Problem of rooms
A rectangle is divided into some smaller rectangles.Each two adjacent rectangles share a door which connects them.Prove that we can start from one of the small rectangles and pass them all without crossing a rectangle more than once.
discrete-mathematics graph-theory recreational-mathematics problem-solving
asked Jan 1 at 21:04
Ali FaryadrasAli Faryadras
211
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Ali Faryadras is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
A rectangle is divided into some smaller rectangles.Each two adjacent rectangles share a door which connects them.Prove that we can start from one of the small rectangles and pass them all without crossing a rectangle more than once.
discrete-mathematics graph-theory recreational-mathematics problem-solving
asked Jan 1 at 21:04
Ali FaryadrasAli Faryadras
211
New contributor
Ali Faryadras is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
What have you tried so far?
– user3482749
Jan 1 at 21:05
@user3482749 I tried to solve this problem by Induction.
– Ali Faryadras
Jan 1 at 21:09
3
And? What progress did you make?
– user3482749
Jan 1 at 21:29
add a comment |
A rectangle is divided into some smaller rectangles.Each two adjacent rectangles share a door which connects them.Prove that we can start from one of the small rectangles and pass them all without crossing a rectangle more than once.
discrete-mathematics graph-theory recreational-mathematics problem-solving
asked Jan 1 at 21:04
Ali FaryadrasAli Faryadras
211
New contributor
Ali Faryadras is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
A rectangle is divided into some smaller rectangles.Each two adjacent rectangles share a door which connects them.Prove that we can start from one of the small rectangles and pass them all without crossing a rectangle more than once.
discrete-mathematics graph-theory recreational-mathematics problem-solving
discrete-mathematics graph-theory recreational-mathematics problem-solving
asked Jan 1 at 21:04
Ali FaryadrasAli Faryadras
211
New contributor
Ali Faryadras is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 1 at 21:04
Ali FaryadrasAli Faryadras
211
New contributor
Ali Faryadras is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 1 at 21:04
Ali FaryadrasAli Faryadras
211
New contributor
Ali Faryadras is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 1 at 21:04
Ali FaryadrasAli Faryadras
211
asked Jan 1 at 21:04
Ali FaryadrasAli Faryadras
211
211
New contributor
Ali Faryadras is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ali Faryadras is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ali Faryadras is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
What have you tried so far?
– user3482749
Jan 1 at 21:05
@user3482749 I tried to solve this problem by Induction.
– Ali Faryadras
Jan 1 at 21:09
3
And? What progress did you make?
– user3482749
Jan 1 at 21:29
add a comment |
1
What have you tried so far?
– user3482749
Jan 1 at 21:05
@user3482749 I tried to solve this problem by Induction.
– Ali Faryadras
Jan 1 at 21:09
3
And? What progress did you make?
– user3482749
Jan 1 at 21:29
1
1
What have you tried so far?
– user3482749
Jan 1 at 21:05
What have you tried so far?
– user3482749
Jan 1 at 21:05
@user3482749 I tried to solve this problem by Induction.
– Ali Faryadras
Jan 1 at 21:09
@user3482749 I tried to solve this problem by Induction.
– Ali Faryadras
Jan 1 at 21:09
3
3
And? What progress did you make?
– user3482749
Jan 1 at 21:29
And? What progress did you make?
– user3482749
Jan 1 at 21:29
add a comment |
2 Answers
2
active
oldest
votes
Here's a tip in the right direction. The main difficulty of an inductive argument would be the following hypothetical case:
Here, it is not obvious how to adapt the pre-existing path to to include this new rectangle. So, why not make the inductive hypothesis be that there exists a path which never exits a rectangle, $r$, from the same "side" that $r$ was entered from, avoiding this problem altogether. With that inductive hypothesis, here’s the case work:
Hopefully everything is clear now?
answered Jan 2 at 6:06
Zachary HunterZachary Hunter
54110
New contributor
Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Do you have another idea without using induction?
