iammisc wrote:Combuster wrote:Nice attempt, but the answer to 4 turned out to be 4.

I don't get it. Did you not write the riddle correctly? Or are you trying to use my algorithm to solve the riddle for 4 because for 4 the answer would be 1 not 4, according to my method.
My method, however, works for your riddle.
Your method works for the three provided cases, but not the general one, and hence not for the riddle in question.
Three things:
1) I had to translate it. In its original language there are several more occurrences where #letters = number/2
2) Your method is a general solution for any such problem. Given 3 points, you need a second degree polynomial to cover them all. You generated an equation of the same degree. Basically, any two properties that are not colinear can be used to generate a fit for this specific riddle. You could even now think of a different parameter and include it to make a third degree equation that would cover the 4-4 case and be exceptionally complex.
basically, you could fit ax²+bx+c to the riddle. (you constructed an au + bv + c, where u and v are not of the same order, just as x and x² aren't)
3) The solution is too complex. Given the circumstances of the riddle, even the most unintellectual person has to be able to solve it by hard, hence generating the solution must be very simple (once you know it, that is), and common people suck at maths
So once again, kudo's for the attempt, but I hope this explains why it wasn't good enough.