Documentation>C API
Agglomerative Information Bottleneck (AIB)
Brian Fulkerson
Andrea Vedaldi

aib.h implemens the Agglomerative Information Bottleneck (AIB) algorithm as first described in [28] .

AIB takes a discrete valued feature \(x\) and a label \(c\) and gradually compresses \(x\) by iteratively merging values which minimize the loss in mutual information \(I(x,c)\).

While the algorithm is equivalent to the one described in [28] , it has some speedups that enable handling much larger datasets. Let N be the number of feature values and C the number of labels. The algorithm of [28] is \(O(N^2)\) in space and \(O(C N^3)\) in time. This algorithm is \(O(N)\) space and \(O(C N^2)\) time in common cases ( \(O(C N^3)\) in the worst case).


Given a discrete feature \(x \in \mathcal{X} = \{x_1,\dots,x_N\}\) and a category label \(c = 1,\dots,C\) with joint probability \(p(x,c)\), AIB computes a compressed feature \([x]_{ij}\) by merging two values \(x_i\) and \(x_j\). Among all the pairs \(ij\), AIB chooses the one that yields the smallest loss in the mutual information

\[ D_{ij} = I(x,c) - I([x]_{ij},c) = \sum_c p(x_i) \log \frac{p(x_i,c)}{p(x_i)p(c)} + \sum_c p(x_i) \log \frac{p(x_i,c)}{p(x_i)p(c)} - \sum_c (p(x_i)+p(x_j)) \log \frac {p(x_i,c)+p(x_i,c)}{(p(x_i)+p(x_j))p(c)} \]

AIB iterates this procedure until the desired level of compression is achieved.

Algorithm details

Computing \(D_{ij}\) requires \(O(C)\) operations. For example, in standard AIB we need to calculate

\[ D_{ij} = I(x,c) - I([x]_{ij},c) = \sum_c p(x_i) \log \frac{p(x_i,c)}{p(x_i)p(c)} + \sum_c p(x_i) \log \frac{p(x_i,c)}{p(x_i)p(c)} - \sum_c (p(x_i)+p(x_j)) \log \frac {p(x_i,c)+p(x_i,c)}{(p(x_i)+p(x_j))p(c)} \]

Thus in a basic implementation of AIB, finding the optimal pair \(ij\) of feature values requires \(O(CN^2)\) operations in total. In order to join all the \(N\) values, we repeat this procedure \(O(N)\) times, yielding \(O(N^3 C)\) time and \(O(1)\) space complexity (this does not account for the space need to store the input).

The complexity can be improved by reusing computations. For instance, we can store the matrix \(D = [ D_{ij} ]\) (which requires \(O(N^2)\) space). Then, after joining \(ij\), all of the matrix D except the rows and columns (the matrix is symmetric) of indexes i and j is unchanged. These two rows and columns are deleted and a new row and column, whose computation requires \(O(NC)\) operations, are added for the merged value \(x_{ij}\). Finding the minimal element of the matrix still requires \(O(N^2)\) operations, so the complexity of this algorithm is \(O(N^2C + N^3)\) time and \(O(N^2)\) space.

We can obtain a much better expected complexity as follows. First, instead of storing the whole matrix D, we store the smallest element (index and value) of each row as \((q_i, D_i)\) (notice that this is also the best element of each column since D is symmetric). This requires \(O(N)\) space and finding the minimal element of the matrix requires \(O(N)\) operations. After joining \(ij\), we have to efficiently update this representation. This is done as follows:

  • The entries \((q_i,D_i)\) and \((q_j,D_j)\) are deleted.
  • A new entry \((q_{ij},D_{ij})\) for the joint value \(x_{ij}\) is added. This requires \(O(CN)\) operations.
  • We test which other entries \((q_{k},D_{k})\) need to be updated. Recall that \((q_{k},D_{k})\) means that, before the merge, the value closest to \(x_k\) was \(x_{q_k}\) at a distance \(D_k\). Then
    • If \(q_k \not = i\), \(q_k \not = j\) and \(D_{k,ij} \geq D_k\), then \(q_k\) is still the closest element and we do not do anything.
    • If \(q_k \not = i\), \(q_k \not = j\) and \(D_{k,ij} < D_k\), then the closest element is \(ij\) and we update the entry in constant time.
    • If \(q_k = i\) or \(q_k = j\), then we need to re-compute the closest element in \(O(CN)\) operations.

This algorithm requires only \(O(N)\) space and \(O(\gamma(N) C N^2)\) time, where \(\gamma(N)\) is the expected number of times we fall in the last case. In common cases one has \(\gamma(N) \approx \mathrm{const.}\), so the time saving is significant.