42
\$\begingroup\$

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()). Use of operators for string concatenation and formatting are ok.
  • 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
\$\endgroup\$
  • 2
    \$\begingroup\$ hopefully by "number" you mean "integer" \$\endgroup\$ – ardnew Aug 1 '12 at 4:06
  • \$\begingroup\$ Is it integer division? Also, can we use the characters */+-% in a string? \$\endgroup\$ – beary605 Aug 1 '12 at 5:04
  • \$\begingroup\$ Yes, integer. @beary605 Yes, you can use in a string. Updating question for both. \$\endgroup\$ – Gaffi Aug 1 '12 at 5:08
  • \$\begingroup\$ This would be much more interesting if you banned any function that internally uses *, /, +, -, or % too - i.e. it would force us to write custom printf() and atoi() calls... :-) \$\endgroup\$ – baby-rabbit Aug 1 '12 at 6:18
  • 1
    \$\begingroup\$ @lnkbug, ~0 is just as short... \$\endgroup\$ – Peter Taylor Aug 1 '12 at 17:20

62 Answers 62

15
\$\begingroup\$

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

\$\endgroup\$
  • 1
    \$\begingroup\$ 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. \$\endgroup\$ – Gareth Aug 17 '12 at 11:21
  • 1
    \$\begingroup\$ Maybe ##~3=_3#\i. for 11 characters? \$\endgroup\$ – Gareth Aug 17 '12 at 11:31
  • 1
    \$\begingroup\$ Actually, you can shrink yours down to 10 characters with ##~0 0 1$~. \$\endgroup\$ – Gareth Aug 17 '12 at 11:34
  • 1
    \$\begingroup\$ You can shrink that down using a hook to 3#.}:(#:~$&3) but it's still longer and it doesn't fix the negative number issue. \$\endgroup\$ – Gareth Aug 17 '12 at 11:54
  • 1
    \$\begingroup\$ 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. \$\endgroup\$ – Gareth Aug 17 '12 at 13:43
56
\$\begingroup\$

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).
\$\endgroup\$
  • 22
    \$\begingroup\$ This is the epitome of hardcoding. ++++++++++ \$\endgroup\$ – Joe Z. Sep 14 '13 at 18:45
  • 4
    \$\begingroup\$ 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. \$\endgroup\$ – Joe Z. Sep 14 '13 at 18:46
  • 15
    \$\begingroup\$ If my boss asked me to compute a division by three without - + / * % I would think he's crazy. \$\endgroup\$ – Mikaël Mayer Jan 2 '14 at 12:35
  • \$\begingroup\$ And yet, a[b] is a syntactic sugar for *(a + b), which does the addition. \$\endgroup\$ – Konrad Borowski Jan 2 '14 at 14:17
  • 11
    \$\begingroup\$ @NicolasBarbulesco There is a size restriction on Stack Exchange answers. \$\endgroup\$ – Timtech Jan 3 '14 at 23:17
38
\$\begingroup\$

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}
\$\endgroup\$
  • 1
    \$\begingroup\$ 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. \$\endgroup\$ – kernigh May 25 '14 at 15:57
26
\$\begingroup\$

Mathematica, 13 chars

Mean@{#,0,0}&
\$\endgroup\$
  • \$\begingroup\$ This is wicked :D I guess you could save the & and use a plain variable (others here do that too). \$\endgroup\$ – Yves Klett Jan 3 '14 at 10:20
  • 2
    \$\begingroup\$ @YvesKlett: Mean is inherently wicked. \$\endgroup\$ – GuitarPicker Nov 7 '16 at 17:33
18
\$\begingroup\$

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!)

\$\endgroup\$
  • \$\begingroup\$ +1 but it doesn't work with negative values \$\endgroup\$ – Francesco Casula Mar 6 '14 at 9:28
  • \$\begingroup\$ Uh, question... how does this count? It uses the - negation operator. \$\endgroup\$ – Patrick Roberts Feb 13 '16 at 0:07
  • \$\begingroup\$ @Patrick I took the specification to mean no subtraction -- if you want, you can replace -~ with parseInt() \$\endgroup\$ – Casey Chu Feb 13 '16 at 22:22
  • \$\begingroup\$ @CaseyChu the value returned by -~prompt() is one greater than parseInt(prompt()). Not sure how you'd deal with that. \$\endgroup\$ – Patrick Roberts Feb 13 '16 at 22:35
  • \$\begingroup\$ alert(Array(parseInt(prompt())).slice(1).join().replace(/,,,/g,1).length) \$\endgroup\$ – Casey Chu Feb 13 '16 at 22:37
15
\$\begingroup\$

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
\$\endgroup\$
  • 2
    \$\begingroup\$ 10/3 equals 3, not 4. \$\endgroup\$ – Joel Cornett Aug 1 '12 at 11:14
  • \$\begingroup\$ Doing print" -"[x<0]+len(range(2,abs(x),3))`` will shave it down to 39 chars \$\endgroup\$ – Joel Cornett Aug 4 '12 at 19:37
  • \$\begingroup\$ golfexchange's comment formatting is messing it up. on the above I used backticks to enclose len() as shorthand for repr() \$\endgroup\$ – Joel Cornett Aug 4 '12 at 19:38
  • \$\begingroup\$ 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. \$\endgroup\$ – grc Aug 5 '12 at 1:31
  • 2
    \$\begingroup\$ Pretend it's Python 3 ;) I like the single char slicing btw. \$\endgroup\$ – Joel Cornett Aug 5 '12 at 1:47
10
\$\begingroup\$

Haskell, 90 106

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, ...].

