PS3: FirstClass Fun
 Due: 5pm Saturday 20 February. (But you should try to complete as much of it as possible before 5pm Fri 19 Februrary, while help is easily available.)
 Notes:
 This is a long assignment. Start soon!
 You should be able to do Problems 1 and 2 now.
 Problem 2 involves reading a paper that will be somewhat challenging for many students. Start reading the paper right away.
 You might be able to do some of Problems 3, 4, and 5 now, but will be able to do them all after the Tue. Feb 16 class.
 The problems needn’t be done in order. Feel free to jump around.
 The problem set is worth 125 points.
 We’re asking you to keep track of the time estimates for each problem/subproblem.
 This is a long assignment. Start soon!
 Submission:
 In the yourFullName CS251 Spring 2016 Folder that you created for PS1, create a Google Doc named yourFullName CS251 PS3.
 For each problem and subproblem, please indicate at the beginning of each writeup approximately how long that problem took you to solve and write up.
 For Problems 1 and 2, include all answers in your PS3 google doc. Please format your evaluation derivations in Problem 1 so that they’re easy to read. Format the derivation using the fixedwidth Courier New font (can use a small font size if that helps).
 For Problems 3 through 5:
 Copy the contents of
yourAccountNameps3functions.rkt
to the Google Doc. Format the definitions using the fixedwidth Courier New font (can use a small font size if that helps).  Drop a copy of your
yourAccountNameps3functions.rkt
in your~/cs251/drop/ps03
drop folder on cs.wellesley.edu.
 Copy the contents of
1. Wacky Lists (20 points)
This problem shows that functions can be used to implement data structures like pairs and lists. Consider the following alternatives to Racket’s usual builtin cons
, car
, cdr
, null
, and null?
:
(define kons (λ (x y) (λ (f) (f #f x y))))
(define kar (λ (k) (k (λ (b f s) f))))
(define kdr (λ (k) (k (λ (b f s) s))))
(define knull (λ (f) (f #t 0 0)))
(define knull? (λ (k) (k (λ (b f s) b))))

(2 points) Use the smallstep substitution model (i.e., using ⇒) to show the evaluation of the value bound to the name
p
by the following declaration:(define p (kons 3 4))
 (12 points) Use the smallstep substitution model (i.e., using ⇒) to show the evalulation of each of the following expressions.
(kar p)
(kdr p)
(knull? p)
(knull? knull)

(3 points) The following
sumto
function uses helper functionssum
anddownfrom
defined in terms of the listlike entities involvingkons
and friends. Does it actually calculate the sum of the integers from 1 ton
(inclusive)? Explain why or why not.(define (sumto n) (sum (downfrom n))) (define (sum nums) (if (knull? nums) 0 (+ (kar nums) (sum (kdr nums))))) ; sum was accidentally ksum before and has been corrected (define (downfrom n) (if (<= n 0) knull (kons n (downfrom ( n 1)))))
 (3 points) Can we replace all instances of
cons
/car
/cdr
/null
/null?
in Racket bykons
/kar
/kdr
/knull
/knull?
? Are there any ways in whichkons
/kar
/kdr
/knull
/knull?
do not behave likecons
/car
/cdr
/null
/null?
2. Backus’s Paper (30 points)
This problem is about John Backus’s 1977 Turing Award Lecture: Can Programming be Liberated from the von Neumann Style? A Functional Style and its Algebra of Programs. His paper can be found here.
You should begin this problem by reading Sections 1–11 and 15–16 of this paper. (Although Sections 12–14 are very interesting, they require more time than I want you to spend on this problem.)
Section 11.2 introduces the details of the FP language. Backus uses many notations that may be unfamiliar to you. For example:

p_{1} → e_{1}; … ; p_{n} → e_{n}; e_{n+1} is similar to the Racket expression
(if
p_{1}e_{1}
…
(if
p_{n}e_{n}
e_{n+1}
)
…)
. 
⟨e_{1}, …, e_{n}⟩ denotes the sequence of the n values of the expressions e_{1}, … e_{n}. φ denotes the empty sequence. Because FP is dynamically typed, such sequences can represent both tuples and lists from Python and OCaml.

