Thread: How Good Are You At Reasoning?

Originally Posted by Xei

4. The point is that it's a circular path; you don't have to start at any fuel stop, you can choose. The question is to prove that there is always some station or other that you can start from, and get all the way round (running out of fuel at the very end). And I promise that it is always possible.

So.... this.

Originally Posted by tommo

For example in number 4, thinking like a mathematician you would say "Okay well if we start at the stop (which could be in the middle or whatever) which has the most fuel, we can suddenly make it the rest of the way".
But thinking in IRL terms, you just start out at the start of the mountain.

So yes I know it is always possible. But it requires you knowing the amount of fuel in each stop, and choosing the one with the most fuel to start at.

Ok....say there are 10 stops over 100km. Stop 1 (wherever you start) must have the most fuel, otherwise you won't make it.
If it does have the most fuel, you will make it.

Basically write this in to a maths equation to prove it

Stops and km's of fuel in each stop
0--10-20-30-40-50-60-70-80-90-100km
60-10-5---5--5---5--5--5---0--0--0

Start (0) will get you to 60km by itself, but you're picking up stop 2, 3, 4, 5, 6, 7 and 8 along the way.
Giving you 100km of fuel by stop 8.

I'd actually enjoy knowing how to put this in to an equation. I should get around to learning one day.
But I don't know if this question really tests reasoning.... Maybe it does but it seemed pretty obvious straight away that it's possible.
As long as you're not thinking about it as a real life situation. Which is what most people do.

Originally Posted by Xei

5. To clarify the 'pairs of towns' thing; you've got six towns, A, B, C, D, E, F. It just means that between any two towns, for instance B and F, there is either a bus service or a train line. I'm not sure what you mean by taking a bus between each town; every single connection between towns could be a train line.

You don't need to travel to a town to find out what services it has. You know everything about the towns and which services are between them. The question is to prove that for whatever the situation is, it'll be possible to find three towns all connected by the same type of service, in a triangle.

Again, it's definitely possible to prove.

Fine, take either all buses or all trains

Ok, math thinking again....

You have six towns and only two different modes of transport.
There may be 1 train 5 buses, you can go between 3 towns on buses,
2 trains, 4 buses, you can go between 3 towns on buses
3 trains, 3 buses, you can between 3 town on either,
etc. vice versa

Again, can't really figure out how to get this in to an equation.

Originally Posted by Xei

Lol... I don't get why 'just take buses' is such a popular 'answer'.

I think it is possibly because the question is fairly simple.
I mean.... if you know there are buses or trains, take the bus to a town with a bus, take that bus to another town with a bus that will take you back to the first town.

When I said 4 and 5 didn't require maths knowledge, I wasn't trying to mislead anybody. Honest, they really don't, they're plain old reasoning problems.

There are no equations involved; there's no method I've learned for doing these questions, they're novel problems. The point is to devise your own method.


4.

Originally Posted by tommo

But I don't know if this question really tests reasoning.... Maybe it does but it seemed pretty obvious straight away that it's possible.

One of the first things you learn to do in a maths degree is to distrust yourself whenever you think a question is obvious. If something looks obvious, yet you can't actually explain in plain words why exactly it's true, it can't really be said to be obvious at all; and indeed, it will often turn out that you are wrong, or at least that the answer is not as obvious as it seems.

A great example of this is actually your proposed solution, namely, "wherever you start must have the most fuel, otherwise you won't make it". This does seem obvious, but can you actually explain in clear terms why it is the case?

If you can't, start to doubt that it actually is the case. See if you can come up with an example where you don't start from the stop with most fuel.


5.

Ok, math thinking again....

You have six towns and only two different modes of transport.
There may be 1 train 5 buses, you can go between 3 towns on buses,
2 trains, 4 buses, you can go between 3 towns on buses
3 trains, 3 buses, you can between 3 town on either,
etc. vice versa

I think the only way to solve this one is to actually draw it and mess around.

