A math puzzle - no numbers

[turtleback]

Here is a little math puzzle some might find amusing.  Try to find a path, without picking up your pencil, that goes through every doorway exactly once, if possible.  If it is possible, solve it.  If it is not possible, prove it.  This is a reformulation of a classic math problem that was first solved by the Swiss mathematician Leonhard Euler (pronounced oiler).

Hint:  think about where you start and stop and the duality of odd and even.

[Golfingdad]

Here is a little math puzzle some might find amusing.  Try to find a path, without picking up your pencil, that goes through every doorway exactly once, if possible.  If it is possible, solve it.  If it is not possible, prove it.  This is a reformulation of a classic math problem that was first solved by the Swiss mathematician Leonhard Euler (pronounced oiler).   [URL=http://thesandtrap.com/content/type/61/id/111279/] [/URL] Hint:  think about where you start and stop and the duality of odd and even.

Maybe I'm under thinking this here, but I'm gonna say it's not possible. There is more than one rectangle with an odd number of doors (3 rectangles have 5 doors). One wouldn't be a problem because you could start inside of it, but you can't start inside of the other two, so once you go through the fifth door in either of those rooms you're trapped inside with no way out.

[iacas]

Maybe I'm under thinking this here, but I'm gonna say it's not possible. There is more than one rectangle with an odd number of doors (3 rectangles have 5 doors). One wouldn't be a problem because you could start inside of it, but you can't start inside of the other two, so once you go through the fifth door in either of those rooms you're trapped inside with no way out.

I think you were close but you forgot that you could both start and finish with an entry/exit that didn't have to be "reversed."

If you enter a room, you have to exit the room, unless it's the last room you enter. You can also exit a room without a paired re-entry if it's the room in which you start.

So that's two rooms for which we're allowed to have an odd number of doors. All the other rooms require entry/exit in pairs (2 or 4, since there are no six-door rooms).

Unfortunately, we have three rooms with 5 doors and only two rooms with 4 doors.

I conclude that it not possible.

[Golfingdad]

I think you were close but you forgot that you could both start and finish with an entry/exit that didn't have to be "reversed." If you enter a room, you have to exit the room, unless it's the last room you enter. You can also exit a room without a paired re-entry if it's the room in which you start. So that's two rooms for which we're allowed to have an odd number of doors. All the other rooms require entry/exit in pairs (2 or 4, since there are no six-door rooms). Unfortunately, we have three rooms with 5 doors and only two rooms with 4 doors. I conclude that it not possible.

Oops ... Right. We can start in an odd doors Room and get out and finish in an odd doored room cuz we don't have to leave ... But, alas, the third room presents the problem. I wonder how much partial credit Id have received in turtles class??

[iacas]

Oops ... Right. We can start in an odd doors

Room and get out and finish in an odd doored room cuz we don't have to leave ... But, alas, the third room presents the problem.

I wonder how much partial credit Id have received in turtles class??

Erik J. Barzeski — ⛳ I knock a ball. It goes in a gopher hole. 🏌🏼‍♂️
Director of Instruction Golf Evolution • Owner, The Sand Trap .com • Author, Lowest Score Wins
Golf Digest "Best Young Teachers in America" 2016-17 & "Best in State" 2017-20 • WNY Section PGA Teacher of the Year 2019  

Check Out: New Topics | TST Blog | Golf Terms | Instructional Content | Analyzr | LSW | Instructional Droplets

[14ledo81]

Just subscribing to make sure I get the answer.

I find math problems interesting (not saying I'm good at them).

[turtleback]

You guys are just too damn smart.  This is a variation on what is known as the Bridges of Konigsburg problem.  It was originally formulated around the town of Konigsburg and its bridges (duh!).  The question was whether one could travel a path that took them over every bridge exactly once.

This led to the development of a branch of topology known as graph theory (which has NOTHING to do with cartesian graphs).

On the last day of school I used to pass out copies of the puzzle and tell them that I would give my car to any student who could do it.  Even though the prize made it obvious (to most, lol) that it was not going to be solved, the optimism of 7th or 8th graders is unbounded that they would be able to do it.

I also offered a (Iesser) prize for anyone who thought it was impossible and could convince me that they knew why - as Drew almost did but as Erik nailed.  So next time I see you , Erik, remind me that I owe you a Math pencil.

I agree with Erik about the partial credit.  If you do everything right on the Hubble telescope but someone forgets to convert English measurements to metric you do not get "partial" success.  You get a telescope that doesn't work.

