CS计算机代考程序代写 SQL scheme python interpreter CS 61A Structure and Interpretation of Computer Programs

CS 61A Structure and Interpretation of Computer Programs
Fall 2018 Final Solutions

INSTRUCTIONS

• You have 3 hours to complete the exam.

• The exam is closed book, closed notes, closed computer, closed calculator, except three hand-written 8.5″ × 11″
crib sheet of your own creation and the official CS 61A midterm 1, midterm 2, and final study guides.

• Mark your answers on the exam itself. We will not grade answers written on scratch paper.

Last name

First name

Student ID number

CalCentral email ( )

TA

Name of the person to your left

Name of the person to your right

All the work on this exam is my own.
(please sign)

POLICIES & CLARIFICATIONS

• If you need to use the restroom, bring your phone and exam to the front of the room.

• You may use built-in Python functions that do not require import, such as min, max, pow, len, abs, sum, next,
iter, list, tuple, map, filter, zip, all, and any.

• You may not use example functions defined on your study guides unless a problem clearly states you can.

• For fill-in-the-blank coding problems, we will only grade work written in the provided blanks. You may only
write one Python statement per blank line, and it must be indented to the level that the blank is indented.

• Unless otherwise specified, you are allowed to reference functions defined in previous parts of the same question.

• You may use the Tree, Link, and BTree classes defined on Page 2 (left column) of the Midterm 2 Study Guide.

http://berkeley.edu

2

1. (12 points) High Quality Air

For each of the expressions in the table below, write the output displayed by the interactive Python interpreter
when the expression is evaluated. The output may have multiple lines. The first row is completed for you.

• If an error occurs, write Error, but include all output displayed before the error.

• To display a function value, write Function.

• To display an iterator value, write Iterator.

• If an expression would take forever to evaluate, write Forever.

The interactive interpreter displays the contents of the repr string of the value of a successfully evaluated
expression, unless it is None.

Assume that you have started python3 and executed the code shown on the left first, then you evaluate each
expression on the right in the order shown. Expressions evaluated by the interpreter have a cumulative effect.

from operator import sub

z = (lambda x: lambda y: 2 * (y-x))(3)

def breath(f, count=1):
if count > 1:

print(count)
count += 1
return lambda x, y: f(x+1, y)

class Day:
aqi = 10
def __init__(self, aqi=0):

if aqi > self.aqi:
self.aqi = aqi

self.n = []
def mask(self, limit):

def f(aqi):
if aqi > limit:

self.n.append(aqi-limit)
return self.mask(aqi)

return f

class Week(Day):
aqi = 50

m, t = Day(), Week(199)
t.mask(200)(100)(150)(160)
Day.aqi = 140
t.aqi = 160

Expression Output
print(None) None

print(print(None), print)
None
None Function

z(4) 2

breath(breath(sub))(5, 3) 4

[Day().aqi, m.aqi] [140, 140]

[Week.aqi, t.aqi] [50, 160]

t.n [50, 10]

Name: 3

2. (8 points) Diagram Horror

def get(out):
out.pop()
out = scary(lambda movie: out)
return lambda: [out]

def scary(movie):
out.append(movie)
return movie(5)[:1]

out = [6]
get([7, 8])()

Fill in the environment diagram that results from ex-
ecuting the code on the left until the entire program
is finished, an error occurs, or all frames are filled.
You may not need to use all of the spaces or frames.
A complete answer will:

• Add all missing names and parent annotations
to all local frames.

• Add all missing values created or referenced
during execution.

• Show the return value for each local frame.

• Use box-and-pointer diagrams for lists
and tuples.

Global get

f2: ___________ [parent=____________]

Return Value

f3: ___________ [parent=____________]

Return Value

func get(out) [parent=Global]

scary func scary(movie) [parent=Global]

f1: ___________ [parent=____________]

Return Value

f4: ___________ [parent=____________]

Return Value

list
0
6

1
out

get Global

out

list
0
7

scary Global

movie func λ(movie) [parent=f1]

λ f1

movie 5

list
0
7

func λ() [parent=f1]

λ f1

list
0

4

3. (16 points) Gainz

Definition. A sequence is near increasing if each element beyond the second is larger than all elements
preceding its previous element. That is, element i must be larger than elements i− 2, i− 3, i− 4, etc.

(a) (3 pt) Implement is_near, which takes a sequence s and returns whether its elements form a near increasing
sequence.

def is_near(s):
“””Return whether s is a near increasing sequence.

