Q16

[noah]

16. At the moment I can't think of a way of slickly solving this question through an inference. We also don't have any previous examples that definitively show more than four pairs of cities connected. With M restricted, as well as the H-T restriction and the PT effect, we can expect that seven and eight are out of reach (indeed, there are only 13 available slots, so 7 pairs), but let's try to fill up the board:

H: give it P and V (it can't connect with M or T). Since P and V both connect to H, they must also connect with T.

So far:
/ /
V / T V T
P ? H P H
H M P T V

(sorry, I know the alignment is a bit off)

Now, moving on to M, let's give that P (we could have given it T or V, it doesn't really matter), and give it's connection an M.

P already connects with M, H and T and it can't connect with V because of an original constraint in the game. Nothing else to do.

T already connects with P and V, and it can't connect with H, so nothing else to do. (though it could be connected to M as well, meaning that P does not).

V connects with T and H, and cannot connect with M nor P (M already has filled its one slot with P, and P can't connect with V because P connects with T.

/ / M /
V / T V T
P P H P H
H M P T V
Count up the letters: 10. So five pairs. I recognize that it's not pretty, but it can be done quickly.

Wait! I just figured out a quick way to approach #16. Since there are 13 open slots, the maximum number of pairs might be 6, and then because of the PT --> ~ PV rule, we have to eliminate one pair. If I was short on time, I might pull the trigger at that point, if I had some in the bank, I would probably spell out the scenario to make sure I'm not missing any restrictions.

[mgalligan04]

I'm not fully understanding why MV cannot be a pair? I keep getting six. This is what I am doing, where am I going wrong?

PT(means that we cannot have PV)
MP, MT, MV (cannot have HM)
HP (cannot have HT)
TV

Thanks,
MG

[noah]

I'm not fully understanding why MV cannot be a pair?

MV can be a pair, it's just that I chose MP. Remember what the first rule says about how many cities M connects with...

I keep getting six. This is what I am doing, where am I going wrong?

PT(means that we cannot have PV)
MP, MT, MV (cannot have HM)
HP (cannot have HT)
TV

Doesn't the part in bold violate the first rule of the game?

[tzyc]

Hi,
I think you started doing the trial and error with filling all slots for H (since it asks the "at most how many..."), but I found you thought of another way to solve the problem written in diagram page.
On that solution you said there are 12 slopts open and because of the Pt→~Pv constraint, eliminate one pair so 10 slots available and 5 pairs is the answer. But I think your first diagram already eliminated one slot by the constraint (/ on the top of P). Why then you need to eliminate 2 slots again? (and why eliminated 2 slots instead of just one?)
And...
Here says there are 14 slots open instead of 12...and actually when I counted I got 13 slots open so I'm kind of confused...did you use the same diagram for the solutions?
Sorry if it is little things but I found the first solution you made was terrific so I'd like to use the same idea when I run out of time and wanted to make it clear.
Thanks.

[noah]

Hi,
I think you started doing the trial and error with filling all slots for H (since it asks the "at most how many..."), but I found you thought of another way to solve the problem written in diagram page.
On that solution you said there are 12 slopts open and because of the Pt→~Pv constraint, eliminate one pair so 10 slots available and 5 pairs is the answer. But I think your first diagram already eliminated one slot by the constraint (/ on the top of P). Why then you need to eliminate 2 slots again? (and why eliminated 2 slots instead of just one?)
And...
Here says there are 14 slots open instead of 12...and actually when I counted I got 13 slots open so I'm kind of confused...did you use the same diagram for the solutions?
Sorry if it is little things but I found the first solution you made was terrific so I'd like to use the same idea when I run out of time and wanted to make it clear.
Thanks.

Thanks for pointing out I should have said 13. Fixed that above.

The other elimination you needed to do was because either P only goes to either T or V, but not both. Yes, we already eliminated a slot on P, but we also have to account for whichever of T and V doesn't get matched with P. That make sense?

[Nina]

16. At the moment I can't think of a way of slickly solving this question through an inference. We also don't have any previous examples that definitively show more than four pairs of cities connected. With M restricted, as well as the H-T restriction and the PT effect, we can expect that seven and eight are out of reach (indeed, there are only 13 available slots, so 7 pairs), but let's try to fill up the board:


So far:
/ /
V / T V T
P ? H P H
H M P T V

(sorry, I know the alignment is a bit off)


Count up the letters: 10. So five pairs. I recognize that it's not pretty, but it can be done quickly.

hey noah,

so far i understand all. but may i have a super basic math question? when you count the letters, do you also include the bold ones? if i don't include those bold letters, i only count up as 9 in total:
V T V T
P ? H P H

but if i include those bold ones, it obviously exceeds 10.
so...could you explain more detailedly how would you count these letters in this diagram?

Thanks a lot!

[noah]

hey noah,

so far i understand all. but may i have a super basic math question? when you count the letters, do you also include the bold ones? if i don't include those bold letters, i only count up as 9 in total:
V T V T
P ? H P H

but if i include those bold ones, it obviously exceeds 10.
so...could you explain more detailedly how would you count these letters in this diagram?

