Programming Challenge #2

[Eric]

We're off, we're off to programming challenge number two. (Runs and hides for being so ridiculous)

Anywho, yes it's the second one. Already. (cough Wrighty, cough).

In case you are unaware of what the Programming Challenges are, I've laid out the details in Programming Challenges Start. This address will also serve as the index for all the challenges, so later on when there are more, just look there.


First Order of Business: Ranking of Challenge #1
Well, I've decided that both of Wrighty and Chantry get the number one spot. Here's my reasoning:

• Wrighty: Yours was fairly good and simple, but you made one slight mistake in that you kept searching for more factors of a factor after one had already been found.
  
• Chantry: Yours got complex with the usage of the strings, which I know you explained was due to the necessity of specifying an array length. However, as I said, you could have just used an ArrayList and done toArray, and made yours incredibly simple (and rather elegant I might add) . However, your sub-factor search was correct, and you also even went up by twos in order to save time.

Points:
Wrighty(1): 13
Chantry(1): 10

Programming Challenge Number 2:
The moment you've all been waiting for (especially Wrighty), the number two challenge. For this one I've decided to mix string manipulation with number calculation.

The challenge this time is to calculate a mathematical expression using reverse order of operations. Write a function that takes in the mathematical expression in string format, and returns it in number format. You can use anything tools available in your programming language aside from something like "eval()" which returns the answer to an expression.

Let me go over how this will work:
() - Parentheses remain the first, due to their nature of allowing exceptions.
+&- - Plus and minus go second
*&/ - Multiplying and dividing go third
^ - Exponents go last

So if you received a statement like this:
"(5 * 7) + 4 * 9 - 3 ^ 2 + 1"

It would break down to:
"35 + 4 * 9 - 3 ^ 2 + 1"

Then:
"39 * 6 ^ 3"

Then:
"234 ^ 3"

Then:
12812904

Have fun programming!

[Wrighty]

hmmm this seems fun... Shouldn't be too hard to use..

I know that the first code I make will be far too long and clumsy. I'll then have to cut it down!

Thanks Eric!

[Me]

Do you want us to deal with nested brackets?

Like (2 + (5 * 2 - (3 * 5)) -(5 / 3))

[Eric]

coolcoolcool said:

Do you want us to deal with nested brackets?

Like (2 + (5 * 2 - (3 * 5)) -(5 / 3))

Yes. Here's a hint:
For brackets, call the function that you write itself to calculate them.

[Me]

I know how to deal with them .

I'm using ruby though ... and I have never written more then 5 lines of code in it before. This should be interesting, seeing as I have to look up every single string manipulation I want to do.

Recursion is much easier in this one.

---

Ok, this is driving me insane. Any reason you see that this would happen?
equation = equation.replace(/(\d+?)(\+|\-)(\d+?)/gi,((RegExp.$2 == "+") ? RegExp.$1 + RegExp.$3 : RegExp.$1 - RegExp.$3)); // replaces with 0
alert((RegExp.$2 == "+") ? RegExp.$1 + RegExp.$3 : RegExp.$1 - RegExp.$3); // alerts -1

Yes, I switched to Javascript.

[Eric]

coolcoolcool said:

I know how to deal with them .

I'm using ruby though ... and I have never written more then 5 lines of code in it before. This should be interesting, seeing as I have to look up every single string manipulation I want to do.

Recursion is much easier in this one.

---

Ok, this is driving me insane. Any reason you see that this would happen?
equation = equation.replace(/(\d+?)(\+|\-)(\d+?)/gi,((RegExp.$2 == "+") ? RegExp.$1 + RegExp.$3 : RegExp.$1 - RegExp.$3)); // replaces with 0
alert((RegExp.$2 == "+") ? RegExp.$1 + RegExp.$3 : RegExp.$1 - RegExp.$3); // alerts -1

Yes, I switched to Javascript.

Javascript and the RegExp object are really funny sometimes. You might try using (RegExp Obj.).test.

Another option is to declare the matches as variables immediately after the check.

[Eric]

I've finished writing the basic one using RegExp, though I might try writing another one using characters, or perhaps some form of syntax trees.

