Mathematical expression translator - Printable Version +- Qbasicnews.com (http://qbasicnews.com/newforum) +-- Forum: QbasicNews.Com (http://qbasicnews.com/newforum/forum-3.html) +--- Forum: Challenges (http://qbasicnews.com/newforum/forum-10.html) +--- Thread: Mathematical expression translator (/thread-10109.html) |
Mathematical expression translator - Frontrunner - 04-21-2008 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 Re: Mathematical expression translator - wildcard - 04-21-2008 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. Re: Mathematical expression translator - Frontrunner - 04-22-2008 That's right, it is quite specific but a general parser would be too easy for a challenge : Cheers, Frontrunner Re: Mathematical expression translator - wildcard - 04-22-2008 Hopefully I can get some time to have a go, seems straight forward enough but will see. Re: Mathematical expression translator - Frontrunner - 04-22-2008 I am looking forward to see your contribution! Cheers, Frontrunner Re: Mathematical expression translator - Ralph - 04-23-2008 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. Re: Mathematical expression translator - Frontrunner - 04-23-2008 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 Re: Mathematical expression translator - Ralph - 04-23-2008 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! Re: Mathematical expression translator - Frontrunner - 04-24-2008 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 Re: Mathematical expression translator - wildcard - 04-24-2008 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 |