It looks like you're new here. If you want to get involved, click one of these buttons!

- 140.7K All Categories
- 104.4K Programming Languages
- 6.4K Assembler Developer
- 1.8K Basic
- 39.7K C and C++
- 4.2K C#
- 7.9K Delphi and Kylix
- 4 Haskell
- 9.6K Java
- 4.1K Pascal
- 1.3K Perl
- 1.9K PHP
- 505 Python
- 48 Ruby
- 4.3K VB.NET
- 1.6K VBA
- 20.8K Visual Basic
- 2.6K Game programming
- 309 Console programming
- 88 DirectX Game dev
- 1 Minecraft
- 109 Newbie Game Programmers
- 2 Oculus Rift
- 8.9K Applications
- 1.8K Computer Graphics
- 726 Computer Hardware
- 3.4K Database & SQL
- 520 Electronics development
- 1.6K Matlab
- 627 Sound & Music
- 254 XML Development
- 3.3K Classifieds
- 189 Co-operative Projects
- 178 For sale
- 189 FreeLance Software City
- 1.9K Jobs Available
- 599 Jobs Wanted
- 201 Wanted
- 2.9K Microsoft .NET
- 1.7K ASP.NET
- 1.1K .NET General
- 3K Miscellaneous
- 3 Join the Team
- 2 User Profiles
- 353 Comments on this site
- 59 Computer Emulators
- 1.8K General programming
- 159 New programming languages
- 597 Off topic board
- 164 Mobile & Wireless
- 38 Android
- 124 Palm Pilot
- 335 Multimedia
- 151 Demo programming
- 184 MP3 programming
- 0 Bash scripts
- 16 Cloud Computing
- 52 FreeBSD
- 1.7K LINUX programming
- 366 MS-DOS
- 0 Shell scripting
- 319 Windows CE & Pocket PC
- 4.1K Windows programming
- 884 Software Development
- 403 Algorithms
- 67 Object Orientation
- 85 Project Management
- 88 Quality & Testing
- 233 Security
- 7.5K WEB-Development
- 1.8K Active Server Pages
- 61 AJAX
- 2 Bootstrap Themes
- 55 CGI Development
- 19 ColdFusion
- 222 Flash development
- 1.4K HTML & WEB-Design
- 1.4K Internet Development
- 2.2K JavaScript
- 33 JQuery
- 285 WEB Servers
- 112 WEB-Services / SOAP

Welcome to the new platform of Programmer's Heaven! We apologize for the inconvenience caused, if you visited us from a broken link of the previous version. The main reason to move to a new platform is to provide more effective and collaborative experience to you all. Please feel free to experience the new platform and use its exciting features. Contact us for any issue that you need to get clarified. We are more than happy to help you.

Matt Jacobs
Posts: **2**Member

in Algorithms

Hi,

I need an algorithm that will overlay a spiral on a square array. Basically, an x/y position is stored with respect to this square array, and if given an offset, the position will be moved in a clockwise spiral direction.

eg. a 3x3 square array will have the spiral pattern as:

012

783

654

saying the position was 2,0 then giving an offset of 3 will move it to position 5, giving it an offset of 6 will move it to position 8 etc

I've wracked my brains trying to figure this one out, and I can't seem to find any algorithms out there on the net. Any help would really be appreciated.

I need an algorithm that will overlay a spiral on a square array. Basically, an x/y position is stored with respect to this square array, and if given an offset, the position will be moved in a clockwise spiral direction.

eg. a 3x3 square array will have the spiral pattern as:

012

783

654

saying the position was 2,0 then giving an offset of 3 will move it to position 5, giving it an offset of 6 will move it to position 8 etc

I've wracked my brains trying to figure this one out, and I can't seem to find any algorithms out there on the net. Any help would really be appreciated.

About & Contact / Terms of use / Privacy statement / Publisher: Lars Hagelin

Programmers Heaven articles / Programmers Heaven files / Programmers Heaven uploaded content / Programmers Heaven C Sharp ebook / Operated by CommunityHeaven LLC

© 1997-2013 Programmersheaven.com - All rights reserved.

## Comments

675Member[b][red]This message was edited by Josh Code at 2002-9-9 18:12:41[/red][/b][hr]

: Hi,

:

: I need an algorithm that will overlay a spiral on a square array. Basically, an x/y position is stored with respect to this square array, and if given an offset, the position will be moved in a clockwise spiral direction.

:

: eg. a 3x3 square array will have the spiral pattern as:

: 012

: 783

: 654

:

: saying the position was 2,0 then giving an offset of 3 will move it to position 5, giving it an offset of 6 will move it to position 8 etc

:

: I've wracked my brains trying to figure this one out, and I can't seem to find any algorithms out there on the net. Any help would really be appreciated.

