I have a very large NoSQL database. Each item in the database is assigned a uniformly distributed random value between 0 and 1. This database is so large that performing a COUNT on queries does not yield acceptable performance, but I'd like to use the random values to estimate COUNT.
The idea is this:
Run a query and order the query by the random value. Random values are indexed, so it's fast.
Grab the lowest N values, and see how big the largest value is, say R.
Estimate COUNT as N / R
The question is two-fold:
Is N / R the best way to estimate COUNT? Maybe it should be (N+1)/R? Maybe we could look at the other values (average, variance, etc), and not just the largest value to get a better estimate?
What is the error margin on this estimated value of COUNT?
Note: I thought about posting this in the math stack exchange, but given this is for databases, I thought it would be more appropriate here.
This actually would be better on math or statistics stack exchange.
The reasonable estimate is that if R is large and x is your order statistic, then R is approximately n / x - 1. About 95% of the time the error will be within 2 R / sqrt(n) of this. So looking at the 100th element will estimate the right answer to within about 20%. Looking at the 10,000th element will estimate it to within about 2%. And the millionth element will get you the right answer to within about 0.2%.
To see this, start with the fact that the n'th order statistic has a Beta distribution with parameters 𝛼 = n and β = R + 1 - n. Which means that the mean value of the n'th smallest value out of R values is n/(R+1). And its variance is 𝛼β / ((𝛼 + β)^2 (𝛼 + β + 1)). If we assume that R is much larger than n, then this is approximately n R / R^3 = n / R^2. Which means that our standard deviation is sqrt(n) / R.
If x is our order statistic, this means that (n / x) - 1 is a reasonable estimate of R. And how much is it off by? Well, we can use the tangent line approximation. The function (n / x) - 1 has a derivative of - n / x^2 Its derivative at x = n/(R+1) is therefore (R + 1)^2 / n. Which for large R is roughly R^2 / n. Stick in our standard deviation of sqrt(n) / R and we come up with an error proportional to R / sqrt(n). Since a 95% confidence interval would be 2 standard deviations, you probably will have an error of around 2 R / sqrt(n).
Related
I was reading about static array queries and this is what I found:
Minimum Queries: There is an O(nlogn) time preprocessing method after which we can answer any minimum query in O(1) time.
The idea is to precalculate all values of min(a, b) where b - a + 1 (the length of the range) is a power of two. The number of precalculated values is O(nlogn), because there are O(logn) range lengths that are powers of two.
The values can be calculated efficiently using the recursive formula:
min(a,b) = min(min(a, a + w - 1), min(a + w, b))
where b-a+1 is a power of two and w = (b - a + 1) / 2
What is meant by the part quoted above? Why do we calculate the minimum only for certain lengths?
What is the idea and the intuition behind it? What does the logic do?
It's kind of a hunch that it must be related to something about a binary tree because we're considering just the powers of two for the lengths.
This structure is referred to an RMQ, a Range Minimum Query. It achieves its O(1) queries by exploiting the associativity and commutativity of the min operation (that is, min(x,y) = min(y,x) and min(x,y,z) = min(x,min(y,z)). The other property that min has is that min(x,x) = x, and more importantly, min(x,y,z) = min(min(x,y),min(y,z))
If you have all the mins for every subarray of length of every power of 2 (hence the n log n memory), you can compute the range min(l-r) by taking the min of the largest power of 2, starting at l, that doesn't overshoot r, with the min of the largest power of 2 ending at r that doesn't undershoot l. The idea here is as follows:
arr=[a,b,c,d,e,f,g,h]
We calculate our RMQ to have mins as follows:
of length 1: [min(a), min(b), min(c), etc]
of length 2: [min(a,b), min(b,c), min(c,d), min(d,e), etc]
of length 4: [min(a,b,c,d}, min(b,c,d,e), min(c,d,e,f), etc]
To take the min from 1 to 6, we want the range min of length 4 starting at 1 (since 8 would go past our right index) and take the min of that with the range min of length 4 ending at 6. So we take these queries from the array of length 4, and take the min of
min(of length 4[1], of length 4[2]) and that's our answer.
We have an array consisting of each entry as a tuple of two integers. Let the array be A = [(a1, b1), (a2, b2), .... , (an, bn)]. Now we have multiple queries where we are given an integer x, we need to find the maximum value of ai + |x - bi| for 1 <= i <= n.
I understand this can be easily achieved in O(n) time complexity for each query but I am looking for something faster than that, probably O(log n) for each query. I can preprocess the array in O(n) time, but the queries should be done faster than O(n).
