cs1001.py , Tel Aviv University, Spring 2019 Recitation 8¶

We reviewed some properties of prime numbers and used them for primality testing. We reviewed the Diffie-Hellman protocol for finding a shared secret key and also tried to crack it. Then, we discussed Linked lists.

Takeaways:¶

• The probabilistic function is_prime, that uses Fermat's primality test, can be used to detect primes quickly and efficiently, but has a (very small) probability of error. Its time complexity is $O(n^3)$, where $n$ is the number of bits of the input.

• The DH protocol relies on two main principles: the following equality $(g^{a}\mod p)^b \mod p = g^{ab} \mod p$ and the (believed) hardness of the discrete log problem (given $g,p$, and $x = g^{a} \mod p$ finding $a$ is hard). Make sure you understand the details of the protocol.

• OOP allows us to create classes at will and to define complex objects. Remember that the most important thing is to find a good inner representation of the object so that you will have an easy time working with it.

• Important properties of Linked lists:

• Not stored continuously in memory.
• Allows for deletion and insertion after a given element in $O(1)$ time.
• Accessing the $i$'th element takes $O(i)$ time.
• Search takes $O(n)$ time (no random access $\implies$ no $O(\log{n})$ search).

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

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

Primes and Diffie-Hellman¶

Fermat's little theorem¶

Fermat's little theorem states that if $p$ is prime and $1 < a < p$, then $a^{p-1} (\textrm{mod}\ p) \equiv 1$

Compositeness witnesses¶

A witness is a piece of evidence that can be produced in order to prove a claim. In our case, the problem we are tackling is deciding whether a given number $m$ is prime or composite. We now describe three types of witnesses for compositeness:

• A clear witness for compositeness can be a (not necessarily prime) factor of $m$. That is, a number $1 < b < m$ such that $b ~|~ m$ ($b$ divides $m$). Since $b$ is a non-trivial factor of $m$, $m$ is clearly composite.
• Another, less trivial witness of compositeness is a GCD witness, that is a number $1 < b < m$ such that $\mathrm{gcd}(m,b) > 1$. Think, why is $b$ a witness of compositeness? Let $\mathrm{gcd}(m,b) = r$, then by definition $r ~|~ n$. As $r \leq b < m$ we get that $r$ is a factor of $m$ and thus $m$ is composite.
• Fermat's primality test defines another set of witness of compositeness - a number $1 < a < m$ is a Fermat witness if $a^{m-1} (\textrm{mod}\ m) \not\equiv 1$

Why go through all of this work? Our tactic would be to randomly draw a number $b \in {1,\ldots, m - 1}$ and hope that $b$ is some witness of compositeness. Clearly, we'd like our witness pool to be as large as possible.

Now, if $m$ is a composite number, let $\mathrm{FACT}_m, \mathrm{GCD}_m, \mathrm{FERM}_m$ be the set of prime factors, GCD witnesses and Fermat witnesses for $m$'s compositeness. It is not hard to show that $$\mathrm{FACT}_m \subseteq \mathrm{GCD}_m \subseteq \mathrm{FERM}_m$$

But the real strength of Fermat's primality test comes from the fact that if $m$ is composite, then apart from very rare cases (where $n$ is a Carmichael number) it holds that $|\mathrm{FERM}_m| \geq m/2$. That is - a random number is a Fermat witness w.p. at least $1/2$.

A side note (for reference only) - Carmichael numbers are exactly the composite numbers $m$ where $\mathrm{GCD}_m = \mathrm{FERM}_m$

Every factor of a composite number is a Fermat's witness¶

Let $m$ be a composite number and write $m = ab$ for some $a,b>1$. We claim that $a$ is a Fermat witness. To see this, assume towards contradiction that $a^{m-1} (\textrm{mod}\ m) \equiv 1$, i.e. $a^{m-1} = c\cdot m + 1= c \cdot a \cdot b + 1$ for some $c \geq 1$.

Rearrange the above to get $a(a^{m-2} - c\cdot b) = 1$. However, $a > 1$ and $(a^{m-2} - c\cdot b) \in \mathbb{Z}$, a contradiction.

Primality test using Fermat's witness¶

We can use Fermat's little theorem in order test whether a given number $m$ is prime. That is, we can test whether we find a Fermat's witness $a\in\mathrm{FERM}_m$ for compositeness. Note that if the number has $n$ bits than testing all possible $a$-s will require $O(2^n)$ iterations (a lot!).

