Implement division

Question

Implement a division algorithm in your favourite language which handles integer division. It need only handle positive numbers - but bonus points if it handles negative and mixed-sign division, too. Results are rounded down for fractional results.

The program may not contain the /, \, div or similar operators. It must be a routine which does not use native division capabilities of the language.

You only need to handle up to 32-bit division.

Input

Take two inputs on stdin separated by new lines or spaces (your choice)

740 
2

Output

In this case, the output would be 370.

I'll be looking for innovative solutions which do not use repeated subtraction. The solution which is the shortest and takes the least time.

Answer 1

Python - 73 chars

Takes comma separated input, eg 740,2

from math import*
x,y=input()
z=int(exp(log(x)-log(y)))
print(z*y+y<=x)+z

Answer 2

Perl, 15 characters

print<>*<>**-1

Answer 3

JavaScript - 36 characters

p=prompt;alert(p()*Math.pow(p(),-1))

Answer 4

JavaScript, 61

A=Array,P=prompt,P((','+A(+P())).split(','+A(+P())).length-1)

This makes a string the length of the dividend ,,,,,, (6) and splits on the divisor ,,, (3), resulting in an array of length 3: ['', '', ''], whose length I then subtract one from. Definitely not the fastest, but hopefully interesting nonetheless!

Answer 5

Python - 72 chars

Takes comma separated input, eg 740,2

x,y=input();z=0
for i in range(32)[::-1]:z+=(1<<i)*(y<<i<=x-z*y)
print z

Answer 6

Python - 41 chars

Takes comma separated input, eg 740,2

x,y=input();z=x
while y*z>x:z-=1 
print z

Answer 7

Mathematica: 34 chars

Solves symbolically the equation (x a == b)

Solve[x#[[1]]==#[[2]],x]&@Input[]

Answer 8

Yabasic - 17 characters

input a,b:?a*b^-1

Answer 9

PHP - 82 characters (buggy)

$i=fgets(STDIN);$j=fgets(STDIN);$k=1;while(($a=$j*$k)<$i)$k++;echo($a>$i?--$k:$k);

This is a very simple solution, however - it doesn't handle fractions or different signs (would jump into an infinite loop). I won't go into detail in this one, it is fairly simple.

Input is in stdin, separated by a new line.

PHP - 141 characters (full)

$i*=$r=($i=fgets(STDIN))<0?-1:1;$j*=$s=($j=fgets(STDIN))<0?-1:1;$k=0;$l=1;while(($a=$j*$k)!=$i){if($a>$i)$k-=($l>>=2)*2;$k+=$l;}echo$k*$r*$s;

Input and output same as the previous one.

Yes, this is almost twice the size of the previous one, but it:

• handles fractions correctly
• handles signs correctly
• won't ever go into an infinite loop, UNLESS the second parameter is 0 - but that is division by zero - invalid input

Re-format and explanation:

$i *= $r = ($i = fgets(STDIN)) < 0 ? -1 : 1;
$j *= $s = ($j = fgets(STDIN)) < 0 ? -1 : 1;
                                    // First, in the parentheses, $i is set to
                                    // GET variable i, then $r is set to -1 or
                                    // 1, depending whether $i is negative or
                                    // not - finally, $i multiplied by $r ef-
                                    // fectively resulting in $i being the ab-
                                    // solute value of itself, but keeping the
                                    // sign in $r.
                                    // The same is then done to $j, the sign
                                    // is kept in $s.

$k = 0;                             // $k will be the result in the end.

$l = 1;                             // $l is used in the loop - it is added to
                                    // $k as long as $j*$k (the divisor times
                                    // the result so far) is less than $i (the
                                    // divided number).

while(($a = $j * $k) != $i){        // Main loop - it is executed until $j*$k
                                    // equals $i - that is, until a result is
                                    // found. Because a/b=c, c*b=a.
                                    // At the same time, $a is set to $j*$k,
                                    // to conserve space and time.

    if($a > $i)                     // If $a is greater than $i, last step
        $k -= ($l >>= 2) * 2;       // (add $l) is undone by subtracting $l
                                    // from $k, and then dividing $l by two
                                    // (by a bitwise right shift by 1) for
                                    // handling fractional results.
                                    // It might seem that using ($l>>=2)*2 here
                                    // is unnecessary - but by compressing the
                                    // two commands ($k-=$l and $l>>=2) into 1
                                    // means that curly braces are not needed:
                                    //
                                    // if($a>$i)$k-=($l>>=2)*2;
                                    //
                                    // vs.
                                    //
                                    // if($a>$i){$k-=$l;$l>>=2;}

    $k += $l;                       // Finally, $k is incremented by $l and
                                    // the while loop loops again.
}

echo $k * $r * $s;                  // To get the correct result, $k has to be
                                    // multiplied by $r and $s, keeping signs
                                    // that were removed in the beginning.

