2D Array Middle Point

4

Given dimensions of a 2D array, except 1xn, that results in an odd number of elements, find the midpoint index.

Example: Input is 3x5, representing the 2D array

[0,0] [0,1] [0,2] [0,3] [0,4]

[1,0] [1,1] [1,2] [1,3] [1,4]

[2,0] [2,1] [2,2] [2,3] [2,4]

The midpoint index is [1,2]

Input

3

5

Output

1 , 2

Rules

Standard code-golf rules, fewest bytes wins
Either full program or function is permitted
When input is either 1xn or the corresponding array would have an even number of elements, output must be [0,0]
nx1 should be treated like any other array

Noir Antares

Posted 2018-07-04T08:38:22.857

Reputation: 469

6Most challenges on PPCG try to avoid special cases or error cases, instead guaranteeing that the input will be valid. It's your call to make, but I find that error handling is usually boilerplate that makes solutions longer without making them more interesting. – Esolanging Fruit – 2018-07-04T09:01:09.757

2

We have a Sandbox for Proposed Challenges.

– user202729 – 2018-07-04T09:07:15.737

@EsolangingFruit The challenge here is to handle the invalid inputs, otherwise it would trivially be ⁽ⁿ⁻²⁾⁄₂. – Adám – 2018-07-04T09:24:08.167

5@Adám A challenge where the entire point is to detect and handle invalid inputs doesn't seem like a very interesting challenge to me. – Esolanging Fruit – 2018-07-04T09:29:22.570

4@EsolangingFruit There's nothing inherently wrong with challenges that a specific user happens to find uninteresting. Personally, I think having to catch such specific cases makes for some interesting golfing material. – Adám – 2018-07-04T09:31:28.377

1

@NoirAntares I've edited your challenge to make it unambiguous. Feel free to edit or roll back If my rewording does not match your intentions.

– Adám – 2018-07-04T09:33:01.970

@JoKing Well, [⁽ⁿ⁻²⁾⁄₂,0], yeah. – Adám – 2018-07-04T09:50:21.517

3output must be [0,0] any reason behind that? How does that part add anything to computing the midpoint? Obviously, 1xn arrays or arrays with an even dimension don't have [0,0] as their midpoints. – JungHwan Min – 2018-07-04T13:57:04.100

Answers

4

APL (Dyalog Classic), 20 19 15 13 bytes

⌊÷∘2×2|×/×1≠⊃

ngn

Posted 2018-07-04T08:38:22.857

Reputation: 11 449

4

Japt, 11 bytes

My first (somewhat) serious Japt submission! Any help is appreciated, I think ~10 bytes or even less may be possible. Saved 4 bytes thanks to Shaggy!

£ÎÉ©Ueu)*Xz

How it works

Slightly outdated.

£eu *UÎ>1 *Xz – Full program.
£             – For each X of the input array.
          *Xz – Multiply floor(X/2) by the truth value of:
 eu           – Are both numbers odd? And...
    *UÎ>1     – Is the initial element greater than 1?
              – Implicit output to STDOUT.

Mr. Xcoder

Posted 2018-07-04T08:38:22.857

Reputation: 39 774

3

05AB1E, 9 8 bytes

Saved 1 byte thanks to Mr. Xcoder (remove D)

¬≠*ÉP*;ï

Try it online! or as a Test Suite

Explanation

¬≠*       # multiply the input by its first value falsified
   ÉP*    # multiply the input by the product of the results values' oddness
      ;   # divide by 2
       ï  # convert to int

Emigna

Posted 2018-07-04T08:38:22.857

Reputation: 50 798

Nice! I had ¬≠sPÉ*s2÷* for 10 bytes. Taking some inspiration from you though, would ¬≠*PÉ*2÷ work for 8 bytes? Alternatively, ¬≠*PÉ*;ï, of course.

– Mr. Xcoder – 2018-07-04T11:34:42.597

@Mr.Xcoder: Of course. I don't need the D here :/ Thanks! – Emigna – 2018-07-04T11:45:47.697

3

Japt, 24 16 bytes

-8 bytes saved thanks to @Shaggy

Neu ©UÉ?Nmz :2ÆT

Luis felipe De jesus Munoz

Posted 2018-07-04T08:38:22.857

Reputation: 9 639

1

You can get this down to 16 bytes with a few tricks, mainly making use of the N variable.

