辅导案例-CS61A

  • August 12, 2020

Exam: CS61A Summer 2020 Midterm Name: Solution Key Email: example_key secure Point breakdown q1: 1.0/1 Score: Total: 1.0 Reskeletonized solution follows ================================================== def cat(password, limit): ” Write a higher-order function `cat` that returns a one-argument\n funct¶ ion `attempt`. Every time `attempt` is called, it checks to see if its argument\¶ n matches the password at the corresponding index.\n\n If the password ent¶ irely matches, return a success string. If more than `limit`\n number of inco¶ rrect hacks are attempted, you should return an error string.\n For details, ¶ see the doctest.\n\n\n Note: to comment out a blank that covers an entire lin¶ e, just put down ‘unnecessary’ (with quotes)\n\n >>> hacker = cat([1,2], 2)\n¶ >>> hacker(1)\n >>> hacker(2)\n ‘Successfully unlocked!’\n >>> hack¶ er = cat([1,2], 1)\n >>> hacker(1)\n >>> hacker(3) # used up attempts to g¶ ain access\n >>> hacker(2) # correct attempt to gain access, but already lock¶ ed\n ‘The safe is now inaccessible!’\n >>> hacker = cat([1,2], 2)\n >>>¶ hacker(1)\n >>> hacker(3) # 1 attempt left to gain access\n >>> hacker(2)¶ # correct attempt to gain access\n ‘Successfully unlocked!’\n ” num_incorrect = 0 index = 0 def attempt(digit): nonlocal num_incorrect nonlocal index if (num_incorrect >= limit): return ‘The safe is now inaccessible!’ if (password[index] == digit): index += 1 if (index == len(password)): return ‘Successfully unlocked!’ else: num_incorrect += 1 return attempt ================================================== Original code follows ================================================== def cat(password, limit): “”” Write a higher-order function `cat` that returns a one-argument function `attempt`. Every time `attempt` is called, it checks to see if its ¶ argument matches the password at the corresponding index. If the password entirely matches, return a success string. If more than `lim¶ it` number of incorrect hacks are attempted, you should return an error string. For details, see the doctest. Note: to comment out a blank that covers an entire line, just put down ‘unne¶ cessary’ (with quotes) >>> hacker = cat([1,2], 2) >>> hacker(1) >>> hacker(2) ‘Successfully unlocked!’ >>> hacker = cat([1,2], 1) >>> hacker(1) >>> hacker(3) # used up attempts to gain access >>> hacker(2) # correct attempt to gain access, but already locked ‘The safe is now inaccessible!’ >>> hacker = cat([1,2], 2) >>> hacker(1) >>> hacker(3) # 1 attempt left to gain access >>> hacker(2) # correct attempt to gain access ‘Successfully unlocked!’ “”” num_incorrect = 0 index = 0 def attempt(digit): nonlocal num_incorrect nonlocal index if num_incorrect >= limit: return ‘The safe is now inaccessible!’ if password[index] == digit: index += 1 if index == len(password): return “Successfully unlocked!” else: num_incorrect += 1 return attempt ================================================== schedule Point breakdown q2: 1.0/1 Score: Total: 1.0 Reskeletonized solution follows ================================================== def schedule(galaxy, sum_to, max_digit): ‘\n A \’galaxy\’ is a string which contains either digits or \’?\’s.\n\n ¶ A \’completion\’ of a galaxy is a string that is the same as galaxy, except\n¶ with digits replacing each of the \’?\’s.\n\n Your task in this question ¶ is to find all completions of the given `galaxy`\n that use digits up to `max¶ _digit`, and whose digits sum to `sum_to`.\n\n Note 1: the function int can b¶ e used to convert a string to an integer and str\n can be used to convert¶ an integer to a string as such:\n\n >>> int(“5”)\n 5\n >>>¶ str(5)\n \’5\’\n\n Note 2: Indexing and slicing can be used on string¶ s as well as on lists.