– Ali Faryadras
Jan 2 at 9:28
induction is by far most natural to me, are you still having difficulty?
– Zachary Hunter
Jan 2 at 14:22
Have you tried anything since?
– Zachary Hunter
Jan 3 at 3:20
please try to provide further guidance.
– Ali Faryadras
2 days ago
I've drawn the cases, did you understand what my inductive proposition is?
– Zachary Hunter
2 days ago
|
show 1 more comment
Convert the rectangular structure to a graph, where each door becomes a vertex and each room becomes an edge. Since each door connects exactly two rooms, each vertex is therefore of degree two. Since no vertex is of odd degree, a Eulerian cycle is possible, i.e. each room/edge/bridge can be visited without repetition.
answered yesterday
GarryBGarryB
824
Since rooms will generally have more than two doors, this requires thinking about "edges" that connect more than two "vertices". That is definitely not a standard concept of graphs, so the standard theorems (like the one you're appealing to) will not work.
– Henning Makholm
yesterday
And "a Eulerian cycle is possible" is too strong a conclusion anyway. There are easy room layouts that will only allow for an Eulerian trail (which may end in a different room from the one it started at).
– Henning Makholm
yesterday
Yes, I was thinking of a simple room structure which produced a simple graph.
– GarryB
10 hours ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Here's a tip in the right direction. The main difficulty of an inductive argument would be the following hypothetical case:
Here, it is not obvious how to adapt the pre-existing path to to include this new rectangle. So, why not make the inductive hypothesis be that there exists a path which never exits a rectangle, $r$, from the same "side" that $r$ was entered from, avoiding this problem altogether. With that inductive hypothesis, here’s the case work:
Hopefully everything is clear now?
answered Jan 2 at 6:06
Zachary HunterZachary Hunter
54110
New contributor
Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Do you have another idea without using induction?
– Ali Faryadras
Jan 2 at 9:28
induction is by far most natural to me, are you still having difficulty?
– Zachary Hunter
Jan 2 at 14:22
Have you tried anything since?
– Zachary Hunter
Jan 3 at 3:20
please try to provide further guidance.
– Ali Faryadras
2 days ago
I've drawn the cases, did you understand what my inductive proposition is?
– Zachary Hunter
2 days ago
|
show 1 more comment
Here's a tip in the right direction. The main difficulty of an inductive argument would be the following hypothetical case:
Here, it is not obvious how to adapt the pre-existing path to to include this new rectangle. So, why not make the inductive hypothesis be that there exists a path which never exits a rectangle, $r$, from the same "side" that $r$ was entered from, avoiding this problem altogether. With that inductive hypothesis, here’s the case work:
Hopefully everything is clear now?
answered Jan 2 at 6:06
Zachary HunterZachary Hunter
54110
New contributor
Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Do you have another idea without using induction?
– Ali Faryadras
Jan 2 at 9:28
induction is by far most natural to me, are you still having difficulty?
– Zachary Hunter
Jan 2 at 14:22
Have you tried anything since?
– Zachary Hunter
Jan 3 at 3:20
please try to provide further guidance.
– Ali Faryadras
2 days ago
I've drawn the cases, did you understand what my inductive proposition is?
– Zachary Hunter
2 days ago
|
show 1 more comment
Here's a tip in the right direction. The main difficulty of an inductive argument would be the following hypothetical case:
Here, it is not obvious how to adapt the pre-existing path to to include this new rectangle. So, why not make the inductive hypothesis be that there exists a path which never exits a rectangle, $r$, from the same "side" that $r$ was entered from, avoiding this problem altogether. With that inductive hypothesis, here’s the case work:
Hopefully everything is clear now?