The symbol ⊥ (pronounced “bottom”) denotes the value of an expression that doesn’t terminate (i.e., it loops infinitely) or terminates with an error.

If f is a function and x is an object (atom or sequence of objects), then f : x denotes the result of applying f to x.

[f_{1}, …, f_{n}] is a functional form denoting a sequence of n functions, f_{1} through f_{n}. The application rule for this functional form is [f_{1}, …, f_{n}] : x = ⟨f_{1} : x, … , f_{n} : x⟩ — i.e., the result of applying a sequence of n functions to an object x is an nelement sequence consisting of the results of applying each of the functions in the function sequence to x.
Consult Lyn if you have trouble understanding Backus’s notation.
Answer the following questions about Backus’s paper. Your answers should be concise but informative.

(2 points) One of the reasons this paper is wellknown is that in it Backus coined the term “von Neumann bottleneck”. Describe what this is and its relevance to the paper.

(2 points) Many programming languages have at least two syntactic categories: expressions and statements. Backus claims that expressions are good but statements are bad. Explain his claim.

(3 points) In Sections 6, 7, and 9 of the paper, Backus discusses three problems/defects with von Neumann languages. Summarize them.

(3 points) What are applicative languages and how do they address the three problems/defects mentioned by Backus for von Neumann languages?

(2 points) The FP language Backus introduces in Section 11 does not support abstraction expressions like Racket’s
lambda
. Why did Backus make this decision in FP? 
(6 points) Backus wrote this paper long before the development of Java and Python. Based on his paper, how do you think he would evaluate these two languages?

(12 points) Consider the following FP definition:
Def F ≡ α/+ ◦ αα× ◦ αdistl ◦ distr ◦ [id, id]
What is the value of F⟨2, 3, 5⟩? Show the evaluation of this expression in algebralike steps.
3. Higherorder List Functions (43 points)
In this and the following problems you will revisit some functions from PS2 Problem 4, as well as some new ones. However, rather than expressing them as recursions, you will express them in terms of higherorderlistoperators.
Notes:

For problems 3 through 5, you should use Dr. Racket to create a single file named
yourAccountNameps3functions.rkt
that contains all the functions (including helper functions) that you define for this problem. 
In your definitions, you are not allowed to use recursion anywhere. (The one exception is the
insertsrec
helper function you are given in Problem 3l.) 
In your definitions, unless otherwise instructed, you should not introduce any new named helper functions, but you can (1) liberally use anonymous functions and (2) use functions you defined in previous parts in later parts.

(3 points) Using Racket’s
map
, define a functionmapremainder
that takes two arguments (an integerdivisor
and a listints
of integers) and returns an integer list the same length asints
in which every element is remainder of dividing the corresponding element ofints
bydivisor
.> (mapremainder 2 (list 16 23 42 57 64 100)) '(0 1 0 1 0 0) > (mapremainder 3 (list 16 23 42 57 64 100)) '(1 2 0 0 1 1) > (mapremainder 5 (list 16 23 42 57 64 100)) '(1 3 2 2 4 0) > (mapremainder 17 (list 16 23 42 57 64 100)) '(16 6 8 6 13 15)

(3 points) Using Racket’s
filter
, define a functionfilterdivisibleby
that takes two arguments (an integerdivisor
and a listints
of integers) and returns a new integer list containing all the elements ofints
that are divisible bydivisor
. Usedivisibleby?
from PS2 Problem 4 to determine divisibility.> (filterdivisibleby 2 (list 16 23 42 57 64 100)) '(16 42 64 100) > (filterdivisibleby 3 (list 16 23 42 57 64 100)) '(42 57) > (filterdivisibleby 4 (list 16 23 42 57 64 100)) '(16 64 100) > (filterdivisibleby 5 (list 16 23 42 57 64 100)) '(100) > (filterdivisibleby 17 (list 16 23 42 57 64 100)) '()

(3 points) Using Racket’s
foldr
, define a functioncontainsmultiple?
that takes an integerm
and a list of integersns
that returns#t
ifm
evenly divides at least one element of the integer listns
; otherwise it returns#f
. Usedivisibleby?
from above to determine divisibility.> (containsmultiple? 5 (list 8 10 14)) #t > (containsmultiple? 3 (list 8 10 14)) #f > (containsmultiple? 5 null) #f

