
(->> [ 53, 47, 33, 87, 81, 3, 17] f
     (map {#{true} 1, #{false} 2, #{true false} 0}))
; (0 0 0 1 0 2 0)

answered Feb 14, 2018 at 1:56

NikoNyrh

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Stack Exchange Network

The modulo parity party

Example

Rules

Test cases

15 Answers 15

Python 2, 68 67 66 bytes

Jelly, 9 bytes

How it works

MATL, 12 bytes

Explanation

Mathematica, 57 49 48 bytes

C (gcc), 118 114 97 92 91 86 bytes

Red, 101 bytes

JavaScript (ES6), 46 bytes

J, 27 20 bytes

Explanation:

Ruby, 58 56 bytes

Perl, 38 bytes

Clean, 95 65 63 bytes

Perl 5, 71 69 + 1 (`-a`) = 70 bytes

Pari/GP, 42 bytes

Java 8, 91 89 bytes

Clojure, 82 bytes

Your Answer

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code-golf
array
number-theory
or ask your own question.

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The modulo parity party

Example

Rules

Test cases

15 Answers 15

Python 2, 68 67 66 bytes

Jelly, 9 bytes

How it works

MATL, 12 bytes

Explanation

Mathematica, 57 49 48 bytes

C (gcc), 118 114 97 92 91 86 bytes

Red, 101 bytes

JavaScript (ES6), 46 bytes

J, 27 20 bytes

Explanation:

Ruby, 58 56 bytes

Perl, 38 bytes

Clean, 95 65 63 bytes

Perl 5, 71 69 + 1 (-a) = 70 bytes

Pari/GP, 42 bytes

Java 8, 91 89 bytes

Clojure, 82 bytes

Your Answer

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Not the answer you're looking for? Browse other questions tagged code-golfarraynumber-theory or ask your own question.

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Hot Network Questions

Perl 5, 71 69 + 1 (`-a`) = 70 bytes

Not the answer you're looking for? Browse other questions tagged
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array
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