\$\endgroup\$
  • 3
    \$\begingroup\$ o.O i never thought of that second solution. I might be able to implement something like that in python \$\endgroup\$ – acolyte Aug 2 '12 at 19:19
8
\$\begingroup\$

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);
    }
}
\$\endgroup\$
  • 2
    \$\begingroup\$ About 5 years later.. You can save 1 byte by removing the space from string[] g, turning it into string[]g \$\endgroup\$ – Metoniem Feb 8 '17 at 14:44
  • \$\begingroup\$ 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? \$\endgroup\$ – Jonathan Frech 15 mins ago
6
\$\begingroup\$

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:

  • abuses array access to implement addition.
  • 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
\$\endgroup\$
6
\$\begingroup\$

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))
\$\endgroup\$
  • 1
    \$\begingroup\$ @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). \$\endgroup\$ – Inkbug Aug 3 '12 at 10:02
  • 1
    \$\begingroup\$ 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. \$\endgroup\$ – Wug Aug 3 '12 at 14:46
  • 1
    \$\begingroup\$ @Sam Also, ~~ converts to an integer, as opposed to +. \$\endgroup\$ – Inkbug Aug 5 '12 at 4:51
  • 1
    \$\begingroup\$ @Wug It is codegolf, thus it's not about efficiency unless specified in the task. \$\endgroup\$ – defhlt Aug 5 '12 at 5:56
  • 3
    \$\begingroup\$ +1 for taking advantage of different bases. It's one of my favorite JavaScript golf techniques. \$\endgroup\$ – DocMax Aug 26 '12 at 3:32
6
\$\begingroup\$

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;}
\$\endgroup\$
  • \$\begingroup\$ If unary negate is permitted, x+3 is -~-~-~x. \$\endgroup\$ – ugoren Nov 23 '12 at 9:11
  • \$\begingroup\$ 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. \$\endgroup\$ – Kaslai Nov 24 '12 at 7:59
6
\$\begingroup\$

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
\$\endgroup\$
  • 2
    \$\begingroup\$ Hey @memowe, nice work! You could save a few more chars (4) by doing $_=3x pop;say s|333||g. \$\endgroup\$ – Dom Hastings Jan 2 '14 at 10:43
  • 2
    \$\begingroup\$ When the input is 0, 1, or 2, then it prints an empty string. If it needs to print 0, then it needs 3 more characters (25 total): '$_=3x pop;say s|333||g||0. Slow with large numbers like 99999999, and doesn't work with negative numbers. \$\endgroup\$ – kernigh May 25 '14 at 16:04
5
\$\begingroup\$

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.
\$\endgroup\$
  • 1
    \$\begingroup\$ 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? \$\endgroup\$ – shiona Aug 1 '12 at 9:28
  • 1
    \$\begingroup\$ C will assume int if unspecified. \$\endgroup\$ – Wug Aug 3 '12 at 14:48
5
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ +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. \$\endgroup\$ – Gaffi Aug 1 '12 at 13:47
  • \$\begingroup\$ For those who care about floor division in python: python-history.blogspot.com/2010/08/… \$\endgroup\$ – Matt Aug 1 '12 at 14:22
  • \$\begingroup\$ what's with the multiplication and subtraction in your second solution? \$\endgroup\$ – boothby Aug 1 '12 at 15:40
  • \$\begingroup\$ @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 \$\endgroup\$ – Matt Aug 1 '12 at 16:34
  • \$\begingroup\$ I'm not understanding the difference between floor division and truncating decimals. Does this have to with negative division? \$\endgroup\$ – Joel Cornett Aug 1 '12 at 19:12
5
\$\begingroup\$

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.

\$\endgroup\$
5
\$\begingroup\$

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).

\$\endgroup\$
  • \$\begingroup\$ Trying to understand the code,somebody shed some lights? Thanks \$\endgroup\$ – Cong Hui Oct 21 '13 at 19:15
  • \$\begingroup\$ @Chui, added explanation. \$\endgroup\$ – ugoren Oct 22 '13 at 11:07
  • \$\begingroup\$ @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 \$\endgroup\$ – Cong Hui Oct 22 '13 at 21:23
5
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ 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.) \$\endgroup\$ – Kevin Cruijssen Nov 7 '16 at 14:19
  • 1
    \$\begingroup\$ @KevinCruijssen So nostalgic to edit my first post after so long. Wasn't too sure why I thought ~2 was better back then. \$\endgroup\$ – Vectorized Nov 7 '16 at 21:41
  • 2
    \$\begingroup\$ @KevinCruijssen well, the challenge says you're not allowed to use -, but I don't know if that counts for unary negation. \$\endgroup\$ – FlipTack Feb 8 '17 at 7:35
  • \$\begingroup\$ @FlipTack Ah you're completely right. In that case forget I ever said it. :) \$\endgroup\$ – Kevin Cruijssen Feb 8 '17 at 7:47
4
\$\begingroup\$

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)

\$\endgroup\$
4
\$\begingroup\$

Ruby (43 22 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.

