# Divide a number by 3 without using *, /, +, -, % operators [closed]

Quoting this question on SO (Spoiler alert!):

This question has been asked in an Oracle interview.

How would you divide a number by 3 without using *, /, +, -, %, operators?

The number may be signed or unsigned.

The task is solvable, but see if you can write the shortest code.

## Rules:

• Perform the required integer division (/3)
• Do not use the non-text-based operators *, /, +, -, or % (or their equivalents, such as __div__ or add()). This also applies to incrementing and decrementing operators, like i++ or i--. Use of operators for string concatenation and formatting are OK. Using these characters for different operators, such as unary - operator for negative numbers, or * to represent a pointer in C is OK as well.
• Input value can be arbitrarily large (whatever your system can handle), both positive and negative
• Input can be on STDIN or ARGV or entered any other way
• Create the shortest code you can to do the above
• How should the result be rounded when positive? How when negative? Jun 22, 2019 at 22:37

# C, 167503724710

Here's my solution to the problem. I admit it is unlikely to win a strict code golf competition, but it doesn't use any tricks to indirectly call built-in division functionality, it is written in portable C (as the original Stack Overflow question asked for), it works perfectly for negative numbers, and the code is exceptionally clear and explicit.

My program is the output of the following script:

#!/usr/bin/env python3
import sys

# 71
sys.stdout.write('''#include <stdint.h>
#include <stdio.h>
int32_t div_by_3(int32_t input){''')

# 39 * 2**32
for i in range(-2**31, 2**31):
# 18 + 11 + 10 = 39
sys.stdout.write('if(input==%11d)return%10d;' % (i, i / 3))

# 95
sys.stdout.write(r'''return 7;}int main(int c,char**v){int32_t n=atoi(a[1]);printf("%d / 3 = %d\n",n, div_by_3(n));}''')


Character count: 71 + 39 * 2**32 + 95 = 167503724710

### Benchmarks

It was asked how long this would take and how much memory it would use, so here are some benchmarks:

• Script execution time — Running ./test.py | pv --buffer-size=1M --average-rate > /dev/null for about 30 seconds gives a rate of about 14.8 MB/s. The rate of output can reasonably be assumed to be roughly constant, so the running time to completion should be about 167503724710 B / (14.8 * 1048576 B/s) ≈ 10794 s.
• Compilation time — The TCC compiler claims to compile C code at 29.6 MB/s, which makes for a compilation time of 167503724710 B / (29.6 * 1048576 B/s) ≈ 5397 s. (Of course this can run in a pipeline with the script.)
• Size of compiled code — I tried estimating it using ./test.py | tcc -c - -o /dev/stdout | pv --buffer-size=1M --average-rate > /dev/null, but it seems tcc doesn't output anything until it reads the entire source file in.
• Memory usage to run — Since the algorithm is linear (and tcc doesn't optimize across lines), the memory overhead should be only a few kilobytes (apart from the code itself, of course).
• This is the epitome of hardcoding. ++++++++++ Sep 14, 2013 at 18:45
• That being said, I'm sure if you gave them a 160 GB source file and asked them to compile and test it, they'd look at you like you were crazy. Sep 14, 2013 at 18:46
• If my boss asked me to compute a division by three without - + / * % I would think he's crazy. Jan 2, 2014 at 12:35
• And yet, a[b] is a syntactic sugar for *(a + b), which does the addition. Jan 2, 2014 at 14:17
• @NicolasBarbulesco There is a size restriction on Stack Exchange answers. Jan 3, 2014 at 23:17

# Ruby 28

b=->n{n.to_s(3).chop.to_i 3}


To divide by 3 we just need to remove the trailing zero in base 3 number: 120 -> 11110 -> 1111 -> 40

Works with negatives:

ice distantstar:~/virt/golf [349:1]% ruby ./div3.rb
666
222
ice distantstar:~/virt/golf [349]% ruby ./div3.rb
-15
-5


