Compare positions of integers in this sequence

Question

A001057 is one way to represent an integer as a natural number. It lists them according to the following pattern:

0, 1, -1, 2, -2, 3, -3, 4, -4, ...

In this challenge, you'll take two distinct integers as input and return which is at a higher index in this enumeration.

Take -2 and 1 as an example: -2 comes after 1, so it would be returned.

You may also choose to instead return the smaller integer, or one of two consistent values (such as 0 or 1) to indicate which is higher or lower. Truthy/falsy values as output are not allowed.

Test cases:

 0   1       1
 1   0       1
 0  -1      -1
-1   0      -1
-1   1      -1
 1  -1      -1
 1  -2      -2
 1   2       2
-1  -2      -2
-1   2       2

Answer 1

Python, 20 bytes

lambda x,y:x*x>y*y-y

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Checks whether \$x^2>y^2-y \$.

Outputs True/False for whether x comes after y in the listing. Note that we're not asked handle the case x==y.

To see that this works, note that each of the x*x value in the table is greater than all the y*y-y values to the left of it, but none to the right of it.

       0  1  -1  2  -2  3  -3   4  -4 ...
x*x:   0  1   1  4   4  9   9  16  16 ...
y*y-y: 0  0   2  2   6  6  12  12  20 ...

The idea is that the entries come in pairs +n, -n (except 0 once), so comparing squares works to check if one input has a bigger absolute value than the other. Then, changing \$y^2\$ to \$y^2-y\$ in the second entry, makes +n "lose" to -n, but is a small enough shift compared to the difference between consecutive squares that it doesn't mess up comparisons between different squares.

Answer 2

J, 8 bytes

>&(-3^|)

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Or verbose: (x - 3 ^ abs(x)) > (y - 3 ^ abs(y)). This maps the inputs to the following sequence to compare them in:

 0  1 _1  2  _2   3  _3   4  _4    5   _5
_1 _2 _4 _7 _11 _24 _30 _77 _85 _238 _248

If sorted input as output is allowed, then 7 byte /:*.@%: is another fun way to sort /: the input list: *. generates (length, angle) pairs of the square root %: of x (which results in complex numbers for negative values):

 0       0      0
 1       1      0
_1       1 1.5708
 2 1.41421      0
_2 1.41421 1.5708
 3 1.73205      0
_3 1.73205 1.5708

Answer 3

Husk, 7 bytes

F>m€Θݱ

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Returns 1 if the second is larger than the first, 0 otherwise

Answer 4

Desmos, 21 bytes

f(x,y)=\{xx-x<yy,0\}

(Newline is required)

Edit: Just noticed that xnor's answer is very similar to mines, but I found this independently.

Function is \$f(x,y)\$, and it returns \$0\$ if \$x\$ has a higher index, otherwise returns \$1\$.

This seems to work, but I just randomly stumbled upon it, so I can't explain how it works.

Try It On Desmos!

Try It On Desmos! - Prettified

Answer 5

Vyxal, 5 bytes

²?Ḣ+⇧

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Takes input as a list and returns [1,0] for first being bigger, [0,1] for second being bigger.

²     # [x^2, y^2]
 ?Ḣ   # [y]
   +  # [x^2+y, y^2]
    ⇧ # Grade up

Answer 6

Python 3, 29 26 bytes

Returns True if a is a at a higher index than b, False otherwise.

lambda a,b:a*a+(a<0)/2>b*b

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Answer 7

Jelly, 5 bytes

²Ḥ_ṠỤ

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-1 byte thanks to ovs
-1 byte thanks to Unrelated String

Returns [1, 2] if the first integer comes first and [2, 1] for the second integer.

²Ḥ_ṠỤ    Main Link; accepts a list
²        x ** 2
 Ḥ       2 * (x ** 2)
   _Ṡ    2 * (x ** 2) - sign(x)
     Ụ   Grade up

 0 => 0
 1 => 1
-1 => 3
 2 => 7
-2 => 9
 3 => 17
-3 => 19
...

Answer 8

Vyxal, 7 bytes

ȧ4*?±-⇧

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Oh yes, jelly port. Very fun.

⟨0|1⟩ for second integer being yes, ⟨1|0⟩ for first integer being yes.

