You have a meter stick placed so that the 0cm marker is on the left, and the 100cm marker is on the right.
99 ants are placed on every cm marker from 1 to 99.
The ant at cm 1 is facing to the right. The ant at cm 99 is facing left.
For the 97 ants in between, a coin is flipped to determine if that ant is facing left or right: heads, the ant faces left, tails the ant faces right.
Ants only walk in a straight line at 1cm/s in the direction they are heading. If an ant bumps into another ant, both ants immediately turn around and start walking in the opposite direction.
If an ant walks to either end of the meter stick, it falls off.
1) Prove that, eventually, all ants will fall off?
2) What is the maximum time it will take for all ants to fall off?
1. Horribly rough, but based on the way it works, because the ants are expected to keep moving till they fall off, Ant 1/99 will guarantee fall off after once bounce, 2/8 after 2 bounces etc.
It doesn't really matter which directions the other ants are facing because 1/99 are facing in and guarantees the next ant bounces and does a chain reaction before falling.
2. I want to say you calculate by determining the expected movements of ant 49 which is supposed to fall off last. I assume it bounces 49 or 48 times with the pairings, so add up 48cm + 47 till 1 for distance traveled in cm and then adjust for seconds/minutes/hours?
I'm sure someone else would get the actual accurate values or something.
Geez, these puzzles are like much more complex than elementary school math. Anything over elementary school math and it ceases to be fun. (ignoring higher level math tools that are just expansions of basic BEDMAS).
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The only math puzzle I distinctively remember was from like grade 3 in a "problem solving" math question. It popped up again in grade 6, so when I was the only person to answer it correctly.
I remember it distinctly because I think we were given like 20 minutes to do it and if you finished early, you went for recess. I handed in my paper 3-4 minutes after we started (teacher actually scolded me in front of the class because she thought I gave up and wanted extra recess) and I was the only one to get it correctly. To "fix" her error in scolding me, she gave me the "honor" of presenting my response to the whole class.
You have a machine that requires exactly 8 litres of liquid or something catastrophic will happen (trap, spaceship explodes, whatever). You have two oddly shaped containers that hold 5 and 7 litres exactly. These are the only two containers that are available that can hold the liquid and you have access to unlimited liquid.
How do you get the exact 8 litres using these two containers?
1. Horribly rough, but based on the way it works, because the ants are expected to keep moving till they fall off, Ant 1/99 will guarantee fall off after once bounce, 2/8 after 2 bounces etc.
It doesn't really matter which directions the other ants are facing because 1/99 are facing in and guarantees the next ant bounces and does a chain reaction before falling.
2. I want to say you calculate by determining the expected movements of ant 49 which is supposed to fall off last. I assume it bounces 49 or 48 times with the pairings, so add up 48cm + 47 till 1 for distance traveled in cm and then adjust for seconds/minutes/hours?
I'm sure someone else would get the actual accurate values or something.
Geez, these puzzles are like much more complex than elementary school math. Anything over elementary school math and it ceases to be fun. (ignoring higher level math tools that are just expansions of basic BEDMAS).
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The only math puzzle I distinctively remember was from like grade 3 in a "problem solving" math question. It popped up again in grade 6, so when I was the only person to answer it correctly.
I remember it distinctly because I think we were given like 20 minutes to do it and if you finished early, you went for recess. I handed in my paper 3-4 minutes after we started (teacher actually scolded me in front of the class because she thought I gave up and wanted extra recess) and I was the only one to get it correctly. To "fix" her error in scolding me, she gave me the "honor" of presenting my response to the whole class.
You have a machine that requires exactly 8 litres of liquid or something catastrophic will happen (trap, spaceship explodes, whatever). You have two oddly shaped containers that hold 5 and 7 litres exactly. These are the only two containers that are available that can hold the liquid and you have access to unlimited liquid.
How do you get the exact 8 litres using these two containers?
I get the gist of your answer to #1 and it will work. However, there is a much more elegant answer that will make #2 a lot easier too!
