Whenever you have 10 W of power, you also have 5 W (\(5 W \leq 10 W\))
The requirement that the feasibility relation map is monotone says that if \(x' \leq_X x\) and \(y \leq_Y y'\), then \(\phi(x,y) \leq_{Bool} \phi(x',y')\)
If x can be obtained from y, something easier to obtain than x can also be obtained from y
If x can be obtained from y, then x can be obtained from something harder to obtain than y
This chapter should make the following table clear:
Bool- | |
---|---|
Bool- | |
Bool- |
Bool is a quantale, meaning it has all joins and a closure operation, \(\mathbb{B}\times\mathbb{B}\xrightarrow{\Rightarrow}\mathbb{B}\). The closure operation must satisfy \(b \land c \leq d\) iff \(b \leq (c \Rightarrow d)\). It is the fact that Bool is a quantale that makes this chapter work; we could alternatively use Cost and obtain the cost of obtaining \(x\) from \(y\), not just whether or not it’s possible.
A feasibility relation for preorder \(X\) given preorder \(Y\)
A monotone map \(X^{op}\times Y \xrightarrow{\phi} \mathbf{Bool}\)
Also denoted \(X \overset{\phi}\nrightarrow Y\)
Given \(x \in X, y \in Y\), if \(\phi(x,y)=true\) then we say x can be obtained from y
Suppose we have the preorders \(X:=\boxed{monoid\rightarrow category \leftarrow preorder}, Y:=\boxed{nothing \rightarrow thisBook}\)
Draw the Hasse diagram for the preorder \(X^{op} \times Y\)
Write a profunctor \(X \overset{\phi}\nrightarrow Y\)
True means “my aunt can explain an x given y"
Interpret the fact that the preimage of true is an upper set in \(X^{op}\times Y\)
Let \(\phi((M,B))=\phi((P,B))=True\) else \(False\)
The preimage of true being an upper set is a consequence of monotone maps to Bool. The domain orders combinations by feasibility (\(x\leq y\) means x is easier than y), and the preimage being an upper set says that if my aunt can explain \(x\) given \(y\), then she can do something easier than \(x\) given \(y\) and can explain \(x\) with something with more explanatory power than \(y\).
Show that the Boolean \(\Rightarrow\) satsifies the requirements of a closure operator.
This boils down to the tautology that \((a \land v) \implies w\) iff \(a \implies (v \implies w)\), where the last \(\implies\) comes from \(\multimap\) and the others come from \(\leq\).
A \(\mathcal{V}\) functor must have \(\mathcal{V}\) categories for domain and codomain, so the \(\mathcal{V}\) of a \(\mathcal{V}\) profunctor must be considered as enriched in itself.
Matrix algorithm for computing distance matrix of profunctor: \(d_X * M_\phi * d_Y\)
A \(\mathcal{V}\) profunctor, denoted \(\mathcal{X}\overset{\phi}{\nrightarrow} \mathcal{Y}\) - where \(\mathcal{V}=(V,\leq,I,\otimes)\) is a (unital commutative) quantale, and \(\mathcal{X},\mathcal{Y}\) are \(\mathcal{V}\) categories.
A \(\mathcal{V}\) functor \(\mathcal{X}^{op}\times Y \xrightarrow{\phi} \mathcal{V}\)
Bool-profunctors and their interpretation as bridges
Let’s consider Bool-profunctors. Recall a preorder (Bool-category) can be drawn as a Hasse diagram.
A Bool-profunctor \(X \overset{\phi}{\nrightarrow} Y\) can look like this:
With bridges coming from the profunctor, one can now use both paths to get from points in \(X\) to points in \(Y\).
There is a path from N to e, so \(\phi(N,e)=\)\(true\) but \(\phi(W,d)=\)\(false\).
We could put a box around both preorders and define a new preorder, called the collage.
Cost profunctors and their interpretation as bridges.
Cost categories are Lawvere metric spaces and can be represented by weighted graphs
This is a depiction of a Cost-profunctor
The distance from a point in the left to a point in the right will run through the left, across a bridge, then through through the right.
\(\phi(B,x)=\)\(11\),\(\phi(A,z)=\)\(20\),\(\phi(C,y)=\)\(17\)
Is it true that a Bool-profunctor is exactly the same as a feasibility relation?
Monotone maps are Bool-functors between Bool-categories, so the definitions line up
Fill out the Cost-matrix associated with Example 4.13 NOCARD
\(\phi\) | x | y | z |
---|---|---|---|
A | 17 | 21 | 20 |
B | 11 | 15 | 14 |
C | 14 | 18 | 17 |
D | 12 | 9 | 15 |
Show that a \(\mathcal{V}\) profunctor is the same as a function \(Ob(\mathcal{X})\times Ob(\mathcal{Y}) \xrightarrow{\phi} V\) such that, \(\forall x,x' \in \mathcal{X}, y,y' \in \mathcal{Y}\), the following holds in \(\mathcal{V}\):
\(\mathcal{X}(x',x)\otimes \phi(x,y) \otimes \mathcal{Y}(y,y') \leq \phi(x',y')\)
A \(\mathcal{V}\) profunctor must be a function satisfying the following constraint, according to the \(\mathcal{V}\) functor definition:
\(Z((x,y),(x',y')) \leq\) \(\mathcal{V}(\phi((x,y)),\phi((x',y')))\)
where \(Z = \mathcal{X}^{op}\times \mathcal{Y}\)
Unpacking the definition of a product \(\mathcal{V}\) category, we obtain
\(\mathcal{X}^{op}(x,x') \otimes \mathcal{Y}(y,y') \leq \mathcal{V}(\phi((x,y)),\phi((x',y')))\)
And applying opposite category definition: \(\mathcal{X}(x',x) \otimes \mathcal{Y}(y,y') \leq \mathcal{V}(\phi((x,y)),\phi((x',y')))\)
Noting the definition of \(\multimap\) for a \(\mathcal{V}\) category enriched in itself:
\(\mathcal{V}(v,w)=v\multimap w\), so now we have: \(\mathcal{X}(x',x) \otimes \mathcal{Y}(y,y') \leq \phi((x,y)) \multimap \phi((x',y'))\)
From the constraint of a hom-element of a symmetric monoidal preorder \(\mathcal{V}\), i.e. \(a \leq (v \multimap w)\) iff \((a \otimes v) \leq w\), we see that the first case pattern matches with:
\(a \mapsto\) \(\mathcal{X}(x',x) \otimes \mathcal{Y}(y,y')\)
\(v \mapsto\) \(\phi((x,y))\)
\(w \mapsto\) \(\phi((x',y'))\)
So using the iff we can rewrite as \((a \otimes v) \leq w\), and use the commutativity of \(\otimes\) to obtain the desired expression.
Each box in a codesign problem has a list of preorders on the LHS and another list on the right.
The box represents a feasibility relation: \(Load \times Velocity \nrightarrow Torque \times Speed \times \$\)
Consider the feasibility relation
\(T:=\boxed{mean\rightarrow nice}\)
\(E:=\boxed{boring\rightarrow funny}\)
\(Money:=\boxed{100 K\rightarrow 500 K \rightarrow 1M}\)
A possible feasibility relation is here:
This says, e.g., a nice but boring movie costs $500,000 to produce.
We can infer that we can also make a mean/boring movie with what we can produce nice/boring movie.
The node \((nice,funny)\) has no arrow from it in Example 4.18a. What does this mean?
It is not feasible, for any amount of money listed, to produce a nice and funny movie.