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 1         


        

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();
}
}

Answer 3

Another way to do it, this time in C#:

int number = 9;
var position = new { x = -1, y = 0 };
var directions = new [] { 
    new { x = 1, y = 0 },
    new { x = 0, y = 1 },
    new { x = -1, y = 0 },
    new { x = 0, y = -1 }
};

var sequence = (
    from n in Enumerable.Range(1, number)
    from o in Enumerable.Repeat(n, n != number ? 2 : 1)
    select o
).Reverse().ToList();

var result = new int[number,number];

for (int i = 0, current = 1; i < sequence.Count; i++)
{
    var direction = directions[i % directions.Length];      

    for (int j = 0; j < sequence[i]; j++, current++)
    {
        position = new {
            x = position.x + direction.x,
            y = position.y + direction.y
        };

        result[position.y, position.x] = current;
    }
}

Answer 4

I have solved your problem using C++. I don't know if it will be helpful for you. But posting it. If it works for you it will be a pleasure.

Here is the Code:

    #include<iostream>
    #include<string.h>
    using namespace std;

    bool valid(int n,int r,int c)
    {
        if(r>=1 && r<=n && c>=1 && c<=n)
            return true;
        return false;
    }


    int main()
    {
        pair<int,int>d1,d2,d3,d4,temp;
        d1 = make_pair(0,1);
        d2 = make_pair(1,0);
        d3 = make_pair(0,-1);
        d4 = make_pair(-1,0);
        /**********************direction******************************/

        int n, i, j, counter=1, newR = 1, newC = 0, direction = 4;
        bool changeDir=true;
        /**************************variables*************************/

        cin>>n;
        int arr[n+1][n+1];
        int visited[n+1][n+1];
        /*************************arrays********************************/

        memset(visited,0,sizeof(visited));
        memset(arr,0,sizeof(arr));
        /***************initializing the array**************************/

        while(counter<=n*n)
        {
            if(direction==1 && changeDir)
            {
                temp = make_pair(d2.first,d2.second);
                direction=2;
                changeDir=false;
            }
            else if(direction==2&& changeDir)
            {
                temp = make_pair(d3.first,d3.second);
                direction=3;
                changeDir=false;
            }
            else if(direction==3&& changeDir)
            {
                temp = make_pair(d4.first,d4.second);
                direction=4;
                changeDir=false;
            }
            else if(direction==4&& changeDir)
            {
                temp = make_pair(d1.first,d1.second);
                direction=1;
                changeDir=false;
            }
            while(counter<=(n*n) && !changeDir)
            {
                newR =newR+temp.first;
                newC=newC+temp.second;
                if(valid(n,newR,newC) && !visited[newR][newC])
                {
                    arr[newR][newC]=counter;
                    visited[newR][newC]=1;
                    counter++;
                }
                else
                {
                    newR-=temp.first;
                    newC-=temp.second;
                    changeDir=true;
                    break;
                }
            }
        }
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                if(arr[i][j]<10)
                    cout<<0;
                cout<<arr[i][j]<<" ";
            }
            cout<<endl;
        }
        return 0;
    }

Here is the output where N=5:

01 02 03 04 05 
16 17 18 19 06 
15 24 25 20 07 
14 23 22 21 08 
13 12 11 10 09

Thank you.

Answer 5

I found a way. Now I've to improve it a bit, especially I've to find a cleaner way to build "fdisp". n = 5

dim = n
pos = (0, -1)
fdisp = []
squares = n % 2 == 0 and n / 2 or n / 2 + 1

for _ in range(squares):
    pos = (pos[0], pos[1] + 1)
    fdisp.append(pos)

    fdisp += [(pos[0],pos[1]+i) for i in range(1, dim)]
    pos = fdisp[-1]
    fdisp += [(pos[0]+i,pos[1]) for i in range(1, dim)]
    pos = fdisp[-1]
    fdisp += [(pos[0],pos[1]-i) for i in range(1, dim)]
    pos = fdisp[-1]
    fdisp += [(pos[0]-i,pos[1]) for i in range(1, dim - 1)]
    pos = fdisp[-1]
    dim = dim - 2

matrix = [[0] * n for i in range(n)]

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)