[Guest]

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Hello, This thing you see is 25 boxes .... 5x5 ordered.

Aim: Is to mark every square once

Restictions : You can only move horizontally or vertically . so this means you can go up , down , right or left only . you cant move cross style . the "X" marked square cant be used . you dont have to or lets say you can't use it...

Solution: No I havent seen any yet , my hope is to see one answer here ...

Thanks for reading my question .

Edited October 1, 2009 by puzzlemussler

[Guest]

I've been working on this for 1 month . I really can't find solution at least share your opinions please . maybe you think what I miss and I think what you miss...

[Guest]

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Hello, This thing you see is 25 boxes .... 5x5 ordered.

Aim: Is to mark every square once

Restictions : You can only move horizontally or vertically . so this means you can go up , down , right or left only . you cant move cross style . the "X" marked square cant be used . you dont have to or lets say you can't use it...

Solution: No I havent seen any yet , my hope is to see one answer here ...

Thanks for reading my question .

impossible. The solitary space in the bottom left corner creates the issue. Whether you start with that square or end with that square it can't be done. These puzzles are typically solvable every time (as long as you don't trap "yourself") or never.

Edited October 1, 2009 by Shadax

[Guest]

Wow I've talked with few mathematicians from other forums and they explained me . Well they said this " lets look like this , you start moving from "X" and try to get back there with filling all boxes , but you have 24 box options but you can only fill odd number boxes since 24 is even you can only fill 23 boxes and 1 remains empty everytime" so this question is just silly impossible thing I guess which has been bugging me for 1 month

[Guest]

If you start out by panting every other square black, so it looks like a chess board, then there will be 13 black ones and 12 white squares. The X is on a white square so there are only 11 white squares that can be marked. If you mark a back square the next square you mark must be white, or you just marked a white square the next square you mark must be back. This means that at any point in time you can only have marked one more back square then white. You need to mark 13 black and 11 white so it is imposable. QED

[Guest]

This one looks a lot like another puzzle... where there's a killer in the room X and regular people in every other room. The exit for the killer to flee is in the top corner of the building. The killer needed to dispatch of everyone in the house before leaving and was able to enter rooms where no one had been murdered... Anyway, the thing is, it depends on how the problem is stated. In this case it is not a 5x5 grid, but a 5x4+4 (being the last four the bottom row). Also, I believe it unfair to call it "the hardest puzzle ever" or something like that, because it's just an impossible task that's being given. It would be the same to tell: Fin a way in which 2 dollars sum up $1,000,000 USD with Lincoln's face in them... Anyway, stated as it is, I think we better move on to the next riddle.

[Guest]

If you start out by panting every other square black, so it looks like a chess board, then there will be 13 black ones and 12 white squares. The X is on a white square so there are only 11 white squares that can be marked. If you mark a back square the next square you mark must be white, or you just marked a white square the next square you mark must be back. This means that at any point in time you can only have marked one more back square then white. You need to mark 13 black and 11 white so it is imposable. QED

Perfect Explanation. So it is impossible.

[Guest]

How is this impossible? Perhaps I'm doing it wrong? I followed the instructions. My spoiler explains how I moved up, down, left, right and only touched each square one time and didn't use the X square.

Start off in square to right of X. Move all the way right. All the way up. All the way left. Down 2. Right, Up, Right, Down, Right, Up, Right, Down, Down, All the way left, Down.

[Guest]

How is this impossible? Perhaps I'm doing it wrong? I followed the instructions. My spoiler explains how I moved up, down, left, right and only touched each square one time and didn't use the X square.

Start off in square to right of X. Move all the way right. All the way up. All the way left. Down 2. Right, Up, Right, Down, Right, Up, Right, Down, Down, All the way left, Down.

You can't cross over squares you have already marked.

