Bitflip and negate

Question

Given an integer, make an expression that produces it from 0 using unary negation - and bitwise complement ~ (~n = -n-1), with the operators applied right to left.

...
-3  = ~-~-~0
-2  = ~-~0
-1  = ~0
 0  = 0
 1  = -~0
 2  = -~-~0
 3  = -~-~-~0
...

Your expression must be as short as possible, which means no redundant parts of ~~, --, -0, or 00. Output or print the expression as a string or a sequence of characters.

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Answer 1

Python, 32 bytes

lambda x:("-~"*abs(x))[x<0:]+"0"

Anonymous lambda function. Given an integer x writes "-~" abs(x) times and removes the first char if x is negative, then a zero is added to the end.

Answer 2

JavaScript (ES6), 33 31 bytes

f=x=>x<0?"~"+f(~x):x&&"-"+f(-x)

Recursion < built-ins < loops (at least in this case). Basically unevaluates the input:

• if it's less than 0, flip it and add a ~ to the string;
• if it's more than 0, negate it and add a - to the string;
• if it's exactly 0, return 0.

Takes advantage of this pattern:

 0         = 0
-1 = ~( 0) = ~0
+1 = -(-1) = -~0
-2 = ~(+1) = ~-~0
+2 = -(-2) = -~-~0
-3 = ~(+2) = ~-~-~0
+3 = -(-3) = -~-~-~0
etc.

Answer 3

Pyth, 14 13 12 Bytes

_<>0Q+0sm"~-

-2 Bytes thanks to @StevenH.

test suite

Decided to try out Pyth, so i translated my python answer to it. Any help welcome!

Explanation:

_<>0Q+0sm"~-     
        m"~-     # Map "~-" onto the input (= a list of n times "~-").
       s         # Join the list to a string.
     +0          # Add "0" in front. 
 <>0Q            # Slice off the last char if the input is negative.
_                # Reverse the whole thing.

Answer 4

C, 46 bytes

m(x){putchar(x?x<0?126:45:48);x&&m(-x-(x<0));}

Unlike most (all?) other answers, this one outputs the operators ~ and - one by one.

Answer 5

05AB1E, 14 13 bytes

Ä„-~×¹0‹i¦}0J

Explanation

 „-~           # the string "-~"
Ä   ×          # repeated abs(input) times
     ¹0‹i¦}    # if input is negative, remove the first char
           0J  # join with 0

Try it online!

Answer 6

Retina, 19 17 bytes

Replace the number with unary, with a zero on the end. Replace each 1 with -~. Remove double negative if there is one.

\d+
$*10
1
-~
--

Try it online

All test cases at once (slightly modified program to support multiple test cases)

Answer 7

Perl 38 35 33 (23 + 1 for -p) 24

s/\d+/"-~"x$&.0/e;s;--;

-13 thanks to Dada

Answer 8

Haskell, 41 bytes

f n=['-'|n>0]++(tail$[1..abs n]>>"-~")++"0"

f n|n<0=tail$f(-n)|x<-[1..n]>>"-~"=x++"0"

Thanks to nimi for 3 bytes

Answer 9

Dyalog APL, 18 bytes

'0',⍨0∘>↓'-~'⍴⍨2×|

'0',⍨ character zero appended to

0∘> negativeness (i.e. 1 for numbers under 0; 0 for zero and up)

↓ dropped from

'-~'⍴⍨ the string "~-" cyclically reshaped to length

2× two times

| the absolute value

+ plus

0∘< positiveness (i.e 1 for numbers over 0)

TryAPL online!

Answer 10

V, 21 bytes

/ä
é
D@"ña-~ñá0kgJó--

Try it online!

V has very limited number support, and it actually has no concept of negative numbers. This means in order to support negatives (or even 0), we have to use some hacky workarounds.

Explanation:

/ä                  "Move forward to the first digit
é                   "And enter a newline
D                   "Delete this number, into register '"'
 @"                 "That number times:
   ñ   ñ            "Repeat the following:
    a               "  Append the string:
     -~             "  '-~'
        á0          "Append a 0
          k         "Move up a line
           gJ       "And join these two lines together
             ó--    "Remove the text '--', if it exists

Answer 11

JavaScript (ES6), 39 37 bytes

x=>"-~".repeat(x<0?-x:x).slice(x<0)+0

Saved 2 bytes thanks to @Neil

Answer 12

Ruby, 34 bytes

->x{("-~"*x.abs+?0)[2[0<=>x]..-1]}

Answer 13

Jelly, 10 bytes

A⁾-~ẋḊẋ¡N0

This is a full program. Try it online!