\n\n >>> \’evocative\'[3]\n \’c\’\n >¶ >> \’evocative\'[3:]\n \’cative\’\n >>> \’evocative\'[:6]\n ¶ \’evocat\’\n >>> \’evocative\'[3:6]\n \’cat\’\n\n\n >>> schedu¶ le(\’?????\’, 25, 5)\n [\’55555\’]\n >>> schedule(\’???\’, 5, 2)\n [\’1¶ 22\’, \’212\’, \’221\’]\n >>> schedule(\’?2??11?\’, 5, 3)\n [\’0200111\’, ¶ \’0201110\’, \’0210110\’, \’1200110\’]\n ‘ def schedule_helper(galaxy, sum_sofar, index): if ((index >= len(galaxy)) and (sum_sofar == sum_to)): return [galaxy] elif ((sum_sofar > sum_to) or (index >= len(galaxy))): return [] elif (galaxy[index] != ‘?’): return schedule_helper(galaxy, (sum_sofar + int(galaxy[index])), (in¶ dex + 1)) ans = [] for x in range((max_digit + 1)): modified_galaxy = ((galaxy[:index] + str(x)) + galaxy[(index + 1):])¶ ans += schedule_helper(modified_galaxy, (sum_sofar + x), (index + 1)¶ ) return ans return schedule_helper(galaxy, 0, 0) ================================================== Original code follows ================================================== def schedule(galaxy, sum_to, max_digit): “”” A ‘galaxy’ is a string which contains either digits or ‘?’s. A ‘completion’ of a galaxy is a string that is the same as galaxy, except with digits replacing each of the ‘?’s. Your task in this question is to find all completions of the given `galaxy` that use digits up to `max_digit`, and whose digits sum to `sum_to`. Note 1: the function int can be used to convert a string to an integer and s¶ tr can be used to convert an integer to a string as such: >>> int(“5”) 5 >>> str(5) ‘5’ Note 2: Indexing and slicing can be used on strings as well as on lists. >>> ‘evocative'[3] ‘c’ >>> ‘evocative'[3:] ‘cative’ >>> ‘evocative'[:6] ‘evocat’ >>> ‘evocative'[3:6] ‘cat’ >>> schedule(‘?????’, 25, 5) [‘55555’] >>> schedule(‘???’, 5, 2) [‘122’, ‘212’, ‘221’] >>> schedule(‘?2??11?’, 5, 3) [‘0200111’, ‘0201110’, ‘0210110’, ‘1200110’] “”” def schedule_helper(galaxy, sum_sofar, index): if index >= len(galaxy) and sum_sofar == sum_to: return [galaxy] elif sum_sofar > sum_to or index >= len(galaxy): return [] elif galaxy[index] != ‘?’: return schedule_helper(galaxy, sum_sofar + int(galaxy[index]), index¶ + 1) ans = [] for x in range(max_digit + 1): modified_galaxy = galaxy[:index] + str(x) + galaxy[index + 1:] ans += schedule_helper(modified_galaxy, sum_sofar + x, index + 1) return ans return schedule_helper(galaxy, 0, 0) ================================================== consume Point breakdown q3: 1.0/1 Score: Total: 1.0 Reskeletonized solution follows ================================================== ‘\nLet a `painting` be a self-referential function that\n – takes in one inte¶ ger\n – returns two values, another painting and well as an integer\n\nFor an¶ example see the function `identity_painting` below.\n\nYou have two tasks in th¶ is assignment, to implement the functions `microscope`\nand `plush`. Both have t¶ heir behavior defined by their doctests.\n\nIt is not necessary to implement `mi¶ croscope` correctly to get the points for\n`plush`. However, the ok test cases f¶ or `plush` will fail if you have not correctly\nimplemented `microscope`.\n’ def identity_painting(x): return (identity_painting, x) def microscope(a=0, s=1): ‘\n This function returns a painting function that processes a sequence\n¶ of integers, and returns the alternating sum of all integers seen thus\n ¶ far (see doctest for an example).\n\n >>> painting_a = microscope()\n >>> ¶ painting_b, x = painting_a(2)\n >>> x # 2\n¶ 2\n >>> painting_c, x = painting_b(8)\n >>> x ¶ # 2 – 8\n -6\n >>> painting_d, x = painting_c(12)\n >>> x ¶ # 2 – 8 + 12\n 6\n >>> painting_e, x = pain¶ ting_d(30)\n >>> x # 2 – 8 + 12 – 30\n -¶ 24\n >>> painting_b_again, x = painting_a(100)\n >>> x ¶ # 100 [note that we are using painting_a not painting_d here]\n ¶ 100\n ‘ def painting(x): return (microscope((a + (s * x)), (- s)), (a + (s * x))) return painting def plush(painting, items): ‘\n The function `plush` takes in a `painting` and a nonempty list of `it¶ ems` and\n runs the given `painting` on each of the `items` in turn, returnin¶ g the final\n numeric result.