现在开始刷cs61a 求队友讨论作业+记录进度

lilirr

对了我没有参加hog test optional的问题我都没有做
之前我朋友给我发了fa18的optional的题，两道题卡了我一天，我最后连答案都没看懂
溜了溜了

lilirr

贴一下我的weix吧有小伙伴有问题哒可以私戳我

您好！
本帖隐藏的内容需要积分高于 350 才可浏览
您当前积分为 0。
使用VIP即刻解锁阅读权限或查看其他获取积分的方式

游客，您好！
本帖隐藏的内容需要积分高于 350 才可浏览
您当前积分为 0。
VIP即刻解锁阅读权限或查看其他获取积分的方式
Unlock interview details and practice with AI
Curated Interview Questions from Top Companies

GuoJSeven

楼主你好，我刚刚准备开始学，请问这个课程网站是不是要科学上网，如果是的话，推荐一下梯子吧

lilirr

GuoJSeven 发表于 2019-7-11 10:30
楼主你好，我刚刚准备开始学，请问这个课程网站是不是要科学上网，如果是的话，推荐一下梯子吧

我用的boomcloud 百度搜索他的网站很稳定我选的套餐是45元/30G/年的

lilirr

我今天把lab02写了前段时间进度有些慢最近要加快速度了
现在把lab的答案贴上来

lilirr

lab02.py

"""Lab 2: Lambda Expressions and Higher Order Functions"""

# Lambda Functions

def lambda_curry2(func):
"""
Returns a Curried version of a two-argument function FUNC.
>>> from operator import add
>>> curried_add = lambda_curry2(add)
>>> add_three = curried_add(3)
>>> add_three(5)
8
"""
def curry2(x):
      def curry3(y):
         return func(x,y)
      return curry3
return curry2

lilirr

lab02_extra.py

""" Optional problems for lab02 """

from lab02 import *

# Higher Order Functions

def compose1(f, g):
"""Return the composition function which given x, computes f(g(x)).

>>> add_one = lambda x: x + 1       # adds one to x
>>> square = lambda x: x**2
>>> a1 = compose1(square, add_one) # (x + 1)^2
>>> a1(4)
25
>>> mul_three = lambda x: x * 3    # multiplies 3 to x
>>> a2 = compose1(mul_three, a1) # ((x + 1)^2) * 3
>>> a2(4)
75
>>> a2(5)
108
"""
return lambda x: f(g(x))

def composite_identity(f, g):
"""
Return a function with one parameter x that returns True if f(g(x)) is
equal to g(f(x)). You can assume the result of g(x) is a valid input for f
and vice versa.

>>> add_one = lambda x: x + 1       # adds one to x
>>> square = lambda x: x**2
>>> b1 = composite_identity(square, add_one)
>>> b1(0)                         # (0 + 1)^2 == 0^2 + 1
True
>>> b1(4)                         # (4 + 1)^2 != 4^2 + 1
False
"""
def is_true(x):
      return compose1(f,g)(x) == compose1(g,f)(x)
return is_true

def count_cond(condition):
"""Returns a function with one parameter N that counts all the numbers from
1 to N that satisfy the two-argument predicate function CONDITION.

>>> count_factors = count_cond(lambda n, i: n % i == 0)
>>> count_factors(2) # 1, 2
2
>>> count_factors(4) # 1, 2, 4
3
>>> count_factors(12)  # 1, 2, 3, 4, 6, 12
6

>>> is_prime = lambda n, i: count_factors(i) == 2
>>> count_primes = count_cond(is_prime)
>>> count_primes(2) # 2
1
>>> count_primes(3) # 2, 3
2
>>> count_primes(4) # 2, 3
2
>>> count_primes(5) # 2, 3, 5
3
>>> count_primes(20) # 2, 3, 5, 7, 11, 13, 17, 19
8
"""
def count_num(n):
      i, sum = 1, 0
      while  i<= n:
         if condition(n,i):
            sum +=1
         i +=1
      return sum
return count_num

def cycle(f1, f2, f3):
"""Returns a function that is itself a higher-order function.

