23
\$\begingroup\$

In this task, you have to write a program, that computes the prime factors of a number. The input is a natural number 1 < n < 2^32. The output is a list of the prime factors of the number in the following format. Exponents must be omitted if they are 1. Only output prime numbers. (Assuming the input is 131784):

131784 = 2^3 * 3 * 17^2 * 19

Using the same amount of whitespace is not required; whitespace may be inserted wherever appropriate. Your program should complete in less then 10 minutes for any input. The program with the shortest amount of characters wins.

\$\endgroup\$
9
  • 11
    \$\begingroup\$ Bonus points if your program can factor 6857599914349403977654744967172758179904114264612947326127169976133296980951450542789808884504301075550786464802304019795402754670660318614966266413770127 in less than 73 days! \$\endgroup\$
    – Joey Adams
    Apr 7 '11 at 2:08
  • 1
    \$\begingroup\$ @Joey Adams: The factorization starts out with 17*71*113*997*313597... \$\endgroup\$
    – FUZxxl
    Apr 9 '11 at 15:48
  • 3
    \$\begingroup\$ @FUZxxl: I think you made a mistake copying the number. It's the product of two large primes. \$\endgroup\$
    – Joey Adams
    Apr 9 '11 at 17:50
  • \$\begingroup\$ @Joey Can we use Shor's Algorithm? \$\endgroup\$ May 3 '11 at 3:30
  • 25
    \$\begingroup\$ @Joey I accidentally spilled some coffee over my quantum computer, and my friend is using his to "hack into the US Government" or something unimportant, so, no. :( \$\endgroup\$ May 3 '11 at 5:59

25 Answers 25

12
\$\begingroup\$

SageMath, 31 Bytes

N=input()
print N,"=",factor(N)

Test case: 83891573479027823458394579234582347590825792034579235923475902312344444 Outputs:

83891573479027823458394579234582347590825792034579235923475902312344444 = 2^2 * 3^2 * 89395597 * 98966790508447596609239 * 263396636003096040031295425789508274613

\$\endgroup\$
1
  • \$\begingroup\$ Congrats on winning a challenge with your very first post, nice work! And welcome to the site! \$\endgroup\$
    – DJMcMayhem
    Sep 20 '16 at 17:10
8
\$\begingroup\$

Ruby 1.9, 74 70 characters

#!ruby -plrmathn
$_+=?=+$_.to_i.prime_division.map{|a|a[0,a[1]]*?^}*?*

Edits:

  • (74 -> 70) Just use the exponent as slice length instead of explicitly checking for exponent > 1
\$\endgroup\$
0
7
\$\begingroup\$

Perl 5.10, 73 88

perl -pe '$_=`factor $_`;s%( \d+)\K\1+%-1-length($&)/length$1%ge;y, -,*^,;s;\D+;=;'

Takes input number from standard input. Will compute factors for multiple inputs if provided.

Counted as a difference to perl -e. 5.10 is needed for the \K regex metacharacter.

\$\endgroup\$
6
  • \$\begingroup\$ +1 for using factor. \$\endgroup\$
    – st0le
    Apr 7 '11 at 4:41
  • \$\begingroup\$ Shouldn't you count the p option? \$\endgroup\$
    – Joey
    Apr 7 '11 at 7:52
  • \$\begingroup\$ @Joey indeed I should. Sorry about that. Fixing. \$\endgroup\$
    – J B
    Apr 7 '11 at 15:06
  • \$\begingroup\$ Haven’t tested this, but instead of split/\D/,~factor $_~;$_="@_"; could you write $_=~factor $_~;s/\D/ /g;? (Of course replace ~ with the backtick.) \$\endgroup\$
    – Timwi
    Apr 9 '11 at 0:35
  • \$\begingroup\$ You mean $_=`factor $_`;s/\D/ /g;? Dual backtick encasing helps. \$\endgroup\$ Apr 9 '11 at 11:37
5
\$\begingroup\$

OCaml, 201 characters

A direct imperative translation of the best Python code:

let(%)s d=if!d>1then Printf.printf"%s%d"s!d
let f n=let x,d,e,s=ref n,ref 1,ref 0,ref"="in""%x;while!d<65536do
incr d;e:=0;while!x mod!d=0do x:=!x/ !d;incr e
done;if!e>0then(!s%d;"^"%e;s:="*")done;!s%x

