cs61a 2021 fall lab 5
网址 https://inst.eecs.berkeley.edu/~cs61a/fa21/lab/lab05/#required-questions
problem1:
def factors_list(n):
"""Return a list containing all the numbers that divide `n` evenly, except
for the number itself. Make sure the list is in ascending order.
>>> factors_list(6)
[1, 2, 3]
>>> factors_list(8)
[1, 2, 4]
>>> factors_list(28)
[1, 2, 4, 7, 14]
"""
all_factors = []
for i in range(1, int(n / 2) + 1):
if n % i == 0:
all_factors += [i]
return all_factors
problem2:
用递归，这种一层一层的很适合用递归
def flatten(s):
"""Returns a flattened version of list s.
>>> flatten([1, 2, 3]) # normal list
[1, 2, 3]
>>> x = [1, [2, 3], 4] # deep list
>>> flatten(x)
[1, 2, 3, 4]
>>> x # Ensure x is not mutated
[1, [2, 3], 4]
>>> x = [[1, [1, 1]], 1, [1, 1]] # deep list
>>> flatten(x)
[1, 1, 1, 1, 1, 1]
>>> x
[[1, [1, 1]], 1, [1, 1]]
"""
"*** YOUR CODE HERE ***"
if not s :
return []
elif type(s[0] ) == list:
return flatten(s[0] + flatten(s[1:]))
else:
return [s[0]] + flatten(s[1:])
problem3:
注意三和四问题不能破坏abstraction barrier
def distance(city_a, city_b):
"""
>>> city_a = make_city('city_a', 0, 1)
>>> city_b = make_city('city_b', 0, 2)
>>> distance(city_a, city_b)
1.0
>>> city_c = make_city('city_c', 6.5, 12)
>>> city_d = make_city('city_d', 2.5, 15)
>>> distance(city_c, city_d)
5.0
"""
"*** YOUR CODE HERE ***"
return round(sqrt((get_lat(city_a) - get_lat(city_b))**2 + (get_lon(city_a) - get_lon(city_b))**2), 2)
problem4:
def closer_city(lat, lon, city_a, city_b):
"""
Returns the name of either city_a or city_b, whichever is closest to
coordinate (lat, lon). If the two cities are the same distance away
from the coordinate, consider city_b to be the closer city.
>>> berkeley = make_city('Berkeley', 37.87, 112.26)
>>> stanford = make_city('Stanford', 34.05, 118.25)
>>> closer_city(38.33, 121.44, berkeley, stanford)
'Stanford'
>>> bucharest = make_city('Bucharest', 44.43, 26.10)
>>> vienna = make_city('Vienna', 48.20, 16.37)
>>> closer_city(41.29, 174.78, bucharest, vienna)
'Bucharest'
"""
"*** YOUR CODE HERE ***"
now = make_city('now', lat, lon)
if distance(now, city_a) <= distance(now, city_b):
return get_name(city_a)
else:
return get_name(city_b)
problem6:
7和6也要注意不要破坏abstraction barrier
def berry_finder(t):
"""Returns True if t contains a node with the value 'berry' and
False otherwise.
>>> scrat = tree('berry')
>>> berry_finder(scrat)
True
>>> sproul = tree('roots', [tree('branch1', [tree('leaf'), tree('berry')]), tree('branch2')])
>>> berry_finder(sproul)
True
>>> numbers = tree(1, [tree(2), tree(3, [tree(4), tree(5)]), tree(6, [tree(7)])])
>>> berry_finder(numbers)
False
>>> t = tree(1, [tree('berry',[tree('not berry')])])
>>> berry_finder(t)
True
"""
"*** YOUR CODE HERE ***"
if label(t) == 'berry':
return True
for b in branches(t):
if berry_finder(b):
return True
return False
problem7：
def sprout_leaves(t, leaves):
"""Sprout new leaves containing the data in leaves at each leaf in
the original tree t and return the resulting tree.
>>> t1 = tree(1, [tree(2), tree(3)])
>>> print_tree(t1)
1
2
3
>>> new1 = sprout_leaves(t1, [4, 5])
>>> print_tree(new1)
1
2
4
5
3
4
5
>>> t2 = tree(1, [tree(2, [tree(3)])])
>>> print_tree(t2)
1
2
3
>>> new2 = sprout_leaves(t2, [6, 1, 2])
>>> print_tree(new2)
1
2
3
6
1
2
"""
"*** YOUR CODE HERE ***"
# if is_leaf(t):
# #直接加一个叶子
# pass
# for b in branches(t):
# if is_leaf(b):
# #加一个叶子
# pass
# else:
# sprout_leaves(b,leaves)
# return t
if is_leaf(t):
return tree(label(t), [tree(leaf) for leaf in leaves])
return tree(label(t), [sprout_leaves(s, leaves) for s in branches(t)])
全部代码
def factors_list(n):
"""Return a list containing all the numbers that divide `n` evenly, except
for the number itself. Make sure the list is in ascending order.
