"That was why I have been repeatedly asking for a proper definition of the problem"
Let us simply things a bit.
Problem definition:
Let n be an integer. Can we compute n^2 in polynomial time for any n?
thank you for your time. but if we do assume pythagorean tripples...is the formula in np or not?
"... If a and b are fixed, then I can write a2+b2 even in constant time.
mmm! ok. so the pythagoras formula cannot be verified in poly time (for all values) and therefor not in np. correct? if correct then i think i understand now. many thanks!
sorry, i did not see the rest of your post. was using a small screen phone. now using laptop... ok. so your saying pythagoras theorem/formula in probably not in NP because some input values cannot be written in finite time. makes sense. am i on the same page with you now?
Great answers guys!
If I understand (I hope I do!) What your saying is that the input I.e in the pythagoras theorem or formula a^2 + b^2 = c^2 . Say your given a, b and c. Verifyng c in polynomial time is not possible for large a and B. is large i.e close to infin
Np-hard: say problem A is np-HARD. ﹰﹰproblem ﹰB is also np-hard if problem A can reduce to problem B.
So when I asked if pythagoras theorem (finding the third side...) is np-hard, I was really asking if there exists an np-hard problem that reduces to the pythagoras theorem.
Aha! Now we are getting somewhere. The question now is : is pythagoras np-hard? If not then why not?
Sorry .. a bit of a lazy typer. Pythagoras theorem. A^2 +b^ = c^2
Going back to the pythagoras example. Given two sides of a right angled triangle A and B. VERIFYING that some given number C is a solution to the third side can be done in polynomial time - just plug it in the formula A^2 + B^2 = C^2. On the other hand, FINDING C can also be done in polynomial...