Solve challenge 47 in Python

I will also implement it in Rust but prototyping is just easier and
faster in Python.

data/set6c47.py

@@ -1,4 +1,6 @@
+import sys
 from math import log, ceil, floor
+from fractions import Fraction
 
 
 e = 3
@@ -22,6 +24,7 @@ def add_padding(m: int, k: int) -> int:
         r.append(0xff)
     r.append(0x0)
     r += m.to_bytes(from_len, 'big')
+    assert(len(r) == k)
     r = int.from_bytes(r, byteorder='big')
     return r
 
@@ -52,7 +55,6 @@ def decrypt(c: int) -> int:
 
 def test():
     m = int.from_bytes(b"kick it, CC", byteorder='big')
-    m = 129852745126415640677073731
     assert(m == 129852745126415640677073731)
 
     k = bytes_needed(n)
@@ -74,6 +76,7 @@ def main():
 
     k = bytes_needed(n)
     m = int.from_bytes(b"kick it, CC", byteorder='big')
+    m_orig = m
     c = encrypt(add_padding(m, k))
     assert(m == remove_padding(decrypt(c), k))
 
@@ -86,7 +89,8 @@ def main():
 
     # Step 2: Searching for PKCS conforming messages.
     while True:
-        print(f"{i=}\n{s=}\n{m=}")
+        # print("========")
+        # print(f"{i=} {s=} {m=}")
         if i == 1:
             # Step 2.a: Starting the search.
             s = ceil(n / (3 * B))
@@ -99,12 +103,13 @@ def main():
             # Step 2.c: Searching with one interval left.
             a, b = m[0]
             found = False
-            r = ceil(2 * ((b * s - 2 * B) / n))
+            r = ceil(Fraction(2 * (b * s - 2 * B), n))
             while not found:
-                s_lower = ceil((2 * B + r * n) / b)
-                s_upper = ceil((3 * B + r * n) / a)
-                assert(s_lower < s_upper + 1)
-                for s in range(s_lower, s_upper):
+                s_lower = Fraction(2 * B + r * n, b)
+                s_upper = Fraction(3 * B + r * n, a)
+                s_lower_ceil = ceil(s_lower)
+                s_upper_ceil = ceil(s_upper)
+                for s in range(s_lower_ceil, s_upper_ceil):
                     if oracle(c * pow(s, e, n) % n):
                         found = True
                         break
@@ -115,25 +120,32 @@ def main():
         # Step 3: Narrowing the set of solutions.
         m_new = []
         for (a, b) in m:
-            lower = ceil((a * s - 3 * B + 1) / n)
-            upper = ceil((b * s - 2 * B) / n)
-            for r in range(lower, upper):
-                a_new = max(a, ceil((2 * B + r * n) / s))
-                b_new = min(b, floor((3 * B - 1 + r * n) / s))
+            lower = Fraction((a * s - 3 * B + 1), n)
+            upper = Fraction((b * s - 2 * B), n)
+            lower_ceil = ceil(lower)
+            upper_ceil = ceil(upper)
+
+            # If upper is an integer we have to increment it to include it into
+            # the range.
+            if upper == upper_ceil:
+                upper_ceil += 1
+
+            for r in range(lower_ceil, upper_ceil):
+                a_new = max(a, ceil(Fraction(2 * B + r * n, s)))
+                b_new = min(b, floor(Fraction(3 * B - 1 + r * n, s)))
                 m_new.append((a_new, b_new))
         m = m_new
 
         # Step 4: Computing the solutions.
         if len(m) == 1 and m[0][0] == m[0][1]:
-            m = a * pow(s0, -1, n) % n
+            m = m[0][0]
             break
         i = i + 1
 
-    print(m)
-
-
-
-
+    assert(m == 5300541194335152988749892502228755547482451611528547105226896651010982723)
+    m = remove_padding(m, k)
+    assert(m == m_orig)
+    print("[okay] Challenge 47: Bleichenbacher's PKCS 1.5 Padding Oracle (Simple Case)")
 
 
 main()