word problem with distance

"shortest" relative to the other 5 choices? or shortest within the choice A? If it's shortest distance within choice A, there could be infinite number of layouts in 2D space with 4 loci's to consider?

how can i do this without a map? i'm more confused than ever before.
I, too, have not taken the time to fully solve it yet.

But I beg you to answer my question about context, which I've been wishing for since I first saw the problem. Where did it come from, and why did you expect to be able to solve it? I'm looking for indications of what methods might be expected, other than raw intelligence. (My guess is still that it is essentially an intelligence test, looking for people who can invent their own methods for solving a non-standard problem. If so, then you are free to give up, and perhaps most of us will!)
 
"shortest" relative to the other 5 choices? or shortest within the choice A? If it's shortest distance within choice A, there could be infinite number of layouts in 2D space with 4 loci's to consider?

how can i do this without a map? i'm more confused than ever before.
Yes, you need to think about choice A as its own problem, which makes it quite different from the normal multiple choice problem. As we have said, it is possible to “solve” A, and therefore A is not the correct answer. Weird. I am 90% sure that this is a basic problem in topology,, but virtually no one who is not a trained mathematician studies topology.
 
i give up. can someone just plainly show how they got the answer? i don't even know what branch of math this would be under other than just word problems.
Where did get this problem? What class/book?

Consider another answer: XABCBCX
Is there a layout where this is the shortest route? No. Why? Because for any layout going back to B instead of returning directly to X makes the route longer. This is the type of reasoning you need to use to conclude that one of the provided choices can't be the shortest route. You'll probably need to use geometry (maybe the fact that in any triangle the sum of 2 sides is always greater than the 3rd).
 
Where did get this problem? What class/book?

Consider another answer: XABCBCX
Is there a layout where this is the shortest route? No. Why? Because for any layout going back to B instead of returning directly to X makes the route longer. This is the type of reasoning you need to use to conclude that one of the provided choices can't be the shortest route. You'll probably need to use geometry (maybe the fact that in any triangle the sum of 2 sides is always greater than the 3rd).
Correction: Because for any layout looping back to B before returning to X from C makes the route longer.
 
this is so poorly worded problem. this is why i hate tests. they get overly convoluted in order to try to trip up students. :(
 
this is so poorly worded problem. this is why i hate tests. they get overly convoluted in order to try to trip up students. :(
The only problem I see is "depending on" vs "independent of".

Look at E: XABCBAX.

Let's try to figure out the layout.
XABC does not tell us much.
After C we are done and can go back to X. But before X we visit B and A. Why? The shortest way from C to X is the straight road the problem tells us must exist. And yet we visit B and A. The reason is the B and A lie on the straight road from C to X.
So, the layout suggested by the answer is X----A----B----C.
Now, is the proposed answer in fact the shortest route? Yes: we need to get to C, then back to X and we do it in a shortest way possible (straight line). We visit A and B along the way, without adding anything to the final distance. That's it. This proves that E _can_ be the shortest route. Therefore, we need to look at another answer to either prove that it can serve as the shortest route for some layout, or that it can't.
 
i think i sort of figured it out. let me reiterate, i think this is freaking poorly worded problem. i don't think i even still understand what it's asking. there's no way they're asking me to think through all the permutations. even if i were to just look at simple line (24 permutations) and a simple square alone (24 permutations) map/layout or whatever the heck they are calling this and then try the 5 choices against all the possible permutations of maps that would be 240 permutations.

here's the thing, choice c doesn't come out to be "not shortest" aka longest. choices A, D and E tie for longest path. at 6. but choice c has that odd diagonal path from c to x that can't be avoided unlike other answer choices. I have no idea how choice B could not be the shortest when the distance is 4 the shortest in this scenario.
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i think i sort of figured it out. let me reiterate, i think this is freaking poorly worded problem. i don't think i even still understand what it's asking. there's no way they're asking me to think through all the permutations. even if i were to just look at simple line (24 permutations) and a simple square alone (24 permutations) map/layout or whatever the heck they are calling this and then try the 5 choices against all the possible permutations of maps that would be 240 permutations.

