We continued with another recurssion question. 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.

- 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.

In [1]:

```
from IPython.core.interactiveshell import InteractiveShell
InteractiveShell.ast_node_interactivity = "all"
```

The $N$ queens problem is to determine how many possibilities are there to legally place $N$ queens on an $N$-by-$N$ chess board. Legally means no queen threatens another queen.

Solution: We build the solution incrementally, column by column. We maintain a partial solution (implemented as a list), which is initially empty.

The function $legal(partial, i)$ is given:

- a partial solution placing the first $len(partial)$ queens on the leftmost columns
- a positive integer $0\leq i<N$

It returns True if it is legal to place a queen in the next column on row $i$, given the partial placement $partial$

In [10]:

```
def legal(partial, i):
''' Can we place a queen in the next column in row i ? '''
# any queens in the same row to the left ?
left = [j for j in partial if j==i]
# diagonal up - left
diag_up = [j for j in partial if j - partial.index(j) == i - len(partial)]
# diagonal down - left
diag_down = [j for j in partial if j + partial.index(j) == i + len(partial)]
res = ( left == diag_up == diag_down == [])
return res
```

The solution:

In [11]:

```
def queens (n, show = True):
''' how many ways to place n queens on an nXn board ? '''
partial = [] # list representing partial placement of queens
return queens_rec (n, partial, show)
def queens_rec (n, partial, show):
''' Given a list representing partial placement of queens ,
can we legally extend it ? '''
if len(partial) == n: # all n queens are placed legally
return 1
else:
cnt = 0
for i in range(n):
# try to place a queen in row i of the next column
if legal(partial, i):
cnt += queens_rec(n, partial + [i], show)
return cnt
```

In [12]:

```
print(queens(0))
print(queens(1))
print(queens(2))
print(queens(3))
print(queens(10))
```

Some intuition: assume we can find a solution for placing $k<N$ queens, how do we expand the solution to $k+1$? queens_rec returns the number of possible legal placements for $N$ queens, where $k$ are already placed at the leftmost columns and there are $N-k$ queens left to place. The recursive idea: Legally place queen number $(k+1)$ and recursively solve the problem, when there is one less queen to place.

Note that the complexity is $O(N!)$ (greater than $O(2^N)$)

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

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 $r ~|~ m$. 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 $\mathrm{GCD}_m = \mathrm{FERM}_m$

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.

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 [9]:

```
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 [14]:

```
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
```

- 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
- One operation of $b/2$ in $O(1)$ time
- 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 - $O(n^3)$

First, notice that if the function says that an imput number $m$ is not prime, then it is true. The function can make a mistake only is the case where a number $m$ is not prime, and is excidentally categorized by the function as prime. This can happen if all $100$ $a$'s that the function tried 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 $(\frac{1}{2})^{100}$ (this is extremely low).

In the rare case where we get a Carmichael number as input we have no guarantee on the performance of the test. In the following example we test whether the Carmichael number $N$ (which is composite) is prime or not using our test.

The number $N$ comes courtesy of Wikipedia

In [15]:

```
# p is a large prime:
p = 29674495668685510550154174642905332730771991799853043350995075531276838753171770199594238596428121188033664754218345562493168782883
# N is a Carmichael number (and is obviously composite):
N = p*(313*(p-1)+1)*(353*(p-1) + 1)
is_prime(N)
```

Out[15]:

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 [16]:

```
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 [17]:

```
prob_prime(2, 10**4)
prob_prime(3, 10**4)
```

Out[17]:

Out[17]:

In [18]:

```
prob_prime(100, 10**4)
```

Out[18]:

In [19]:

```
prob_prime(200, 10**4)
```

Out[19]:

Diffie Hellman from lecture

In [8]:

```
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
```

In [10]:

```
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 [21]:

```
import random
p = find_prime(10)
print(p)
g, a, b, x, y, key = DH_exchange(p)
g, a, b, x, y, key
```

Out[21]:

In [22]:

```
print(pow(g, a, p))
print(pow(x, b, p))
```

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 [24]:

```
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 [28]:

```
g, a, b, x, y, key = DH_exchange(p)
print(a)
crack_DH(p, g, x)
```

Out[28]:

In [30]:

```
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)
```

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 [31]:

```
a
```

Out[31]:

In [32]:

```
a//100000/60/60/24/365
```

Out[32]:

In [ ]:

```
```