The No-Straight Maze
$begingroup$
Consider the following maze.
You can walk on the black lines, and your aim is to go from the green at the maze's bottom to the red on the left side. However, each time you reach an intersection of three or more black paths (spokes), you must turn 90 degrees either direction, rather than continuing straight.
Find a valid path through the maze, or prove that no such solution exists.
- It's not a lateral-thinking puzzle; the solution is not a trick.
- If you arrive at a corner, simply follow the path.
- You cannot suddenly turn around and walk the other way, but you may retrace your steps otherwise.
- It's a puzzle of my own creation, and I already have the solution.
- The missing line in the middle of the maze is intentional.
logical-deduction no-computers mazes
$endgroup$
add a comment |
$begingroup$
Consider the following maze.
You can walk on the black lines, and your aim is to go from the green at the maze's bottom to the red on the left side. However, each time you reach an intersection of three or more black paths (spokes), you must turn 90 degrees either direction, rather than continuing straight.
Find a valid path through the maze, or prove that no such solution exists.
- It's not a lateral-thinking puzzle; the solution is not a trick.
- If you arrive at a corner, simply follow the path.
- You cannot suddenly turn around and walk the other way, but you may retrace your steps otherwise.
- It's a puzzle of my own creation, and I already have the solution.
- The missing line in the middle of the maze is intentional.
logical-deduction no-computers mazes
$endgroup$
add a comment |
$begingroup$
Consider the following maze.
You can walk on the black lines, and your aim is to go from the green at the maze's bottom to the red on the left side. However, each time you reach an intersection of three or more black paths (spokes), you must turn 90 degrees either direction, rather than continuing straight.
Find a valid path through the maze, or prove that no such solution exists.
- It's not a lateral-thinking puzzle; the solution is not a trick.
- If you arrive at a corner, simply follow the path.
- You cannot suddenly turn around and walk the other way, but you may retrace your steps otherwise.
- It's a puzzle of my own creation, and I already have the solution.
- The missing line in the middle of the maze is intentional.
logical-deduction no-computers mazes
$endgroup$
Consider the following maze.
You can walk on the black lines, and your aim is to go from the green at the maze's bottom to the red on the left side. However, each time you reach an intersection of three or more black paths (spokes), you must turn 90 degrees either direction, rather than continuing straight.
Find a valid path through the maze, or prove that no such solution exists.
- It's not a lateral-thinking puzzle; the solution is not a trick.
- If you arrive at a corner, simply follow the path.
- You cannot suddenly turn around and walk the other way, but you may retrace your steps otherwise.
- It's a puzzle of my own creation, and I already have the solution.
- The missing line in the middle of the maze is intentional.
logical-deduction no-computers mazes
logical-deduction no-computers mazes
asked Feb 25 at 6:44
ZanyGZanyG
1,153421
1,153421
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Since we must always turn 90 degrees, every odd-numbered step we take is always vertical and every even-numbered step is horizontal. Since the last step has to be horizontal, it means our total path has to be an even number of steps. In other words, once we've reached the red squares, we have made the same amount of vertical and horizontal steps.
After each odd-numbered vertical step, we are an odd number of rows away from the starting point. After each even-numbered step, we are an even number of rows away. The same applies for horizontal steps.
In the end, we have taken the same number of steps both horizontally and vertically (i.e. either both are even or both are odd). We are 4 columns away from the starting point, so we must have taken an even number of horizontal steps. However, we are 5 rows away from the starting point, so we must have taken an odd number of vertical steps. This is impossible, so we can conclude that there is no such path through the maze.
Note though that this only works because this particular maze has no straight moves available (i.e. intersections where you can't turn but only go forward). For example, using the same rules, if you remove the vertical line directly below the 2x1 block...
...there actually is a solution.
$endgroup$
$begingroup$
Yes, much better. It's the answer that should be the most obvious thing. Thank you.
