Although CS 60 will introduce graphics in the coming weeks, this week's "graphical" programming refers to mathematical graphs, i.e., connection networks of nodes and edges that might represent human relationships, physical links, or hierarchical structures. Trees and even lists are special types of graphs: trees are graphs for which each node has at most 1 parent node, but any number of children nodes. Lists are graphs in which each node has at most 1 parent node and at most 1 child node, i.e., a purely linear structure.
All of the problems this week may be done in a pair, in you wish. Certainly it's OK to work on them individually, too. If you do work in a pair, make sure you follow the pair-programming practices as noted on the CS 60 syllabus.
Whether individual or pair, place all of your functions for this week in a file named hw2pr1.rkt. There is a (small) starter file by that name linked below.
On average, each problem this week is worth 10 points The earlier ones are a bit less; the later ones, a bit more... .
Just a reminder that these are important! 25-30% of the assignment's points will be based on these. A few details:
Be sure to submit your hw2pr1.rkt file to the submission site!
Start with the hw2pr1.rkt starter file -- it has placeholders for the functions you will need.
Some of these problems have a component of open-ended algorithm design, as well as reinforcement of this week's themes. Feel free to seek out help in discussions with other students, the CS 60 tutors, or to chat with Zach or Colleen any time about any of these problems... .
Using higher-order functions
For the first three functions,
superreverse, indivisible, and
lotto-winner, see if you can use higher-order
functions (filter, map, foldr,
and sort, along with lambda) instead of using
recursion explicitly, i.e., instead of calling
superreverse from within superreverse's
definition, and so on... . This isn't a requirement,
but it will help familiarize the ideas of those HOFs (and
will be more concise, codewise).
As with the previous homeworks, be sure to devise and include not only the provided tests in your submitted hw2pr1.rkt file, but at least one more test for each of the homework functions. Writing the tests will ensure you're confident of what each function should do and asks you to consider the space of possibilities for the function's behavior. This case analysis is a crucial skill in software design and development -- some would say the crucial skill... .
In addition, be sure to include a comment just above
the test you create in order to indicate it. For example,
Feel free to use this test for indivisible, below!
;; This is an additional test I've included:
(check-expect (indivisible 1 (range 1 10)) '())
This problem asks you to write two functions. First, implement (superreverse
L), whose functionality is identical to what it was in
hw1. Again, assume that the input L
will be a list that
contains zero or more lists (and only lists) as elements.
The output of (superreverse L) should be a list similar
to L, but with all of its top-level sublists reversed.
Two example tests:
Hint: the reverse function is built-in to Racket. Use it, along with one of
Racket's higher-order functions... .
(check-expect (superreverse '( (1 2 3) (4 5 6) (#\k #\o #\o #\l) (#\a #\m) ))
'( (3 2 1) (6 5 4) (#\l #\o #\o #\k) (#\m #\a) ) )
(check-expect (superreverse '( (1 2 3) (4 5 6 (7 8) 9 ) ))
'( (3 2 1) (9 (7 8) 6 5 4) ) )
Second, implement (indivisible k L), where k
is a positive
integer and L is a list of positive integers.
Then, indivisible should return a list consisting of all of the
elements of L that are not
evenly divisible by k. Those elements
should appear in the same order as they do in L.
For instance,
You might consider using both an anonymous function and
a higher-order function for indivisible.
(check-expect (indivisible 3 '( 2 3 4 5 6 7 8 9 10 )) '(2 4 5 7 8 10))
This problem is challenging! We have a starting point for this one, along with suggestions to help make progress... .
For this problem, write
in which the winning-numbers input will be a list
of distinct positive integers representing a set of
winning lottery numbers, e.g., '(2 3 15 32 42).
The list-of-tickets input will be a nonempty list of "lottery
tickets" that were purchased. Each "lottery ticket" will, in turn,
be a list whose first element is a symbol (the name of the person
who purchased the ticket) and whose rest is a list of the lottery
numbers they played. The number of lottery numbers will be equal to the length
of winning-numbers.
For example, here is a possible list-of-tickets:
(define (lotto-winner list-of-tickets winning-numbers)
Note: in class we looked at a helper function named
matches. However, there may have been a pair of
parens missing from that definition. Here it is for reference,
along with a couple of tests:
'( (alice 2 4 16 33 42) (bob 3 4 5 6 7) (cdrthecat 3 15 16 41 42) )
It's important that the inputs to a lambda-expression
be in a list, i.e., in parentheses; otherwise, Racket will
add list structure to the inputs!
(define (matches L W)
(length (filter (lambda (x) (member x W)) L)))
(check-expect (matches '(ace 2 3 4) '(3 42 2)) 2)
(check-expect (matches '(ace 2 3 4) '(8 4 5)) 1)
The lotto-winner function should output a list of two
things: first, the name (symbol) of the person who matched the
most numbers and, second, the number of numbers they matched.