For example,

# f 4294967292;;
4294967292=2^2*3^2*7*11*31*151*331- : unit = ()

(note that I've omitted outputting the final endline.) Just for fun, at 213 characters, a purely functional version, thoroughly obfuscated through liberal use of operators:

let(%)s d=if d>1then Printf.printf"%s%d"s d
let f x=let s=ref"="in""%x;let rec(@)x d=if d=65536then!s%x else
let rec(^)x e=if x/d*d<x then x,e else x/d^e+1in
let x,e=x^0in if e>0then(!s%d;"^"%e;s:="*");x@d+1in x@2
\$\endgroup\$
5
\$\begingroup\$

Python, 140 135 133 chars

M=N=input()
s=''
f=1
while f<4**8:
 f+=1;e=0
 while N%f<1:e+=1;N/=f
 if e:s+='*%d'%f+'^%d'%e*(e>1)
print M,'=',(s+'*%d'%N*(N>1))[1:]
\$\endgroup\$
2
  • \$\begingroup\$ I think the output requires some more spaces, e.g. ' * %d'... And two more things: 65536 == 4**8; Line 7: if e:s+='*%d'%f+'^%d'%e*(e>1) \$\endgroup\$ Apr 7 '11 at 13:48
  • \$\begingroup\$ @BlaXpirit: "same amount of whitespace is not required". Thanks for the other two, I'll incorporate them. \$\endgroup\$ Apr 7 '11 at 16:56
5
\$\begingroup\$

J, 72

(":*/f),'=',([,'*',])/(":"0~.f),.(('^',":)`(''"0)@.(=&1))"0+/(=/~.)f=.q:161784

Typical J. Two characters to do most of the work, sixty characters to present it.

Edit: Fixed the character count.

\$\endgroup\$
3
  • 2
    \$\begingroup\$ This doesn't look like 62 characters to me. Even when assuming 161784 is your input, it's still 72 characters. \$\endgroup\$
    – Ventero
    Apr 8 '11 at 16:27
  • \$\begingroup\$ Wouldn't it be shorter with |: __ q: y? \$\endgroup\$
    – Eelvex
    Apr 9 '11 at 12:24
  • 2
    \$\begingroup\$ @Ventero: typical JB. Two hours to golf the damned thing, fifteen seconds to mess up the character count. \$\endgroup\$
    – J B
    Apr 16 '11 at 19:41
5
\$\begingroup\$

J, 53 52 characters

This solution takes the rplc trick from the solution of randomra but comes up with some original ideas, too.

":,'=',(":@{.,'^','*',~":@#)/.~@q:}:@rplc'^1*';'*'"_

In non-tacit notation, this function becomes

f =: 3 : 0
(": y) , '=' , }: (g/.~ q: y) rplc '^1*' ; '*'
)

where g is defined as

g =: 3 : 0
": {. y) , '^' , (": # y) , '*'
)
  • q: y is the vector of prime factors of y. For instance, q: 60 yields 2 2 3 5.
  • x u/. y applies u to y keyed by x, that is, u is applied to vectors of elements of y for which the entries in x are equal. This is a bit complex to explain, but in the special case y u/. y or u/.~ y, u is applied to each vector of distinct elements in y, where each element is repeated for as often as it appears in y. For instance, </.~ 1 2 1 2 3 1 2 2 3 yields

    ┌─────┬───────┬───┐
    │1 1 1│2 2 2 2│3 3│
    └─────┴───────┴───┘
    
  • # y is the tally of y, that is, the number of items iny.

  • ": y formats y as a string.
  • x , y appends x and y.
  • {. y is the head y, that is, its first item.
  • Thus, (": {. y), '^' , (": # y) , '*' formats a vector of n repetitions of a number k into a string of the form k ^ n *. This phrase in tacit notation is :@{.,'^','*',~":@#, which we pass to the adverb /. described further above.
  • x rplc y is the library function replace characters. y has the form a ; b and every instance of string a in x is replaced by b. x is ravelled (that is, reshaped such that it has rank 1) before operation takes place, which is used here. This code replaces ^1* with * as to comply with the mandated output format.
  • }: y is the curtail of y, that is, all but its last item. This is used to remove the trailing *.
\$\endgroup\$
2
  • \$\begingroup\$ Couldn't you save a lot of work by using __ q:? Try it online! \$\endgroup\$
    – Adám
    May 4 '17 at 10:46
  • \$\begingroup\$ @Adám Indeed, good idea! \$\endgroup\$
    – FUZxxl
    May 4 '17 at 13:12
4
\$\begingroup\$