The only time I ever gave partial credit was for answers that were correct, but not in the required form (e.g., 49/231 when it should have been 7/33).  I also based almost all of the grade on in-class quizes and tests, with very little credit for HW or notes or other stuff.  My philosophy was simple - if you cannot get correct answers to problems in a controlled setting, ie., in front of me in my classroom, then you do not know the material.  And I was not going to puff up grades, and lie to the kids and parents, when they cannot get correct answer by giving away a bunch of credit for things that have nothing to do with showing competency in the subject at hand.  I had colleagues who had grading system under which a kid could fail every test and still get a C or B.  Craziness, IMO.

[Golfingdad]

[SPOILER=OT: don't read if you don't want it spoiled] You guys are just too damn smart.  This is a variation on what is known as the Bridges of Konigsburg problem.  It was originally formulated around the town of Konigsburg and its bridges (duh!).  The question was whether one could travel a path that took them over every bridge exactly once. This led to the development of a branch of topology known as graph theory (which has NOTHING to do with cartesian graphs).  On the last day of school I used to pass out copies of the puzzle and tell them that I would give my car to any student who could do it.  Even though the prize made it obvious (to most, lol) that it was not going to be solved, the optimism of 7th or 8th graders is unbounded that they would be able to do it. I also offered a (Iesser) prize for anyone who thought it was impossible and could convince me that they knew why - as Drew almost did but as Erik nailed.  So next time I see you , Erik, remind me that I owe you a Math pencil.  ;-) [/SPOILER] I agree with Erik about the partial credit.  If you do everything right on the Hubble telescope but someone forgets to convert English measurements to metric you do not get "partial" success.  You get a telescope that doesn't work. The only time I ever gave partial credit was for answers that were correct, but not in the required form (e.g., 49/231 when it should have been 7/33).  I also based almost all of the grade on in-class quizes and tests, with very little credit for HW or notes or other stuff.  My philosophy was simple - if you cannot get correct answers to problems in a controlled setting, ie., in front of me in my classroom, then you do not know the material.  And I was not going to puff up grades, and lie to the kids and parents, when they cannot get correct answer by giving away a bunch of credit for things that have nothing to do with showing competency in the subject at hand.  I had colleagues who had grading system under which a kid could fail every test and still get a C or B.  Craziness, IMO.

Fair enough. Although, I'd argue that my explanation clearly showed I was thinking along the right track, which does show some competency in the subject. As opposed to somebody saying "no it can't be done" without any explanation of their thought process. As far as you know, they just guessed. For the record, though, in real life I never argued for anything. I got whatever grade I was given and moved on.

[iacas]

I also offered a (Iesser) prize for anyone who thought it was impossible and could convince me that they knew why - as Drew almost did but as Erik nailed.  So next time I see you , Erik, remind me that I owe you a Math pencil.

My wife teaches 7th grade math as you may remember, so I have plenty of math pencils if I need 'em! :D But I'll hold you to that anyway! Bwah ha ha ha ha!

[newtogolf]

My wife teaches 7th grade math as you may remember, so I have plenty of math pencils if I need 'em! :D But I'll hold you to that anyway! Bwah ha ha ha ha!

I thought @turtleback was going to give you his car

[turtleback]

I thought @turtleback was going to give you his car

Nah, you have to actually draw the path to get my car.  LOL

[iacas]

Nah, you have to actually draw the path to get my car.  LOL

Can we draw the shape on a 3D surface? Because I'm pretty sure I could do it then…

What kind of car do you drive again? :)

[turtleback]

Can we draw the shape on a 3D surface? Because I'm pretty sure I could do it then…

What kind of car do you drive again? :)

Nope.  3D frequently makes things easier in topology.

A great example is what I grew up calling the 4-color conjecture and which now is the 4-color theorem.  It says that any map drawn on a planar surface can be colored in such a way that no 2 contiguous areas are the same color and not more than 4 colors are used.  This turned out to be a very very hard proof that ended up finally being accomplished with the aid of computers.

OTOH there was a related problem where we look at the same thing but maps on a torus (donut-shaped).  It turned out that this one was MUCH easier and was proved way before the 4-color theorem was.  BTW, on a torus a map can require up to 7 colors, as it turns out.

The whole 3D/2D thing provides some other cool stuff.  I'm sure everyone has seen some of the cool stuff surrounding moebius strips where we create the paradox of a surface that has 2-sides locally but only one side globally.  At the beginning of the year I'd always tell my students that by the end of the year I would show them one-sided paper.  And then on one of those lazy days right before a break I would show them one-sided paper (moebius strips) and we would do some fun things, drawing a line down the middle to prove it was one-sided, cutting the strip down the middle to see what would happen, cutting it a third of the way all the way around to see what happen.

You don't need to know what kind of car I drive because you ain't never gonna win it.