Let's say I have a simple system
a + b - c = 0
a > 0
b > 0
c > 0
and let's say that I'm 'blinded'--meaning that I can't explicitly take any two values and perform anything other than an equality comparison (this means that I can't compute piece-wise or branching functions, nor can I explicitly check for x > y).
If somebody gives me three example points a, b, c, is there any way I can determine whether my system is satisfied? I had originally thought to attempt to derive a single system of equalities using Lagrangian multipliers that will have a solution iff a, b, and c satisfy the original system; however, it seems I'm too retarded to get anything other than a single point (where my goal is a system) from the method.
Any thoughts?
>>7637675
No clue what the hell you are talking about. What do you mean by three example points.
>>7637790
Say I have a hypothesis of a = 1, b = 2, c = 3.
I want to be able to verify that the above example points satisfy my system without the use of branching, piece-wise functions, or comparison operators other than = or !=.
>>7637922
Okay, I think I see what your saying. So your elements don't start off positive, you have to prove they are positive, correct?
If so that's clearly impossible. You haven't defined a way to show anything is greater than anything, since you have ruled out using standard methods.
You need to have an axiom that says, "We know x>y when [some quality] is satisfied." And you don't have that.
>>7637675
You have 3 unknowns and one equation.
This means that there are an infinite number of solutions. Two of the three are free variables.
>>7638141
It's not that x > y isn't defined. It's just that you can't explicitly rely on that definition. All the properties of ordering on the real number line still hold, for example. The challenge is in determining a relative state computationally without relying on branches or conditional logic.
>>7638175
Right. There are an infinite number of solutions.
The meat of the problem at hand relies on, given a possible solution, verifying it without explicitly relying on '>' or '<' operators.
>>7638178
No, you don't get it. How would anyone go about proving the ordering? You need a condition for the proof. In math we say that if there is an injective function from A->B than |A|=<|B|. The natural ordering on the real line springs from this condition.
So what condition would you allow? How would I show for example 1<2. Give me some necessary and/or sufficient conditions to prove this.
You don't have any (from what I can tell, unless you count something trivial like x+1=y => x<y). Until you get some it will be impossible.
If you have a-b-c=y for example
If two other systems are satisfied then the other two systems should interect your system at different variables and should intersect each other at the variable not intersected by the original equation.
These three examples satisfy your system of equations.
>>7638322
Intersect***
>>7638322
I think I see where you're going with this; however, I can't tell if you've quite made it to the point yet.
Given generic values proposed for a, b, c, how do apply this to determine whether or not they satisfy the original system?
