# [2021-Fall] Hw04 of CS61A of UCB

## Mobiles

### Q2: Weights

Implement the planet data abstraction by completing the planet constructor and the size selector so that a planet is represented using a two-element list where the first element is the string 'planet' and the second element is its size.

def planet(size):
"""Construct a planet of some size."""
assert size > 0
return ['planet', size]

def size(w):
"""Select the size of a planet."""
assert is_planet(w), 'must call size on a planet'
return w


### Q3: Balanced

Implement the balanced function, which returns whether m is a balanced mobile. A mobile is balanced if both of the following conditions are met:

1. The torque applied by its left arm is equal to that applied by its right arm. The torque of the left arm is the length of the left rod multiplied by the total weight hanging from that rod. Likewise for the right. For example, if the left arm has a length of 5, and there is a mobile hanging at the end of the left arm of weight 10, the torque on the left side of our mobile is 50.
2. Each of the mobiles hanging at the end of its arms is balanced.

Planets themselves are balanced, as there is nothing hanging off of them.

1. 它是 planet, 根据描述可以知道本身是平衡的
2. 它是 arm, 并且 total_weight(left_arm) == total_weight(right_arm). 并且它的所有子树都要符合这个平衡的条件. 注意不要漏掉对子树的判断

def balanced(m):
"""Return whether m is balanced.

>>> t, u, v = examples()
>>> balanced(t)
True
>>> balanced(v)
True
>>> w = mobile(arm(3, t), arm(2, u))
>>> balanced(w)
False
>>> balanced(mobile(arm(1, v), arm(1, w)))
False
>>> balanced(mobile(arm(1, w), arm(1, v)))
False
>>> from construct_check import check
>>> # checking for abstraction barrier violations by banning indexing
>>> check(HW_SOURCE_FILE, 'balanced', ['Index'])
True
"""
# planet is balanced
if is_planet(m):
return True

left_arm, right_arm = left(m), right(m)

# end(...arm) will is a mobile or a planet
left_val = length(left_arm) * total_weight(end(left_arm))
right_val = length(right_arm) * total_weight(end(right_arm))
return left_val == right_val and balanced(end(left_arm)) and balanced(end(right_arm))


### Q4: Totals

Implement totals_tree, which takes in a mobile or planet and returns a tree whose root is the total weight of the input. For a planet, totals_tree should return a leaf. For a mobile, totals_tree should return a tree whose label is that mobile's total weight, and whose branches are totals_trees for the ends of its arms. As a reminder, the description of the tree data abstraction can be found here.

def totals_tree(m):
"""Return a tree representing the mobile with its total weight at the root.

>>> t, u, v = examples()
>>> print_tree(totals_tree(t))
3
2
1
>>> print_tree(totals_tree(u))
6
1
5
3
2
>>> print_tree(totals_tree(v))
9
3
2
1
6
1
5
3
2
>>> from construct_check import check
>>> # checking for abstraction barrier violations by banning indexing
>>> check(HW_SOURCE_FILE, 'totals_tree', ['Index'])
True
"""
if is_planet(m):
return tree(size(m))

return tree(total_weight(m), [totals_tree(i) for i in [end(left(m)), end(right(m))]])


## More trees

### Q5: Replace Loki at Leaf

Define replace_loki_at_leaf, which takes a tree t and a value lokis_replacement. replace_loki_at_leaf returns a new tree that's the same as t except that every leaf label equal to "loki" has been replaced with lokis_replacement.

If you want to learn about the Norse mythology referenced in this problem, you can read about it here.

def replace_loki_at_leaf(t, lokis_replacement):
"""Returns a new tree where every leaf value equal to "loki" has
been replaced with lokis_replacement.

>>> yggdrasil = tree('odin',
...                  [tree('balder',
...                        [tree('loki'),
...                         tree('freya')]),
...                   tree('frigg',
...                        [tree('loki')]),
...                   tree('loki',
...                        [tree('sif'),
...                         tree('loki')]),
...                   tree('loki')])
>>> laerad = copy_tree(yggdrasil) # copy yggdrasil for testing purposes
>>> print_tree(replace_loki_at_leaf(yggdrasil, 'freya'))
odin
balder
freya
freya
frigg
freya
loki
sif
freya
freya
>>> laerad == yggdrasil # Make sure original tree is unmodified
True
"""
if is_leaf(t):
if label(t) == 'loki':
return tree(lokis_replacement)
return t
return tree(label(t), [replace_loki_at_leaf(b, lokis_replacement) for b in branches(t)])


### Q6: Has Path

Write a function has_path that takes in a tree t and a string word. It returns True if there is a path that starts from the root where the entries along the path spell out the word, and False otherwise. (This data structure is called a trie, and it has a lot of cool applications, such as autocomplete). You may assume that every node's labelis exactly one character.