>>> is_near([]) and is_near([1]) and is_near([1, 2]) and is_near(range(10))
True
>>> is_near([4, 2]) and is_near((1, 4, 2, 5)) and is_near((1, 2, 4, 3))
True
>>> is_near((3, 2, 1)) # 1 <= 3 False >>> is_near([1, 4, 2, 3, 5]) # 3 <= 4 False >>> is_near([1, 4, 2, 5, 3]) # 3 <= 4 False >>> is_near([1, 2, 4, 2, 5]) # 2 <= 2 False """ return all([s[i] > max(s[:i-1]) for i in range(2, len(s))])

(b) (6 pt) Implement fast_near, which takes an iterable value and returns whether its elements form a near
increasing sequence. fast_near must run in Θ(n) time and Θ(1) space (not including the input itself) for an
iterable input with n elements. Assume that s has a finite number of elements. You may not call is_near.

def fast_near(s):
“””Return whether s is a near increasing sequence.

>>> fast_near([2, 5, 3, 6, 6, 7, 7, 9, 8])
True
“””
t, s = iter(s), None # Do not refer to s below this line.

try:

largest, last = next(t), next(t)

except StopIteration:

return True

for x in t:

if x <= largest: return False largest, last = max(largest, last), x return True Name: 5 Alternative Definition. (Equivalent to the one on the previous page, but stated in a more useful way for the problem below.) A sequence is near increasing if each element but the last two is smaller than all elements following its subsequent element. That is, element i must be smaller than elements i + 2, i + 3, i + 4, etc. (c) (6 pt) Implement near, which takes a non-negative integer n and returns the largest near increasing sequence of digits within n as an integer. The arguments smallest and d are part of the implementation; you must determine their purpose. You may not call is_near or fast_near. You may not use any values except integers and booleans (True and False) in your solution (no lists, strings, etc.). def near(n, smallest=10, d=10): """Return the longest sequence of near-increasing digits in n. >>> near(123)
123
>>> near(153)
153
>>> near(1523)
153
>>> near(15123)
1123
>>> near(11111111)
11
>>> near(985357)
557
>>> near(14735476)
143576
>>> near(812348567)
1234567
“””

if n == 0:

return 0

no = near(n//10, smallest, d)

if smallest > n % 10:

yes = 10 * near(n//10, d, min(d, n%10)) + n%10

# OR yes = 10 * near(n//10, min(smallest, d), n%10) + n%10

return max(yes, no)

return no

(d) (1 pt) What is the largest possible integer that could ever be returned from the near function? Note: In
general, integers in Python can be arbitrarily large.

112233445566778899

6

4. (11 points) Tree Time

Definition. A runt node is a node in a tree whose label is smaller than all of the labels of its siblings. A
sibling is another node that shares the same parent. A node with no siblings is a runt node.

(a) (7 pt) Implement runts, which takes a Tree instance t in which every label is different and returns a list
of the labels of all runt nodes in t, in any order. Also implement apply_to_nodes, which returns nothing
and is part of the implementation. Do not mutate any tree. The Tree class is on the Midterm 2 Guide.

def runts(t):
“””Return a list of the labels of all smallest siblings in any order.

>>> sorted(runts(Tree(9, [Tree(3), Tree(4, [Tree(5, [Tree(6)]), Tree(7)]), Tree(2)])))
[2, 5, 6, 9]
“””
result = [f]

def g(node):

if not node.is_leaf():

result.append(min([b.label for b in node.branches]))

apply_to_nodes(g, t)

return [t.label] + result

def apply_to_nodes(f, t):
“””Apply a function f to each node in a Tree instance t.”””

f(t)

for b in t.branches:

apply_to_nodes(f, b)

(b) (4 pt) Implement max_label, which takes a Tree t and returns its largest label. Do not mutate any tree.

def max_label(t):
“””Return the largest label in t.

>>> max_label(Tree(4, [Tree(5), Tree(3, [Tree(6, [Tree(1), Tree(2)])])]))
6
“””
def f(node):

nonlocal t

t = max(t, node, key=lambda n: n.label)

apply_to_nodes(f, t)

return t.label

Name: 7

5. (9 points) Run, Program, Run

Implement runs, a Scheme procedure that takes a list of integers s and returns a list of non-empty lists of
integers t. Together, the lists in t should contain all elements of s in order. The first element in each list in t
must be less than the last element in the previous list, if there is one. The rest of the elements in each list in
t must be greater than or equal to the previous element.

Also implement and use next-run in your solution, which takes a non-empty list of integers s and returns a
pair of lists: the longest non-decreasing prefix of s and the rest of s. Use the provided pair data abstraction.
Your implementation should be correct even if the pair implementation were to change.

