2D Array Middle Point

Question

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

• Esolanging Fruit: Most 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.
• user202729: We have a Sandbox for Proposed Challenges.
• Esolanging Fruit: @Adám A challenge where the entire point is to detect and handle invalid inputs doesn't seem like a very interesting challenge to me.
• Adám: @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.
• JungHwan Min: output 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.

Answer 1

APL (Dyalog Classic), (削除) 20 (削除ここまで) (削除) 19 (削除ここまで) (削除) 15 (削除ここまで) 13 bytes

×ばつ1≠⊃

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Answer 2

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

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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Slightly outdated.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.

Answer 3

APL (Dyalog Unicode), 23 bytes ^SBCS

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

{0::0 0⋄÷∘÷@0⌽ ×ばつ⍵-1}

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{...} "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

×ばつ 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)

Answer 4

05AB1E, (削除) 9 (削除ここまで) 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

• Mr. Xcoder: 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.
• Emigna: @Mr.Xcoder: Of course. I don't need the D here :/ Thanks!

Answer 5

Japt, (削除) 24 (削除ここまで) 16 bytes

-8 bytes saved thanks to @Shaggy

Neu ©UÉ?Nmz :2ÆT

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• Shaggy: You can get this down to 16 bytes with a few tricks, mainly making use of the N variable.
• Shaggy: Feel free to ping me in chat if you've any questions.

Answer 6

Perl 6, (削除) 42 (削除ここまで) 39 bytes

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

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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

• Adám: Seems to fail on 7×1: Try it online!
• Jo King: I'm confused. Why is 1xn expected to fail, but nx1 is fine?
• Adám: Because OP says so.

Answer 7

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]

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---

Original version, 32 bytes

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

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

Try it online!

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ドル\,ドル which means that \$w-1\$ is not equal to \0ドル\$. 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ドル\$ with a standard division. For \$h\,ドル we use a bitwise shift instead because it's still odd.

• Neil: w=>h=>w&h&w>1?[h>>1,w>>1]:[0,0]?
• Arnauld: @Neil This is a bit more straightforward indeed. Thanks!

Answer 8

Haskell, (削除) 46 (削除ここまで) 42 bytes

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

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

Answer 9

Jelly, 10 bytes

^{This confused me :p}

×ばつ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.

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How?

×ばつ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
 ×ばつ - multiply (vectorises) [0,0] [9,6] [0,0] [9,0]

Answer 10

Jelly, 11 bytes

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

×ばつḢn1Ɗ

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Alternative, 11 bytes

ḷ/&g×ばつ:2

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Answer 11

Julia 0.6, (削除) 28 (削除ここまで) 27 bytes

x\y=x&y&1&(x>1)*[x÷2,y÷2]

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(-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).

• Jo King: I don't know Julia, but could you assign to a one byte operator instead, like +?
• Sundar R: 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.

Answer 12

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")

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not the smallest language but submission for practicing :)

• Sundar R: 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 R: Also, this doesn't return 0,0 when x is 1, which is one of the question's requirements
• Sundar R: 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.)
• Jo King: 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

Answer 13

Rust, 42 bytes

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

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Answer 14

R, (削除) 38 (削除ここまで) (削除) 29 (削除ここまで) 28 bytes

An extra byte removed thanks to @Guiseppe

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

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

• Giuseppe: !i instead of i<2?
• MickyT: @Giuseppe good call, thanks

Answer 15

Python 2, 39 bytes

-2 bytes thanks to ovs

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

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• ovs: 39 bytes

Answer 16

Python 2, (削除) 43 (削除ここまで) 41 bytes

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

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No rocket science here.

-2 with thanks to @MrXcoder and @Emigna

• Mr. Xcoder: [a*b%2*(a!=1)] should work for 42 bytes.
• Emigna: @Mr.Xcoder: or even a>1.