Homework/HW4.md

Benjamin's Solution

Looks like Ben opted to literally translate the Prog nonterminal into Haskell. This is perfectly valid: if anything, it's a more faithful translation of the syntax. However, as you probably know, the definition is isomorphic to a list:

data Prog = Empty | Stmt Stmt Prog
data List Stmt = Nil | Cons Stmt (List Stmt)

So, you could've defined Prog to be [Stmt]. There are a few small advantages to this, mostly in terms of using Haskell as a metalanguage. Here's a line from my interpreter for this language:

stmt (Loop p) = prog (cycle p) `catchError` (const $ return ())

In order to implement the looping behavior of Loop, I used Haskell's built-in cycle :: [a] -> [a] function. This function infinitely repeats a list:

cycle [1,2,3] = [1,2,3,1,2,3,1,2,3,...]

The more compact list notation is another advantage to this approach.

I appreciated the following comment in the code:

- Starting register A at x, set R to zero, and add to R until A passes y (my approach above)
- Same as above, but use y-x as the boundary instead of y, and adjust accordingly
- Don't use a do loop, and instead hardcode out y-x additions of x+1 each time

What about a closed form approach? Something like (b-a+1)(b+a)/2. If * and / were O(1) operations, this would make the whole summation O(1).

Owen's Solution

Looks like Owen had some trouble with this homework assignment. From what I can tell, the code should not compile; there are a few things here that can be improved.

Allowing Expr to be Int

I'm looking at this line of code:

-- since there is already an int type in Haskell, I think I can consider it already encoded
data Expr = Int

Unfortunately, this isn't quite right. This creates a constructor (value!) Int of type Expr. This value Int has nothing to do with the type Int (integers). Thus, Expr can't contain integers like you intended. The proper syntax would be something like:

data Expr = Lit Int 

Unlike the previous version, this will create a constructor called Lit with type Int -> Expr. This time, the Int refers to the type Int, and so, Lit 3 will represent an expression that contains the number 3.

Defining Reg

Haskell's type is a way to create a type alias. So, if you have a type you use a lot (perhaps Maybe String), you can give it another name:

type MyType = Maybe String

Then, Maybe String and MyType mean the same thing. For defining Reg, you want to use data:

data Reg = A | B | C

Using undefined or _holes

It's hard to get the whole program compiling in one go. Haskell supports placeholders, which can be used anywhere, and will compile (mostly) no matter what. Of course, they'll either crash at runtime (undefined) or not really compile (_holes). However, if you have something like:

while = -- can't quite do this yet

sumFromTo = ... -- but now there's a parse error!

You can use undefined:

while = undefined

This way, there's no parse error, and GHC goes on to check the rest of your program. Holes are useful when you're in the middle of a complicated expression:

foldr _help 1 [1,2,3]

This will almost compile, except that at the very end, Haskell will tell you the type of the expression you need to fill in for _help. It will also suggest what you can put for _help!

Parenthesization

This is a small nitpick. Instead of writing RegStore A (x), you would write RegStore A x. In this case, you probably wanted RegStore A (Lit x) (you do need parentheses here!).

Making Type Errors Impossible

I'm looking at this part of Part 2:

data Stmt = RegStore Reg IntExpr
	| RegStore Reg BoolExpr
	| Contitional IntExpr Prog Prog
	| Contitional BoolExpr Prog Prog
	| Loop Prog
	| BreakLoop

You were on the right track!

I think this should make it so that I have the full functionality of the previous setting but now I have one version for each type of expr?

You're exactly right. You can represent any program from Part 1 using this syntax. However, the point was: you don't need to! Programs in the form Conditional IntExpr Prog Prog are specifcally broken! You can't use a number as a condition for if/then/else, and IntExpr is guaranteed to produce a number! The same is true for RegStore Reg BoolExpr: you aren't allowed to store booleans into registers! See this part of the homework description:

Note that although the expression language may produce both booleans and integers, the registers may only contain integers.

So, it is in fact sufficient to leave out the duplicate constructors you have:

data Stmt = RegStore Reg IntExpr
	| Contitional BoolExpr Prog Prog
	| Loop Prog
	| BreakLoop

Hope all this helps!

Jaspal's Solution

Notably different in Jaspal's solution from the others in this group is that they used strings for registers instead of a data type definition. I think this was allowed by this homework assignment; however, I think that using a data type is better for our purposes.

The issue is, if your register is a string, it can be any string. You can have register "A", register "a", register "Hello World", and so on. But our language only has three registers: A, B, and R. If we define a data type as follows:

data Reg = A | B | R

Then a value of type Reg can only be either A, B, or R. This means that we can't represent programs like

Hello := 5

Which is good!

This solution also uses variables to encode programs. I think that's a nice touch! It makes some expressions less bulky than they otherwise would be.

It doesn't look like the code compiles; here are a few things you can do to bring it closer to a working state.

Type Names and Constructor Names

Type names (Expr, Int, Bool) all have to start with an uppercase letter. Thus, you should change the following line:

data expr

To the following:

data Expr

And so on.

The exception to this rule is type variables for when things are polymorphic. For instance, we have

data Maybe a = Nothing | Just a

Here, a is lowercase, but tht's because a is actually a placeholder for an arbitrary type. You can tell because it also occurs on the left side of the = : data Maybe a.

Constructor names also have to start with an uppercase letter. So, you would have:

data Stmt
  = Eql Reg Expr 
  | IfElse Expr Prog Prog 
  | Do Prog 
  | Break 

String Notation

I have made a case that you shouldn't be using strings for this homework. However, you'll probably use them in the future. So, I want to point out: 'A' does not mean the string "A". It means a single character, like 'A' in C. To create a String in Haskell, you want to use double quotes: "A".

Other than those two things, it looks like this should work!

My Solution

My solution to Part 1 is pretty much identical to everyone else's; I opted to use Prog = [Stmt] instead of defining a data type, but other than that, not much is different.