answered Jan 2 at 6:06
Zachary HunterZachary Hunter
54110
New contributor
Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Here's a tip in the right direction. The main difficulty of an inductive argument would be the following hypothetical case:
Here, it is not obvious how to adapt the pre-existing path to to include this new rectangle. So, why not make the inductive hypothesis be that there exists a path which never exits a rectangle, $r$, from the same "side" that $r$ was entered from, avoiding this problem altogether. With that inductive hypothesis, here’s the case work:
Hopefully everything is clear now?
answered Jan 2 at 6:06
Zachary HunterZachary Hunter
54110
New contributor
Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
answered Jan 2 at 6:06
Zachary HunterZachary Hunter
54110
New contributor
Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 2 at 6:06
Zachary HunterZachary Hunter
54110
answered Jan 2 at 6:06
Zachary HunterZachary Hunter
54110
54110
New contributor
Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Do you have another idea without using induction?
– Ali Faryadras
Jan 2 at 9:28
induction is by far most natural to me, are you still having difficulty?
– Zachary Hunter
Jan 2 at 14:22
Have you tried anything since?
– Zachary Hunter
Jan 3 at 3:20
please try to provide further guidance.
– Ali Faryadras
2 days ago
I've drawn the cases, did you understand what my inductive proposition is?
– Zachary Hunter
2 days ago
|
show 1 more comment
Do you have another idea without using induction?
– Ali Faryadras
Jan 2 at 9:28
induction is by far most natural to me, are you still having difficulty?
– Zachary Hunter
Jan 2 at 14:22
Have you tried anything since?
– Zachary Hunter
Jan 3 at 3:20
please try to provide further guidance.
– Ali Faryadras
2 days ago
I've drawn the cases, did you understand what my inductive proposition is?
– Zachary Hunter
2 days ago
Do you have another idea without using induction?
– Ali Faryadras
Jan 2 at 9:28
Do you have another idea without using induction?
– Ali Faryadras
Jan 2 at 9:28
induction is by far most natural to me, are you still having difficulty?
– Zachary Hunter
Jan 2 at 14:22
induction is by far most natural to me, are you still having difficulty?
– Zachary Hunter
Jan 2 at 14:22
Have you tried anything since?
– Zachary Hunter
Jan 3 at 3:20
Have you tried anything since?
– Zachary Hunter
Jan 3 at 3:20
please try to provide further guidance.
– Ali Faryadras
2 days ago
please try to provide further guidance.
– Ali Faryadras
2 days ago
I've drawn the cases, did you understand what my inductive proposition is?
– Zachary Hunter
2 days ago
I've drawn the cases, did you understand what my inductive proposition is?
– Zachary Hunter
2 days ago
|
show 1 more comment
Convert the rectangular structure to a graph, where each door becomes a vertex and each room becomes an edge. Since each door connects exactly two rooms, each vertex is therefore of degree two. Since no vertex is of odd degree, a Eulerian cycle is possible, i.e. each room/edge/bridge can be visited without repetition.
answered yesterday
GarryBGarryB
824
Since rooms will generally have more than two doors, this requires thinking about "edges" that connect more than two "vertices". That is definitely not a standard concept of graphs, so the standard theorems (like the one you're appealing to) will not work.
– Henning Makholm
yesterday
And "a Eulerian cycle is possible" is too strong a conclusion anyway. There are easy room layouts that will only allow for an Eulerian trail (which may end in a different room from the one it started at).
– Henning Makholm
yesterday
Yes, I was thinking of a simple room structure which produced a simple graph.
– GarryB
10 hours ago
add a comment |
Convert the rectangular structure to a graph, where each door becomes a vertex and each room becomes an edge. Since each door connects exactly two rooms, each vertex is therefore of degree two. Since no vertex is of odd degree, a Eulerian cycle is possible, i.e. each room/edge/bridge can be visited without repetition.
answered yesterday
GarryBGarryB
824
Since rooms will generally have more than two doors, this requires thinking about "edges" that connect more than two "vertices". That is definitely not a standard concept of graphs, so the standard theorems (like the one you're appealing to) will not work.
