Looping in a spiral

Question

A friend was in need of an algorithm that would let him loop through the elements of an NxM matrix (N and M are odd). I came up with a solution, but I wanted to see if my fellow SO'ers could come up with a better solution.

I'm posting my solution as an answer to this question.

Example Output:

For a 3x3 matrix, the output should be:

(0, 0) (1, 0) (1, 1) (0, 1) (-1, 1) (-1, 0) (-1, -1) (0, -1) (1, -1)

Furthermore, the algorithm should support non-square matrices, so for example for a 5x3 matrix, the output should be:

(0, 0) (1, 0) (1, 1) (0, 1) (-1, 1) (-1, 0) (-1, -1) (0, -1) (1, -1) (2, -1) (2, 0) (2, 1) (-2, 1) (-2, 0) (-2, -1)

Answer 1

C++ anyone? Quick translation from python, posted for completeness

void Spiral( int X, int Y){
    int x,y,dx,dy;
    x = y = dx =0;
    dy = -1;
    int t = std::max(X,Y);
    int maxI = t*t;
    for(int i =0; i < maxI; i++){
        if ((-X/2 <= x) && (x <= X/2) && (-Y/2 <= y) && (y <= Y/2)){
            // DO STUFF...
        }
        if( (x == y) || ((x < 0) && (x == -y)) || ((x > 0) && (x == 1-y))){
            t = dx;
            dx = -dy;
            dy = t;
        }
        x += dx;
        y += dy;
    }
}

Answer 2

I am sharing this code which I designed for a different purpose; it is about finding the Column number "X", and the row number "Y" of array element @ spiral index "index". This function takes the width "w" and height "h" of the matrix, and the required "index". Of course, this function can be used to produce the same required output. I think it is the fastest possible method (as it jumps over cells instead of scanning them).

    rec BuildSpiralIndex(long w, long h, long index = -1)
    {  
        long count = 0 , x = -1,  y = -1, dir = 1, phase=0, pos = 0,                            length = 0, totallength = 0;
        bool isVertical = false;
        if(index>=(w*h)) return null;

        do 
        {                
            isVertical = (count % 2) != 0;
            length = (isVertical ? h : w) - count/2 - count%2 ;
            totallength += length;
            count++;
        } while(totallength<index);

        count--; w--; h--;
        phase = (count / 4); pos = (count%4);
        x = (pos > 1 ? phase : w - phase);
        y = ((pos == 1 || pos == 2) ? h - phase : phase) + (1 * (pos == 3 ? 1 : 0));
        dir = pos > 1 ? -1 : 1;
        if (isVertical) y -= (totallength - index - 1) * dir;
        else x -= (totallength - index -1) * dir;
        return new rec { X = x, Y = y };
    }

Answer 3

This is my very very bad solution, made from bare minimum knowledge of Java. Here I have to place units on a field in a spiral. Units cannot be placed on top of other units or on mountains or in the ocean.

To be clear. This is not a good solution. This is a very bad solution added for the fun of other people to laugh at how bad it can be done

private void unitPlacementAlgorithm(Position p, Unit u){
    int i = p.getRow();
    int j = p.getColumn();

    int iCounter = 1;
    int jCounter = 0;

    if (getUnitAt(p) == null) {
            unitMap.put(p, u);
    } else {
        iWhileLoop(i, j, iCounter, jCounter, -1, u);
    }