\n\n For example, on the items [1, 2, 3, 4, 5¶ ] with the painting microscope\n we return 1 – 2 + 3 – 4 + 5 = 3\n\n >>> p¶ lush(microscope(), [1, 2, 3, 4, 5])\n 3\n >>> plush(microscope(), [4000])\¶ n 4000\n >>> plush(microscope(), [2, 90])\n -88\n >>> plush(identity¶ _painting, [2, 90])\n 90\n ‘ (painting, x) = painting(items[0]) if (len(items) == 1): return x return plush(painting, items[1:]) ================================================== Original code follows ================================================== “”” Let a `painting` be a self-referential function that – takes in one integer – returns two values, another painting and well as an integer For an example see the function `identity_painting` below. You have two tasks in this assignment, to implement the functions `microscope` and `plush`. Both have their behavior defined by their doctests. It is not necessary to implement `microscope` correctly to get the points for `plush`. However, the ok test cases for `plush` will fail if you have not correc¶ tly implemented `microscope`. “”” def identity_painting(x): return identity_painting, x def microscope(a=0, s=1): “”” This function returns a painting function that processes a sequence of integers, and returns the alternating sum of all integers seen thus far (see doctest for an example). >>> painting_a = microscope() >>> painting_b, x = painting_a(2) >>> x # 2 2 >>> painting_c, x = painting_b(8) >>> x # 2 – 8 -6 >>> painting_d, x = painting_c(12) >>> x # 2 – 8 + 12 6 >>> painting_e, x = painting_d(30) >>> x # 2 – 8 + 12 – 30 -24 >>> painting_b_again, x = painting_a(100) >>> x # 100 [note that we are using painti¶ ng_a not painting_d here] 100 “”” def painting(x): return microscope(a + s * x, -s), a + s * x return painting def plush(painting, items): “”” The function `plush` takes in a `painting` and a nonempty list of `items` an¶ d runs the given `painting` on each of the `items` in turn, returning the fina¶ l numeric result. For example, on the items [1, 2, 3, 4, 5] with the painting microscope we return 1 – 2 + 3 – 4 + 5 = 3 >>> plush(microscope(), [1, 2, 3, 4, 5]) 3 >>> plush(microscope(), [4000]) 4000 >>> plush(microscope(), [2, 90]) -88 >>> plush(identity_painting, [2, 90]) 90 “”” painting, x = painting(items[0]) if len(items) == 1: return x return plush(painting, items[1:]) ================================================== exact_copy Point breakdown q4: 1.0/1 Score: Total: 1.0 Reskeletonized solution follows ================================================== def lemon(xv): ‘\n A lemon-copy is a perfect replica of a nested list\’s box-and-pointer¶ structure.\n If an environment diagram were drawn out, the two should be¶ entirely\n separate but identical.\n\n A `xv` is a list that only con¶ tains ints and other lists.\n\n The function `lemon` generates a lemon-copy o¶ f the given list `xv`.\n\n Note: The `isinstance` function takes in a value a¶ nd a type and determines\n whether the value is of the given type. So\n\n¶ >>> isinstance(“abc”, str)\n True\n >>> isinstance(“abc”, ¶ list)\n False\n\n Here\’s an example, where lemon_y = lemon(y)\n\n\n ¶ +—–+—–+ +—–+—–+—–+\n ¶ | | | | | | |\n ¶ | + | +————-> | 200 | 300 | + |\n y +——¶ ———-> | | | | | | | | |\n ¶ +—–+—–+ +–> +—–+—–+—–+\n lemon_y +-+ ¶ | | ^ |\n | +–¶ ————–+ | |\n | ¶ +———–+\n |\n | +¶ —–+—–+ +—–+—–+—–+\n | | ¶ | | | | | |\n +——-> | + | +-¶ ————> | 200 | 300 | + |\n | | | | ¶ | | | | |\n +—–+—–+ +¶ –> +—–+—–+—–+\n | | ¶ ^ |\n +—————-+ | ¶ |\n +———–+\¶ n\n >>> x = [200, 300]\n >>> x.append(x)\n >>> y = [x, x] ¶ # this is the `y` from the doctests\n >>> lemon_y = lemon(y) # this is t¶ he `lemon_y` from the doctests\n >>> # check