>>> def add1(x):
...    return x + 1
>>> def times2(x):
...    return x * 2
>>> def add3(x):
...    return x + 3
>>> my_cycle = cycle(add1, times2, add3)
>>> identity = my_cycle(0)
>>> identity(5)
5
>>> add_one_then_double = my_cycle(2)
>>> add_one_then_double(1)
4
>>> do_all_functions = my_cycle(3)
>>> do_all_functions(2)
9
>>> do_more_than_a_cycle = my_cycle(4)
>>> do_more_than_a_cycle(2)
10
>>> do_two_cycles = my_cycle(6)
>>> do_two_cycles(1)
19
"""
"*** YOUR CODE HERE ***"
def take_arg_n(n):
      def take_arg_x(x):
         sum_res,i = x,0
         ori_ls = [f1,f2,f3]
         func_ls = ori_ls*(n//3) + ori_ls[0:n%3]
         if n == 0:
            return x
         else:
            while i<n:
                  sum_res = func_ls[i](sum_res)
                  i +=1
            return sum_res
      return take_arg_x
return take_arg_n

lilirr

我现在把hw03的代码po上来，ps我为了赶进度没有做extra hw 之后也一样

lilirr

HW_SOURCE_FILE = 'hw03.py'

#############
# Questions #
#############

def has_seven(k):
"""Returns True if at least one of the digits of k is a 7, False otherwise.

>>> has_seven(3)
False
>>> has_seven(7)
True
>>> has_seven(2734)
True
>>> has_seven(2634)
False
>>> has_seven(734)
True
>>> has_seven(7777)
True
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'has_seven',
...    ['Assign', 'AugAssign'])
True
"""
"*** YOUR CODE HERE ***"
if k<10:
      return k%10 ==7
else:HW_SOURCE_FILE = 'hw03.py'

#############
# Questions #
#############

def has_seven(k):
"""Returns True if at least one of the digits of k is a 7, False otherwise.

>>> has_seven(3)
False
>>> has_seven(7)
True
>>> has_seven(2734)
True
>>> has_seven(2634)
False
>>> has_seven(734)
True
>>> has_seven(7777)
True
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'has_seven',
...    ['Assign', 'AugAssign'])
True
"""
"*** YOUR CODE HERE ***"
if k<10:
      return k%10 ==7
else:
      return k%10 == 7 or has_seven(k//10)

def summation(n, term):

"""Return the sum of the first n terms in the sequence defined by term.
Implement using recursion!

>>> summation(5, lambda x: x * x * x) # 1^3 + 2^3 + 3^3 + 4^3 + 5^3
225
>>> summation(9, lambda x: x + 1) # 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
54
>>> summation(5, lambda x: 2**x) # 2^1 + 2^2 + 2^3 + 2^4 + 2^5
62
>>> # Do not use while/for loops!
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'summation',
...    ['While', 'For'])
True
"""
assert n >= 1
"*** YOUR CODE HERE ***"
if n ==1:
      return term (1) #base case
else:
      return term(n)+summation(n-1,term)

def square(x):
return x * x

def identity(x):
return x

triple = lambda x: 3 * x

increment = lambda x: x + 1

add = lambda x, y: x + y

mul = lambda x, y: x * y

def accumulate(combiner, base, n, term):
"""Return the result of combining the first n terms in a sequence and base.
The terms to be combined are term(1), term(2), ..., term(n).  combiner is a
two-argument commutative function.

