Easy Logic game: What should be the strategy?

Question

Try this out
http://content.boredmuch.com/game/07183679.swf

Rules are simple:
When you start, you will get a n x n board on which few black balls are placed. To start with, you can click on "any" unoccupied cell and then you can move left- right - up - down creating a trail. The objective is to "occupy" all the blank cells. Figure out rest of the things your self .

Question:
What should you think before you place the "first" ball inside the board, so that you can win the game?

I am asking this in math section since it involves little bit of math :)

Answer 1

I beat Level 1.



I hope Level 2 is impossible because, if it's not, I'm probably just stupid.



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Edit:

Wooo I just beat Level 2. Started with second from the left, second from the bottom.



Thanks for posting this. Now I won't get any work done for the next week or so.



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I beat Level 6. I'm sort of getting the hang of this, yet I can't really pinpoint my strategy.

Spoiler for Level 6: http://i44.tinypic.com/1zv795v.jpg



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And only a few minutes later, I beat Levels 7, 8, & 9 within only about 15 minutes. As I said, I think I have a strategy, I just can't exactly pinpoint it. This game definitely messes with your head a little.

Spoiler for Level 7: http://i42.tinypic.com/otf5o7.jpg

Spoiler for Level 8: http://i39.tinypic.com/13ztax.jpg

Spoiler for Level 9: http://i44.tinypic.com/mcye4j.jpg



Maybe someone can look at these screenshots and figure out what the strategy is to win.

Answer 2

I was thinking that a solution would have to start from the middle of an empty block of 9 (if one existed) but when I got to level 6 my theory crashed when I found the solution started from (5,5) then it crashed again on level 7 with (3,4) and again on level 8 with (2,5), I won't give full solutions cause that would be like telling you the end of a movie before you'd watched it.

Then I realised that this is the same as finding a Hamiltionian path, but it's more difficult than that because no paths are allowed to cross either, and so far no one has ever found a general set of rules to show the existence of a hamiltonian path in a graph so I'd say there is no general set of rules for finding a solution to this problem either, only conditions that would simplify the brute force method. But there are exceptions to this in some specific cases like the n-cube and the puzzle called 'the knights tour' so this problem might be one of those exceptions even though it is more complicated.

I seem to have hit a brick wall on level 9 and I shouldn't really be doing this cause I have an exam on friday but I will undoubtedly get up in the morning and have another go cause I'm addicted now and won't be able to rest until I finish it, damn you!

,.,.,

Answer 3

n•n board; 5•5 board. I will try to explain this according to how my brain matter deciphers logical patterned puzzles. The two dots in Line 1, Row 1, & Line 1, Row 2 establish a 2 to 3 & a 3 to 2 pattern. In other words, all moves must be 2 then 3 and/or 3 then 2 trails. Level one upper row third block is the logical starting point since the first two blocks are already a completed 2 pattern.

Answer 4

Are you looking for something that inputs the locations of the black squares and finds the solution through some special calculation?



Finding it by brute force seems easy.



Brute force could also be simplified by systematically eliminating positions (without certain changes) that would result in a "division" of the path, but that would just potentially save a bit of processing time.



It's easier to determine locations where you must begin or must end... and sometimes it is obvious that you cannot begin there (because every path out results in you blocking some open spaces in)... but not every level has ones that are like that, and you probably need to evaluate an entire path to really know whether you start or end there sometimes.

Answer 5

WOW

REALLY AWESOME GAME.

I TOTALLY LOVE IT.

THIS GAME BEATS UR PREVIOUS GRAVITY GAME.



Right after u posted the Q, i did level 1 and 2.

And today i did level 3 and 4

For some reason I’m stuck @ level 5.



Vikram did you finished all levels?

I mean, are all levels possible?

I've been trying level 5 since today morning and it's 6.15pm now. Yet i couldn't figure-out. Just say whether all levels are possible or not. I mean there are no glitches in the game, right?

Gosh!! This is driving me mad.



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For those who care

Level 1 => http://i579.photobucket.com/albums/ss239/Nino_ops/Level1.jpg

Level 2 => http://i579.photobucket.com/albums/ss239/Nino_ops/Level2.jpg

Level 3 => http://i579.photobucket.com/albums/ss239/Nino_ops/Level3.jpg

Level 4 => http://i579.photobucket.com/albums/ss239/Nino_ops/Level4.jpg



+ + +

EDIT =>

Of-course NOT, don’t give us the solutions.



