3.7.2 Remove code randomizing which side odd elements end up on

This could still be gamed. For [1, 2, 3], the options were ([1], [2, 3]) or
([1, 2], [3]). This means 2 would always have the maximum round count, and
thus this is still game-able. There's no point to keeping its complexity
accordingly when the algorithm is as efficient as it is.

While a proper random could be used to satisfy 3.7.2, it'd break the
expected determinism.

crypto/multiexp/src/batch.rs

@@ -22,13 +22,9 @@ where
 /// A batch verifier intended to verify a series of statements are each equivalent to zero.
 #[allow(clippy::type_complexity)]
 #[derive(Clone, Zeroize)]
-pub struct BatchVerifier<Id: Copy + Zeroize, G: Group + Zeroize>
+pub struct BatchVerifier<Id: Copy + Zeroize, G: Group + Zeroize>(Zeroizing<Vec<(Id, Vec<(G::Scalar, G)>)>>)
 where
-  <G as Group>::Scalar: PrimeFieldBits + Zeroize,
+  <G as Group>::Scalar: PrimeFieldBits + Zeroize;
-{
-  split: u64,
-  statements: Zeroizing<Vec<(Id, Vec<(G::Scalar, G)>)>>,
-}
 
 impl<Id: Copy + Zeroize, G: Group + Zeroize> BatchVerifier<Id, G>
 where
@@ -37,7 +33,7 @@ where
   /// Create a new batch verifier, expected to verify the following amount of statements.
   /// This is a size hint and is not required to be accurate.
   pub fn new(capacity: usize) -> BatchVerifier<Id, G> {
-    BatchVerifier { split: 0, statements: Zeroizing::new(Vec::with_capacity(capacity)) }
+    BatchVerifier(Zeroizing::new(Vec::with_capacity(capacity)))
   }
 
   /// Queue a statement for batch verification.
@@ -47,15 +43,8 @@ where
     id: Id,
     pairs: I,
   ) {
-    // If this is the first time we're queueing a value, grab a u64 (AKA 64 bits) to determine
-    // which side to handle odd splits with during blame (preventing malicious actors from gaming
-    // the system by maximizing the round length)
-    if self.statements.len() == 0 {
-      self.split = rng.next_u64();
-    }
-
     // Define a unique scalar factor for this set of variables so individual items can't overlap
-    let u = if self.statements.is_empty() {
+    let u = if self.0.is_empty() {
       G::Scalar::one()
     } else {
       let mut weight;
@@ -98,20 +87,20 @@ where
     };
 
     self
-      .statements
+      .0
       .push((id, pairs.into_iter().map(|(scalar, point)| (scalar * u, point)).collect()));
   }
 
   /// Perform batch verification, returning a boolean of if the statements equaled zero.
   #[must_use]
   pub fn verify(&self) -> bool {
-    multiexp(&flat(&self.statements)).is_identity().into()
+    multiexp(&flat(&self.0)).is_identity().into()
   }
 
   /// Perform batch verification in variable time.
   #[must_use]
   pub fn verify_vartime(&self) -> bool {
-    multiexp_vartime(&flat(&self.statements)).is_identity().into()
+    multiexp_vartime(&flat(&self.0)).is_identity().into()
   }
 
   /// Perform a binary search to identify which statement does not equal 0, returning None if all
@@ -119,30 +108,9 @@ where
   /// multiple are invalid.
   // A constant time variant may be beneficial for robust protocols
   pub fn blame_vartime(&self) -> Option<Id> {
-    let mut slice = self.statements.as_slice();
+    let mut slice = self.0.as_slice();
-    let mut split_side = self.split;
-
     while slice.len() > 1 {
-      let mut split = slice.len() / 2;
+      let split = slice.len() / 2;
-      // If there's an odd number of elements, this can be gamed to increase the amount of rounds
-      // For [0, 1, 2], where 2 is invalid, this will take three rounds
-      // ([0], [1, 2]), then ([1], [2]), before just 2
-      // If 1 and 2 were valid, this would've only taken 2 rounds to complete
-      // To prevent this from being gamed, if there's an odd number of elements, randomize which
-      // side the split occurs on
-
-      // This does risk breaking determinism
-      // The concern is if the select split point causes different paths to be taken when multiple
-      // invalid elements exist
-      // While the split point may move an element from the right to the left, always choosing the
-      // left side (if it's invalid) means this will still always return the left-most,
-      // invalid element
-
-      if slice.len() % 2 == 1 {
-        split += usize::try_from(split_side & 1).unwrap();
-        split_side >>= 1;
-      }
-
       if multiexp_vartime(&flat(&slice[.. split])).is_identity().into() {
         slice = &slice[split ..];
       } else {