# Ruby, 6045

Alternatively, w/o using base conversion:

d=->n{x=n.abs;r=(0..1.0/0).step(3).take(x).index x;n>0?r:-r}

d=->n{(r=1.step(n.abs,3).to_a.size);n>0?r:-r}

• The alternate without base conversion has the banned / operator where Float::INFINITY became 1.0/0. With Ruby 2.1, one may golf (0..1.0/0).step(3) into 0.step(p,3), removing the /. The bigger problem is that -r uses - to negate. It costs 5 characters to change -r to ~r.pred, abusing Integer#pred to subtract 1 without the subtraction operator. May 25, 2014 at 15:57

# Mathematica, 13 chars

Mean@{#,0,0}&

• This is wicked :D I guess you could save the & and use a plain variable (others here do that too). Jan 3, 2014 at 10:20
• @YvesKlett: Mean is inherently wicked. Nov 7, 2016 at 17:33

## JavaScript, 56

alert(Array(-~prompt()).join().replace(/,,,/g,1).length)


Makes a string of length n of repeating ,s and replaces ,,, with 1. Then, it measures the string's resulting length. (Hopefully unary - is allowed!)

• +1 but it doesn't work with negative values Mar 6, 2014 at 9:28
• Uh, question... how does this count? It uses the - negation operator. Feb 13, 2016 at 0:07
• @Patrick I took the specification to mean no subtraction -- if you want, you can replace -~ with parseInt() Feb 13, 2016 at 22:22
• @CaseyChu the value returned by -~prompt() is one greater than parseInt(prompt()). Not sure how you'd deal with that. Feb 13, 2016 at 22:35
• alert(Array(parseInt(prompt())).slice(1).join().replace(/,,,/g,1).length) Feb 13, 2016 at 22:37

## Python, 41 38

print"-"[x:]+len(xrange(2,abs(x),3))


xrange seems to be able to handle large numbers (I think the limit is the same as for a long in C) almost instantly.

>>> x = -72
-24

>>> x = 9223372036854775806
3074457345618258602

• 10/3 equals 3, not 4. Aug 1, 2012 at 11:14
• Doing print" -"[x<0]+len(range(2,abs(x),3)) will shave it down to 39 chars Aug 4, 2012 at 19:37
• golfexchange's comment formatting is messing it up. on the above I used backticks to enclose len() as shorthand for repr() Aug 4, 2012 at 19:38
• I've updated it. I can't use range, because it will actually create the list. xrange just pretends to, so it is able to handle huge numbers without wasting time/memory.
– grc
Aug 5, 2012 at 1:31
• Pretend it's Python 3 ;) I like the single char slicing btw. Aug 5, 2012 at 1:47

# J, 45 44 10 chars

".,&'r3'":

Works with negatives:

".,&'r3'": 15
5
".,&'r3'": _9
_3
".,&'r3'": 3e99
1e99


": - format as text

,&'r3' - append r3 to the end

". - execute the string, e.g. 15r3

• It works if you do 3 3 3 #: 9. It looks like you need to know how long your ternary number will be. _3]\i. is also a possible starting point for something, but I don't know if it would be shorter than your solution here. The problem with #_3]\i. as it stands is that it always rounds up instead of down. Aug 17, 2012 at 11:21
• Maybe ##~3=_3#\i. for 11 characters? Aug 17, 2012 at 11:31
• Actually, you can shrink yours down to 10 characters with ##~0 0 1$~. Aug 17, 2012 at 11:34 • You can shrink that down using a hook to 3#.}:(#:~$&3) but it's still longer and it doesn't fix the negative number issue. Aug 17, 2012 at 11:54
• Yes, you can use either the Power function ^: or Agenda @. for an if or if...else replacement. In this case you might be able to use @. with two verbs connected with a '' character (a gerund in J-speak) to select one or the other based on a condition. Aug 17, 2012 at 13:43