$endgroup$
– Overmind
Feb 26 at 12:11
add a comment |
$begingroup$
Here is an answer that I think is slightly easier to understand:
I have marked all the intersections with a white or a black dot in the manner of a checkerboard. Whatever path you travel, the intersections you visit will alternate black and white. Since you also have to alternate your direction of travel at every intersection between east/west and north/south, it is easy to see that when you arrive at any black intersection, your next move is east/west, and at a white intersection your next move is north/south. The only way to reach the red block is west from a white intersection, which you will therefore never be able to do.
To hammer the point home, from the restrictions above it follows that each edge can only be traversed in one direction. In the following picture I have marked each edge with the travel direction. As you can see, the edge to the red block has the arrow going in the wrong direction.
$endgroup$
add a comment |
$begingroup$
The path through the maze is:
non-existent.
Some notes about this -
The design of the maze is such that you'll basically be turning after following any line.
The missing segment is largely irrelevant.
Each step north, including the first, requires the next step to be east or west. Likewise for any step south.
You have to travel a net distance north of 5 steps.
After 1 step north + 1 step east/west, you're an ODD number of steps east or west of the starting position.
After two steps north + 1 east/west, you'll be an EVEN number of steps east or west of the start.
...
On a turn that leaves you 5 steps north, no matter how you get there, you'll be an even number of steps from the origin and about to have to turn east or west. The end state would require you to be an odd (3) number of steps west, so you will never be able to exit there.
$endgroup$
$begingroup$
After checking the parity of the cells it appears you are correct.
$endgroup$
– Overmind
Feb 26 at 11:47
add a comment |
$begingroup$
This seems to work, don't know the math of odds evens up down but i ended up on the red.
$endgroup$
$begingroup$
Sorry didn't turn at the first one but if you do it is actually easier
$endgroup$
– Josh Lewis
Feb 25 at 14:56
1
$begingroup$
If you do turn at the first intersection (as you are supposed to), it won't be easier. It will be impossible, as the accepted answer shows. Are you sure you didn't go straight ahead at the last intersection?
$endgroup$
– Jaap Scherphuis
Feb 25 at 14:59
2
$begingroup$
This makes an invalid move at the first intersection by not turning. Please update your answer to indicate a correct solution. If it is actually easier, it sounds like you already have a solution in mind: please post that.
$endgroup$
– doppelgreener
Feb 25 at 15:29
add a comment |
$begingroup$
I think I have it...the key is the missing line which makes a couple of intersections of less than 3 black lines....
$endgroup$
6
$begingroup$
The arrow clearly is pointing to an intersection of 3 spokes. Take a look at the other answers to see what others have done.
$endgroup$
– Rubio♦
Feb 25 at 22:35
add a comment |
Your Answer
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since we must always turn 90 degrees, every odd-numbered step we take is always vertical and every even-numbered step is horizontal. Since the last step has to be horizontal, it means our total path has to be an even number of steps. In other words, once we've reached the red squares, we have made the same amount of vertical and horizontal steps.
After each odd-numbered vertical step, we are an odd number of rows away from the starting point. After each even-numbered step, we are an even number of rows away. The same applies for horizontal steps.
In the end, we have taken the same number of steps both horizontally and vertically (i.e. either both are even or both are odd). We are 4 columns away from the starting point, so we must have taken an even number of horizontal steps. However, we are 5 rows away from the starting point, so we must have taken an odd number of vertical steps. This is impossible, so we can conclude that there is no such path through the maze.
Note though that this only works because this particular maze has no straight moves available (i.e. intersections where you can't turn but only go forward). For example, using the same rules, if you remove the vertical line directly below the 2x1 block...
...there actually is a solution.
$endgroup$
$begingroup$
Yes, much better. It's the answer that should be the most obvious thing. Thank you.
$endgroup$
– Overmind
Feb 26 at 12:11
add a comment |
$begingroup$
Since we must always turn 90 degrees, every odd-numbered step we take is always vertical and every even-numbered step is horizontal. Since the last step has to be horizontal, it means our total path has to be an even number of steps. In other words, once we've reached the red squares, we have made the same amount of vertical and horizontal steps.
After each odd-numbered vertical step, we are an odd number of rows away from the starting point. After each even-numbered step, we are an even number of rows away. The same applies for horizontal steps.