– Shaggy – 2018-07-04T14:13:34.287

Feel free to ping me in chat if you've any questions.

– Shaggy – 2018-07-04T14:44:09.180

2

APL (Dyalog Unicode), 23 bytes^SBCS

Anonymous prefix function. Takes rows,columns as right argument.

{0::0 0⋄÷∘÷@0⌽¨⍨.5×⍵-1}

{…} "dfn"; ⍵ is right argument (rightmost letter of Greek alphabet).

0:: if any error happens:

0 0 return [0,0]

⋄ now try:

⍵-1 right argument minus 1

.5× multiply by a half

⌽¨⍨ rotate each by itself (this errors on non-integers)

÷∘÷@0 reciprocal of reciprocal at position 0 (this errors when the first coordinate is 0)

Adám

Posted 2018-07-04T08:38:22.857

Reputation: 37 779

2

JavaScript (ES6), 31 bytes

Saved 1 byte thanks to @Neil

Takes the dimensions in currying syntax (w)(h).

w=>h=>w&h&w>1?[h>>1,w>>1]:[0,0]

Original version, 32 bytes

Takes the dimensions in currying syntax (w)(h).

w=>h=>w*h&!!--w?[h>>1,w/2]:[0,0]

How?

We want to test whether both \$w\$ and \$h\$ are odd, which is true if and only if \$w\times h\$ is odd. We also want to make sure that \$w\$ is not equal to \$1\$, which means that \$w-1\$ is not equal to \$0\$. We can merge both tests into:

w * h & !!--w

If this test passes, we know that \$w\$ was odd and is now even (since it was decremented), so we can divide it by \$2\$ with a standard division. For \$h\$, we use a bitwise shift instead because it's still odd.

Arnauld

Posted 2018-07-04T08:38:22.857

Reputation: 111 334

1w=>h=>w&h&w>1?[h>>1,w>>1]:[0,0]? – Neil – 2018-07-04T13:51:41.563

@Neil This is a bit more straightforward indeed. Thanks! – Arnauld – 2018-07-04T13:59:22.610

2

Perl 6, 42 39 bytes

{(^$^a X ^$^b)[$b-1&&$a*$b%2&&$a*$b/2]}

Anonymous code block that returns a list of two integers.

Explanation:

{                                     }  # Anonymous code block
 (^$^a X ^$^b)   # Cross product of the two ranges, basically the flattened 2D array
              [                      ] # Get the element at
                            &&$a*$b/2    # Midpoint of the array
                     $a*$b%2             # If the numbers are odd
               $b-1&&                    # And the second element is not 1
                                        # Else the 0th element, which is 0,0

Jo King

Posted 2018-07-04T08:38:22.857

Reputation: 38 234

Seems to fail on 7×1: Try it online!

– Adám – 2018-07-04T09:51:24.103

I'm confused. Why is 1xn expected to fail, but nx1 is fine? – Jo King – 2018-07-04T10:13:01.303

2Because OP says so. – Adám – 2018-07-04T10:19:26.890

2

Haskell, 46 42 bytes

x%y|odd$x*y,x>1=(`div`2)<$>[x,y]|1<2=[0,0]

EDIT: -4 bytes thanks to a trick from Arnauld's answer.

Amphibological

Posted 2018-07-04T08:38:22.857

Reputation: 1 394

1

Julia 0.6, 28 27 bytes

x\y=x&y&1&(x>1)*[x÷2,y÷2]

(-1 byte thanks to Jo King.)

Explanation:

x\y - define an operator instead of a function, to save bytes
x&y&1 - check that both inputs are odd.
&(x>1) - and that the first input is not 1
* - the result of the above checks is 0 or 1, multiply that with:
[x÷2,y÷2] - ÷ is integer division, so here for odd numbers gives (n-1)/2 i.e. the middle index. So this forms a vector of the two middle indices (which might be made [0, 0] if any of the previous checks fail).

sundar - Reinstate Monica

Posted 2018-07-04T08:38:22.857

Reputation: 5 296

I don't know Julia, but could you assign to a one byte operator instead, like +?

– Jo King – 2018-07-04T13:09:18.767

2

Short answer: Yes thank you! Long answer: Your comment finally prompted me to check why those work on TIO but not on my system (I'd assumed it was just version differences). Turns out I can assign to those if I haven't used them yet in that session. Makes some confusing behaviour around this make sense now.

