# Homework 5

*Due by 11:59pm on Monday, 3/28*

## Instructions

Download hw05.zip. Inside the archive, you will find a file called hw05.py, along with a copy of the OK autograder.

**Submission:** When you are done, submit with ```
python3 ok
--submit
```

. You may submit more than once before the deadline; only the
final submission will be scored. See Lab 0 for instructions on submitting
assignments.

**Using OK:** If you have any questions about using OK, please
refer to this guide.

## Assorted Problems You've Seen Before

The following problems were extra from this week's lab. If you've already done them, feel free to re-use your solutions.

### Question 1: Cumulative Sum

Write a function `cumulative_sum`

that returns a new `Tree`

, where each entry is the sum of all entries in the corresponding subtree of the old `Tree`

.

```
def cumulative_sum(t):
"""Return a new Tree, where each entry is the sum of all entries in the
corresponding subtree of t.
>>> t = Tree(1, [Tree(3, [Tree(5)]), Tree(7)])
>>> cumulative = cumulative_sum(t)
>>> t
Tree(1, [Tree(3, [Tree(5)]), Tree(7)])
>>> cumulative
Tree(16, [Tree(8, [Tree(5)]), Tree(7)])
>>> cumulative_sum(Tree(1))
Tree(1)
"""
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

`python3 ok -q cumulative_sum`

### Question 2: Link to List

Write a function `link_to_list`

that converts a given `Link`

to a
Python list.

```
def link_to_list(link):
"""Takes a Link and returns a Python list with the same elements.
>>> link = Link(1, Link(2, Link(3, Link(4))))
>>> link_to_list(link)
[1, 2, 3, 4]
>>> link_to_list(Link.empty)
[]
"""
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

`python3 ok -q link_to_list`

### Question 3: Reverse

Implement `reverse`

, which takes a linked list `link`

and returns a linked list
containing the elements of `link`

in reverse order. The original `link`

should be
unchanged.

```
def reverse(link):
"""Returns a Link that is the reverse of the original.
>>> print_link(reverse(Link(1)))
<1>
>>> link = Link(1, Link(2, Link(3)))
>>> new = reverse(link)
>>> print_link(new)
<3 2 1>
>>> print_link(link)
<1 2 3>
"""
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

`python3 ok -q reverse`

### Question 4: Cycles

The `Link`

class can represent lists with cycles. That is, a list may
contain itself as a sublist.

```
>>> s = Link(1, Link(2, Link(3)))
>>> s.rest.rest.rest = s
>>> s.rest.rest.rest.rest.rest.first
3
```

Implement `has_cycle`

,that returns whether its argument, a `Link`

instance, contains a cycle.

Hint: Iterate through the linked list and try keeping track of which`Link`

objects you've already seen.

```
def has_cycle(link):
"""Return whether link contains a cycle.
>>> s = Link(1, Link(2, Link(3)))
>>> s.rest.rest.rest = s
>>> has_cycle(s)
True
>>> t = Link(1, Link(2, Link(3)))
>>> has_cycle(t)
False
>>> u = Link(2, Link(2, Link(2)))
>>> has_cycle(u)
False
"""
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

`python3 ok -q has_cycle`

**Extra for experts**: Implement `has_cycle`

with only constant space. (If
you followed the hint above, you will use linear space.) The solution is short
(less than 20 lines of code), but requires a clever idea. Try to discover the
solution yourself before asking around:

```
def has_cycle_constant(link):
"""Return whether link contains a cycle.
>>> s = Link(1, Link(2, Link(3)))
>>> s.rest.rest.rest = s
>>> has_cycle_constant(s)
True
>>> t = Link(1, Link(2, Link(3)))
>>> has_cycle_constant(t)
False
"""
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

`python3 ok -q has_cycle_constant`

## Binary Search Trees

The following problems deal with *binary trees*, in which each node has
at most two children. We further modify the definition of tree to allow
*empty trees*, which we haven't dealt with up to now. So here, we define
a binary tree as either

- An empty tree, or
- A label and two children, called the left and right child respectively, both of which are binary trees. The left and right children, therefore, are the roots of the left and right subtrees.