Instead, we will try 100 random $a$-s in the range and see if one works as a Fermat's witness.

In :
import random

def is_prime(m, show_witness=False):

""" probabilistic test for m's compositeness """''

for i in range(0, 100):
a = random.randint(1, m - 1) # a is a random integer in [1..m-1]
if pow(a, m - 1, m) != 1:
if show_witness:  # caller wishes to see a witness
print(m, "is composite", "\n", a, "is a witness, i=", i + 1)
return False

return True

For $a,b,c$ of at most $n$ bits each, time complexity of modpower is $O(n^3)$

In :
def modpower(a, b, c):
""" computes a**b modulo c, using iterated squaring """
result = 1
while b > 0: # while b is nonzero
if b % 2 == 1: # b is odd
result = (result * a) % c
a = (a*a) % c
b = b // 2
return result

Runtime analysis:¶

• The main loop runs over $b$, dividing $b \to b/2$ at each iteration, so it runs $O(n)$ times.
• In each iteration we do:
• One operation of $b~\%~2$ in $O(1)$ time (check least significant bit of $b$)
• One operation of $b~//~2$ in $O(1)$ time (snip $b$'s least significant bit)
• At most two multiplication and two modulu operations
• Multiplication of two $n$ bit numbers runs in time $O(n^2)$
• Modulu can be implemented by addition, division and multiplication: $a \textrm{ mod } b = a - (a // b) b$ and division runs in time $O(n^2)$ same as multiplication
• Finally, the modulu operation keeps all numbers at most $n$ bits, thus the running time does not increase with each iteration
• In total - $O(n^3)$

The probability of error:¶

First, notice that if the function says that an input number $m$ is not prime, then it is true. Stated the other way around, if $m$ is prime, the function is always right.

The function can make a mistake only in the case where a number $m$ is not prime, and is accidentally categorized by the function as prime. This can happen if all $100$ $a$'s that the function had drawn were not witnesses.

A quick computation shows that if $m$ is not a Charmichael number then at least $\frac{1}{2}$ of all possible $a$s are witnesses, so in almost all cases the probability for error is $(\frac{1}{2})^{100}$ (which is negligible).

Testing the prime number theorem: For a large n, a number of n bits is prime with a prob. of O(1/n)¶

We decide on the size of the sample (to avoid testing all possible $2^{n-1}$ numbers of $n$ bits) and test whether each number we sample is prime. Then we divide the number of primes with the size of the sample.

In :
def prob_prime(n, sample):
cnt = 0
for i in range(sample):
m = random.randint(2**(n-1), 2**n - 1)
cnt += is_prime(m)
return cnt / sample
In :
prob_prime(2, 10**4)
prob_prime(3, 10**4)
Out:
1.0
Out:
0.5057
In :
prob_prime(100, 10**4)
Out:
0.0148
In :
prob_prime(200, 10**4)
Out:
0.0081

Diffie Hellman from lecture The protocol as code¶

In :
def DH_exchange(p):
""" generates a shared DH key """
g = random.randint(1, p - 1)
a = random.randint(1, p - 1) #Alice's  secret
b = random.randint(1, p - 1) #Bob's  secret
x = pow(g, a, p)
y = pow(g, b, p)
key_A = pow(y, a, p)
key_B = pow(x, b, p)
#the next line is different from lecture
return g, a, b, x, y, key_A #key_A=key_B

Find a prime number¶

In :
def find_prime(n):
""" find random n-bit long prime """
while(True):
candidate = random.randrange(2**(n-1), 2**n)
if is_prime(candidate):
return candidate

Demostration:

In :
import random
p = find_prime(10)
print(p)
g, a, b, x, y, key = DH_exchange(p)
g, a, b, x, y, key
557
Out:
(247, 88, 353, 367, 2, 206)
In :
print(pow(g, a, p))
print(pow(x, b, p))
367
206

Crack the Diffie Hellman code¶

There is no known way to find $a$ efficiently, so we try the naive one: iterating over all $a$-s and cheking whether the equation $g^a \mod p = x$ holds for them.

If we found $a'$ that satisfies the condition but is not the original $a$, does it matter?