1. 什么是这道题的 base case ? 我们要记住从根结点出发到当前所在结点一路经过的结点拼起来的字符串是什么. 举例来说, 在一开始的时候我们在根结点, 我们的字符串是 label(t). 也就是 h. 当我们到了根结点的第一个子结点的时候, 我们就得到了 h + i = hi. 为了实现这样的功能我们可以另外写个递归函数. 然后在 has_path 里面调用即可. 一个提高效率的方法是如果我们当前得到的字符串不是目标字符串的一部分(开头位置), 那么我们显然没有必要继续到更深的子树里面寻找. 这也是递归的剪枝操作.
2. 怎么把问题分解为更简单的子问题 ? 我们可以使用 python 提供的 any 函数, 只要任意一个子树上存在这样的路径即可.
def has_path(t, word):
"""Return whether there is a path in a tree where the entries along the path
spell out a particular word.

>>> greetings = tree('h', [tree('i'),
...                        tree('e', [tree('l', [tree('l', [tree('o')])]),
...                                   tree('y')])])
>>> print_tree(greetings)
h
i
e
l
l
o
y
>>> has_path(greetings, 'h')
True
>>> has_path(greetings, 'i')
False
>>> has_path(greetings, 'hi')
True
>>> has_path(greetings, 'hello')
True
>>> has_path(greetings, 'hey')
True
>>> has_path(greetings, 'bye')
False
>>> has_path(greetings, 'hint')
False
"""
assert len(word) > 0, 'no path for empty word.'
def helper(t, cur):
if cur == word:
return True
if cur not in word:
return False
return any([helper(b, cur + label(b)) for b in branches(t)])
return helper(t, label(t))


## Trees

### Q7: Preorder

Define the function preorder, which takes in a tree as an argument and returns a list of all the entries in the tree in the order that print_tree would print them.

The following diagram shows the order that the nodes would get printed, with the arrows representing function calls.

def preorder(t):
"""Return a list of the entries in this tree in the order that they
would be visited by a preorder traversal (see problem description).

>>> numbers = tree(1, [tree(2), tree(3, [tree(4), tree(5)]), tree(6, [tree(7)])])
>>> preorder(numbers)
[1, 2, 3, 4, 5, 6, 7]
>>> preorder(tree(2, [tree(4, [tree(6)])]))
[2, 4, 6]
"""
result = []
def helper(t):
if t is not None:
result.append(label(t))
for b in branches(t):
helper(b)
helper(t)
return result


## Data Abstraction

### Q8: Interval Abstraction

Acknowledgements. This interval arithmetic example is based on a classic problem from 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 measurements from physical devices) with known precision, so that when computations are done with such approximate quantities the results will be numbers of known precision. For example, if a measured quantity x lies between two numbers a and b, Alyssa would like her system to use this range in computations involving x.

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 also an interval, one that represents the range of the result.

Alyssa suggests the existence of an abstraction 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 create the interval using data abstraction. Using this constructor and the appropriate selectors, she defines the following operations:

Alyssa's program is incomplete because she has not specified the implementation of the interval abstraction. She has implemented the constructor for you; fill in the implementation of the selectors.

def interval(a, b):
"""Construct an interval from a to b."""
assert a <= b, 'Lower bound cannot be greater than upper bound'
return [a, b]

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

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


### Q9: Interval Arithmetic

After implementing the abstraction, Alyssa decided to implement a few interval arithmetic functions.

This is her current implementation for interval multiplication. Unfortunately there are some data abstraction violations, so your task is to fix this code before someone sets it on fire.

def mul_interval(x, y):
"""Return the interval that contains the product of any value in x and any
value in y."""
p1 = x * y
p2 = x * y
p3 = x * y
p4 = x * y
return [min(p1, p2, p3, p4), max(p1, p2, p3, p4)]


def mul_interval(x, y):
"""Return the interval that contains the product of any value in x and any
value in y."""
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))


#### Interval Subtraction

Using a similar approach as mul_interval and add_interval, define a subtraction function for intervals. If you find yourself repeating code, see if you can reuse functions that have already been implemented.

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


#### Interval Division

Alyssa implements division below by multiplying by the reciprocal of y. A 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 assertstatement 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."""
assert lower_bound(y) > 0, "AssertionError!"
reciprocal_y = interval(1 / upper_bound(y), 1 / lower_bound(y))
return mul_interval(x, reciprocal_y)


### Q10: Par Diff

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 theformula 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 par2(r1, r2):
one = interval(1, 1)
rep_r1 = div_interval(one, r1)
rep_r2 = div_interval(one, r2)


Lem points out that Alyssa's program gives different answers for the two ways of computing. Find two intervals r1 and r2 that demonstrate the difference in behavior between par1 and par2 when passed into each of the two functions.

Demonstrate that Lem is right. Investigate the behavior of the system on a variety of arithmetic expressions. Make some intervals r1 and r2, and show that par1 and par2 can give different results.

def check_par():
"""Return two intervals that give different results for parallel resistors.

>>> r1, r2 = check_par()
>>> x = par1(r1, r2)
>>> y = par2(r1, r2)
>>> lower_bound(x) != lower_bound(y) or upper_bound(x) != upper_bound(y)
True
"""
r1 = interval(5, 7)
r2 = interval(5, 7)
return r1, r2

posted @ 2022-02-20 15:18  MartinLwx  阅读(1686)  评论(0编辑  收藏  举报