Drawing one-liner

Question

CodeDrawing one-liner

Teaser

Behold this formidable drawing:

Can you draw this in a single stroke? Give it a try.

Can you do this one, now:

Give it a try.

How it works

These "make this drawing with one pen stroke" problems are graph-theory problems with a rather simple solution. For each vertex, compute its degree. If all vertices have an even degree or if only two vertices have an odd degree, this is possible. In any other case, it is impossible. This is the same as finding an Eulerian path in the graph. This is also related to the very famous seven bridges of Königsberg puzzle.

For the first drawing, the degrees are

   2
 /   \
4-----4
|\   /|
|  4  |
|/   \|
3-----3

so we are in the second possible case, as all numbers are even except for the two 3s in the bottom. What is more, when we have odd numbers, we have to start in one of them and finish in the other. When everything is even, we can start wherever we want but we will finish where we started.

For the second drawing it is impossible, as there are too many odd degrees:

     2
   /   \
  5-----5
 /|\   /|\
2 |  4  | 2
 \|/   \|/
  5-----5
   \   /
     2

Task

Given the degrees of the edges as input, decide if a corresponding drawing could be done in a single pen stroke. Output Truthy if such a feat is possible and output Falsy otherwise.

Input

The degrees of the vertices in any sensible format, such as

• an array of integers like [2,4,4,4,3,3] or [2,5,5,2,4,2,5,5,2]
• separate arguments to a function, like f(2,4,4,4,3,3) or f(2,5,5,2,4,2,5,5,2)

Output

A Truthy value if all numbers are even or exactly two of them are odd, Falsy otherwise.

Test cases

Truthy

[2, 4, 4, 3, 3]
[2, 2]
[1, 2, 3, 4, 6, 8]
[2, 4, 2, 4, 2, 4, 2, 4]
[8, 6, 8, 2, 2, 3, 11]
[1, 3]
[8]

Falsy

[1, 2, 3, 4, 5, 6]
[2, 3]
[9]
[7, 2, 2, 2]
[4, 2, 7]
[3, 3, 3, 3]
[1, 2, 3, 1, 1]
[1, 1, 4, 1, 1]
[7, 7, 7, 7, 7, 7, 7, 7]

---

This is code-golf so shortest submission in bytes, wins! If you liked this challenge, consider upvoting it... And happy golfing!

Answer 1

Python, 31 bytes

lambda l:sum(n%2for n in l)|2<3

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We use k|2<3 to check that k is either 0 or 2. This works because the bit operation |2 sets the bit for place-value 2, so to get a result that's 2 or less, there must be no other bits set.

Answer 2

Wolfram Language (Mathematica), 25 bytes

(a=Tr@Mod[#,2])==0||a==2&

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Answer 3

05AB1E, 5 4 bytes

Thanks to 05AB1E's weird boolification system, any number but 1 is falsy.

-1 Thanks to Jonathan Allan

ÉO1α

Try it online!

Explanation

É     Vectorizing n % 2
 O    Sum the resulting list
  1α  |n - 1|

Answer 4

Charcoal, 8 bytes

⁼¹↔⊖Σ﹪Ａ²

Try it online! Link is to verbose version of code. Port of @ovs' 05AB1E answer. Explanation:

      Ａ     Input array
     ﹪ ²    Vectorised modulo literal 2
    Σ       Sum
   ⊖        Decremented
  ↔         Absolute
⁼¹          Equals literal 1

There are other ways to check for 0 or 2 for the same byte count, such as Not(Decremented(Absolute(Decremented(...)))), Equals(2, BitwiseOr(2, ...)), or Count("02", Cast(...)) (careful not to use "20" of course).

Answer 5

Retina 0.8.2, 18 bytes

M`[13579]\b
^[02]$

Try it online! Link includes test cases. Explanation:

M`[13579]\b

Count the number of odd numbers.

^[02]$

Compare to 0 or 2.

