# Recitation 9¶

We discussed two data structures - Binary Search Trees and Hash Tables.

#### Takeaways:¶

• Important properties of Binary Search Trees:
• Insert and find take $O(h)$ time where $h$ is the height of the tree.
• When a tree containing $n$ nodes is balanced, $h = O(\log{n})$, in the worst case (totally unbalanced), $h=O(n)$
• Many methods in this class are implemented using recursion.
• Important properties of Hash Tables:
• Hash tables can be useful for many algorithms, including memoization.
• Insert and find operation run in $O(1)$ on average, but $O(n)$ in the worst case (where $n$ is the number of elements in the table)
• Make sure you understand the complexity analysis for hash tables (see the links below).

#### Code for printing several outputs in one cell (not part of the recitation):¶

In [26]:
from IPython.core.interactiveshell import InteractiveShell
InteractiveShell.ast_node_interactivity = "all"

In [27]:
## This file contains functions for the representation of binary trees.
## used in class Binary_search_tree's __repr__
## Written by a former student in the course - thanks to Amitai Cohen
## No need to fully understand this code

def printree(t, bykey = True):
"""Print a textual representation of t
bykey=True: show keys instead of values"""
#for row in trepr(t, bykey):
#        print(row)
return trepr(t, bykey)

def trepr(t, bykey = False):
"""Return a list of textual representations of the levels in t
bykey=True: show keys instead of values"""
if t==None:
return ["#"]

thistr = str(t.key) if bykey else str(t.val)

return conc(trepr(t.left,bykey), thistr, trepr(t.right,bykey))

def conc(left,root,right):
"""Return a concatenation of textual represantations of
a root node, its left node, and its right node
root is a string, and left and right are lists of strings"""

lwid = len(left[-1])
rwid = len(right[-1])
rootwid = len(root)

result = [(lwid+1)*" " + root + (rwid+1)*" "]

ls = leftspace(left[0])
rs = rightspace(right[0])
result.append(ls*" " + (lwid-ls)*"_" + "/" + rootwid*" " + "|" + rs*"_" + (rwid-rs)*" ")

for i in range(max(len(left),len(right))):
row = ""
if i<len(left):
row += left[i]
else:
row += lwid*" "

row += (rootwid+2)*" "

if i<len(right):
row += right[i]
else:
row += rwid*" "

result.append(row)

return result

def leftspace(row):
"""helper for conc"""
#row is the first row of a left node
#returns the index of where the second whitespace starts
i = len(row)-1
while row[i]==" ":
i-=1
return i+1

def rightspace(row):
"""helper for conc"""
#row is the first row of a right node
#returns the index of where the first whitespace ends
i = 0
while row[i]==" ":
i+=1
return i


## Binary Search Trees¶

Recall the (recursive) definition of a binary tree:

• A binary tree is an empty tree (a tree which contains no nodes), or
• Is composed of a root node, a Binary Tree called the left subtree of the root and a Binary Tree called the right subtree of the root

A binary search tree is a binary tree whose nodes have values, with the additional property that if $v$ is a node then all nodes in the left subtree of $v$ have keys smaller than $v.key$ and all those in the right subtree of $v$ have keys larger than $v.key$.

Binary search trees support operations such as insertion, deletion, search and many more.

In [28]:
class Tree_node():
def __init__(self, key, val):
self.key = key
self.val = val
self.left = None
self.right = None

def __repr__(self):
return "(" + str(self.key) + ":" + str(self.val) + ")"

class Binary_search_tree():

def __init__(self):
self.root = None

def __repr__(self): #no need to understand the implementation of this one
out = ""
for row in printree(self.root): #need printree.py file
out = out + row + "\n"
return out

def lookup(self, key):
''' return node with key, uses recursion '''

def lookup_rec(node, key):
if node == None:
return None
elif key == node.key:
return node
elif key < node.key:
return lookup_rec(node.left, key)
else:
return lookup_rec(node.right, key)

return lookup_rec(self.root, key)

def insert(self, key, val):
''' insert node with key,val into tree, uses recursion '''