PHP, 112

echo$n=$_GET[0],'=';$c=0;for($i=2;;){if($n%$i<1){$c++;$n/=$i;}else{if($c){echo"$i^$c*";}$c=0;if(++$i>$n)break;}}

118

echo $n=$_GET[0],'=';for($i=2;;){if(!($n%$i)){++$a[$i];$n/=$i;}else{if($a[$i])echo "$i^$a[$i]*";$i++;if($i>$n)break;}}
\$\endgroup\$
3
\$\begingroup\$

Python 119 Chars

M=N=input()
i=1
s=""
while N>1:
 i+=1;c=0
 while N%i<1:c+=1;N/=i
 if c:s+=" * %d"%i+['','^%d'%c][c>1]
print M,'=',s[3:]
\$\endgroup\$
3
  • 1
    \$\begingroup\$ That's what I tried first, but it is too slow for big primes, like 4294967291. \$\endgroup\$ Apr 7 '11 at 5:56
  • \$\begingroup\$ @Keith The question allows upto 10 minutes. Will this take more than 10 minutes for the worst case? \$\endgroup\$
    – fR0DDY
    Apr 7 '11 at 7:01
  • 2
    \$\begingroup\$ It took 32 minutes on my machine for that number. \$\endgroup\$ Apr 7 '11 at 16:54
3
\$\begingroup\$

JavaScript, 124 122 119

for(s='',i=2,o=p=prompt();i<o;i++){for(n=0;!(p%i);n++)p/=i;n?s+=i+(n-1?'^'+n:'')+'*':0}alert(s.substring(0,s.length-1))
\$\endgroup\$
3
\$\begingroup\$

Perl, 78

use ntheory":all";say join" * ",map{(join"^",@$_)=~s/\^1$//r}factor_exp(shift)

It uses the s///r feature of Perl 5.14 to elide the ^1s. 81 characters to run in a loop:

perl -Mntheory=:all -nE 'chomp;say join" * ",map{(join"^",@$_)=~s/\^1$//r}factor_exp($_);'
\$\endgroup\$
1
  • \$\begingroup\$ You can leave out the spaces if you like. This would save two characters. Nice solution! \$\endgroup\$
    – FUZxxl
    Nov 1 '14 at 10:08
2
\$\begingroup\$

PHP, 236 characters

$f[$n=$c=$argv[1]]++;echo"$n=";while($c){$c=0;foreach($f as$k=>$n)for($r=~~($k/2);$r>1;$r--){if($k%$r==0){unset($f[$k]);$f[$r]++;$f[$k/$r]++;$c=1;break;}}}foreach($f as$k=>$n)if(--$n)$f[$k]="$k^".++$n;else$f[$k]=$k;echo implode("*",$f);

Output for 131784: 2^3*3*17^2*19

Completes all numbers within a few seconds while testing.

4294967296=2^32
Time: 0.000168

Input was never specified, so I chose to call it using command line arguments.

php factorize.php 4294967296
\$\endgroup\$
2
\$\begingroup\$

Scala 374:

def f(i:Int,c:Int=2):List[Int]=if(i==c)List(i)else 
if(i%c==0)c::f(i/c,c)else f(i,c+1)
val r=f(readInt)
class A(val v:Int,val c:Int,val l:List[(Int,Int)])
def g(a:A,i:Int)=if(a.v==i)new A(a.v,a.c+1,a.l)else new A(i,1,(a.v,a.c)::a.l)
val a=(new A(r.head,1,Nil:List[(Int,Int)])/:(r.tail:+0))((a,i)=>g(a,i))
a.l.map(p=>if(p._2==1)p._1 else p._1+"^"+p._2).mkString("", "*", "")

ungolfed:

def factorize (i: Int, c: Int = 2) : List [Int] = {
  if (i == c) List (i) else 
    if (i % c == 0) c :: f (i/c, c) else 
      f (i, c+1)
}
val r = factorize (readInt)
class A (val value: Int, val count: Int, val list: List [(Int, Int)])
def g (a: A, i: Int) = 
  if (a.value == i) 
    new A (a.value, a.count + 1, a.list) else 
    new A (i, 1, (a.value, a.count) :: a.list)
val a = (new A (r.head, 1, Nil: List[(Int,Int)]) /: (r.tail :+ 0)) ((a, i) => g (a, i))
a.l.map (p => if (p._2 == 1) p._1 else
  p._1 + "^" + p._2).mkString ("", "*", "")
\$\endgroup\$
2
\$\begingroup\$