>>> accumulate(add, 0, 5, identity)  # 0 + 1 + 2 + 3 + 4 + 5
15
>>> accumulate(add, 11, 5, identity) # 11 + 1 + 2 + 3 + 4 + 5
26
>>> accumulate(add, 11, 0, identity) # 11
11
>>> accumulate(add, 11, 3, square) # 11 + 1^2 + 2^2 + 3^2
25
>>> accumulate(mul, 2, 3, square) # 2 * 1^2 * 2^2 * 3^2
72
"""
"*** YOUR CODE HERE ***"
if n == 0:
      return base
else:
      return combiner(term(n),accumulate(combiner,base,n-1,term)) #一定要一个代表n情况的表达式和之后的循环
      #除了tree recursion一般很少两个base case的，所以当写了两个bc看能不能合并否则可能报错

def summation_using_accumulate(n, term):
"""Returns the sum of term(1) + ... + term(n). The implementation
uses accumulate.

>>> summation_using_accumulate(5, square)
55
>>> summation_using_accumulate(5, triple)
45
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'summation_using_accumulate',
...    ['Recursion', 'For', 'While'])
True
"""
"*** YOUR CODE HERE ***"

return accumulate(add,0,n,term)

def product_using_accumulate(n, term):
"""An implementation of product using accumulate.

>>> product_using_accumulate(4, square)
576
>>> product_using_accumulate(6, triple)
524880
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'product_using_accumulate',
...    ['Recursion', 'For', 'While'])
True
"""
"*** YOUR CODE HERE ***"
return accumulate(mul,1,n,term)

def filtered_accumulate(combiner, base, pred, n, term):
"""Return the result of combining the terms in a sequence of N terms
that satisfy the predicate pred. combiner is a two-argument function.
If v1, v2, ..., vk are the values in term(1), term(2), ..., term(N)
that satisfy pred, then the result is
      base combiner v1 combiner v2 ... combiner vk
(treating combiner as if it were a binary operator, like +). The
implementation uses accumulate.

>>> filtered_accumulate(add, 0, lambda x: True, 5, identity)  # 0 + 1 + 2 + 3 + 4 + 5
15
>>> filtered_accumulate(add, 11, lambda x: False, 5, identity) # 11
11
>>> filtered_accumulate(add, 0, odd, 5, identity) # 0 + 1 + 3 + 5
9
>>> filtered_accumulate(mul, 1, greater_than_5, 5, square)  # 1 * 9 * 16 * 25
3600
>>> # Do not use while/for loops or recursion
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'filtered_accumulate',
...    ['While', 'For', 'Recursion'])
True
"""
def combine_if(x, y):
      "*** YOUR CODE HERE ***"
      if pred(x):
         return combiner(x,y)
      else:
         return y
return accumulate(combine_if, base, n, term)

def odd(x):
return x % 2 == 1

def greater_than_5(x):
return x > 5

def make_repeater(f, n):
"""Return the function that computes the nth application of f.

>>> add_three = make_repeater(increment, 3)
>>> add_three(5)
8
>>> make_repeater(triple, 5)(1) # 3 * 3 * 3 * 3 * 3 * 1
243
>>> make_repeater(square, 2)(5) # square(square(5))
625
>>> make_repeater(square, 4)(5) # square(square(square(square(5))))
152587890625
>>> make_repeater(square, 0)(5)
5
"""
"*** YOUR CODE HERE ***"
'''第一个解法'''
'''if n == 0:
      return identity
elif n ==1:
      return f
else:
      a = f
      i = 1
      while i < n:
         a = compose1(a,f)
         i = i+1
      return a'''
'''第二个解法'''
return accumulate(compose1,lambda x: x,n,lambda x:f)#注意一定是lambdax:f，如果只是f在term(n)即f(n)的时候返回的一个int 返回的f而非f(x)说明x对f没有作用，即n对increment等函数没有影响，返回的只是一个函数而已

def compose1(f, g):
"""Return a function h, such that h(x) = f(g(x))."""
def h(x):
      return f(g(x))
return h

###################
# Extra Questions #
###################

quine = """
"*** YOUR CODE HERE ***"
"""

def zero(f):
return lambda x: x

def successor(n):
return lambda f: lambda x: f(n(f)(x))

def one(f):
"""Church numeral 1: same as successor(zero)"""
"*** YOUR CODE HERE ***"

def two(f):
"""Church numeral 2: same as successor(successor(zero))"""
"*** YOUR CODE HERE ***"

three = successor(two)

def church_to_int(n):
"""Convert the Church numeral n to a Python integer.