For the above example data,
should pass, because 'cdrthecat matched three numbers (3, 15, 42), whereas
'bob matched only one number (3) and 'alice matched
two numbers (32, 42).
If there is more than one equally-good ticket, you may return either one -- but
do return only one!
(check-expect (lotto-winner
'( (alice 2 4 16 33 42) (bob 3 4 5 6 7) (cdrthecat 3 15 16 41 42) )
'(2 3 15 32 42))
'(cdrthecat 3))
Again, if this problem seems hard to start or is going awry, consider our lotto starting point and suggestions on how to proceed... .
The next few problems in this assignment ask you to implement functions that manipulate binary search trees (BSTs) in a variety of ways. You might recall from class that the representation of our binary search tree that we are using is either null? (the empty list) or a list of three elements: first, the root of the tree; second, the left-hand subtree; and third, the right-hand subtree. Also, all leaves within a left-hand subtree are strictly less than the value of the root. All elements within a right-hand subtree are strictly greater than the root. Finally, no value may be repeated in a BST. We will use binary search trees of only integers for the following three problems.
For example,
binds the label BT1 to the binary search tree shown. Its root
is 42 and the root of its left subtree is 20 (and so
on...).
(define BT1 '( 42
(20 (7 (1 () ()) (8 () ())) (31 () (41 () ())))
(100 (60 () ()) ()) ) )
Note: you're welcome to use raw recursion for these problems -- and we encourage you to do so. Higher-order functions are OK, too, but may not be as natural in this context.
Code from class This is the file of tree-based example functions from class - at the bottom are the find-max, find, and insert functions, good starting points for these problems.
... what are these tree-structure things? If you're feeling uncertain about trees, you might try Prof. Colleen Lewis's Racket-trees YouTube channel!
First, write a function (find-min BT)
whose input will always be a
non-empty binary search tree and whose output
is the value of the smallest node in that binary search tree.
For instance,
In a comment in your code, explain the big-O running time of
the height and find-min functions in the
best-case and the worst-case for binary
search trees with N nodes.
(check-expect (find-min BT1) 1) ;; using the tree defined above
Hint: Compare the find-max function we wrote in
class... .
Next, write a function (height BT)
whose input is a binary
search tree and whose output
is the length of the longest path from the root of BT
to any one of its bottom-level nodes, i.e.,
the height of the binary search tree.
The height of the empty binary search tree is 0.
For instance,
(check-expect (height BT1) 4) ;; using the tree defined above
(check-expect (height '()) 0)
Hint: max is built-in to Racket - it will help here!
Write a function (flatten-tree BT) whose input is any binary search tree and whose output is a list of all of the elements, in order, of the input. However, for this function you should not use the flatten function we wrote in class, because it would require a subsequent call to sort (and pay its penalty in big-O run time...). Instead, use the recursive structure of the tree. (Note that you can call flatten-tree recursively on the left and right subtrees. Where will the root go?)
In a comment
with your code, explain what the big-O running time of your flatten-tree
function is in the worst case.
For example,
(check-expect (flatten-tree BT1) '(1 7 8 20 31 41 42 60 100)) ;; using the tree defined above
Hint: You'll want to append three lists here: what three lists? Two of
them will involve calling flatten-tree recursively... .
The final binary search tree function this week
is to write a function
(delete value BT)
whose input
value is an integer and whose input
BT is a binary search tree. If value
does not appear in BT, then
a binary search tree identical to BT is returned.
On the other hand, if value
does appear in BT, then a tree similar to BT
is returned, except with
the node value deleted -- and other adjustments
made, as necesary, to ensure that the result is a valid
binary search tree. The next paragraph describes these
adjustments.
If the value to delete has zero children, it is
straightforward to delete.
Similarly, if it has only one non-empty child, it is replaced by that
child.
When the value to be deleted has two
non-empty children, however, it is not clear which of its children (or
descendants) are to take its place. For the
sake of this problem,
the node that should take value's place should be
the smallest value in BT
that is greater than value. (Use your
find-min function!)
Here are two examples:
(check-expect (delete 20 BT1)
'(42
(31 (7 (1 () ()) (8 () ())) (41 () ()))
(100 (60 () ()) ())))
(check-expect (delete 42 BT1)
'(60
(20 (7 (1 ()()) (8 ()())) (31 () (41 () ())))
(100 ()())))
Hint #1: As with the insert example, which is included at the bottom of this file of examples from Tuesday's class, this is challenging because you need to ensure your code always returns a complete binary search tree! If you're not deleting the right-hand branch, for example, you'll need to make sure it stays as the right-hand branch!
