ISBN: 978-3748190486 Paperback: 220 pages
We give a brief overview of popular hash functions and hash tables that are widely used in probabilistic data structures.
We study the most well-known known probabilistic data structures that are used to answer approximate membership queries.
We learn data structures that allow to estimate the number of unique elements in a dataset.
We discuss solutions for common problems in streaming applications, such as finding frequency of some element, filtering most frequent elements in the stream, detecting trending elements, etc.
We study data structures to estimate quantiles and percentiles in a data stream using only one pass through the data.
We learn how to effciently solve the nearest neighbor problem for large datasets using sub-linear in time solutions.
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Edition | Location | Original text | Corrected text |
---|---|---|---|
1st, 2019 | p. 31 | "... by bitwise XOR, however ..." | "... by bitwise AND, however ..." |
1st, 2019 | p. 62, Example 3.1 | "... of unique elements is 23 GB." | "... of unique elements is 2.3 GB." |
1st, 2019 | p. 65, Algorithm 3.2 | $$\mathbf{Add}(e, LINEARCOUNTER)$$ | $$\mathbf{Add}(x, LINEARCOUNTER)$$ |
1st, 2019 | p. 66, Example 3.3 | $$V = \frac{9}{16} = 0.5625$$ | $$V = \frac{7}{16} = 0.4375$$ |
1st, 2019 | p. 66, Example 3.3 | $$n \approx - 16 \cdot \ln{0.5625} \approx 9.206$$ | $$n \approx - 16 \cdot \ln{0.4375} \approx 13.22$$ |
1st, 2019 | p. 67 | ... a fixed constant \(O(N)\), where \(N\) is the total number of elements, including duplicates. Thus, the algorithm has \(O(N)\) time complexity. | ... a fixed constant, and thus the algorithm has \(O(N)\) time complexity, where \(N\) is the total number of elements, including duplicates. |
1st, 2019 | p. 81, Formula 3.12 | $$\hat n \approx \alpha_{m}\cdot m^{2} \cdot \left(\sum_{j=0}^{m-1}2^{-COUNTER[j]}\right)$$ | $$\hat n \approx \alpha_{m}\cdot m^{2} \cdot \left(\sum_{j=0}^{m-1}2^{-COUNTER[j]}\right)^{-1}$$ |
1st, 2019 | p. 96, Example 4.2 | "... 500 billion ..." | "... 500 million ..." |
1st, 2019 | p. 135, Example 5.3 | "... remaining empty buffer \(B_{2}\)." | "... remaining empty buffer \(B_{4}\)." |
1st, 2019 | p. 173, Example 6.8 | $$c(d_1, d_4) = \frac{255 \cdot 0 + 0 \cdot 191 + 0 \cdot 255}{255 \cdot 255} = 0$$ | $$c(d_1, d_4) = \frac{255 \cdot 0 + 0 \cdot 191 + 0 \cdot 255}{\sqrt{255^{2} + 0^2 + 0^2} \cdot \sqrt{0^{2} + 191^2 + 255^2}} = 0$$ |