>>> church_to_int(zero)
0
>>> church_to_int(one)
1
>>> church_to_int(two)
2
>>> church_to_int(three)
3
"""
"*** YOUR CODE HERE ***"

def add_church(m, n):
"""Return the Church numeral for m + n, for Church numerals m and n.

>>> church_to_int(add_church(two, three))
5
"""
"*** YOUR CODE HERE ***"

def mul_church(m, n):
"""Return the Church numeral for m * n, for Church numerals m and n.

>>> four = successor(three)
>>> church_to_int(mul_church(two, three))
6
>>> church_to_int(mul_church(three, four))
12
"""
"*** YOUR CODE HERE ***"

def pow_church(m, n):
"""Return the Church numeral m ** n, for Church numerals m and n.

>>> church_to_int(pow_church(two, three))
8
>>> church_to_int(pow_church(three, two))
9
"""
"*** YOUR CODE HERE ***"

return k%10 == 7 or has_seven(k//10)

def summation(n, term):

"""Return the sum of the first n terms in the sequence defined by term.
Implement using recursion!

>>> summation(5, lambda x: x * x * x) # 1^3 + 2^3 + 3^3 + 4^3 + 5^3
225
>>> summation(9, lambda x: x + 1) # 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
54
>>> summation(5, lambda x: 2**x) # 2^1 + 2^2 + 2^3 + 2^4 + 2^5
62
>>> # Do not use while/for loops!
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'summation',
...    ['While', 'For'])
True
"""
assert n >= 1
"*** YOUR CODE HERE ***"
if n ==1:
      return term (1) #base case
else:
      return term(n)+summation(n-1,term)

def square(x):
return x * x

def identity(x):
return x

triple = lambda x: 3 * x

increment = lambda x: x + 1

add = lambda x, y: x + y

mul = lambda x, y: x * y

def accumulate(combiner, base, n, term):
"""Return the result of combining the first n terms in a sequence and base.
The terms to be combined are term(1), term(2), ..., term(n).  combiner is a
two-argument commutative function.

>>> accumulate(add, 0, 5, identity)  # 0 + 1 + 2 + 3 + 4 + 5
15
>>> accumulate(add, 11, 5, identity) # 11 + 1 + 2 + 3 + 4 + 5
26
>>> accumulate(add, 11, 0, identity) # 11
11
>>> accumulate(add, 11, 3, square) # 11 + 1^2 + 2^2 + 3^2
25
>>> accumulate(mul, 2, 3, square) # 2 * 1^2 * 2^2 * 3^2
72
"""
"*** YOUR CODE HERE ***"
if n == 0:
      return base
else:
      return combiner(term(n),accumulate(combiner,base,n-1,term)) #一定要一个代表n情况的表达式和之后的循环
      #除了tree recursion一般很少两个base case的，所以当写了两个bc看能不能合并否则可能报错

def summation_using_accumulate(n, term):
"""Returns the sum of term(1) + ... + term(n). The implementation
uses accumulate.

>>> summation_using_accumulate(5, square)
55
>>> summation_using_accumulate(5, triple)
45
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'summation_using_accumulate',
...    ['Recursion', 'For', 'While'])
True
"""
"*** YOUR CODE HERE ***"

return accumulate(add,0,n,term)

def product_using_accumulate(n, term):
"""An implementation of product using accumulate.

>>> product_using_accumulate(4, square)
576
>>> product_using_accumulate(6, triple)
524880
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'product_using_accumulate',
...    ['Recursion', 'For', 'While'])
True
"""
"*** YOUR CODE HERE ***"
return accumulate(mul,1,n,term)

def filtered_accumulate(combiner, base, pred, n, term):
"""Return the result of combining the terms in a sequence of N terms
that satisfy the predicate pred. combiner is a two-argument function.
If v1, v2, ..., vk are the values in term(1), term(2), ..., term(N)
that satisfy pred, then the result is
      base combiner v1 combiner v2 ... combiner vk
(treating combiner as if it were a binary operator, like +). The
implementation uses accumulate.