:

Here is a way you could do it:

[b]Variables[/b]

let c = a counter variable (integer) that counts the value of the number stored in the array at a given point

let x,y = the coordinates of the current point being being processed in the array

let dx,dy = the change in x and y

let dir = the direction of motion (up, down, left, right). This is used to set the dx and dy values

let s = the length of the sides of a particullar "square" within the array. I am talking about the smaller "squares" that would be inside the parimiters of the array that approach the centre of the square. I hope this makes some sence.

let z = a counter for the loop along a straight line on one of the sprial's sides

[b]code segment[/b]

s = the length of the square array (3 for the example you provided)

dir = 0

// direction is initially right

x = 0

y = 0

// top left corner of the array

c = 0

// start the spiral at 0

Loop while the x,y is not the coordinates of the centre of the array

case dir of

0: dx = 1

dy = 0

1: dx = 0

dy = 1

2: dx = -1

dy = 0

3: dx = 0

dy = -1

s = s -1

dir = -1

end of case statement

// now the direction would be updated

add 1 to dir

z = 0

loop while z<s

set array[x,y] to c

if z=0 and dir=1 then subtract 1 from s

add dx to x

add dy to y

add 1 to z

add 1 to c

end of z-loop

end of coordinate loop

I haven't tested that but try that algorithm out. If it doesn't work, tell me about it.

- Spam

0 · Vote Down Vote Up · Share on Facebook89Membertransform ordinary matrix positions to spiral positions.

123 position(p) 123

456 -> 894

789 765

Matrix position p is (row*n+col) where n is the sidelength of the square matrix. position(p) will give the

To move the value from position 6 (2,1) spiralwise 3 steps, get the new position from position(6+3)

I tried this last night and it works for all matrixes with n>0. Sorry I couldn't put this in proper algorithm terms.

' n is the sidelength of the matrix

n=

For t = 1 To n: position(t) = t: Next

turnfactor(0) = -1 / n: turnfactor(1) = n

toggle = 0

ds = n

cnt = n - 1

cnt_start = n - 1

For t = n + 1 To n * n

position(t) = position(t - 1) + ds

cnt = cnt - 1

If cnt = 0 Then

ds = ds * turnfactor(toggle)

cnt_start = cnt_start - toggle

cnt = cnt_start

toggle = toggle Xor 1

End If

Next

- Spam

0 · Vote Down Vote Up · Share on Facebook1Member: transform ordinary matrix positions to spiral positions.

:

: 123 position(p) 123

: 456 -> 894

: 789 765

:

: Matrix position p is (row*n+col) where n is the sidelength of the

: square matrix. position(p) will give the

:

: To move the value from position 6 (2,1) spiralwise 3 steps, get the

: new position from position(6+3)

:

: I tried this last night and it works for all matrixes with n>0.

: Sorry I couldn't put this in proper algorithm terms.

:

: ' n is the sidelength of the matrix

: n=

: For t = 1 To n: position(t) = t: Next

: turnfactor(0) = -1 / n: turnfactor(1) = n

: toggle = 0

: ds = n

: cnt = n - 1

: cnt_start = n - 1

:

: For t = n + 1 To n * n

: position(t) = position(t - 1) + ds

: cnt = cnt - 1

: If cnt = 0 Then

: ds = ds * turnfactor(toggle)

: cnt_start = cnt_start - toggle

: cnt = cnt_start

: toggle = toggle Xor 1

: End If

: Next

:

:

Here is a Java version of the above algorithm. However, I put

the array in reverse order for my own purpose.

private static int[] spiralArray(int dimension){

int numberOfItem = dimension*dimension;

int[] spiralArr = new int[numberOfItem];

byte toggle = 0;

int ds = dimension;

int cnt = dimension - 1;

int cntStart = dimension - 1;

for (int i = 1; i <= dimension; i++){

spiralArr[numberOfItem-i] = i;

}

for (int i = numberOfItem - dimension ; i > 0 ; i--){

spiralArr[i - 1] = spiralArr[i] + ds;

cnt = cnt - 1;

if (cnt == 0){

ds = (int)(ds * turnFactor(toggle, dimension));

cntStart = cntStart - toggle;

cnt = cntStart;

toggle = (byte)(toggle^1);

}

}

return spiralArr;

}

/**

*

* @param arg

* @param n

* @return

*/

private static float turnFactor(int arg, float n){

if (arg == 0){

return -1/n;

}else {

return n;

}

}

- Spam

0 · Vote Down Vote Up · Share on Facebook