The time complexity of crack_DH is $O(2^nn^3)$

In :
def crack_DH(p, g, x):
''' find secret "a" that satisfies g**a%p == x
Not feasible for large p '''
for a in range(1, p - 1):
if a % 100000 == 0:
print(a) #just to estimate running time
if pow(g, a, p) == x:
return a
return None #should never get here
In :
#g, a, b, x, y, key = DH_exchange(p)
print(a)
crack_DH(p, g, x)
88
Out:
88

Cracking DH with different values of $a$ (different private keys)¶

The algorithm crack_DH can return a different private key ($a$) than the one chosen by Alice, i.e. - $crack\_DH(p,g,x) = a' \neq a$, however, in this case we have, by definition of the cracking algorithm: $g^{a'} \textrm{ mod } p = g^{a} \textrm{ mod } p$, thus:

$$y^{a'} \textrm{ mod } p = \left(g^{b}\right)^{a'} \textrm{ mod } p = \left(g^{a'}\right)^{b} \textrm{ mod } p = \left(g^{a} \textrm{ mod } p \right)^{b} \textrm{ mod } p = g^{ab} \textrm{ mod } p$$

I.e. - we can still compute the shared secret!

Trying to crack the protocol with a 100 bit prime¶

In :
import random
p = find_prime(100)
print(p)
g, a, b, x, y, key = DH_exchange(p)
print(g, a, b, x, y, key)

crack_DH(p, g, x)
870752849128025769690613494727
556076925348494470910111020503 563975178376164531572554628719 591095820882694542257093278496 683629518338524578223612745129 622210352872361792591098681438 318844905364032174750094749602
100000
200000
300000
400000
500000
600000
---------------------------------------------------------------------------
KeyboardInterrupt                         Traceback (most recent call last)
<ipython-input-26-6b3d13779c00> in <module>
5 print(g, a, b, x, y, key)
6
----> 7 crack_DH(p, g, x)

<ipython-input-17-9a63fe3af2e8> in crack_DH(p, g, x)
3         Not feasible for large p '''
4     for a in range(1, p - 1):
----> 5         if a % 100000 == 0:
6             print(a) #just to estimate running time
7         if pow(g, a, p) == x:

KeyboardInterrupt:

Analyzing the nubmer of years it will take to crack the protocol if $a$ is found at the end (assuming iterating over 100000 $a$s takes a second)

In :
a
Out:
563975178376164531572554628719
In :
a//100000/60/60/24/365
Out:
1.7883535590314707e+17

A linked list is a linear data structure (i.e. - nodes can be accessed one after the other). The list is composed of nodes, where each node contains a value and a "pointer" to the next node in the list.

Linked lists support operations such as insertion, deletion, search and many more.

In :
class Node():
def __init__(self, val):
self.value = val
self.next = None

def __repr__(self):
#return  str(self.value)

#for today's recitation, we print the id of self as well
return str(self.value) + "(" + str(id(self))+ ")"

def __init__(self, seq=None):
self.next = None
self.len = 0
if seq != None:
for x in seq[::-1]:

def __repr__(self):
out = ""
p = self.next
while  p != None :
out += str(p) + ", " # str(p) envokes __repr__ of class Node
p = p.next
return "[" + out[:-2] + "]"

def __len__(self):
''' called when using Python's len() '''
return self.len

''' add node with value val at the list head '''
p = self
tmp = p.next
p.next = Node(val)
p.next.next = tmp
self.len += 1

''' add node with value val at the list tail '''
p = self
while (p.next != None):
p = p.next
p.next = Node(val)
self.len += 1

def insert(self, loc, val):
''' add node with value val after location 0<=loc<len of the list '''
assert 0 <= loc < len(self)
p = self
for i in range(0, loc):
p = p.next
tmp = p.next
p.next = Node(val)
p.next.next = tmp
self.len += 1

def __getitem__(self, loc):
''' called when using L[i] for reading
return node at location 0<=loc<len '''
assert 0 <= loc < len(self)
p = self.next
for i in range(0, loc):
p = p.next
return p

def __setitem__(self, loc, val):
''' called when using L[loc]=val for writing
assigns val to node at location 0<=loc<len '''
assert 0 <= loc < len(self)
p = self.next
for i in range(0, loc):
p = p.next
p.value = val
return None