J, 74 chars

f=.3 :0
(":y),'=',' '-.~('^1 ';'')rplc~}:,,&' *'"1(,'^'&,)&":/"{|:__ q:y
)

   f 131784
131784=2^3*3*17^2*19

64 chars with input in variable x:

   x=.131784

   (":x),'=',' '-.~('^1 ';'')rplc~}:,,&' *'"1(,'^'&,)&":/"{|:__ q:x
131784=2^3*3*17^2*19
\$\endgroup\$
11
  • \$\begingroup\$ If you manage to turn this into a tacit definition, you can avoid escaping all the quotes. You could also use a 3 : 0 defintion. \$\endgroup\$
    – FUZxxl
    Feb 8 '15 at 15:53
  • \$\begingroup\$ @FUZxxl I expected I can just put in the unescaped string in the 3 : 0 version but it didn't work somewhy. I might try tacit later though. This is the 3:0 I tried: pastebin.com/rmTVAk4j. \$\endgroup\$
    – randomra
    Feb 8 '15 at 15:57
  • \$\begingroup\$ It should work. I don't see why. Did you name your argument y as you are supposed to? \$\endgroup\$
    – FUZxxl
    Feb 8 '15 at 15:59
  • \$\begingroup\$ @FUZxxl This is the 3:0 I tried: pastebin.com/rmTVAk4j. \$\endgroup\$
    – randomra
    Feb 8 '15 at 16:00
  • \$\begingroup\$ The 3:0 you tried doesn't exactly match the one-liner you provide. It uses '' instead of a: in one place. Maybe that's the difference? \$\endgroup\$
    – FUZxxl
    Feb 8 '15 at 16:04
2
\$\begingroup\$

Jelly, 16 bytes (non-competing on request of OP)

³”=³ÆFḟ€1j€”^j”*

One of my first Jelly answers, so can definitely be golfed (especially ³”=³)..

Try it online.

Explanation:

³                 # Push the first argument
 ”=               # Push string "="
   ³ÆF            # Get the prime factor-exponent pairs of the first argument
      ḟ€1         # Remove all 1s from each pair
         j€”^     # Join each pair by "^"
             j”*  # Join the pair of strings by "*"
                  # (implicitly join the entire 'stack' together)
                  # (which is output implicitly as result)
\$\endgroup\$
5
  • \$\begingroup\$ Disqualified as your language was created after this task was published. \$\endgroup\$
    – FUZxxl
    May 13 '19 at 11:38
  • \$\begingroup\$ @FUZxxl Since mid-2017 non-competing isn't in the meta anymore, unless the challenge explicitly states that languages should be older than the time of posting. But if you as the one who posted the challenge chooses to not allow languages newer than your challenge post-date, I will edit my answers to add the explicit (non-competing). :) \$\endgroup\$ May 13 '19 at 11:47
  • \$\begingroup\$ I believe the site consensus that was in place when this challenge was posted should define the rules for answers. Everything else (i.e. rules that change after the challenge was posted) would be unfair. Please mark your answers as non competing. \$\endgroup\$
    – FUZxxl
    May 13 '19 at 11:55
  • \$\begingroup\$ @FUZxxl I've marked my answers as non-competing, as requested. \$\endgroup\$ May 13 '19 at 12:02
  • \$\begingroup\$ Thank you for your help. \$\endgroup\$
    – FUZxxl
    May 13 '19 at 12:13
2
\$\begingroup\$

Java 10, 109 108 bytes (lambda function) (non-competing on request of OP)

n->{var r=n+"=";for(int i=1,f;i++<n;r+=f<1?"":(f<2?i:i+"^"+f)+(n>1?"*":""))for(f=0;n%i<1;n/=i)f++;return r;}

Try it online.