def insert_rec(node, key, val):
if key == node.key:
node.val = val     # update the val for this key
elif key < node.key:
if node.left == None:
node.left = Tree_node(key, val)
else:
insert_rec(node.left, key, val)
else: #key > node.key:
if node.right == None:
node.right = Tree_node(key, val)
else:
insert_rec(node.right, key, val)
return

if self.root == None: #empty tree
self.root = Tree_node(key, val)
else:
insert_rec(self.root, key, val)

def minimum(self):
''' return node with minimal key '''
if self.root == None:
return None
node = self.root
left = node.left
while left != None:
node = left
left = node.left
return node

def depth(self):
''' return depth of tree, uses recursion'''
def depth_rec(node):
if node == None:
return -1
else:
return 1 + max(depth_rec(node.left), depth_rec(node.right))

return depth_rec(self.root)

def size(self):
''' return number of nodes in tree, uses recursion '''
def size_rec(node):
if node == None:
return 0
else:
return 1 + size_rec(node.left) + size_rec(node.right)

return size_rec(self.root)

def inorder(self):
'''prints the keys of the tree in a sorted order'''
def inorder_rec(node):
if node == None:
return
inorder_rec(node.left)
print(node.key)
inorder_rec(node.right)

inorder_rec(self.root)

t = Binary_search_tree()
t.insert(4,'a')
t.insert(2,'a')
t.insert(6,'a')
t.insert(1,'a')
t.insert(3,'a')
t.insert(5,'a')
t.insert(7,'a')
t
t.inorder()

Out[28]:
              4
______/ |______
2               6
__/ |__         __/ |__
1       3       5       7
/ |     / |     / |     / |
#   #   #   #   #   #   #   #
1
2
3
4
5
6
7


Claim: An inorder traversal of a binary search tree prints the keys in ascending order.

Proof, by complete induction on the size of tree:

• Base - for $n = 1$ the claim is trivial
• Assume the claim holds for all $i \leq n$
• For a tree of size $n+1$ with root $v$ consider the following:
• Both the right and the left subtree of $v$ have size at most $n$
• By induction, all keys in $v.left$ are printed in ascending order (and they are all smaller than $v.key$)
• Next, $v.key$ is printed
• Finally, by induction, all keys in $v.right$ are printed in ascending order (and they are all larger than $v.key$)
• Thus, the keys of the tree (which has size $n+1$) are printed in ascending order

Question: Assume $N = 2^n - 1$ for some $n$, find a method of inserting the numbers $[1,...,N]$ to a BST such that it is completely balanced (i.e. - it is a complete tree).

Solution: As we usually do with trees, we give a recursive solution. Our input will be a node and a first and last index to enter into the tree rooted in the node. We start with the root, $first = 1, last = 2^n - 1$

• Base case: If $first = last$ then we simply need to create a root with no sons labeled $first$
• Otherwise, we find the mid point $mid = (first + last - 1 ) // 2$. We recursively insert $first,...,mid$ into the left son of the node, $mid + 1$ into the current node and $mid + 2, ..., last$ into the right son of the node.
In [30]:
class Tree_node():
def __init__(self, key, val):
self.key = key
self.val = val
self.left = None
self.right = None

def __repr__(self):
return "(" + str(self.key) + ":" + str(self.val) + ")"

class Binary_search_tree():

def __init__(self):
self.root = None

def __repr__(self): #no need to understand the implementation of this one
out = ""
for row in printree(self.root): #need printree.py file
out = out + row + "\n"
return out

def lookup(self, key):
''' return node with key, uses recursion '''

def lookup_rec(node, key):
if node == None:
return None
elif key == node.key:
return node
elif key < node.key:
return lookup_rec(node.left, key)
else:
return lookup_rec(node.right, key)

return lookup_rec(self.root, key)

def insert(self, key, val):
''' insert node with key,val into tree, uses recursion '''

def insert_rec(node, key, val):
if key == node.key:
node.val = val     # update the val for this key
elif key < node.key:
if node.left == None:
node.left = Tree_node(key, val)
else:
insert_rec(node.left, key, val)
else: #key > node.key:
if node.right == None:
node.right = Tree_node(key, val)
else:
insert_rec(node.right, key, val)
return

if self.root == None: #empty tree
self.root = Tree_node(key, val)
else:
insert_rec(self.root, key, val)

