Position item in an array larger than 3D
Re: Position item in an array larger than 3D
Hello everyone
This is a little experiment and possibly useful for 3D graphics with only whole numbers.
This example shows the vertices (X, Y, Z) of a 3D cube.
1 = 1,1,1
2 = 1,10,1
3 = 10,1,1
4 = 10,10,1
5 = 1,1,10
6 = 1,10,10
7 = 10,1,10
8 = 10,10,10
With this system we need to create a 3D array (x, y, z) = 24 items
':::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
dim as integer a,b,c,ro,co,le, p
cls
ro = 10
co = 10
le = 10
input "row position = ";a
input "col position = ";b
input "level position = ";c
p = ((a  1) * co) + b + ((ro * co) * (c  1))'position
print "Position = ";p
sleep
end
'::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Now the same vertices, obtained with this application:
1 = 1
2 = 10
3 = 91
4 = 100
5 = 901
6 = 910
7 = 991
8 = 1000
As we see, in this case, we only need to create a 1D array (P) = 8 elements
To regain the coordinates of the vertices, we just have to break down the numbers of previous positions and get back the coordinates of the vertices.
':::::::::::::::::::::::::::::::::::::::::::::::::::::
dim as integer c,co,d,f,h,l,p,r
cls
'input "qty blocs = ";d
input "qty rows (rows <= cols) = "; r
input "qty cols (cols >=rows) = ";c
input "qty lev = ";l
input "find position = ";p
h = l
'calculate dimension/block
if p mod (r*c*l) <> 0 then
d = int(p/(r*c*l))+1
else
d = int(p/(r*c*l))
end if
'::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
'calculate level
l = int(p/(r*c))+1
if p mod (r*c) = 0 then l = l1
':::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
'calculate col
if p < c then co = p
if p >= c then co = p mod c
if p mod c = 0 then co = c
'::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
'calculate row
f = ((pco((r*c)*(l1)))/c)+1
'::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
print "blo = ";d;" row = ";f; " col = ";co; " lev = ";l((d1)*h)
sleep
end
'::::::::::::::::::::::::::::::::::::::::::::::::::::
This same example applied to a sphere, we would need only 1/3 of a 3D array.
I wish you like this idea.!
Regards
This is a little experiment and possibly useful for 3D graphics with only whole numbers.
This example shows the vertices (X, Y, Z) of a 3D cube.
1 = 1,1,1
2 = 1,10,1
3 = 10,1,1
4 = 10,10,1
5 = 1,1,10
6 = 1,10,10
7 = 10,1,10
8 = 10,10,10
With this system we need to create a 3D array (x, y, z) = 24 items
':::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
dim as integer a,b,c,ro,co,le, p
cls
ro = 10
co = 10
le = 10
input "row position = ";a
input "col position = ";b
input "level position = ";c
p = ((a  1) * co) + b + ((ro * co) * (c  1))'position
print "Position = ";p
sleep
end
'::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Now the same vertices, obtained with this application:
1 = 1
2 = 10
3 = 91
4 = 100
5 = 901
6 = 910
7 = 991
8 = 1000
As we see, in this case, we only need to create a 1D array (P) = 8 elements
To regain the coordinates of the vertices, we just have to break down the numbers of previous positions and get back the coordinates of the vertices.
':::::::::::::::::::::::::::::::::::::::::::::::::::::
dim as integer c,co,d,f,h,l,p,r
cls
'input "qty blocs = ";d
input "qty rows (rows <= cols) = "; r
input "qty cols (cols >=rows) = ";c
input "qty lev = ";l
input "find position = ";p
h = l
'calculate dimension/block
if p mod (r*c*l) <> 0 then
d = int(p/(r*c*l))+1
else
d = int(p/(r*c*l))
end if
'::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
'calculate level
l = int(p/(r*c))+1
if p mod (r*c) = 0 then l = l1
':::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
'calculate col
if p < c then co = p
if p >= c then co = p mod c
if p mod c = 0 then co = c
'::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
'calculate row
f = ((pco((r*c)*(l1)))/c)+1
'::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
print "blo = ";d;" row = ";f; " col = ";co; " lev = ";l((d1)*h)
sleep
end
'::::::::::::::::::::::::::::::::::::::::::::::::::::
This same example applied to a sphere, we would need only 1/3 of a 3D array.
I wish you like this idea.!
Regards
Re: Position item in an array larger than 3D
Hi all:
I've done some work on this subject.
Written in Spanish.
If someone wants to see, is in this link:
http://es.slideshare.net/ZkrouhnZyx/mat ... onbloques
Regards
I've done some work on this subject.
Written in Spanish.