Here's a my explanation which is the same as a couple posts above:

You're 5x5 grid can be looked at like a chess board. Black, white, black, white, etc. This would leave 13 black squares and 12 white squares (13+12=25 squares, if you decide to flip this possibility, the explanation remains the same except with too many white squares vs. too many black, and where the X lies would just be an opposite color, it doesn't matter).

Example:

In this pattern, the X would lie on a white square, which means now only 11 white squares are available. Because we cannot move diagonally, the pattern for marking off squares would have to be Black, white, black, white, black, etc. Or you can start with a white square, it doesn't matter. The fact is that since there are TWO fewer white squares then there are black, that means you would HAVE to at some point cross black TWICE, which is impossible since no two black squares are in sequence (except for diagonally).

Edited October 1, 2009 by Shadax

[Guest]

Actualy it is possible (although the answer may not be what was intended). Everyone is making an assumption that is not in the restrictions.

To solve the problem simply go up from the x to the top square, then go right 1, then down to the bottom square, then right 1, then up to the top square, then right 1 then down to ***just below the bottom square then over to below the first row. then up to the top!

So you see it works if you go outside the 5x5 grid which was not forbiden in the rules.

[Guest]

Actualy it is possible (although the answer may not be what was intended). Everyone is making an assumption that is not in the restrictions.

To solve the problem simply go up from the x to the top square, then go right 1, then down to the bottom square, then right 1, then up to the top square, then right 1 then down to ***just below the bottom square then over to below the first row. then up to the top!

So you see it works if you go outside the 5x5 grid which was not forbiden in the rules.

It is then now longer a 5x5 grid which pretty much outlines the boundaries.

Furthermore, if that's the case then we can jump squares, teleport, go under squares etc...

I guess when a puzzle is impossible per obvious rules we just have to get creative

Edited October 1, 2009 by Shadax

[Guest]

I Strongly agree with Newbie, The whole reason for this puzzle is to open up your mind instead of Keeping it locked inside a Box all the time. In other words you have to think outside the box.

Nowhere does it say to stay in the Grid.

this is the only solution for this puzzle otherwise you will try and try without Success.

Well said Newbie!!!

[Guest]

I Strongly agree with Newbie, The whole reason for this puzzle is to open up your mind instead of Keeping it locked inside a Box all the time. In other words you have to think outside the box.

Nowhere does it say to stay in the Grid.

this is the only solution for this puzzle otherwise you will try and try without Success.

Well said Newbie!!!

Right but the initial intention is to solve a puzzle with specific boundaries to ensure a challenge and logical way to prove it possible or impossible. Otherwise the purpose is defeated, no challenge equals no critical thinking. If we handled every puzzle by finding an ambiguous loophole and went wild with it then we can solve any puzzle very, very easily...

I guess the flowchart has a point that asks "Is it possible?" and the "Nope" cloud points to "Get creative!"

[Guest]

Right but the initial intention is to solve a puzzle with specific boundaries to ensure a challenge and logical way to prove it possible or impossible. Otherwise the purpose is defeated, no challenge equals no critical thinking. If we handled every puzzle by finding an ambiguous loophole and went wild with it then we can solve any puzzle very, very easily...

I guess the flowchart has a point that asks "Is it possible?" and the "Nope" cloud points to "Get creative!"

I see the point you are trying to make and dont think I didn't get Creative but The Final Conclusion of the puzzle is that it's a Dead End!!

It sounds like you think that there is a solution, if or when you figure it out let me know and I will say Sorry I was Wrong, But until then you can Try and figure it out with no Success!

I will not waste any more of my time on this puzzle Because there is no point other than thinking outside the box.

Good Luck Shadax!!

[Guest]

Impossible!

[Guest]

You can't cross over squares you have already marked.

Here's a my explanation which is the same as a couple posts above:

You're 5x5 grid can be looked at like a chess board. Black, white, black, white, etc. This would leave 13 black squares and 12 white squares (13+12=25 squares, if you decide to flip this possibility, the explanation remains the same except with too many white squares vs. too many black, and where the X lies would just be an opposite color, it doesn't matter).