What you're talking about currently is not actually what I described; there are not 6 different services in total. There are lines between each PAIR of towns. Like this.



The key to solving these problems is to first break it down into its absolute simplest terms. A boiled down version of this question would be, "if you colour each line either red or blue, do you always have to draw a triangle of one colour?".

Alright, let's see if we can pull the 'ol modus ponens and find some answers.

Originally Posted by Xei

1. I draw N circles on a piece of paper. Prove that they divide the paper into no more than N(N - 1) + 2 regions.

Originally Posted by Xei

2. x is any number on the number line such that x + 1/x is a whole number. Prove x^n + 1/x^n is also a whole number, for any positive whole number n.

I interpret this as,

x ∈ ℤ, where ƒ(x) = [x + (1/x)], and ƒ(x) ∈ ℤ.

ƒ₂(x) = [x^n + (1/x^n)], where ƒ₂(x) ∈ ℕ only when n ⊂ x.

I'm not sure the first function should look like [(x+1)/x] or [x+(1/x)].

If I were to take the sum of the geometric series which results from expanding ƒ₂(x) while keeping x static at some value, namely 2, then the series doesn't have a common ratio and does diverge to an arbitrary whole number.

Originally Posted by Xei

3. We have the set of consecutive numbers from 1 to some even number, for instance, {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

Now we take one more than half of the numbers from this set, for instance, {2, 3, 4, 7, 9, 10}.

a) Prove that in this new set, there will be two numbers with no common factors (for instance, 4 and 9).

b) Prove that in this new set, one number will divide another (for instance, 3 and 9).

B ⊂ A and A ∪ B and B ∩ A. If both sets are communicative with respect to the other, and each set contains elements of the other, then every valid operation for the set A will be valid for the set B. The number of valid operations depends on the total number of elements taken from the consecutive sequence set to the subset B and is proportional to limiting value of ½+1 number of elements in the subset.

Originally Posted by Xei

4. You wish to drive around a mountain. There are fuel stops at various places on the road, and the total amount of fuel in them is exactly enough to make the journey once. Prove that it's possible to make the journey in a car with an empty fuel tank from some fuel stop.

Do you mean the car or the fuel stations when you say "them"? The gas stations would need to be at equivalent distances around the mountain and the surface of the terrain would need to be consistent for this to be possible.

Originally Posted by Xei

5. There are six towns. Between each pair of towns is either a bus service or a train service, but never both. Prove that you can travel through three towns in a loop (without visiting any others) with only one type of transport.

Does this have to be true for five towns?

There will always be at least one more bus than train station or vice versa between all the towns, and no that doesn't have to be true for five towns.

1. It's quite easy for 3, but can you work out how to prove it for 6 circles and 32 regions?

Also the solution requires a brief explanation of why you can't create more regions.

2. x is not generally an integer.

3. I guess you're using some more general theorem I haven't encountered yet. I don't really understand your answer. How does your solution fail when the set has less than n + 1 elements?

4. Fuel stations. If you add up all of the fuel in each station, there's just enough to make the journey once. It's possible for any position of the fuel stops around the route; the question is to prove this.

5. I make it 15 different services between towns, so I accept that there will always be at least one more of one type of transport. I don't see how the conclusion follows (i.e. that you will always be able to travel in a loop through three towns via only one means of transport).

Originally Posted by Xei

When I said 4 and 5 didn't require maths knowledge, I wasn't trying to mislead anybody. Honest, they really don't, they're plain old reasoning problems.

Oh, I couldn't even remember you saying that.

Originally Posted by Xei

4.
One of the first things you learn to do in a maths degree is to distrust yourself whenever you think a question is obvious. If something looks obvious, yet you can't actually explain in plain words why exactly it's true, it can't really be said to be obvious at all; and indeed, it will often turn out that you are wrong, or at least that the answer is not as obvious as it seems.

