**Due by 4pm on Monday, 2/6**

You can grab a template for this homework either by downloading the file from the calendar or by running the following command in terminal on one of the school computers (the dot is significant: it denotes the current directory)

cp ~cs61a/lib/hw/hw3.py .

**Readings.** Chapter 2.1, 2.2

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.** Alyssa implements division by multiplying 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)

**Q3.** Using reasoning analogous to Alyssa's, define a subtraction function,
`sub_interval` for
intervals. Add a doctest.

**Q4.** In passing, Ben also cryptically comments, "By testing the signs of the
endpoints of the intervals, it is possible to break mul_interval into nine
cases, only one of which requires more than two multiplications." Write a
fast multiplication function, `mul_interval_fast`, using Ben's suggestion.
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.

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.

**Q6.** 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.

**Q7.** 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 hw3.py file you submit.

**Q8.** 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. Make sure that your implementation returns the smallest such interval, one that does not suffer from the multiple references problem.

*Hint*: the derivative f'(t) = 2 * a * t + b, and so the extreme point of the
quadratic is -b/(2*a).

**Q9.** Extra for those who'd like to work ahead (we'll ask it for real in
HW #4). Write three similar functions, each of which takes as an argument
a sequence
of intervals and returns the sum of the square of each interval that does not
contain 0.

sum_nonzero_with_for(seq)uses a for statement containing an if statement.sum_nonzero_with_map_filter_reduce(seq)uses map and filter and reduce.sum_nonzero_with_generator_reduce(seq)uses generator expression and reduce.

**Q10.** Extra for experts: 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 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.