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https://issues.apache.org/jira/browse/DATAFU-21?page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel&focusedCommentId=13944325#comment-13944325
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jian wang edited comment on DATAFU-21 at 7/9/14 6:05 AM:
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Some investigation updates:
Based on the theories from paper:
http://utopia.duth.gr/~pefraimi/research/data/2007EncOfAlg.pdf, I plan to
associate each item with a key X(j) = 1 - pow(U, 1/w(j)), U is a random
variable between (0,1). Then follow the thought of Random Sort, we sort the
items in ascending order based on X(j) and select the smallest k = p * n items.
Also as simple random sampling algorithm, we could also consider the
possibility of rejecting items applying Maurer's lemma and accepting items
applying Bernstein's lemma.
Apply Maurer's lemma:
we would like to find 0<q1<1, so that we reject items whose key is greater
than q1.
let Y(j) = 1 if (X(j) < q1)
= 0 otherwise
{Y(j), j = 1 to n} are independent random variables.
E(Y(j)) = Pr(X(j) < q1) * 1 + Pr(X(j) >= q1) * 0
= Pr(1 - pow( U, 1/w(j) ) < q1)
= Pr(1 - q1 < pow( U, 1/w(j) ))
= Pr(pow(1 - q1, w(j) ) < U) = 1 - pow( 1 - q1, w(j) )
E(Y(j) ^ 2) = E(Y(j)) = 1 - pow(1 - q1, w(j) )
set Y = sum(Y(j), j = 1 to n),
Q1 = sum( pow( 1-q1, w(j) ) , j = 1 to n)
E(Y) = sum(E(Y(j))) = n - sum( pow( 1-q1, w(j) ) ) = n - Q1
apply Maurer's lemma with t = (1 - p) * n - sum( pow(1 - q1, w(j) ) ) = (1 - p)
* n - Q1, since t > 0, Q1 < (1 - p) * n. Solving the inequality, I get
abs( Q1 - (1 - p) * n - log(err) ) >= sqrt( log(err) ^ 2 - 2 * p * n *
log(err) ) (1)
we could get q1 by solving (1)
Apply Berstein's lemma:
similar to applying Maurer's lemma, we could get a q2 so that we could accept
item whose key is smaller than q2, 0 <= q2 <= 1.
let Z(j) = 1 if X(j) < q2,
= 0 if X(j) >= q2
{Z(j), j = 1 to n} are independent random variables.
E(Z(j)) = Pr(X(j) < q2) * 1 + Pr(X(j) >= q2) * 0
= Pr(1 - pow( U, 1/w(j) ) < q2)
= 1 - pow(1 - q2, w(j) )
E(Z(j) ^ 2) = E(Z(j))
Z(j) - E(Z(j)) <= 1 - E(Z(j)) = pow(1 - q2, w(j) ) <= 1 = M
theta(j) ^ 2 = E(Z(j) ^ 2) - E(Z(j)) ^ 2 <= E(Z(j) ^ 2) = 1 - pow(1 - q2, w(j) )
set Z = sum(Z(j), j = 1 to n)
Q2 = sum( pow(1 - q2, w(j) ) )
E(Z) = sum(E(Z(j)), j = 1 to n) = n - sum( pow(1 - q2, w(j) ) ) = n - Q2
apply Berstein's lemma with t = sum( pow(1 - q2, w(j) ) ) - (1 - p) * n = Q2 -
(1 - p) * n, since t > 0, Q2 > (1 - p) * n, I get
abs(Q2 - n * (1 - p) - 2 / 3 * log(err) ) =
Q2 - n * (1 - p) - 2 / 3 * log(err) <= 2 / 3 * sqrt( log(err) * log(err) -
9 * n * p / 2 * log(err) ) (2)
we could get q2 by solving (2)
Questions:
(1) Please help comment on the above approach. Do they overall make sense?
(2) I am stuck in getting q1 and q2 by solving (1) and (2) respectively. Would
like to seek some advice on it.
Use Newton method to solve (1):
abs( Q1 - (1 - p) * n - log(err) ) >= sqrt( log(err) ^ 2 - 2 * p * n *
log(err) ) = F(n, p, err) (1)
Q1 = sum( pow( 1-q1, w(j) ) , j = 1 to n), 0 < q1 < 1
Q1' = sum( -1 * w(j) * (1 - q1) ^ (w(j) - 1) ), 0 < q1 < 1, Q1' means the 1st
derivative of Q1.
Remove the less-than inequality of (1), our target is to get an approximate q1
that makes abs( Q1 - (1 - p) * n - log(err) ) close to F(n, p, err), we name
it as q1_t.
Set function f(q1) = abs( Q1 - (1 - p) * n - log(err) ) - F(n, p, err). Use
Newton–Raphson method to get q1_t so that f(q1_t) is close to 0.
Use Newton to solve (2):
Q2 - n * (1 - p) - 2 / 3 * log(err) <= 2 / 3 * sqrt( log(err) *
log(err) - 9 * n * p / 2 * log(err) ) (2)
Q2 = sum( pow(1 - q2, w(j) ) )
Q2' = sum( -1 * w(j) * (1 - q2) ^ (w(j) - 1) ), 0 < q2 < 1, Q2' means the 1st
derivative of Q2.