Answer 6

Turing Machine But Way Worse, 1245 bytes

0 0 0 1 1 0 0
0 1 0 1 2 0 0
1 2 1 1 3 0 0
1 3 1 1 4 0 0
0 4 0 1 5 0 0
1 4 1 1 5 0 0
0 5 0 1 6 0 0
1 5 1 1 6 0 0
0 6 0 1 7 0 0
1 6 1 1 7 0 0
0 7 0 1 0 0 0
1 7 1 1 8 0 0
0 8 0 1 9 0 0
0 9 0 1 a 0 0
1 a 1 1 b 0 0
1 b 1 1 c 0 0
0 c 0 1 d 0 0
1 c 1 1 d 0 0
0 d 0 1 e 0 0
1 d 1 1 e 0 0
0 e 0 1 f 0 0
1 e 1 1 f 0 0
0 f 0 1 0 0 0
1 f 1 1 8 0 0
0 b 0 1 k 0 0
0 3 0 1 4 0 0
0 g 0 1 h 0 0
0 h 0 1 i 0 0
1 i 1 1 j 0 0
1 j 1 1 k 0 0
0 k 0 1 l 0 0
1 k 1 1 l 0 0
0 l 0 1 m 0 0
1 l 1 1 m 0 0
0 m 0 1 n 0 0
1 m 1 1 n 0 0
0 n 0 1 g 0 0
1 n 1 1 o 0 0
0 o 0 1 p 0 0
0 p 0 1 q 0 0
1 q 1 1 r 0 0
1 r 1 1 s 0 0
0 s 0 1 t 0 0
1 s 1 1 t 0 0
0 t 0 1 u 0 0
1 t 1 1 u 0 0
0 u 0 1 v 0 0
1 u 1 1 v 0 0
0 v 0 1 g 0 0
1 v 1 1 o 0 0
0 r 0 1 A 0 0
0 j 0 1 k 0 0
0 w 0 1 x 0 0
0 x 0 1 y 0 0
1 y 1 1 z 0 0
1 z 1 1 A 0 0
0 A 0 1 B 0 0
1 A 1 1 B 0 0
0 B 0 1 C 0 0
1 B 1 1 C 0 0
0 C 0 1 D 0 0
1 C 1 1 D 0 0
0 D 0 1 w 0 0
1 D 1 1 E 0 0
0 E 0 1 F 0 0
0 F 0 1 G 0 0
1 G 1 1 H 0 0
1 H 1 1 I 0 0
0 I 0 1 J 0 0
1 I 1 1 J 0 0
0 J 0 1 K 0 0
1 J 1 1 K 0 0
0 K 0 1 L 0 0
1 K 1 1 L 0 0
0 L 0 1 w 0 0
1 L 1 1 E 0 0
0 H 0 1 0 0 1
0 z 0 1 A 0 0
0 2 1 1 M 0 0
0 y 1 1 M 0 0
0 q 1 1 M 0 0
0 M 1 1 N 0 0
0 N 0 1 O 0 0
0 O 0 1 P 0 0
0 P 0 1 Q 0 0
0 Q 1 1 0 1 1
0 a 0 1 0 0 1
0 i 0 1 0 0 1
0 G 0 1 0 0 1

Try it online!

Input: A list like n, n, n, n, ...
Output: (blank) for falsy, 1 for truthy

Answer 7

05AB1E, 6 5 bytes

-1 byte thanks to a'_'.

ÉO<ÄΘ

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Explanation

É       is the number odd (vectorizes over the input list)
 O      sum (number of odd values)
  <     decrement (number of odd values - 1)
   Ä    absolute value (|number of odd values - 1|)
    Θ   equal to 1 (|number of odd values - 1| == 1)

Answer 8

JavaScript (ES6), 29 bytes

a=>a.map(x=>m>>=x&1,m=5)&&m&1

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How?

We start with the bitmask \$m=101_2=5_{10}\$ and right-shift it by one position for each odd value in the input array. The final result is given by the parity of \$m\$.

Answer 9

Bash + Unix utilities, 28 bytes

dc<<<`grep -c [13579]$`d2-*p

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Input is on stdin: one number per line.

Output is on stdout: 0 is truthy, any non-zero value is falsy.

How it works:

• grep counts the number n of lines ending with an odd digit;
• dc then computes and prints n*(n-2), which is 0 iff n equals 0 or 2.

Answer 10

C++ (gcc), 73 64 bytes

I don't know why I keep trying.

int f(int*a,int s){int o=1;for(;s;)o+=a[--s]%2;return o%2&&o<4;}

Commented :

int f(int*a,int s){ // function definition

  int o=1;      // odd counter starts at 1 because then parity of truthy inputs is one

  for(;s;)          // Use size arg as counter and iterate backwards (Thanks @MitchellSpector)
    o+=a[--s]%2;    // check for odd, add parity

  return o%2&&o<4;  // Valid values for odd are 0 and 2 (or rather 1 and 3)
                    // Solve with parity and range check
}

TIO-k6qqj7ln

Answer 11

Excel, 50 bytes

=ISNUMBER(FIND(SUMPRODUCT(--(MOD(A:A,2)=1)),"02"))

MOD(A:A,2)=1 to find odd number SUMPRODUCT() to count them FIND( ,"02") to check if 0 or 2 ISNUMBER() to converts to TRUE/FALSE

Feels very clunky, hope to refine...

Answer 12

Python 3, 34 bytes

lambda a:sum(n%2for n in a)in(0,2)

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Answer 13

W d, 8 7 bytes

♦æ╥└¬y÷

Uncompressed:

2mJ1-z1=

Explanation

2m       % Modulo the list by 2, auto-vectorizes.
  J      % Sum the list.
   1-    % Decrement by 1.
     z   % Absolute value.
      1= % Is this 1?

Answer 14

Stax, 8 bytes

This method is shorter in Stax.

ƒ╟⌡≡ù₧╬)

Run and debug it

Explanation

{2%m       Map modulo 2 over the input
    |+     Sum the resulting list
      02\  In the list [0, 2]:
         # How many times does this number appear?

Answer 15

Keg, 13 bytes

÷⑷2%⑸⅀:0=$2=+

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Simply using everyone else's algorithm!