As a result of this definition, most recursion on binary trees uses the empty tree as a base case.

Our implementation of binary trees is the following:

```
# Tree definition
class BinTree:
empty = ()
def __init__(self, label, left=empty, right=empty):
self.label = label
self.left = left
self.right = right
def __repr__(self):
if self.left == self.empty and self.right == self.empty:
return 'BinTree({})'.format(repr(self.label))
left = 'BinTree.empty' if self.left == self.empty else repr(self.left)
right = 'BinTree.empty' if self.right == self.empty else repr(self.right)
return 'BinTree({}, {}, {})'.format(repr(self.label), left, right)
def binTree(tupleTree):
"""A convenience method for succinctly constructing binary trees. The
empty tuple or list stands for BinTree.empty; (V,) or [V] stands
for BinTree(V); (V, T1, T2) or [V, T1, T2] stands for
BinTree(V, binTree(T1), binTree(T2)).
>>> binTree((3,
... (1, (), [2]),
... (6, [5], [7])))
BinTree(3, BinTree(1, BinTree.empty, BinTree(2)), BinTree(6, BinTree(5), BinTree(7)))
"""
if len(tupleTree) == 0:
return BinTree.empty
elif len(tupleTree) == 1:
return BinTree(tupleTree[0])
else:
return BinTree(tupleTree[0], binTree(tupleTree[1]), binTree(tupleTree[2]))
```

We also included a function `print_bintree`

, which prints out a string
representation of a tree:

```
>>> print_bintree(binTree( (1, (3, (), [2]), (4, [5], [6])) ))
-1-
/ \
3 4
\ / \
2 5 6
```

A *binary search tree* (or *BST*) is a binary tree that obeys the following
constraint:

- All labels in the left subtree have a value less than or equal to the label of
*v*, and - All labels in the right subtree have a value greater than or equal to the
label of
*v*. - This is true for its children subtree and their grandchildren (and so on) as well.

Thus, these are BSTs:

```
--4-- 1 4 4
/ \ \ / /
2 6 2 3 1
/ \ / \ / \
1 3 5 3 2 3
\ / /
4 1 2
```

These are binary trees, but not BSTs:

```
--6-- 4
/ \ \
2 4 3
/ \ / \
1 3 5 2
\
1
```

### Question 5: Min

Define the function `bst_min`

, which takes in a binary search tree and returns
the min of all of the labels of its nodes. Be sure to take
advantage of the properties of the binary search tree to get an efficient
solution.

You may assume the input is a non-empty `BinTree`

.

```
def bst_min(b):
r""" Returns the min of all the values of each node in b.
Calling bst_min on the following tree should return 1:
4
/ \
2 6
/ /
1 5
>>> t1 = binTree([4, [2, [1], []], [6, [5], []]])
>>> bst_min(t1)
1
>>> t2 = binTree([4])
>>> bst_min(t2)
4
>>> t3 = binTree([4, [2], []])
>>> bst_min(t3)
2
>>> t4 = binTree([4, [], [6]])
>>> bst_min(t4)
4
>>> t5 = binTree([9, [6, [5], [7]], []])
>>> bst_min(t5)
5
>>> # Illegal BST, but if your solution is right, it should NOT
>>> # print 100.
>>> t6 = binTree([9, [6, [5], [7]], [-100]])
>>> bst_min(t6)
5
"""
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

`python3 ok -q bst_min`

### Question 6: Max

Define the function `bst_max`

, which takes in a binary search tree and returns
the max of all of the labels of its nodes. Be sure to take
advantage of the properties of the binary search tree to get an efficient
solution.

You may assume the input is a non-empty `BinTree`

.

