Due by 11:59 PM on (that is, the end of) Tuesday, 7/3
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Readings. All problems in this homework can be solved with the subset of Python 3 introduced in sections 1-2.2 of the lecture notes.
If you would like, you can use the template file hw4.py.
To copy this file to your lab account you can run the command:
cp ~cs61a/lib/hw/hw04/hw4.py .
to copy it into your current directory.
This interval arithmetic example is based on SICP, Section 2.1.4.
Alyssa P. Hacker is designing a system to help people solve engineering problems. One feature she wants to provide in her system is the ability to manipulate inexact quantities, such as measured parameters of physical devices, with known precision, so that when computations are done with such approximate quantities the results will be numbers of known precision.
Alyssa's idea is to implement interval arithmetic as a set of arithmetic operations for combining intervals, objects that represent the range of possible values of an inexact quantity. The result of adding, subracting, multiplying, or dividing two intervals is itself an interval, representing the range of the result.
Alyssa postulates the existence of an abstract object called an interval that has two endpoints: a lower bound and an upper bound. She also presumes that, given the endpoints of an interval, she can construct the interval using the data constructor make_interval. Using the constructor and selectors, she defines the following operations:
def str_interval(x):
"""Return a string representation of interval x.
>>> str_interval(make_interval(-1, 2))
'-1 to 2'
"""
return '{0} to {1}'.format(lower_bound(x), upper_bound(x))
def add_interval(x, y):
"""Return an interval that contains the sum of any value in interval x and
any value in interval y.
>>> str_interval(add_interval(make_interval(-1, 2), make_interval(4, 8)))
'3 to 10'
"""
lower = lower_bound(x) + lower_bound(y)
upper = upper_bound(x) + upper_bound(y)
return make_interval(lower, upper)
def mul_interval(x, y):
"""Return the interval that contains the product of any value in x and any
value in y.
>>> str_interval(mul_interval(make_interval(-1, 2), make_interval(4, 8)))
'-8 to 16'
"""
p1 = lower_bound(x) * lower_bound(y)
p2 = lower_bound(x) * upper_bound(y)
p3 = upper_bound(x) * lower_bound(y)
p4 = upper_bound(x) * upper_bound(y)
return make_interval(min(p1, p2, p3, p4), max(p1, p2, p3, p4))
Q1. Alyssa's program is incomplete because she has not specified the implementation of the interval abstraction. Define the constructor make_interval and selectors lower_bound and upper_bound in terms of two-element tuples.
Q2. Write sum_intervals, which takes a sequence (tuple) of intervals and returns the interval representing their sum. Add a doctest.
Q3. Alyssa implements division of two intervals x and y by multiplying x by the reciprocal of y. Ben Bitdiddle, an expert systems programmer, looks over Alyssa's shoulder and comments that it is not clear what it means to divide by an interval that spans zero. Add an assert statement to Alyssa's code to ensure that no such interval is used as a divisor:
def div_interval(x, y):
"""Return the interval that contains the quotient of any value in
x divided by any value in y.
Division is implemented as the multiplication of x by the reciprocal
of y.
>>> str_interval(div_interval(make_interval(-1, 2), make_interval(4, 8)))
'-0.25 to 0.5'
"""
"*** YOUR CODE HERE ***"
reciprocal_y = make_interval(1/upper_bound(y), 1/lower_bound(y))
return mul_interval(x, reciprocal_y)
Hint: It might help to write a local function spans_zero which returns True if the interval includes zero.
Q4. Using reasoning analogous to Alyssa's, define a subtraction function, sub_interval for intervals. Add a doctest.
Q5. After debugging her program, Alyssa shows it to a potential user, who complains that her program solves the wrong problem. He wants a program that can deal with numbers represented as a center value and an additive tolerance; for example, he wants to work with intervals such as 3.5 +/- 0.15 rather than 3.35 to 3.65. Alyssa returns to her desk and fixes this problem by supplying an alternate constructor and alternate selectors in terms of the existing ones:
def make_center_width(c, w):
"""Construct an interval from center and width."""
return make_interval(c - w, c + w)
def center(x):
"""Return the center of interval x."""
return (upper_bound(x) + lower_bound(x)) / 2
def width(x):
"""Return the width of interval x."""
return (upper_bound(x) - lower_bound(x)) / 2
Unfortunately, most of Alyssa's users are engineers. Real engineering situations usually involve measurements with only a small uncertainty, measured as the ratio of the width of the interval to the midpoint of the interval. Engineers usually specify percentage tolerances on the parameters of devices. For example an engineer might say a value is 5 with a percentage tolerance of 10% which would mean that the values can range from 4.5 to 5.5, since 10% of 5 is .5 and so we have (5 - 10% of 5) through (5 + 10% of 5).