– Henning Makholm
yesterday
And "a Eulerian cycle is possible" is too strong a conclusion anyway. There are easy room layouts that will only allow for an Eulerian trail (which may end in a different room from the one it started at).
– Henning Makholm
yesterday
Yes, I was thinking of a simple room structure which produced a simple graph.
– GarryB
10 hours ago
add a comment |
Convert the rectangular structure to a graph, where each door becomes a vertex and each room becomes an edge. Since each door connects exactly two rooms, each vertex is therefore of degree two. Since no vertex is of odd degree, a Eulerian cycle is possible, i.e. each room/edge/bridge can be visited without repetition.
answered yesterday
GarryBGarryB
824
Convert the rectangular structure to a graph, where each door becomes a vertex and each room becomes an edge. Since each door connects exactly two rooms, each vertex is therefore of degree two. Since no vertex is of odd degree, a Eulerian cycle is possible, i.e. each room/edge/bridge can be visited without repetition.
answered yesterday
GarryBGarryB
824
answered yesterday
GarryBGarryB
824
answered yesterday
GarryBGarryB
824
answered yesterday
GarryBGarryB
824
824
Since rooms will generally have more than two doors, this requires thinking about "edges" that connect more than two "vertices". That is definitely not a standard concept of graphs, so the standard theorems (like the one you're appealing to) will not work.
– Henning Makholm
yesterday
And "a Eulerian cycle is possible" is too strong a conclusion anyway. There are easy room layouts that will only allow for an Eulerian trail (which may end in a different room from the one it started at).
– Henning Makholm
yesterday
Yes, I was thinking of a simple room structure which produced a simple graph.
– GarryB
10 hours ago
add a comment |
Since rooms will generally have more than two doors, this requires thinking about "edges" that connect more than two "vertices". That is definitely not a standard concept of graphs, so the standard theorems (like the one you're appealing to) will not work.
– Henning Makholm
yesterday
And "a Eulerian cycle is possible" is too strong a conclusion anyway. There are easy room layouts that will only allow for an Eulerian trail (which may end in a different room from the one it started at).
– Henning Makholm
yesterday
Yes, I was thinking of a simple room structure which produced a simple graph.
– GarryB
10 hours ago
Since rooms will generally have more than two doors, this requires thinking about "edges" that connect more than two "vertices". That is definitely not a standard concept of graphs, so the standard theorems (like the one you're appealing to) will not work.
– Henning Makholm
yesterday
Since rooms will generally have more than two doors, this requires thinking about "edges" that connect more than two "vertices". That is definitely not a standard concept of graphs, so the standard theorems (like the one you're appealing to) will not work.
– Henning Makholm
yesterday
And "a Eulerian cycle is possible" is too strong a conclusion anyway. There are easy room layouts that will only allow for an Eulerian trail (which may end in a different room from the one it started at).
– Henning Makholm
yesterday
And "a Eulerian cycle is possible" is too strong a conclusion anyway. There are easy room layouts that will only allow for an Eulerian trail (which may end in a different room from the one it started at).
– Henning Makholm
yesterday
Yes, I was thinking of a simple room structure which produced a simple graph.
– GarryB
10 hours ago
Yes, I was thinking of a simple room structure which produced a simple graph.
– GarryB
10 hours ago
add a comment |
Ali Faryadras is a new contributor. Be nice, and check out our Code of Conduct.
Ali Faryadras is a new contributor. Be nice, and check out our Code of Conduct.
Ali Faryadras is a new contributor. Be nice, and check out our Code of Conduct.
Ali Faryadras is a new contributor. Be nice, and check out our Code of Conduct.
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1
What have you tried so far?
– user3482749
Jan 1 at 21:05
@user3482749 I tried to solve this problem by Induction.
– Ali Faryadras
Jan 1 at 21:09
3
And? What progress did you make?
– user3482749
Jan 1 at 21:29