d n=snd.head.dropWhile((/=n).fst)$zip([0..]>>=ν)([0..]>>=replicate 3>>=ν);ν q=[negate q,q]  Creates an infinite (lazy) lookup list [(0,0),(0,0),(-1,0),(1,0),(-2,0),(2,0),(-3,-1),(3,1), ...], trims all elements that don't match n (/= is inequality in Haskell) and returns the first which does. This gets much simpler if there are no negative numbers: ## 25 27 (([0..]>>=replicate 3)!!)  simply returns the nth element of the list [0,0,0,1,1,1,2, ...]. • o.O i never thought of that second solution. I might be able to implement something like that in python Aug 2, 2012 at 19:19 ## C#, 232 bytes My first code golf... And since there wasn't any C# and I wanted to try a different method not tried here, thought I would give it a shot. Like some others here, only non-negative numbers. class l:System.Collections.Generic.List<int>{}class p{static void Main(string[] g){int n=int.Parse(g[0]);l b,a=new l();b=new l();while(a.Count<n)a.Add(1);while(a.Count>2){a.RemoveRange(0,3);b.Add(1);}System.Console.Write(b.Count);}}  ## Ungolfed class l : System.Collections.Generic.List<int> { } class p { static void Main(string[] g) { int n = int.Parse(g[0]); l b, a = new l(); b = new l(); while (a.Count < n) a.Add(1); while (a.Count > 2) { a.RemoveRange(0, 3); b.Add(1); } System.Console.Write(b.Count); } }  • About 5 years later.. You can save 1 byte by removing the space from string[] g, turning it into string[]g Feb 8, 2017 at 14:44 • Quoting "Do not use the non-text-based operators *, /, +, -, or % (or their equivalents, such as div or add()" -- are you not using an equivalent -- .Add? Jun 20, 2019 at 21:22 • @JonathanFrech That add method does not work on two numbers, it only adds a value to a collection Aug 30, 2019 at 3:07 ## Perl (26 22) $_=3x pop;say s|333||g


This version (ab)uses Perl's regex engine. It reads a number as the last command line argument (pop) and builds a string of 3s of this length ("3" x $number). The regex substitution operator (s///, here written with different delimitiers because of the puzzle's rules and with a global flag) substitues three characters by the empty string and returns the number of substitutions, which is the input number integer-divided by three. It could even be written without 3, but the above version looks funnier. $ perl -E '$_=3x pop;say s|333||g' 42 14  • Hey @memowe, nice work! You could save a few more chars (4) by doing $_=3x pop;say s|333||g. Jan 2, 2014 at 10:43
• When the input is 0, 1, or 2, then it prints an empty string. If it needs to print 0, then it needs 3 more characters (25 total): '$_=3x pop;say s|333||g||0. Slow with large numbers like 99999999, and doesn't work with negative numbers. May 25, 2014 at 16:04 • Use -p on the commandline, and you can do: $_=3x$_;$_=0|s|...||g for a total of 22, including coverage of the 0, 1, or 2 inputs. Jun 20, 2019 at 23:06

## C, 160 chars

Character by character long division solution using lookup tables, i.e. without string atoi() or printf() to convert between base 10 strings and integers.

Output will sometimes include a leading zero - part of it's charm.

main(int n,char**a){
char*s=a[1],*x=0;
if(*s==45)s=&s[1];
for(;*s;s=&s[1])n=&x[*s&15],x="036"[(int)x],*s=&x["000111222333"[n]&3],x="012012012012"[n]&3;
puts(a[1]);
}


Note:

• compiles with clang 4.0, other compilers may barf.

Testing:

./a.out -6            -2
./a.out -5            -1
./a.out -4            -1
./a.out -3            -1
./a.out -2            -0
./a.out -1            -0
./a.out 0             0
./a.out 1             0
./a.out 2             0
./a.out 3             1
./a.out 4             1
./a.out 5             1
./a.out 6             2
./a.out 42            14
./a.out 2011          0670


# Python 42

int(' -'[x<0]+str(len(range(2,abs(x),3))))


Since every solution posted here that Ive checked truncates decimals here is my solution that does that.

# Python 50 51

int(' -'[x<0]+str(len(range([2,0][x<0],abs(x),3))))


Since python does floor division, here is my solution that implements that.

