## database: in DHIS2, what is the largest number of data elements that a numerator can contain to form an indicator?

We recently updated the family planning dataset to collect data on 6 different age groups based on clients' preferred methods across all family planning acceptors (both new and ongoing). So when calculating the acceptance rate, we have to add more than 228 individual data elements in the numerator to make the indicator.

We only succeeded in completing 108 data items. Anything beyond this, we couldn't save. What are you not doing well? Is there a way to avoid it, for example using subtotals instead of individual data elements? What is the largest number of data elements that a numerator can contain to form an indicator in DHIS2?

## database: what is the largest number of data elements that a numerator can contain to form an indicator?

We recently updated the family planning dataset to collect data on 6 different age groups based on clients' preferred methods across all family planning acceptors (both new and ongoing). So when calculating the acceptance rate, we have to add more than 228 individual data elements in the numerator to make the indicator. We only succeeded in completing 108 data items. Anything beyond this, we couldn't save. What are you not doing well? Is there a way to avoid it, for example using subtotals instead of individual data elements? What is the largest number of data elements that a numerator can contain to form an indicator in DHIS2?

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This publication has been edited by LiteForexOfficial: Today, 10:15 a.m.

## Create a function with foreach that receives a list of dates, returns what is the largest date among them

Largest public list (list data)
{
List r = new List ();

``````foreach (DateTime item in data)
{
if (item > item) //Aqui é onde quero ver se a data é maior ou não
{
}
}
``````

}

## Reference Request – Dependency Range: What is the size of the largest sub-collection of random variables that is statistically independent?

Leave $$X_1, ldots, X_p$$ be random variables in the same space. Define your range of dependency, denoted $$range (X_1, ldots, X_p)$$ as the largest non-negative integer $$k$$ such that there is a subcollection of $$k$$ outside $$p$$ random variables that are statistically independent by pairs (that is, their joint distribution function factors as a product of their individual distribution functions). Of course, each singleton $${X_i }$$ forms an independent collection so $$1 le range (X_1, ldots, X_p) le p$$, and the upper limit is reached if $$X_1, ldots, X_p$$ they are independent to begin with.

Question. Is the above notion of rank defined above being studied in the literature? Is there any interesting quantity property? Are they the known representatives of this concept?

## What is an algorithm for finding the largest white space rectangle in an image?

Subtract $$254.9999999$$ of all pixel values, so black corresponds to $$-254.9999999$$ and white to $$0.0000001$$.

Find the 2D subrectangle with maximum sum. There is standard $$O (n ^ 3)$$time algorithms for this problem (for example https://stackoverflow.com/q/19064133/781723, https://www.geeksforgeeks.org/maximum-sum-rectangle-in-a-2d-matrix-dp- 27 /). As long as you have no more than 10,000,000 pixels in the image, this will be the solution to your problem (since each white pixel contributes another $$0.0000001$$ to the sum).

Proof that this finds the optimal solution: Only all-white subrectangles have a positive sum (if you have a single non-white pixel, then you contribute a negative value of $$-0.9999999$$ or less than the sum, which cannot be exceeded by any number of white pixels); and the larger the completely white subrectangle, the greater its sum.

## algorithms – Find the largest sum of elements \$ k \$ below the threshold

You can use the encounter function in the middle to reduce the execution time to $$O (n ^ { lceil k / 2 rceil})$$.

For simplicity let me assume that $$k$$ even.

The idea is the following:

• Divide $$A$$ in two parts
• For each part, compute an ordered list of sums of $$k / 2$$ part elements
• For each $$k / 2$$-sum in the first part, use binary search to find the best $$k / 2$$-sum in the second part that conforms to the restrictions.

As stated, there are several problems with the idea:

1. The optimal solution is assumed to contain exactly $$k / 2$$ addends of each part.
2. Run on time $$O (n ^ {k / 2} log n)$$.
3. Uses $$O (n ^ {/ 2})$$ memory.

To handle the first difficulty, there are several options. We could repeat the whole algorithm $$sqrt {k}$$ times. Alternatively, there may be deterministic ways to achieve the same goal. Here is a hybrid solution:

• Random partition $$A$$ inside $$k ^ 3$$ parts $$P_1, ldots, P_ {k ^ 3}$$ (something substantially larger than $$k ^ 2$$ would work). With high probability, each element of the optimal solution is in its own part.
• Consider all possible partitions of $$A$$ in two parts of the form $$P_1, ldots, P_i$$ Y $$P_ {i + 1}, ldots, P_ {k ^ 3}$$. One of these will contain exactly $$k / 2$$ elements of the optimal solution.

To handle the second difficulty, we need to be a little more careful in implementation. By merging (the mergesort family subroutine), it should be possible to calculate the $$k / 2$$-sums of each part in $$O (n ^ {/ 2})$$. The final step can be implemented in $$O (n ^ {/ 2})$$ using a classic two-pointer technique (the first pointer goes up in the first half, the second pointer goes down in the second half).

There are tricks to reduce the required memory of $$O (n ^ {/ 2})$$ to $$O (n ^ {/ 4})$$: Divide each part into two sub parts. You can easily review everything $$k / 2$$-sum in the first part considering all pairs of $$k / 4$$-sums in its subparts. The binary search in the second part can be implemented using the two pointer technique.

## Probable Probability: The probability that the third largest of 39 nine-sided dice equals 8?

How can I calculate the probability that the third greatest of 39 nine-sided dice equals 8?
I only see three results:
(two nines, one eight, the rest is less than 8)
(one nine, two eights, the rest is less than 8)
(three eights, the rest is less than 8)

How would this be done mathematically?

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