– sundar - Reinstate Monica – 2018-07-04T13:44:29.090

1

Jelly, 10 bytes

^{This confused me :p}

:2×Ḃ_ỊḢ$PƲ

A monadic link accepting a list of the dimensions returning a list of the middle-indices or [0,0] in the special-cases.

Try it online!

How?

:2×Ḃ_ỊḢ$PƲ - Link: list of dimension lengths        e.g.  [19,12]  or  [19,13]  or  [1,19]  or  [19,1]
:2        - integer divide by two (gets middle indices)   [9,6]         [9,6]       [0,9]       [9,0]
        Ʋ - last four links as a monad:
   Ḃ      -   bit (n%2) (of the input)                    [1,0]         [1,1]       [1,1]       [1,1]
       $  -   last two links as a monad (of the input):
     Ị    -     insignificant? (abs(n)<=1)                [0,0]         [0,0]       [1,0]       [0,1]
      Ḣ   -     head                                      0             0           1           0
    _     -   subtract (vectorises)                       [1,0]         [1,1]       [0,0]       [1,1]
       P  -   product                                     0             1           0           1
  ×       - multiply (vectorises)                         [0,0]         [9,6]       [0,0]       [9,0]

Jonathan Allan

Posted 2018-07-04T08:38:22.857

Reputation: 67 804

1

Jelly, 11 bytes

^{_{Ugh. Waaaaaaaay too long!}}

:2×ḂẠ$×Ḣn1Ɗ

Alternative, 11 bytes

ḷ/>1×ḂẠ$×:2

Mr. Xcoder

Posted 2018-07-04T08:38:22.857

Reputation: 39 774

1

Lua, 77 bytes

x,y=...f=math.floor print((x%2==1 and y%2==1)and f(x/2)..","..f(y/2)or "0,0")

not the smallest language but submission for practicing :)

Lycea

Posted 2018-07-04T08:38:22.857

Reputation: 141

I don't know much Lua (although I wish I did!), but I believe you can replace the oddness check x%2==1 and y%2==1 with x%2+y%2>1 Try it online!

– sundar - Reinstate Monica – 2018-07-04T14:30:11.360

2Also, this doesn't return 0,0 when x is 1, which is one of the question's requirements – sundar - Reinstate Monica – 2018-07-04T14:32:34.733

You can extend the condition to check that too, here's my attempt: Try it online! (also changed x%2+y%2 to x*y%2 after seeing Arnauld's answer.)

– sundar - Reinstate Monica – 2018-07-04T14:45:19.570

It's been a while since I've done anything in Lua, but it has the // operator yes? So you don't need the floor function – Jo King – 2018-07-04T14:49:42.227

1

Rust, 42 bytes

|w,h|if w>1&&w*h%2>0{(w/2,h/2)}else{(0,0)}

Konrad Borowski

Posted 2018-07-04T08:38:22.857

Reputation: 11 185

1

R, 38 29 28 bytes

An extra byte removed thanks to @Guiseppe

(i=scan()-1)/2*!any(i%%2,!i)

Get the input, subtract 1 store in i and divide by 2 and multiply by not any evens or 0's in i

MickyT

Posted 2018-07-04T08:38:22.857

Reputation: 11 735

!i instead of i<2? – Giuseppe – 2018-07-04T22:15:49.377

@Giuseppe good call, thanks – MickyT – 2018-07-04T23:57:18.547

0

Python 2, 39 bytes

-2 bytes thanks to ovs

lambda a,b:[0,a/2,0,b/2][a>1==a*b%2::2]

Dead Possum

Posted 2018-07-04T08:38:22.857

Reputation: 3 256

39 bytes – ovs – 2018-07-04T11:12:25.520

0

Python 2, 43 41 bytes

lambda a,b:((0,0),(a/2,b/2))[a*b%2*(a>1)]

No rocket science here.

-2 with thanks to @MrXcoder and @Enigma

ElPedro

Posted 2018-07-04T08:38:22.857

Reputation: 5 301

1[a*b%2*(a!=1)] should work for 42 bytes. – Mr. Xcoder – 2018-07-04T13:01:04.690

2@Mr.Xcoder: or even a>1. – Emigna – 2018-07-04T13:13:03.090

Asked: 2018-07-04T08:38:22.857

Viewed: 672 times

Active: 2018-07-24T15:28:15.047