that lemon_y has the same struct¶ ure as y\n >>> len(lemon_y)\n 2\n >>> lemon_y[0] is lemon_y[1]\n Tru¶ e\n >>> len(lemon_y[0])\n 3\n >>> lemon_y[0][0]\n 200\n >>> lemon¶ _y[0][1]\n 300\n >>> lemon_y[0][2] is lemon_y[0]\n True\n >>> # chec¶ k that lemon_y and y have no list objects in common\n >>> lemon_y is y\n F¶ alse\n >>> lemon_y[0] is y[0]\n False\n ‘ lemon_lookup = [] def helper(xv): if isinstance(xv, int): return xv for old_new in lemon_lookup: if (old_new[0] is xv): return old_new[1] new_xv = [] lemon_lookup.append((xv, new_xv)) for element in xv: new_xv.append(helper(element)) return new_xv return helper(xv) ================================================== Original code follows ================================================== def lemon(xv): “”” A lemon-copy is a perfect replica of a nested list’s box-and-pointer structu¶ re. If an environment diagram were drawn out, the two should be entirely separate but identical. A `xv` is a list that only contains ints and other lists. The function `lemon` generates a lemon-copy of the given list `xv`. Note: The `isinstance` function takes in a value and a type and determines whether the value is of the given type. So >>> isinstance(“abc”, str) True >>> isinstance(“abc”, list) False Here’s an example, where lemon_y = lemon(y) +—–+—–+ +—–+—–+—–+ | | | | | | | | + | +————-> | 200 | 300 | + | y +—————-> | | | | | | | | | +—–+—–+ +–> +—–+—–+—–+ lemon_y +-+ | | ^ | | +—————-+ | | | +———–+ | | +—–+—–+ +—–+—–+—–+ | | | | | | | | +——-> | + | +————-> | 200 | 300 | + | | | | | | | | | | +—–+—–+ +–> +—–+—–+—–+ | | ^ | +—————-+ | | +———–+ >>> x = [200, 300] >>> x.append(x) >>> y = [x, x] # this is the `y` from the doctests >>> lemon_y = lemon(y) # this is the `lemon_y` from the doctests >>> # check that lemon_y has the same structure as y >>> len(lemon_y) 2 >>> lemon_y[0] is lemon_y[1] True >>> len(lemon_y[0]) 3 >>> lemon_y[0][0] 200 >>> lemon_y[0][1] 300 >>> lemon_y[0][2] is lemon_y[0] True >>> # check that lemon_y and y have no list objects in common >>> lemon_y is y False >>> lemon_y[0] is y[0] False “”” lemon_lookup = [] def helper(xv): if isinstance(xv, int): return xv for old_new in lemon_lookup: if old_new[0] is xv: return old_new[1] new_xv = [] lemon_lookup.append((xv, new_xv)) for element in xv: new_xv.append(helper(element)) return new_xv return helper(xv) ================================================== nth_repeating_seq Point breakdown q5: 1.0/1 Score: Total: 1.0 Reskeletonized solution follows ================================================== def subsaltshaker(disk): “\n A ‘saltshaker’ is a sequence of digits of length `d` composed entirel¶ y of the digit `d`. Examples include\n 1\n 4444\n 7777777\n¶ \n Note that `1 <= d <= 9`; there are no 0-length saltshakers.\n\n Your ta¶ sk is to implement the `subsaltshaker` function, which takes in an integer `disk¶ ` and returns\n whether `disk` contains a saltshaker as a consecutive sub¶ integer of its digits.\n\n >>> subsaltshaker(2233) # 22 counts\n True\n ¶ >>> subsaltshaker(2444423) # 4444 counts\n True\n >>> subsaltshaker(82223¶ ) # 22 counts even if it appears as part of 222\n True\n >>> subsaltshaker¶ (234562) # 2…2 does not count if the 2s are not consecutive\n False\n >>¶ > subsaltshaker(1) # 1 counts\n True\n >>> subsaltshaker(498729879871) # 1¶ counts\n True\n >>> subsaltshaker(149872987987) # 1 counts\n True\n ¶ >>> subsaltshaker(4445555) # no saltshakers in this number\n False\n >>> ¶ subsaltshaker(20) # no saltshakers in this number\n False\n ” current_digit = (disk % 10) count = 0 while (disk != 0): last = (disk % 10) if (current_digit == last): count += 1 else: count = 1 current_digit = last if (count == current_digit): return True disk = (disk // 10) return False ================================================== Original code follows ================================================== def subsaltshaker(disk): “”” A ‘saltshaker’ is a sequence of digits of length `d` composed entirely of th¶ e digit `d`. Examples include 1 4444 7777777 Note that `1 <= d <= 9`; there are no 0-length saltshakers. Your task is to implement the `subsaltshaker` function, which takes in an in¶ teger `disk` and returns whether `disk` contains a saltshaker as a consecutive subinteger of its ¶ digits. >>> subsaltshaker(2233) # 22 counts True >>> subsaltshaker(2444423) # 4444 counts True >>> subsaltshaker(82223) # 22 counts even if it appears as part of 222 True >>> subsaltshaker(234562) # 2…2 does not count if the 2s are not consecuti¶ ve False >>> subsaltshaker(1) # 1 counts True >>> subsaltshaker(498729879871) # 1 counts True >>> subsaltshaker(149872987987) # 1 counts True >>> subsaltshaker(4445555) # no saltshakers in this number False >>> subsaltshaker(20) # no saltshakers in this number False “”” current_digit = disk % 10 count = 0 while disk != 0: last = disk % 10 if current_digit == last: count += 1 else: count = 1 current_digit = last if count == current_digit: return True disk = disk // 10 return False ================================================== copycat Point breakdown q6: 1.0/1 Score: Total: 1.0 Reskeletonized solution follows ================================================== def copycat(lst1, lst2): “\n Write a function `copycat` that takes in two lists.\n `lst1` i¶ s a list of strings\n `lst2` is a list of integers\n\n It returns a ne¶ w list where every element from `lst1` is copied the\n number of times as the¶ corresponding element in `lst2`. If the number\n of times to be copied is ne¶ gative (-k), then it removes the previous\n k elements added.\n\n Note 1: ¶ `lst1` and `lst2` do not have to be the same length, simply ignore\n any extr¶ a elements in the longer list.\n\n Note 2: you can assume that you will never¶ be asked to delete more\n elements than exist\n\n\n >>> copycat([‘a’, ‘b’¶ , ‘c’], [1, 2, 3])\n [‘a’, ‘b’, ‘b’, ‘c’, ‘c’, ‘c’]\n >>> copycat([‘a’, ‘b¶ ‘, ‘c’], [3])\n [‘a’, ‘a’, ‘a’]\n >>> copycat([‘a’, ‘b’, ‘c’], [0, 2, 0])\¶ n [‘b’, ‘b’]\n >>> copycat([], [1,2,3])\n []\n >>> copycat([‘a’, ‘b’¶ , ‘c’], [1, -1, 3])\n [‘c’, ‘c’, ‘c’]\n ” def copycat_helper(lst1, lst2, lst_so_far): if ((len(lst1) == 0) or (len(lst2) == 0)): return lst_so_far if (lst2[0] >= 0): lst_so_far = (lst_so_far + [lst1[0] for _ in range(lst2[0])]) else: lst_so_far = lst_so_far[:lst2[0]] return copycat_helper(lst1[1:], lst2[1:], lst_so_far) return copycat_helper(lst1, lst2, []) ================================================== Original code follows ================================================== def copycat(lst1, lst2): “”” Write a function `copycat` that takes in two lists. `lst1` is a list of strings `lst2` is a list of integers It returns a new list where every element from `lst1` is copied the number of times as the corresponding element in `lst2`. If the number of times to be copied is negative (-k), then it removes the previous k elements added. Note 1: `lst1` and `lst2` do not have to be the same length, simply ignore any extra elements in the longer list. Note 2: you can assume that you will never be asked to delete more elements than exist >>> copycat([‘a’, ‘b’, ‘c’], [1, 2, 3]) [‘a’, ‘b’, ‘b’, ‘c’, ‘c’, ‘c’] >>> copycat([‘a’, ‘b’, ‘c’], [3]) [‘a’, ‘a’, ‘a’] >>> copycat([‘a’, ‘b’, ‘c’], [0, 2, 0]) [‘b’, ‘b’] >>> copycat([], [1,2,3]) [] >>> copycat([‘a’, ‘b’, ‘c’], [1, -1, 3]) [‘c’, ‘c’, ‘c’] “”” def copycat_helper(lst1, lst2, lst_so_far): if len(lst1) == 0 or len(lst2) == 0: return lst_so_far if lst2[0] >= 0: lst_so_far = lst_so_far + [lst1[0] for _ in range(lst2[0])] else: lst_so_far = lst_so_far[:lst2[0]] return copycat_helper(lst1[1:], lst2[1:], lst_so_far) return copycat_helper(lst1, lst2, []) ================================================== flatmap_tree Point breakdown q7: 1.0/1 Score: Total: 1.0 Reskeletonized solution follows ================================================== def village(apple, t): ‘\n The `village` operation takes\n a function `apple` that maps a¶ n integer to a tree where\n every label is an integer.