## Unwrapping A Spiral

### June 1, 2010

This exercise appeared on Proggit a while ago. The task is to enumerate the elements of a matrix in spiral order. For instance, consider the matrix:

` 1 2 3 4`

5 6 7 8

9 10 11 12

13 14 15 16

17 18 19 20

The spiral starts across the first row, yielding 1, 2, 3, and 4. Then the spiral turns right and runs down the right column, yielding 8, 12, 16, and 20. The spiral turns right again and runs across the bottom row, from right to left, yielding 19, 18, and 17. Then up the first column with 13, 9, 5, right with 6 and 7, down with 11 and 15, right to 14, and up to 10. Thus, unwrapping the given matrix in a spiral gives the list of elements 1, 2, 3, 4, 8, 12, 16, 20, 19, 18, 17, 13, 9, 5, 6, 7, 11, 15, 14, and 10.

Your task is to write a function to unwrap spirals. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

Pages: 1 2

[...] Praxis – Unwrapping A Spiral By Remco Niemeijer In today’s Programming Praxis exercise our task is to write a function that walks a 2-dimensional array in a [...]

My Haskell solution (see http://bonsaicode.wordpress.com/2010/06/01/programming-praxis-unwrapping-a-spiral/ for a version with comments):

hm, implemented the reverse thing in common lisp, from an spiral list to an matrix:

(defparameter *spiral* ‘(1 2 3 4 8 12 16 20 19 18 17 13 9 5 6 7 11 15 14 10))

(defun spiral-unwind (spiral)

(let* ((partion ‘())

(delta (maplist

#'(lambda (x)

(if (> (length x) 1)

(let ((delta (- (second x) (first x))))

(if (and partion (= delta (car (car partion))))

(setf (car partion) (cons delta (car partion)))

(push (list delta) partion))

delta)

0))

spiral))

(dimension-x (+ 1 (length (first (reverse partion)))))

(dimension-y (+ 1 (length (second (reverse partion)))))

(matrix (make-array (list dimension-y dimension-x)))

(x 0) (y 0))

(loop

for elem in spiral

for d in delta

do (progn

(setf (aref matrix y x) elem)

(cond ((= d 1) (incf x))

((= d -1) (decf x))

((= d (- dimension-y 1)) (incf y))

(t (decf y)))))

matrix))

A few minutes of playing around yielded this faintly obscure Python version…

Ooops. Left a line in that was unnecessary.

My Python version:

‘matrix’ is a list of lists in row major order. The idea is to append the top row of the matrix to the spiral. Then rotate the rest of the matrix (rows 2 through n) ccw, so that the right-most column is now the top row. Repeat until the matrix is empty.

I like yours better, Mike. I bow to your tidy Python-fu.

Oops. Line 18 should be:

My Java solution can be seen here. I must say mine seems overly verbose in comparison to the solutions showed here. Awesome!

A Haskell solution that generates a list of indices (starting from 0, matrix stored by rows).

A recursive and an iterative solution in Common Lisp along the same lines as the Haskell solution.

If we do not want to allocate an intermediate list with indices we could do something like:

a=[[1,2,3,4],

[5,6,7,8],

[9,10,11,12],

[13,14,15,16],

[17,18,19,20]]

def unwrap(matrix):

rows=len(matrix)

cols=len(matrix[0])

counter=0

curr_row=0

curr_col=0

while (counter<rows*cols):

j=curr_col

while (j<(cols-curr_col)):

print matrix[curr_row][j],

counter=counter+1

j=j+1

curr_col=j-1

i=curr_row+1

while (i=(cols-curr_col-1)):

print matrix[curr_row][j],

counter=counter+1

j=j-1

curr_col=j+1

i=curr_row-1

while (i>(rows-curr_row-1)):

print matrix[i][curr_col],

counter=counter+1

i=i-1

curr_row=i+1

curr_col=curr_col+1

unwrap(a)

——————————————————————————————

This is a very simple code written in python…very easy to understand!!

Python code using generators.

A quick c/c++ hack:

def spiral(matrix):

while len(matrix):

matrix = horizontal(matrix)

if (len(matrix)):

matrix = vertical(matrix)

for i in range(len(matrix)):

matrix[i].reverse()

matrix.reverse()

def vertical(matrix):

lastCol = len(matrix[0])

for i in range(len(matrix)):

listM.append(matrix[i][lastCol-1])

del(matrix[i][lastCol-1])

return matrix

def horizontal(matrix):

for i in range(len(matrix[0])):

listM.append(matrix[0][i])

return matrix[1:len(matrix)]

`listM =[]`

matrixM = [ [1,2,3,4],

[5,6,7,8],

[9,10,11,12],

[13,14,15,16],

[17,18,19,20]

]

spiral(matrixM)

print listM

My python code:

My C implementation

http://codepad.org/LhPOJjmq

:)

Little bit modified

http://codepad.org/dpnlp8zz

Thanks :)

import java.util.Iterator;

public class Unwrapp implements Iterator, Iterable

{

private enum Direction { EAST, SOUTH, WEST, NORTH }

private int row, col, minRow, maxRow, minCol, maxCol;

private T matrix[][];

private Direction d;

public Unwrapp(T aMatrix[][]) {

matrix = aMatrix;

row = col = 0;

minRow = 0; maxRow = matrix.length – 1;

minCol = 0; maxCol = matrix[0].length – 1;

d = Direction.EAST;

}

public Iterator iterator() { return this; }

public void remove() { throw new UnsupportedOperationException(); }

public boolean hasNext() {

return (d == Direction.EAST && col <= maxCol)

|| (d == Direction.SOUTH && row = minCol)

|| (d == Direction.NORTH && row >= minRow);

}

public T next() {

T x = matrix[row][col];

switch (d) {

case EAST:

if (col == maxCol) { ++minRow; ++row; d = Direction.SOUTH; }

else ++col;

break;

case SOUTH:

if (row == maxRow) { –maxCol; –col; d = Direction.WEST; }

else ++row;

break;

case WEST:

if (col == minCol) { –maxRow; –row; d = Direction.NORTH; }

else –col;

break;

case NORTH:

if (row == minRow) { ++minCol; ++col; d = Direction.EAST; }

else –row;

break;

}

return x;

}

public static void main(String args[]) {

Integer matrix[][] = {

{1, 2, 3, 4},

{5, 6, 7, 8},

{9, 10, 11, 12},

{13, 14, 15, 16},

{17, 18, 19, 20}

};

for (Integer i : new Unwrapp(matrix)) System.out.print(” ” + i);

System.out.println();

}

}

Unwrapp.java

/home/fabio/Desktop/Unwrapp.java

Here is an R version, with shortness of code traded for memory/performance:

I want program to print reverse spiral program. e.g If a matrix of 4×4, instead of statig A(oo), I want to print revesre spiral order starting from D(33).

Similar to @mikes solution, but in Perl