Java 6+, 181 bytes (full program)

class M{public static void main(String[]a){long n=new Long(a[0]),i=1,f;String r=n+"=";for(;i++<n;r+=f<1?"":(f<2?i:i+"^"+f)+(n>1?"*":""))for(f=0;n%i<1;n/=i)f++;System.out.print(r);}}

Try it online.

-1 byte thanks to @ceilingcat.

Explanation:

n->{                // Method with integer parameter and String return-type
  var r=n+"=";      //  Result-String, starting at the input with an appended "="
  for(int i=1,f;i++<n;
                    //  Loop in the range [2, n]
      r+=           //    After every iteration: append the following to the result-String:
        f<1?        //     If the factor `f` is 0:
         ""         //      Append nothing
        :           //     Else:
         (f<2?      //      If the factor `f` is 1:
           i        //       Append the current prime `i`
          :         //      Else:
           i+"^"+f) //       Append the current prime `i` with it's factor `f`
         +(n>1?     //      And if we're not done yet:
            "*"     //       Also append a "*"
           :        //      Else:
            ""))    //       Append nothing more
    for(f=0;        //   Reset the factor `f` to 0
        n%i<1;      //   Loop as long as `n` is divisible by `i`
      n/=i)         //    Divide `n` by `i`
      f++;          //    Increase the factor `f` by 1
  return r;}        //  Return the result-String
\$\endgroup\$
4
  • \$\begingroup\$ @ceilingcat Thanks! \$\endgroup\$ May 10 '19 at 6:29
  • \$\begingroup\$ Disqualified as Java 10 was created after this task was published. \$\endgroup\$
    – FUZxxl
    May 13 '19 at 11:38
  • \$\begingroup\$ @FUZxxl I've marked the Java 10 lambda as non-competing, and added a Java 6 program, which was released in December 2006. \$\endgroup\$ May 13 '19 at 12:01
  • \$\begingroup\$ Okay, cool. That works for me! \$\endgroup\$
    – FUZxxl
    May 13 '19 at 12:14
2
\$\begingroup\$

Japt, 28 27 26 bytes

-1 byte thanks to Shaggy

+'=+Uk ü ®ÊÉ?ZÌ+'^+Zl:ZÃq*

Try it

\$\endgroup\$
7
  • \$\begingroup\$ Disqualified as your language was created after this task was published. \$\endgroup\$
    – FUZxxl
    May 13 '19 at 11:36
  • 1
    \$\begingroup\$ @FUZxxl Using a newer language on an older challenge is allowed \$\endgroup\$
    – Oliver
    May 13 '19 at 14:28
  • \$\begingroup\$ It was not allowed back when the challenge was posted. I consider it to be unfair to amend the rules of a challenge after the challenge has been posted, so languages published after this challenge remain illegal. \$\endgroup\$
    – FUZxxl
    May 13 '19 at 14:30
  • 1
    \$\begingroup\$ @FUZxxl You don't have to accept my answer, but I am allowed to answer it regardless. \$\endgroup\$
    – Oliver
    May 13 '19 at 14:31
  • 1
    \$\begingroup\$ 24 bytes \$\endgroup\$
    – Shaggy
    May 14 '19 at 16:56
1
\$\begingroup\$

Powershell, 113 97 bytes

Inspired by Joey's answer. It's a slow but short.

param($x)(2..$x|%{for(;!($x%$_)){$_
$x/=$_}}|group|%{$_.Name+"^"+$_.Count-replace'\^1$'})-join'*'

Explained test script:

$f = {

param($x)               # let $x stores a input number > 0
(2..$x|%{               # loop from 2 to initial input number
    for(;!($x%$_)){     # loop while remainder is 0
        $_              # push a current value to a pipe
        $x/=$_          # let $x is $x/$_ (new $x uses in for condition only)
    }
}|group|%{              # group all values
    $_.Name+"^"+$_.Count-replace'\^1$'  # format and remove last ^1
})-join'*'              # make string with *

}

&$f 2
&$f 126
&$f 129
&$f 86240
#&$f 7775460

Output:

2
2*3^2*7
3*43
2^5*5*7^2*11
\$\endgroup\$
1
\$\begingroup\$

05AB1E, 22 20 bytes (non-competing on request of OP)

ÐfsÓ0Køε1K'^ý}'*ý'=ý

-2 bytes thanks to @Emigna.