In the end, we have taken the same number of steps both horizontally and vertically (i.e. either both are even or both are odd). We are 4 columns away from the starting point, so we must have taken an even number of horizontal steps. However, we are 5 rows away from the starting point, so we must have taken an odd number of vertical steps. This is impossible, so we can conclude that there is no such path through the maze.
Note though that this only works because this particular maze has no straight moves available (i.e. intersections where you can't turn but only go forward). For example, using the same rules, if you remove the vertical line directly below the 2x1 block...
...there actually is a solution.
$endgroup$
$begingroup$
Yes, much better. It's the answer that should be the most obvious thing. Thank you.
$endgroup$
– Overmind
Feb 26 at 12:11
add a comment |
$begingroup$
Since we must always turn 90 degrees, every odd-numbered step we take is always vertical and every even-numbered step is horizontal. Since the last step has to be horizontal, it means our total path has to be an even number of steps. In other words, once we've reached the red squares, we have made the same amount of vertical and horizontal steps.
After each odd-numbered vertical step, we are an odd number of rows away from the starting point. After each even-numbered step, we are an even number of rows away. The same applies for horizontal steps.
In the end, we have taken the same number of steps both horizontally and vertically (i.e. either both are even or both are odd). We are 4 columns away from the starting point, so we must have taken an even number of horizontal steps. However, we are 5 rows away from the starting point, so we must have taken an odd number of vertical steps. This is impossible, so we can conclude that there is no such path through the maze.
Note though that this only works because this particular maze has no straight moves available (i.e. intersections where you can't turn but only go forward). For example, using the same rules, if you remove the vertical line directly below the 2x1 block...
...there actually is a solution.
$endgroup$
Since we must always turn 90 degrees, every odd-numbered step we take is always vertical and every even-numbered step is horizontal. Since the last step has to be horizontal, it means our total path has to be an even number of steps. In other words, once we've reached the red squares, we have made the same amount of vertical and horizontal steps.
After each odd-numbered vertical step, we are an odd number of rows away from the starting point. After each even-numbered step, we are an even number of rows away. The same applies for horizontal steps.
In the end, we have taken the same number of steps both horizontally and vertically (i.e. either both are even or both are odd). We are 4 columns away from the starting point, so we must have taken an even number of horizontal steps. However, we are 5 rows away from the starting point, so we must have taken an odd number of vertical steps. This is impossible, so we can conclude that there is no such path through the maze.
Note though that this only works because this particular maze has no straight moves available (i.e. intersections where you can't turn but only go forward). For example, using the same rules, if you remove the vertical line directly below the 2x1 block...
...there actually is a solution.
edited Feb 26 at 12:09
answered Feb 25 at 7:06
jafejafe
24.6k472246
24.6k472246
$begingroup$
Yes, much better. It's the answer that should be the most obvious thing. Thank you.
$endgroup$
– Overmind
Feb 26 at 12:11
add a comment |
$begingroup$
Yes, much better. It's the answer that should be the most obvious thing. Thank you.
$endgroup$
– Overmind
Feb 26 at 12:11
$begingroup$
Yes, much better. It's the answer that should be the most obvious thing. Thank you.
$endgroup$
– Overmind
Feb 26 at 12:11
$begingroup$
Yes, much better. It's the answer that should be the most obvious thing. Thank you.
$endgroup$
– Overmind
Feb 26 at 12:11
add a comment |
$begingroup$
Here is an answer that I think is slightly easier to understand:
I have marked all the intersections with a white or a black dot in the manner of a checkerboard. Whatever path you travel, the intersections you visit will alternate black and white. Since you also have to alternate your direction of travel at every intersection between east/west and north/south, it is easy to see that when you arrive at any black intersection, your next move is east/west, and at a white intersection your next move is north/south. The only way to reach the red block is west from a white intersection, which you will therefore never be able to do.
To hammer the point home, from the restrictions above it follows that each edge can only be traversed in one direction. In the following picture I have marked each edge with the travel direction. As you can see, the edge to the red block has the arrow going in the wrong direction.