I've heard the liquid problem before, so will let someone else solve it who hasn't seen it. To me, it is a relative of the bridge riddle that Scorch posted above.
You have a machine that requires exactly 8 litres of liquid or something catastrophic will happen (trap, spaceship explodes, whatever). You have two oddly shaped containers that hold 5 and 7 litres exactly. These are the only two containers that are available that can hold the liquid and you have access to unlimited liquid.
How do you get the exact 8 litres using these two containers?
Spoiler!
- Fill the smaller jug and pour the liquid into the larger jug.
- Fill the smaller jug again.
- Pour the contents of the smaller jug into the larger jug until it is full. You now have 3 litres of liquid in the smaller jug.
- Pour those 3 litres into the machine, then fill the small jug again and pour contents into the machine.
Here's a puzzle that I saw the other day and found quite interesting. The video also contains the solution, so pause after the question is given to you if you want to try to figure it out on your own.
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You have a meter stick placed so that the 0cm marker is on the left, and the 100cm marker is on the right.
99 ants are placed on every cm marker from 1 to 99.
The ant at cm 1 is facing to the right. The ant at cm 99 is facing left.
For the 97 ants in between, a coin is flipped to determine if that ant is facing left or right: heads, the ant faces left, tails the ant faces right.
Ants only walk in a straight line at 1cm/s in the direction they are heading. If an ant bumps into another ant, both ants immediately turn around and start walking in the opposite direction.
If an ant walks to either end of the meter stick, it falls off.
1) Prove that, eventually, all ants will fall off?
2) What is the maximum time it will take for all ants to fall off?
Spoiler!
1 is easy an end ant will either colide turn around and walk off or continue to walk off the other side and each ant will eventually become an end ant
My gut feel for 2 is that the answer should be 99 or 100 seconds as at some point in time all of the ants will be facing the same direction so the maximum distance the first any can travel is 99 and a colliding ant can travel is all the way the other way. So .5 seconds to create a collision and 98.5 seconds back off the other side. That leaves the collisions in middle but I don’t think they matter as they will resolve themselves faster than the time it takes for mr 98.5. And any colloidions with 98.5 will occur before it gets to half way. So I think 99 is correct But I can’t prove the internal collisions don’t matter.
Thinking about collisions further each ant hitting another ant will be deflected and each other returning to its original position and will take 1 second so you have in the opposite order with 2 less ants. Then it would take 49 seconds to eliminate ants + 50 seconds to walk off. I think this is correct so the general solution is (N-1) seconds for eliminating / sorting ants and then (N+1)/2 to walk off in whatever direction so N seconds where N is the number of ants
So if we have a 4cm/3ants. 1R 2L 3L you get 1 second and you have 1L 2L 2R and then everyone falls off in 2 seconds. 3 seconds total.
With 5 we have 8 scenarios to look at 1R 2L 3L 4L 5L and 1R 2L 3R 4L 5L and 1R 2L 3R 4R 5L and 1R 2L 3L 4R 5L but half are symmetrical.
With 1 you have after 1 second 1L 2R 2L 3L 4L after 2 seconds 1L, 2L, 3R,3L after 3 seconds you have 1L, 2L, 4R and in two more seconds everyone dies for 5.
The second case after 1 second 1L, 2R, 3L, 4L, 5R after 2 seconds 2L, 3R,3L after 3 seconds you have 1L, 2L 4R and death in 5 tot
The 3rd you have 1L, 2R, 4R,4L, 5R then 3R,3L, 5R and 3 seconds to death.
Pretty sure I’m right
Each collision sets two ants on their final death path and these collisions occur every second. The maximum distance is being in the Center.
Also I think maximum is the wrong word I believe in all conditions it will take N seconds to clear the ruler provided the two outer ants face in. Without that condition there would be an N/2 seconds case.
And there’s an easier way, collisions don’t occur the ants just walking past is indistinguishable from a deflection. You can see 1R and 5L walk through every time.
