Question

4. In LSH, remember that we divide signature columns to b bands where there are r elements in each band for each column. Consider two columns C and C2 of length m=br. Suppose that for each i = 1,.., m, the probability that C ) = C2() is equal to s independently from other elements. Note that s is in fact the similarity of the two columns. (a) Show that the probability that the two columns are equal in one band is s". (b) Show that the probability that none of the bands are equal is (1 - s"). (c) Conclude that the probability that the columns are chosen as a candidate pair (i.e., at least one of the bands match) is 1-(1 - s")

Answer 1

ANSWER::

4Q)

given that

Suppose that for each i = 1, .., m, the probability that $\dpi{100} c_{1}(i)=c_{2}(i)$ is equal to s independently from other elements.

(A) Show that the probability that the two columns are equal in one band is $\dpi{100} s^r$ .

The bond of  $\dpi{100} r$ elements $\dpi{100} x_{1},x_{2},x_{3},....x_{r}.$

the two column will be equal if each elements are equal in both column.

We Know that,

$\dpi{100} P({C_{1}(x_{1})=C_{2}(x_{1})}\cap {C_{1}(x_{2})=C_{2}(x_{2})}.......\cap {C_{1}(x_{r})=C_{2}(x_{r})}$

Where ,

$\dpi{100} P({C_{1}(x_{1})=C_{2}(x_{1})})\ *\ P(C_{1}(x_{2})=C_{2}(x_{2}))*...................\\\Rightarrow \ s\ * \ s\ *\ s\ *................\ * \ s\\\ \ \ \Rightarrow s^{r}$

$\dpi{100} \therefore \ the \ probality\ that\ the\ two\ column\ are\ equal \ in \ one\ band\ is\ s^r$

(B) Show that the probaility that none of the bands are equals is $\dpi{100} (1-s^{r})^b$

given that ,

bands = $\dpi{100} y_{1},y_{2},......y_{b}$

$\dpi{100} p(C_{1}(y_{1})=C_{2}(y_{1}))\ =\ s^r\\\Rightarrow \ p(C_{1}(y_{1})\neq \ C_{2}(y_{1}))\ =\ 1-s^r$

hence ,

$\dpi{100} \Rightarrow \ p(C_{1}(y_{1})\neq \ C_{2}(y_{1}))\ \ *\ p(C_{1}(y_{2})\neq \ C_{2}(y_{2}))\\substute\ values\ in\ above\ equation\\\Rightarrow \ \ \ (1-s^r)\ * \ (1-s^r)\ * \ .......\\\Rightarrow \ \ \ (1-s^r)^b.$

$\dpi{100} \therefore \ \ the\ probaility\ that\ none\ of \ the\ bands\ are\ equals\ is\ (1-s^r)^b.$

(C) Conclude that the probability that the columns are chosen as candidate pair.

Given that

for the candidate pair we are chosen the we are above two bands.

$\Rightarrow p(atleast\ one\ of\ the\ band\ )\\\Rightarrow 1-p(none\ of\ the\ band)\\ substute \ above\ value\ in\ equation\\\Rightarrow 1-(1-s^r)^b.$

$\therefore \ \ the\ cloumn \are\ chosen\ as\ a \candidate\ pair \ is\ 1-(1-s^r)^b.$

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