4
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Related:

Challenge:

Given a grid, with an ID starting at the center and spiraling out, what is the ID given a position in the fewest number of bytes?

Grid:

+---------------+---------------+---------------+---------------+---------------+ | id: 20 | id: 19 | id: 18 | id: 17 | id: 16 | | pos: (-2, -2) | pos: (-1, -2) | pos: (0, -2) | pos: (1, -2) | pos: (2, -2) | +---------------+---------------+---------------+---------------+---------------+ | id: 21 | id: 6 | id: 5 | id: 4 | id: 15 | | pos: (-2, -1) | pos: (-1, -1) | pos: (0, -1) | pos: (1, -1) | pos: (2, -1) | +---------------+---------------+---------------+---------------+---------------+ | id: 22 | id: 7 | id: 0 | id: 3 | id: 14 | | pos: (-2, 0) | pos: (-1, 0) | pos: (0, 0) | pos: (1, 0) | pos: (2, 0) | +---------------+---------------+---------------+---------------+---------------+ | id: 23 | id: 8 | id: 1 | id: 2 | id: 13 | | pos: (-2, 1) | pos: (-1, 1) | pos: (0, 1) | pos: (1, 1) | pos: (2, 1) | +---------------+---------------+---------------+---------------+---------------+ | id: 24 | id: 9 | id: 10 | id: 11 | id: 12 | | pos: (-2, 2) | pos: (-1, 2) | pos: (0, 2) | pos: (1, 2) | pos: (2, 2) | +---------------+---------------+---------------+---------------+---------------+

Tests:

f(0, 0) = 0 f(1, 1) = 2 f(0, -1) = 5 f(2, 0) = 14 f(-2, -2) = 20 f(x, y) = id

Notes:

  • Grid will always be square (height == width)
  • Grid can be infinite in size
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  • \$\begingroup\$ How can the grid be both square and infinite in size? \$\endgroup\$ Commented Mar 27, 2018 at 23:00
  • \$\begingroup\$ Sorry, just trying to say it can be bigger than the example grid provided. \$\endgroup\$ Commented Mar 27, 2018 at 23:02
  • \$\begingroup\$ Sorta odd you chose to make the bottom right 1,1 and the top left -1,-1. Then again I guess Java does this in java.awt.* \$\endgroup\$ Commented Mar 27, 2018 at 23:04
  • \$\begingroup\$ @MagicOctopusUrn - makes you think a little more? It's just the grid setup I've been using in my codebase. I didn't want to use the same old grid in my game. \$\endgroup\$ Commented Mar 27, 2018 at 23:06
  • 2
    \$\begingroup\$ The last case should output 20, not 24 \$\endgroup\$ Commented Mar 27, 2018 at 23:46

4 Answers 4

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1
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JavaScript (ES6), 64 bytes

Takes input in currying syntax (x)(y).

x=>y=>(d=(A=Math.abs)(A(x)-A(y))+A(x)+A(y))*d+(d-x+y)*(-y>x||-1)

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Heavily inspired by this answer from math.stackexchange.com.

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1
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JavaScript (Node.js), 58 bytes

y=>x=>(m=Math.max)(4*x*x-m(x+y,3*x-y),4*y*y+m(-x-y,x-3*y))

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Use the formula from my previous answer, with some edit (because Javascript doesn't vectorize).

I came up with the formula myself. Basically, knowing that the formula must be quadratic in each 1/4 grid section, I find out the formula mostly by trial-and-error and then find a way to fit those together.

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0
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MATL, 15 bytes

|sQtE1YLqwG+X{)

Try it online! Or verify all test cases.

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0
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J, 37 bytes

(g@[-+<.-+[+[)>.(g=:4**:)@]-+>.]+]+-~

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My first J answer! Use the formula from my Javascript answer.

  1. Precedence rules are weird. (at least I don't need to memorize too much, but it's counterintuitive)
  2. More than half of the characters are BF keywords.
  3. -+> and :).
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