Last edited by GGG; 07-06-2022 at 09:12 PM.
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I think we have a winner. GGG, I appreciate you walking through your thought process.
Here's the concise solution:
Spoiler!
As GGG discovered, if we assume ants are all identical (indistinguishable from each other) then the puzzle's set up is exactly the same as if the ants just walk past each other instead of reversing direction when they meet.
1) If all ants just walk past each other, they will all walk off the edge.
2) The maximum time (and, indeed, the only time) it takes for all ants to fall off is 99 seconds, the time for ants 1 and 99 (furthest from the edge) to walk the remaining length of the meter stick.
My comments on your solution in bold:
Quote:
Originally Posted by GGG
Spoiler!
1 is easy an end ant will either colide turn around and walk off or continue to walk off the other side and each ant will eventually become an end ant
Yes, 1 is deliberately easy, but how you solve 1 affects how easy 2 is
My gut feel for 2 is that the answer should be 99 or 100 seconds as at some point in time all of the ants will be facing the same direction so the maximum distance the first any can travel is 99 and a colliding ant can travel is all the way the other way. So .5 seconds to create a collision and 98.5 seconds back off the other side. That leaves the collisions in middle but I don’t think they matter as they will resolve themselves faster than the time it takes for mr 98.5. And any colloidions with 98.5 will occur before it gets to half way. So I think 99 is correct But I can’t prove the internal collisions don’t matter.
A good start, but I think the statement "at some point in time all of the ants will be facing the same direction" isn't necessarily true.
Thinking about collisions further each ant hitting another ant will be deflected and each other returning to its original position and will take 1 second so you have in the opposite order with 2 less ants.
I don't think it's guaranteed that you have 2 less ants after a collision causes two ants to return to their original position.
So if we have a 4cm/3ants. 1R 2L 3L you get 1 second and you have 1L 2L 2R and then everyone falls off in 2 seconds. 3 seconds total.
With 5 we have 8 scenarios to look at 1R 2L 3L 4L 5L and 1R 2L 3R 4L 5L and 1R 2L 3R 4R 5L and 1R 2L 3L 4R 5L but half are symmetrical.
With 1 you have after 1 second 1L 2R 2L 3L 4L after 2 seconds 1L, 2L, 3R,3L after 3 seconds you have 1L, 2L, 4R and in two more seconds everyone dies for 5.
The second case after 1 second 1L, 2R, 3L, 4L, 5R after 2 seconds 2L, 3R,3L after 3 seconds you have 1L, 2L 4R and death in 5 tot
The 3rd you have 1L, 2R, 4R,4L, 5R then 3R,3L, 5R and 3 seconds to death.
These are great steps! Reduce the problem to a more manageable size so that you can see what is going on.
Also I think maximum is the wrong word I believe in all conditions it will take N seconds to clear the ruler provided the two outer ants face in. Without that condition there would be an N/2 seconds case.
I used "maximum" in the problem wording deliberately so that I don't reveal that the time taken is the same regardless. That would be too much of a hint.
And there’s an easier way, collisions don’t occur the ants just walking past is indistinguishable from a deflection. You can see 1R and 5L walk through every time.
And this is the elegant solution. Congrats on getting to it!
I think we have a winner. GGG, I appreciate you walking through your thought process.
Here's the concise solution:
Spoiler!
As GGG discovered, if we assume ants are all identical (indistinguishable from each other) then the puzzle's set up is exactly the same as if the ants just walk past each other instead of reversing direction when they meet.
1) If all ants just walk past each other, they will all walk off the edge.
2) The maximum time (and, indeed, the only time) it takes for all ants to fall off is 99 seconds, the time for ants 1 and 99 (furthest from the edge) to walk the remaining length of the meter stick.
My comments on your solution in bold:
Spoiler!