Try it online.

Explanation:

Ð                # Triplicate the (implicit) input-integer
 f               # Pop and push all prime factors (without counting duplicates)
  s              # Swap to take the input again
   Ó             # Get all prime exponents
    0K           # Remove all 0s from the exponents list
      ø          # Zip it with the prime factors, creating pairs
       ε         # Map each pair to:
        1K       #  Remove all 1s from the pair
        '^ý     '#  And then join by "^"
       }'*ý     '# After the map: join the string/integers by "*"
           '=ý  '# And join the stack by "=" (with the input we triplicated at the start)
                 # (after which the result is output implicitly)
\$\endgroup\$
3
  • \$\begingroup\$ 1K should work instead of `≠iy in the loop. \$\endgroup\$
    – Emigna
    May 10 '19 at 8:39
  • \$\begingroup\$ @Emigna Ah lol.. I actually do that in my Jelly answer I just posted. Not sure why I didn't think of it earlier here. :) \$\endgroup\$ May 10 '19 at 8:49
  • \$\begingroup\$ Disqualified as your language was created after this task was published. \$\endgroup\$
    – FUZxxl
    May 13 '19 at 11:37
1
\$\begingroup\$

APL(NARS), 66 chars, 132 bytes

{(⍕⍵),'=',3↓∊{m←' * ',⍕↑⍵⋄1=w←2⊃⍵:m⋄m,'^',⍕w}¨v,¨+/¨{k=⍵}¨v←∪k←π⍵}

test and comment:

  f←{(⍕⍵),'=',3↓∊{m←' * ',⍕↑⍵⋄1=w←2⊃⍵:m⋄m,'^',⍕w}¨v,¨+/¨{k=⍵}¨v←∪k←π⍵}
  f 131784
131784=2^3 * 3 * 17^2 * 19
  f 2
2=2
  f (2*32)
4294967296=2^32

{(⍕⍵),'=',3↓∊{m←' * ',⍕↑⍵⋄1=w←2⊃⍵:m⋄m,'^',⍕w}¨v,¨+/¨{k=⍵}¨v←∪k←π⍵}
k←π⍵      find the factors with repetition of ⍵ and assign that array to k example for 12 k is 2 2 3
v←∪       gets from k unique elements and put them in array v
+/¨{k=⍵}¨ for each element of v count how many time it appear in k (it is array exponents)
v,¨       make array of couples from element of v (factors unique) and the array above (exponents unique)
∊{m←' * ',⍕↑⍵⋄1=w←2⊃⍵:m⋄m,'^',⍕w}¨ pretty print the array of couples factor exponent as array chars
3↓                                 but not the first 3 chars
(⍕⍵),'='  but print first the argument and '=' in char format

if someone has many time with these primitives, know them very well them, for me it is possible that the code is clearer of comments... so code more clear than comments, comments unuseful...

\$\endgroup\$
1
\$\begingroup\$

QBasic, 156 characters

INPUT n#
?n#;"=";
f=2
1c=0
WHILE n#/f=INT(n#/f)
n#=n#/f
c=c+1
WEND
IF c THEN?f;
IF c>1THEN?"^";c;
IF c*(n#>1)THEN?"*";
f=f+1
IF f*f<=n#GOTO 1
IF n#>1THEN?n#

When run at Archive.org, this takes about 2½ minutes to finish for an input of 4294967291, the largest prime in the specified range.

Ungolfed, with some remarks

This is the version I used to get the timing information:

CLS
INPUT num#
PRINT num#; "=";
factor = 2

startTime# = TIMER

DO
    count = 0
    WHILE num# / factor = INT(num# / factor)
        num# = num# / factor
        count = count + 1
    WEND
    IF count > 0 THEN PRINT factor;
    IF count > 1 THEN PRINT "^"; count;
    IF count > 0 AND num# > 1 THEN PRINT "*";
    factor = factor + 1
LOOP WHILE factor * factor <= num#

IF n# > 1 THEN PRINT n# ELSE PRINT

time# = TIMER - startTime#
PRINT "Search took"; INT(time# / 60); "minutes and";
PRINT INT(time# mod 60); "seconds"

The program uses the standard approach of trial division of each factor between \$2\$ and \$\sqrt n\$. It prints a * after each successful factor if the remaining \$n\$ is still greater than 1. If the factor becomes greater than \$\sqrt n\$ while \$n > 1\$, then \$n\$ is prime and we output it after exiting the loop.