here's the thing, choice c doesn't come out to be "not shortest" aka longest. choices A, D and E tie for longest path. at 6. but choice c has that odd diagonal path from c to x that can't be avoided unlike other answer choices. I have no idea how choice B could not be the shortest when the distance is 4 the shortest in this scenario.
View attachment 28312
Did you read and did you understand the explanation for E in my post #26?
 
this is so poorly worded problem. this is why i hate tests. they get overly convoluted in order to try to trip up students. :(
Once again I need to ask you: Where did this question come from? If it is a test for a class, that is one thing; if it is something like a contest or admission test, that is entirely different. It may well, in that case, be precisely intended to separate the best thinkers from the ordinary, and therefore to trip up the unwary. But if it is for a course, then it should use ideas that were taught, even though they may be hidden.
 
Once again I need to ask you: Where did this question come from? If it is a test for a class, that is one thing; if it is something like a contest or admission test, that is entirely different. It may well, in that case, be precisely intended to separate the best thinkers from the ordinary, and therefore to trip up the unwary. But if it is for a course, then it should use ideas that were taught, even though they may be hidden.
it would be the latter situation in this case. they're just being devilishly tricky.
 
it would be the latter situation in this case. they're just being devilishly tricky.
It bothers me when people stubbornly refuse to answer direct questions about the source of a problem.

I searched for the problem and found that it's #48 in the "TSA Oxford 2009 Section 1 - Thinking Skills Assessment". Clearly it is more or less the sort of thing I suspected. Of course it's tricky, and of course it isn't a routine problem. And clearly you aren't ready to take it. But that's not shameful; so why hide it?

I also found the answer key, which says the answer is C. Now, at one point I had almost convinced myself that all of them could in fact be the shortest; but that one was the most subtle. The layout that almost solves it looks like this:

B / | A --- X | \ | C

or with B or C inline with A and X. Can you see the shorter route than XAXBCX? And can you see how knowledge of the triangle inequality helps in the proof?

It's also worth noting that rather than actually proving that this can't be the shortest route for any layout, you can use counterexamples to eliminate all the other answers, and go with that. Of course, I'm not happy doing that (and perhaps others whom they should want to give good scores to would feel the same); that's my main concern about this problem.

While we're getting this out in the open, let's make it easier for others to find this discussion by searching. Here is the problem, in text rather than image form:




48 Starting from its depot in town X, a delivery lorry is to make deliveries to three towns A, Band C, in any order which the driver chooses and finally to return to X.​
Straight roads connect each town to every other town.​
Depending on the layout of the towns, which of the following could NOT be the shortest journey for him to take?​
A XAXBCBX​
B XACBX​
C XAXBCX​
D XBABXCX​
E XABCBAX​
 
i understand triangle inequality but how that can be applied to this, i still don't get. how did you know to put the x in the middle and not say a,b, or c? is that layout the same to be used against all 5 choices? so the objective is to prove that c can't be the shortest possible path of the 5 choices isn't it?
 
i understand triangle inequality but how that can be applied to this, i still don't get. how did you know to put the x in the middle and not say a,b, or c? is that layout the same to be used against all 5 choices? so the objective is to prove that c can't be the shortest possible path of the 5 choices isn't it?
Did you read any of my posts???
Each answer corresponds to a different layout. Please reread my posts if you missed them. Especially the one going over "E".
The 5 choices are independent, the objective is to find the one (yes, it's C) that can't be the shortest route for any of infinitely many layouts.
 
I searched for the problem and found that it's #48 in the "TSA Oxford 2009 Section 1 - Thinking Skills Assessment".
...
I also found the answer key, which says the answer is C.
Great research!