How it works

A⁾-~ẋḊẋ¡N0  Main link. Argument: n

A           Take the absolute value of n.
 ⁾-~ẋ       Repeat the string "-~" that many times. Result: s
       ¡    Conditional application:
     Ḋ        Dequeue; remove the first element of s...
      ẋ N     if s, repeated -n times, is non-empty.
         0  Print the previous return value. Set the return value to 0.
            (implicit) Print the final return value.

Answer 14

Java 7, 95 79 bytes

79 bytes:

String s(int x){String t=x<0?"~":"";while((x<0?++x:x--)!=0)t+="-~";return t+0;}

Ungolfed:

String s(int x) {
    String t = x<0 ? "~" : "";
    while((x<0 ? ++x : x--) != 0)
        t += "-~";
    return t+0;
}

Old version (95 bytes):

String s(int x){return new String(new char[x<0?-x:x]).replace("\0","-~").substring(x<0?1:0)+0;}

Usage:

class A {
    public static void main(String[]a) {
        System.out.println(s(-3));
        System.out.println(s(-2));
        System.out.println(s(-1));
        System.out.println(s(0));
        System.out.println(s(1));
        System.out.println(s(2));
        System.out.println(s(3));
    }
    static String s(int x){String t=x<0?"~":"";while((x<0?++x:x--)!=0)t+="-~";return t+0;}
}

Try it here!

Output:

~-~-~0
~-~0
~0
0
-~0
-~-~0
-~-~-~0

Answer 15

EXCEL: 55 33 bytes

=REPT("-~",IF(A1>0,A1,ABS(A1)-1))&"0"

Input is in the form of putting a number in the A1 cell. Formula can go anywhere except A1.

Answer 16

CJam, 18 14 bytes

Took some inspiration from Emigna's answer to save 4 bytes.

li_z"-~"*\0<>0

Try it online! (As a linefeed-separated test suite.)

Explanation

li      e# Read input and convert to integer N.
_z      e# Duplicate and get |N|.
"-~"*   e# Repeat this string |N| times.
\0<     e# Use the other copy of N to check if it's negative.
>       e# If so, discard the first '-'.
0       e# Put a 0 at the end.

Answer 17

T-SQL, 87 bytes

select substring(replicate('-~',abs(x)),case when x<0then 2 else 1 end,x*x+1)+'0'from #

The x*x+1 condition in substring is sufficient, since x^2+1>=2*abs(x) for all x.

As usually in SQL, the input is stored in a table:

create table # (x int)

insert into # values (0)
insert into # values (1)
insert into # values (-1)
insert into # values (2)
insert into # values (-2)

Answer 18

Vim - 31 keystrokes

First vim golf, prolly missed a ton of stuff.

`i-~<esc>:s/-\~-/\~-/dwp<left>ii<esc><left>d$@"<esc>a0`

Answer 19

Matlab, 61 bytes

x=input('');A=repmat('-~',1,abs(x));disp([A((x<0)+1:end) 48])

Answer 20

Pyke, 14 13 bytes

X,"-~"*Q0<>0+

Try it here!

Answer 21

Perl 6, 25 bytes

{substr '-~'x.abs~0,0>$_}

Explanation:

{
  substr
    # string repeat ｢-~｣ by the absolute value of the input
    '-~' x .abs

    # concatenate 0 to that
    ~ 0

    ,

    # ignore the first character of the string if it is negative
    0 > $_
}

Answer 22

Jelly, 14 12 bytes

-2 bytes thanks to @Dennis (return 0 rather than concatenate "0", making this a full program only.)

0>‘
A⁾-~ẋṫÇ0

Test it at TryItOnline

How?

0>‘      - link 1 takes an argument, the input
0>       - greater than 0? 1 if true 0 if false
  ‘      - increment

A⁾-~ẋṫÇ0 - main link takes an argument, the input
      Ç  - y = result of previous link as a monad
A        - x = absolute value of input
 ⁾-~     - the string "-~"
    ẋ    - repeat the sting x times
     ṫ   - tail repeatedString[y:] (y will be 1 or 2, Jelly lists are 1-based)
       0 - implicit print then return 0

Answer 23

><>, 18 + 3 = 22 bytes

:?!n0$-:0):1go-
-~

Try it online! +3 bytes for the ​ -v flag to initialise the stack with the input. If assuming that STDIN is empty is okay, then the following is a byte shorter:

:?!ni*:0):1go-
-~

The program keeps flipping the input n as necessary until it reaches 0, after which it errors out.