Input integer is in the variable x.

Tested in Python 2.7 but I suspect it works in 3 as well.

• +1 For offering both alternatives to the negative value situation. Since there are already so many answers, I'll not be adjusting the spec to exclude one or the other option, though I would personally agree that -3 is the correct answer to -10/3. Aug 1, 2012 at 13:47
• For those who care about floor division in python: python-history.blogspot.com/2010/08/…
– Matt
Aug 1, 2012 at 14:22
• what's with the multiplication and subtraction in your second solution? Aug 1, 2012 at 15:40
• @boothby The second solution implements floor division. I wanted to do range(0,abs(x),3) for negative numbers and range(2,abs(x),3) for positive numbers. In order to do that I had range(2... then i subtracted 2 when x is negative. X<0 is True when x is negative, (True)*2 == 2
– Matt
Aug 1, 2012 at 16:34
• I'm not understanding the difference between floor division and truncating decimals. Does this have to with negative division? Aug 1, 2012 at 19:12

## JavaScript, 55

alert(parseInt((~~prompt()).toString(3).slice(0,-1),3))


If one can't use -1, then here is a version replacing it with ~0 (thanks Peter Taylor!).

alert(parseInt((~~prompt()).toString(3).slice(0,~0),3))

• @ArtemIce One ~ is a Bitwise operator which inverts the bits of the operand (first converting it to a number). This is the shortest way to convert a string to a number (as far as I know). Aug 3, 2012 at 10:02
• I feel like using string parsing/conversion is cheating, since its a) a very complicated and expensive process compared to bitwise operations, b) uses the forbidden operators internally, and c) would take up waaaaay more characters than a homerolled solution. Kind of like how people get grumpy when you use the built in sorts when asked to implement a quicksort.
– Wug
Aug 3, 2012 at 14:46
• @Sam Also, ~~ converts to an integer, as opposed to +. Aug 5, 2012 at 4:51
• @Wug It is codegolf, thus it's not about efficiency unless specified in the task. Aug 5, 2012 at 5:56
• +1 for taking advantage of different bases. It's one of my favorite JavaScript golf techniques. Aug 26, 2012 at 3:32

# C 83 characters

The number to divide is passed in through stdin, and it returns it as the exit code from main() (%ERRORLEVEL% in CMD). This code abuses some versions of MinGW in that when optimizations aren't on, it treats the last assignment value as a return statement. It can probably be reduced a bit. Supports all numbers that can fit in to an int

If unary negate (-) is not permitted: (129)

I(unsigned a){a=a&1?I(a>>1)<<1:a|1;}main(a,b,c){scanf("%i",&b);a=b;a=a<0?a:I(~a);for(c=0;a<~1;a=I(I(I(a))))c=I(c);b=b<0?I(~c):c;}


If unary negate IS permitted: (123)

I(unsigned a){a=a&1?I(a>>1)<<1:a|1;}main(a,b,c){scanf("%i",&b);a=b;a=a<0?a:-a;for(c=0;a<~1;a=I(I(I(a))))c=I(c);b=b<0?-c:c;}


# EDIT: ugoren pointed out to me that -~ is an increment...

83 Characters if unary negate is permitted :D

main(a,b,c){scanf("%i",&b);a=b;a=a<0?a:-a;for(c=0;a<~1;a=-~-~-~a)c=-~c;b=b<0?-c:c;}

• If unary negate is permitted, x+3 is -~-~-~x. Nov 23, 2012 at 9:11
• Thank you for that. I don't know why that never occurred to me. I guess I didn't realize you could stack unaries so gratuitously hehe. Nov 24, 2012 at 7:59

# C, 139 chars

t;A(a,b){return a?A((a&b)<<1,a^b):b;}main(int n,char**a){n=atoi(a[1]);for(n=A(n,n<0?2:1);n&~3;t=A(n>>2,t),n=A(n>>2,n&3));printf("%d\n",t);}


Run with number as command line argument

• Handles both negative and positive numbers

Testing:

 ./a.out -6            -2
./a.out -5            -1
./a.out -4            -1
./a.out -3            -1
./a.out -2            0
./a.out -1            0
./a.out 0             0
./a.out 1             0
./a.out 2             0
./a.out 3             1
./a.out 4             1
./a.out 5             1
./a.out 6             2
./a.out 42            14
./a.out 2011          670


Edits:

• saved 10 chars by shuffling addition (A) to remove local variables.
• Nicely done. I tried my best at bit twiddling and got to 239. I just can't get my head around your A, my function just checks the bit i in number n. Does C standard allow omitting type declarations or is that some compiler thing? Aug 1, 2012 at 9:28
• C will assume int if unspecified.
– Wug
Aug 3, 2012 at 14:48

## ZSH — 31 20/21

echo {2..x..3}|wc -w


For negative numbers:

echo {-2..x..3}|wc -w


### With negative numbers (ZSH + bc) — 62 61

I probably shouldn't give two programs as my answer, so here's one that works for any sign of number:

echo 'obase=10;ibase=3;'echo 'obase=3;x'|bc|sed 's/.$//'|bc  This uses the same base conversion trick as Artem Ice's answer. ## C, 81 73 chars Supports non-negative numbers only. char*x,*i; main(){ for(scanf("%d",&x);x>2;x=&x[~2])i=&i[1]; printf("%d",i); }  The idea is to use pointer arithemtic. The number is read into the pointer x, which doesn't point anywhere. &x[~2] = &x[-3] = x-3 is used to subtract 3. This is repeated as long as the number is above 2. i counts the number of times this is done (&i[1] = i+1). • Trying to understand the code,somebody shed some lights? Thanks Oct 21, 2013 at 19:15 • @Chui, added explanation. Oct 22, 2013 at 11:07 • @ugoren as far as I understand, shouldn't printf("%d") print out the memory address pointer i holds in Hex? why it is printing out an integer? or char* i was initialized to point to memory address 0 by default? Thanks Oct 22, 2013 at 21:23 ## Java 86 79 Assume the integer is in y: Converts to a string in base 3, removes the last character ( right shift ">>" in base 3 ), then converts back to integer. Works for negative numbers. If the number, y, is < 3 or > -3, then it gives 0. System.out.print(~2<y&y<3?0:Long.valueOf(Long.toString(y,3).split(".$")[0],3));


First time posting on code golf. =) So can't comment yet.

Thx Kevin Cruijssen for the tips.

• I know it's been more than two years ago, but you can golf a few parts: && to &, and 2x Integer to Long. (Also, why do you use ~2 instead of just -3? They are the same byte-count.) Nov 7, 2016 at 14:19
• @KevinCruijssen So nostalgic to edit my first post after so long. Wasn't too sure why I thought ~2 was better back then. Nov 7, 2016 at 21:41
• @KevinCruijssen well, the challenge says you're not allowed to use -, but I don't know if that counts for unary negation. Feb 8, 2017 at 7:35
• @FlipTack Ah you're completely right. In that case forget I ever said it. :) Feb 8, 2017 at 7:47

# Python2.6 (29)(71)(57)(52)(43)

z=len(range(2,abs(x),3))
print (z,-z)[x<0]


print len(range(2,input(),3))


Edit - Just realized that we have to handle negative integers too. Will fix that later

Edit2 - Fixed

Edit3 - Saved 5 chars by following Joel Cornett's advice

Edit4 - Since input doesn't have to be necessarily be from STDIN or ARGV, saved 9 chars by not taking any input from stdin

• Go ahead with abs() Aug 3, 2012 at 7:53
• shorter to do print z if x==abs(x) else -z Aug 3, 2012 at 7:53
• better yet, print (z,-z)[x<0] Aug 3, 2012 at 7:54
• @ArtemIce thanks, only realized I could use that after reading another answer above. Aug 3, 2012 at 7:57
• @JoelCornett humm, didn't know about that, thanks Aug 3, 2012 at 7:58

# Javascript, 47 29

Uses eval to dynamically generate a /. Uses + only for string concatenation, not addition.

alert(eval(prompt()+"\57"+3))


EDIT: Used "\57" instead of String.fromCharCode(47)

• -1 for alert(eval(prompt()+"\573"))? Jun 24, 2019 at 4:33

## Ruby (4322 17)

Not only golf, but elegance also :)

p Rational gets,3


Output will be like (41/1). If it must be integer then we must add .to_i to result, and if we change to_i to to_f then we will can get output for floats also.