(3 points) Using Racket’s
foldr
, define a functionallcontainmultiple?
that takes an integern
and a list of lists of integersnss
(pronounced “enziz”) and returns#t
if each list of integers innss
contains at least one integer that is a multiple ofn
; otherwise it returns#f
. Usecontainsmultiple?
in your definition ofallcontainmultiple?
.> (allcontainmultiple? 5 (list (list 17 10 2) (list 25) (list 3 8 5))) #t > (allcontainmultiple? 2 (list (list 17 10 2) (list 25) (list 3 8 5))) #f > (allcontainmultiple? 3 null) #t ; said to be "vacuously true"; there is no counterexample!

(3 points) Using Racket’s
foldr
, define a functionsnoc
that takes listys
and a valuex
, and returns the new list that results from addingx
to the end ofys
.> (snoc (list 7 2 5) 4) '(7 2 5 4) > (snoc (list) 4) '(4)

(3 points) Using Racket’s
foldr
, define a functionmyappend
that takes two lists,xs
andys
, and returns the new list that contains all the elements ofxs
followed by all the elements ofys
.> (myappend (list 7 2 5) (list 4 6)) '(7 2 5 4 6) > (myappend (list) (list 4 6)) '(4 6) > (myappend (list 7 2 5) (list)) '(7 2 5) > (myappend (list) (list)) '()
Note: You may not use Racket’s
append
in your definition. 
(3 points) Using Racket’s
foldr
, define a functionappendall
that takes a list of listsxss
and returns a new list that contains all the elements of the sublists ofxss
in their relative order.> (appendall (list (list 1 2) (list 3) (list 4 5 6))) '(1 2 3 4 5 6) > (appendall (list (list 1 2) (list 3))) '(1 2 3) > (appendall (list (list 1 2))) '(1 2) > (appendall (list)) '() > (appendall (list (list (list 1 2) (list 3 4 5)) (list (list 6)) (list (list 7 8) (list) (list 9)))) '((1 2) (3 4 5) (6) (7 8) () (9))
Note: You may use
append
ormyappend
in your definition. 
(3 points) Using Racket’s
map
, define a functionmapcons
that takes any valuex
and an nelement listys
and returns an nelement list of all pairs'(x . y)
wherey
ranges over the elements ofys
. The pair'(x . y)
should have the same relative position in the resulting list asy
has inys
.> (mapcons 17 (list 8 5 42 23)) '((17 . 8) (17 . 5) (17 . 42) (17 . 23)) > (mapcons 3 (list (list 1 6 2) (list 4 5) (list) (list 9 6 8 7))) '((3 1 6 2) (3 4 5) (3) (3 9 6 8 7)) > (mapcons 42 null) '()

(5 points) Using Racket’s
foldr
, define a functionmycartesianproduct
that takes two listsxs
andys
and returns a list of all pairs'(x . y)
wherex
ranges over the elements ofxs
andy
ranges over the elements ofys
. The pairs should be sorted first by thex
entry (relative to the order inxs
) and then by they
entry (relative to the order inys
).> (mycartesianproduct (list 1 2) (list "a" "b" "c")) ; yes, Racket has string values '((1 . "a") (1 . "b") (1 . "c") (2 . "a") (2 . "b") (2 . "c")) > (mycartesianproduct (list 2 1) (list "a" "b" "c")) '((2 . "a") (2 . "b") (2 . "c") (1 . "a") (1 . "b") (1 . "c")) > (mycartesianproduct (list "c" "b" "a") (list 2 1)) '(("c" . 2) ("c" . 1) ("b" . 2) ("b" . 1) ("a" . 2) ("a" . 1)) > (mycartesianproduct (list "a" "b") (list 2 1)) '(("a" . 2) ("a" . 1) ("b" . 2) ("b" . 1)) > (mycartesianproduct (list 1) (list "a")) '((1 . "a")) > (mycartesianproduct null (list "a" "b" "c")) '()
Note: You may use
mapcons
andappend
ormyappend
in your definition. 
(4 points) Using Racket’s
foldr
, define a functionmyreverse
that takes a listxs
and returns a new list whose elements are the elements ofxs
in reverse order. You may not use the builtinreverse
function.> (myreverse (list 1 2 3 4)) '(4 3 2 1) > (myreverse (list 1)) '(1) > (myreverse (list)) '()
Note:
 We ask you to name your function
myreverse
because Racket already provides the same function namedreverse
(which you cannot use, of course).  You may use
snoc
orappend
ormyappend
in your definition.
 We ask you to name your function

