Logic Challenge #30 - The Processors Game

[noah]

Please post your explanations here! When we post the next one, we'll figure out who wrote the best explanation and reward that person with jewels ("jewels" here being defined as "$200 off one of our courses, a free Strategy Guide, or an LSAT t-shirt (yes, those exist)."

Good luck!

[Kurst]

Thanks for posting another game, Noah! These are always fun. My explanation for the processors game is in this PDF:

processors.pdf

(128.82 KiB) Downloaded 586 times

[cinderellarose14]

Fun game!!
Do we have to post the explanations in pdf format? I was just wondering.

[Kurst]

You do not have to post explanations as PDFs.

[noah]

Fun game!!
Do we have to post the explanations in pdf format? I was just wondering.

As i.kurst mentions, you can simply write it in. If you have diagrams that are easier to show in a word doc (or pdf), e-mail that to us and we'll load it up to your post for you.

[cinderellarose14]

Here is my explanation! It took me a while to write, and this is my first time trying doing such a thing... Hopefully it is not too confusing.

Thanks,
ZLeitter

[interestedintacos]

I'm not trying to win the contest; I just wanted to point out that both explanations are missing a real explanation for how we get going. You have to explain why it's the case that we get 1/8, 2/7, 3/6 and 4/5.

In case anyone missed that, here's why. Basically you know that you will have 4 equal sums (the sums of R, S, T and U will be equal) so whatever total number you have, it will be divisible by 4.

Since we know we have numbers 1-8, to get the sum total--which we'll end up dividing by 4--we have to add each number from 1-8, i.e. 1+2+3+4+5+6+7+8, which is 36. 36 divided by 4 = 9, so each set will have a total of 9. Then it should be clear that our sets will come out to be 1/8, 2/7, 3/6 and 4/5.

Another point: I used RSTU as the base (and put the possible number sets on top) and found that to be more effective in seeing the possible arrangements:

__ __ __ ___
__ __ __ __
R S T U

One more thing: I did this game without working out templates, but I found the last question to require them. It's tricky because it isn't just asking for another rule that would preclude 3/6 from going into T (which answer choice C would satisfy); it's looking for a rule that would do that AND be consistent in other ways with the original possibilities. C is incorrect because it precludes some other original possibilities, even though it still precludes 3/6 from going to T.

Trying to solve that question by putting 3/6 into T and finding which rule would be violated doesn't work (because you will be left between B and C), nor does just looking at the choices and finding which one would preclude 3/6 from going into T, which again can leave you with C. I think the most effective way is simply applying the choices to the templates. I worked out 4 templates based on the only 4 possibilities for placement of T and U.

[cinderellarose14]

For question 7 I did the same as you by applying each answer choice to all the possibilities. However I thought it would be fun to also think about a way to not use that method. As I stated in the explanation it is tricky to eliminate the (C) choice due to the reasons you mentioned (so second step thinking is required here), however even if you only have (B) and (C) left, it is still obvious why (C) needs to be eliminated (cause it adds a rule so it does not replace).

Thanks for pointing out that the distribution was not explained clearly enough - I meant to explain it but did not do as nearly a good job as you did. Kudos!!

[Kurst]

You have to explain why it's the case that we get 1/8, 2/7, 3/6 and 4/5.

In case anyone missed that, here's why. Basically you know that you will have 4 equal sums (the sums of R, S, T and U will be equal) so whatever total number you have, it will be divisible by 4.

Since we know we have numbers 1-8, to get the sum total--which we'll end up dividing by 4--we have to add each number from 1-8, i.e. 1+2+3+4+5+6+7+8, which is 36. 36 divided by 4 = 9, so each set will have a total of 9. Then it should be clear that our sets will come out to be 1/8, 2/7, 3/6 and 4/5.

This is an excellent explanation of how the processors must be combined. Thanks tacos! My explanation regarding this aspect of the game was perhaps too succinct.

I used RSTU as the base (and put the possible number sets on top) and found that to be more effective in seeing the possible arrangements

When motherboards are used as the base, how do you represent the third rule?

[interestedintacos]

For the third rule I realized first off that 1/8 can't be assigned to T. If U is going to have at least one processor higher than T's, T can't have 8 since it's the highest processor.

Knowing from the other rule that T also can't have 3/6 I realized that there are only 2 possibilities for T. I put the 2 possibilities in the slots above T separated with a slash:
7/ 5
2 /4

I then, again using slashes, showed the corresponding possibilities for U. On the left side of the slash for U I put 1/8--to indicate that if the T had 2/7, U would have to have 1/8. So that's the first possibility. The other side of the slash leaves U with two additional possibilities (as well as the first, which is still valid), so I also represented that (with a dash to separate the additional 2 because using another slash would be confusing).

From this basic diagram I went ahead and actually created the 4 templates (again with RSTU as the base) for the last question. Otherwise the templates weren't necessary.