• Works without the require rational line on Ruby 1.9.3. Omitting the parentheses saves one more char. Dec 3, 2012 at 20:53

# TI-Basic, 8 bytes

Winner? :)

int(mean({Ans,0,0


P.S. Rounds towards infinity for negative numbers (see here for why). To round to zero instead, replace int( with iPart( for no byte change.

Test cases

-4:prgmDIVIDE
-2
11:prgmDIVIDE
3
109:prgmDIVIDE
36


# Python 2.x, 5453 51

print' -'[x<0],len(range(*(2,-2,x,x,3,-3)[x<0::2]))

Where _ is the dividend and is entered as such.

>>> x=-19
>>> print' -'[x<0],len(range(*(2,-2,x,x,3,-3)[x<0::2]))
- 6


Note: Not sure if using the interactive interpreter is allowed, but according to the OP: "Input can be on STDIN or ARGV or entered any other way"

Edit: Now for python 3 (works in 2.x, but prints a tuple). Works with negatives.

• Works in python 3 as well? Aug 1, 2012 at 9:42
• Doesn't have to be subscriptable; having __len__ is enough. Aug 1, 2012 at 10:08
• len(range(100,1000)) gives 900 in 3.2.3 on linux. Aug 1, 2012 at 10:09
• This doesn't work for negative numbers. And len(xrange(0,_,3)) is shorter and massively faster anyway.
– grc
Aug 1, 2012 at 10:20
• @Mechanicalsnail: Point taken. I concede. It does work on 3. Aug 1, 2012 at 11:04

## C++, 191

With main and includes, its 246, without main and includes, it's only 178. Newlines count as 1 character. Treats all numbers as unsigned. I don't get warnings for having main return an unsigned int so its fair game.

My first ever codegolf submission.

#include<iostream>
#define R return
typedef unsigned int U;U a(U x,U y){R y?a(x^y,(x|y^x^y)<<1):x;}U d(U i){if(i==3)R 1;U t=i&3,r=i>>=2;t=a(t,i&3);while(i>>=2)t=a(t,i&3),r=a(r,i);R r&&t?a(r,d(t)):0;}U main(){U i;std::cin>>i,std::cout<<d(i);R 0;}


uses shifts to divide number by 4 repeatedly, and calculates sum (which converges to 1/3)

Pseudocode:

// typedefs and #defines for brevity

function a(x, y):
magically add x and y using recursion and bitwise things
return x+y.

function d(x):
if x = 3:
return 1.
variable total, remainder
until x is zero:
remainder = x mod 4
x = x / 4
total = total + x
if total and remainder both zero:
return 0.
else:
return a(total, d(remainder)).


As an aside, I could eliminate the main method by naming d main and making it take a char ** and using the programs return value as the output. It will return the number of command line arguments divided by three, rounded down. This brings its length to the advertised 191:

#define R return
typedef unsigned int U;U a(U x,U y){R y?a(x^y,(x|y^x^y)<<1):x;}U main(U i,char**q){if(i==3)R 1;U t=i&3,r=i>>=2;t=a(t,i&3);while(i>>=2)t=a(t,i&3),r=a(r,i);R r&&t?a(r,d(t)):0;}


## Golfscript - 13 chars

~3base);3base

• doesn't seem to handle negative input Aug 19, 2012 at 18:05
• @r.e.s. s/seem to // :(. I'll have to have a think about it Aug 19, 2012 at 22:30

## PowerShell 57 or 46

In 57 characters using % as the PowerShell foreach operator, not modulo. This solution can accept positive or negative integers.

