Matrix and algorithm "spiral" - algorithm

Question

i wanted ask if there some algorithm ready, that allowed me to do this: i have a matrix m (col) x n (row) with m x n elements. I want give position to this element starting from center and rotating as a spiral, for example, for a matrix 3x3 i have 9 elements so defined:
5 6 7
4 9 8
3 2 1
or for una matrix 4 x 3 i have 12 elements, do defined:
8 9 10 1
7 12 11 2
6 5 4 3
or again, a matrix 5x2 i have 10 elements so defined:
3 4
7 8
10 9
6 5
2 1
etc.
I have solved basically defining a array of integer of m x n elements and loading manually the value, but in generel to me like that matrix maked from algorithm automatically.
Thanks to who can help me to find something so, thanks very much.
UPDATE
This code, do exactely about i want have, but not is in delphi; just only i need that start from 1 and not from 0. Important for me is that it is valid for any matrics m x n. Who help me to translate it in delphi?
(defun spiral (rows columns)
(do ((N (* rows columns))
(spiral (make-array (list rows columns) :initial-element nil))
(dx 1) (dy 0) (x 0) (y 0)
(i 0 (1+ i)))
((= i N) spiral)
(setf (aref spiral y x) i)
(let ((nx (+ x dx)) (ny (+ y dy)))
(cond
((and (< -1 nx columns)
(< -1 ny rows)
(null (aref spiral ny nx)))
(setf x nx
y ny))
(t (psetf dx (- dy)
dy dx)
(setf x (+ x dx)
y (+ y dy)))))))
> (pprint (spiral 6 6))
#2A ((0 1 2 3 4 5)
(19 20 21 22 23 6)
(18 31 32 33 24 7)
(17 30 35 34 25 8)
(16 29 28 27 26 9)
(15 14 13 12 11 10))
> (pprint (spiral 5 3))
#2A ((0 1 2)
(11 12 3)
(10 13 4)
(9 14 5)
(8 7 6))
Thanks again very much.

Answer 1

Based on the classic spiral algorithm. supporting non-square matrix:
program SpiralMatrix;
{$APPTYPE CONSOLE}
uses
SysUtils;
type
TMatrix = array of array of Integer;
procedure PrintMatrix(const a: TMatrix);
var
i, j: Integer;
begin
for i := 0 to Length(a) - 1 do
begin
for j := 0 to Length(a[0]) - 1 do
Write(Format('%3d', [a[i, j]]));
Writeln;
end;
end;
var
spiral: TMatrix;
i, m, n: Integer;
row, col, dx, dy,
dirChanges, visits, temp: Integer;
begin
m := 3; // columns
n := 3; // rows
SetLength(spiral, n, m);
row := 0;
col := 0;
dx := 1;
dy := 0;
dirChanges := 0;
visits := m;
for i := 0 to n * m - 1 do
begin
spiral[row, col] := i + 1;
Dec(visits);
if visits = 0 then
begin
visits := m * (dirChanges mod 2) + n * ((dirChanges + 1) mod 2) - (dirChanges div 2) - 1;
temp := dx;
dx := -dy;
dy := temp;
Inc(dirChanges);
end;
Inc(col, dx);
Inc(row, dy);
end;
PrintMatrix(spiral);
Readln;
end.
3 x 3:
1 2 3
8 9 4
7 6 5
4 x 3:
1 2 3 4
10 11 12 5
9 8 7 6
2 x 5:
1 2
10 3
9 4
8 5
7 6

