gatoatigrado
2010-07-27 05:27:25
This is wrong because the problem statement says there may be multiple elements with the smallest radius and they should all have value 0.
Alexey Romanov
2010-07-27 06:04:56
ah, sorry, I didn't get the differentiation between value and radius. Thanks.
gatoatigrado
2010-07-27 06:18:31
A:
What you wrote is also true if there are no elements with the smallest radius. If this is desired, you are correct; if not, you need to add a clause to that effect:
(\forall e1 \in S. \forall e2 \in S. Radius e1 <= Radius e2 --> Value e1 = 0) \and (\exists e1 \in S. \forall e2 \in S. Radius e1 <= Radius e2)
Alexey Romanov
2010-07-27 06:03:33
This has the same problem as the original. This statement asserts that the value of every element is 0.
luqui
2010-07-27 06:25:33
Don't vote him down b/c you think he's making mistakes! Vote downs should be reserved for cases where answers are opinion b/s unrelated to the question i.m.h.o. Voted back up :)
gatoatigrado
2010-07-27 06:27:09
Well y.h.o. differs from m.h.o. I want to draw the attention of future readers to the correct answers.
luqui
2010-07-27 06:31:46
+1
A:
Just to clarify with parentheses, what you wrote is usually taken to mean:
\forall e1 \in S. (\forall e2 \in S. (Radius e1 <= Radius e2 --> Value e1 = 0))
This statement asserts that the value of every element is 0. Here's how: Pick an arbitrary e1, now pick e2 = e1, and we have: Radius e1 <= Radius e1 --> Value e1 = 0. Since the antecedent (thing before the -->) is true, we have Value e1 = 0. And since we made no assumptions about e1, we have forall e \in S. Value e = 0.
The problem is that your parentheses are off.
\forall e1 \in S. (\forall e2 \in S. Radius e1 <= Radius e2) --> Value e1 = 0
In order for the antecedent to be true now, e1's radius has to be less than or equal to every (as opposed to any) other radius, which seems like what you intended.
luqui
2010-07-27 06:30:43