def minimum(self):
''' return node with minimal key '''
if self.root == None:
return None
node = self.root
left = node.left
while left != None:
node = left
left = node.left
return node

def depth(self):
''' return depth of tree, uses recursion'''
def depth_rec(node):
if node == None:
return -1
else:
return 1 + max(depth_rec(node.left), depth_rec(node.right))

return depth_rec(self.root)

def size(self):
''' return number of nodes in tree, uses recursion '''
def size_rec(node):
if node == None:
return 0
else:
return 1 + size_rec(node.left) + size_rec(node.right)

return size_rec(self.root)

def inorder(self):
'''prints the keys of the tree in a sorted order'''
def inorder_rec(node):
if node == None:
return
inorder_rec(node.left)
print(node.key)
inorder_rec(node.right)

inorder_rec(self.root)

def insert_balanced(self, n):
'''inserts the numbers 1, ... , 2**n - 1 into a full BST'''
def insert_balanced_rec(node, first, last):
if first == last:
node.key = first
node.val = first
else:
mid = (first + last - 1) // 2
node.key = mid + 1
node.val = mid + 1
node.left = Tree_node(0,0)
insert_balanced_rec(node.left, first, mid)
node.right = Tree_node(0,0)
insert_balanced_rec(node.right, mid + 2, last)

self.root = Tree_node(0,0)
insert_balanced_rec(self.root, 1, 2**n - 1)

t = Binary_search_tree()
t.insert(4,'a')
t.insert(2,'a')
t.insert(6,'a')
t.insert(1,'a')
t.insert(3,'a')
t.insert(5,'a')
t.insert(7,'a')

#t.inorder()

t = Binary_search_tree()
t.insert_balanced(4)
printree(t.root)

Out[30]:
['                              8                                    ',
'               ______________/ |________________                   ',
'              4                                 12                 ',
'       ______/ |______                  _______/  |_______         ',
'      2               6               10                  14       ',
'   __/ |__         __/ |__         __/  |__            __/  |__    ',
'  1       3       5       7       9        11        13        15  ',
' / |     / |     / |     / |     / |      /  |      /  |      /  | ',
'#   #   #   #   #   #   #   #   #   #    #    #    #    #    #    #']

## Hash Tables¶

We wish to have a data structure that implements the operations: insert, search and delete in expected $O(1)$ time.

Summarizing the insert and search complexity of the data structures that we have seen already:

implementation insert search delete
Python list O(1) always at the end O(n) O(n)
Python ordered list O(n) O(log n) O(n)
Linked list O(1) always at the start O(n) O(1) given the node before the one to delete
Sorted linked list O(n) O(n) O(1) given the node before the one to delete
Unbalanced Binary Search Tree O(n) O(n) O(n)
Balanced Binary Search Tree O(log n) O(log n) O(log n)

Please read the following summary on the various data structures mentioned in class.

A detailed summary on the complexity of insert/search operations using hash tables can be found here. Make sure you read it.

### Exercise:¶

Given a string $st$ of length $n$ and a small integer $\ell$, write a function that checks whether there is a substring in $st$ of length $\ell$ that appears more than once.

#### Solution #1: Naive¶

The complexity is $O(\ell(n-\ell)^2)$. There $O((n-\ell)^2)$ iterations (make sure you undersand why) and in each iteration we perform operations in $O(\ell)$ time.

In [ ]:
def repeat_naive(st, l):
for i in range(len(st)-l+1):
for j in range(i+1,len(st)-l+1):
if st[i:i+l]==st[j:j+l]:
return True
return False

repeat_naive("hello", 1)
repeat_naive("hello"*10, 45)
repeat_naive("hello"*10, 46)


Let's test our algorithm with by generating a random string of a given size.

