Creating Distinct Sums

Question

You should write a program or function which receives an integers as input and outputs or returns two integers whose sum is the first one.

There is one further requirement: no number can be part of the output for two different inputs.

Details

• You should be able to handle inputs for at least the range -32768 .. 32767 (inclusive).
• If your datatype can't handle arbitrary whole numbers, that is fine but your algorithm should work for arbitrary large and small numbers in theory.

Examples

Each block shows a part of a correct or incorrect solution in the format of input => output.

1 => 6 -5
2 => -2 4
15 => 20 -5

Incorrect, as `-5` is used in two outputs.

-5 => -15 10
0 => 0 0
1 => 5 6
2 => -5 7

Incorrect, as `5 + 6` isn't `1`.

-1 => -1 0
0 => 6 -6
2 => 1 1

Can be correct if other outputs doesn't collide.

This is code golf so the shortest entry wins.

Answer 1

Pyth, 8 bytes

_J^Q3+QJ

Demonstration. Equivalent to the Python 2 code:

Q=input()
J=Q**3
print -J
print Q+J

So, the output has form (-n**3, n+n**3)

Some outputs:

-5 (125, -130)
-4 (64, -68)
-3 (27, -30)
-2 (8, -10)
-1 (1, -2)
 0 (0, 0)
 1 (-1, 2)
 2 (-8, 10)
 3 (-27, 30)
 4 (-64, 68)
 5 (-125, 130)

These are distinct because cubes are far enough spaced that adding n to n**3 is not enough to cross the gap to the next cube :n**3 < n+n**3 < (n+1)**3 for positive n, and symmetrically for negative n.

Answer 2

Snowman 0.1.0, 101 chars

}vg0aa@@*45,eQ.:?}0AaG0`NdE`;:?}1;bI%10sB%nM2np`*`%.*#NaBna!*+#@~%@0nG\]:.;:;bI~0-NdEnMtSsP" "sP.tSsP

Input on STDIN, space-separated output on STDOUT.

This uses the same method as isaacg's answer.

Commented version with newlines, for "readability":

}vg0aa          // get input, take the first char
@@*45,eQ.       // check if it's a 45 (ASCII for -) (we also discard the 0 here)
// this is an if-else
:               // (if)
  ?}0AaG        // remove first char of input (the negative sign)
  0`NdE`        // store a -1 in variable e, set active vars to beg
;
:               // (else)
  ?}1           // store a 1 in variable e, set active vars to beg
;bI             // active variables are now guaranteed to be beg
%10sB           // parse input as number (from-base with base 10)
%nM             // multiply by either 1 or -1, as stored in var e earlier
2np`*`          // raise to the power of 2 (and discard the 2)
%.              // now we have the original number in b, its square in d, and
                //   active vars are bdg
*#NaBna!*+#     // add abs(input number) to the square (without modifying the
                //   input variable, by juggling around permavars)
@~%@0nG\]       // active vars are now abcfh, and we have (0>n) in c (where n is
                //   the input number)
:.;:;bI         // if n is negative, swap d (n^2) and g (n^2+n)
~0-NdEnM        // multiply d by -1 (d is n^2 if n is positive, n^2+n otherwise)
tSsP            // print d
" "sP           // print a space
.tSsP           // print g

Commentary on the very first Snowman solution on PPCG: I think my design goal of making my language as confusing as possible has been achieved.

This actually could have been a lot shorter, but I'm an idiot and forgot to implement negative numbers for string -> number parsing. So I had to manually check whether there was a - as the first character and remove it if so.

Answer 3

Pyth, 15 11 bytes

4 bytes thanks to @Jakube

*RQ,hJ.aQ_J

Demonstration.

This maps as follows:

0  -> 0, 0
1  -> 2, -1
-1 -> -2, 1
2  -> 6, -4
-2 -> -6, 4

And so, on, always involving n^2 and n^2 + n, plus or minus.

Answer 4

APL, 15 bytes

{(-⍵*3)(⍵+⍵*3)}

This creates an unnamed monadic function that returns the pair -n^3 (-⍵*3), n+n^3 (⍵+⍵*3).

You can try it online.

Answer 5

Pyth - 11 10 bytes

Just multiplies by 10e10 and -10e10+1 Thanks to @xnor for showing me that I could use CG for the number.

*CGQ_*tCGQ

Try it online here.

Answer 6

O, 17 15 9 bytes

Uses some new features of O.

Q3^.Q+p_p

Older Version

[i#.Z3^*\Z3^)_*]o

Answer 7

Python 3, 29 27

Edit: doesn't meet the requirement in the 2nd "Details" bullet point

Bonus: it works from -99998 to 99998 inclusive

---

lambda n:[99999*n,-99998*n]

This creates an anonymous function*, which you can use by enclosing in brackets and then placing the argument in brackets afterwards like this:

(lambda n:[99999*n,-99998*n])(arg)

*Thanks to @vioz- for suggesting this.

---

Example input / output:

>>> (lambda n:[99999*n,-99998*n])(1)
[99999, -99998]
>>> (lambda n:[99999*n,-99998*n])(2)
[199998, -199996]
>>> (lambda n:[99999*n,-99998*n])(0)
[0, 0]
>>> (lambda n:[99999*n,-99998*n])(-1)
[-99999, 99998]
>>> (lambda n:[99999*n,-99998*n])(-2)
[-199998, 199996]
>>> (lambda n:[99999*n,-99998*n])(65536)
[6553534464, -6553468928]

Answer 8

Haskell, 16 bytes

I shamelessly copied @xnor's method. There probably isn't much better than this.

f x=(-x^3,x^3+x)