Answer 10

Ruby 1.9, 28 characters

(?a*a+?b).split(?a*b).size-1

Rest of division, 21 characters

?a*a=~/(#{?a*b})\1*$/  

Sample:

a = 756
b = 20
print (?a*a+?b).split(?a*b).size-1  # => 37
print ?a*a=~/(#{?a*b})\1*$/         # => 16

For Ruby 1.8:

a = 756
b = 20
print ('a'*a+'b').split('a'*b).size-1  # => 37
print 'a'*a=~/(#{'a'*b})\1*$/          # => 16

Answer 11

Python, 70

Something crazy I just thought (using comma separated input):

from cmath import*
x,y=input()
print round(tan(polar(y+x*1j)[1]).real)

If you accept small float precision errors, the round function can be dropped.

Answer 12

APL (6)

⌊*-/⍟⎕

/ is not division here, but foldr. i.e, F/a b c is a F (b F c). If I can't use foldr because it's called /, it can be done in 9 characters:

⌊*(⍟⎕)-⍟⎕

Explanation:

• ⎕: input()
• ⍟⎕: map(log, input())
• -/⍟⎕: foldr1(sub, map(log, input()))
• *-/⍟⎕: exp(foldr1(sub, map(log, input())))
• ⌊*-/⍟⎕: floor(exp(foldr1(sub, map(log, input()))))

Answer 13

PHP, 55 characters

<?$a=explode(" ",fgets(STDIN));echo$a[0]*pow($a[1],-1);

Output (740/2): http://codepad.viper-7.com/ucTlcq

Answer 14

Haskell, 96 characters

main=getLine>>=print.d.map read.words
d[x,y]=pred.snd.head.filter((>x).fst)$map(\n->(n*y,n))[0..]

Input is on a single line.

The code just searches for the answer by taking the divisor d and multiplying it against all integers n >= 0. Let m be the dividend. The largest n such that n * d <= m is picked to be the answer. The code actually picks the least n such that n * d > m and subtracts 1 from it because I can take the first element from such a list. In the other case, I would have to take the last, but it's hard work to take the last element from an infinite list. Well, the list can be proven to be finite, but Haskell does not know better when performing the filter, so it continues to filter indefinitately.

Answer 15

Scala 77

def d(a:Int,b:Int,c:Int=0):Int=if(b<=a)d(a-b,b,c+1)else c
d(readInt,readInt)

Answer 16

Common Lisp, 42 charaacters

(1-(loop as x to(read)by(read)counting t))

Accepts space or line-separated input

Answer 17

Python, 37

Step 1. Convert to unary.

Step 2. Unary division algorithm.

print('1'*input()).count('1'*input())

Answer 18

Python, 25 characters

print input()*input()**-1

Answer 19

Python, 46 bytes

Nobody had posted the boring subtraction solution, so I could not resist doing it.

a,b=input()
i=0
while a>=b:a-=b;i+=1
print i

Answer 20

Smalltalk, Squeak 4.x flavour

define this binary message in Integer:

% d 
    | i |
    d <= self or: [^0].
    i := self highBit - d highBit.
    d << i <= self or: [i := i - 1].
    ^1 << i + (self - (d << i) % d)

Once golfed, this quotient is still long (88 chars):

%d|i n|d<=(n:=self)or:[^0].i:=n highBit-d highBit.d<<i<=n or:[i:=i-1].^1<<i+(n-(d<<i)%d)

But it's reasonnably fast:

[0 to: 1000 do: [:n |
    1 to: 1000 do: [:d |
        self assert: (n//d) = (n%d)]].
] timeToRun.

-> 127 ms on my modest mac mini (8 MOp/s)

Compared to regular division:

[0 to: 1000 do: [:n |
    1 to: 1000 do: [:d |
        self assert: (n//d) = (n//d)]].
] timeToRun.

-> 31 ms, it's just 4 times slower

I don't count the chars to read stdin or write stdout, Squeak was not designed for scripting.

FileStream stdout nextPutAll:
    FileStream stdin nextLine asNumber%FileStream stdin nextLine asNumber;
    cr

Of course, more stupid repeated subtraction

%d self>d and:[^0].^self-d%d+1

or plain stupid enumeration

%d^(0to:self)findLast:[:q|q*d<=self]

could work too, but are not really interesting

Answer 21

#include <stdio.h>
#include <string.h>
#include <math.h>


main()
{
   int i,j,ans;
   i=740;
   j=2;

   ans = pow(10,log10(i) - log10(j));
   printf("\nThe answer is %d",ans);
}