>>> filtered_accumulate(add, 0, lambda x: True, 5, identity)  # 0 + 1 + 2 + 3 + 4 + 5
15
>>> filtered_accumulate(add, 11, lambda x: False, 5, identity) # 11
11
>>> filtered_accumulate(add, 0, odd, 5, identity) # 0 + 1 + 3 + 5
9
>>> filtered_accumulate(mul, 1, greater_than_5, 5, square)  # 1 * 9 * 16 * 25
3600
>>> # Do not use while/for loops or recursion
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'filtered_accumulate',
...    ['While', 'For', 'Recursion'])
True
"""
def combine_if(x, y):
      "*** YOUR CODE HERE ***"
      if pred(x):
         return combiner(x,y)
      else:
         return y
return accumulate(combine_if, base, n, term)

def odd(x):
return x % 2 == 1

def greater_than_5(x):
return x > 5

def make_repeater(f, n):
"""Return the function that computes the nth application of f.

>>> add_three = make_repeater(increment, 3)
>>> add_three(5)
8
>>> make_repeater(triple, 5)(1) # 3 * 3 * 3 * 3 * 3 * 1
243
>>> make_repeater(square, 2)(5) # square(square(5))
625
>>> make_repeater(square, 4)(5) # square(square(square(square(5))))
152587890625
>>> make_repeater(square, 0)(5)
5
"""
"*** YOUR CODE HERE ***"
'''第一个解法'''
'''if n == 0:
      return identity
elif n ==1:
      return f
else:
      a = f
      i = 1
      while i < n:
         a = compose1(a,f)
         i = i+1
      return a'''
'''第二个解法'''
return accumulate(compose1,lambda x: x,n,lambda x:f)#注意一定是lambdax:f，如果只是f在term(n)即f(n)的时候返回的一个int 返回的f而非f(x)说明x对f没有作用，即n对increment等函数没有影响，返回的只是一个函数而已

def compose1(f, g):
"""Return a function h, such that h(x) = f(g(x))."""
def h(x):
      return f(g(x))
return h

###################
# Extra Questions #
###################

quine = """
"*** YOUR CODE HERE ***"
"""

def zero(f):
return lambda x: x

def successor(n):
return lambda f: lambda x: f(n(f)(x))

def one(f):
"""Church numeral 1: same as successor(zero)"""
"*** YOUR CODE HERE ***"

def two(f):
"""Church numeral 2: same as successor(successor(zero))"""
"*** YOUR CODE HERE ***"

three = successor(two)

def church_to_int(n):
"""Convert the Church numeral n to a Python integer.

>>> church_to_int(zero)
0
>>> church_to_int(one)
1
>>> church_to_int(two)
2
>>> church_to_int(three)
3
"""
"*** YOUR CODE HERE ***"

def add_church(m, n):
"""Return the Church numeral for m + n, for Church numerals m and n.

>>> church_to_int(add_church(two, three))
5
"""
"*** YOUR CODE HERE ***"

def mul_church(m, n):
"""Return the Church numeral for m * n, for Church numerals m and n.

>>> four = successor(three)
>>> church_to_int(mul_church(two, three))
6
>>> church_to_int(mul_church(three, four))
12
"""
"*** YOUR CODE HERE ***"

def pow_church(m, n):
"""Return the Church numeral m ** n, for Church numerals m and n.

>>> church_to_int(pow_church(two, three))
8
>>> church_to_int(pow_church(three, two))
9
"""
"*** YOUR CODE HERE ***"

GuoJSeven

lilirr 发表于 2019-7-13 15:11
我用的boomcloud 百度搜索他的网站很稳定我选的套餐是45元/30G/年的

okok谢谢啦

现在开始刷cs61a 求队友讨论作业+记录进度

浏览过的版块