Example:

In this pattern, the X would lie on a white square, which means now only 11 white squares are available. Because we cannot move diagonally, the pattern for marking off squares would have to be Black, white, black, white, black, etc. Or you can start with a white square, it doesn't matter. The fact is that since there are TWO fewer white squares then there are black, that means you would HAVE to at some point cross black TWICE, which is impossible since no two black squares are in sequence (except for diagonally).

How am I going over squares I already marked? I touched each square only one time and went at 90 degree angels. Where does it say I can't do it that way?

[Guest]

I Have found the answer!

So, everyone is simply assuming that once you go into a square it is marked. However the restrictions don't make that so. So...

*Shrug*

-Hex

Just my two cents.

[Guest]

How am I going over squares I already marked? I touched each square only one time and went at 90 degree angels. Where does it say I can't do it that way?

If you label the squares 1-25 (l to r, top to bottom), your solution runs over squares 10, 15 and 20 twice.

[Guest]

Actualy it is possible (although the answer may not be what was intended). Everyone is making an assumption that is not in the restrictions.

To solve the problem simply go up from the x to the top square, then go right 1, then down to the bottom square, then right 1, then up to the top square, then right 1 then down to ***just below the bottom square then over to below the first row. then up to the top!

So you see it works if you go outside the 5x5 grid which was not forbiden in the rules.

Thanks for wasting our time reading that post. You are basically saying that the question the poster asked is a joke or a lateral thinking puzzle or that he wasted a month on a totaly worthless question. The question is either solvable as written or you can give proof that it is not solvable, but dont give a "gotcha" answer like "I used magic, and it is solved, see!" It is obvious to everyone that you can not go outside the grid.

Edited October 2, 2009 by Chewbacca

[Guest]

I figured this out like the postman delivering only on roads - nodes/links problem.

treat each square as a node, and each side (that you use to travel) as a road.

start anywhere and only use roads once to cover all nodes without repeats.

now at school we did some work on this, and if you have n nodes there are certain rules to follow (n being large enough >3 I think)

give each node a number relating to it's number of roads either into or out of (l). These will either be odd or even

(e.g. - bottom left node only has one road, next one up has 3, corner nodes have 2, edge nodes have 3, and middle nodes have 4 (up, down, left, right))

in an n body problem:

- if all nodes have l=even then it's possible starting anywhere

- if one node has l=odd and all others l=even then it's possible as long as you start or finish at the one odd node

- if precisely 2 nodes have an odd value for l then you must start at one of these and finish at the other

and finaly

- if you have more than 3 nodes with l=odd then the problem becomes impossible.

in this case you have 8 middle nodes with l=4, 11 side nodes with l=3, 1 middle node with l=3, 1 side node with l = 2, 3 corner nodes with l=2, one corner node with l=1

there are 13 nodes with l=odd, therefore impossible.

[Guest]

I figured this out like the postman delivering only on roads - nodes/links problem.

treat each square as a node, and each side (that you use to travel) as a road.

start anywhere and only use roads once to cover all nodes without repeats.

now at school we did some work on this, and if you have n nodes there are certain rules to follow (n being large enough >3 I think)

give each node a number relating to it's number of roads either into or out of (l). These will either be odd or even

(e.g. - bottom left node only has one road, next one up has 3, corner nodes have 2, edge nodes have 3, and middle nodes have 4 (up, down, left, right))

in an n body problem:

- if all nodes have l=even then it's possible starting anywhere

- if one node has l=odd and all others l=even then it's possible as long as you start or finish at the one odd node

- if precisely 2 nodes have an odd value for l then you must start at one of these and finish at the other

and finaly

- if you have more than 3 nodes with l=odd then the problem becomes impossible.

in this case you have 8 middle nodes with l=4, 11 side nodes with l=3, 1 middle node with l=3, 1 side node with l = 2, 3 corner nodes with l=2, one corner node with l=1

there are 13 nodes with l=odd, therefore impossible.