I quite enjoyed the puzzle and was annoyed at the simplicity and elegance in the end. I was satisfied with my poorly explained intuitive solution about sorting (it works in the symmetrical case and the concept of collisions sort ants work. Essentially it’s similar to ants passing each other means less ant passes until the system is done just much tougher to explain) and was just checking my work for the 5 case when it clicked.
Prove that there exists two irrational numbers, a and b, such that a^b (a to the power of b) is rational.
As a reminder, rational numbers are numbers that can be expressed as a ratio of two integers (like 1/3, or 12). Irrational numbers are numbers that aren't rational.
Prove that there exists two irrational numbers, a and b, such that a^b (a to the power of b) is rational.
As a reminder, rational numbers are numbers that can be expressed as a ratio of two integers (like 1/3, or 12). Irrational numbers are numbers that aren't rational.
Spoiler!
I was working on this as a laws of exponents problem and was messing around with 'e'. I suspect there is a general proof here but one example proves that it does exist, and e to the power of ln(2) equals 2, and both e and ln(2) are irrational.
I vaguely remember proving than ln(2) is irrational, but that was probably 15 years ago, so I'm going to go with, "it can be shown that ln(2) and e are irrational" and 2 is obviously rational. Neither have terminating or repeating decimals.
Last edited by bizaro86; 07-08-2022 at 10:42 PM.
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I was working on this as a laws of exponents problem and was messing around with 'e'. I suspect there is a general proof here but one example proves that it does exist, and e to the power of ln(2) equals 2, and both e and ln(2) are irrational.
I vaguely remember proving than ln(2) is irrational, but that was probably 15 years ago, so I'm going to go with, "it can be shown that ln(2) and e are irrational" and 2 is obviously rational. Neither have terminating or repeating decimals.
That wasn't what I was looking for but I think it's a great solution. Simple and immediately obvious. As you say, you do have to prove that the one term is irrational - and if we are being pedantic, you should have to prove that both terms are irrational. One isn't too hard, but the other is a bit more difficult.
Funny that I first heard this problem in my first year algebra course over 30 years ago, and I've seen it pop up every now and then, but never saw your solution.
Let me amend the problem a little. Prove that there exists two irrational, non-transcendental numbers (so no pi or e, for example) a and b such that a^b is rational.
That wasn't what I was looking for but I think it's a great solution. Simple and immediately obvious. As you say, you do have to prove that the one term is irrational - and if we are being pedantic, you should have to prove that both terms are irrational. One isn't too hard, but the other is a bit more difficult.
Funny that I first heard this problem in my first year algebra course over 30 years ago, and I've seen it pop up every now and then, but never saw your solution.
Let me amend the problem a little. Prove that there exists two irrational, non-transcendental numbers (so no pi or e, for example) a and b such that a^b is rational.
Spoiler!
Thanks! I agree that to consider the problem complete a proof that those are both irrational is necessary, but I didn't feel like it, and probably wouldn't be able to do that any more
A very slightly modified version of my solution works for your amended problem as well.
A=sqrt(2)
B= log-sqrt(2)(3) ie a logarithm of base sqrt(2) of 3
Which by the same rules of exponents/logarithms equals 3.
It is fairly trivial to prove that sqrt(2) is irrational - I haven't done that for 15 years either but recall the rough outline. If you assume it is rational in the form a/b and that a/b is fully reduced then a and b can't both be even, and by squaring both sides and moving the denominator you get 2b^2=a^2. That means a is even so b can't be even.
But if a^2 is even then a is even, so its a multiple of 2, and if you sub 2x for a (as its an even number) into the previous you get 2b^2=4x^2 which means that b must be even and is a contradiction.
I'm quite confident that logarithm is irrational as well but leave that as an exercise to the reader
Thanks! I agree that to consider the problem complete a proof that those are both irrational is necessary, but I didn't feel like it, and probably wouldn't be able to do that any more
A very slightly modified version of my solution works for your amended problem as well.
A=sqrt(2)
B= log-sqrt(2)(3) ie a logarithm of base sqrt(2) of 3
Which by the same rules of exponents/logarithms equals 3.
