*Learn paging/partitioning mathematical equations and how to apply them in JavaScript and .NET C# code.*

The **Paging** or **Partitioning** concept is used in many fields. When you have a set of items and you want to divide them equally between some sort of containers or groups, you are thinking of paging or partitioning but may be you don’t recognize it, yet…

The main goals of this story are to:

Explain some mathematical equations which can make it easy for you to implement the paging or partitioning concept.

Provide some code examples of how to implement paging.

## Disclaimer

If you are expecting to find explanations for the paging concept on specific applications like operating systems memory management or file system,… then you are reading the wrong story.

Whenever “

**Page**” or “**Paging**” is mentioned, “**Partition**” and “**Partitioning**” would be valid. Therefore, for brevity, I will use “Page” and “Paging”.

## Back to Basics

The best way to explain paging is to apply to an example. Let’s assume that we have a collection of 10 items, any kind of items, maybe bottles, pencils, bananas… whatever makes you happy.

Now, we want to divide these **10 **items into groups where each group contains **3 **items.

If we do this manually, we will have the items distributed as in the image below.

This was easy, right?

Let’s explain what we have in the table above.

The

**First**item, holding**0**as its Zero-Index, would be found at the**0**Zero-Index position, inside the page of**0**Zero-Index.The

**Second**item, holding**1**as its Zero-Index, would be found at the**1**Zero-Index position, inside the page of**0**Zero-Index.The

**Third**item, holding**2**as its Zero-Index, would be found at the**2**Zero-Index position, inside the page of**0**Zero-Index.The

**Fourth**item, holding**3**as its Zero-Index, would be found at the**0**Zero-Index position, inside the page of**1**Zero-Index.The

**Fifth**item, holding**4**as its Zero-Index, would be found at the**1**Zero-Index position, inside the page of**2**Zero-Index.And so on,…

This time you were able to do it manually as the items count is not that big, but, this is not always the case. Therefore, we need to analyze the example above and come up with some mathematical formulas.

## The Formula

Doing some analysis, we can come up with the following formula:

Let me explain; when you divide the **Item Zero-Index** on the **Page Size**, you get two things:

The

**result**which would represent the**Page Zero-Index.**The

**remainder**which would represent the**Item Zero-Index Per Page**.

Don’t you believe me? ok, I will show you…

## Moment of Truth

Let’s apply the formula on our example and see where it goes from there.

I think now you believe me. The formula is proved to be working as charm.

## Further Analysis

Having the formula proved, let’s do some further analysis.

The formula could be transformed into the shape below:

Item Zero-Index= (Page Zero-Index*Page Size) + (Item Zero-Index Per Page)

This means that if we have a value for **Page Zero-Index,** and a value for **Page Size,** and we need to know the **Zero-Index** of the **first item** and the **last item** on this page, we can use the equation above as follows.

**First Item Zero-Index**= (Page Zero-Index * Page Size) + Minimum (Item Zero-Index Per Page)= (Page Zero-Index * Page Size) + 0**= (Page Zero-Index * Page Size)**

**Last Item Zero-Index**= (Page Zero-Index * Page Size) + Maximum (Item Zero-Index Per Page)**= (Page Zero-Index *** Page Size) + (Page Size — 1)**

**But note** that if the calculated **Last Item Zero-Index** is greater than the index of the last item in the whole collection, then take the smaller number which is the index of the last item in the whole collection.

Again, still you don’t believe me?

To verify these equations, let’s apply on our example:

On the first page, (first item Zero-Index = 0

*3 = 0) and (last item Zero-Index = (0*3) + (3–1) = 2)On the second page, (first item Zero-Index = 1

*3 = 3) and (last item Zero-Index = (1*3) + (3–1) = 5)On the third page, (first item Zero-Index = 2

*3 = 6) and (last item Zero-Index = (2*3) + (3–1) = 8)On the fourth page, (first item Zero-Index = 3

*3 = 9) and (last item Zero-Index = (3*3) + (3–1) = 11 which is greater than the max available item index (9), therefore, last item Zero-Index = 9)

## Let’s See Some Code

Having all the equations proved, let’s see how to apply them in code. I will provide two implementations, one in **JavaScript**, and the other one in **.NET C#**.

### JavaScript Code

Applying the equations in the form of an extension function to the **JavaScript Array**.

Now, executing this code

Would return

And, executing this code

Would return

### .NET C# Code

Applying the equations in the form of an extension method to the **.NET C# List**.

Now, executing this code

Would return

And, executing this code

Would return

## Final Words

That’s it. These are the most important equations you would need to know about to apply paging in any situation.

You can also use the code samples to apply in your own project using whatever language or framework you are using.

That’s it, hope you found reading this article as interesting as I found writing it.

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