Matthew

> >**Puzzle 206.** Suppose you are trying to feed hungry children with the loaves of bread you bought in the previous puzzle, and you can feed at most three children with each loaf of bread. Describe this using a feasibility relation \\(\Phi : \mathbb{N} \nrightarrow \lbrace 0,1,2\rbrace \\). Here \\(\mathbb{N}\\) is the set of natural numbers \\( \lbrace 0,1,2,3,\dots \rbrace \\) with its usual ordering.

>

> \[

\Psi(x, y) =

\begin{cases}

\texttt{true} & \mbox{if } (x \leq 0 \mbox{ and } y = 0) \mbox{ or } (x \leq 3 \mbox{ and } y = 1) \mbox{ or } (x \leq 6 \mbox{ and } y = 2) \\\\

\texttt{false} & \mbox{otherwise.}

\end{cases}

\]

> I can see a conjoint that gives rise to your answer.

Thanks for doing this. Guidance like this is really helpful as a newb. Really appreciate it.

I think the conjoint is pretty obvious from the feasibility relation.

First set \\(G(y) =3x\\):

\[ \text{y loaves feeds G(y) children} \]

Then \\(\check{G}(x,y) = \text{true iff } x \le G(y) \\).

Feasibily relation is saying "If we have y loaves, we can feed at most G(y) children."

> The companion, if it exists at all, is less intuitive. If it doesn't exist do you see how we'd go about proving it doesn't?

I think I see why this whole business is confusing. I can give a companion solution using a different G(y) but I do not see how we can make a companion using the function described above nor can I prove it. I can plug it in using definitions and can see it is impossible to give a viable companion solution tho. Using a different G(y) or to make notation more applicable, using a new function F(x), we can do the following:

First set \\(F(x) = \left\lceil x/3 \right\rceil\\):

\[\text{x children need F(x) loaves of bread} \]

Then \\(\hat{F}(x,y) = \text{true iff } F(x) \le y \\):

\[

\Psi(x, y) =

\begin{cases}

\texttt{true} & \mbox{if } (0 \leq x \mbox{ and } 0 \leq y) \mbox{ or } (3 \leq x \mbox{ and } 1 \leq y) \mbox{ or } (6 \leq x \mbox{ and } 2 \leq y ) \\\\

\texttt{false} & \mbox{otherwise.}

\end{cases}

\]

In this case, feasibility relation is saying "if we have at least x children, we need at least F(x) loaves of bread."

Since F(x) and G(y) are adjoints, shouldn't \\(\hat{F}(x,y) = \check{G}(x,y)\\)?

> >**Puzzle 206.** Suppose you are trying to feed hungry children with the loaves of bread you bought in the previous puzzle, and you can feed at most three children with each loaf of bread. Describe this using a feasibility relation \\(\Phi : \mathbb{N} \nrightarrow \lbrace 0,1,2\rbrace \\). Here \\(\mathbb{N}\\) is the set of natural numbers \\( \lbrace 0,1,2,3,\dots \rbrace \\) with its usual ordering.

>

> \[

\Psi(x, y) =

\begin{cases}

\texttt{true} & \mbox{if } (x \leq 0 \mbox{ and } y = 0) \mbox{ or } (x \leq 3 \mbox{ and } y = 1) \mbox{ or } (x \leq 6 \mbox{ and } y = 2) \\\\

\texttt{false} & \mbox{otherwise.}

\end{cases}

\]

> I can see a conjoint that gives rise to your answer.

Thanks for doing this. Guidance like this is really helpful as a newb. Really appreciate it.

I think the conjoint is pretty obvious from the feasibility relation.

First set \\(G(y) =3x\\):

\[ \text{y loaves feeds G(y) children} \]

Then \\(\check{G}(x,y) = \text{true iff } x \le G(y) \\).

Feasibily relation is saying "If we have y loaves, we can feed at most G(y) children."

> The companion, if it exists at all, is less intuitive. If it doesn't exist do you see how we'd go about proving it doesn't?

I think I see why this whole business is confusing. I can give a companion solution using a different G(y) but I do not see how we can make a companion using the function described above nor can I prove it. I can plug it in using definitions and can see it is impossible to give a viable companion solution tho. Using a different G(y) or to make notation more applicable, using a new function F(x), we can do the following:

First set \\(F(x) = \left\lceil x/3 \right\rceil\\):

\[\text{x children need F(x) loaves of bread} \]

Then \\(\hat{F}(x,y) = \text{true iff } F(x) \le y \\):

\[

\Psi(x, y) =

\begin{cases}

\texttt{true} & \mbox{if } (0 \leq x \mbox{ and } 0 \leq y) \mbox{ or } (3 \leq x \mbox{ and } 1 \leq y) \mbox{ or } (6 \leq x \mbox{ and } 2 \leq y ) \\\\

\texttt{false} & \mbox{otherwise.}

\end{cases}

\]

In this case, feasibility relation is saying "if we have at least x children, we need at least F(x) loaves of bread."

Since F(x) and G(y) are adjoints, shouldn't \\(\hat{F}(x,y) = \check{G}(x,y)\\)?