square puzzle solution

Question

Question: given an integer number n, print the numbers from 1 up to n² like this:

n = 4

result is:

01 02 03 04
12 13 14 05
11 16 15 06
10 09 08 07

How do you solve it (apart from the solution provided in the link below)?

http://www.programmersheaven.com/mb/CandCPP/81986/81986/problem-in-making-ap-c++-program/?S=B20000

I'm looking in another direction. So far, I'm trying to figure out if I could obtain the ordered list of positions I have to fill in.

Here's what Im looking into: is there a way to obtain the "fdisp" so as to solve the problem that way, instead of "walk" in the matrix?

matrix = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]
n = len(matrix)

# final disposition wrote by hand: how to get it for arbitrary n?
fdisp = [(0,0), (0,1), (0,2), (0,3), (1,3), (2,3), (3,3), (3,2),
         (3,1), (3,0), (2,0), (1,0), (1,1), (1,2), (2,2), (2,1)]

for val,i in enumerate(fdisp):
    matrix[i[0]][i[1]] = val + 1

def show_matrix(matrix, n):
    for i,l in enumerate(matrix):
        for j in range(n):
            print "%d\t" % matrix[i][j],
        print

show_matrix(matrix, n)

Answer 1

Here's a different approach. It relies on spotting that the movements you make cycle between: right, down, left, up, right, .... Further, the number of times you move goes: 3 right, 3 down, 3 left, 2 up, 2 right, 1 down, 1 left. So without further ado, I will code this up in Python.

First, I will use some itertools and some numpy:

from itertools import chain, cycle, imap, izip, repeat
from numpy import array

The directions cycle between: right, down, left, up, right, ...:

directions = cycle(array(v) for v in ((0,1),(1,0),(0,-1),(-1,0)))

(I'm using numpy's arrays here so I can easily add directions together. Tuples don't add nicely.)

Next, the number of times I move counts down from n-1 to 1, repeating each number twice, and the first number three times:

countdown = chain((n-1,), *imap(repeat, range(n-1,0,-1), repeat(2)))

So now my sequence of directions can be created by repeating each successive direction by the paired number in countdown:

dirseq = chain(*imap(repeat, directions, countdown))

To get my sequence of indices, I can just sum this sequence, but (AFAIK) Python does not provide such a method, so let's quickly throw one together:

def sumseq(seq, start=0):
  v = start
  yield v
  for s in seq:
    v += s
    yield v

Now to generate the original array, I can do the following:

a = array(((0,)*n,)*n) # n-by-n array of zeroes
for i, v in enumerate(sumseq(dirseq, array((0,0)))):
  a[v[0], v[1]] = i+1
print a

Which, for n = 4, gives:

[[ 1  2  3  4]
 [12 13 14  5]
 [11 16 15  6]
 [10  9  8  7]]

and, for n = 5, gives:

[[ 1  2  3  4  5]
 [16 17 18 19  6]
 [15 24 25 20  7]
 [14 23 22 21  8]
 [13 12 11 10  9]]

This approach can be generalised to rectangular grids; I leave this as an exercise for the reader ;)

Answer 2

Though your example is in python and this is in Java, I think you should be able to follow the logic:

public class SquareTest {

public static void main(String[] args) {
    SquareTest squareTest = new SquareTest(4);
    System.out.println(squareTest);
}

private int squareSize;
private int[][] numberSquare;
private int currentX;
private int currentY;
private Direction currentDirection;

private enum Direction {
    LEFT_TO_RIGHT, RIGHT_TO_LEFT, TOP_TO_BOTTOM, BOTTOM_TO_TOP;
};

public SquareTest(int squareSize) {
    this.squareSize = squareSize;
    numberSquare = new int[squareSize][squareSize];
    currentY = 0;
    currentX = 0;
    currentDirection = Direction.LEFT_TO_RIGHT;
    constructSquare();
}

private void constructSquare() {
    for (int i = 0; i < squareSize * squareSize; i = i + 1) {
        numberSquare[currentY][currentX] = i + 1;
        if (Direction.LEFT_TO_RIGHT.equals(currentDirection)) {
            travelLeftToRight();
        } else if (Direction.RIGHT_TO_LEFT.equals(currentDirection)) {
            travelRightToLeft();
        } else if (Direction.TOP_TO_BOTTOM.equals(currentDirection)) {
            travelTopToBottom();
        } else {
            travelBottomToTop();
        }
    }
}

private void travelLeftToRight() {
    if (currentX + 1 == squareSize || numberSquare[currentY][currentX + 1] != 0) {
        currentY = currentY + 1;
        currentDirection = Direction.TOP_TO_BOTTOM;
    } else {
        currentX = currentX + 1;
    }
}

private void travelRightToLeft() {
    if (currentX - 1 < 0 || numberSquare[currentY][currentX - 1] != 0) {
        currentY = currentY - 1;
        currentDirection = Direction.BOTTOM_TO_TOP;
    } else {
        currentX = currentX - 1;
    }
}

private void travelTopToBottom() {
    if (currentY + 1 == squareSize || numberSquare[currentY + 1][currentX] != 0) {
        currentX = currentX - 1;
        currentDirection = Direction.RIGHT_TO_LEFT;
    } else {
        currentY = currentY + 1;
    }
}

private void travelBottomToTop() {
    if (currentY - 1 < 0 || numberSquare[currentY - 1][currentX] != 0) {
        currentX = currentX + 1;
        currentDirection = Direction.LEFT_TO_RIGHT;
    } else {
        currentY = currentY - 1;
    }
}

@Override
public String toString() {
    StringBuilder builder = new StringBuilder();
    for (int i = 0; i < squareSize; i = i + 1) {
        for (int j = 0; j < squareSize; j = j + 1) {
            builder.append(numberSquare[i][j]);
            builder.append(" ");
        }
        builder.append("\n");
    }

    return builder.toString();
}
}