## Spiral Wrapping

### October 14, 2014

Today’s exercise appears regularly on lists of interview questions. We’ve done something similar in the past, but since it’s so common we’ll do it again.

Given a matrix, print a list of elements of the matrix. Start in the top-right corner of the matrix, move left across the top row, then down the left column, then across the bottom row, then up the right column to the element below the top row, then left, then down, then right, and so on, collecting the elements of the matrix as it goes. For instance, with this matrix

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20the elements are collected in order 4, 3, 2, 1, 5, 9, 13, 17, 18, 19, 20, 16, 12, 8, 7, 6, 10, 14, 15, 11.

Your task is to write a program to unwrap the elements of a matrix in the indicated spiral. 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.

(Sorry it is not a complete solution)

You can define a bijection between both representations (enumeration spaces) with O(1) cost (per cell).

Then, you can transform using whatever “path function” from one space to another one (eg. take only numbers inside some circle; numbers into a line; …).

The next `coords` function transform from spiral enumeration to coordinates; to solve the problem we need to code the inverse of that function.

Of course, this is a generalized solution.

(argh! I can’t edit my wrong comment, sorry!)

“to solve the problem we need to code the inverse of that function”

no, only `spiral -> coordinates -> cell` then, `coords == spiral -> coordinates` then:

Haskell: http://codepad.org/C9LValE7

A functional, recursive approach.

Similar Python solution as for the previously posted problem.

Allowing any rectangle of integer coordinates between an included point b and and an excluded point e, vertical real axis growing down, horizontal imaginary axis growing right, left and right turn specified as the positive and negative imaginary units, stopping when area vanishes. Risking inexact arithmetic, but both Gambit-C (gsi) and Guile appear sensible.

(define (direction b e t)

(let ((dh (if (< (magnitude (- e (+ b +i))) (magnitude (- e b))) +i -i))

(dv (if (< (magnitude (- e (+ b +1))) (magnitude (- e b))) +1 -1)))

(if (< (magnitude (- (* t dh) dv)) (magnitude (- (* t dv) dh))) dh dv)))

(define (empty? b e)

(or (zero? (- (real-part b) (real-part e)))

(zero? (- (imag-part b) (imag-part e)))))

`(define (spirally b e t)`

(let loop ((p b)

(d (direction b e t))

(b b)

(e e)

(r '()))

(cond

((empty? b e) (reverse r))

((empty? p e)

(let ((b (+ (- p d) (* d t)))

(e (+ e (- b p d))))

(loop b (* d t) b e r)))

(else (loop (+ p d) d b e (cons p r))))))

Mapping coordinates to values that give the example matrix in row-major order between (0,0) inclusive and (5,4) exclusive.

(define (M p) (inexact->exact (+ (* 4 (real-part p)) (imag-part p) 1)))

(define (C r k) (make-rectangular r k))

(define (test exp obs)

(display "expected: ") (display exp) (newline)

(display "observed: ") (display obs) (newline) (newline))

; From top right, left turns

(test '(4 3 2 1 5 9 13 17 18 19 20 16 12 8 7 6 10 14 15 11)

(map M (spirally (C 0 3) (C 5 -1) +i)))

; From top right, right turns

(test '(4 8 12 16 20 19 18 17 13 9 5 1 2 3 7 11 15 14 10 6)

(map M (spirally (C 0 3) (C 5 -1) -i)))

`; Down last column, up next column`

(test '(4 8 12 16 20 21 17 13 8 5)

(map M (spirally (C 0 3) (C 5 5) +i)))

Observing some expected results.

$ gsi spiral.scm

expected: (4 3 2 1 5 9 13 17 18 19 20 16 12 8 7 6 10 14 15 11)

observed: (4 3 2 1 5 9 13 17 18 19 20 16 12 8 7 6 10 14 15 11)

expected: (4 8 12 16 20 19 18 17 13 9 5 1 2 3 7 11 15 14 10 6)

observed: (4 8 12 16 20 19 18 17 13 9 5 1 2 3 7 11 15 14 10 6)

expected: (4 8 12 16 20 21 17 13 8 5)

observed: (4 8 12 16 20 21 17 13 9 5)

Er, that one discrepancy is a typo in the expected value :)

A mechanical translation of my original Scheme to C99 (only printing the values of the extended matrix M).

Hm. Forgot to return from main. There’s always something.

Again observing the expected, as described in my original Scheme entry (the task matrix spirally as specified, then clockwise, and then the last column down and the next column up).

recursive scheme solution using matrix rotation

http://repl.it/18y

Here’s a J solution:

And here’s the sample output;

#include

using namespace std;

#define SIZE 5

void SPiralArray(int size,int array[][SIZE])

{

int a = size/2*2 + 1;

int y = a/2,x =a/2;

//for(int i = 1; i=1; i–)

{

if(x=y)

{

array[y][x] = i;

x++;

}

else if(x>a-y-1 && x>y)

{

array[y][x] = i;

y++;

}

else if(x>a-y-1 && x<= y)

{

array[y][x] = i;

x–;

}

else if(x<=a-y-1&& x<y)

{

array[y][x] = i;

y–;

}

}

}

int main(int argc, char *argv[])

{

int array[SIZE][SIZE];

for( int i = 0; i<SIZE; i++)

{

for(int j = 0; j<SIZE; j++)

array[i][j] = 0;

}

SPiralArray(SIZE, array);

for( int i = 0; i<SIZE; i++)

{

for(int j = 0; j<SIZE; j++)

{

cout << array[i][j] << "\t";

}

cout<<endl;

}

return 0;

}

It is really weird, some of the codes that I posted are missing. It should be like this way:

void SPiralArray(int size,int array[][SIZE])

{

int a = size/2*2 + 1;

int y = a/2,x =a/2;

for(int i =size*size; i>=1; i–)

{……}

}