Any kind of help would be appreciated.
It seems to be way too easy to over-think this.
For n = 1, the function is v-shaped with a minimum of a1 at b1, with slopes of -1 and 1, respectively - let's call these values ac and bc (for combined).
For an additional pair (ai, bi), one of the pairs may dominate the other (|bc - bi| ≤ |ac - ai), which may then be ignored.
Otherwise, the falling slope of the combination will be from the pair with the larger b, the rising slope from the other.
The minimum will be between the individual b, closer to the b of the pair with the larger a, the distance being half the difference between the (absolute value of the) "coordinate" differences, the minimum value that amount higher.
The main catch is that neither needs to be an integer - the only alternative being exactly in the middle between two integers.
(Ending up with the falling slope from max ai + bi, and the rising slope of max ai - bi.)
I have been trying this question on hackerearth practice which requires below work done.
PROBLEM
Given an integer n which signifies a sequence of n numbers from {0,1,2,3,4,5.......n-2,n-1}
We are provided m ranges in form of (L,R) such that (0<=L<=n-1)(0<=R<=n-1)
if(L <= R) (L,R) signifies numbers {L,L+1,L+2,L+3.......R-1,R} from above sequence
else (L,R) signifies numbers {R,R+1,R+2,.......n-2,n-1} & {0,1,2,3....L-1,L} ie numbers wrap around
example
n = 5 ie {0,1,2,3,4}
(0,3) signifies {0,1,2,3}
(3,0) signifies {3,4,0}
(3,2) signifies {3,4,0,1,2}
Now we have to select ONE (only one) number from each range without repeating any selection. We have to tell is it possible to select one number from each(and every) range without repetition.
Example test case
n = 5// numbers {0,1,2,3,4}
// ranges m in number //
0 0 ie {0}
1 2 ie {1,2}
2 3 ie {2,3}
4 4 ie {4}
4 0 ie {4,0}
Answer is "NO" it's not possible.
Because we cannot select any number from range 4 0 because if we select 4 from it we could not be able to select from range 4 4 and if select 0 from it we would not be able to select from 0 0
My approaches -
1) it can be done in O(N*M) using recurrsion checking all possibilitie of selection from each range and side by side using hash map to record our selections.
2) I was trying it in order n or m ie linear order .Problem lack editorial explanation .Only a code is mentioned in the editorial without comments and explanation . I m not able to get the codelinear solution code by someone which passes all test cases and got accepted.
I am not able to understand the logic/algo used in the code and why is it working?
Please suggest ANY linear method and logic behind it because problem has these constraints
1 <= N<= 10^9
1 <= M <= 10^5
0 <= L, R < N
which demands a linear or nlogn solution as i guess??
The code in the editorial can also be seen here http://ideone.com/5Xb6xw
Warning --After looking The code I found the code is using n and m interchangebly So i would like to mention the input format for the problem.
INPUT FORMAT
The first line contains test cases, tc, followed by two integers N,M- the first one depicting the number of countries on the globe, the second one depicting the number of ranges his girlfriend has given him. After which, the next M lines will have two integers describing the range, X and Y. If (X <= Y), then range covers countries [X,X+1... Y] else range covers [X,X+1,.... N-1,0,1..., Y].
Output Format
Print "YES" if it is possible to do so, print "NO", if it is not.
There are two components to the editorial solution.
Linear-time reduction to a problem on ordinary intervals
Assume to avoid trivial cases that the number of input intervals is less than n.
The first is to reduce the problem to one where the intervals don't wrap around as follows. Given an interval [L, R], if L ≤ R, then emit two intervals [L, R] and [L + n, R + n]; if L > R, emit [L, R + n]. The easy direction of the reduction is showing that, if the original problem has a solution, then the reduced problem has a solution. For [L, R] with L ≤ R assigned a number k, assign k to [L, R] and k + n to [L + n, R + n]. For [L, R] with L > R, assign whichever of k, k + n belongs to [L, R + n]. Except for the dual assignment of k and k + n for intervals [L, R] and [L + n, R + n] respectively, each interval gets its own residue class mod n, so the assignments do not conflict.
