Matthew wrote:

> John mentioned that one of the problems has *neither* a conjoint nor a companion [....]

I think I might have been wrong about that.

My intuition for companions and conjoints wasn't so good at first: witness my surprise in [comment #9](https://forum.azimuthproject.org/discussion/comment/20337/#Comment_20337)! But I think that surprise wound up improving my intuition.

So, let me try to explain my intuition for this stuff. It's easiest with an example:

> **Puzzle 207.** Suppose you buy loaves of bread and then use them to feed hungry children. Compose the feasibility relation \\(\Psi : \{0,1,2\} \nrightarrow [0,\infty) \\) from Puzzle 205 and the feasibility relation \\(\Phi : \mathbb{N} \nrightarrow \lbrace 0,1,2\rbrace \\) from Puzzle 206 to get a feasibility relation \\(\Psi \Phi : \mathbb{N} \nrightarrow [0,\infty) \\) describing how many children you can feed for a certain amount of money (given the fact that you plan to buy at most two loaves).

Suppose we want \\(\Psi\\) to be the companion of some monotone function \\(F \colon \{0,1,2\} \nrightarrow [0,\infty) \\).

\\(F\\) is a function that takes a number of loaves of bread and gives an amount of money. So, the obvious first guess is a function \\(F\\) such that

\\(x\\) loaves of bread costs \\(F(x)\\) dollars.

\\(\hat{F}\\) will then take this function and 'pad it out' to give a feasibility relation. If you have _more than_ \\(F(x)\\) dollars, it's still feasible to buy \\(x\\) loaves of bread. And if you have \\(F(x)\\) dollars, it's feasible to buy _less than_ \\(x\\) loaves of bread. So the feasibility relation \\(\hat{F}\\) is true for more pairs \\( (x,y) \\) than those for which \\(F(x) = y\\).

In fact, it's true exactly for pairs with \\(F(x) \le y\\)!

This isn't hard to see. The idea of a companion is that

\[ \hat{F}(x,y) = \mathbf{Bool}(F(x), y ) \]

but if you unravel all the jargon and notation, this just means

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

So, to find a function \\(F\\) whose companion is \\(\Psi\\), we need a function such that

You can buy \\(x\\) loaves of bread for \\(y\\) dollars iff \\(F(x) \le y\\).

But to repeat myself: if we're seeking such a function, the obvious first guess is a function such that

\\(x\\) loaves of bread costs \\(F(x)\\) dollars.

For Puzzle 207, I think such a function exists.

> John mentioned that one of the problems has *neither* a conjoint nor a companion [....]

I think I might have been wrong about that.

My intuition for companions and conjoints wasn't so good at first: witness my surprise in [comment #9](https://forum.azimuthproject.org/discussion/comment/20337/#Comment_20337)! But I think that surprise wound up improving my intuition.

So, let me try to explain my intuition for this stuff. It's easiest with an example:

> **Puzzle 207.** Suppose you buy loaves of bread and then use them to feed hungry children. Compose the feasibility relation \\(\Psi : \{0,1,2\} \nrightarrow [0,\infty) \\) from Puzzle 205 and the feasibility relation \\(\Phi : \mathbb{N} \nrightarrow \lbrace 0,1,2\rbrace \\) from Puzzle 206 to get a feasibility relation \\(\Psi \Phi : \mathbb{N} \nrightarrow [0,\infty) \\) describing how many children you can feed for a certain amount of money (given the fact that you plan to buy at most two loaves).

Suppose we want \\(\Psi\\) to be the companion of some monotone function \\(F \colon \{0,1,2\} \nrightarrow [0,\infty) \\).

\\(F\\) is a function that takes a number of loaves of bread and gives an amount of money. So, the obvious first guess is a function \\(F\\) such that

\\(\hat{F}\\) will then take this function and 'pad it out' to give a feasibility relation. If you have _more than_ \\(F(x)\\) dollars, it's still feasible to buy \\(x\\) loaves of bread. And if you have \\(F(x)\\) dollars, it's feasible to buy _less than_ \\(x\\) loaves of bread. So the feasibility relation \\(\hat{F}\\) is true for more pairs \\( (x,y) \\) than those for which \\(F(x) = y\\).

In fact, it's true exactly for pairs with \\(F(x) \le y\\)!

This isn't hard to see. The idea of a companion is that

\[ \hat{F}(x,y) = \mathbf{Bool}(F(x), y ) \]

but if you unravel all the jargon and notation, this just means

So, to find a function \\(F\\) whose companion is \\(\Psi\\), we need a function such that

But to repeat myself: if we're seeking such a function, the obvious first guess is a function such that

For Puzzle 207, I think such a function exists.