>>> factors_list(6)
[1, 2, 3]
>>> factors_list(8)
[1, 2, 4]
>>> factors_list(28)
[1, 2, 4, 7, 14]
"""
all_factors = []
for i in range(1, int(n / 2) + 1):
if n % i == 0:
all_factors += [i]
return all_factors
def flatten(s):
"""Returns a flattened version of list s.
>>> flatten([1, 2, 3]) # normal list
[1, 2, 3]
>>> x = [1, [2, 3], 4] # deep list
>>> flatten(x)
[1, 2, 3, 4]
>>> x # Ensure x is not mutated
[1, [2, 3], 4]
>>> x = [[1, [1, 1]], 1, [1, 1]] # deep list
>>> flatten(x)
[1, 1, 1, 1, 1, 1]
>>> x
[[1, [1, 1]], 1, [1, 1]]
"""
"*** YOUR CODE HERE ***"
if not s :
return []
elif type(s[0] ) == list:
return flatten(s[0] + flatten(s[1:]))
else:
return [s[0]] + flatten(s[1:])
from math import sqrt
def distance(city_a, city_b):
"""
>>> city_a = make_city('city_a', 0, 1)
>>> city_b = make_city('city_b', 0, 2)
>>> distance(city_a, city_b)
1.0
>>> city_c = make_city('city_c', 6.5, 12)
>>> city_d = make_city('city_d', 2.5, 15)
>>> distance(city_c, city_d)
5.0
"""
"*** YOUR CODE HERE ***"
return round(sqrt((get_lat(city_a) - get_lat(city_b))**2 + (get_lon(city_a) - get_lon(city_b))**2), 2)
def closer_city(lat, lon, city_a, city_b):
"""
Returns the name of either city_a or city_b, whichever is closest to
coordinate (lat, lon). If the two cities are the same distance away
from the coordinate, consider city_b to be the closer city.
>>> berkeley = make_city('Berkeley', 37.87, 112.26)
>>> stanford = make_city('Stanford', 34.05, 118.25)
>>> closer_city(38.33, 121.44, berkeley, stanford)
'Stanford'
>>> bucharest = make_city('Bucharest', 44.43, 26.10)
>>> vienna = make_city('Vienna', 48.20, 16.37)
>>> closer_city(41.29, 174.78, bucharest, vienna)
'Bucharest'
"""
"*** YOUR CODE HERE ***"
now = make_city('now', lat, lon)
if distance(now, city_a) <= distance(now, city_b):
return get_name(city_a)
else:
return get_name(city_b)
def check_city_abstraction():
"""
There's nothing for you to do for this function, it's just here for the extra doctest
>>> change_abstraction(True)
>>> city_a = make_city('city_a', 0, 1)
>>> city_b = make_city('city_b', 0, 2)
>>> distance(city_a, city_b)
1.0
>>> city_c = make_city('city_c', 6.5, 12)
>>> city_d = make_city('city_d', 2.5, 15)
>>> distance(city_c, city_d)
5.0
>>> berkeley = make_city('Berkeley', 37.87, 112.26)
>>> stanford = make_city('Stanford', 34.05, 118.25)
>>> closer_city(38.33, 121.44, berkeley, stanford)
'Stanford'
>>> bucharest = make_city('Bucharest', 44.43, 26.10)
>>> vienna = make_city('Vienna', 48.20, 16.37)
>>> closer_city(41.29, 174.78, bucharest, vienna)
'Bucharest'
>>> change_abstraction(False)
"""
# Treat all the following code as being behind an abstraction layer,
# you shouldn't need to look at it.
def make_city(name, lat, lon):
"""
>>> city = make_city('Berkeley', 0, 1)
>>> get_name(city)
'Berkeley'
>>> get_lat(city)
0
>>> get_lon(city)
1
"""
if change_abstraction.changed:
return {"name": name, "lat": lat, "lon": lon}
else:
return [name, lat, lon]
def get_name(city):
"""
>>> city = make_city('Berkeley', 0, 1)
>>> get_name(city)
'Berkeley'
"""
if change_abstraction.changed:
return city["name"]
else:
return city[0]
def get_lat(city):
"""
>>> city = make_city('Berkeley', 0, 1)
>>> get_lat(city)
0
"""
if change_abstraction.changed:
return city["lat"]
else:
return city[1]
def get_lon(city):
"""
>>> city = make_city('Berkeley', 0, 1)
>>> get_lon(city)
1
"""
if change_abstraction.changed:
return city["lon"]
else:
return city[2]
def berry_finder(t):
"""Returns True if t contains a node with the value 'berry' and
False otherwise.