(5 points) Assume that the elements of a list are indexed starting with 0. Using Racket’s
foldr
, define a functionalts
that takes a listxs
and returns a twoelement list of of lists, the first of which has all the evenindexed elements (in the same relative order as inxs
) and the second of which has all the evenindexed elements (in the same relative order as inxs
).> (alts (list 7 5 4 6 9 2 8 3)) '((7 4 9 8) (5 6 2 3)) > (alts (list 5 4 6 9 2 8 3)) '((5 6 2 3) (4 9 8)) > (alts (list 4 6 9 2 8 3)) '((4 9 8) (6 2 3)) > (alts (list 3)) '((3) ()) > (alts null) '(() ())
Note: There is no need to treat evenlength and oddlength cases differently, nor is there any need to treat the singleton list specially.

(5 points) Assume you are supplied with the following recursive version of the
inserts
function from PS2 Problem 4:(define (insertsrec x ys) (if (null? ys) (list (list x)) (cons (cons x ys) (mapcons (car ys) (insertsrec x (cdr ys))))))
Using Racket’s
foldr
, define a functionmypermutations
that takes as its single argument a listxs
of distinct elements (i.e., no duplicates) and returns a list of all the permutations of the elements ofxs
. The order of the permutations does not matter.> (mypermutations null) '(()) > (mypermutations (list 4)) '((4)) > (mypermutations (list 3 4)) '((3 4) (4 3)) ; order doesn't matter > (mypermutations (list 2 3 4)) '((2 3 4) (3 2 4) (3 4 2) (2 4 3) (4 2 3) (4 3 2)) > (mypermutations (list 1 2 3 4)) '((1 2 3 4) (2 1 3 4) (2 3 1 4) (2 3 4 1) (1 3 2 4) (3 1 2 4) (3 2 1 4) (3 2 4 1) (1 3 4 2) (3 1 4 2) (3 4 1 2) (3 4 2 1) (1 2 4 3) (2 1 4 3) (2 4 1 3) (2 4 3 1) (1 4 2 3) (4 1 2 3) (4 2 1 3) (4 2 3 1) (1 4 3 2) (4 1 3 2) (4 3 1 2) (4 3 2 1))
Note: It is helpful to use
appendall
,map
, andinsertsrec
in your solution.
4. forall?
, exists?
, find
, and zip
(20 points)
Here are some listprocessing functions that are not built in to Racket, but are handy in many situations:
(define (forall? pred xs)
(if (null? xs)
#t
(and (pred (car xs))
(forall? pred (cdr xs)))))
(define (exists? pred xs)
(if (null? xs)
#f
(or (pred (car xs))
(exists? pred (cdr xs)))))
(define (find pred notfound xs)
(if (null? xs)
notfound
(if (pred (car xs))
(car xs)
(find pred notfound (cdr xs)))))
(define (zip xs ys)
(if (or (null? xs) (null? ys))
null
(cons (cons (car xs) (car ys))
(zip (cdr xs) (cdr ys)))))
forall?
, exists?
, and find
are higherorder list functions involving a predicate.
forall?
returns#t
if the predicate is true on all elements of the list, and otherwise returns#f
.exists?
returns#t
if the predicate is true on at least one element of the list, and otherwise returns#f
.
find
returns the first element of the list for which the predicate is true. If there is no such element, it returns the value supplied as thenotfound
argument.> (forall? (λ (x) (> x 0)) (list 7 2 5 4 6)) #t > (forall? (λ (x) (> x 0)) (list 7 2 5 4 6)) #f > (exists? (λ (x) (< x 0)) (list 7 2 5 4 6)) #t > (exists? (λ (x) (< x 0)) (list 7 2 5 4 6)) #f > (find (λ (x) (< x 0)) #f (list 7 2 5 4 6)) 5 > (find (λ (x) (< x 0)) #f (list 7 2 5 4 6)) #f
The zip
function is not higher order, but combines two lists by pairing (using cons
) the corresponding elements of the two lists. If the lists do not have the same length, zip
returns a list of pairs whose length is the length of the shorter of the two input lists:
> (zip (list 1 2 3) (list "a" "b" "c"))
'((1 . "a") (2 . "b") (3 . "c"))
> (zip (list 1 2 3 4 5) (list "a" "b" "c"))
'((1 . "a") (2 . "b") (3 . "c"))
> (zip (list 1 2 3) (list "a" "b" "c" "d" "e"))
'((1 . "a") (2 . "b") (3 . "c"))