Conversely, the hard direction of the reduction (if the original problem has no solution, then the reduced problem has no solution) is proved using Hall's marriage theorem. By Hall's criterion, an unsolvable original problem has, for some k, a set of k input intervals whose union has size less than k. We argue first that there exists such a set of input intervals whose union is a (circular) interval (which by assumption isn't all of 0..n-1). Decompose the union into the set of maximal (circular) intervals that comprise it. Each input interval is contained in exactly one of these intervals. By an averaging argument, some maximal (circular) interval contains more input intervals than its size. We finish by "lifting" this counterexample to the reduced problem. Given the maximal (circular) interval [L*, R*], we lift it to the ordinary interval [L*, R*] if L* ≤ R*, or [L*, R* + n] if L* > R*. Do likewise with the circular intervals contained in this interval. It is tedious but straightforward to show that this lifted counterexample satisfies Hall's criterion, which implies that the reduced problem has no solution.
O(m log m)-time solution for ordinary intervals
This is a sweep-line algorithm. Sort the intervals by lower endpoint and scan them in that order. We imagine that the sweep line moves from lower endpoint to lower endpoint. Maintain the set of intervals that intersect the sweep line and have not been assigned a number, sorted by upper endpoint. When the sweep line is about to move, assign the numbers between the old and new positions to the intervals in the set, preferentially to the ones whose upper endpoint is the lowest. The correctness of this strategy should be clear: the intervals that could be assigned a number but are passed over have at least as many options (in the sense of being a superset) as the intervals that are assigned, so we never make a choice that we have cause to regret.
In the chapter about Value Iteration algorithm to calculate optimal policy for MDPs, there is an algorithm:
function Value-Iteration(mdp,ε) returns a utility function
inputs: mdp, an MDP with states S, actions A(s), transition model P(s'|s,a),
rewards R(s), discount γ
ε, the maximum error allowed in the utility of any state
local variables: U, U', vectors of utilities for states in S, initially zero
δ, the maximum change in the utility of any state in an iteration
repeat
U ← U'; δ ← 0
for each state s in S do
U'[s] ← R(s) + γ max(a in A(s)) ∑ over s' (P(s'|s,a) U[s'])
if |U'[s] - U[s]| > δ then δ ← |U'[s] - U[s]|
until δ < ε(1-γ)/γ
return U
(I apologize for the formatting, but I need 10 rep to post picture and $latex formatting$ doesn't seem to work here.)
and also a chapter earlier there was a statement:
A discount factor of γ is equivalent to an interest rate of (1/γ) − 1.
Could anyone explain to me what does the interest rate (1/γ)-1 mean? How did they get it? Why is it used in the termination condition in the algorithm above?
The reward at t-1 is considered discounted by a factor gamma (y). That is to say, old = y x new. So new = (1/y) * old, and new - old = ((1/y) - 1) * old. That is your interest rate.
I am not so sure why it is used in the termination condition. The value of epsilon is arbitrary, anyway.
In fact, I believe this termination criterion is very bad. It does not work when y = 1. When y = 0, then the iteration should stop immediately, since it is enough to estimate perfect values. When y = 1, many iterations are necessary.
I want to write a C program that will calculate a series:
1/x + 1/2*x^2 + 1/3*x^3 + 1/4*x^4 + ...
up to five decimal places.
The program will take x as input and print the f(x) (value of series) up to five decimal places. Can you help me?
For evaluating a polynomial, Horner form generally has better numerical stability than expanded form See http://reference.wolfram.com/legacy/v5/Add-onsLinks/StandardPackages/Algebra/Horner.html
If first term was a typo then try (((((1/4 )* x + 1/3) * x ) + 1/2) * x + 1) * x
Else if first term is really 1/x (((((1/4 )* x + 1/3) * x ) + 1/2) * x*x + 1/x
Of course, you still have to analyze convergence and numerical stability as developped in Eric Postpischil answer.
Last thing, does the serie you submited as example really converge to a finite value for some x???
In order to know that the sum you have calculated is within a desired distance to the limit of the series, you need to demonstrate that the sources of error are less than the desired distance.
When evaluating a series numerically, there are two sources of error. One is the limitations of numerical calculation, such as floating-point rounding. The other is the sum of the remaining terms, which have not been added into the partial sum.
The numerical error depends on the calculations done. For each series you want to evaluate, a custom analysis of the error must be performed. For the sample series you show, a crude but sufficient bound on the numerical error could like be calculated without too much effort. Is this the series you are primarily interested in, or are there others?
The sum of the remaining terms also requires a custom analysis. Often, given a series, we can find an expression that can be proven to be at least as large as the sum of all remaining terms but that is more easily calculated.
After you have established bounds on these two errors, you could sum terms of the series until the sum of the two bounds is less than the desired distance.