It is fairly trivial to prove that sqrt(2) is irrational - I haven't done that for 15 years either but recall the rough outline. If you assume it is rational in the form a/b and that a/b is fully reduced then a and b can't both be even, and by squaring both sides and moving the denominator you get 2b^2=a^2. That means a is even so b can't be even.
But if a^2 is even then a is even, so its a multiple of 2, and if you sub 2x for a (as its an even number) into the previous you get 2b^2=4x^2 which means that b must be even and is a contradiction.
I'm quite confident that logarithm is irrational as well but leave that as an exercise to the reader
Haha, the old "exercise to the reader" argument. I remember in undergrad someone put together a list of different proof types. Along with common ones like "proof by induction" or "proof by contradiction" were "back of the textbook proof" and "proof by intimidation".
I think your method works, but you do have to prove that that one item is irrational. It certainly looks irrational, but how do you know for sure?
Maybe, to help further narrow the scope, I'll amend the problem one last time to say there exists 2 irrational algebraic numbers a and b such that a^b is rational.
Algebraic numbers are numbers that are solutions to P(x)=0 where P(x) is any polynomial in x with rational coefficients.
Haha, the old "exercise to the reader" argument. I remember in undergrad someone put together a list of different proof types. Along with common ones like "proof by induction" or "proof by contradiction" were "back of the textbook proof" and "proof by intimidation".
I think your method works, but you do have to prove that that one item is irrational. It certainly looks irrational, but how do you know for sure?
Maybe, to help further narrow the scope, I'll amend the problem one last time to say there exists 2 irrational algebraic numbers a and b such that a^b is rational.
Algebraic numbers are numbers that are solutions to P(x)=0 where P(x) is any polynomial in x with rational coefficients.
Spoiler!
I figured that would be the objection here (and its valid, obviously). Since I'm terrible at proving things but my proof of sqrt(2) being irrational didn't illicit a correction I'm going to take it as accepted and start from there.
Sqrt(2) is irrational. After screwing around with exponents of sqrt(2) I have a solution which conveniently uses that proof as the only one I need for both terms, which eliminates any future proving things from the problem.
Sqrt(2)^2=2 (rational)
Therefore
Sqrt(2)^sqrt(2)^sqrt(2)=2 (rational)
Sqrt(2)^sqrt(2) is either rational or irrational.
If rational:
Then a=b=sqrt(2)
If irrational
Then a= sqrt(2)^sqrt(2)
B=sqrt(2)
One of the two must be true which I believe proves the original contention that at least one value of a/b exist.
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I don’t like this problem isn’t there a wide variety of sets of numbers that prove the general case of there is at least one irrational number that satisfies the solution and then the debate is just proving the initials conditions are irrational?
Going from memory the construct is something like root 3^ log 4 = 2 and then you prove the construct isn’t true. Anyway I dislike mathematical proofs….
I figured that would be the objection here (and its valid, obviously). Since I'm terrible at proving things but my proof of sqrt(2) being irrational didn't illicit a correction I'm going to take it as accepted and start from there.
Sqrt(2) is irrational. After screwing around with exponents of sqrt(2) I have a solution which conveniently uses that proof as the only one I need for both terms, which eliminates any future proving things from the problem.
Sqrt(2)^2=2 (rational)
Therefore
Sqrt(2)^sqrt(2)^sqrt(2)=2 (rational)
Sqrt(2)^sqrt(2) is either rational or irrational.
If rational:
Then a=b=sqrt(2)
If irrational
Then a= sqrt(2)^sqrt(2)
B=sqrt(2)
One of the two must be true which I believe proves the original contention that at least one value of a/b exist.
And we have another winner!
Yes, you've come up with the solution I was looking for.
Spoiler!
I really like this solution because you don't actually have to determine whether sqrt(2)^sqrt(2) is rational or not. You just use its existence to come up with a rational number, either directly or indirectly.
