CMSI 585: Programming Language Foundations: Homework #4 Answers

(9.1.5) Going with the book’s “we use long function names” style:

evalRoman ⟦h t u⟧ = evalHundreds ⟦h⟧ + evalTens ⟦t⟧ + evalUnits ⟦u⟧
evalHundreds ⟦ε⟧ = 0
evalHundreds ⟦C⟧ = 100
evalHundreds ⟦CC⟧ = 200
evalHundreds ⟦CCC⟧ = 300
evalHundreds ⟦CD⟧ = 400
evalTens ⟦lt⟧ = evalLowTens ⟦lt⟧
evalTens ⟦XL⟧ = 40
evalTens ⟦L lt⟧ = 50 + evalLowTens ⟦lt⟧
evalTens ⟦XC⟧ = 90
evalLowTens ⟦ε⟧ = 0
evalLowTens ⟦lt X⟧ = evalLowTens ⟦lt⟧ + 10 
evalUnits ⟦lu⟧ = evalLowUnits ⟦lu⟧
evalUnits ⟦IV⟧ = 4
evalUnits ⟦V lu⟧ = 5 + evalLowUnits ⟦lu⟧
evalUnits ⟦IX⟧ = 9
evalLowUnits ⟦ε⟧ = 0
evalLowUnits ⟦lu I⟧ = evalLowUnits ⟦lu⟧ + 1

(9.2.6) Syntax: Add a new clause to the Answer syntax category:
```
Answer ::= M+ | = | +/- | sqr
```
Semantics: Add a new clause to the calculate semantic function:
```
calculate ⟦sqr⟧ (a,op,d,m) = (a,op,times(d,d),m)
```
(9.3.1 a,b) Part (a) using the book's notation:
```
execute ⟦repeat C until E⟧ = loop
    where loop sto = if evaluate ⟦E⟧ sto' then sto' else loop sto'
        where sto' = execute ⟦C⟧ sto
```
A more conventional and concise approach:

$\mathscr{C}[\![\mathtt{repeat}\;c\;\mathtt{until}\;e]\!] = \mathit{r}\; \mathsf{where}\;r\;\sigma = \mathsf{let}\;\sigma' = \mathscr{C}\,c\,\sigma\;\mathsf{in}\; \mathsf{if}\;\mathscr{E}\,e\,\sigma'\;\mathsf{then}\;\sigma'\;\mathsf{else}\;r\;\sigma'$

If you want to go with the fix point style:

$\mathscr{C}[\![\mathtt{repeat}\;c\;\mathtt{until}\;e]\!] = \mathit{fix}\; \lambda r. \lambda \sigma. \mathsf{let}\;\sigma' = \mathscr{C}\,c\,\sigma\;\mathsf{in}\; \mathsf{if}\;\mathscr{E}\,e\,\sigma'\;\mathsf{then}\;\sigma'\;\mathsf{else}\;r\;\sigma'$

Part (b), also using the book’s notation:
```
evaluate ⟦if E1 then E2 else E3⟧ =
    if evaluate ⟦E1⟧ then evaluate ⟦E2⟧ else evaluate ⟦E3⟧
```
In class, I mentioned you could skip the proof of semantic equivalence that was requested in Part (b), so that will not be graded.
(9.3.9) The book uses a pretty lame syntax for variables. As mentioned in the helper, it is much nicer to use the grammar I used in my notes on Logic. So here’s my concrete syntax:
```
var      → "p" | "q" | "r" | var "′"
formula  → "true" | "false" | var
         | "¬" forumula
         | "(" formula "∧" formula ")"
         | "(" formula "∨" formula ")"
         | "(" formula "⊃" formula ")"
         | "(" formula "≡" formula ")"
```
The abstract syntax is:
```
v: Var
f: Formula = true | false | ¬f | f ∧ f | f ∨ f | f ⊃ f | f ≡ f
```
The semantics requires formulas to be evaluated against an interpretation which is the logician’s word for a “memory” or “store” that maps variables to truth values. This domain looks like:

$\mathsf{Interpretation} = \mathsf{Var} \rightarrow \mathbb{B}$

and the semantic function for evaluating formulas is:

$\mathscr{E}: \mathsf{Formula} \rightarrow \mathsf{Interpretation} \rightarrow \mathbb{B}$

Defined like so:

$\begin{array}{l} \mathscr{E} [\![ \mathtt{true} ]\!] \varphi = T \\ \mathscr{E} [\![ \mathtt{false} ]\!] \varphi = F \\ \mathscr{E} [\![ p ]\!] \varphi = \varphi\;p \\ \mathscr{E} [\![ \neg p ]\!] \varphi = \neg\mathscr{E} [\![ p ]\!]\varphi \\ \mathscr{E} [\![ p \wedge q ]\!] \varphi = \mathscr{E} [\![ p ]\!]\varphi \wedge \mathscr{E} [\![ q ]\!]\varphi \\ \mathscr{E} [\![ p \vee q ]\!] \varphi = \mathscr{E} [\![ p ]\!]\varphi \vee \mathscr{E} [\![ q ]\!]\varphi \\ \mathscr{E} [\![ p \supset q ]\!] \varphi = \mathscr{E} [\![ p ]\!]\varphi \supset \mathscr{E} [\![ q ]\!]\varphi \\ \mathscr{E} [\![ p \equiv q ]\!] \varphi = \mathscr{E} [\![ p ]\!]\varphi \equiv \mathscr{E} [\![ q ]\!]\varphi \end{array}$
(10.1.2) It’s just:

$\mathsf{Tree} = \mathbb{N} \cup (\mathsf{Tree} \times \mathsf{Tree})$