Remove the less-than inequality of (2), our target is to get an approximate q2
that makes f(q2) = Q2 - n * (1 - p) - 2 / 3 * log(err) - 2 / 3 * sqrt(
log(err) * log(err) - 9 * n * p / 2 * log(err) ) is close to 0.
was (Author: king821221):
Some investigation updates:
Based on the theories from paper:
http://utopia.duth.gr/~pefraimi/research/data/2007EncOfAlg.pdf, I plan to
associate each item with a key X(j) = 1 - pow(U, 1/w(j)), U is a random
variable between (0,1). Then follow the thought of Random Sort, we sort the
items in ascending order based on X(j) and select the smallest k = p * n items.
Also as simple random sampling algorithm, we could also consider the
possibility of rejecting items applying Maurer's lemma and accepting items
applying Bernstein's lemma.
Apply Maurer's lemma:
we would like to find 0<q1<1, so that we reject items whose key is greater
than q1.
let Y(j) = 1 if (X(j) < q1)
= 0 otherwise
{Y(j), j = 1 to n} are independent random variables.
E(Y(j)) = Pr(X(j) < q1) * 1 + Pr(X(j) >= q1) * 0
= Pr(1 - pow( U, 1/w(j) ) < q1)
= Pr(1 - q1 < pow( U, 1/w(j) ))
= Pr(pow(1 - q1, w(j) ) < U) = 1 - pow( 1 - q1, w(j) )
E(Y(j) ^ 2) = E(Y(j)) = 1 - pow(1 - q1, w(j) )
set Y = sum(Y(j), j = 1 to n),
Q1 = sum( pow( 1-q1, w(j) ) , j = 1 to n)
E(Y) = sum(E(Y(j))) = n - sum( pow( 1-q1, w(j) ) ) = n - Q1
apply Maurer's lemma with t = (1 - p) * n - sum( pow(1 - q1, w(j) ) ) = (1 - p)
* n - Q1, since t > 0, Q1 < (1 - p) * n. Solving the inequality, I get
abs( Q1 - (1 - p) * n - log(err) ) >= sqrt( log(err) ^ 2 - 2 * p * n *
log(err) ) (1)
we could get q1 by solving (1)
Apply Berstein's lemma:
similar to applying Maurer's lemma, we could get a q2 so that we could accept
item whose key is smaller than q2, 0 <= q2 <= 1.
let Z(j) = 1 if X(j) < q2,
= 0 if X(j) >= q2
{Z(j), j = 1 to n} are independent random variables.
E(Z(j)) = Pr(X(j) < q2) * 1 + Pr(X(j) >= q2) * 0
= Pr(1 - pow( U, 1/w(j) ) < q2)
= 1 - pow(1 - q2, w(j) )
E(Z(j) ^ 2) = E(Z(j))
Z(j) - E(Z(j)) <= 1 - E(Z(j)) = pow(1 - q2, w(j) ) <= 1 = M
theta(j) ^ 2 = E(Z(j) ^ 2) - E(Z(j)) ^ 2 <= E(Z(j) ^ 2) = 1 - pow(1 - q2, w(j) )
set Z = sum(Z(j), j = 1 to n)
Q2 = sum( pow(1 - q2, w(j) ) )
E(Z) = sum(E(Z(j)), j = 1 to n) = n - sum( pow(1 - q2, w(j) ) ) = n - Q2
apply Berstein's lemma with t = sum( pow(1 - q2, w(j) ) ) - (1 - p) * n = Q2 -
(1 - p) * n, I get
abs(Q2 - n * (1 - p) - 2 / 3 * log(err) ) <= 2 / 3 * sqrt( log(err) *
log(err) - 9 * n * p / 2 * log(err) ) (2)
we could get q2 by solving (2)
Questions:
(1) Please help comment on the above approach. Do they overall make sense?
(2) I am stuck in getting q1 and q2 by solving (1) and (2) respectively. Would
like to seek some advice on it.
Some thoughts on how to resolve this, eg: solve (1)
abs( Q1 - (1 - p) * n - log(err) ) >= sqrt( log(err) ^ 2 - 2 * p * n *
log(err) ) = F(n, p, err) (1)
Q1 = sum( pow( 1-q1, w(j) ) , j = 1 to n), 0 < q1 < 1
Q1' = sum( -1 * w(j) * (1 - q1) ^ (w(j) - 1) ), 0 < q1 < 1, Q1' means the 1st
derivative of Q1.
Remove the less-than inequality of (1), our target is to get an approximate q1
that makes abs( Q1 - (1 - p) * n - log(err) ) close to F(n, p, err), we name
it as q1_t.
Set function f(q1) = abs( Q1 - (1 - p) * n - log(err) ) - F(n, p, err). Use
Newton–Raphson method to get q1_t so that f(q1_t) is close to 0.
> Probability weighted sampling without reservoir
> -----------------------------------------------
>
> Key: DATAFU-21
> URL: https://issues.apache.org/jira/browse/DATAFU-21
> Project: DataFu
> Issue Type: New Feature
> Environment: Mac OS, Linux
> Reporter: jian wang
> Assignee: jian wang
>
> This issue is used to track investigation on finding a weighted sampler
> without using internal reservoir.
> At present, the SimpleRandomSample has implemented a good
> acceptance-rejection sampling algo on probability random sampling. The
> weighted sampler could utilize the simple random sample with slight
> modification.
> One slight modification is: the present simple random sample generates a
> uniform random number lies between (0, 1) as the random variable to accept or
> reject an item. The weighted sample may generate this random variable based
> on the item's weight and this random number still lies between (0, 1) and
> each item's random variable remain independent between each other.
> Need further think and experiment the correctness of this solution and how to
> implement it in an effective way.
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