Dizzy integer enumeration

Question

Your challenge today is to output a given term of a sequence enumerating all of the integers. The sequence is as follows: If we have a 0-indexed function generating the sequence f(n) and ceil(x) is the ceiling function, then f(0) = 0; abs(f(n)) = ceil(n/2); sign(f(n)) is positive when n and ceil(n/2) are either both even or both odd.

To help understand this sequence, the first few terms are as follows: 0 1 -1 -2 2 3 -3 -4 4 5 -5 -6 6 7 -7...

Your task is to write a program to that takes an integer n and outputs the nth term of the sequence. Input may be 0 or 1-indexed only.

Test cases (0-indexed):

0  =>  0
1  =>  1
2  => -1
3  => -2
4  =>  2
5  =>  3

This is code-golf, fewest bytes wins!

Answer 1

Python 2, 26 24 bytes

lambda x:-~x/2/(1-(x&2))

Try it online!

Answer 2

SOGL V0.12, 8 6 bytes

I».»⌡±

Try it Here! or try the first couple numbers (changed a bit so it'd work)
0-indexed.

Explanation:

I       increment the input
 »      floor divide by 2
  .     push the original input
   »    floor divide by 2
    ⌡   that many times
     ±    negate

Or simpler:

(input + 1) // 2 negated input // 2 times
        I     »     ±      .     »    ⌡

Answer 3

JavaScript (ES6), 18 bytes

1-indexed.

n=>n/(++n&2||-2)|0

Demo

let f =

n=>n/(++n&2||-2)|0

for(n = 1; n < 20; n++) {
  console.log(n + ' --> ' + f(n))
}

Answer 4

Jelly, 6 bytes

HµĊN⁸¡

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Uses dzaima's algorithm.

-1 thanks to Jonathan Allan.

Answer 5

C, 25 bytes

f(n){return~n/2*~-(n&2);}

Answer 6

Haskell, 26 bytes

f x=div(x+1)2*(-1)^div x 2

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The other Haskell answers seem to be overcomplicating things… ^^

Answer 7

Pyke, 6 bytes

heQeV_

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Uses dzaima's approach... Beats Ties Jelly!

Explanation

h      - Increment the input, which is implicit at the beginning.
 e     - Floor halve.
  Q    - Push the input.
   e   - Floor halve.
    V_ - Apply repeatedly (V), ^ times, using negation (_).
       - Output implicitly.

The hex-encoded bytes equivalent would be: 68 65 51 65 56 5F.

Answer 8

05AB1E, 6 bytes

;DîsF(

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Uses dzaima's algorithm.

Answer 9

Python 2, 21 bytes

lambda x:-x/2*~-(x&2)

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Answer 10

Java 8, 15 bytes

n->~n/2*~-(n&2)

EDIT: Is Java really the shortest of the non-golfing languages?! o.Ô

Explanation:

Try it here.

I'll use the table below as reference of what's happening.

1. ~n is equal to -n-1.
2. Since integer division in Java automatically floors on positive integers and ceils on negative integers, ~n/2 will result in the sequence 0,-1,-1,-2,-2,-3,-3,-4,-4,-5,-5,...
3. n&2 will result in either 0 or 2, in the sequence 0,0,2,2,0,0,2,2,0,0,2,...
4. ~-x is equal to (x-1), so ~-(n&2) (((n&2)-1)) results in the sequence -1,-1,1,1,-1,-1,1,1,-1,-1,1,...
5. Multiplying the two sequences of ~n/2 and ~-(n&2) gives is the correct sequence asked in the challenge: 0,1,-1,-2,2,3,-3,-4,4,5,-5,...