$endgroup$
add a comment |
$begingroup$
Here is an answer that I think is slightly easier to understand:
I have marked all the intersections with a white or a black dot in the manner of a checkerboard. Whatever path you travel, the intersections you visit will alternate black and white. Since you also have to alternate your direction of travel at every intersection between east/west and north/south, it is easy to see that when you arrive at any black intersection, your next move is east/west, and at a white intersection your next move is north/south. The only way to reach the red block is west from a white intersection, which you will therefore never be able to do.
To hammer the point home, from the restrictions above it follows that each edge can only be traversed in one direction. In the following picture I have marked each edge with the travel direction. As you can see, the edge to the red block has the arrow going in the wrong direction.
$endgroup$
add a comment |
$begingroup$
Here is an answer that I think is slightly easier to understand:
I have marked all the intersections with a white or a black dot in the manner of a checkerboard. Whatever path you travel, the intersections you visit will alternate black and white. Since you also have to alternate your direction of travel at every intersection between east/west and north/south, it is easy to see that when you arrive at any black intersection, your next move is east/west, and at a white intersection your next move is north/south. The only way to reach the red block is west from a white intersection, which you will therefore never be able to do.
To hammer the point home, from the restrictions above it follows that each edge can only be traversed in one direction. In the following picture I have marked each edge with the travel direction. As you can see, the edge to the red block has the arrow going in the wrong direction.
$endgroup$
Here is an answer that I think is slightly easier to understand:
I have marked all the intersections with a white or a black dot in the manner of a checkerboard. Whatever path you travel, the intersections you visit will alternate black and white. Since you also have to alternate your direction of travel at every intersection between east/west and north/south, it is easy to see that when you arrive at any black intersection, your next move is east/west, and at a white intersection your next move is north/south. The only way to reach the red block is west from a white intersection, which you will therefore never be able to do.
To hammer the point home, from the restrictions above it follows that each edge can only be traversed in one direction. In the following picture I have marked each edge with the travel direction. As you can see, the edge to the red block has the arrow going in the wrong direction.
edited Feb 25 at 16:29
answered Feb 25 at 15:19
Jaap ScherphuisJaap Scherphuis
16.4k12772
16.4k12772
add a comment |
add a comment |
$begingroup$
The path through the maze is:
non-existent.
Some notes about this -
The design of the maze is such that you'll basically be turning after following any line.
The missing segment is largely irrelevant.
Each step north, including the first, requires the next step to be east or west. Likewise for any step south.
You have to travel a net distance north of 5 steps.
After 1 step north + 1 step east/west, you're an ODD number of steps east or west of the starting position.
After two steps north + 1 east/west, you'll be an EVEN number of steps east or west of the start.
...
On a turn that leaves you 5 steps north, no matter how you get there, you'll be an even number of steps from the origin and about to have to turn east or west. The end state would require you to be an odd (3) number of steps west, so you will never be able to exit there.
$endgroup$
$begingroup$
After checking the parity of the cells it appears you are correct.
$endgroup$
– Overmind
Feb 26 at 11:47
add a comment |
$begingroup$
The path through the maze is:
non-existent.
Some notes about this -
The design of the maze is such that you'll basically be turning after following any line.
The missing segment is largely irrelevant.
Each step north, including the first, requires the next step to be east or west. Likewise for any step south.
You have to travel a net distance north of 5 steps.
After 1 step north + 1 step east/west, you're an ODD number of steps east or west of the starting position.
After two steps north + 1 east/west, you'll be an EVEN number of steps east or west of the start.
...
On a turn that leaves you 5 steps north, no matter how you get there, you'll be an even number of steps from the origin and about to have to turn east or west. The end state would require you to be an odd (3) number of steps west, so you will never be able to exit there.
$endgroup$
$begingroup$
After checking the parity of the cells it appears you are correct.
$endgroup$
– Overmind
Feb 26 at 11:47
add a comment |
$begingroup$
The path through the maze is:
non-existent.
Some notes about this -
The design of the maze is such that you'll basically be turning after following any line.
The missing segment is largely irrelevant.