\n a tre¶ e `t` whose labels are all integers\n\n And applies `apple` to every label in¶ `t`.\n\n To recombine this tree of trees into a a single tree,\n simp¶ ly copy all its branches to each of the leaves\n of the new tree.\n\n ¶ For example, if we have\n apple(x) = tree(x, [tree(x + 1), tree(x + 2)])\¶ n and\n t = 10\n / 20 ¶ 30\n\n We should get the output\n\n village(apple, t)\n = ¶ 10\n / ¶ / 11 12\n / ¶ \\ / 20 30 20 30\n ¶ / \\ / \\ / \\ / 21 22 31 32 21 22 31 ¶ 32\n >>> t = tree(10, [tree(20), tree(30)])\n >>> apple = lambda x: tree(x¶ , [tree(x + 1), tree(x + 2)])\n >>> print_tree(village(apple, t))\n 10\n ¶ 11\n 20\n 21\n 22\n 30\n 31\n ¶ 32\n 12\n 20\n 21\n 22\n 30\n ¶ 31\n 32\n ‘ def graft(t, bs): ‘\n Grafts the given branches `bs` onto each leaf\n of the¶ given tree `t`, returning a new tree.\n ‘ if is_leaf(t): return tree(label(t), bs) new_branches = [graft(b, bs) for b in branches(t)] return tree(label(t), new_branches) base_t = apple(label(t)) bs = [village(apple, b) for b in branches(t)] return graft(base_t, bs) def tree(label, branches=[]): ‘Construct a tree with the given label value and a list of branches.’ 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.’ return tree[0] def branches(tree): ‘Return the list of branches of the given tree.’ return tree[1:] def is_tree(tree): ‘Returns True if the given tree is a tree, and False otherwise.’ 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\n ¶ otherwise.\n " return (not branches(tree)) def print_tree(t, indent=0): 'Print a representation of this tree in which each node is\n indented by ¶ two spaces times its depth from the entry.\n ' print(((' ' * indent) + str(label(t)))) for b in branches(t): print_tree(b, (indent + 1)) ================================================== Original code follows ================================================== def village(apple, t): """ The `village` operation takes a function `apple` that maps an integer to a tree where every label is an integer. a tree `t` whose labels are all integers And applies `apple` to every label in `t`. To recombine this tree of trees into a a single tree, simply copy all its branches to each of the leaves of the new tree. For example, if we have apple(x) = tree(x, [tree(x + 1), tree(x + 2)]) and t = 10 / \ 20 30 We should get the output village(apple, t) = 10 / \ / \ 11 12 / \ / \ 20 30 20 30 / \ / \ / \ / \ 21 22 31 32 21 22 31 32 >>> t = tree(10, [tree(20), tree(30)]) >>> apple = lambda x: tree(x, [tree(x + 1), tree(x + 2)]) >>> print_tree(village(apple, t)) 10 11 20 21 22 30 31 32 12 20 21 22 30 31 32 “”” def graft(t, bs): “”” Grafts the given branches `bs` onto each leaf of the given tree `t`, returning a new tree. “”” if is_leaf(t): return tree(label(t), bs) new_branches = [graft(b, bs) for b in branches(t)] return tree(label(t), new_branches) base_t = apple(label(t)) bs = [village(apple, b) for b in branches(t)] return graft(base_t, bs) def tree(label, branches=[]): “””Construct a tree with the given label value and a list of branches.””” 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.””” return tree[0] def branches(tree): “””Return the list of branches of the given tree.””” return tree[1:] def is_tree(tree): “””Returns True if the given tree is a tree, and False otherwise.””” 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 entry. """ print(' ' * indent + str(label(t))) for b in branches(t): print_tree(b, indent + 1) ==================================================

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