>>> scrat = tree('berry')
>>> berry_finder(scrat)
True
>>> sproul = tree('roots', [tree('branch1', [tree('leaf'), tree('berry')]), tree('branch2')])
>>> berry_finder(sproul)
True
>>> numbers = tree(1, [tree(2), tree(3, [tree(4), tree(5)]), tree(6, [tree(7)])])
>>> berry_finder(numbers)
False
>>> t = tree(1, [tree('berry',[tree('not berry')])])
>>> berry_finder(t)
True
"""
"*** YOUR CODE HERE ***"
if label(t) == 'berry':
return True
for b in branches(t):
if berry_finder(b):
return True
return False
def sprout_leaves(t, leaves):
"""Sprout new leaves containing the data in leaves at each leaf in
the original tree t and return the resulting tree.
>>> t1 = tree(1, [tree(2), tree(3)])
>>> print_tree(t1)
1
2
3
>>> new1 = sprout_leaves(t1, [4, 5])
>>> print_tree(new1)
1
2
4
5
3
4
5
>>> t2 = tree(1, [tree(2, [tree(3)])])
>>> print_tree(t2)
1
2
3
>>> new2 = sprout_leaves(t2, [6, 1, 2])
>>> print_tree(new2)
1
2
3
6
1
2
"""
"*** YOUR CODE HERE ***"
# if is_leaf(t):
# #直接加一个叶子
# pass
# for b in branches(t):
# if is_leaf(b):
# #加一个叶子
# pass
# else:
# sprout_leaves(b,leaves)
# return t
if is_leaf(t):
return tree(label(t), [tree(leaf) for leaf in leaves])
return tree(label(t), [sprout_leaves(s, leaves) for s in branches(t)])
# Abstraction tests for sprout_leaves and berry_finder
def check_abstraction():
"""
There's nothing for you to do for this function, it's just here for the extra doctest
>>> change_abstraction(True)
>>> scrat = tree('berry')
>>> berry_finder(scrat)
True
>>> sproul = tree('roots', [tree('branch1', [tree('leaf'), tree('berry')]), tree('branch2')])
>>> berry_finder(sproul)
True
>>> numbers = tree(1, [tree(2), tree(3, [tree(4), tree(5)]), tree(6, [tree(7)])])
>>> berry_finder(numbers)
False
>>> t = tree(1, [tree('berry',[tree('not berry')])])
>>> berry_finder(t)
True
>>> t1 = tree(1, [tree(2), tree(3)])
>>> print_tree(t1)
1
2
3
>>> new1 = sprout_leaves(t1, [4, 5])
>>> print_tree(new1)
1
2
4
5
3
4
5
>>> t2 = tree(1, [tree(2, [tree(3)])])
>>> print_tree(t2)
1
2
3
>>> new2 = sprout_leaves(t2, [6, 1, 2])
>>> print_tree(new2)
1
2
3
6
1
2
>>> change_abstraction(False)
"""
def change_abstraction(change):
"""
For testing purposes.
>>> change_abstraction(True)
>>> change_abstraction.changed
True
"""
change_abstraction.changed = change
change_abstraction.changed = False
# Tree ADT
def tree(label, branches=[]):
"""Construct a tree with the given label value and a list of branches."""
if change_abstraction.changed:
for branch in branches:
assert is_tree(branch), 'branches must be trees'
return {'label': label, 'branches': list(branches)}
else:
for branch in branches:
assert is_tree(branch), 'branches must be trees'
return [label] + list(branches)
def label(tree):
"""Return the label value of a tree."""
if change_abstraction.changed:
return tree['label']
else:
return tree[0]
def branches(tree):
"""Return the list of branches of the given tree."""
if change_abstraction.changed:
return tree['branches']
else:
return tree[1:]
def is_tree(tree):
"""Returns True if the given tree is a tree, and False otherwise."""
if change_abstraction.changed:
if type(tree) != dict or len(tree) != 2:
return False
for branch in branches(tree):
if not is_tree(branch):
return False
return True
else:
if type(tree) != list or len(tree) < 1:
return False
for branch in branches(tree):
if not is_tree(branch):
return False
return True
def is_leaf(tree):
"""Returns True if the given tree's list of branches is empty, and False
otherwise.
"""
return not branches(tree)
def print_tree(t, indent=0):
"""Print a representation of this tree in which each node is
indented by two spaces times its depth from the root.
>>> print_tree(tree(1))
1
>>> print_tree(tree(1, [tree(2)]))
1
2
>>> numbers = tree(1, [tree(2), tree(3, [tree(4), tree(5)]), tree(6, [tree(7)])])
>>> print_tree(numbers)
1
2
3
4
5
6
7
"""
print(' ' * indent + str(label(t)))
for b in branches(t):
print_tree(b, indent + 1)
def copy_tree(t):
"""Returns a copy of t. Only for testing purposes.
>>> t = tree(5)
>>> copy = copy_tree(t)
>>> t = tree(6)
>>> print_tree(copy)
5
"""
return tree(label(t), [copy_tree(b) for b in branches(t)])