[Loop]
:?!n      If n is 0, output it as a num. If this happens then the stack is now
          empty, and the next subtraction fails
0$-       Subtract n from 0
:0)       Push (n > 0)
:1go      Output the char at (n>0, 1) which is a char from the second line
-         Subtract, overall updating n -> -n-(n>0)

Answer 24

PseudoD, 688 579 521 bytes

utilizar mate.pseudo
utilizar entsal.pseudo
adquirir n
adquirir a
adquirir r
adquirir i
fijar n a llamar LeerPalabra finargs
si son iguales n y CERO
escribir {0}
salir
fin
fijar a a llamar ValorAbsoluto n finargs
fijar i a CERO
si comparar Importar.Ent.Comparar n < CERO
fijar r a {~}
sino
fijar r a {-}
fin
mientras comparar Importar.Ent.Comparar i < a
escribir r finargs
si son iguales r y {~}
fijar r a {-}
Importar.Ent.Sumar i UNO i
sino
fijar r a {~}
fin
finbucle
si son iguales r y {~}
escribir {~}
fin
escribir {0}

Explain:

Read a number from STDIN;
If the number is zero (0); Then:
    Writes 0 to STDOUT and exits;
End If;
If the number is less than zero (0); Then:
    Set the fill character to "~";
Else:
    Set the fill character to "-";
End If;
For i = 0; While i is less than abs(number); do:
    Write the fill character to STDOUT;
    If the fill character is "~":
        Set the fill character to "-"
        Increment i by one
    Else:
        Set the fill character to "~"
    End if;
End for;
If the fill character is "~"; Then:
    Write "~" to STDOUT;
End If;
Write "0" to STDOUT

Answer 25

Octave, 51 bytes

x=input('');[("-~"'*[1:abs(x)>0])((x<0)+1:end),'0']

At first blatantly copying the Matlab approach by @pajonk and then modifying some details, rewriting as an "outer product" between a vector of ones and the characters "-~" and abusing on-the-fly-indexing (or what it could be called) lets us save some bytes. It still pains me slightly that I can't get the index expression to take fewer bytes.

Octave allows a(i1)(i2) or even (...)(i1)(i2) for indexing where Matlab would want us to store variables in between the indexings.

((x<0)+1:end)

is far too long to describe "skip first if". There must be a better way.

Answer 26

PHP, 61 bytes

if(0>$n=$argv[1]){echo"~";$n=~$n;}echo str_repeat("-~",$n),0;

Answer 27

PHP, 58 bytes

<?=((0<$a=$argv[1])?'-':'').str_pad('0',2*abs($a),'~-',0);

Answer 28

Labyrinth, 25 bytes

`?+#~.
.  ; 6
54_"#2
  @!

Try it online!

Explanation

I really like the control flow in this one. The IP runs in a figure 8 (or actually a ∞, I guess) through the code to reduce the input slowly to 0 while printing the corresponding characters.

The code starts in the upper left corner going right. The ` doesn't do anything right now. ? reads the input and + adds it to the implicit zero below. Of course that doesn't do anything either, but when we run over this code again, ? will push a zero (because we're at EOF), and + will then get rid of that zero.

Next the # pushes the stack depth, simply to ensure that there's a positive value on the stack to make the IP turn south, and ; discards it again.

The " is a no-op and acts as the main branch of the code. There are three cases to distinguish:

• If the current value is positive, the IP turns right (west) and completes one round of the left loop:

  _45.`?+
  _45      Push 45.
     .     Print as character '-'.
      `    Negate the current value (thereby applying the unary minus).
       ?+  Does nothing.
  

• If the current value is negative, the IP turns left (east) and the following code is run:

  #26.~
  #        Push stack depth, 1.
   26      Turn it into a 126.
     .     Print as character '~'.
      ~    Bitwise NOT of the current value (applying the ~).
  

  Note that these two will alternate (since both change the sign of the input) until the input value is reduced to zero. At that point...

• When the current value is zero, the IP simply keeps moving south, and executes the ! and then turns west onto the @. ! prints the 0 and @ terminates the program.

Answer 29

GolfScript

, 30 24 20 bytes

• Saved 6 bytes thanks to xnor.
• Saved 4 bytes thanks to Dennis.

~."-~"\abs*\0<{(;}*0

Input: -5

Output: -5 = ~-~-~-~-~0

Explanation

~.     # Input to integer and duplicate
"-~"   # We shall output a repetition of this string
\abs   # Move the input onto the stack and computes abs
*      # Multiply "-~" for abs(input) times
\      # Copy onto the stack the input
0<     # Is it less than 0?
{(;}*  # Yes: remove first '-' from the output
0      # Push 0

Try it online!