# [2021-Fall] Hw05 of CS61A of UCB

### Q1: Generate Permutations

Given a sequence of unique elements, a permutation of the sequence is a list containing the elements of the sequence in some arbitrary order. For example, [2, 1, 3], [1, 3, 2], and [3, 2, 1] are some of the permutations of the sequence [1, 2, 3].

Implement gen_perms, a generator function that takes in a sequence seq and returns a generator that yields all permutations of seq. For this question, assume that seq will not be empty.

Permutations may be yielded in any order. Note that the doctests test whether you are yielding all possible permutations, but not in any particular order. The built-in sorted function takes in an iterable object and returns a list containing the elements of the iterable in non-decreasing order.

Hint: If you had the permutations of all the elements in seq not including the first element, how could you use that to generate the permutations of the full seq?

Hint: Remember, it's possible to loop over generator objects because generators are iterators!

def gen_perms(seq):
if len(seq) <= 1:
yield seq
...


def gen_perms(seq):
"""Generates all permutations of the given sequence. Each permutation is a
list of the elements in SEQ in a different order. The permutations may be
yielded in any order.

>>> perms = gen_perms()
>>> type(perms)
<class 'generator'>
>>> next(perms)

>>> try: #this piece of code prints "No more permutations!" if calling next would cause an error
...     next(perms)
... except StopIteration:
...     print('No more permutations!')
No more permutations!
>>> sorted(gen_perms([1, 2, 3])) # Returns a sorted list containing elements of the generator
[[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]
>>> sorted(gen_perms((10, 20, 30)))
[[10, 20, 30], [10, 30, 20], [20, 10, 30], [20, 30, 10], [30, 10, 20], [30, 20, 10]]
>>> sorted(gen_perms("ab"))
[['a', 'b'], ['b', 'a']]
"""
# This problem requires list type, see example above
if type(seq) != list:
seq = list(seq)
# base case
if len(seq) <= 1:
yield seq
else:
# iterate every permutation in the generator
for perm in gen_perms(seq[1:]):
# enumerate every position for insertation
for i in range(len(seq)):
yield perm[:i] + seq[:1] + perm[i:]


### Q2: Yield Paths

Define a generator function path_yielder which takes in a tree t, a value value, and returns a generator object which yields each path from the root of t to a node that has label value.

Each path should be represented as a list of the labels along that path in the tree. You may yield the paths in any order.

We have provided a skeleton for you. You do not need to use this skeleton, but if your implementation diverges significantly from it, you might want to think about how you can get it to fit the skeleton.

def path_yielder(t, value):
for _______________ in _________________:
for _______________ in _________________:


def path_yielder(t, value):
"""Yields all possible paths from the root of t to a node with the label
value as a list.

>>> t1 = tree(1, [tree(2, [tree(3), tree(4, [tree(6)]), tree(5)]), tree(5)])
>>> print_tree(t1)
1
2
3
4
6
5
5
>>> next(path_yielder(t1, 6))
[1, 2, 4, 6]
>>> path_to_5 = path_yielder(t1, 5)
>>> sorted(list(path_to_5))
[[1, 2, 5], [1, 5]]

>>> t2 = tree(0, [tree(2, [t1])])
>>> print_tree(t2)
0
2
1
2
3
4
6
5
5
>>> path_to_2 = path_yielder(t2, 2)
>>> sorted(list(path_to_2))
[[0, 2], [0, 2, 1, 2]]
"""
if label(t) == value:
yield [label(t)]
for b in branches(t):
for path in path_yielder(b, value):
yield [label(t)] + path


### Q3: 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


### Q4: Generate Preorder

Similarly to preorder in Question 3, define the function generate_preorder, which takes in a tree as an argument and now instead yields the entries in the tree in the order that print_tree would print them.

Hint: How can you modify your implementation of preorder to yield from your recursive calls instead of returning them?

def generate_preorder(t):
"""Yield 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)])])
>>> gen = generate_preorder(numbers)
>>> next(gen)
1
>>> list(gen)
[2, 3, 4, 5, 6, 7]
"""
yield label(t)
for b in branches(t):
yield from generate_preorder(b)


### Q5: Remainder Generator

Like functions, generators can also be higher-order. For this problem, we will be writing remainders_generator, which yields a series of generator objects.

remainders_generator takes in an integer m, and yields m different generators. The first generator is a generator of multiples of m, i.e. numbers where the remainder is 0. The second is a generator of natural numbers with remainder 1 when divided by m. The last generator yields natural numbers with remainder m - 1 when divided by m.

Hint: To create a generator of infinite natural numbers, you can call the naturals function that's provided in the starter code.

Hint: Consider defining an inner generator function. Each yielded generator varies only in that the elements of each generator have a particular remainder when divided by m. What does that tell you about the argument(s) that the inner function should take in?

def helper(i, m):
num = 1         # loop variable
while True:
if num % m == i:
yield num
num += 1
# you can give it a test
>>> it = helper(2, 3)
>>> next(it)
2				# 2 % 3 == 2
>>> next(it)
5				# 5 % 3 == 2
>>> next(it)
8				# 8 % 3 == 2


def remainders_generator(m):
"""
Yields m generators. The ith yielded generator yields natural numbers whose
remainder is i when divided by m.

>>> import types
>>> [isinstance(gen, types.GeneratorType) for gen in remainders_generator(5)]
[True, True, True, True, True]
>>> remainders_four = remainders_generator(4)
>>> for i in range(4):
...     print("First 3 natural numbers with remainder {0} when divided by 4:".format(i))
...     gen = next(remainders_four)
...     for _ in range(3):
...         print(next(gen))
First 3 natural numbers with remainder 0 when divided by 4:
4
8
12
First 3 natural numbers with remainder 1 when divided by 4:
1
5
9
First 3 natural numbers with remainder 2 when divided by 4:
2
6
10
First 3 natural numbers with remainder 3 when divided by 4:
3
7
11
"""
def helper(i, m):
num = 1         # loop variable
while True:
if num % m == i:
yield num
num += 1
for i in range(m):
yield helper(i, m)

posted @ 2022-02-24 00:00  MartinLwx  阅读(3026)  评论(0编辑  收藏  举报