Hint #2: This one also has many cases. Here is only one
of those cases, when the element to be deleted is less than the
root:
Note here that the correct tree to return in this case has
exactly the same root (it's not what we're deleting) and
exactly the same right-hand subtree (it's not affected).
The left-hand subtree, however, may need to be changed,
and we're doing that (recursively)... .
Thanks to Emily for suggesting this!
(if (< value rt)
(list rt (delete value L) R)
;; otherwise, probably check some other cases...!
This problem boils down to handling all of the (many!) different cases in appropriate ways... .
These three problems involve either the use-it-or-lose-it approach to problem solving (discussed in the second lecture this week) or graphs of general structure -- or both!
Code from class: Three well-commented examples of UIOLI are at the top of the example functions from class - you might use these as a starting point... .
Aargh! What is this UIOLI approach? If you're feeling uncertain about use-it-or-lose-it, you might try Prof. Colleen Lewis's Racket-based UIOLI YouTube channel!
For this problem write a scheme function
whose first input total is a nonnegative integer and whose
second input coin-list is a (possibly empty) list of
coin values that you currently have. The make-change
function should output #t if some combination of
your coin values sum up to the desired total.
If the total is 0, then the result
should be #t (since you can always use no coins at all).
What should the result be if total is less than 0?
(define (make-change total coin-list)
If no combination of coin values sums up to total, then make-change should return #f. For this problem, you have one coin of each value that appears in the list coin-list: imagine a pocketful of coins of those values. Later on in the term we'll tackle the more general problem of having as many coins of each value as possible.
We won't limit coins to U.S. denominations: any nonnegative values
will be allowed. For example,
The use-it-or-lose-it approach will help here! For each coin in the list,
consider "asking" (recursively), whether the appropriate problem
can be solved with it and without it... .
(check-expect (make-change 4 '(2 2)) #t)
(check-expect (make-change 4 '(2)) #f)
(check-expect (make-change 0 '(1 4 6 15 25 29 54)) #t)
(check-expect (make-change 29 '(1 4 6 15 25 29 54)) #t)
(check-expect (make-change 11 '(1 4 6 15 25 29 54)) #t)
(check-expect (make-change 76 '(1 4 6 15 25 29 54)) #t)
(check-expect (make-change 9 '(1 4 6 15 25 29 54)) #f)
(check-expect (make-change 77 '(1 4 6 15 25 29 54)) #f)
Write a function
which takes in
(define (gchildren N n G)
;; Graph1
(define Graph1 '((a b) (b c) (c d) (d e) (c f) (e g) (e h) (f x) (x y) (y z) (z b)) )
(check-expect (gchildren 0 'a Graph1) '(a))
(check-expect (gchildren 1 'a Graph1) '(b))
(check-expect (gchildren 2 'a Graph1) '(c))
;; these tests use sortsym to avoid ambiguity in node order...
(define (sym<? s1 s2) (string<? (symbol->string s1) (symbol->string s2)))
(define (sortsym L) (sort L sym<?)) ;; will sort a list of symbols
(check-expect (sortsym (gchildren 1 'c Graph1)) '(d f))
(check-expect (sortsym (gchildren 2 'c Graph1)) '(e x))
(check-expect (sortsym (gchildren 3 'a Graph1)) '(d f))
(check-expect (sortsym (gchildren 3 'c Graph1)) '(g h y))
You will likely want to use the helper funciton
(kids n G) - a function that returns
the list of children of node n in graph G: it can
be used to do most of the work of gchildren.
This problem does not really fall into the use-it-or-lose-it pattern, since "it" is already provided (the input n). Rather, it is an extension of the gkids function we wrote in class -- here, to handle any number of generations of descendants!
Write two related functions -- the first is (min-dist a b G), which takes in two nodes a and b and a directed, positively-weighted graph G. Graph G will be represented as a list of edges: (src dst weight). See Gt, below, for an example.
Your min-dist should return the smallest distance that is required to travel from a to b through graph G. Every node should be considered 0 away from itself. However, if there is no path from a to b in G, for different nodes a and b, your function should return the value 42000000, which will be bigger than any feasible path in our graphs. (The problem with using the value +inf.0 is that it is floating-point, which changes check-expect's behavior.)
Just to emphasize - be sure to use 42000000 to represent the "infinite" distance between two nodes for which there is no connecting path.
For example,
To help sanity-check your tests, here is a picture of Gt:
(define Gt '((e b 100) (a b 25) (e a 42) (a c 7) (a d 13) (a e 15)
(b c 10) (b d 5) (d e 2) (b e 100) (c d 1) (c e 7)) )
(check-expect (min-dist 'a 'e Gt) 10)
(check-expect (min-dist 'e 'b Gt) 67)
(check-expect (min-dist 'd 'd Gt) 0)
(check-expect (min-dist 'f 'a Gt) 42000000)

The second function, an extension of the first, is min-path:
whose inputs are the same as min-dist's.
