Can a non-constant function V exist such that for every N-tuple t there exists a weight w such that V(t, w) is at least as great as V(j, w) for every other N-tuple j?
The idea for an application of such a function is for devising a system in which every "applicant" (represented by an N-tuple) has at least one instance (represented by a weight) where they are a "best choice." In short, it would be useful in a "no losers" system.
If there is nothing more, define V(j,w) = - | w - ||j|| |, where for a given t you will choose w = ||t||.
No, the maximum principle restricts such cases; it must be a constant function.
It sounds to me like there may be formulation mishap in the question; forall t, j where t =/= j, exists w such that V(t,w)
If there is nothing more, define V(j,w) = - | w - ||j|| |, where for a given t you will choose w = ||t||.
I'll try to clarify with the example that lead me to think about this:
Let's say that we have some pants that we would wish to buy. We are able to numerically judge the quality of each in addition to the price of each.
Now let's weight our decision of which pair that we would like to buy such that a weight of 1 represents only caring about the quality and a weight of 0 represents only caring about the price. Weights between 1 and 0 behave as expected.
Now, if we were to "slide" the weight from 0 to 1, the "best choice" (maximum of function V) would change from pair to pair. However, this does not mean that every pair would have a weight at which it would be the best choice. Pairs of pants that would never be the best choice can be discarded from the set of pants we are considering.
The question now comes to: is there a system for weighting the properties of possible choices such that each item we are considering has a weight at which it would be the best choice?
I worded it with "at least as great" because multiple items might "tie" for best at a weight.
By the wording "non-constant" I mean that V(t, w) =/= c.
Additionally, I am now considering this question under the restriction that both the weight and the elements of the N-tuples are limited to the interval [0, 1].
As I said, if there is nothing more - your pdf formula requirement is satisfied by the above function!
Be careful of Nominal type data. It is arbitrary and caries little information. In other words, you are attempting to project an n-dimensional manifold onto the unit interval [0,1]. This will greatly restrict the ability to have maximas and minimas of any great degree or complexity.
What if we consider a different V for each N? I.e., one V might be applicable for 2-tuples and another for 3-tuples.
In this way, we only care to define a V regarding tuples of the size applicable in a given system.
Such a V (that fails the above formula) for N=2 could be: V(t, w) = wt_1 + (1 - w)t_2
For the sake of this conversation, let's just assume N to always be 2.
How about assigning a binary range for n-tuples, where each variable is unique? You then have a function of n-variables that take on either a zero or one iff they meet the criteria. So, in the end you will have a function with so many variables being either zero or one, then you may sum the ones, which would imply something desirable and the total sum of each category would correspond to the Nominal rank or degree of significance. In other words, you could have say ten variables defining the function F, and say fifteen categories, so, you would say at category Macy's the jeans satisfy 6 out of the ten possible variables (descriptors or relators). Alright, then at Marshall's there are only three variables assigned unity, cause only three conditions are satisfied. In this case, Macy's wins the prize... Maybe this configuration would work?
@Luisiana: good thinking. However, the primary goal here is to develop a system where, for example, every single department store could win depending on the weight selected. Before we do so, I would like to prove that a function for managing the weighting of values in such a way could even exist. Your method could find use in scoring individual fields however (e.g. ranking jean quality based on a sum of 1s or 0s).
@Alex: I will look at the reducibility of our two models. If it's a fit, then I guess such a "loserless" system could not exist; otherwise, then it's still a project to find V!
If I now understand you correctly, what you're doing is determining the so called Pareto optimal plane (or Pareto optimal set). Your "weight" is actually the "time" in the bezier-sense. The Pareto optimal set in the case of 2-tuples is a line in a 2D plane.
To generalise this for N-tuples of any N, you must either determine the precise path the Pareto optimal set (line) will have, so that you can vary on the interval [0,1] and touch all theses points. The alternative is to have (N-1)-dimensional weight factors; where the weight for the N^th tuple element is 1 - Sum(i=1..N-1; f_i). Thus, we have an n-tuple of metrics (pant quality, pant kost) and an n-tuple of factors that sum to 1. These tuples can be pair-wise multiplied and summed.
All of this as in: iff I've now understood you; Pareto-stuff
@Philip: interesting. I will have to give that a look.
@everybody: just in case there is any confusion, when considering a set of N-tuples for a best choice (choice being governed by V and a weight), I mean a set containing all tuples with the same size (where size = N, a constant to be determined by the application), not the set of all tuples of any size.
Also, I have an idea for V that seems to satisfy the above constraint, as well as using a w that actually behaves as a weight. I will do some more work on verifying that it is defined for all values and post a PDF later.