(-join(1..(Read-Host)|%{1})-replace111,0-replace1).Length


In 46 characters if * is allowed as the string repetition operator, not multiply. This option requires positive integers as input values.

("1"*(Read-Host)-replace111,0-replace1).Length


# R

These only work with positive integers:

max(sapply(split(1:x,1:3), length))
# Gives a warning that should be ignored


Or:

min(table(rep(1:3, x)[1:x]))


Or:

length((1:x)[seq(3,x,3)])


Or:

sum(rep(1,x)[seq(3,x,3)])


[[EDIT]] And an ugly one:

trunc(sum(rep(0.3333333333, x)))


[[EDIT2]] Plus probably the best one - inspired by the matlab code above by Elliot G:

length(seq(1,x,3))

• I wanted to implement the same idea as in your EDIT2 but it does not work for negative numbers: wrong sign in 'by' argument  Nov 9, 2016 at 15:59

# SmileBASIC, 5851 36 bytes (no mathematical functions!)

INPUT N
BGANIM.,4,-3,N
WAIT?BGROT(0)


Explanation:

INPUT N           'get input
BGANIM 0,"R",-3,N 'smoothly rotate background layer 0 by N degrees over 3 frames
WAIT              'wait 1 frame
PRINT BGROT(0)    'display angle of layer 0


The program moves the background layer smoothly over 3 frames, and then gets the angle after 1 frame, when it has traveled 1/3 of its total distance.

### Float division version, 38 bytes:

INPUT N
BGANIM.,7,-3,N
WAIT?BGVAR(0,7)


Explanation:

INPUT N           'input
BGANIM 0,"V",-3,N 'smoothly change layer 0's internal variable to N over 3 frames
WAIT              'wait 1 frame
PRINT BGVAR(0,7)  'display layer 0's internal variable


Works with the full set of positive and negative integers

f n=sum[sum$1:[-2|n<0]|i<-[3,6..abs n]]  First creates a list 1's or (-1)'s (depending on the sign of the input) for every third integer from 0 up until the input n. abs(n) for negative numbers inclusive. e.g n=8 -> [0,3,6] It then returns the sum of this list. • Doesn't work on negative numbers (-3/3 is -1 not 1). Jun 23, 2019 at 12:27 • Nice you made it work on negative numbers but you can't use /, read the specification. Jun 23, 2019 at 12:45 • Oh gosh, you got me twice. All fixed ;) Jun 23, 2019 at 13:00 • Nice! Btw You can get 39 chars with f n=sum[sum$1:[-2|n<0]|i<-[3,6..abs n]] Jun 23, 2019 at 21:58

# Python 3, 72 bytes

f=lambda x:x and s(x>>1,f(x>>1))
s=lambda x,y:y and s(x^y,(~x&y)<<1)or x


Try it online!

A bitwise solution consisting of two recursive functions, one to calculate third, the other to subtract. Because multiplying and dividing by 2 is easy to do bitwise, we can use the fact:
$$\X/3=X/2-(X/2)/3\$$
to recursively find the third.

It does produce some weird rounding patterns for non-multiples of three.

0 -> 0 (0)
1 -> 0 (0.333...)
2 -> 1 (0.666...)
3 -> 1 (1)
4 -> 1 (1.333...)
5 -> 1 (1.666...)
6 -> 2 (2)
7 -> 2 (2.333...)
8 -> 3 (2.666...)
9 -> 3 (3)
10 -> 4 (3.333...)
11 -> 4 (3.666...)
12 -> 4 (4)
13 -> 4 (4.333...)
14 -> 5 (4.666...)
15 -> 5 (5)


# Clojure, 87; works with negatives; based on lazyseqs

(defn d[n](def r(nth(apply interleave(repeat 3(range)))(Math/abs n)))(if(> n 0)r(- r)))


Ungolfed:

(defn d [n]
(let [r (nth (->> (range) (repeat 3) (apply interleave))
(Math/abs n))]
(if (pos? n)
r
(- r))))
`