}

private void iWhileLoop(int i, int j, int iCounter, int jCounter, int fortegn, Unit u){
    if(iCounter == 3) {
        for(int k = 0; k < 3; k++) {
            if(k == 2) { //This was added to make the looping stop after 9 units
                System.out.println("There is no more room around the city");
                return; 
            }
            i--;

            if (getUnitAt(new Position(i, j)) == null 
                && !(getTileAt(new Position(i, j)).getTypeString().equals(GameConstants.OCEANS)) 
                && !(getTileAt(new Position(i, j)).getTypeString().equals(GameConstants.MOUNTAINS))) {
                    unitMap.put(new Position(i, j), u);
                    return;
            }
            iCounter--;
        }
    }

    while (iCounter > 0) {
        if (fortegn > 0) {
            i++;
        } else {
            i--;
        }

        if (getUnitAt(new Position(i, j)) == null 
            && !(getTileAt(new Position(i, j)).getTypeString().equals(GameConstants.OCEANS)) 
            && !(getTileAt(new Position(i, j)).getTypeString().equals(GameConstants.MOUNTAINS))) {
                unitMap.put(new Position(i, j), u);
                return;
        }
        iCounter--;
        jCounter++;
    }
    fortegn *= -1;
    jWhileLoop(i, j, iCounter, jCounter, fortegn, u);
}

private void jWhileLoop(int i, int j, int iCounter, int jCounter,
        int fortegn, Unit u) {
    while (jCounter > 0) {
        if (fortegn > 0) {
            j++;
        } else {
            j--;
        }

        if (getUnitAt(new Position(i, j)) == null 
            && !(getTileAt(new Position(i, j)).getTypeString().equals(GameConstants.OCEANS)) 
            && !(getTileAt(new Position(i, j)).getTypeString().equals(GameConstants.MOUNTAINS))) {
                unitMap.put(new Position(i, j), u);
                return;

        }
        jCounter--;
        iCounter++;
        if (jCounter == 0) {
            iCounter++;
        }

    }
    iWhileLoop(i, j, iCounter, jCounter, fortegn, u);
}

Cudos to anyone who can actually read this

Bonus question: What is the running time of this "algorithm"? :P

Answer 4

I recently had a similar challenge where I had to create a 2D array and use a spiral matrix algorithm to sort and print the results. This C# code will work with a N,N 2D array. It is verbose for clarity and can likely be re-factored to fit your needs.

//CREATE A NEW MATRIX OF SIZE 4 ROWS BY 4 COLUMNS - SCALE MATRIX SIZE HERE
SpiralMatrix SM = new SpiralMatrix(4, 4);
string myData = SM.Read();


public class SpiralMatrix
{
    //LETS BUILD A NEW MATRIX EVERY TIME WE INSTANTIATE OUR CLASS
    public SpiralMatrix(int Rows, int Cols)
    {
        Matrix = new String[Rows, Cols];

        int pos = 1;
        for(int r = 0; r<Rows; r++){
            for (int c = 0; c < Cols; c++)
            {
                //POPULATE THE MATRIX WITH THE CORRECT ROW,COL COORDINATE
                Matrix[r, c] = pos.ToString();
                pos++;
            }
        }
    }

    //READ MATRIX
    public string Read()
    {
        int Row = 0;
        int Col = 0;

        string S = "";
        bool isDone = false;

        //CHECK tO SEE IF POSITION ZERO IS AVAILABLE
        if(PosAvailable(Row, Col)){
            S = ConsumePos(Row, Col);
        }


        //START READING SPIRAL
        //THIS BLOCK READS A FULL CYCLE OF RIGHT,DOWN,LEFT,UP EVERY ITERATION
        while(!isDone)
        {
            bool goNext = false;

            //READ ALL RIGHT SPACES ON THIS PATH PROGRESSION
            while (PosAvailable(Row, Col+1))
            {
                //Is ReadRight Avail
                Col++;
                S += ConsumePos(Row, Col);
                goNext = true;
            }

            //READ ALL DOWN SPACES ON THIS PATH PROGRESSION
            while(PosAvailable(Row+1, Col)){
                //Is ReadDown Avail
                Row++;
                S += ConsumePos(Row, Col);
                goNext = true;
            }

            //READ ALL LEFT SPACES ON THIS PATH PROGRESSION
            while(PosAvailable(Row, Col-1)){
                //Is ReadLeft Avail
                Col--;
                S += ConsumePos(Row, Col);
                goNext = true;
            }