A great example of this is actually your proposed solution, namely, "wherever you start must have the most fuel, otherwise you won't make it". This does seem obvious, but can you actually explain in clear terms why it is the case?

If you can't, start to doubt that it actually is the case. See if you can come up with an example where you don't start from the stop with most fuel.

Yes, of course it is possible. But it won't be possible every time, I already stated that (if you start at one with 1km of fuel and need to go 2km).
Whereas to keep it simple, all you have to know is which pump has the most fuel.

Maybe I should have said "wherever you start must have the most fuel, otherwise you are less certain to make it".
I can't explain in clear terms why this is so. I can picture it in my head but can't grasp it fully.
If you have 100km's of fuel, distributed around a 100km circle you still have 100km's of fuel.
You could also start at a pump which has less than the most full pump if the distance to pump 2 is shorter than the amount of km's you get from pump 1.

See this is where I thought some algorithm would come in handy, coz you could figure out the least fuel you would need from each pump, given their distances apart etc.
It's tangential to the question, but still.

You seem to be changing the question though. You just said "prove that it's possible", at first. And I did.

Originally Posted by Xei

5.
I think the only way to solve this one is to actually draw it and mess around.

What you're talking about currently is not actually what I described; there are not 6 different services in total. There are lines between each PAIR of towns. Like this.



The key to solving these problems is to first break it down into its absolute simplest terms. A boiled down version of this question would be, "if you colour each line either red or blue, do you always have to draw a triangle of one colour?".

Ok, what you should have said is "There is a line (or, service, whatever) connecting each town to each other town". That's much more clear. Or just put the picture up first.

You did it here again, btw
First you said "Prove that you can travel through three towns in a loop (without visiting any others) with only one type of transport."
Now you're saying "do you always have to draw a triangle of one colour?" i.e Is there always a way between 3 towns using one mode of transport.

Originally Posted by tommo

Oh, I couldn't even remember you saying that.


Yes, of course it is possible. But it won't be possible every time, I already stated that (if you start at one with 1km of fuel and need to go 2km).
Whereas to keep it simple, all you have to know is which pump has the most fuel.

Maybe I should have said "wherever you start must have the most fuel, otherwise you are less certain to make it".
I can't explain in clear terms why this is so. I can picture it in my head but can't grasp it fully.
If you have 100km's of fuel, distributed around a 100km circle you still have 100km's of fuel.
You could also start at a pump which has less than the most full pump if the distance to pump 2 is shorter than the amount of km's you get from pump 1.

See this is where I thought some algorithm would come in handy, coz you could figure out the least fuel you would need from each pump, given their distances apart etc.
It's tangential to the question, but still.

You seem to be changing the question though. You just said "prove that it's possible", at first. And I did.

The question is to prove that it's always possible.

Obviously it's possible in some circumstances. For instance, there could simply be one stop with all of the fuel. The question is to prove it can be done in all circumstances, for any given arrangement of fuel pumps.

Are you saying that if you start from the stop with most fuel, you are certain to make it..? Is that the basis of your argument?

Ok, what you should have said is "There is a line (or, service, whatever) connecting each town to each other town". That's much more clear. Or just put the picture up first.

I simply copied the question from its source. It isn't ambiguous. There's a service between each pair of towns... what else can that mean?

I didn't put the picture up because working out what the picture looks like is part of the question. I only put it up now because lots of people did not have the right picture in mind.

You did it here again, btw
First you said "Prove that you can travel through three towns in a loop (without visiting any others) with only one type of transport."
Now you're saying "do you always have to draw a triangle of one colour?" i.e Is there always a way between 3 towns using one mode of transport.

It don't understand what you think the difference is.

The two formulations are equivalent, it's just that the latter is boiled down to the absolute basics of what you need to prove, minus all of the superfluous information about travelling and things.

Originally Posted by Xei

The question is to prove that it's always possible.

Obviously it's possible in some circumstances. For instance, there could simply be one stop with all of the fuel. The question is to prove it can be done in all circumstances, for any given arrangement of fuel pumps.

