diff --git a/qkd.py b/00_qkd.py
similarity index 100%
rename from qkd.py
rename to 00_qkd.py
diff --git a/computer.py b/01_computer.py
similarity index 100%
rename from computer.py
rename to 01_computer.py
diff --git a/02_cirq.py b/02_cirq.py
new file mode 100644
index 0000000..f40d918
--- /dev/null
+++ b/02_cirq.py
@@ -0,0 +1,18 @@
+import cirq
+
+# Pick a qubit.
+qubit = cirq.GridQubit(0, 0)
+
+# Create a circuit
+circuit = cirq.Circuit.from_ops(
+    cirq.X(qubit)**0.5,  # Square root of NOT.
+    cirq.measure(qubit, key='m')  # Measurement.
+)
+print("Circuit:")
+print(circuit)
+
+# Simulate the circuit several times.
+simulator = cirq.Simulator()
+result = simulator.run(circuit, repetitions=20)
+print("Results:")
+print(result)
\ No newline at end of file
diff --git a/03_shors.md b/03_shors.md
new file mode 100644
index 0000000..6de8fb4
--- /dev/null
+++ b/03_shors.md
@@ -0,0 +1,29 @@
+RSA relies on:
+
+```
+N = p*q
+```
+
+where N is a big number and `p` and `q` are large prime number.
+
+So, say we have `N`. Make a guess `g`. `g` doesn't have to be a factor of N,
+it can be a number that shares factors with N, e.g.
+
+```
+N = g * h
+
+N = a * b  # 4 = 2 * 2
+g = a * c  # 6 = 2 * 3
+gcd(N, g) = a
+
+g^(p/2) +- 1
+
+A, B -> no common factors
+A, B -> A^p = m*B + 1
+7, 15 -> 7^2 = 3*15 + 1
+         7^3 = 22*15 + 13
+         160*15 + 1
+
+```
+
+
diff --git a/03_shors.py b/03_shors.py
new file mode 100644
index 0000000..e5d5226
--- /dev/null
+++ b/03_shors.py
@@ -0,0 +1,84 @@
+"""
+RSA is an asymmetric encr algo that relies on a public key and a private key. 
+Public key encrypts message M to a cyphertext C:
+  
+  C = M^e % N
+
+e is part of the public key and is implementation specific, chosen
+but frequently chosen to be 3 (bad) or 17 or 65537 (a number that has a 
+lot of 0s, makes "computer maths easy).
+
+N is also part of the public key - N = p*q, where p and q are (large) prime 
+numbers.
+
+Decrypting the message:
+
+  M = C^d % N
+
+d is the part of the private key (private exponent) derived using p and q,
+such that:
+
+  e*d=1%((p-1)(q-1))
+
+To break the encryption, if one gets to factor N to p and q, d can be 
+derviced from there
+
+"""
+import sys
+
+def d(p, q, e=3):
+    """ TODO: Doesn't make sense!?"""
+    return 1%((p-1)*(q-1))
+
+def gcd(a, b):
+    if b == 0:
+        return a
+    return gcd(b, a % b) 
+
+
+def trick(A, B, p_max=10):
+    """
+    If A and B are mutually prime (gcd is 1):
+    
+    A^p = m*B + 1
+
+    for some p and m. 
+    
+    returns p and m 
+    """
+    if gcd(A,B) != 1:
+        print("ERROR: A, B not mutually prime!")
+        return None, None
+    for p in range(2, p_max):
+        m, found = 1, False
+        lhs, rhs = A**p, m*B + 1
+        while lhs > rhs:
+            rhs = m*B + 1
+            print("({}, {}): {} =? {}".format(p, m, lhs, rhs))
+            if lhs == rhs:
+                found = True
+                break
+            m += 1
+        if found:
+            print('===== FOUND')
+            print("p={}, m={}".format(p, m))
+            return p, m
+    print("Not found up to {}".format(p_max))
+    return None, None
+
+
+
+# (g^p/2 + 1) * (g^p/2 -1) = m*N --- this is almost like a*b~=N
+# because of m*N, the terms on LHS might not be the factors but
+# multiple of factors, i.e. a*factor, b*factor
+# e.g. 7^4/2 + 1 = 50; 7^4/2 - 1 = 48; which are not factors of 15
+# gcd(50, 15) = 5, gcd(48, 15) = 3
+
+# Trick #2:
+# g^x = m_1*N + r
+# g^(x+p) = m_2*N + r
+# [if g^p = m_1*N + 1 => expanding g^(x+p)=g^x*x^p=(m_1*N+1)*(m_2*N+r)=...=(m_1*m_2 + 3m_1+m_2)*N +r ~= something*N+r 
+
+if __name__ == '__main__':
+    trick(int(sys.argv[1]), int(sys.argv[2]))
+