Suppose we have m N-tuples, say t1, ..., tm, then we can construct a non-constant function V as follows:
V(t1,1)=2, V(ti,1)=1, for i1,
V(t2,2)=2, V(ti,2)=1, for i2,
....
V(tm,m)=2, V(ti,m)=1, for im,
V(ti,n)=0, for n>m.
The above funcition could be what you want to have, suppose all weights are natural numbers. Simiarly, we can construct lots of such functions.
**this is an answer from a friend****
@Zhiming: it is true that your method defines a function that meets the requirements, although I am not sure where that particular function would be especially useful.
@Alex: yes, you definitely misread something. Have you taken a look at the PDF I posted a few posts up? It defines the proposition quite clearly.
By this point, we have found a number of functions which prove that such a function can exist. Now, it is a matter of finding one that (1) uses a weight that behaves like a weight and (2) is applicable in a "loserless" system.
To be more clear on that last point, a "loserful" system is one in which a participant may exist that is "always a loser" (never the best choice), regardless of the value of the weight. A "loserless" system is just the contrary, allowing every participant at least opportunity to be the best choice.
A special case is the game with 2D nxn grid. any point (i,j) moves along a path of grids to the right-up point (n,n). There exists the short path with a decision sequence w ( \in {up, down, right, left}* . The V((x,y),w) is the corresponding the number of steps to the up-point (the reverse of the problem).
Generalize the problem, is the N-verctor space, from any given point, there exists a shortest path (from the point "t" to the orginal point (0, 0, ... 0)) with a choice of w as the path .
@Zhiming: very interesting. It appears that such functions as the one I have been looking for have applications in many different problems. Maybe we should give further examination into these functions for more than just my "loserless system" problem.
@everyone: in any case, I have attached a short PDF describing (briefly and informally) my "loserless system" problem and my function for V.
I agree with the example of Zhiming Liu. More generaly, to answer the question of the existence of V, if your N-tuples form a countable set, by definition there exists an injective function f to the set of integers. Thus you can define the following function V :
V(t,w) = 1 if f(t)=w and 0 otherwise
Of course there is nothing constructive here. If your ground set is finite of size B, a natural injective function f is to write each N-tuples as a number in base B.
Notice that these function V ensures that for each t, there exists an integer w (is it what you call a weight ?) such that V(t,w) > V(t',w) for all t' t, that is we have a strict inequality. If you only require the condition "at least as great as" the existence of V is trivial, considering only {0,1} for weights :
- pick one particular N-tuple t0
- define V(t0,0) = 1 and V(t,0) = 0 for any t t0
-define V(t0,1)=0 and V(t,1)=1 for any t t0
Here we are.
As you said, Christophe, the existance of V is trivial. In fact, the more we look at this problem, the more we find rather simple solutions to it. What I like about this conversation, however, this the various way that we are all approaching our solutions.
What if we require that V be a continuous function and that V(t, w) be strictly greater than V(j, w) for at least one w? This makes our piecewise and absolute value solutions invalid. My initial thoughts on this would be a cleverly composed sin-based function.
Well here's a related problem: Devise a continuous function f : [0, 1]^3 \to [0, 1] such that for all (x, y) exists t such that f(x, y, t) > f(x', y', t).
Sorry to sound Bourbaki-like but an N-tuple of what? Integers? Elements ?
Currently, I am considering N-tuples of real numbers in the interval [0, 1]. However, this certainly could be adapted for addressing many other considerations.
For example, given that we are considering 4-tuples of dishes in a 4-course meal, the function V could value the meal given a weight between healthiness and taste. (Of course, this would simply be reduced to 4-tuples of 2-tuples of a health metric and a taste metric.)
@everybody: regarding the consideration that V must be continuous, proving this seems almost trivial. Regarding the consideration that t must be strictly greater then j (as opposed to at least as great as), proving this seems to require proving that a function could exist that maps 1D-space to nD-space (1D-space of the weight to the nD-space of all N-tuples accepted by V).
@Ville: is this 1D to nD mapping the well-known problem you were referring to?
Yes, exactly :) Perhaps I can spoil it now since actually constructing such a function is rather hard: This is called the Hilbert curve (at least in the 2d case). After that, it's a simple matter to construct the actual function.
Bryan, if the weigth depend on t, what you real want if a two variable function f(x,y) such that the function y \mapsto f(x,y) reaches maximum in y=x. The obvious choice is a Dirac centered in x ie f(x,y)=delta(x-y). If you demand continuity differentiability etc just replace delta by a suitable approximation. If x,y are in [0,1]^2 that any thumbtack like function will do; eg f(x,y)=exp(-|x-y|)....
The most simple function V must be a quadratic form. Think about a concave paraboloid, which reaches its maximum in t. Potentially, there exist many matrices of that quadratic form depending on the parameter w.
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