Well, that's not what it said.
So, wherever they are, choosing a certain starting point will enable you to complete the course.
But as I said, I don't know how to prove this, I just know that it's possible.
It seems like some maths is required for this.
The best way I can explain is to give examples like I did before, but obviously that doesn't suffice.

Originally Posted by Xei

Are you saying that if you start from the stop with most fuel, you are certain to make it..? Is that the basis of your argument?

No.
If the one with the most fuel has a huge distance, you may not make it.
But if you start at the one after it and collect all the smaller amounts of fuel (which will be closer together) you will make it.

I can't see how it's a difficult question, it's just difficult to explain why.
There's always 100km's of fuel. You just have to choose the correct one to start at.
Which would differ every time depending on amount of fuel and distance between each one.

Originally Posted by Xei

I simply copied the question from its source. It isn't ambiguous. There's a service between each pair of towns... what else can that mean?

What I thought at first.

Originally Posted by Xei

I didn't put the picture up because working out what the picture looks like is part of the question. I only put it up now because lots of people did not have the right picture in mind.

Hmmmm, what does that say about the question? Maybe that it is ambiguous.
It's pretty obvious that "There is a line (or, service, whatever) connecting each town to each other town" is infinitely more clear.

Originally Posted by Xei

It don't understand what you think the difference is.

"Prove that you can travel through three towns in a loop (without visiting any others) with only one type of transport."
Means, show that you can travel through three towns in a loop using one mode of transport. So everyone says, "Ok, I'll go through town with only buses."

The other one is saying that you need to prove that you can ALWAYS travel through 3 towns in a loop using one mode of transport, no matter which modes of transport are between each town.

You'll probably need to switch your mind out of mathematician mode to be able to see the difference, but I've tried to explain it clearly.

I guess they were phrased by a mathematician for a mathematician and we have certain ways of speaking. In both questions, you're given the conditions first, so the implication is that the object already objectively exists. You can't choose it. To prove that you can do it, it is therefore required that you prove you can do it for all circumstances which satisfy the conditions, and hence the specific circumstance that you'll have been provided with.

I suppose we are also used to talking about pairwise conditions, but really it isn't ambiguous. In your original interpretation, which involved six lines, if you actually drew it and asked, 'is it true here that every pair of towns is connected, as required?', the answer would have been no. It's an understandable misunderstanding though.

As to question 4 not being difficult but it being hard to explain why... well, the entire question is to explain why, which nobody has come close to yet. The question isn't 'can you do it', you're already told that you can definitely do it. The challenge is to prove the assertion. And like I say, I can speak from plenty of experience: whenever something seems obvious and yet you can't actually explain why it's obvious, you probably don't actually have any justification, and could well be wrong. The human brain is only capable of making very basic inferences; if your contention can't be pinned down to any inferences, it begs the question where your contention is coming from in the first place.

Consider this question (this is just a metaphor, not a puzzle for the thread): you have three dice, A, B, C, (they are cubes), with numbers on each face (not necessarily 1 to 6). Prove that if die A is more likely to beat die B, and die B is more likely to beat die C, then die C can't be more likely to beat die A.

Would you say that this is roughly as obvious as the fuel question?

Pedagogy aside, now that you get the gist of the towns puzzle, I recommend you take a shot at it.

The third problem is eerily related to the classic NP problem, one of the seven greatest problems of the century.

Every proof question is essentially NP. That's one of the reasons why P = NP would be so dramatic; a computer could prove any theorem in polynomial time.

It's probably the hardest question there, certainly the second part.

Originally Posted by Xei

I guess they were phrased by a mathematician for a mathematician and we have certain ways of speaking. In both questions, you're given the conditions first, so the implication is that the object already objectively exists. You can't choose it. To prove that you can do it, it is therefore required that you prove you can do it for all circumstances which satisfy the conditions, and hence the specific circumstance that you'll have been provided with.