(3 points) Using
exists?
, define a functionmember?
that determines if an elementx
appears in a listys
.> (member? 4 (list 7 2 5 4 6)) #t > (member? 3 (list 7 2 5 4 6)) #f > (member? (list 7 8) (list (list 1 2) (list 3 4 5) (list 6) (list 7 8) (list) (list 9))) #t > (member? (list) (list (list 1 2) (list 3 4 5) (list 6) (list 7 8) (list) (list 9))) #t > (member? (list 5 6) (list (list 1 2) (list 3 4 5) (list 6) (list 7 8) (list) (list 9))) #f
Note: Use
equal?
to compare the equality of two values. 
(5 points) Using
forall?
andexists?
, define a functionallcontainmultiplealt?
that is an alternative implementation of theallcontainmultiple?
function from Problem 3.> (allcontainmultiplealt? 5 (list (list 17 10 2) (list 25) (list 3 8 5))) #t > (allcontainmultiplealt? 2 (list (list 17 10 2) (list 25) (list 3 8 5))) #f > (allcontainmultiplealt? 3 null) #t ; said to be "vacuously true"; there is no counterexample!
Note: You may use the
divisible_by
function from above, but not thecontainsmultiple?
function, and you may not define any new helper functions. 
(4 points) An association list is a list of pairs that represents a mapping from key to value. Each pair of key and value is represented by a cons cell, with the key in the
car
and the value in thecdr
. For example, the association list:(list (cons 2 3) (cons 5 1) (cons "mountain" #t))
maps the key
2
to the value3
, the key5
to the value1
, and the key"mountain"
to the value#t
.Using
find
, define a functionlookup
that takes a keyk
and an association listas
and returns:#f
if no mapping with keyk
was not found in the list; and a cons cell whose
car
isk
and whosecdr
is the corresponding value for the shallowest mapping ofk
in the association list.
For example:
> (lookup 1 (list (cons 2 3) (cons 5 1) (cons "mountain" #t))) #f > (lookup 5 (list (cons 2 3) (cons 5 1) (cons "mountain" #t))) '(5 . 1) > (lookup 5 (list (cons 2 3) (cons 5 1) (cons 5 "river"))) '(5 . 1) > (lookup (list 3 5) (list (cons (list 3 5) 2) (cons 5 1))) '((3 5) . 2)
Note: Use
equal?
to test for equality of keys. This will support keys more interesting than just simple values. 
(5 points) Using
forall?
andzip
, define a functionsorted?
that determines if a list of numbersns
is in sorted order from low to high.> (sorted? (list 7 4 2 5 4 6)) #f > (sorted? (list 2 3 3 5 6 7)) #t > (sorted? (list 2)) #t > (sorted? (list)) #t
Note: You will need to have a special case for the empty list.

