Learning from Examples » k-means Clustering | Taskflow QuickStart

# Learning from Examples » k-means Clustering

We study a fundamental clustering problem in unsupervised learning, k-means clustering. We will begin by discussing the problem formulation and then learn how to write a parallel k-means algorithm.

## Problem Formulation

k-means clustering uses centroids, k different randomly-initiated points in the data, and assigns every data point to the nearest centroid. After every point has been assigned, the centroid is moved to the average of all of the points assigned to it. We describe the k-means algorithm in the following steps:

- Step 1: initialize k random centroids
- Step 2: for every data point, find the nearest centroid (L2 distance or other measurements) and assign the point to it
- Step 3: for every centroid, move the centroid to the average of the points assigned to that centroid
- Step 4: go to Step 2 until converged (no more changes in the last few iterations) or maximum iterations reached

The algorithm is illustrated as follows:

A sequential implementation of k-means is described as follows:

    // sequential implementation of k-means on a CPU
    // N: number of points
    // K: number of clusters
    // M: number of iterations
    // px/py: 2D point vector
    void kmeans_seq(
      int N, int K, int M, const std::vector<float>& px, const std::vector<float>& py
    ) {

      std::vector<int> c(K);
      std::vector<float> sx(K), sy(K), mx(K), my(K);

      // initial centroids
      std::copy_n(px.begin(), K, mx.begin());
      std::copy_n(py.begin(), K, my.begin());

      // k-means iteration
      for(int m=0; m<M; m++) {

        // clear the storage
        std::fill_n(sx.begin(), K, 0.0f);
        std::fill_n(sy.begin(), K, 0.0f);
        std::fill_n(c.begin(), K, 0);

        // find the best k (cluster id) for each point
        for(int i=0; i<N; ++i) {
          float x = px[i];
          float y = py[i];
          float best_d = std::numeric_limits<float>::max();
          int best_k = 0;
          for (int k = 0; k < K; ++k) {
            const float d = L2(x, y, mx[k], my[k]);
            if (d < best_d) {
              best_d = d;
              best_k = k;
            }
          }
          sx[best_k] += x;
          sy[best_k] += y;
          c [best_k] += 1;
        }

        // update the centroid
        for(int k=0; k<K; k++) {
          const int count = max(1, c[k]);  // turn 0/0 to 0/1
          mx[k] = sx[k] / count;
          my[k] = sy[k] / count;
        }
      }

      // print the k centroids found
      for(int k=0; k<K; ++k) {
        std::cout << "centroid " << k << ": " << std::setw(10) << mx[k] << ' '
                                              << std::setw(10) << my[k] << '\n';
      }
    }

## Parallel k-means using CPUs

The second step of k-means algorithm, assigning every point to the nearest centroid, is highly parallelizable across individual points. We can create a parallel-for task to run parallel iterations.

    std::vector<int> best_ks(N);  // nearest centroid of each point

    unsigned P = 12;  // 12 partitioned tasks

    // update cluster
    taskflow.for_each_index(0, N, 1, [&](int i){
      float x = px[i];
      float y = py[i];
      float best_d = std::numeric_limits<float>::max();
      int best_k = 0;
      for (int k = 0; k < K; ++k) {
        const float d = L2(x, y, mx[k], my[k]);
        if (d < best_d) {
          best_d = d;
          best_k = k;
        }
      }
      best_ks[i] = best_k;
    });

The third step of moving every centroid to the average of points is also parallelizable across individual centroids. However, since k is typically not large, one task of doing this update is sufficient.

    taskflow.emplace([&](){
      // sum of points
      for(int i=0; i<N; i++) {
        sx[best_ks[i]] += px[i];
        sy[best_ks[i]] += py[i];
        c [best_ks[i]] += 1;
      }

      // average of points
      for(int k=0; k<K; ++k) {
        auto count = max(1, c[k]);  // turn 0/0 to 0/1
        mx[k] = sx[k] / count;
        my[k] = sy[k] / count;
      }
    });

To describe `M` iterations, we create a condition task that loops the second step of the algorithm by `M` times. The return value of zero goes to the first successor which we will connect to the task of the second step later; otherwise, k-means completes.

    taskflow.emplace([m=0, M]() mutable {
      return (m++ < M) ? 0 : 1;
    });

The entire code of CPU-parallel k-means is shown below. Here we use an additional storage, `best_ks`, to record the nearest centroid of a point at an iteration.