I suppose we are also used to talking about pairwise conditions, but really it isn't ambiguous. In your original interpretation, which involved six lines, if you actually drew it and asked, 'is it true here that every pair of towns is connected, as required?', the answer would have been no. It's an understandable misunderstanding though.

The problem was that I automatically put them in 3 pairs.
OO
OO
OO

So I didn't think about "pairs" the way it's meant.

But anyway.... I think you pretty much get why people are confused at these types of questions sometimes.

Originally Posted by Xei

Consider this question (this is just a metaphor, not a puzzle for the thread): you have three dice, A, B, C, (they are cubes), with numbers on each face (not necessarily 1 to 6). Prove that if die A is more likely to beat die B, and die B is more likely to beat die C, then die C can't be more likely to beat die A.

Would you say that this is roughly as obvious as the fuel question?

Not really. At first it seems obvious. But then I quickly realise that I don't know what the numbers are so C could easily beat A.

Originally Posted by Xei

Pedagogy aside, now that you get the gist of the towns puzzle, I recommend you take a shot at it.

5. There are six towns. Between each pair of towns is either a bus service or a train service, but never both. Prove that you can always () travel through three towns in a loop (without visiting any others) with only one type of transport.

Does this have to be true for five towns?

Think it's probably easiest to explain with a diagram.

Say all possible second town's paths (excluding ones leading back to the first town) are trains, but you've taken a bus to the second town. This is the worst situation.
But you can just start instead at one of the other towns, which all have all trains in loops.

Like this.



But then we can eliminate some of those paths to break the loops.
I'll break all of them.



But this just opens up several loops to take buses on again.

I keep confusing myself trying to figure out the possible paths and why there are always loops available.

Basically, as soon as you cut off enough loops (by replacing them with the other type of transport) you have just created loops for the other type of transport.

Originally Posted by Xei

But that's not the only information you have, is it? A beats another die B, and that die B beats C. So surely it's obvious that A beats C?

Yeah but, who knows. What are the rules? Highest number?
Anyway....

Originally Posted by Xei

This is basically just a restatement of what the question is rather than an answer. Of course, for the assertion of the question to be true, it must be true that breaking loops will create another loop. But it's not clear that this will always happen. Maybe for some specially created circumstance you haven't considered yet it will fail. Also, if this argument works, it shouldn't matter how many towns there are; but are you sure it is even true for any number of towns (consider 5)?

Playing with maps is the right general way to go about it, though. I recommend you try taking the birds eye view of the transportation planner, rather than the perspective of the person setting out on a journey.

I realise my answer doesn't suffice. And also that it is better not to think of it like a person taking a journey, but I'd started like that and so went with it. I haven't done anything like this for years and I think my brain is having trouble grasping precisely what I'm thinking. I rewrote that like 5 times coz I kept getting mixed up with my train of thought haha

I will re-attempt it tomorrow, even though I'm getting over it now.

Why the fuck did that post go before yours, and shows me posting it an hour before you?
That's a real puzzle right there.

Why the fuck did that post go before yours, and shows me posting it an hour before you?
That's a real puzzle right there.

Originally Posted by tommo

Not really. At first it seems obvious. But then I quickly realise that I don't know what the numbers are so C could easily beat A.

But that's not the only information you have, is it? A beats another die B, and that die B beats C. So surely it's obvious that A beats C?

Basically, as soon as you cut off enough loops (by replacing them with the other type of transport) you have just created loops for the other type of transport.

This is basically just a restatement of what the question is rather than an answer. Of course, for the assertion of the question to be true, it must be true that breaking loops will create another loop. But it's not clear that this will always happen. Maybe for some specially created circumstance you haven't considered yet it will fail. Also, if this argument works, it shouldn't matter how many towns there are; but are you sure it is even true for any number of towns (consider 5)?

Playing with maps is the right general way to go about it, though. I recommend you try taking the birds eye view of the transportation planner, rather than the perspective of the person setting out on a journey.