Explained

ȧ4*?±-⇧
ȧ4*     # 4 * abs(input) [vectorised] # call this x
   ?±   # sign_of(input) [vectorised - <0 = -1, =0 = 0, >0 = 1] # call this y
     -  # x - y [vectorised element-wise] # call this z
      ⇧ # indexes of z in an order such that they would arrange z to be sorted ascending

Answer 9

Jelly, 4 bytes

Ḥc2Ụ

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Alternative grade-up solution. Uses (2n)C2 = n(2n-1) as the grade key function. This function maps 0, 1, -1, 2, -2, 3, -3, ... to 0, 1, 3, 6, 10, 15, 21, ..., i.e. the triangular numbers.

This allows a shorter APL solution:

APL(Dyalog Unicode), 5 bytes ^SBCS

⍋2!+⍨

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Answer 10

Japt, 5 bytes

Outputs true for the first number being higher or false for the second.

²+V>²

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Answer 11

COW, 183 bytes

oomMMMmoOMMMmoOoomMMMmoOMMMmOomOoMMMmoOmoOmoOMMMMMMmoOMMMmOomOoMMMmoOmoOmoOMMMMMMmoOMMMmOoMOOmOomOomOomOomOomOoMOOMOomoOmoOMOOMoOmoOmoOMOOMoOmoOmoOMOoMMMOOOmooOOOmooOOOmooMMMmoomoOOOM

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Returns the smaller integer. It increases and decreases both, the first to reach 0 (with precedence to the decreasing one) is the smaller.

[0]: n to be decreased   [1]: n   [2]: m to be increased   [3]: m   [4]: n to be increased   [5]: n   [6]: m to be decreased   [7]: m

i=>=>i=>=<<=>>>==>=<<=>>>==>=<    | set up memory cells as above
[                                 | loop until a cell reaches 0
    <<<<<<[->>[+>>[+>>-=*]*]*]=   | [0]-- [2]++ [4]++ [6]--
]>o                               | print the redundant copy of the smaller

moo ]    mOo <    MOo -    OOO *    OOM i
MOO [    moO >    MoO +    MMM =    oom o

Answer 12

Python 2, 30 bytes

lambda x,y:(x*x,x<0)>(y*y,y<0)

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This is one of those irritating ones that can't be shortened with a loop because the loop is only over 2 items.

The abs(x) -> x*x trick was taken from ovs' answer.

Answer 13

APL (Dyalog Unicode), 6 bytes

A port of hyper-neutrino's Jelly answer. Outputs are swapped.

⍋×-4×|

Try it online! g gets the "higher" number according to the output.

Answer 14

Wolfram Language (Mathematica), 19 bytes

Most@*SortBy[-# #&]

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-23 bytes from @att

Answer 15

C (gcc), 20 bytes

f(a,b){a=a*a>b*b-b;}

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Port of xnor's Python answer.

Answer 16

Scala, 37 bytes

_.maxBy(x=>if(x>0)x*2-1 else x.abs*2)

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Very simple, simply takes a list containing the two number, maps each to its index in the sequence, and finds the max according to that.

Answer 17

Jelly, 5 4 bytes

²+ḊỤ

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Terrible port of xnor's Python solution, also stealing from hyper-neutrino's Jelly solution ovs's output format ([1, 2] if y is greater, [2, 1] if x is greater). Takes input as a list [x, y].

²       [x^2, y^2]
 +      plus
  Ḋ     [y]:
²+Ḋ     [x^2 + y, y^2].
   Ụ    Grade up.

Answer 18

Charcoal, 10 bytes

Ｉ⌊Φθ⁼↔ι⌈↔θ

Try it online! Link is to verbose version of code. Takes input as a list of two (actually arbitrary many) integers (which need not be distinct). Explanation: Outputs the element with the greatest absolute value, or the less of two elements tied for absolute value. Explanation:

   θ        Input list
  Φ         Filter where
      ι     Current element
     ↔      Absolute value
    ⁼       Equals
         θ  Input list
        ↔   Vectorised absolute value
       ⌈    Take the maximum
 ⌊          Take the minimum
Ｉ           Cast to string
            Implicitly print

Actually a port of @xnor's Python answer would probably be only 9 bytes but that would be boring.

Answer 19

K (oK), 14 bytes

f:{x*x>-y+y*y}

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