# Assignment 12: Introduction to Monads

Our biggest mistake: using the scary term “monad” rather than “warm fuzzy thing”.

## Assignment

Please complete the problems below. When you are finished, place all of your code in a file named a12.rkt and submit it to Oncourse.

There are an abundance of lecture notes you could use for this assignment.

## Requirements

• You should use the monads.rkt file. You should add it to your C311 directory. This contains all the monad definitions you should need.
• You should also pull down and use the a12-student-tests.rkt test file.
• Do not use set! or any another Racket procedures that perform side effects.
• Unless otherwise specified, you may use bind* to write any program below.

Recall the definition of the maybe monad presented in lecture.

1. The Racket function findf takes a list and a predicate against which to test the list's elements. In our implementation, we will return either the leftmost list element to pass the predicate or (Nothing) if none of the list's elements pass.

> (findf-maybe symbol? '(1 2 c))
(Just c)
> (findf-maybe boolean? '(#f 1 2 c))
(Just #f)
> (findf-maybe number? '(a b c))
(Nothing)

The writer monad provides a mechanism to write data separately from the actual return value. If we use a list to represent these writes, we can use this monad to implement some rather useful functions.

2. The function partition takes a list and a predicate, returning a dotted pair with the values that do not pass the predicate in the first position and the values that do in the second position. Implement this using the writer monad. Do not use bind* syntax for this problem.

> (partition-writer even? '(1 2 3 4 5 6 7 8 9 10))
((1 3 5 7 9) . (2 4 6 8 10))

> (partition-writer odd? '(1 2 3 4 5 6 7 8 9 10))
((2 4 6 8 10) . (1 3 5 7 9))

3. Exponentiation by squaring is a method for quickly raising numbers to integer powers. Here is the definition of power, a function that raises a base x to a power n using this algorithm:

(define power
(lambda (x n)
(cond
[(zero? n) 1]
[(zero? (sub1 n)) x]
[(odd? n) (* x (power x (sub1 n)))]
[(even? n) (let ((nhalf (/ n 2)))
(let ((y (power x nhalf)))
(* y y)))])))

Using the writer monad, implement the powerXpartials procedure, which also takes a base and an exponent. It should return the answer as a natural value, along with each partial result computed along the way.

> (powerXpartials 2 6)
(64 . (2 4 8))

> (powerXpartials 3 5)
(243 . (3 9 81))

> (powerXpartials 5 7)
(78125 . (5 25 125 15625))

Recall from lecture that the state monad uses a state and works with return-state and bind-state.

4. Given a symbol x and a binary tree of symbols tr (i.e. a tree with symbols at the leaves), via a preorder walk replace every occurrence of x with the number of xs that have been seen so far.

> ((replace-with-count 'o '(a o (t o (e o t ((n . m) . o) . f) . t) . r)) 0)
((a 0 (t 1 (e 2 t ((n . m) . 3) . f) . t) . r) . 4)

> ((replace-with-count 'o '(((h (i s . o) . a) o s o e . n) . m)) 0)
(((h (i s . 0) . a) 1 s 2 e . n) . 3)

> ((replace-with-count 'o '(o (h (o s . o) . o) . o)) 1)
((1 (h (2 s . 3) . 4) . 5) . 6)

One of the neat things about monadic code is that it can reveal the underlying structure in the code that uses them. This enables you to parameterize your code over the monad. You can then drop in a different monad and monadic operation, and get different behavior as a result. We'll do that here.

You'll use the following traverse in the next three problems.

> (define traverse
(lambda (return bind f)
(letrec
((trav
(lambda (tree)
(cond
[(pair? tree)
(do bind
(a <- (trav (car tree)))
(d <- (trav (cdr tree)))
(return (cons a d)))]
[else (f tree)]))))
trav)))

5. The reciprocal of a number n is computed by (/ 1 n). Note that 0 has no reciprocal. Implement reciprocal using the maybe monad, returning any value computed and (Nothing) when 0 is provided.