I don't think you have to use all of the 'roads.' You just need to enter all 'intersections' (nodes). So your reasoning would need to be tweaked. Consider the OP without the Xed square. There will be >3 nodes with l=odd, but the problem is possible.

[Guest]

While I accept that a node argument might be "tweaked," as ljb has stated, the black-white argument is much simpler and has been used for a variety of similar problems. Note how the problem is soluble whenever the numbers of ranks and files are even, and never when the elided square, wherever positioned on an odd by odd board, is of opposite parity to a corner square.

A similar problem was mentioned by Martin Gardner regarding thw Knight's Tour problem on a standard chessboard, in which two diagonally opposite corner squares were cut out.

I fully agree with an earlier post regarding the unnecessary hyperbole regarding the difficulty of the problem.

[Gmaster479]

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Hello, This thing you see is 25 boxes .... 5x5 ordered.

Aim: Is to mark every square once

Restictions : You can only move horizontally or vertically . so this means you can go up , down , right or left only . you cant move cross style . the "X" marked square cant be used . you dont have to or lets say you can't use it...

Solution: No I havent seen any yet , my hope is to see one answer here ...

Thanks for reading my question .

Okay there has been a lot of discussion on this thread and for good reason. This problem is a larger and slightly modified version of one if the first logic problems I knew of. It is the one with 9 dots that have to be connected with 4 straight lines without moving the pencil from the paper as seen

to "think outside the box" and therefore, go outside the 5x5 box that you are required to mark and then re-enter the box to finish it. This can be done many different ways. I personally started at the 'X', went to the left and went all the way up, then went down the columns as though covering a 5x4 box, then left and re-entered to get the bottom three squares that I missed

There is no other way to do it. Congratulations to those who got this by thinking in this way. I am happy to see a new twist on such a classic problem

[Guest]

Okay there has been a lot of discussion on this thread and for good reason. This problem is a larger and slightly modified version of one if the first logic problems I knew of. It is the one with 9 dots that have to be connected with 4 straight lines without moving the pencil from the paper as seen

to "think outside the box" and therefore, go outside the 5x5 box that you are required to mark and then re-enter the box to finish it. This can be done many different ways. I personally started at the 'X', went to the left and went all the way up, then went down the columns as though covering a 5x4 box, then left and re-entered to get the bottom three squares that I missed

There is no other way to do it. Congratulations to those who got this by thinking in this way. I am happy to see a new twist on such a classic problem

Nonsense. The solution to the nine dot problem always was, in my view, excellent. The so-called "lateral thinking" regarding the current problem is a cheap trivialization of a problem which deserves, and gets, a sensible answer.

Edited October 3, 2009 by jerbil

[Guest]

Wow, I feel dumb. I solved this one (er, found out there was no solution), by looking at the bottom right hand corner of the grid and seeing that the starting point would have to be in either one of the two S's:

(E stands for end, S stands for start)

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|_|_|_|S|_|

|E|X|_|_|S|

because that portion of the grid has to be either

(U stands for up-down, L stands for left-right, C stands for corner)

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|_|_|U|U|U|

|E|X|C|L|C|

or

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|_|_|U|U|U|

|E|X|C|C|U|

and I also filled in

|C|_|_|_|C|

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|_|_|_|S|_|

|E|X|_|_|S|

And then I eliminated all the possibilities, which took maybe five or ten minutes of just playing around.

I wish I had thought up the white/black squares solution, which is waaaaaaaaaaay cooler

[Guest]

By logic, I repeat, BY LOGIC and following the restrictions, this is not really a hard puzzle. It's just impossible. That does not necesarilly make it hard. It's like saying, find a way that 0 times 0 equals 1. Is that hard? Not necessarily. It's just impossible.

If it was possible, it wouldn't be hard.

By getting outside the box you would jump, or skip squares.

Of course, I could be wrong.