(define (min-path a b G)
The output of min-path should be a list whose
Note that because of this output format, you will want to cons the best distance onto the list of nodes in the best path. Keep this structure in mind (dist-cons-path) to avoid lots of extra-parens bugs!!!
Here are some examples using Gt from above:
in which Gt is the same graph used in testing min-dist.
(check-expect (min-path 'd 'd Gt) '(0 d))
(check-expect (min-path 'a 'z Gt) '(42000000)) ;; z's not there!
(check-expect (min-path 'a 'e Gt) '(10 a c d e))
(check-expect (min-path 'e 'b Gt) '(67 e a b))
Hint: It's helpful to start with the reachable (reach) code from class and consider how to change the return values appropriately from booleans to numeric quantities... !
What if there are two different paths, each of which is the same (minimal) distance? In this, case you can break ties arbitrarily in returning a single solution. We will, however, only test your code on unique minimal paths -- that way there will be only one answer to worry about... .
Head back to CodingBat to practice some java
I finished the java problems!
under hw2pr2.txt within the submission site. Again,
don't place that in the "partner-situation" textbox; rather,
click "edit" and place it in the assignment-submission textbox.
Be sure to submit hw2pr1.rkt to the submission site; if you've completed the Java problems, submit that note under hw2pr2.txt. If you're interested in more trees, Racket, and a very cool application (20 Q's), check out the optional extra-credit below!
This completely optional problem asks you to implement the game of 20 questions. It is worth up to +10 points.
Submission Please submit this extra credit as well-labeled functions at the bottom of your hw2pr1.rkt file... .
First, to give you a sense of the game of 20 questions,
here is an example of a run in Scheme. The user's input is in
blue
> (twenty-questions)
Is it bigger than a breadbox? (y/n) y
Is it a computer scientist? (y/n) n
What was it? a chemist
What's a question distinguishing a computer scientist and a chemist? Does it titrate?
And what's the answer for a chemist? (y/n) y
Play again? (y/n) y
Is it bigger than a breadbox? (y/n) y
Does it titrate? (y/n) y
Is it a chemist? (y/n) n
What was it? Wally Wart
What's a question distinguishing a chemist and Wally Wart? Is it concrete?
And what's the answer for Wally Wart? (y/n) y
Play again? (y/n) y
Is it bigger than a breadbox? (y/n) y
Does it titrate? (y/n) y
Is it concrete? (y/n) y
Is it Wally Wart? (y/n) y
I got it!
Play again? (y/n) n
Features
Your twenty questions should allow users to
Representation
For consistency, your game should always start with the
question Is it bigger than a
breadbox? and should have the objects
spam and a computer scientists
available as guesses if the user answers the initial
question with a n or a y,
respectively.
You are welcome to represent the game's data however you like. However as a guide to one possible
representation (the one used for the example program demoed in class), you might
consider a tree in which internal nodes are questions and leaf nodes are objects, e.g.,
Strings are likely the simplest data structure for questions and objects. The function
string-append, along with the other string functions noted on the Scheme reference card may be of use.
(define initialTree '("Is it bigger than a breadbox?" "spam" "a computer scientist"))
Possible approach to 20Q...
As with the data structure, the choice here is yours. As one possible suggestion
to get you started (if you'd like), consider the decomposition of the demo version,
which included
Input:
Here are two functions demonstrating how to use (read-line) to
grab a line of user input.
The first tests the resulting value (which will
be a string). The second is a more general-purpose question-asking function:
(define (askyn Q) ;; asks a yes/no question Q and returns #t or #f
(begin
(display Q)
(display " (y/n) ")
(if (equal? "y" (read-line))
#t
#f)
))
(define (ask Q)
(begin
(display Q)
(read-line)))
Hints: (feel free to disregard!)
The binary tree used in this game is slightly different than the numeric
ones used earlier in the HW. In this case, the leaves are not empty lists;
rather, they contain possible objects; internal nodes contain questions.
After the first round
in the example above, the tree's structure would be
Note that, by our convention, "no" traverses to the left and "yes" to the right.
("Is it bigger than a breadbox?" "spam" ("Do you titrate?" "a computer scientist" "a chemist"))
A crucial facet of the game is that (tq-once tree) must return
a valid twenty-questions tree from all of the different conditional branches it handles.
That returned tree will be augmented by an additional question and an additional object if
the previous run did not guess the object correctly. If it did guess the object
correctly, the original tree will be returned.
The tq-continuer function will want to give a name to the value of that new tree
(using let is one way to do this). It will then ask the user whether they would
like to continue and use that new tree as appropriate.
As with any large functional program, the key is to break up the tasks into manageable
chunks, write functions that handle those pieces, and then compose them into a solution.
Write a number of helper functions that will keep your code succinct and straightforward.