If someone wants to see, is in this link:
http://es.slideshare.net/ZkrouhnZyx/mat ... onbloques
Regards
Re: Position item in an array larger than 3D
Hi all:
I've done some work on this subject.
Written in English.
If someone wants to see, is in this link:
http://es.slideshare.net/ZkrouhnZyx/mathblocks
Regard
I've done some work on this subject.
Written in English.
If someone wants to see, is in this link:
http://es.slideshare.net/ZkrouhnZyx/mathblocks
Regard
Re: Position item in an array larger than 3D
Hi, all:
Question:
Any number in few DIM's all we can transform?
(R = Rows  C = Cols  L = Level)
Example: 27
27 = 3L * 3R * 3C
27 = 1R * 27C
27 = 3R * 9C
Decomposing the number:
27  3
9  3
3  3
1 
27 = 3 ^ 3
Anyone know any way to solve this question?
Question:
Any number in few DIM's all we can transform?
(R = Rows  C = Cols  L = Level)
Example: 27
27 = 3L * 3R * 3C
27 = 1R * 27C
27 = 3R * 9C
Decomposing the number:
27  3
9  3
3  3
1 
27 = 3 ^ 3
Anyone know any way to solve this question?
Re: Position item in an array larger than 3D
HI, all:
Well, I have not found time to find how all forms (dimensions) possible.
But if the number of combinations.
Just finding all divisors of the number searching and combining them with themselves.
We can only use its divisors, if we use other numbers, the result is wrong.
Prime numbers = 1
Demonstration:
Example: 729
729 243 81 27 9 3 1
Combinations = 6
Example: 27
27 9 3 1
Combinations = 3
Example: 25
25 5 1
Combinations = 2
Example: 29 <<< >>> prime number = 1
29 1
Combinations = 1
Similarly, the value of the rows, columns, levels, dimensions, will be all the divisors of the number to find.
Regards
Well, I have not found time to find how all forms (dimensions) possible.
But if the number of combinations.
Just finding all divisors of the number searching and combining them with themselves.
We can only use its divisors, if we use other numbers, the result is wrong.
Prime numbers = 1
Demonstration:
Code: Select all
DIM AS INTEGER J,N,Z
CLS
INPUT "Number = ";Z
CLS
J = 1
FOR N = Z TO 1 STEP  1
IF Z MOD N = 0 THEN PRINT N;:J = J + 1
NEXT N
PRINT
PRINT "Combinations = ";J
SLEEP
END
Example: 729
729 243 81 27 9 3 1
Combinations = 6
Example: 27
27 9 3 1
Combinations = 3
Example: 25
25 5 1
Combinations = 2
Example: 29 <<< >>> prime number = 1
29 1
Combinations = 1
Similarly, the value of the rows, columns, levels, dimensions, will be all the divisors of the number to find.
Regards
Last edited by lrcvs on Jun 15, 2015 19:49, edited 2 times in total.

 Posts: 2958
 Joined: Jun 02, 2015 16:24
Re: Position item in an array larger than 3D
Hi,
I dont understand what is made for? Can you make a demo of drawing some 3D with your notation?
I dont understand what is made for? Can you make a demo of drawing some 3D with your notation?
Re: Position item in an array larger than 3D
Hi,Tourist Trap:
Ok!
It's simple and complex to explain.
Look to understand this before we made up here or what makes this calculation.
It is seeking the position of an element in any Dimension / Array (2D, 3D, 4D, 5D, 6D, .....)
Similarly, we can transform any Dimension / 1D Array.
On the other hand, knowing the position, we can look at that Block / Dimension / Array, Row, Column, Level, it is an element of an array.
See:
http://es.slideshare.net/ZkrouhnZyx/mathblocks
Regards
Excuse my English!
Ok!
It's simple and complex to explain.
Look to understand this before we made up here or what makes this calculation.
It is seeking the position of an element in any Dimension / Array (2D, 3D, 4D, 5D, 6D, .....)
Similarly, we can transform any Dimension / 1D Array.
On the other hand, knowing the position, we can look at that Block / Dimension / Array, Row, Column, Level, it is an element of an array.
See:
http://es.slideshare.net/ZkrouhnZyx/mathblocks
Regards
Excuse my English!

 Posts: 2958
 Joined: Jun 02, 2015 16:24
Re: Position item in an array larger than 3D
I've watched the slideshow, that was my first action. It doesn't do 3D :)
However from your last explanation I tend to think that you want to change a "n"D array in a 1D one. Is it true?
Why not chaining the dimensions so like this :
[quote]2D version
x1 y1
x2 y2
...
xn yn
1D version
x1 x2 .... xn y1 y2 .... yn[/quote]
But I still dont see why you would do that. Maybe I'm totally wrong.
