# Find the position within an array larger than 3D (New Features)

Hey, hey, hey, STOP.

I have found my error in 4D.

It is a error of conception dimendional.

Well, let's let W = (5 * 4 * 3 * 2) = 120 elements, Ok!
w = (x * y * z * "D")
x = 1D
y = x * y = 2D
z = x * y * z = 3D
D = number of times that repeats the block x, y, z

Let's see:
w = (5 * 4 * 3 * 2) = 120
We see that "2" is the number of times
Now comes my error of conception,
I have assumed that 4D was an extension of 3D, that was my error of conception.

As we see from my calculation / deduction wrong in my post:
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We have now in the 4D = 2.
I Understand That at level 3, the last layer,
and has an element Each extension of 2 more items of the fourth dimension.
So I think 20 elements of the last level * 2 = 40 elements more extensions.
I still think if 3D = 60 items + 40 items in the fouth dimension = 100 items.
That item is not the 120 Mathematically speaking.

This is a ERROR!!!
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... Then the fourth dimension (4D) are the times you repeat the block: x, y, z.

Now, by extension ...
we know what they mean the fifth, sixth, seventh, eighth,..... dimensions:

An example:
w = (2 * 2 * 2 * 2 * 2) = (a b c d e)
a * b * c = 2 * 2 * 2 = 8 = 3D starting block

8 * d (d as we have said is the number of times to repeat xyz) then 8 * 2 = 16 elements to the fourth dimension

we ...

16 * e ("e" also indicates the number of times to repeat the previous block, xyz + d)
16 * 2 = 32 total elements in W (2 * 2 * 2 * 2 * 2)

I believe that I have understood the concept of multidimensional array.

Thank you all for your interest in explaining the concepts of multidimensional array.