Move with_huffman_tree logic to TaprootBuilder
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@ -191,24 +191,9 @@ pub struct TaprootSpendInfo {
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}
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}
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impl TaprootSpendInfo {
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impl TaprootSpendInfo {
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/// Create a new [`TaprootSpendInfo`] from a list of script(with default script version) and
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/// Create a new [`TaprootSpendInfo`] from a list of script(with default script version).
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/// weights of satisfaction for that script. The weights represent the probability of
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/// each branch being taken. If probabilities/weights for each condition are known,
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/// constructing the tree as a Huffman tree is the optimal way to minimize average
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/// case satisfaction cost. This function takes input an iterator of tuple(u64, &Script)
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/// where usize represents the satisfaction weights of the branch.
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/// For example, [(3, S1), (2, S2), (5, S3)] would construct a TapTree that has optimal
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/// satisfaction weight when probability for S1 is 30%, S2 is 20% and S3 is 50%.
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///
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///
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/// # Errors:
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/// See [`TaprootBuilder::with_huffman_tree`] for more detailed documentation
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///
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/// - When the optimal huffman tree has a depth more than 128
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/// - If the provided list of script weights is empty
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///
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/// # Edge Cases:
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/// - If the script weight calculations overflow, a sub-optimal tree may be generated. This
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/// should not happen unless you are dealing with billions of branches with weights close to
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/// 2^32.
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pub fn with_huffman_tree<C, I>(
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pub fn with_huffman_tree<C, I>(
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secp: &Secp256k1<C>,
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secp: &Secp256k1<C>,
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internal_key: UntweakedPublicKey,
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internal_key: UntweakedPublicKey,
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@ -218,29 +203,7 @@ impl TaprootSpendInfo {
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I: IntoIterator<Item=(u32, Script)>,
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I: IntoIterator<Item=(u32, Script)>,
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C: secp256k1::Verification,
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C: secp256k1::Verification,
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{
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{
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let mut node_weights = BinaryHeap::<(Reverse<u64>, NodeInfo)>::new();
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TaprootBuilder::with_huffman_tree(script_weights)?.finalize(secp, internal_key)
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for (p, leaf) in script_weights {
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node_weights.push((Reverse(p as u64), NodeInfo::new_leaf_with_ver(leaf, LeafVersion::TapScript)));
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}
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if node_weights.is_empty() {
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return Err(TaprootBuilderError::IncompleteTree);
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}
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while node_weights.len() > 1 {
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// Combine the last two elements and insert a new node
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let (p1, s1) = node_weights.pop().expect("len must be at least two");
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let (p2, s2) = node_weights.pop().expect("len must be at least two");
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// Insert the sum of first two in the tree as a new node
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// N.B.: p1 + p2 can not practically saturate as you would need to have 2**32 max u32s
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// from the input to overflow. However, saturating is a reasonable behavior here as
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// huffman tree construction would treat all such elements as "very likely".
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let p = Reverse(p1.0.saturating_add(p2.0));
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node_weights.push((p, NodeInfo::combine(s1, s2)?));
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}
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// Every iteration of the loop reduces the node_weights.len() by exactly 1
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// Therefore, the loop will eventually terminate with exactly 1 element
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debug_assert!(node_weights.len() == 1);
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let node = node_weights.pop().expect("huffman tree algorithm is broken").1;
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return Ok(Self::from_node_info(secp, internal_key, node));
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}
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}
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/// Create a new key spend with internal key and proided merkle root.
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/// Create a new key spend with internal key and proided merkle root.
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@ -396,6 +359,56 @@ impl TaprootBuilder {
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pub fn new() -> Self {
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pub fn new() -> Self {
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TaprootBuilder { branch: vec![] }
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TaprootBuilder { branch: vec![] }
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}
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}
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/// Create a new [`TaprootBuilder`] from a list of script(with default script version) and
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/// weights of satisfaction for that script. The weights represent the probability of
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/// each branch being taken. If probabilities/weights for each condition are known,
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/// constructing the tree as a Huffman tree is the optimal way to minimize average
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/// case satisfaction cost. This function takes input an iterator of tuple(u64, &Script)
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/// where usize represents the satisfaction weights of the branch.
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/// For example, [(3, S1), (2, S2), (5, S3)] would construct a TapTree that has optimal
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/// satisfaction weight when probability for S1 is 30%, S2 is 20% and S3 is 50%.
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///
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/// # Errors:
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///
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/// - When the optimal huffman tree has a depth more than 128
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/// - If the provided list of script weights is empty
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///
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/// # Edge Cases:
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/// - If the script weight calculations overflow, a sub-optimal tree may be generated. This
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/// should not happen unless you are dealing with billions of branches with weights close to
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/// 2^32.
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pub fn with_huffman_tree<I>(
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script_weights: I,
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) -> Result<Self, TaprootBuilderError>
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where
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I: IntoIterator<Item=(u32, Script)>,
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{
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let mut node_weights = BinaryHeap::<(Reverse<u64>, NodeInfo)>::new();
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for (p, leaf) in script_weights {
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node_weights.push((Reverse(p as u64), NodeInfo::new_leaf_with_ver(leaf, LeafVersion::TapScript)));
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}
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if node_weights.is_empty() {
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return Err(TaprootBuilderError::IncompleteTree);
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}
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while node_weights.len() > 1 {
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// Combine the last two elements and insert a new node
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let (p1, s1) = node_weights.pop().expect("len must be at least two");
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let (p2, s2) = node_weights.pop().expect("len must be at least two");
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// Insert the sum of first two in the tree as a new node
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// N.B.: p1 + p2 can not practically saturate as you would need to have 2**32 max u32s
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// from the input to overflow. However, saturating is a reasonable behavior here as
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// huffman tree construction would treat all such elements as "very likely".
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let p = Reverse(p1.0.saturating_add(p2.0));
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node_weights.push((p, NodeInfo::combine(s1, s2)?));
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}
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// Every iteration of the loop reduces the node_weights.len() by exactly 1
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// Therefore, the loop will eventually terminate with exactly 1 element
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debug_assert!(node_weights.len() == 1);
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let node = node_weights.pop().expect("huffman tree algorithm is broken").1;
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Ok(TaprootBuilder{branch: vec![Some(node)]})
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}
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/// Just like [`TaprootBuilder::add_leaf`] but allows to specify script version
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/// Just like [`TaprootBuilder::add_leaf`] but allows to specify script version
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pub fn add_leaf_with_ver(
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pub fn add_leaf_with_ver(
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self,
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self,
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