Time: Thu 10/12/17 3 pm
They are naturally sustainable! Oh well, think about how trees
are conceptually related to pairs.
Define the terminologies around trees--tree, root, label, branch, leaf (the language we use may vary in different places).
Consider the following implementation:
def tree(label, branches=[]):
return [label] + branches
def label(tree):
return tree[0]
def branches(tree):
return tree[1:]There are multiple ways to get started on playing with such tree structure! As a few examples:
Fill in the blanks of the implementation of sixty_ones below, a function that takes a Link instance representing a sequence of integers and returns the number of times that 6 and 1 appear consecutively.
def sixty_ones(s):
"""
Return the number of times that 1 directly follows 6 in linked list s.
>>> once = Link(4, Link(6, Link(1, Link(6, Link(0, Link(1))))))
>>> twice = Link(1, Link(6, Link(1, once)))
>>> thrice = Link(6, twice)
>>> apply_to_all(sixty_ones, [Link.empty, once, twice, thrice])
[0, 1, 2, 3]
"""
if :
return 0
elif :
return 1 +
else:
return Fill in the blanks of the implementation of no_eleven below, a function that returns a list of all distinct length-n lists of ones and sixes in which 1 and 1 do not appear consecutively.
def no_eleven(n):
"""
Return a list of lists of 1’s and 6’s that do not contain 1 after 1.
>>> no_eleven(2)
[[6, 6], [6, 1], [1, 6]]
>>> no_eleven(3)
[[6, 6, 6], [6, 6, 1], [6, 1, 6], [1, 6, 6], [1, 6, 1]]
>>> no_eleven(4)[:4] # first half
[[6, 6, 6, 6], [6, 6, 6, 1], [6, 6, 1, 6], [6, 1, 6, 6]]
>>> no_eleven(4)[4:] # second half
[[6, 1, 6, 1], [1, 6, 6, 6], [1, 6, 6, 1], [1, 6, 1, 6]]
"""
if n == 0:
return
elif n == 1:
return
else:
a, b = no_eleven(), no_eleven()
return [ for s in a] + [ for s in b]This question uses the following tree data abstraction.
def tree(entry, children=[]):
return [entry, children]
def entry(tree):
return tree[0]
def children(tree):
return tree[1]Define the function track_lineage that takes in a tree of strings family_tree and a string name. Assume that there is a unique node with entry name. track_lineage returns a list with the entries of the parent and grandparent of that node. If the node with entry name does not have a parent or grandparent, return None for that element in the list. See the doctests for details. Do not violate abstraction barriers. You may only use the lines provided. You may not need to fill all the lines.
def track_lineage(family_tree, name):
"""
Return the entries of the parent and grandparent of the node with entry name in family_tree.
>>> t = tree('Tytos', [
... tree('Tywin', [
... tree('Cersei'), tree('Jaime'), tree('Tyrion')
... ]),
... tree('Kevan', [
... tree('Lancel'), tree('Martyn'), tree('Willem')
... ])])
>>> track_lineage(t, 'Tywin')
['Tytos', None]
>>> track_lineage(t, 'Tytos')
[None, None]
"""
def tracker(t, p, gp):
if
for c in children(t):
return tracker(, , )Assuming that track_lineage works correctly, define the function are cousins that takes in a tree of strings family_tree and two strings name1 and name2 and returns True if the node with entry name1 and the node with entry name2 are cousins in family_tree. Assume that there are unique nodes with entries name1 and name2 in family_tree. See the doctests for details. Two nodes are cousins if they have the same grandparent but different parents.
You may only use the lines provided. You may not need to fill all the lines.
def are_cousins(family_tree, name1, name2):
"""Return True if a node with entry name1 is a cousin of a node with entry name2 in family_tree.
>>> are_cousins(t, ‘Kevan’, ‘Tytos’) # same tree as before False
>>> are_cousins(t, ‘Cersei’, ‘Lancel’)
True
>>> are_cousins(t, ‘Jaime’, ‘Lancel’)
True
>>> are_cousins(t, ‘Jaime’, ‘Tyrion’)
False
"""
A GrootTree g is a binary tree that has an attribute parent. Its parent is the GrootTree in which g is a branch. If a GrootTree instance is not a branch of any other GrootTree instance, then its parent is BinaryTree.empty.
BinaryTree.empty should not have a parent attribute. Assume that every GrootTree instance is a branch of at most one other GrootTree instance and not a branch of any other kind of tree.
Fill in the blanks below so that the parent attribute is set correctly. You may not need to use all of the lines. Indentation is allowed. You should not include any assert statements. Using your solution, the doctests for fib_groot should pass.
class GrootTree(BinaryTree):
"""A binary tree with a parent."""
def __init__(self, entry, left=BinaryTree.empty, right=BinaryTree.empty):
BinaryTree.__init__(self, entry, left, right)
def fib_groot(n):
"""
Return a Fibonacci GrootTree.
>>> t = fib_groot(3)
>>> t.entry
2
>>> t.parent.is_empty
True
>>> t.left.parent.entry
2
>>> t.right.left.parent.right.parent.entry
1
"""
if n == 0 or n == 1:
return GrootTree(n)
else:
left, right = fib_groot(n - 2), fib_groot(n - 1)
return GrootTree(left.entry + right.entry, left, right)Fill in the blanks of the implementation of paths, a function that takes two arguments: a GrootTree instance g and a list s. It returns the number of paths through g whose entries are the elements of s. A path through a GrootTree can extend either to a branch or its parent.
You may assume that the GrootTree class is implemented correctly and that the list s is non-empty.
There are two paths for [2, 1, 2, 1, 0], one path for [2, 1, 0, 1, 0] in the tree below:
2 +-------+ 1 1 +---+ 0 1
def paths(g, s):
"""
The number of paths through g with entries s.
>>> t = fib_groot(3)
>>> paths(t, [1])
0
>>> paths(t, [2])
1
>>> paths(t, [2, 1, 2, 1, 0])
2
>>> paths(t, [2, 1, 0, 1, 0])
1
>>> paths(t, [2, 1, 2, 1, 2, 1])
8
"""
if g is BinaryTree.empty :
return 0
elif :
return 1
else:
xs = []
return sum([ for x in xs])