My method was, to think about all the possible 'bend-ways' around the black dots and come-up with a path. first 2 was easy, next was bit hard, 5th is REALLY hard.

And I don’t wanna skip levels either.

I’ll come back as soon as i got any method for the 5th one.



BTW, do you think that any method for any level is unique?



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After like another hour of work I got the solution for level 5

I had to go from block by block to get the solution.



Level 5 => http://i579.photobucket.com/albums/ss239/Nino_ops/Level5.jpg



+ + + + + +

OK I woke-up early and i started to do the rest form level 7.

Yesterday i did level 6.

And since there are already images for level 6 7 8 and 9,

There’s no need to post mine. LOL.

Answer 6

Vikram, after trying level 7 for 20 mins, I just wrote a small program to solve this which was done in less than an hour. The only heuristic that I found easy to use is that you cannot form more than one contiguous region of open blocks while traversing the path. Any way, here is the program for level 7 -- you can modify the variable blk and set mx, my and it will solve any size board for you. I wont post the solutions as they are already posted below. It prints the ball locations in the path in reverse order. And it takes the computer less than a second to solve ... time to get ready for the robot overlords :-)

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import java.util.ArrayList;

public class Board {

ArrayList b;

enum direction {up, down, left, right};

int x, y;

int cx, cy;

int open;

int min_open=9999;

static final int mx = 7, my = 7;

static final int [][] blk = {{5,1}, {3,2}, {2,2}, {2,4}, {4,5}};



public static void main(String[] args) {

int i,j;

Board b = new Board(mx,my);

for(i=0;i
b.set(blk[i][0],blk[i][1],true);



boolean found = false;

for(i=0;i
for(j=0;j
if(b.get(i,j)==false) {

b.setInit(i, j);

if(findPath(b,i,j)) {

found = true;

break;

} else {

b.set(i,j,false);

}

}

if(!found)

System.out.println("No path found");

}



static boolean findPath(Board b, int i, int j) {

Board.direction [] alldir = Board.direction.values();

b.set(i,j,true);

int bx = b.cx, by = b.cy;

int idx;

for(idx=0;idx
if(b.openDir(alldir[idx], false)) {

if(b.goDir(alldir[idx]))

return true;

if(findPath(b,b.cx,b.cy)) {

System.out.println(i+ " "+j);

return true;

}

b.backDir(bx, by);

}

}



b.backDir(i,j);

return false;

}





Board(int xx, int yy) {

x = xx; y = yy;

cx = 0; cy = 0;

initBoard();

}



void initBoard() {

if(b ==null)

b = new ArrayList(y);

else

b.clear();



open = x*y;



for(int i=0;i
b.add(i, new boolean[x]);

for(int j=0;j
b.get(i)[j] = false;

}

}



void set(int xx, int yy, boolean v) {

if(xx
if(b.get(yy)[xx] != v) {

b.get(yy)[xx] = v;

open = v?open-1:open+1;

min_open = (open
}

}

}



boolean get(int xx, int yy) {

return b.get(yy)[xx];

}



boolean setInit(int xx, int yy) {

if(get(xx,yy) == false) {

cx = xx; cy = yy;

set(xx,yy,true);

return true;

}

return false;

}



boolean openDir(direction d, boolean put) {

int nx = (d == direction.left)? cx-1:(d == direction.right)?cx+1:cx;

int ny = (d == direction.up)? cy-1: (d== direction.down)?cy+1:cy;



if(nx < x && ny < y && nx > -1 && ny > -1) {

if(get(nx,ny) == false) {

if(put) {

cx = nx; cy = ny;

set(cx,cy,true);

}

return true;

}

}



return false;

}



boolean goDir(direction d) {

while(openDir(d,true));

return (open==0);

}



void backDir(int bx,int by) {

if(bx != cx) {

int len = java.lang.Math.abs(bx-cx)+1;

int stx = (bx>cx)?cx:bx;

for(int i=stx;i<(stx+len);i++) {

set(i,cy,false);

}

cx = bx;

} else if(by != cy){

int len = java.lang.Math.abs(by-cy)+1;

int sty = (by>cy)?cy:by;

for(int i=sty;i<(sty+len);i++) {

set(cx,i,false);

}

cy = by;

}

set(cx,cy,true);

}

}

Answer 7

Place the first ball on a corner