I'm gonna start knocking these out in a few days if you guys are done playing...

Originally Posted by Xei

6. In the UK you can drink at 18. There are four people drinking at the bar. One is 21, one is 16, one is drinking coke, one is drinking beer. What do I have to check to make sure nobody is breaking the law?

The 16 year old is on one side of the bar drinking coke. On the other side of the bar is the 21 year old barmaid, the landlady and the landlord. The landord is drinking the beer. The barmaid, landlord and landlady have to be at least 18 to be on the serving side of the bar.

There wasn't anything clever about the way the situation was described. There are literally four people sat at the bar with drinks, that's all. You know for sure that one is 21, another is 16, another is drinking coke, and another is drinking beer. The question is to determine the minimum amount of information about drinks and ages I would need to obtain to be sure that nobody is breaking the law.

It isn't supposed to be very hard at all. The point is that it's coupled with the cards question one so try that too.

Originally Posted by PhilosopherStoned

I'm gonna start knocking these out in a few days if you guys are done playing...

I look forwards to it. And no appealing to arcane theorems please!

I started reading around some of the basics of graph theory a couple of days after I set that towns puzzle (which I knew nothing at all of beforehand) and it turns out that one of the most famous theorems has this as a simple consequence (which, thinking about it, is no doubt what inspired the tutor to set this as an introductory puzzle for undergrads in the first place). Can't remember the name but it was to do with graphs with every vertex joined (which is what the towns are) having subgraphs with every vertex joined (which is what a triangle is) when you colour the edges.

When a few of them have been knocked down (which none have so far), I'll put up some really awesome riddles I've found recently which are definitely pure reasoning riddles with nothing to do with mathematics.

Well technically you'd also need to check the age of the person serving the drinks, and whether the bar has a valid license to sell alcohol. And that it wasn't violating various health and safety laws. And....

It's funny because the reasoning behind puzzles like these are by design capable of being demonstrated from many facets of mathematical understanding.

Originally Posted by Xei

1. Along the right lines, but doesn't check out with the actual numbers (2 is non-prime but tails for example).

This brought me joy to an otherwise dull day.

Sorry tommo, didn't see the posts. To be fair you did put them in a silly place.

Originally Posted by tommo

Yeah but, who knows. What are the rules? Highest number?

If it were the highest number on the die that determined the winner then A would indeed always beat C. But as I imagine you guessed, it's not true. It seems very obvious, but if you relied on this intuition, and a reason that you 'know but can't put into words', you'd be incorrect. It is of course normally true. But there are various dice for which it is not true. For instance,

die A has sides: 2, 2, 4, 4, 9, 9
die B has sides: 1, 1, 6, 6, 8, 8
die C has sides: 3, 3, 5, 5, 7, 7.

You can work out the probabilities of A beating B and so on (any pair of numbers from the two dice is equally likely to come up, so just count the ones which win). You find that the probability that A rolls a higher number than B is 5/9 (55.55%), the probability that B rolls a higher number than C is 5/9, and the probability that C rolls a higher number than A is 5/9.

I realise my answer doesn't suffice. And also that it is better not to think of it like a person taking a journey, but I'd started like that and so went with it. I haven't done anything like this for years and I think my brain is having trouble grasping precisely what I'm thinking. I rewrote that like 5 times coz I kept getting mixed up with my train of thought haha

I will re-attempt it tomorrow, even though I'm getting over it now.

Like all of these things, the answer is hard to get, but very easy to understand. It can be explained in two or three short sentences, I think.

Originally Posted by Dreams4free

This brought me joy to an otherwise dull day.

I saw what you did there.

Even numbers are never prime. Sure you could tout 2 as a so called 'counter-example' but that would just be cherry-picking, wouldn't it.

Clearly I'd been studying too hard and was actually thinking in Z[√-5]... thankfully what I said does actually make sense in context. Very well spotted though.