Thanks a lot!

Thanks for your question. Looking back, I realized there was a mistake in my post.

Here's what you should get to:

/ / M /
V / T V T
P P H P H
H M P T V

And the bold letters are the labels for the diagram, so don't count those.

Tell me if you have any more questions about this.

[Reijovem.36]

Following a second play of this game I realized that the simplest way, for me, to attack q.16 was to use the sketch I drew in q.12

*P-T-V
With:
*H connected to P
*M connected to P
*V connected to H

I then looked at each element individually to see whether or not it was possible to form an additional connection

It was not.

[csunnerberg13]

So I understood the question and the setup for it - I actually ended up with basically the same diagram that was posted by Noah, except that I matched M with T instead of with P but that part doesn't matter. What I'm confused with is how you end up with 5 pairs...once I got my picture, I figured I needed to list out all the connections made and then that was the number I would have. So I listed out the possibilities: HV, HP, MT, HT, TP, TV - there are 6, though, not 5...what am I missing here? I would never be confident enough math-wise to just say "oh 10 letters, must be 5 pairs" - I would still go by my matchings that I can see exist so I'm confused about how to end up at 5.

[noah]

So I understood the question and the setup for it - I actually ended up with basically the same diagram that was posted by Noah, except that I matched M with T instead of with P but that part doesn't matter. What I'm confused with is how you end up with 5 pairs...once I got my picture, I figured I needed to list out all the connections made and then that was the number I would have. So I listed out the possibilities: HV, HP, MT, HT, TP, TV - there are 6, though, not 5...what am I missing here? I would never be confident enough math-wise to just say "oh 10 letters, must be 5 pairs" - I would still go by my matchings that I can see exist so I'm confused about how to end up at 5.

When this sort of moment happens, you know that you've broken a rule somewhere. I ran through the pairs and see that you've violated one (if you don't see it after another pass, tell me). Perhaps you're overlooking that these pairs would all need to be connected at the same time.

Tell me when you figure it out--or if it you're just about to kill me because I haven't told you!

[yuhefan]

Hi Noah,

I think I sort of solved this in a strange way, so I'm not sure if I got it out of pure luck or I can actually count on it...

I started with H:

HP --> PT --> - PV
HV --> TV

Then with M, either MP, MV, or MT, but M can only connect to 1.

So counting the possible pairs, I came up with 5.

[noah]

Hi Noah,

I think I sort of solved this in a strange way, so I'm not sure if I got it out of pure luck or I can actually count on it...

I started with H:

HP --> PT --> - PV
HV --> TV

Then with M, either MP, MV, or MT, but M can only connect to 1.

So counting the possible pairs, I came up with 5.

From a quick glance, it looks like you tried a scenario and it worked out to 5. Here, that worked. Elsewhere it's dicey to just try one scenario as another might lead to more--but sometimes that's the only way to wade through it.

[rachellewrx]

I got stuck on this one during timed PT. However, I managed to figure out a quicker way to come up with the answer by using math.

So the simpler version would be: 10-3-1-1=5. Explanation is as below.

10: Setting aside all the restrictions, if we randomly choose 2 cities out of five, we could get 10 different combinations. I don't know how to attach the math symbol, but it's something we learned in math class in Sequencing & Combination chapter. Well, imagine a letter C with 2 on the upper right corner and 4 on the lower right corner. Yeah, that's it.

-3: M could've paired with 4 other cities if it weren't for the restrictions. Now with the first restriction, it can only pair with one city. So we lose 3 possible pairs.

-1: H and T cannot be together. We lose this pair.

-1: P can either be with T or V, or neither. But never both. So we lose another pair.

Hence, 5.

I hope it makes sense. Or, maybe I came up with the correct answer by luck. Any feedback would be appreciated.

Please pardon my english. Non-native speaker here.

[ZIYAOW681]

I got stuck on this one during timed PT. However, I managed to figure out a quicker way to come up with the answer by using math.

So the simpler version would be: 10-3-1-1=5. Explanation is as below.

10: Setting aside all the restrictions, if we randomly choose 2 cities out of five, we could get 10 different combinations. I don't know how to attach the math symbol, but it's something we learned in math class in Sequencing & Combination chapter. Well, imagine a letter C with 2 on the upper right corner and 4 on the lower right corner. Yeah, that's it.

-3: M could've paired with 4 other cities if it weren't for the restrictions. Now with the first restriction, it can only pair with one city. So we lose 3 possible pairs.

-1: H and T cannot be together. We lose this pair.

-1: P can either be with T or V, or neither. But never both. So we lose another pair.

Hence, 5.

I hope it makes sense. Or, maybe I came up with the correct answer by luck. Any feedback would be appreciated.

Please pardon my english. Non-native speaker here.

I did it in exactly the same way.
And with Q12 we can confidently eliminate A and confirm B is the correct one.