Mathematical expression translator

[Frontrunner]

Hi all...

My challenge is to write a routine which will translate correctly (in all circumstances) pow functions as done in C.

The winner we judged on two things.
1) Translations must be correct
2) Efficiency

Example:
Basic
r1 = (-c + (SQR(b(x) ^ 2 - (4 * a * c)))) / (2 * a)

C
r1 = (-c + (sqrt(pow(b(x), 2) - (4 * a * c)))) / (2 * a)

Good luck!
Fronrunner

[wildcard]

Sounds like an interesting, if quite specific, challenge. Its too bad I'm busy at the moment as I have always wanted to get around to writing parsers.

[Frontrunner]

That's right, it is quite specific but a general parser would be too easy for a challenge

Cheers,
Frontrunner

[wildcard]

Hopefully I can get some time to have a go, seems straight forward enough but will see.

[Frontrunner]

I am looking forward to see your contribution!

Cheers,
Frontrunner

[Ralph]

Could you please be patient with me and tell me what POW stands for?  The only meaning I have for those three letters at present is, Prisoner Of War, and, I'm sure, that is not what is intended. 

[Frontrunner]

Hi Ralph,

Sure I will try to explain you what POW means:

Like in qbasic there is in C a mathematical function to compute the power exponent.
In qbasic we use ^ but in C it is called POW.

I will give some more example in both C and bqasic so you can see the differences.

Example in C
  printf ("7 ^ 3 = %lf\n", pow (7,3));
  printf ("4.73 ^ 12 = %lf\n", pow (4.73,12));
  printf ("32.01 ^ 1.54 = %lf\n", pow (32.01,1.54));

Example in qbasic
  print "7 ^ 3 = ", 7 ^ 3
  print "4.73 ^ 3 = ", 4.73 ^ 12
  print "32.01 ^ 1.54 = ", 32.01 ^ 1.54

Both should output something like this:
7 ^ 3 = 343.000000
4.73 ^ 12 = 125410439.217423
32.01 ^ 1.54 = 208.036691

Please forget the C syntax you see in the example.
A correct translation (according to this challenge) should translate print 3 ^ 7 to print pow(3,7).

Now this looks easy but things start to get more complicated when translating
print (10 ^ 2) - 5 + (21 * (3 - 4 ^ 6) * 2 + 10 - 5) ^ 2 + 3 - (3 + 7) + 99
To
print pow(10,2))-5+pow((21*(3-pow(4,6))*2+10-5),2)+3-(3+7)+99)

I hope that helped a little.

Kind regards,
Frontrunner

[Ralph]

Frontrunner:

Yes, I now understand what your challenge is all about; translating a qb expression with a power expression into its equivalent code in C. 

I am not into C, so, I can not compete.  But, I know that others will!

[Frontrunner]

Hi Ralph,

You are to some part right!
But please forget about the C part, as long as the POW and the ^ operator are being translated correctly.
I am not here on a qbasic forum  for C coding 

Cheers,
Frontrunner

[wildcard]

I'm trying to dust off my head and have a go, I know its about parsing correcting but am confused by your original example:

r1 = (-b + (SQR(b(x) ^ 2 - (4 * a * c)))) / (2 * a)

Is b and b(x) an integer and an arrary respectively or am I miss reading, I've been out of maths and coding for too long

[Frontrunner]

Thank you for pointing me to the confusion.
I have changed the original example 
b is indeed an array.

But you have seen it right that the challenge is all about an parsing correction!

Sorry for the confusion.

Kind regards,
Frontrunner

[Ralph]

There is nothing as good for explaining as one or more examples!  Could you post at least one, good example of a math expression and the qbasic equivalent code that you would want?  That would help a lot!!!

[wildcard]

Frontrunner: No problem, haven't got to using it yet though still working on the simpler bracket-less parser.

Ralph: The challenge is to write a converter/parser that changes x ^ y to pow(x,y) however in the context of a more complex equation such as posted.

[Ralph]

Thanks, Wildcard.  I've never tried things like this before.  At first thought, it would seem that the algorithm would look for each character ^, then, for each, compare the characters going backwards and consider where to stop, put a "(" there, then continue until either a "(" or any mathematical function is found; next, do the same, going forward.  If this is the way to do it, I can see I would flounder in all the considerations I would have to take into account!  It really is too much for me, and definetely, "Not my cup of tea!"

[Frontrunner]

Hi All...

Ralph I see you are understanding the problem quite good and your approach is also good.

As for your reference I will give another tricky example:

Basic
a = -x ^ 3 - 3 * x ^ 2 + 4 * x - (x ^ 3 + x ^ 2 - 2) * (-1)

C
a = -pow(x,3) - 3 * pow(x,2) + 4 * x - (pow(x,3) + pow(x,2) - 2) * (-1)

Kind regards,
Frontrunner