\$\endgroup\$
  • 1
    \$\begingroup\$ Works without the require rational line on Ruby 1.9.3. Omitting the parentheses saves one more char. \$\endgroup\$ – steenslag Dec 3 '12 at 20:53
4
\$\begingroup\$

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
\$\endgroup\$
3
\$\begingroup\$

Python 2.x, 54 53 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.

\$\endgroup\$
  • \$\begingroup\$ Works in python 3 as well? \$\endgroup\$ – Mechanical snail Aug 1 '12 at 9:42
  • \$\begingroup\$ Doesn't have to be subscriptable; having __len__ is enough. \$\endgroup\$ – Mechanical snail Aug 1 '12 at 10:08
  • \$\begingroup\$ len(range(100,1000)) gives 900 in 3.2.3 on linux. \$\endgroup\$ – Mechanical snail Aug 1 '12 at 10:09
  • \$\begingroup\$ This doesn't work for negative numbers. And len(xrange(0,_,3)) is shorter and massively faster anyway. \$\endgroup\$ – grc Aug 1 '12 at 10:20
  • \$\begingroup\$ @Mechanicalsnail: Point taken. I concede. It does work on 3. \$\endgroup\$ – Joel Cornett Aug 1 '12 at 11:04
3
\$\begingroup\$

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;}
\$\endgroup\$
3
\$\begingroup\$

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

\$\endgroup\$
  • \$\begingroup\$ Go ahead with abs() \$\endgroup\$ – defhlt Aug 3 '12 at 7:53
  • \$\begingroup\$ shorter to do print z if x==abs(x) else -z \$\endgroup\$ – Joel Cornett Aug 3 '12 at 7:53
  • \$\begingroup\$ better yet, print (z,-z)[x<0] \$\endgroup\$ – Joel Cornett Aug 3 '12 at 7:54
  • \$\begingroup\$ @ArtemIce thanks, only realized I could use that after reading another answer above. \$\endgroup\$ – elssar Aug 3 '12 at 7:57
  • \$\begingroup\$ @JoelCornett humm, didn't know about that, thanks \$\endgroup\$ – elssar Aug 3 '12 at 7:58
3
\$\begingroup\$

Golfscript - 13 chars

~3base);3base
\$\endgroup\$
  • \$\begingroup\$ doesn't seem to handle negative input \$\endgroup\$ – r.e.s. Aug 19 '12 at 18:05
  • 1
    \$\begingroup\$ @r.e.s. s/seem to // :(. I'll have to have a think about it \$\endgroup\$ – gnibbler Aug 19 '12 at 22:30
3
\$\begingroup\$

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))
\$\endgroup\$
  • \$\begingroup\$ I wanted to implement the same idea as in your EDIT2 but it does not work for negative numbers: wrong sign in 'by' argument \$\endgroup\$ – Andreï Kostyrka Nov 9 '16 at 15:59
3
\$\begingroup\$

SmileBASIC, 58 51 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
\$\endgroup\$
2
\$\begingroup\$

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))))
\$\endgroup\$
2
\$\begingroup\$

Sage Notebook (21)

ZZ(n.digits(3)[1:],3)
\$\endgroup\$
2
\$\begingroup\$

Mathematica 36 51 42 chars

This is easily achieved in base 3.

IntegerDigits[n,3] converts the absolute value of the number to base 3.

Most takes all but the rightmost digit. This "rightward shift" amounts to integer division by 3.

FromDigits converts back to base 10.

Sign restores, if necessary, the sign.

Sign@n*Most@IntegerDigits[n,3]~FromDigits~3
\$\endgroup\$
2
\$\begingroup\$

F# (81)

Using inclusive ranges:

let(^)s x=if s<0 then int("-"+string x)else x
let d m =sign m^[3..3..abs m].Length

Alternatively, using strings:

let (^) s x = if s < 0 then int("-" + string x) else x
let d m =
  let n = abs m
  [0..n]
  |> Seq.map (fun i -> i, String.replicate i "aaa")
  |> Seq.takeWhile (fun (i, s) -> s.Length <= n)
  |> Seq.last
  |> fst
  |> (^) (sign m)

This is verbose but it's better than being forced to use loops in many imperative languages. String.replicate is especially valuable.

\$\endgroup\$
  • \$\begingroup\$ This looks like it uses multiplication * to restore the sign bit. \$\endgroup\$ – kernigh May 25 '14 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.