Each step north, including the first, requires the next step to be east or west. Likewise for any step south.
You have to travel a net distance north of 5 steps.
After 1 step north + 1 step east/west, you're an ODD number of steps east or west of the starting position.
After two steps north + 1 east/west, you'll be an EVEN number of steps east or west of the start.
...
On a turn that leaves you 5 steps north, no matter how you get there, you'll be an even number of steps from the origin and about to have to turn east or west. The end state would require you to be an odd (3) number of steps west, so you will never be able to exit there.
$endgroup$
The path through the maze is:
non-existent.
Some notes about this -
The design of the maze is such that you'll basically be turning after following any line.
The missing segment is largely irrelevant.
Each step north, including the first, requires the next step to be east or west. Likewise for any step south.
You have to travel a net distance north of 5 steps.
After 1 step north + 1 step east/west, you're an ODD number of steps east or west of the starting position.
After two steps north + 1 east/west, you'll be an EVEN number of steps east or west of the start.
...
On a turn that leaves you 5 steps north, no matter how you get there, you'll be an even number of steps from the origin and about to have to turn east or west. The end state would require you to be an odd (3) number of steps west, so you will never be able to exit there.
answered Feb 25 at 7:09
Rubio♦Rubio
30.4k567188
30.4k567188
$begingroup$
After checking the parity of the cells it appears you are correct.
$endgroup$
– Overmind
Feb 26 at 11:47
add a comment |
$begingroup$
After checking the parity of the cells it appears you are correct.
$endgroup$
– Overmind
Feb 26 at 11:47
$begingroup$
After checking the parity of the cells it appears you are correct.
$endgroup$
– Overmind
Feb 26 at 11:47
$begingroup$
After checking the parity of the cells it appears you are correct.
$endgroup$
– Overmind
Feb 26 at 11:47
add a comment |
$begingroup$
This seems to work, don't know the math of odds evens up down but i ended up on the red.
$endgroup$
$begingroup$
Sorry didn't turn at the first one but if you do it is actually easier
$endgroup$
– Josh Lewis
Feb 25 at 14:56
1
$begingroup$
If you do turn at the first intersection (as you are supposed to), it won't be easier. It will be impossible, as the accepted answer shows. Are you sure you didn't go straight ahead at the last intersection?
$endgroup$
– Jaap Scherphuis
Feb 25 at 14:59
2
$begingroup$
This makes an invalid move at the first intersection by not turning. Please update your answer to indicate a correct solution. If it is actually easier, it sounds like you already have a solution in mind: please post that.
$endgroup$
– doppelgreener
Feb 25 at 15:29
add a comment |
$begingroup$
This seems to work, don't know the math of odds evens up down but i ended up on the red.
$endgroup$
$begingroup$
Sorry didn't turn at the first one but if you do it is actually easier
$endgroup$
– Josh Lewis
Feb 25 at 14:56
1
$begingroup$
If you do turn at the first intersection (as you are supposed to), it won't be easier. It will be impossible, as the accepted answer shows. Are you sure you didn't go straight ahead at the last intersection?
$endgroup$
– Jaap Scherphuis
Feb 25 at 14:59
2
$begingroup$
This makes an invalid move at the first intersection by not turning. Please update your answer to indicate a correct solution. If it is actually easier, it sounds like you already have a solution in mind: please post that.
$endgroup$
– doppelgreener
Feb 25 at 15:29
add a comment |
$begingroup$
This seems to work, don't know the math of odds evens up down but i ended up on the red.
$endgroup$
This seems to work, don't know the math of odds evens up down but i ended up on the red.
answered Feb 25 at 14:55
Josh LewisJosh Lewis
1
1
$begingroup$
Sorry didn't turn at the first one but if you do it is actually easier
$endgroup$
– Josh Lewis
Feb 25 at 14:56
1
$begingroup$
If you do turn at the first intersection (as you are supposed to), it won't be easier. It will be impossible, as the accepted answer shows. Are you sure you didn't go straight ahead at the last intersection?
