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clusterlin: permit passing in existing linearization to Linearize
This implements the LIMO algorithm for linearizing by improving an existing linearization. See https://delvingbitcoin.org/t/limo-combining-the-best-parts-of-linearization-search-and-merging for details.
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3 changed files with 53 additions and 9 deletions
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@ -109,7 +109,7 @@ void BenchLinearizePerIterWorstCase(ClusterIndex ntx, benchmark::Bench& bench)
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});
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}
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/** Benchmark for linearization of a trivial linear graph using just ancestor sort.
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/** Benchmark for linearization improvement of a trivial linear graph using just ancestor sort.
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*
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* Its goal is measuring how much time linearization may take without any search iterations.
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*
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@ -124,8 +124,10 @@ void BenchLinearizeNoItersWorstCase(ClusterIndex ntx, benchmark::Bench& bench)
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{
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const auto depgraph = MakeLinearGraph<SetType>(ntx);
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uint64_t rng_seed = 0;
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std::vector<ClusterIndex> old_lin(ntx);
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for (ClusterIndex i = 0; i < ntx; ++i) old_lin[i] = i;
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bench.run([&] {
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Linearize(depgraph, /*max_iterations=*/0, rng_seed++);
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Linearize(depgraph, /*max_iterations=*/0, rng_seed++, old_lin);
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});
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}
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@ -663,23 +663,27 @@ public:
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}
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};
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/** Find a linearization for a cluster.
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/** Find or improve a linearization for a cluster.
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*
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* @param[in] depgraph Dependency graph of the cluster to be linearized.
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* @param[in] max_iterations Upper bound on the number of optimization steps that will be done.
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* @param[in] rng_seed A random number seed to control search order. This prevents peers
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* from predicting exactly which clusters would be hard for us to
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* linearize.
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* @param[in] old_linearization An existing linearization for the cluster (which must be
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* topologically valid), or empty.
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* @return A pair of:
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* - The resulting linearization.
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* - The resulting linearization. It is guaranteed to be at least as
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* good (in the feerate diagram sense) as old_linearization.
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* - A boolean indicating whether the result is guaranteed to be
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* optimal.
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*
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* Complexity: O(N * min(max_iterations + N, 2^N)) where N=depgraph.TxCount().
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*/
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template<typename SetType>
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std::pair<std::vector<ClusterIndex>, bool> Linearize(const DepGraph<SetType>& depgraph, uint64_t max_iterations, uint64_t rng_seed) noexcept
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std::pair<std::vector<ClusterIndex>, bool> Linearize(const DepGraph<SetType>& depgraph, uint64_t max_iterations, uint64_t rng_seed, Span<const ClusterIndex> old_linearization = {}) noexcept
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{
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Assume(old_linearization.empty() || old_linearization.size() == depgraph.TxCount());
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if (depgraph.TxCount() == 0) return {{}, true};
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uint64_t iterations_left = max_iterations;
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@ -690,9 +694,17 @@ std::pair<std::vector<ClusterIndex>, bool> Linearize(const DepGraph<SetType>& de
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linearization.reserve(depgraph.TxCount());
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bool optimal = true;
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/** Chunking of what remains of the old linearization. */
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LinearizationChunking old_chunking(depgraph, old_linearization);
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while (true) {
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// Initialize best as the best remaining ancestor set.
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// Find the highest-feerate prefix of the remainder of old_linearization.
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SetInfo<SetType> best_prefix;
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if (old_chunking.NumChunksLeft()) best_prefix = old_chunking.GetChunk(0);
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// Then initialize best to be either the best remaining ancestor set, or the first chunk.
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auto best = anc_finder.FindCandidateSet();
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if (!best_prefix.feerate.IsEmpty() && best_prefix.feerate >= best.feerate) best = best_prefix;
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// Invoke bounded search to update best, with up to half of our remaining iterations as
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// limit.
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@ -703,6 +715,12 @@ std::pair<std::vector<ClusterIndex>, bool> Linearize(const DepGraph<SetType>& de
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if (iterations_done_now == max_iterations_now) {
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optimal = false;
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// If the search result is not (guaranteed to be) optimal, run intersections to make
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// sure we don't pick something that makes us unable to reach further diagram points
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// of the old linearization.
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if (old_chunking.NumChunksLeft() > 0) {
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best = old_chunking.Intersect(best);
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}
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}
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// Add to output in topological order.
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@ -712,6 +730,9 @@ std::pair<std::vector<ClusterIndex>, bool> Linearize(const DepGraph<SetType>& de
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anc_finder.MarkDone(best.transactions);
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if (anc_finder.AllDone()) break;
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src_finder.MarkDone(best.transactions);
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if (old_chunking.NumChunksLeft() > 0) {
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old_chunking.MarkDone(best.transactions);
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}
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}
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return {std::move(linearization), optimal};
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@ -143,8 +143,9 @@ public:
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/** A simple linearization algorithm.
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*
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* This matches Linearize() in interface and behavior, though with fewer optimizations, and using
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* just SimpleCandidateFinder rather than AncestorCandidateFinder and SearchCandidateFinder.
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* This matches Linearize() in interface and behavior, though with fewer optimizations, lacking
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* the ability to pass in an existing linearization, and using just SimpleCandidateFinder rather
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* than AncestorCandidateFinder and SearchCandidateFinder.
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*/
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template<typename SetType>
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std::pair<std::vector<ClusterIndex>, bool> SimpleLinearize(const DepGraph<SetType>& depgraph, uint64_t max_iterations)
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@ -614,12 +615,32 @@ FUZZ_TARGET(clusterlin_linearize)
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reader >> VARINT(iter_count) >> Using<DepGraphFormatter>(depgraph) >> rng_seed;
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} catch (const std::ios_base::failure&) {}
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// Optionally construct an old linearization for it.
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std::vector<ClusterIndex> old_linearization;
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{
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uint8_t have_old_linearization{0};
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try {
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reader >> have_old_linearization;
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} catch(const std::ios_base::failure&) {}
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if (have_old_linearization & 1) {
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old_linearization = ReadLinearization(depgraph, reader);
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SanityCheck(depgraph, old_linearization);
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}
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}
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// Invoke Linearize().
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iter_count &= 0x7ffff;
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auto [linearization, optimal] = Linearize(depgraph, iter_count, rng_seed);
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auto [linearization, optimal] = Linearize(depgraph, iter_count, rng_seed, old_linearization);
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SanityCheck(depgraph, linearization);
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auto chunking = ChunkLinearization(depgraph, linearization);
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// Linearization must always be as good as the old one, if provided.
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if (!old_linearization.empty()) {
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auto old_chunking = ChunkLinearization(depgraph, old_linearization);
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auto cmp = CompareChunks(chunking, old_chunking);
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assert(cmp >= 0);
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}
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// If the iteration count is sufficiently high, an optimal linearization must be found.
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// Each linearization step can use up to 2^k iterations, with steps k=1..n. That sum is
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// 2 * (2^n - 1)
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