If I were being pedantic, I'd say that a and b have to be different numbers, but it's easy enough to use sqrt(3)^sqrt(2). Proving that sqrt(3) is irrational is only slightly more difficult than proving sqrt(2) is irrational, using the basically the same format as the sqrt(2) proof, but with an added couple steps at the end.
Alice, Bob, and their dog Cujo all start at the same point. Alice and Bob begin walking down the same path in the same direction: Alice walks at 4km/h, Bob at 3km/h. Cujo, meanwhile, runs back and forth on the path between Alice and Bob at 10km/h (ie. he runs until he meets Alice, then immediately turns around and runs until he meets Bob then turns around again to run to Alice etc, always staying on the path).
Assume they all travel at constant speed the entire time.
After 1 hour, Alice is 4km down the path from the start point. Bob is 3km on the path from the start point. Where on the path is Cujo?
I don’t like this problem isn’t there a wide variety of sets of numbers that prove the general case of there is at least one irrational number that satisfies the solution and then the debate is just proving the initials conditions are irrational?
Going from memory the construct is something like root 3^ log 4 = 2 and then you prove the construct isn’t true. Anyway I dislike mathematical proofs….
I probably should have specified algebraic numbers in the original wording. When we got this problem as an assignment in uni, it didn't specify algebraic numbers, but it did say "consider sqrt(2)" as a hint, which led us down the right path for the solution.
If you consider logs, then the issue is, as you say, proving that the log of some value is irrational, which isn't trivial.
I do think math proofs have a certain beauty about them because they can show an assertion is always true and it is airtight. Also, they often reveal something new about the nature of the system you are working in. Brute force proofs are like the ugly stepchild - they prove the result by just checking every possible scenario, and usually don't reveal anything new - just the result (I know the 4-color problem was initially solved this way. Haven't checked if there is a non-brute force proof for that problem).
Yes, you've come up with the solution I was looking for.
Spoiler!
I really like this solution because you don't actually have to determine whether sqrt(2)^sqrt(2) is rational or not. You just use its existence to come up with a rational number, either directly or indirectly.
If I were being pedantic, I'd say that a and b have to be different numbers, but it's easy enough to use sqrt(3)^sqrt(2). Proving that sqrt(3) is irrational is only slightly more difficult than proving sqrt(2) is irrational, using the basically the same format as the sqrt(2) proof, but with an added couple steps at the end.
I think the bold would be enough changes for me to stop playing.
Out of curiosity, I did find proofs that both e and ln(2) are irrational (using Google so doesn't count). It turns out I'm not as good at math as Euler... I still sort of like my first e to the power of ln(2) solution better. The proofs are harder (even though its intuitively obvious both are irrational) but the solution itself doesn't have a logic tree, just one of the exponential laws.
I do sort of agree with ggg about proofs in general, although iirc correctly he (like me) is an engineer. And I know I'm certainly fine with empirical solutions that work. Most of the math I've done in my career is numerical solutions to equations of state, which is basically just complicated guess-and-check So from that perspective trying to solve this analytically was fun, as it's different.
Alice, Bob, and their dog Cujo all start at the same point. Alice and Bob begin walking down the same path in the same direction: Alice walks at 4km/h, Bob at 3km/h. Cujo, meanwhile, runs back and forth on the path between Alice and Bob at 10km/h (ie. he runs until he meets Alice, then immediately turns around and runs until he meets Bob then turns around again to run to Alice etc, always staying on the path).
Assume they all travel at constant speed the entire time.
After 1 hour, Alice is 4km down the path from the start point. Bob is 3km on the path from the start point. Where on the path is Cujo?
Spoiler!
This is like a backwards version of the fly on a train problem. Where two trains are travelling toward each other at different rates and the fly goes back and forth. The question there is how far does the fly travel which you can determine from the flys velocity times the time or by summing the series.
The framing is different here as you start with an infinitely small distance to be travelled in an instantly small time like the paradox that states and journey can never end or begin because you always can only go half of the remaining distance. So I keep getting stuck in this reverse limit.