$endgroup$
– Jaap Scherphuis
Feb 25 at 14:59
2
$begingroup$
This makes an invalid move at the first intersection by not turning. Please update your answer to indicate a correct solution. If it is actually easier, it sounds like you already have a solution in mind: please post that.
$endgroup$
– doppelgreener
Feb 25 at 15:29
add a comment |
$begingroup$
Sorry didn't turn at the first one but if you do it is actually easier
$endgroup$
– Josh Lewis
Feb 25 at 14:56
1
$begingroup$
If you do turn at the first intersection (as you are supposed to), it won't be easier. It will be impossible, as the accepted answer shows. Are you sure you didn't go straight ahead at the last intersection?
$endgroup$
– Jaap Scherphuis
Feb 25 at 14:59
2
$begingroup$
This makes an invalid move at the first intersection by not turning. Please update your answer to indicate a correct solution. If it is actually easier, it sounds like you already have a solution in mind: please post that.
$endgroup$
– doppelgreener
Feb 25 at 15:29
$begingroup$
Sorry didn't turn at the first one but if you do it is actually easier
$endgroup$
– Josh Lewis
Feb 25 at 14:56
$begingroup$
Sorry didn't turn at the first one but if you do it is actually easier
$endgroup$
– Josh Lewis
Feb 25 at 14:56
1
1
$begingroup$
If you do turn at the first intersection (as you are supposed to), it won't be easier. It will be impossible, as the accepted answer shows. Are you sure you didn't go straight ahead at the last intersection?
$endgroup$
– Jaap Scherphuis
Feb 25 at 14:59
$begingroup$
If you do turn at the first intersection (as you are supposed to), it won't be easier. It will be impossible, as the accepted answer shows. Are you sure you didn't go straight ahead at the last intersection?
$endgroup$
– Jaap Scherphuis
Feb 25 at 14:59
2
2
$begingroup$
This makes an invalid move at the first intersection by not turning. Please update your answer to indicate a correct solution. If it is actually easier, it sounds like you already have a solution in mind: please post that.
$endgroup$
– doppelgreener
Feb 25 at 15:29
$begingroup$
This makes an invalid move at the first intersection by not turning. Please update your answer to indicate a correct solution. If it is actually easier, it sounds like you already have a solution in mind: please post that.
$endgroup$
– doppelgreener
Feb 25 at 15:29
add a comment |
$begingroup$
I think I have it...the key is the missing line which makes a couple of intersections of less than 3 black lines....
$endgroup$
6
$begingroup$
The arrow clearly is pointing to an intersection of 3 spokes. Take a look at the other answers to see what others have done.
$endgroup$
– Rubio♦
Feb 25 at 22:35
add a comment |
$begingroup$
I think I have it...the key is the missing line which makes a couple of intersections of less than 3 black lines....
$endgroup$
6
$begingroup$
The arrow clearly is pointing to an intersection of 3 spokes. Take a look at the other answers to see what others have done.
$endgroup$
– Rubio♦
Feb 25 at 22:35
add a comment |
$begingroup$
I think I have it...the key is the missing line which makes a couple of intersections of less than 3 black lines....
$endgroup$
I think I have it...the key is the missing line which makes a couple of intersections of less than 3 black lines....
answered Feb 25 at 22:01
Darren ReddickDarren Reddick
1
1
6
$begingroup$
The arrow clearly is pointing to an intersection of 3 spokes. Take a look at the other answers to see what others have done.
$endgroup$
– Rubio♦
Feb 25 at 22:35
add a comment |
6
$begingroup$
The arrow clearly is pointing to an intersection of 3 spokes. Take a look at the other answers to see what others have done.
$endgroup$
– Rubio♦
Feb 25 at 22:35
6
6
$begingroup$
The arrow clearly is pointing to an intersection of 3 spokes. Take a look at the other answers to see what others have done.
$endgroup$
– Rubio♦
Feb 25 at 22:35
$begingroup$
The arrow clearly is pointing to an intersection of 3 spokes. Take a look at the other answers to see what others have done.
$endgroup$
– Rubio♦
Feb 25 at 22:35
add a comment |
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