;; Return a list of non-decreasing lists that together contain the elements of s.
;; scm> (runs ‘(3 4 7 6 6 8 1 2 5 5 4))
;; ((3 4 7) (6 6 8) (1 2 5 5) (4))
;; scm> (runs ‘(4 3 2 3))
;; ((4) (3) (2 3))
(define (runs s)

(if (null? s) nil

(let ((p (next-run s)))

(cons (first p) (runs (rest p))))))

;; A data abstraction for a pair of a first run and the rest of a list.
(define (pair a b) (lambda (c) (if c a b)))
(define (first p) (p #t))
(define (rest p) (p #f))

;; Return a pair containing the first run in s (a list) and the rest of s (another list).
;; scm> (first (next-run ‘(4 5 1 3 2)))
;; (4 5)
;; scm> (rest (next-run ‘(4 5 1 3 2)))
;; (1 3 2)
(define (next-run s)

(if (or (null? (cdr s))

(< (car (cdr s)) (car s))) (pair (list (car s)) (cdr s)) (begin (define p (next-run (cdr s))) (pair (cons (car s) (first p)) (rest p))))) 8 6. (9 points) Generation Z (a) (4 pt) Implement rev, a generator function that takes a Link instance and yields the elements of that linked list in reverse order. The Link class appears on Page 2 of the Midterm 2 Study Guide. def rev(s): """Yield the elements in Link instance s in reverse order. >>> list(rev(Link(1, Link(2, Link(3)))))
[3, 2, 1]
>>> next(rev(Link(2, Link(3))))
3
“””
if s is not Link.empty:

yield from rev(s.rest)

yield s.first

(b) (2 pt) Using the provided add procedure, define not-three, an infinite stream of all positive integers that
are not evenly divisible by 3. The not-three stream is increasing and begins with 1, 2, 4, 5, 7, 8, 10, 11, 13.

(define (add k s) (cons-stream (+ k (car s)) (add k (cdr-stream s))))

(define not-three (cons-stream 1 (cons-stream 2 (add 3 not-three))))

(c) (3 pt) Implement infix, a Scheme macro that evaluates infix expressions. An infix expression is either a
number or a three-element list containing an infix expression, a procedure, and another infix expression. The
value of a compound infix expression is the value of its second element applied to the values of its first and
third elements. Note: The last line begins with a quasiquote. If you cross out the quasiquote and solve the
problem without using quasiquote or unquote, you can receive up to 2 out of 3 points (not recommended).

;; A macro to evaluate infix expressions.
;; scm> (infix (2 * 3))
;; 6
;; scm> (infix ((1 + 1) * (1 + 2)))
;; 6
;; scm> (infix ((1 + (3 – 2)) * ((2 + 3) + 2)))
;; 14
(define-macro (infix e)

(if (number? e) e

`(,(cadr e) (infix ,(car e)) (infix ,(caddr e)))))

(define (cadr x) (car (cdr x)))
(define (caddr x) (car (cdr (cdr x))))

Name: 9

7. (10 points) SQL of Course

The courses table describes the course name, start time hour (h) and minute (m), and length in minutes (len)
for different lectures. For example, 61A starts at 13:00 and lasts 50 minutes. The locations table describes
the course name and location (loc) of these courses. Assume that each course name appears exactly once in
each table. Write your SQL statements so that they would still be correct if the table contents changed.

CREATE TABLE courses AS
SELECT “1” AS course, 14 AS h, 0 AS m, 80 AS len UNION
SELECT “2” , 13 , 30 , 80 UNION
SELECT “8” , 12 , 30 , 50 UNION
SELECT “10” , 12 , 30 , 110 UNION
SELECT “50AC” , 13 , 30 , 45 UNION
SELECT “61A” , 13 , 0 , 50;

CREATE TABLE locations AS
SELECT “1” AS name, “VLSB” AS loc UNION
SELECT “2” , “Dwinelle” UNION
SELECT “10” , “VLSB” UNION
SELECT “50AC” , “Wheeler” UNION
SELECT “61A” , “Wheeler”;

(a) (2 pt) Select a one-column table that contains the course names of all courses that start before 13:30.

SELECT course FROM courses WHERE h < 13 OR (h=13 AND m<30); 61A 8 10 (b) (4 pt) Select a two-column table with one row per location that contains the location, as well as the shortest length in minutes of any lecture held in that location. SELECT loc, min(len) FROM courses, locations WHERE name=course GROUP BY loc; Dwinelle 80 VLSB 80 Wheeler 45 (c) (4 pt) Select a three-column table where each row describes an earlier course, a later course, and the amount of time in minutes between the end time of the earlier course and the start time of the later course. Only include pairs of courses where the lectures do not overlap in time. Note: There are 60 minutes in an hour. SELECT a.course, b.course, (b.h - a.h) * 60 + b.m - a.m - a.len AS gap FROM courses as a, courses as b WHERE gap > 0

61A 1 10
8 1 40
8 2 10
8 50AC 10

10

8. (0 points) Draw! (Optional) Draw a picture of some function or procedure.