(3 points) It is possible to define alternative versions of
forall?
andexists?
in terms offoldr
, as show below.(define (forallalt? pred xs) (foldr (λ (x subres) (and (pred x) subres)) #t xs)) (define (existsalt? pred xs) (foldr (λ (x subres) (or (pred x) subres)) #f xs)) > (forallalt? (λ (x) (> x 0)) (list 7 2 5 4 6)) #t > (forallalt? (λ (x) (> x 0)) (list 7 2 5 4 6)) #f > (existsalt? (λ (x) (< x 0)) (list 7 2 5 4 6)) #t > (existsalt? (λ (x) (< x 0)) (list 7 2 5 4 6))
However, just because it’s possible to define a function in terms of
foldr
does not mean its a good idea. Give a concrete example of a situation in whichforall?
is better thanforallalt?
.Note: For this problem, it’s critical to understand that
(and e1 e2)
desugars to(if e1 e2 #f)
5. foldrternop
(12 points)
Sometimes it is difficult to express a recursive list accumulation in terms of foldr
because the binary combiner function needs more information from the list than its first element. The following foldrternop
higherorder list function solves this problem by having the combiner function be a ternary (i.e., threeargument) function that takes both the first and rest of the given list in addition to the result of recursively processing the list:
(define (foldrternop ternop nullvalue xs)
(if (null? xs)
nullvalue
(ternop (first xs)
(rest xs)
(foldrternop ternop nullvalue (rest xs)))))
In this problem, you will use foldrternop
to implement two list functions that are very challenging to implement in terms of foldr
(see the extra credit problem below).

(6 points) Using
foldrternop
, define a functioninserts
that takes a valuex
and an nelement listys
and returns an n+1element list of lists showing all ways to insert a single copy ofx
intoys
.> (inserts 3 (list 5 7 1)) '((3 5 7 1) (5 3 7 1) (5 7 3 1) (5 7 1 3)) > (inserts 3 (list 7 1)) '((3 7 1) (7 3 1) ( 7 1 3)) > (inserts 3 (list 1)) '((3 1) (1 3)) > (inserts 3 null) '((3)) > (inserts 3 (list 5 3 1)) '((3 5 3 1) (5 3 3 1) (5 3 3 1) (5 3 1 3))
Notes:

Your definition should have exactly this pattern:
(define (insertsfoldr x ys) (foldrternop {ternarycombiner} {nullvalue} ys))

You may use
mapcons
in your ternarycombiner function.


(6 points) Using
foldrternop
, define a functionsortedalt?
that is an alternative implementation of thesorted?
function from Problem 4.> (sortedalt? (list 7 4 2 5 4 6)) #f > (sortedalt? (list 2 3 3 5 6 7)) #t > (sortedalt? (list 2)) #t > (sortedalt? (list)) #t
Note:

Your definition should have exactly this pattern:
(define (sortedalt? xs) (foldrternop {ternarycombiner} {nullvalue} xs))

Extra Credit: Using foldr
to define inserts
and sorted?
(20 points)
This problem is optional. You should only attempt it after completing all the other problems.
As noted in Problem 5, it is challenging to define inserts
and sorted
in terms of foldr
, but it turns out that it is stil possible to do this.

(8 points) Using
foldr
, define a functioninsertsfoldr
that is an alternative implementation of theinserts
function from Probem 5.> (insertsfoldr 3 (list 5 7 1)) '((3 5 7 1) (5 3 7 1) (5 7 3 1) (5 7 1 3)) > (insertsfoldr 3 (list 7 1)) '((3 7 1) (7 3 1) ( 7 1 3)) > (insertsfoldr 3 (list 1)) '((3 1) (1 3)) > (insertsfoldr 3 null) '((3)) > (insertsfoldr 3 (list 5 3 1)) '((3 5 3 1) (5 3 3 1) (5 3 3 1) (5 3 1 3))
Note:

Your definition should have exactly this pattern:
(define (insertsfoldr x ys) (foldr {binarycombiner} {nullvalue} ys))

You may use
mapcons
in your binarycombiner function.


(12 points) Using
foldr
, define a functionsortedfoldr?
that is an alternative implementation of thesorted?
function from Problem 5.> (sortedfoldr? (list 7 4 2 5 4 6)) #f > (sortedfoldr? (list 2 3 3 5 6 7)) #t > (sortedfoldr? (list 2)) #t > (sortedfoldr? (list)) #t
Note:

Your definition should have exactly this pattern:
(define (insertsfoldr x ys) (cdr (foldr {binarycombiner} (cons {nullvalue1} {nullvalue2}) ys)))
The idea is to accumulate a pair of (1) the first element of the rest of the list (or
#f
if there is none) and (2) a boolean indicating whether the rest of the list is sorted.