```
def bst_max(b):
r""" Returns the max of all the labels in B, assuming B is a BST.
Calling bst_max on the following tree should return 6:
4
/ \
2 6
/ /
1 5
>>> t1 = binTree([4, [2, [1], []], [6, [5], []]])
>>> bst_max(t1)
6
>>> t2 = binTree([4])
>>> bst_max(t2)
4
>>> t3 = binTree([4, [2], []])
>>> bst_max(t3)
4
>>> t4 = binTree([4, [], [6]])
>>> bst_max(t4)
6
>>> t5 = binTree([4, [], [6, [5], [7]]])
>>> bst_max(t5)
7
>>> # Illegal BST, but if your solution is right, it should NOT
>>> # print 100.
>>> t6 = binTree([4, [100], [6, [5], [7]]])
>>> bst_max(t6)
7
"""
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

`python3 ok -q bst_max`

### Question 7: is_valid

Write a function `is_valid_bst`

, which takes a given binary
tree and returns `True`

iff that tree is a valid binary search tree.
That is, for every node, the labels in that node's left children
are less than the label of the node,
and the labels of its right children are greater than that of the node.
You might find the `bst_min`

and `bst_max`

functions helpful.

```
def is_valid_bst(bst):
"""
>>> t1 = binTree([4, [2, [1], []], [6, [5], []]])
>>> is_valid_bst(t1)
True
>>> t2 = binTree([4, [100], [6, [5], [7]]])
>>> is_valid_bst(t2)
False
>>> t3 = binTree([6, [5, [3, [1], [4]], [7]], [8, [7], [9]]])
>>> is_valid_bst(t3)
False
>>> t4 = binTree([1])
>>> is_valid_bst(t4)
True
>>> t5 = binTree([6, [5, [3, [4], [1]], []], [8, [7], [9]]])
>>> is_valid_bst(t5)
False
>>> is_valid_bst(BinTree.empty)
True
"""
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

`python3 ok -q is_valid_bst`

## Python Sets

A set is an **unordered** collection of **distinct** objects that supports
membership testing, union, intersection, and adjunction. The main difference between
sets and lists are 1. they are unordered, and 2. there are no duplicates. Other than
that almost everything is the same.

```
>>> a = [1, 1, 2, 2, 3, 3]
>>> a = set(a)
>>> a # No duplicates
{1, 2, 3}
>>> a = {3, 1, 2}
>>> a # Not necessarily in same order
{1, 2, 3}
```

The Python documentation on sets has more details. One really convenient thing about Python sets is that many operations on sets (adding elements, removing elements, checking membership) run in O(1) (constant) time (usually).

Some of the problems use a utility method called
`timeit`

, which takes a parameterless function as argument, executes it,
and returns the time required to do so. It's a variation on the function
`timeit.timeit`

function in the Python3 library.

### Question 8: Add up

Write the following function so it (usually) runs in O(m) time, where m is
the length of `lst`

.

```
def add_up(n, lst):
"""Returns True if any two non identical elements in lst add up to n.
>>> add_up(100, [1, 2, 3, 4, 5])
False
>>> add_up(7, [1, 2, 3, 4, 2])
True
>>> add_up(10, [5, 5])
False
>>> add_up(151, range(0, 200000, 2))
False
>>> timeit(lambda: add_up(151, range(0, 200000, 2))) < 1.0
True
>>> add_up(50002, range(0, 200000, 2))
True
"""
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

`python3 ok -q add_up`

### Question 9: Missing value

Write the following function so it (usually) runs in O(n) time, where n is
the length of `lst0`

.

```
def missing_val(lst0, lst1):
"""Assuming that LST0 contains all the values in LST1, but LST1 is missing
one value in LST0, return the missing value. The values need not be
numbers.
>>> from random import shuffle
>>> missing_val(range(10), [1, 0, 4, 5, 7, 9, 2, 6, 3])
8
>>> big0 = [str(k) for k in range(15000)]
>>> big1 = [str(k) for k in range(15000) if k != 293 ]
>>> shuffle(big0)
>>> shuffle(big1)
>>> missing_val(big0, big1)
'293'
>>> timeit(lambda: missing_val(big0, big1)) < 1.0
True
"""
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

`python3 ok -q missing_val`