One-to-one correspondence between pairs of integers and the positive integers

Question

It is well-known that there are one-to-one correspondences between pairs of integers and the positive integers. Your task is to write code defining such a correspondence (by defining a pair of functions/programs that are inverses of each other) in your programming language of choice, plus a correctness check (see below) with the smallest number of bytes for the correspondence definition (not taking the correctness check into account).

The solution must include :

• the definition of a function/program f having two integer arguments and returning an integer (that's one direction of the bijection).

• either the definition of a function/program g having one integer argument and returning a pair of integers (might be an array, a list, the concatenation of the two integers separated by something ...) or two functions/programs a and b having an integer argument and returning an integer (that's the other direction).

• an additional code snippet checking that for the f and g (or f and a,b) you defined above, you have g(f(x,y))=(x,y) (or a(f(x,y))=x and b(f(x,y))=y) for any integers x,y in the range -100 < x < 100,-100 < y < 100. Note that f and g have to work for values outside of this range as well.

You may rename a,b,f or g of course. The two solutions do not have to be written in the same language.

Below is a not-optimal-at-all solution in PARI/GP, with 597 characters for the function definitions.

plane_to_line(x,y)={
my(ax,ay,z);
ax=abs(x);
ay=abs(y);
if((ax<=ay)*(y<0),        z=4*y*y-2*y+x+2;);
if((ax<=ay)*(y>=0),       z=4*y*y-2*y-x+2;);
if((ay<=ax)*(x<0),        z=4*x*x    -y+2;);
if((ay<=ax)*(x>=0)*(y<0), z=4*x*x+4*x+y+2;);
if((ay<=ax)*(x>=0)*(y>=0),z=4*x*x-4*x+y+2;);
if((x==0)*(y==0),z=1;);
return(z);
}


line_to_plane(z)={
my(l,d,x,y);
l=floor((1+sqrt(z-1))/2);
d=z-(4*l*l-4*l+2);
if(d<=l,x=l;y=d;);
if((l<d)*(d<=3*l),x=2*l-d;y=l;);
if((3*l<d)*(d<=5*l),x=(-l);y=4*l-d;);
if((5*l<d)*(d<=7*l),x=d-6*l;y=(-l););
if((7*l<d)*(d<8*l) ,x=l;y=d-8*l;);
if(z==1,x=0;y=0;);
return([x,y]);
}

and the correctness-check code :

accu=List([])
m=100;
for(x=-m,m,for(y=-m,m,if(line_to_plane(plane_to_line(x,y))!=[x,y],\
listput(accu,[x,y]);)))
Vec(accu)

Answer 1

MATL, 43 36 bytes

This uses the spiral (1YL) function, which generates a square 2D array of given size with values arranged in an outward spiral. For example, with input 7 it produces

43 44 45 46 47 48 49
42 21 22 23 24 25 26
41 20  7  8  9 10 27
40 19  6  1  2 11 28
39 18  5  4  3 12 29
38 17 16 15 14 13 30
37 36 35 34 33 32 31

The center of the array, which contains 1, corresponds to the tuple [0 0]. The upper left corner corresponds to [-3 -3] etc. So for example f (-3,-3) will be 43 and g (43) will be [-3 -3].

The code generates a 2D array with this spiral matrix, as large as needed to do the conversion. Note that larger sizes always give the same result for the entries already included in smaller sizes.

From Z² to N (18 bytes):

|X>tEQ1YLGb+QZ}3$)

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|X>   % input is a 2-tuple. Take maximum of absolute values
tEQ   % duplicate. Multiply by 2 and increase by 1. This gives necessary size of spiral
1YL   % generate spiral
G     % push input 2-tuple again
b+Q   % bubble up, add, increase by 1. This makes the center correspont to [0 0]
Z}    % split tuple into its values
3$)   % use those two value as indices into the spiral array to obtain result

From N to Z² (25 18 bytes)

Eqt1YLG=2#fhw2/k-q

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Eq      % input is a number. Multiply by 2, add 1. This assures size is enough and odd
t1YL    % duplicate. Generate spiral of that size
G=      % compare each entry with the input value
2#fh    % 2-tuple of row and column indices of matching entry
w2/k-q  % swap. Offset values so that center corresponds to [0 0]

Snippet for checking

Note that G needs to be modified to accomodate the fact that we have don't have a single input. The code is slow, so the link checks tuples with values from -9 to 9 only. For -99 to 99 just replace the first line.

The code tests each tuple with values in the defined range. It does the conversion to a number, then from that number back to a tuple, and then checks if the original and recovered tuple are equal. The results should all be 1, indicating that all comparisons give true.

It takes a while to run.

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-9:9                     % Or use -99:99. But it takes long
HZ^!"@                   % Cartesian power: gives tuples [-9 -9] ... [9 9].
                         % For each such tuple
|X>tEQ1YL@b+QZ}3$)       % Code from Z^2 to N with `G` replaced by `@` (current tuple)
XJ                       % Copy result into clipboard J
Eqt1YLJ=2#fhw2/k-q       % Code from N to Z^2 with `G` replaced by `J`
@!X=                     % Compare original tuple with recovered tuple: are they equal?

Answer 2

JavaScript (ES6), 171 bytes

(x,y)=>(h=x=>parseInt((x^x>>31).toString(2)+(x>>>31),4),h(x)*2+h(y))
x=>(h=x=>parseInt(x.toString(2).replace(/.(?!(..)*$)/g,''),2),i=x=>x<<31>>31^x>>>1,[i(h(x>>>1)),i(h(x))])

Bit twiddling: negative numbers have their bits flipped; each integer is then doubled, and 1 added if it was originally negative; the bits from the integers are then interleaved. The reverse operation deletes alternate bits, divides by 2, and flips all the bits if the value was negative. I could save 3 bytes by limiting myself to 15-bit values instead of 16-bit values.

f=(x,y)=>(h=x=>parseInt((x^x>>31).toString(2)+(x>>>31),4),h(x)*2+h(y))
g=x=>(h=x=>parseInt(x.toString(2).replace(/.(?!(..)*$)/g,''),2),i=x=>x<<31>>31^x>>>1,[i(h(x>>>1)),i(h(x))])
for(i=-100;i<=100;i++)for(j=-100;j<100;j++)if(g(f(i,j))[0]!=i||g(f(i,j))[1]!=j)alert([i,j])

Answer 3

Jelly, 50 48 46 45 43 40 39 bytes

Plane to line (18 17 16 bytes):

AḤ_<0$
+RS+
,Ñç/

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Line to plane (32 30 29 27 24 23 bytes):

HĊN⁸¡
³R‘R’U⁹¡F³ịÇ
ç1,¢

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Explanation:

I'll only explain plane to line, because line to plane is just its opposite.

Firstly, we convert each integer to a natural number, by the function f(x) = 2*|x| - (x<0).

Then, we convert the two natural numbers to another two natural numbers, by the function g(x,y) = (x+y,y).

Finally, we convert them into one natural number, by the function h(x,y) = (x+1)C2 + y