# 61A Homework 4

Due by 11:59pm on Tuesday, 10/8

Submission. See the online submission instructions. We have provided a hw4.py starter file for the questions below.

Readings. Sections 2.1, 2.2, and 2.3 of the online textbook.

Acknowledgements. This interval arithmetic example is based on Structure and Interpretation of Computer Programs, Section 2.1.4.

Introduction. 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 interval. Using the constructor and selectors, she defines the following operations:

```def str_interval(x):
"""Return a string representation of interval x.

>>> str_interval(interval(-1, 2))
'-1 to 2'
"""
return '{0} to {1}'.format(lower_bound(x), upper_bound(x))

"""Return an interval that contains the sum of any value in interval x and
any value in interval y.

'3 to 10'
"""
lower = lower_bound(x) + lower_bound(y)
upper = upper_bound(x) + upper_bound(y)
return 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(interval(-1, 2), 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 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 and selectors in terms of two-element tuples:

```def interval(a, b):
"""Construct an interval from a to b."""

def lower_bound(x):
"""Return the lower bound of interval x."""

def upper_bound(x):
"""Return the upper bound of interval x."""
```

Q2. Alyssa implements division below, 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(interval(-1, 2), interval(4, 8)))
'-0.25 to 0.5'
"""
reciprocal_y = 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 for intervals:

```def sub_interval(x, y):
"""Return the interval that contains the difference between any value in x
and any value in y.

>>> str_interval(sub_interval(interval(-1, 2), interval(4, 8)))
'-9 to -2'
"""
```

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 using Ben's suggestion:

```def mul_interval_fast(x, y):
"""Return the interval that contains the product of any value in x and any
value in y, using as few multiplications as possible.

>>> str_interval(mul_interval_fast(interval(-1, 2), interval(4, 8)))
'-8 to 16'
>>> str_interval(mul_interval_fast(interval(-2, -1), interval(4, 8)))
'-16 to -4'
>>> str_interval(mul_interval_fast(interval(-1, 3), interval(-4, 8)))
'-12 to 24'
>>> str_interval(mul_interval_fast(interval(-1, 2), interval(-8, 4)))
'-16 to 8'
"""
```

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 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:

```def make_center_percent(c, p):
"""Construct an interval from center and percentage tolerance.

>>> str_interval(make_center_percent(2, 50))
'1.0 to 3.0'
"""

def percent(x):
"""Return the percentage tolerance of interval x.

>>> percent(interval(1, 3))
50.0
"""
```

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:

par1(r1, r2) = (r1 * r2) / (r1 + r2), or

par2(r1, r2) = 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):

def par2(r1, r2):
one = interval(1, 1)
rep_r1 = div_interval(one, r1)
rep_r2 = div_interval(one, 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 show that par1 and par2 can give different results. You will get the most insight by using intervals whose width is a small percentage of the center value:

```# These two intervals give different results for parallel resistors:
```

Note: No tests will be run on your solution to this problem.

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? Write a function that returns an explanation string:

```def multiple_references_explanation():
return """The mulitple reference problem..."""
```

Note: No tests will be run on your solution to this problem.

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) is a quadratic function:

`````f(t) = a * t * t + b * t + c``
```

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):

```def quadratic(x, a, b, c):
"""Return the interval that is the range of the quadratic defined by
coefficients a, b, and c, for domain interval x.

>>> str_interval(quadratic(interval(0, 2), -2, 3, -1))
'-3 to 0.125'
>>> str_interval(quadratic(interval(1, 3), 2, -3, 1))
'0 to 10'
"""
```

Q9. 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.

1. Using a for statement containing an if statement.
2. Using map and filter and reduce.
3. Using generator expression and reduce.

Hint: Square is a special case of quadratic, but you can also use the simpler square_interval function below for intervals that do not contain 0:

```def non_zero(x):
"""Return whether x contains 0."""
return lower_bound(x) > 0 or upper_bound(x) < 0

def square_interval(x):
"""Return the interval that contains all squares of values in x, where x
does not contain 0.
"""
assert non_zero(x), 'square_interval is incorrect for x containing 0'
return mul_interval(x, x)

# The first two of these intervals contain 0, but the third does not.
seq = (interval(-1, 2), make_center_width(-1, 2), make_center_percent(-1, 50))

zero = interval(0, 0)

def sum_nonzero_with_for(seq):
"""Returns an interval that is the sum of the squares of the non-zero
intervals in seq, using a for statement.

>>> str_interval(sum_nonzero_with_for(seq))
'0.25 to 2.25'
"""

from functools import reduce
def sum_nonzero_with_map_filter_reduce(seq):
"""Returns an interval that is the sum of the squares of the non-zero
intervals in seq, using using map, filter, and reduce.

>>> str_interval(sum_nonzero_with_map_filter_reduce(seq))
'0.25 to 2.25'
"""

def sum_nonzero_with_generator_reduce(seq):
"""Returns an interval that is the sum of the squares of the non-zero
intervals in seq, using using reduce and a generator expression.

>>> str_interval(sum_nonzero_with_generator_reduce(seq))
'0.25 to 2.25'
"""
```

Q10. (Optional extra for experts) Write a function polynomial that takes an interval x and a tuple of coefficients c, and returns the 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 * 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. Try using Newton's method:

```def polynomial(x, c):
"""Return the interval that is the range of the polynomial defined by
coefficients c, for domain interval x.

>>> str_interval(polynomial(interval(0, 2), (-1, 3, -2)))
'-3 to 0.125'
>>> str_interval(polynomial(interval(1, 3), (1, -3, 2)))
'0 to 10'
>>> str_interval(polynomial(interval(0.5, 2.25), (10, 24, -6, -8, 3)))
'18.0 to 23.0'
"""