            //READ ALL UP SPACES ON THIS PATH PROGRESSION
            while(PosAvailable(Row-1, Col)){
                //Is ReadUp Avail
                Row--;
                S += ConsumePos(Row, Col);
                goNext = true;
            }

            if(!goNext){
                //DONE - SET EXIT LOOP FLAG
                isDone = true;
            }
        }

        return S;
    }

    //DETERMINE IF THE POSITION IS AVAILABLE
    public bool PosAvailable(int Row, int Col)
    {
        //MAKE SURE WE ARE WITHIN THE BOUNDS OF THE ARRAY
        if (Row < Matrix.GetLength(0) && Row >= 0
            && Col < Matrix.GetLength(1) && Col >= 0)
        {
            //CHECK COORDINATE VALUE
            if (Matrix[Row, Col] != ConsumeChar)
                return true;
            else
                return false;
        }
        else
        {
            //WE ARE OUT OF BOUNDS
            return false;
        }
    }

    public string ConsumePos(int Row, int Col)
    {
        string n = Matrix[Row, Col];
        Matrix[Row, Col] = ConsumeChar;
        return n;
    }

    public string ConsumeChar = "X";
    public string[,] Matrix;
}

Answer 5

This is in C.

I happened to choose bad variable names. In the names T == top, L == left, B == bottom, R == right. So, tli is top left i and brj is bottom right j.

#include<stdio.h>

typedef enum {
   TLTOR = 0,
   RTTOB,
   BRTOL,
   LBTOT
} Direction;

int main() {
   int arr[][3] = {{1,2,3},{4,5,6}, {7,8,9}, {10,11,12}};
   int tli = 0, tlj = 0, bri = 3, brj = 2;
   int i;
   Direction d = TLTOR;

   while (tli < bri || tlj < brj) {
     switch (d) {
     case TLTOR:
    for (i = tlj; i <= brj; i++) {
       printf("%d ", arr[tli][i]);
    }
    tli ++;
    d = RTTOB;
    break;
     case RTTOB:
    for (i = tli; i <= bri; i++) {
       printf("%d ", arr[i][brj]);
    }
    brj --;
    d = BRTOL;
    break;
     case BRTOL:
    for (i = brj; i >= tlj; i--) {
       printf("%d ", arr[bri][i]);
    }
    bri --;
        d = LBTOT;
    break;
     case LBTOT:
    for (i = bri; i >= tli; i--) {
       printf("%d ", arr[i][tlj]);
    }
    tlj ++;
        d = TLTOR;
    break;
 }
   }
   if (tli == bri == tlj == brj) {
      printf("%d\n", arr[tli][tlj]);
   }
}

Answer 6

//PHP implementation

function spiral($n) {

    $r = intval((sqrt($n + 1) - 1) / 2) + 1;

    // compute radius : inverse arithmetic sum of 8+16+24+...=
    $p = (8 * $r * ($r - 1)) / 2;
    // compute total point on radius -1 : arithmetic sum of 8+16+24+...

    $en = $r * 2;
    // points by face

    $a = (1 + $n - $p) % ($r * 8);
    // compute de position and shift it so the first is (-r,-r) but (-r+1,-r)
    // so square can connect

    $pos = array(0, 0, $r);
    switch (intval($a / ($r * 2))) {
        // find the face : 0 top, 1 right, 2, bottom, 3 left
        case 0:
            $pos[0] = $a - $r;
            $pos[1] = -$r;
            break;
        case 1:
            $pos[0] = $r;
            $pos[1] = ($a % $en) - $r;
            break;
        case 2:
            $pos[0] = $r - ($a % $en);
            $pos[1] = $r;
            break;
        case 3:
            $pos[0] = -$r;
            $pos[1] = $r - ($a % $en);
            break;
    }
    return $pos;
}

for ($i = 0; $i < 168; $i++) {

    echo '<pre>';
    print_r(spiral($i));
    echo '</pre>';
}