A few extra bytes had to be spent to cope with very large numbers:

QBasic's numbers are either single- or double-precision floating point under the hood, although it casts to integers for certain calculations. Integers greater than \$2^{24}\$ are not guaranteed to be accurately represented in single-precision floating point, so we have to use double-precision for \$n\$ (thus the double-precision sigil on the variable num#). However, since we're only checking factors up to \$\sqrt n < 2^{16}\$, we can use the default single precision for factor.

At first, I had WHILE num# MOD factor = 0 for the inner loop. This worked until I tried numbers close to \$2^{32}\$, at which point it said "Overflow." It seems that the MOD operator casts its operands to integers--specifically, 32-bit signed integers. A value of \$2^{32}-1\$ can fit in a 32-bit unsigned integer, but not a signed one. So MOD wasn't going to work. Neither was my usual golf trick of checking divisibility with a/b=a\b, since the int-div operator \ also casts to integers. Fortunately, I could still use floating point division and then truncate explicitly with the INT function (which seemingly does not cast to integer--??).

Given the issues with single- vs. double-precision, I assumed that squaring factor (single-precision) before comparing it with num# (double) wasn't going to work. But surprisingly, it did. My best guess after some experimentation is that if there's a double-precision value anywhere in an expression, QBasic casts all of the values to double before computing anything. This contrasts with C, for example, where implicit casting only occurs between the operands of a single operator.

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0
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JavaScript, 107

n=prompt()
s=n+'='
c=0
for(i=2;;){if(n%i<1){c++
n/=i}else{if(c)s+=i+'^'+c+'*'
c=0
if(++i>n)break}}
alert(s)

120

n=prompt()
o={2:0}
for(i=2,c=n;i<=c;)!(c%i)?++o[i]?c/=i:0:o[++i]=0
s=n+'='
for(i in o)s+=o[i]?i+'^'+o[i]+'*':''
alert(s)
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4
  • 1
    \$\begingroup\$ Has a trailing * in the output and prints the exponent even if it's 1. \$\endgroup\$
    – Ventero
    Apr 9 '11 at 8:54
  • \$\begingroup\$ no need to downvote. There's nowhere that said that it couldn't print the exponent if it's 1. Also, the trailing * assumes multiplying by 1. If it's that big an issue, I'll fix it. \$\endgroup\$
    – zzzzBov
    Apr 9 '11 at 17:47
  • 1
    \$\begingroup\$ »in the following format« in the task description pretty much implies that an exponent of 1 should not be printed. And no, a trailing * is also against that. If one could choose the output format that freely, then shelling out to factor(1) would be the easiest one. Answers can only reasonably compared if they all solve the same problem. \$\endgroup\$
    – Joey
    Apr 9 '11 at 23:13
  • 3
    \$\begingroup\$ As the creator of this task, I say, that the exponents have to be omitted if 1 and only prime-numbers can be factors. \$\endgroup\$
    – FUZxxl
    Apr 10 '11 at 11:48
0
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PHP, 112 bytes

<?=$a=$argn,"=";for($i=2;1<$a;)$a%$i?$i++:$a/=$i+!++$r[$i];foreach($r as$k=>$v)echo$t?"*":!++$t,$v<2?$k:"$k^$v";

Try it online!

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0
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PHP, 93 bytes

<?=$n=$argn;for($i=2;$n>1;$k&&$p=print($p?"*":"=")."$i^$k",$i++)for($k=0;$n%$i<1;$n/=$i)$k++;

I could do 89 bytes with PHP 5.5 (or later), but that postdates the challenge by more than 2 years:

<?=$n=$argn;for($i=2;$n>1;$k&&$p=print"=*"[$p]."$i^$k",$i++)for($k=0;$n%$i<1;$n/=$i)$k++;

Run as pipe with -nF or try them online.

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0
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Japt, 24 bytes

+'=+Uk ü ËÎ+DÅÊÎç^+DÊÃq*

Try it

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