    // N: number of points
    // K: number of clusters
    // M: number of iterations
    // px/py: 2D point vector
    void kmeans_par(
      int N, int K, int M, cconst std::vector<float>& px, const std::vector<float>& py
    ) {

      unsigned P = 12;  // 12 partitions of the parallel-for graph

      tf::Executor executor;
      tf::Taskflow taskflow("K-Means");

      std::vector<int> c(K), best_ks(N);
      std::vector<float> sx(K), sy(K), mx(K), my(K);

      // initial centroids
      tf::Task init = taskflow.emplace([&](){
        for(int i=0; i<K; ++i) {
          mx[i] = px[i];
          my[i] = py[i];
        }
      }).name("init");

      // clear the storage
      tf::Task clean_up = taskflow.emplace([&](){
        for(int k=0; k<K; ++k) {
          sx[k] = 0.0f;
          sy[k] = 0.0f;
          c [k] = 0;
        }
      }).name("clean_up");

      // update cluster
      tf::Task pf = taskflow.for_each_index(0, N, 1, [&](int i){
        float x = px[i];
        float y = py[i];
        float best_d = std::numeric_limits<float>::max();
        int best_k = 0;
        for (int k = 0; k < K; ++k) {
          const float d = L2(x, y, mx[k], my[k]);
          if (d < best_d) {
            best_d = d;
            best_k = k;
          }
        }
        best_ks[i] = best_k;
      }).name("parallel-for");

      tf::Task update_cluster = taskflow.emplace([&](){
        for(int i=0; i<N; i++) {
          sx[best_ks[i]] += px[i];
          sy[best_ks[i]] += py[i];
          c [best_ks[i]] += 1;
        }

        for(int k=0; k<K; ++k) {
          auto count = max(1, c[k]);  // turn 0/0 to 0/1
          mx[k] = sx[k] / count;
          my[k] = sy[k] / count;
        }
      }).name("update_cluster");

      // convergence check
      tf::Task condition = taskflow.emplace([m=0, M]() mutable {
        return (m++ < M) ? 0 : 1;
      }).name("converged?");

      init.precede(clean_up);

      clean_up.precede(pf);
      pf.precede(update_cluster);

      condition.precede(clean_up)
               .succeed(update_cluster);

      executor.run(taskflow).wait();
    }

The taskflow consists of two parts, a `clean_up` task and a parallel-for graph. The former cleans up the storage `sx`, `sy`, and `c` that are used to average points for new centroids, and the later parallelizes the searching for nearest centroids across individual points using 12 tasks (may vary depending on the machine). If the iteration count is smaller than `M`, the condition task returns 0 to let the execution path go back to `clean_up`. Otherwise, it returns 1 to stop (i.e., no successor tasks at index 1). The taskflow graph is illustrated below:

The scheduler starts with `init`, moves on to `clean_up`, and then enters the parallel-for task `parallel-for` that spawns a subflow of 12 workers to perform parallel iterations. When `parallel-for` completes, it updates the cluster centroids and checks if they have converged through a condition task. If not, the condition task informs the scheduler to go back to `clean_up` and then `parallel-for`; otherwise, it returns a nominal index to stop the scheduler.

## Benchmarking

Based on the discussion above, we compare the runtime of computing various k-means problem sizes between a sequential CPU and parallel CPUs on a machine of 12 Intel i7-8700 CPUs at 3.2 GHz.

N | K | M | CPU Sequential | CPU Parallel
--- | --- | --- | --- | ---
10 | 5 | 10 | 0.14 ms | 77 ms
100 | 10 | 100 | 0.56 ms | 86 ms
1000 | 10 | 1000 | 10 ms | 98 ms
10000 | 10 | 10000 | 1006 ms | 713 ms
100000 | 10 | 100000 | 102483 ms | 49966 ms

When the number of points is larger than 10K, the parallel CPU implementation starts to outperform the sequential CPU implementation.

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