Overview table:

n       ~n      ~n/2    n&2     ~-(n&2)     ~n/2*~-(n&2)
0       -1      0       0       -1          0
1       -2      -1      0       -1          1
2       -3      -1      2       1           -1
3       -4      -2      2       1           -2
4       -5      -2      0       -1          2
5       -6      -3      0       -1          3
6       -7      -3      2       1           -3
7       -8      -4      2       1           -4
8       -9      -4      0       -1          4
9       -10     -5      0       -1          5
10      -11     -5      2       1           -5

Answer 11

Mathematica, 24 bytes

(s=⌈#/2⌉)(-1)^(#+s)&  

-14 bytes from @Misha Lavrov

Answer 12

Haskell, 25 43 42 bytes

((do a<-[0..];[[-a,a],[a,-a]]!!mod a 2)!!)

Try it online! 1-indexed.

Edit: The previous version had the signs in a wrong order, thanks to @Potato44 for pointing out. Fixed for 18 bytes ...

Edit 2: Thanks to BMO for -1 byte!

Answer 13

Python 3, 29 bytes

lambda n:-~n//2*(-1)**(n%4>1)

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Answer 14

Pyth, 9 bytes

_F/hQ2/Q2

Try it here!

Uses dzaima's approach.

Answer 15

Haskell, 36 bytes

g x=x: -x:g(-x-signum x)
((0:g 1)!!)

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Answer 16

Batch, 29 bytes

@cmd/cset/a"%1/2^(%1<<30>>30)

Answer 17

JavaScript (ES6), 18 bytes

f=
n=>n/2^(n<<30>>30)

<input type=number min=0 value=0 oninput=o.textContent=f(this.value)><pre id=o>0

0-indexed.

Answer 18

Javascript, 17 bytes

n=>~n/2*~-(n&2)^0

f=
n=>~n/2*~-(n&2)^0

<input type=number min=0 value=0 oninput=o.textContent=f(this.value)><pre id=o>0

This one is 0 indexed. It's entirely bitwise trickery.

Answer 19

Cubically, 23 bytes

(1-indexed)

FDF'$:7+8/0_0*0-8*7/0%6

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The main difficulty when writing code in Cubically are:

• There is only 1 write-able variable, and
• Get constants is hard.

So, this solution calculate

((((n+1)/2)%2)*2-1)*n/2

where / denotes integer division. That only need 1 temporary variable, and constants 1 and 2.

Explanation:

FDF'$:7+8/0_0*0-8*7/0%6
FDF'                      Set face value of face 0 to 2, and value of memory index 8 (cube is unsolved) to 1 (true = unsolved)
    $                     Read input
     :7                                 input
       +8                                + 1
         /0                        (        ) /2
           _0                     (             ) %2
             *0                  (                  ) *2
               -8                                        -1
                 *7             (                          ) *n
                   /0                                          /2
                     %6   Print

Answer 20

TI-Basic (TI-84 Plus CE), 20 bytes

‾int(‾Ans/2)(1-2remainder(int(Ans/2),2

A full program that is called like 5:prgmNAME.

TI-Basic is a tokenized lanugage, all tokens used here are one byte, except for remainder( which is two. ‾ represents the regative token, which is typed with the (-) key.

Examples:

0:prgmNAME
 => 0
1:prgmNAME
 => 1
2:prgmNAME
 => -1
#etc

Explanation:

‾int(‾Ans/2)(1-2remainder(int(Ans/2),2
‾int(‾Ans/2)                           # -int(-X) is ciel(X), so ciel(Ans/2)
                          int(Ans/2)   # int(X) is floor(X), so floor(Ans/2)
                remainder(int(Ans/2),2 # 1 if floor(Ans/2) is odd else 0
            (1-2remainder(int(Ans/2),2 # -1 if floor(Ans/2) is odd, else 1
_int(_Ans/2)(1-2remainder(int(Ans/2),2 # -ciel(Ans/2) if floor(Ans/2) is odd, else ciel(Ans/2)

Same formula as a Y-var function:

Y1= ‾int(‾X/2)(1-2remainder(int(X/2),2

Answer 21

dc, 16 bytes

1+d2~+2%2*1-r2/*

I am sure there's a way to make 0..1 to -1..1 in dc shorter, but no ideas for now.