Also, I made an edit above. And you really don't have to worry about the no double posting rule, because its allowed by default in my threads

[Ross]

I'm sure I could have found a shorter way to deal with embedded brackets but at least this way works (or seems to anyway ) Oh, and it's coded in PHP.

   function asmdo($x) {
      $x = preg_replace("/(\d+)(\+|-)(\d+)/ei", "$1 $2 $3", $x);
      $x = preg_replace("/(\d+)(\*|\/)(\d+)/ei", "$1 $2 $3", $x);
      $x = preg_replace("/(\d+)\^(\d+)/ei", "pow($1, $2)", $x);
      return $x;
   }
   function basmdo($equation) {
      $y = preg_replace("/[\t\s\r\n]/", "", $equation);
      $e = preg_split("/(\(|\))/", $y, -1, PREG_SPLIT_DELIM_CAPTURE);
      $a = Array();
      $b = 0;
      $c = "";
      foreach($e as $x) {
         switch($x) {
            case '(':
               $b++;
               $a[$b] = "";
            break;
            case ')':
               if($b > 1)
                  $a[$b - 1] .= asmdo($a[$b]);
               else
                  $c .= asmdo($a[$b]);
               $a[$b] = "";
               $b--;
            break;
            default:
               if($b) {
                  $a[$b] .= $x;
               } else {
                  $c .= $x;
               }
            break;
         }
      }
      $z = asmdo($c);
      return $z;
   }

   echo basmdo("(5 * 7) + 4 * 9 - 3 ^ 2 + 1");


Note: The function names are based on BODMAS except with the order of operations reversed. They're not just random letters

[Eric]

Note: The function names are based on BODMAS except with the order of operations reversed. They're not just random letters

I noticed

Thanks for participating!

[Me]

13lade said:

I've finished writing the basic one using RegExp, though I might try writing another one using characters, or perhaps some form of syntax trees.

Also, I made an edit above. And you really don't have to worry about the no double posting rule, because its allowed by default in my threads

Sorry, won't be able to do this one. It is a challenging piece of code for me to write, and with my Track, my Hockey Tournament, and my homework, I can't get it done .

I'll be sure to try the next one though. Oh, and nicely done Ross.

[Eric]

coolcoolcool said:

13lade said:

I've finished writing the basic one using RegExp, though I might try writing another one using characters, or perhaps some form of syntax trees.

Also, I made an edit above. And you really don't have to worry about the no double posting rule, because its allowed by default in my threads

Sorry, won't be able to do this one. It is a challenging piece of code for me to write, and with my Track, my Hockey Tournament, and my homework, I can't get it done .

I'll be sure to try the next one though. Oh, and nicely done Ross.

You can still turn it in late and get points, you just won't be in the running for the best one.

[Eric]

function reverse_oo(expr)
{
   var ex, num1, num2, op;
   while(expr.match(/\s+/))
   {
      expr = expr.replace(/\s+/g, "");
   }
   while(expr.match(/(\(([^()]+)\))/))
   {
      expr = expr.replace(RegExp.$1, reverse_oo(RegExp.$2));
   }
   while(expr.match(/(([0-9.]+)(\+|\-)([0-9.]+))/))
   {
      ex = RegExp.$1;
      num1 = RegExp.$2 * 1;
      op = RegExp.$3;
      num2 = RegExp.$4 * 1;
      if(op == "+")
         num1 += num2;
      else
         num1 -= num2;
      expr = expr.replace(ex, num1);
   }
   while(expr.match(/(([0-9.]+)(\*|\/)([0-9.]+))/))
   {
      ex = RegExp.$1;
      num1 = RegExp.$2 * 1;
      op = RegExp.$3;
      num2 = RegExp.$4 * 1;
      if(op == "*")
         num1 *= num2;
      else
         num1 /= num2;
      expr = expr.replace(ex, num1);
   }
   while(expr.match(/(([0-9.]+)\^([0-9.]+))/))
   {
      ex = RegExp.$1;
      num1 = RegExp.$2 * 1;
      num2 = RegExp.$3 * 1;
      expr = expr.replace(ex, Math.pow(num1, num2));
   }
   return (expr * 1);
}