However from your last explanation I tend to think that you want to change a "n"D array in a 1D one. Is it true?
Why not chaining the dimensions so like this :
[quote]2D version
x1 y1
x2 y2
...
xn yn
1D version
x1 x2 .... xn y1 y2 .... yn[/quote]
But I still dont see why you would do that. Maybe I'm totally wrong.
Re: Position item in an array larger than 3D
HI,
I've watched the slideshow, That was my first action. It does not do 3D :) >>> Ok, Not 3D Graphics
However from your last explanation I Tend To Think That You want to change to "n" D array in a 1D one. Is it true? >>> Is it true
Maybe I'm totally wrong. >>> No, you is not wrong, but may not understand the whole program / project.
Imagine you, you have a block of 3 * 3 * 3 elements (3 rows, 3 columns, 3 levels).
We know that "position number" the element at row 2, column 2 and Level 2, is if we start counting from the top down and from left to right, we see that the search item is located at position 14.
On the other hand, we can reverse the problem, knowing the position and composition of the block.
I've watched the slideshow, That was my first action. It does not do 3D :) >>> Ok, Not 3D Graphics
However from your last explanation I Tend To Think That You want to change to "n" D array in a 1D one. Is it true? >>> Is it true
Maybe I'm totally wrong. >>> No, you is not wrong, but may not understand the whole program / project.
Imagine you, you have a block of 3 * 3 * 3 elements (3 rows, 3 columns, 3 levels).
We know that "position number" the element at row 2, column 2 and Level 2, is if we start counting from the top down and from left to right, we see that the search item is located at position 14.
On the other hand, we can reverse the problem, knowing the position and composition of the block.

 Posts: 2958
 Joined: Jun 02, 2015 16:24
Re: Position item in an array larger than 3D
lrcvs wrote:On the other hand, we can reverse the problem, knowing the position and composition of the block.
Are you sure that we can? How do you know if 7 = 3 + 3 + 1 , or 7 = 2 + 2 + 3 for instance ?
Code: Select all
I I I I I I
I I I I I I
I I I x I I I
I I I x I I I
I x I I x I I I
I x I I x I I I
I x I I x I I x I
\_x_/ \_x_/ \_x_/ == 12
I I I I I I
I I I I I x I
I I I I I x I
I I I I I x I
I I I x I I x I
I I I x I I x I
I I I x I I x I
\_x_/ \_x_/ \_x_/ == 12
Re: Position item in an array larger than 3D
Hi,
Are you sure That we can? Yes !!!
How do you know if: 7 = 3 + 3 + 1, or 7 = 2 + 2 + 3 for instance? <<< Is to Error!
Is very simple...
1) The blocks are composed of: Rows * Columns * Levels * Dims
Tell me how many rows, many columns and how many levels does your block and then tell me what position is the element you want to search for?
2) you indicate drawing I do not understand, give me a block data (rows, columns and levels), tell me that row, column and want to put an item level to know his position.
Are you sure That we can? Yes !!!
How do you know if: 7 = 3 + 3 + 1, or 7 = 2 + 2 + 3 for instance? <<< Is to Error!
Is very simple...
1) The blocks are composed of: Rows * Columns * Levels * Dims
Tell me how many rows, many columns and how many levels does your block and then tell me what position is the element you want to search for?
2) you indicate drawing I do not understand, give me a block data (rows, columns and levels), tell me that row, column and want to put an item level to know his position.
Re: Position item in an array larger than 3D
Hi lrcvs
I remember doing the offsets for multidimensional arrays earlier in this thread (a while ago).
The offsets (for pointer purposes) start at 0, your conversion to a single array (starts at one) will be offsets+1.
So it is easy going from an element in an array to your single dimension substitute element number.
Going the other way, well, that is what you are working on here?
I have done a brute force kinda way, what method would you use to get the 5 dimensional coordinates of a given position in the single dimension equivalent array.
I remember doing the offsets for multidimensional arrays earlier in this thread (a while ago).
The offsets (for pointer purposes) start at 0, your conversion to a single array (starts at one) will be offsets+1.
So it is easy going from an element in an array to your single dimension substitute element number.
Going the other way, well, that is what you are working on here?
I have done a brute force kinda way, what method would you use to get the 5 dimensional coordinates of a given position in the single dimension equivalent array.