> (reciprocal 0)
(Nothing)

> (reciprocal 2)
(Just 1/2)

Using this, we can return a tree of reciprocals, and instead signal failure if the tree contains a 0.

> (define traverse-reciprocal
(traverse return-maybe bind-maybe reciprocal))

> (traverse-reciprocal '((1 . 2) . (3 . (4 . 5))))
(Just ((1 . 1/2) . (1/3 . (1/4 . 1/5))))

> (traverse-reciprocal '((1 . 2) . (0 . (4 . 5))))
(Nothing)

6. Halve. Implement the function halve that, given a number, either will return in the monad half the number, or, if the number is not divisible by two, will instead leave the original number in place, and also log that number (using the writer monad).

> (halve 6)
(3 . ())

> (halve 5)
(5 . (5))

Using this, we can return a tree in which the even numbers have been halved, the odds remain in place, and in which we've logged the odd numbers (which are not cleanly divisible by 2).

> (define traverse-halve
(traverse return-writer bind-writer halve))

> (traverse-halve '((1 . 2) . (3 . (4 . 5))))
(((1 . 1) . (3 . (2 . 5))) . (1 3 5))

7. State/sum. Implement a function state/sum which will, when given a number, return the current state as the value, and add that number to the current state.

> ((state/sum 5) 0)
(0 . 5)

> ((state/sum 2) 0)
(0 . 2)

> ((state/sum 2) 3)
(3 . 5)

Using this, we can return a tree consisting of partial sums of the elements, and in which the state contains the final sum of the tree.

> (define traverse-state/sum
(traverse return-state bind-state state/sum))

> ((traverse-state/sum '((1 . 2) . (3 . (4 . 5)))) 0)
((0 . 1) 3 6 . 10)
15

Take a look in the monads.rkt file for the definition of the continuation monad.

For more examples using the Cont monad, see pp. 16-18 of A Schemer's View of Monads.

The following interpreter is a direct style interpreter resembling what Dan wrote in class:

(define value-of
(lambda (expr env)
(match expr
[(? number?) expr]
[(? boolean?) expr]
[(? symbol?) (apply-env env expr)]
[`(* ,x1 ,x2) (* (value-of x1 env) (value-of x2 env))]
[`(sub1 ,x) (sub1 (value-of x env))]
[`(zero? ,x) (zero? (value-of x env))]
[`(if ,test ,conseq ,alt) (if (value-of test env)
(value-of conseq env)
(value-of alt env))]
[`(capture ,k-id ,body) (call/cc (lambda (k)
(value-of body (extend-env k-id k env))))]
[`(return ,k-exp ,v-exp) ((value-of k-exp env) (value-of v-exp env))]
[`(lambda (,id) ,body) (closure id body env)]
[`(,rator ,rand) (apply-proc (value-of rator env) (value-of rand env))])))

Use the Cont monad to create a monadic value-of, and call it value-of-cps. Provide your own empty-env, apply-env, extend-env, closure, and apply-proc in representations of your choice. Most of the same helpers should work for both value-of and value-of-cps. Here are some tests your interpreter should pass:

> (define fact-5
'((lambda (f)
((f f) 5))
(lambda (f)
(lambda (n)
(if (zero? n)
1
(* n ((f f) (sub1 n))))))))

> (apply-k (lambda (v) v) (value-of-cps fact-5 (empty-env)))
120

> (define capture-fun
'(* 3 (capture q (* 2 (return q 4)))))

> (apply-k (lambda (v) v) (value-of-cps capture-fun (empty-env)))
12

### Just Dessert

Here is the Reverse State monad. Do something interesting with it. You might have to use Lazy Racket. Since it might be in a different language, you should leave your answer commented out at the bottom of your file. Of course, if you want to go forward and backward through time, you'd use a Tardis.