Detect ECB via duplicated chunks instead of Hemming distance as I should have done in the first place.

src/bytes.rs

@@ -95,6 +95,20 @@ impl Bytes {
         )
     }
 
+    pub fn has_duplicated_cycle(&self, block_size: usize) -> bool {
+        let Bytes(v) = self;
+        let chunks: Vec<&[u8]> = v.chunks(block_size).collect();
+        // we could use a hashmap to get O(n) instead of O(n^2)
+        for i in 0..chunks.len() {
+            for j in (i + 1)..chunks.len() {
+                if chunks[i] == chunks[j] {
+                    return true;
+                }
+            }
+        }
+        false
+    }
+
     #[allow(dead_code)]
     pub fn hemming(Bytes(a): &Bytes, Bytes(b): &Bytes) -> u32 {
         let v: Vec<u32> = Iterator::zip(a.iter(), b.iter())

src/set1.rs

@@ -193,17 +193,28 @@ pub fn challenge8() {
         rating
     }
 
+    let expected_index: usize = 132;
     let bytes_vector = read("data/8.txt");
     let ratings: Vec<u32> = bytes_vector.iter().map(|b| rate(&b, 16)).collect();
     let min_rating = ratings.iter().min().unwrap();
     let index = ratings.iter().position(|e| e == min_rating).unwrap();
     let average = ratings.iter().sum::<u32>() as f32 / ratings.len() as f32;
     let partial_cipher = bytes_vector[index].to_hex()[..10].to_string();
-    if index != 132 {
+    if index != expected_index {
         panic!("Regression in challenge 8.");
     }
     println!(
         "[okay] Challenge 8: Cipher {} [{}...] with rating {} (average = {}) is the solution.",
         index, partial_cipher, min_rating, average
     );
+
+    // More elegant solution.
+    let bytes_vector = read("data/8.txt");
+    let ix: Vec<usize> = bytes_vector
+        .iter()
+        .enumerate()
+        .filter(|&(_, bytes)| bytes.has_duplicated_cycle(16))
+        .map(|(i, _)| i)
+        .collect();
+    assert_eq!(ix[0], expected_index);
 }

src/set2.rs

@@ -57,11 +57,9 @@ pub fn challenge11() {
         // Write a function that encrypts data under an unknown key --- that is, a
         // function that generates a random key and encrypts under it.
         let key = Bytes(random_bytes(16));
-
         // Under the hood, have the function append 5-10 bytes (count chosen randomly)
         // before the plaintext and 5-10 bytes after the plaintext.
-        let _padded_data = pad_data(data.to_vec());
+        let padded_data = pad_data(data.to_vec());
-        let padded_data = Bytes(data.to_vec());
         // Now, have the function choose to encrypt under ECB 1/2 the time, and under CBC
         // the other half (just use random IVs each time for CBC). Use rand(2) to decide
         // which to use.
@@ -69,9 +67,8 @@ pub fn challenge11() {
         let (data, encryption_type) = if zero_or_one == 1 {
             (ecb::encrypt(&key, &padded_data), EncryptionType::ECB)
         } else {
-            // let iv = Bytes(random_bytes(16));
+            let iv = Bytes(random_bytes(16));
-            // (cbc::encrypt(&key, &iv, &padded_data), EncryptionType::CBC)
+            (cbc::encrypt(&key, &iv, &padded_data), EncryptionType::CBC)
-            (Bytes(random_bytes(2876)), EncryptionType::CBC)
         };
         (data, encryption_type)
     }
@@ -80,22 +77,7 @@ pub fn challenge11() {
         // Detect the block cipher mode the function is using each time. You should end up
         // with a piece of code that, pointed at a block box that might be encrypting ECB
         // or CBC, tells you which one is happening.
-
+        if data.has_duplicated_cycle(16) {
-        fn rate(Bytes(v): &Bytes, size: usize) -> u32 {
-            let mut rating = 0;
-            let chunks: Vec<&[u8]> = v.chunks(size).collect();
-            let mut count = 0;
-            for i in 0..chunks.len() {
-                for j in (i + 1)..chunks.len() {
-                    rating +=
-                        Bytes::hemming(&Bytes(chunks[i].to_vec()), &Bytes(chunks[j].to_vec()));
-                    count += 1;
-                }
-            }
-            rating / count
-        }
-        let rating = rate(&data, 16);
-        if rating < 50 {
             EncryptionType::ECB
         } else {
             EncryptionType::CBC
@@ -103,6 +85,11 @@ pub fn challenge11() {
     }
 
     fn run_oracle(count: usize) {
+        // I struggled for a while to understand how the detection oracle can
+        // work.  The issue is that if we provide a random text to encrypt, we
+        // won't be able to use duplicated blocks as a method for detecting the
+        // cipher. I don't think there is a way around that, but if we can choose
+        // the plaintext it becomes trivial.
         let text = Bytes::from_utf8("aaaabbbbccccddddaaaabbbbccccddddaaaabbbbccccdddd");
         let mut correct: usize = 0;
         for _ in 0..count {