| 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 |
$$c(d_1, d_4) = \frac{255 \cdot 0 + 0 \cdot 191 + 0 \cdot 255}{255 \cdot 255} = 0$$ |
$$c(d_1, d_4) = \frac{255 \cdot 0 + 0 \cdot 191 + 0 \cdot 255}{\sqrt{255^{2} + 0^2 + 0^2} \cdot \sqrt{0^{2} + 191^2 + 255^2}} = 0$$ |