So if I run the problem backwards such that Bob and Alice are 1 km apart walking home and the gap is closing at 1km/hr. In this scenario it doesn’t matter where the dog starts it just gets smushed after 1 hour when the gap disappears and since 1 hr has elapsed The dog has run 10k. But this means that dog can be anywhere between them.
I don’t think that’s quite right but I can’t resolve the paradox I’m in.
The trick is to play the scenario backwards as you discovered. If you assume the path they take is the y-axis of a graph, then you can plot each of their positions over time (in minutes).
A will be a straight line from (0,0) to (60,4)
B will be a straight line from (0,0) to (60,3)
C will be some sort of line that bounces between A's line and B's line. The slope of line segments that make up C's line will be 10 or -10.
The main point is that C's line is bounded by A and B's line, and converges to (0,0).
If you were to place C anywhere between (60,4) and (60,3), C's line would still converge to (0,0).
So, in fact, C can be anywhere.
We get caught up because the scenario is set up to be deterministic. They all start at the same spot, they move at fixed speeds and follow a specific path. But in the immediate nanosecond time interval after they start, C's path is chaotic - C bounces between A and B an infinite number of times!
A few comments below.
Spoiler!
Quote:
Originally Posted by GGG
This is like a backwards version of the fly on a train problem. Where two trains are travelling toward each other at different rates and the fly goes back and forth. The question there is how far does the fly travel which you can determine from the flys velocity times the time or by summing the series.
Yeah, I was thinking of posting that problem. It's made famous because it was posed (allegedly) to John Von Neumann and he immediately answered. The person said "oh, you figured out the trick!" to which he responded "what trick, I just summed the infinite series!".
I was at a social gathering several years ago, and was chatting with another guy who enjoyed math, so I told him the Von Neumann story. It was loud because there were 20 or so people in a smallish space all talking in their own conversations. When I got to the punchline, there was a sudden lull in the conversations, and in almost perfect silence I said with gusto "I just summed the infinite series!". Immediate laughter.
Quote:
Originally Posted by GGG
The framing is different here as you start with an infinitely small distance to be travelled in an instantly small time like the paradox that states and journey can never end or begin because you always can only go half of the remaining distance. So I keep getting stuck in this reverse limit.
Yup, limits and infinity are hard to grasp.
I remember when I was younger (maybe elementary or junior high?) and I had learned about the math concept of a ray - a line going out to infinity from a point. I thought "What if you were on a train that took the path of a ray. It would go forever. Now what if you were on a train that went in the opposite direction of a ray. You would come from infinity and stop at a point. The conductor of the train would say "We've been travelling forever, but we finally made it". It made me laugh, but also blew my mind because I couldn't wrap my head around the idea of "coming from infinity" .
Quote:
Originally Posted by GGG
So if I run the problem backwards such that Bob and Alice are 1 km apart walking home and the gap is closing at 1km/hr. In this scenario it doesn’t matter where the dog starts it just gets smushed after 1 hour when the gap disappears and since 1 hr has elapsed The dog has run 10k. But this means that dog can be anywhere between them.
I don’t think that’s quite right but I can’t resolve the paradox I’m in.
You were dead on! Laughed at the phrasing of the dog getting smushed - should be in a Sliver thread! - but I love how you got to the right conclusion even though you can't believe it's right because it doesn't mesh with a deterministic real-world scenario. Math is not always real world
My favorite quote is "In theory, there is no difference between theory and reality, but not in reality"
I hope these have been fun problems. I've had several in my head from when I first heard them, but I'm starting to look around for more - hopefully I can keep finding interesting ones to solve that don't require lots of background knowledge. Again, if anyone else has any to share, please do.
Here's a new problem. Let's see if it gets solved before the weekend is over or not.
You have a complete graph on 6 vertices (graph where all vertices are connected to each other) like below
Each edge can be colored either red or blue.
Show that regardless of the coloring, you can always find 3 edges that form a triangle that will have its edges all the same color.