Code: Select all
'================= Offset macros ==============================
#define ub ubound
#define lb lbound
#define _r(arr,n) (ub(arr,n)lb(arr,n)+1)
#define d1(a,n1) (n1lb(a,1))
#define d2(a,n1,n2) (_r(a,2)*d1(a,n1)+(n2lb(a,2)))
#define d3(a,n1,n2,n3) (_r(a,3)*d2(a,n1,n2)+(n3lb(a,3)))
#define d4(a,n1,n2,n3,n4) (_r(a,4)*d3(a,n1,n2,n3)+(n4lb(a,4)))
#define d5(a,n1,n2,n3,n4,n5) (_r(a,5)*d4(a,n1,n2,n3,n4)+(n5lb(a,5)))
#define d6(a,n1,n2,n3,n4,n5,n6) (_r(a,6)*d5(a,n1,n2,n3,n4,n5)+(n6lb(a,6)))
#define d7(a,n1,n2,n3,n4,n5,n6,n7) (_r(a,7)*d6(a,n1,n2,n3,n4,n5,n6)+(n7lb(a,7)))
#define d8(a,n1,n2,n3,n4,n5,n6,n7,n8) (_r(a,8)*d7(a,n1,n2,n3,n4,n5,n6,n7)+(n8lb(a,8)))
'========================================================
dim shared as integer i(1 to 30,1 to 30,1 to 30,1 to 30,1 to 30)
print d5(i,30,30,30,30,30)+1,"Size of one dimensional array" 'added 1 to the max offset
dim as integer position=21000000
print "position required = ";position
'for position 21000000 the offset will be one less
for x1 as integer=1 to 30
for x2 as integer=1 to 30
for x3 as integer=1 to 30
for x4 as integer=1 to 30
for x5 as integer=1 to 30
if d5(i,x1,x2,x3,x4,x5)=position1 then print "Element ";x1,x2,x3,x4,x5:goto fin
next:next:next:next:next
fin:
sleep

 Posts: 2958
 Joined: Jun 02, 2015 16:24
Re: Position item in an array larger than 3D
lrcvs wrote:2) you indicate drawing I do not understand, give me a block data (rows, columns and levels), tell me that row, column and want to put an item level to know his position.
Code: Select all
dim as Integer someArray(1 to 10, 1 to 10, 1 to 10)
someArray(4, 8, 2) = 999
If the position in 3D style is 4, 8, 2 , with a value of 999 stored there... What is the position in your counting system?
Re: Position item in an array larger than 3D
Does it will last long (position=372)!
position = @someArray(4, 8, 2)  @someArray(Lbound(someArray, 1), Lbound(someArray, 2), Lbound(someArray, 3)) + 1
position = @someArray(4, 8, 2)  @someArray(Lbound(someArray, 1), Lbound(someArray, 2), Lbound(someArray, 3)) + 1
Re: Position item in an array larger than 3D
Two methods to retrieve the index.
The second method same as fxm.
The second method same as fxm.
Code: Select all
'================= Offset macros ==============================
#define ub ubound
#define lb lbound
#define _r(arr,n) (ub(arr,n)lb(arr,n)+1)
#define d1(a,n1) (n1lb(a,1))
#define d2(a,n1,n2) (_r(a,2)*d1(a,n1)+(n2lb(a,2)))
#define d3(a,n1,n2,n3) (_r(a,3)*d2(a,n1,n2)+(n3lb(a,3)))
#define d4(a,n1,n2,n3,n4) (_r(a,4)*d3(a,n1,n2,n3)+(n4lb(a,4)))
#define d5(a,n1,n2,n3,n4,n5) (_r(a,5)*d4(a,n1,n2,n3,n4)+(n5lb(a,5)))
#define d6(a,n1,n2,n3,n4,n5,n6) (_r(a,6)*d5(a,n1,n2,n3,n4,n5)+(n6lb(a,6)))
#define d7(a,n1,n2,n3,n4,n5,n6,n7) (_r(a,7)*d6(a,n1,n2,n3,n4,n5,n6)+(n7lb(a,7)))
#define d8(a,n1,n2,n3,n4,n5,n6,n7,n8) (_r(a,8)*d7(a,n1,n2,n3,n4,n5,n6,n7)+(n8lb(a,8)))
'========================================================
dim shared as integer i(1 to 10,1 to 10,1 to 10)
print d3(i,10,10,10)+1," = Size of one dimensional array" 'added 1 to the max offset
print "position of array( 4, 8, 2)= ";1+d3(i,4,8,2)
'==============================================================
'method 2
sub Show(a() as integer,x as integer,y as integer,z as integer)
var k1=lbound(a,1),k2=lbound(a,2),k3=lbound(a,3)
var u1=ubound(a,1),u2=ubound(a,2),u3=ubound(a,3)
print @a(u1,u2,u3)@a(k1,k2,k3)+1;" = Size of one dimensional array"
print "position of array(";x;",";y;",";z;")= "; @a(x,y,z)@a(k1,k2,k3)+1
end sub
print
Show(i(),4,8,2)
sleep
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