def find(self, val):
''' find (first) node with value val in list '''
p = self.next
# loc = 0     # in case we want to return the location
while p != None:
if p.value == val:
return p
else:
p = p.next
#loc=loc+1   # in case we want to return the location
return None

def delete(self, loc):
''' delete element at location 0<=loc<len '''
assert 0 <= loc < len(self)
p = self
for i in range(0, loc):
p = p.next
# p is the element BEFORE loc
p.next = p.next.next
self.len -= 1

def insert_ordered(self, val):
''' assume self is an ordered list,
insert Node with val in order '''
p = self
q = self.next
while q != None and q.value < val:
p = q
q = q.next
newNode = Node(val)
p.next = newNode
newNode.next = q
self.len += 1

def find_ordered(self, val):
''' assume self is an ordered list,
find Node with value val '''
p = self.next
while p != None and p.value < val:
p = p.next
if p != None and p.value == val:
return p
else:
return None

Converting a string to a list

In :
L
Out:
[a(2165820022568), b(2165820022736), c(2165820022624)]

Memory View¶ Adding elements one by one. Try using Python tutor.

In :
l
Out:
[b(2165819893856), a(2165819894528)]
A short summary of methods complexity of Linked lists vs. lists:¶
 Method List Linked list init $O(1)$ $O(1)$ init with $k$ items $O(k)$ $O(k)$ for the rest of the methods we assume that the structure contains $n$ items add at start $O(n)$ $O(1)$ add at end $O(1)$ $O(n)$ $lst[k]$ (get the $k$th item) $O(1)$ $O(k)$ delete $lst[k]$ $O(n-k)$ $O(k)$

Reversing a linked list in $O(1)$ additional memory. See the code and demo here

In :
class Node():
def __init__(self, val):
self.value = val
self.next = None

def __repr__(self):
#return str(self.value)

#for today's recitation, we print the id of self as well
return  str(self.value) + "(" + str(id(self))+ ")"

def __init__(self, seq=None):
self.next = None
self.len = 0
if seq != None:
for x in seq[::-1]:

def __repr__(self):
out = ""
p = self.next
while  p != None :
out += str(p) + ", " # str(p) envokes __repr__ of class Node
p = p.next
return "[" + out[:-2] + "]"

def __len__(self):
''' called when using Python's len() '''
return self.len

''' add node with value val at the list head '''
p = self
tmp = p.next
p.next = Node(val)
p.next.next = tmp
self.len += 1

''' add node with value val at the list tail '''
p = self
while (p.next != None):
p = p.next
p.next = Node(val)
self.len += 1

def insert(self, loc, val):
''' add node with value val after location 0<=loc<len of the list '''
assert 0 <= loc < len(self)
p = self
for i in range(0, loc):
p = p.next
tmp = p.next
p.next = Node(val)
p.next.next = tmp
self.len += 1

def __getitem__(self, loc):
''' called when using L[i] for reading
return node at location 0<=loc<len '''
assert 0 <= loc < len(self)
p = self.next
for i in range(0, loc):
p = p.next
return p

def __setitem__(self, loc, val):
''' called when using L[loc]=val for writing
assigns val to node at location 0<=loc<len '''
assert 0 <= loc < len(self)
p = self.next
for i in range(0, loc):
p = p.next
p.value = val
return None

def find(self, val):
''' find (first) node with value val in list '''
p = self.next
# loc = 0     # in case we want to return the location
while p != None:
if p.value == val:
return p
else:
p = p.next
#loc=loc+1   # in case we want to return the location
return None

def delete(self, loc):
''' delete element at location 0<=loc<len '''
assert 0 <= loc < len(self)
p = self
for i in range(0, loc):
p = p.next
# p is the element BEFORE loc
p.next = p.next.next
self.len -= 1

def insert_ordered(self, val):
''' assume self is an ordered list,
insert Node with val in order '''
p = self
q = self.next
while q != None and q.value < val:
p = q
q = q.next
newNode = Node(val)
p.next = newNode
newNode.next = q
self.len += 1

def find_ordered(self, val):
''' assume self is an ordered list,
find Node with value val '''
p = self.next
while p != None and p.value < val:
p = p.next
if p != None and p.value == val:
return p
else:
return None

def reverse(self):
prev = None
curr = self.next
while curr != None:
tmp = curr.next
curr.next = prev
prev = curr
curr = tmp
self.next = prev