I think I just proved that A is also a good answer. In all dimensions (linear, 2d, 3d,...) AND in all possible layouts, route B ≤ A. So the people at Oxford may be turning away good candidates (unless I'm wrong in the following logic :whistle: )

First, split the routes into individual legs...
A) xa ax xb bc cb bx
B) xa ac cb bx

Put each leg into alphabetical order (eg. write "xa" as "ax"). This doesn't change the distance of each separate leg of the journey.
A) ax ax bx bc bc bx
B) ax ac bc bx

Now "cancel" any pairs that A and B have in common. If A contains "ab" and B also contains "ab" then remove them both. This will reduce the length of both routes by exactly the same amount.
A) (ax) ax bx (bc) bc (bx)
B) (ax) ac (bc) (bx)

A) ax bx bc
B) ac

Clearly, route B which goes direct "ac" is always less than or equal to route A "ax xb bc".

(Yes, I just did the "clearly" thing :ROFLMAO: )
 
Great research!

I think I just proved that A is also a good answer. In all dimensions (linear, 2d, 3d,...) AND in all possible layouts, route B ≤ A. So the people at Oxford may be turning away good candidates (unless I'm wrong in the following logic :whistle: )

First, split the routes into individual legs...
A) xa ax xb bc cb bx
B) xa ac cb bx

Put each leg into alphabetical order (eg. write "xa" as "ax"). This doesn't change the distance of each separate leg of the journey.
A) ax ax bx bc bc bx
B) ax ac bc bx

Now "cancel" any pairs that A and B have in common. If A contains "ab" and B also contains "ab" then remove them both. This will reduce the length of both routes by exactly the same amount.
A) (ax) ax bx (bc) bc (bx)
B) (ax) ac (bc) (bx)

A) ax bx bc
B) ac

Clearly, route B which goes direct "ac" is always less than or equal to route A "ax xb bc".

(Yes, I just did the "clearly" thing :ROFLMAO: )
Why does it matter how the 5 answers relate to each other?
 
Why does it matter how the 5 answers relate to each other?
I don't understand. Maybe there's a flaw in my post that you've spotted? Isn't A a valid answer because it "could NOT be the shortest journey".
 
I don't understand. Maybe there's a flaw in my post that you've spotted? Isn't A a valid answer because it "could NOT be the shortest journey".
I haven't reviewed your post, it's most likely a valid proof, that given a particular layout, route A is not shorter than B. However, in my understanding, this is not what the problem is asking us to do. The layout is not the same for all choices. We need to consider them independently. Is there a layout for which A is the shortest route? Yes. Next. Is there a layout for which B is the shortest route? Yes. Next. Is there a layout for which C is the shortest route? No. C is the correct answer.
 
I haven't reviewed your post, it's most likely a valid proof, that given a particular layout, route A is not shorter than B. However, in my understanding, this is not what the problem is asking us to do. The layout is not the same for all choices. We need to consider them independently. Is there a layout for which A is the shortest route? Yes. Next. Is there a layout for which B is the shortest route? Yes. Next. Is there a layout for which C is the shortest route? No. C is the correct answer.
Ahh, very interesting! I understand your interpretation now, and you're probably correct because it means that the question could be solved in much less time (without, potentially, doing LOTS of comparisons between the options). However, I still can't see why C could not be a shortest route if this is the layout in 2d...

A----X----B----C

C is XAXBCX, however there are some equivalent length routes, like XAXBCBX, XACBX.

Perhaps another part of this puzzle is that if a direct route passes through another location then you HAVE to state that location? So, for the above layout, you couldn't write "CX" since it would have to be written as "CBX" because the lorry passes through B en-route. I was thinking that each letter in the route would just be a delivery stop.
 
Perhaps another part of this puzzle is that if a direct route passes through another location then you HAVE to state that location? So, for the above layout, you couldn't write "CX" since it would have to be written as "CBX" because the lorry passes through B en-route. I was thinking that each letter in the route would just be a delivery stop.
Yes, it should've been CBX, not CX. If we record this route as XAXB... then we are recording any stops along the way, not just deliveries - the second X is not a delivery.
 
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