Answer 16

Japt -!, 7 bytes

èu a1 É

Try it

Answer 17

PHP, 39 43 bytes

for(;$n=$argv[++$i];)$s+=$n%2;echo!($s&-3);

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Answer 18

Perl 5 -pa, 20 bytes

$_=(2|grep$_%2,@F)<3

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Borrows the |2<3 trick from @xnor's Python answer

Answer 19

Japt -!, 6 bytes

èu &-3

Try it

èu &-3        Full program taking in array of degrees U
èu            Amount of odd numbers in array
   &          Bitwise AND
    -3        With -3; -3 has very bit set except of the second bit (1111...11101)
-!            Zero -> True, Other numbers -> False

Answer 20

GolfScript, 11 bytes

{2%+}*(2?(!

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My first go at it, may shrink it a bit. I made it so 1 is true, 0 is false, but if you're okay with the other way around, I can do different things :)

Short explanation; Convert all the elements to mod 2 and sum them. The only valid answers would be 0 or 2. Now, subtract 1 and square them. 0 is -1 is 1. 2 is 1 is 1. 0 and 2 both give you 1. Lastly, subtract 1 and perform a not operation to make all non-0, non-2 values to a 0 and the 0-2s to a 1.

Kind of roundabout, and I'm sure I can cut two characters somewhere.

Answer 21

APL (Dyalog Unicode), 10 bytes^SBCS

0 2∊⍨1⊥2∘|

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Because a challenge needs an APL answer.

How it works

0 2∊⍨1⊥2∘|
       2∘|  ⍝ Modulo 2
     1⊥     ⍝ Sum
0 2∊⍨       ⍝ Is member of [0 2]?

Answer 22

Whitespace, 102 bytes

[S S S N
_Push_0][S N
S _Dupe_0][N
S S N
_Create_Label_LOOP][S N
S _Dupe][S N
S _Dupe][T  N
T   T   _Read_STDIN_as_integer][T   T   T   _Retrieve][S N
S _Dupe_input][N
T   T   S N
_If_neg_Jump_to_Label_DONE][S S S T S N
_Push_2][T  S T T   _Modulo][T  S S S _Add][N
S N
N
_Jump_to_Label_LOOP][N
S S S N
_Create_Label_DONE][S N
N
_Discard][S N
S _Dupe][N
T   S T N
_If_0_Jump_to_Label_TRUTHY][S S S T S N
_Push_2][T  S S T   _Subtract][N
T   S T N
_If_0_Jump_to_Label_TRUTHY][T   N
S T _Print_as_integer][N
N
N
_Exit_program][N
S S T   N
_Create_Label_TRUTHY][S S S T   N
_Push_1][T  N
S T _Print_as_integer]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Since Whitespace inputs one integer at a time, the input should contain a trailing -1 so it knows when to stop reading integers and the input is done.
Outputs 1/0 for truthy/falsey respectively.

Try it online (with raw spaces, tabs and new-lines only).

Explanation in pseudo-code:

Integer n = 0
Start LOOP:
  Integer i = STDIN as input
  If(i < 0):
    Jump to Label DONE
  n = n + (i modulo-2)
  Go to next iteration of LOOP

Label DONE:
  If(n == 0): Jump to Label TRUTHY
  If(n-2 == 0): Jump to Label TRUTHY
  Print 0 as integer to STDOUT
  Exit program

Label TRUTHY:
  Print 1 as integer to STDOUT

Answer 23

MathGolf, 6 bytes

¥Σ(±1=

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Explanation:

¥       # Take modulo-2 on each integer of the (implicit) input-list
 Σ      # Sum those
  (     # Decrease it by 1
   ±    # Take the absolute value
    1=  # And check whether this is equal to 1
        # (after which the entire stack joined together is output implicitly)

Answer 24

Java 8, 28 bytes

s->(s.map(i->i%2).sum()|2)<3

Port of @xnor's Python answer

Try it online.

Explanation:

s->               // Method with IntStream parameter and boolean return-type
  (s.map(i->i%2)  //  Take modulo-2 for each value in the IntStream
    .sum()        //  And them sum those
    |2)           //  Take a bitwise-OR with 2
       <3         //  And check whether that is smaller than 3

Answer 25

C# (Visual C# Interactive Compiler), 26 bytes

p=>(p.Count(n=>n%2>0)|2)<3

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Thanks to answer of @xnor -1B

27 bytes

p=>(p.Count(n=>n%2>0)&-3)<1

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This one was my first attempt, but I made mistake at first so I'm publishing it now when I fixed it.

Answer 26

Chevron, 172 147 bytes

^__>^n
>^n>^t
^t~s>>^s
0>>^o
0>>^i
^i+1>>^i
^t,^i~c>>^c
->+2??^c~^_i
->-3
^n^c>^n
^n~o>>^z
^o+^z>>^o
^__>^n
->+2?^i=^s
->-9
^o-2>>^p
^o*^p>>^o
>^o

Takes numbers from stdin in the form 2 4 4 3 3.

Outputs 0 or -0 for truthy, anything else falsy.