Answer 2

There you go!!! After 30some syntax errors...
On ideone.com, I ran it with some tests and it seems to work fine. I think you can see the output there still and run it yourself...
I put some comments in the code. Enough to understand most of it. The main navigation system is a little bit harder to explain. Briefly, doing a spiral is going in first direction 1 time, second 1 time, third 2 times, fourth 2 times, fifth 3 times, 3, 4, 4, 5, 5, and so on. I use what I called a seed and step to get this behavior.
program test;
var
w, h, m, n, v, d : integer; // Matrix size, then position, then value and direction.
spiral : array of array of integer; // Matrix/spiral itself.
seed, step : integer; // Used to travel the spiral.
begin
readln(h);
readln(w);
setlength(spiral, h, w);
v := w * h; // Value to put in spiral.
m := trunc((h - 1) / 2); // Finding center.
n := trunc((w - 1) / 2);
d := 0; // First direction is right.
seed := 2;
step := 1;
// Travel the spiral.
repeat
// If in the sub-spiral, store value.
if ((m >= 0) and (n >= 0) and (m < h) and (n < w)) then
begin
spiral[m, n] := v;
v := v - 1;
end;
// Move!
case d of
0: n := n + 1;
1: m := m - 1;
2: n := n - 1;
3: m := m + 1;
end;
// Plan trajectory.
step := step - 1;
if step = 0 then
begin
d := (d + 1) mod 4;
seed := seed + 1;
step := trunc(seed / 2);
end;
until v = 0;
// Print the spiral.
for m := 0 to (h - 1) do
begin
for n := 0 to (w - 1) do
begin
write(spiral[m, n], ' ');
end;
writeln();
end;
end.
If you really need that to print text spirals I'll let you align the numbers. Just pad them with spaces.
EDIT:
Was forgetting... In order to make it work on ideone, I put the parameters on 2 lines as input. m, then n.
For example:
5
2
yields
3 4
7 8
10 9
6 5
2 1

Answer 3

Here's the commented JavaScript implementation for what you're trying to accomplish.
// return an array representing a matrix of size MxN COLxROW
function spiralMatrix(M, N) {
var result = new Array(M * N);
var counter = M * N;
// start position
var curCol = Math.floor((M - 1) / 2);
var curRow = Math.floor(N / 2);
// set the center
result[(curRow * M) + curCol] = counter--;
// your possible moves RIGHT, UP, LEFT, DOWN * y axis is flipped
var allMoves = [[1,0], [0,-1], [-1,0], [0,1]];
var curMove = 0;
var moves = 1; // how many times to make current Move, 1,1,2,2,3,3,4,4 etc
// spiral
while(true) {
for(var i = 0; i < moves; i++) {
// move in a spiral outward counter clock-wise direction
curCol += allMoves[curMove][0];
curRow += allMoves[curMove][1];
// naively skips locations that are outside of the matrix bounds
if(curCol >= 0 && curCol < M && curRow >= 0 && curRow < N) {
// set the value and decrement the counter
result[(curRow * M) + curCol] = counter--;
// if we reached the end return the result
if(counter == 0) return result;
}
}
// increment the number of times to move if necessary UP->LEFT and DOWN->RIGHT
if(curMove == 1 || curMove == 3) moves++;
// go to the next move in a circular array fashion
curMove = (curMove + 1) % allMoves.length;
}
}
The code isn't the most efficient, because it walks the spiral naively without first checking if the location it's walking on is valid. It only checks the validity of the current location right before it tries to set the value on it.

Answer 4

Even though the question is already answered, this is an alternative solution (arguably simpler).
The solution is in python (using numpy for bidimendional arrays), but can be easily ported.
The basic idea is to use the fact that the number of steps is known (m*n) as end condition,
and to properly compute the next element of the loop at each iteration:
import numpy as np
def spiral(m, n):
"""Return a spiral numpy array of int with shape (m, n)."""
a = np.empty((m, n), int)
i, i0, i1 = 0, 0, m - 1
j, j0, j1 = 0, 0, n - 1
for k in range(m * n):
a[i, j] = k
if i == i0 and j0 <= j < j1: j += 1
elif j == j1 and i0 <= i < i1: i += 1
elif i == i1 and j0 < j <= j1: j -= 1
elif j == j0 and 1 + i0 < i <= i1: i -= 1
else:
i0 += 1
i1 -= 1
j0 += 1
j1 -= 1
i, j = i0, j0
return a
And here some outputs:
>>> spiral(3,3)
array([[0, 1, 2],
[7, 8, 3],
[6, 5, 4]])
>>> spiral(4,4)
array([[ 0, 1, 2, 3],
[11, 12, 13, 4],
[10, 15, 14, 5],
[ 9, 8, 7, 6]])
>>> spiral(5,4)
array([[ 0, 1, 2, 3],
[13, 14, 15, 4],
[12, 19, 16, 5],
[11, 18, 17, 6],
[10, 9, 8, 7]])
>>> spiral(2,5)
array([[0, 1, 2, 3, 4],
[9, 8, 7, 6, 5]])