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Answer 22

Brain-Flak, 86 74 72 70 bytes

{({}[()]<({}<>([({})]{(<{}([{}]())>)}{}())<>)>)}{}<>{}{<>([{}])(<>)}<>

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Explanation

There are two parts to this code. The first part

({}[()]<({}<>([({})]{(<{}([{}]())>)}{}())<>)>)}{}

does the brawn of the computation. It determines ceil(n/2) and whether or not to negate the output.

To explain how it works I will first explain how one would calculate ceil(n/2). This could be done with the following code

{({}[()]<({}([{}]()))>)}{}

This counts down from n each time it performs a not (([{}]())) on a counter and adds the counter to a result. Since the counter is zero half the time we only increment every other run starting with the first one.

Now I want to also compute the sign of our results. To do this we start another counter. This counter only changes state if the first counter is off. That way we get the desired pattern. We put these two counters on the off stack to ease with moving them around when the time comes.

Now once we have finished that computation our stack looks like this

          parity(n)
ceil(n/2) sign

So we need to do some work to get the intended result this second part does it.

<>{}{<>([{}])(<>)}<>

Answer 23

Perl 5, 32 + 1 (-p) = 33 bytes

$_='-'x(($_+++($b=0|$_/2))%2).$b

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Answer 24

Proton, 23 bytes

n=>-~n//2//(-1)**(n//2)

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Port of Halvard's solution.

Proton, 23 bytes

n=>-~n//2*(-1)**(n%4>1)

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Port of Leaky's solution.

A bit more Protonic, 24 bytes:

n=>-~n//2*(**)(-1,n%4>1)

Answer 25

QBIC, 27 26 bytes

g=(:+1)'\2`~(a-g)%2|?-g\?g

Explanation

g=          set worker var 'g' to
(:+1)           our index (plus one for the ceil() bit)
'\2`            integer divided by 2 (the int div needs a code literal: '..`
~(a-g)%2    IF index - temp result is odd (index 2 minus result 1 = 1)
|?-g        THEN PRINT g negated
\?g         ELSE PRINT g

Answer 26

Clojure 122 bytes

Verbose, even when golfed. I'm going for the sympathy vote here... :-)

Golfed:

(defn d[n](let[x(int(Math/ceil(/ n 2)))y(cond(or(and(even? n)(even? x))(and(odd? n)(odd? x)))(Math/abs x):else(- 0 x))]y))

Ungolfed:

(defn dizzy-integer [n]
  (let [x   (int (Math/ceil (/ n 2)))
        y   (cond
                (or (and (even? n) (even? x))
                    (and (odd? n)  (odd? x))) (Math/abs x)
                :else (- 0 x)) ]
    y))

Answer 27

Excel VBA 32-Bit, ₃₉ 37 Bytes

Anonymous VBE immediate window function that takes input from cell A1 and outputs to the VBE immediate window

?[Sign((-1)^Int(A1/2))*Int((A1+1)/2)]

Restricted to 32-Bit as A^Bis not valid in 64-Bit (A ^B is as close as can be achieved)

Answer 28

Julia 0.6, 16 bytes

It's just the java solution except I need ÷ for integer division.

n->~n÷2*~-(n&2)

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Answer 29

Husk, 9 bytes

!m?I_%2İZ

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1-indexed.

Explanation

!m?I_%2İZ
       İZ List of all integers (builtin) [0,1,-1,2,-2,...]
 m        map to the following function
  ?  %2   if the number is even,
    _     negate it.
   I      otherwise return it
!         find element at input index

Alternate solution, 9 bytes

!⌊½¹¡_⌊½→

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Uses the same equation as the SOGL answer.

Answer 30

Japt, 8 bytes

z°U&2ªJÑ

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