Define a constructor make_center_percent that takes a center and a percentage tolerance and produces the desired interval. You must also define a selector percent that produces the percentage tolerance for a given interval. The center selector is the same as the one shown above. Hint: The percentage tolerance is defined in terms of the interval's center and width. Make use of the two other selectors already provided.
Q6. Write a function quadratic that returns the interval of all values f(t) such that t is in the argument interval x and
f(t) = a * t * t + b * t + c
where a, b, and c are also parameters.
Note: You might have noticed that we're calculating a much larger interval than what is the true range of possible outputs given the input interval x, we explore this problem in questions in the Extra for Experts section that deals with the "Multiple Reference Problem".
Q7. Express the order of growth in time of the following functions, using big-Theta notation.
def one(n):
return n
def two(n):
if n > 10 or n < 0:
return n
return two(n+1)
def three(n):
if n <= 0:
return 0
elif n % 2 == 0:
return 2 + three(n/2)
else:
return 1 + three(n-1)
def four(n):
if n == 0:
return 0
else:
return two(n) + four(n-1)
def five(n):
if n == 0:
return 0
else:
return three(n) + five(n-1)
def six(n):
total = 0
while n >= 0:
total = total + four(n)
n = n-1
return total
Extra for Experts
Note: "Extra for Experts" problems are completely optional. You are encouraged to try these questions. However, these are questions which either we consider more difficult than what we would consider asking you on an exam or focus on ideas we are not concerned with in this course, so please don't be discouraged if you don't solve them.
Q8. After considerable work, Alyssa P. Hacker delivers her finished system. Several years later, after she has forgotten all about it, she gets a frenzied call from an irate user, Lem E. Tweakit. It seems that Lem has noticed that the formula for parallel resistors can be written in two algebraically equivalent ways:
(r1 * r2) / (r1 + r2)
and
1 / (1/r1 + 1/r2)
He has written the following two programs, each of which computes the parallel_resistors formula differently:
def par1(r1, r2):
return div_interval(mul_interval(r1, r2), add_interval(r1, r2))
def par2(r1, r2):
one = make_interval(1, 1)
rep_r1 = div_interval(one, r1)
rep_r2 = div_interval(one, r2)
return div_interval(one, add_interval(rep_r1, rep_r2))
Lem complains that Alyssa's program gives different answers for the two ways of computing. This is a serious complaint.
Demonstrate that Lem is right. Investigate the behavior of the system on a variety of arithmetic expressions. Make some intervals A and B, and use them in computing the expressions A/A and A/B. You will get the most insight by using intervals whose width is a small percentage of the center value.
Q9. Eva Lu Ator, another user, has also noticed the different intervals computed by different but algebraically equivalent expressions. She says that the problem is multiple references to the same interval.
The Multiple References Problem: A formula to compute with intervals using Alyssa's system will produce tighter error bounds if it can be written in such a form that no variable that represents an uncertain number is repeated.
Thus, she says, par2 is a better program for parallel resistances than par1. Is she right? Why? Include an explanation as a string in the hw4.py file you submit.
Q10. Write a new version of quadratic, accurate_quadratic, which takes the multiple references problem into account and returns a tighter interval than our original solution.
Hint: The derivative is
f'(t) = 2 * a * t + b
and so the extreme point of the quadratic is -b/(2*a).
Q11. Write a function polynomial that takes an interval x and a tuple of coefficients c, and returns the (possibly approximate) interval containing all values of f(t) for t in interval x, where:
f(t) = c[k-1] * pow(t, k-1) + c[k-2] * pow(t, k-2) + ... + c[0] * 1
Like accurate_quadratic, your polynomial function should return the smallest such interval, one that does not suffer from the multiple references problem.
Hint: You can approximate this result. Consider using Newton's
method. Feel free to use the code from newton.py for this
question.