In [ ]:
import random
def gen_str(size, alphabet = "abcdefghijklmnopqrstuvwxyz"):
''' Generate a random string of length size over alphabet '''
s=""
for i in range(size):
s += random.choice(alphabet)
return s
rndstr = gen_str(1000)
print(rndstr)
repeat_naive(rndstr, 3)
repeat_naive(rndstr, 10)


For bigger $n$ and $\ell$, this could be quite slow:

In [ ]:
rndstr = gen_str(10000)
repeat_naive(rndstr, 3)
repeat_naive(rndstr, 10)


### The class Hashtable from the lectures¶

In [ ]:
class Hashtable:
def __init__(self, m, hash_func=hash):
""" initial hash table, m empty entries """
##bogus initialization #1:
#self.table = [[]*m]
##bogus initialization #2:
#empty=[]
#self.table = [empty for i in range(m)]

self.table = [ [] for i in range(m)]
self.hash_mod = lambda x: hash_func(x) % m # using python hash function

def __repr__(self):
L = [self.table[i] for i in range(len(self.table))]
return "".join([str(i) + " " + str(L[i]) + "\n" for i in range(len(self.table))])

def find(self, item):
""" returns True if item in hashtable, False otherwise  """
i = self.hash_mod(item)
return item in self.table[i]
#if item in self.table[i]:
#    return True
#else:
#    return False

def insert(self, item):
""" insert an item into table """
i = self.hash_mod(item)
if item not in self.table[i]:
self.table[i].append(item)


#### Solution #2: using the class Hashtable¶

In [ ]:
def repeat_hash1(st, l):
m=len(st)-l+1
htable = Hashtable(m)
for i in range(len(st)-l+1):
if htable.find(st[i:i+l])==False:
htable.insert(st[i:i+l])
else:
return True
return False


The expected (average) complexity is: $O(\ell(n-\ell))$

Creating the table takes $O(n-\ell)$ time, and there are $O(n-\ell)$ iterations, each taking expected $O(\ell)$ time.

The worst case complexity is: $O(\ell(n-\ell)^2)$

Creating the table takes $O(n-\ell)$ time, and the time for executing the loop is $\ell\cdot\sum_{i=0}^{n-\ell}{i}= O(\ell(n-\ell)^2)$

Which native Python data structure is most suitable for this problem?

#### Solution #3: using Python's set implementation¶

In [ ]:
def repeat_hash2(st, l):
htable = set() #Python sets use hash functions for fast lookup
for i in range(len(st)-l+1):
if st[i:i+l] not in htable:
else: return True
return False


#### Competition between the 3 solutions¶

In [ ]:
import time
str_len=1000
st=gen_str(str_len)
l=10
for f in [repeat_naive,repeat_hash1,repeat_hash2]:
t0=time.perf_counter()
res=f(st, l)
t1=time.perf_counter()
print(f.__name__, t1-t0, "found?",res)

In [ ]:
str_len=2000
st=gen_str(str_len)
l=10
for f in [repeat_naive,repeat_hash1,repeat_hash2]:
t0=time.perf_counter()
res=f(st, l)
t1=time.perf_counter()
print(f.__name__, t1-t0, "found?",res)


For a random string of size $n=1000$ and for $l=10$ the running time of repeat_hash2 is the smallest, while the one for repeat_naive is the largest.

When increasing $n$ to 2000, the running time of repeat_naive increases by ~4, while the running time of repeat_hash1, repeat_hash2 increases by ~2.

#### Time spent on creating the table¶

When $st$ is "a"$*1000$, repeat_hash1 is the slowest, since it spends time on creating an empty table of size 991.

In [ ]:
st="a"*1000
l=10
for f in [repeat_naive,repeat_hash1,repeat_hash2]:
t0=time.perf_counter()
res=f(st, l)
t1=time.perf_counter()
print(f.__name__, t1-t0, "found?",res)


### The effect of table size¶

The second solution, with control over the table size

In [ ]:
def repeat_hash1_var_size(st, l, m=0):
if m==0: #default hash table size is ~number of substrings to be inserted
m=len(st)-l+1
htable = Hashtable(m)
for i in range(len(st)-l+1):
if htable.find(st[i:i+l])==False:
htable.insert(st[i:i+l])
else:
return True
return False


#### Comparing tables sizes¶

In [ ]:
import time
str_len=1000
st=gen_str(str_len)
l=10
print("str_len=",str_len, "repeating substring len=",l)
for m in [1, 10, 100, 1000, 1500, 10000, 100000]:
t0=time.perf_counter()
res=repeat_hash1_var_size(st, l, m)
t1=time.perf_counter()
print(t1-t0, "found?",res, "table size=",m)


### Summary¶

Make sure you read the following summary that includes a detailed explanation on the experiments.