I prefer my method of using RSTU as a base because I could see the relationships better that way--the 4 possibilities for T and U are clearly displayed, and it's obvious that R and S simply get the leftovers.

The only problem was representing the fact that the second possibility (when T is 4/5) leaves open all 3 options to U and not just the 2 that are excluded when T is 2/7.

[noah]

Great discussion! The numbering issue is interesting. If someone didn't get it, the game is unbearable. Once you do get it, it seems so obvious that it doesn't even seem worth discussing.

It's awesome that we're already seeing some divergent solution styles. I'll be interested to see if someone has a radically different approach.

[noah]

I just finished reading the two pdf explanations and I have to say I was thoroughly impressed. I'm supposed to choose the best, and frankly, that's really not possible. The approaches were both great.

i.kurst used a strange but effective diagram that treats teh game as an assignment game (and I intentionally wrote this to try to "trick" players into thinking this way, but apparently, it is quite possible to do this). He also noted a few places to improve timing.

Cindereallrose treated the game as a numbered ordering game, as expected, but she exhausted the possibilities - usually not a great method, but it worked well here. She also pointed out a few nice time-saving tips as well.

Honestly, these were stellar explanations, and both authors should be proud!

[efcaley]

I see I'm a little late in the game, but I just stumbled upon the site today. I posted as an attachment,

processor game explanation.pdf

(22.87 KiB) Downloaded 290 times

but if you can't download that, here is my attempt at an explanation:

The solution to this game relies on the inferences made from the rules given.

The most important inference comes from rule #1: Since all boards must have the same processing capacity, we know that each must have (8+7+6+5+4+3+2+1)/8 = 9 gHz capacity.

From this, we can infer that each board will carry a pair consisting of (8,1), (7,2), (6,3), or (5,4).

Rule #2 and Rule #3 combine to determine which pairs can occupy boards T and U:

Rule #3 states that U must have at least one chip with greater processing power than both those assigned to T

and

Rule #2 states that T cannot contain the (6,3) pair. This allows for three initial setups:

U = (6,3), (7,2), or (8,1)
T= (5,4) or (7,2)

A

7/8 8/7 5 6
2/1 1/2 4 3
R S T U

B

7/6 6/7 5 7
2/3 3/2 4 2
R S T U

The third setup can be split into two, since T can hold either the (7,2) or (5,4) pairs:

C

6/5 5/6 7 8
3/4 4/3 2 1
R S T U

D

7/6 6/7 5 8
2/3 3/2 4 1
R S T U
From this, there are only 8 possible solutions to the game, 2 for each setup. The solutions to questions 1-6 are arrived at simply by comparing the setup diagrams.

1) (B) Which of the following is an acceptable partial list of processors and the motherboards for which they are used?

T can only contain the pairs (5,4) or (7,2), therefore (A), (C), and (E) are incorrect. (D) places (7,2) into T and (6,3) into U, which violates Rule #3, and so is incorrect. (B) satisfies solution
diagram D above and is the correct answer.

2) (B) What is the smallest processor that can be assigned to T?

From our setup diagrams, T must contain either (7,2) or (5,4). Therefore 2 is the smallest processor which can be placed in T, and (B) is correct.

3) (D) Which of the following is a complete and accurate list of boards that could be constructed with a pair of processors that differ in power by one gigahertz?

The only pair of chips which can be used which differ by 1 gHz are (5,4). From the diagrams R, S, and T can each hold the (5,4) pair, therefore (D) is correct.

4) (C) If both S and T have processors that are consecutively-sized with at least one of the processors of the other board, how many different assignments of all eight processors are possible?

From the setup diagrams, (6,3) and (5,4) occupy S and T, respectively, in two potential solution sets (B and C); and (6,3) and (7,2) occupy S and T, respectively, in one solution set (D). Therefore, there are three potential solutions and (C) is correct.

5) (D) If R is assigned processor 6, which of the following must be true?

If R is assigned processor 6, then there are three possible solution sets:

R (6,3) S (8,1) T (5,4) U (7,2)
R (6,3) S (5,4) T (7,2) U (8,1)
R (6,3) S (7,2) T (5,4) U (8,1)

The second case demonstrates that (A) and (B) are incorrect. In the first case, (C) is incorrect. The first and third cases prove (E) wrong. Therefore, (D) is correct.

6) (D) Each of the following could be the boards to which processors 6 and 7 are assigned, though not necessarily in the order listed, EXCEPT:

From the diagrams, 6 and 7 can occupy each of the following pairs of boards:

R, S; R, T; S, T; U,R; U, S.

(D) contains the only unlisted pair and is therefore correct.

7) (B) Which of the following, if substituted for the rule that T cannot be assigned processor 6 would have the same effect on the assignment of processors to boards?

This question requires that a rule be established which creates the same restrictions as rule #2.

(B) provides these restrictions: the first portion of the rule covers setup C (2 < 3,4,5,6) and the
second portion covers A,B, and D (6,7,8 >4,5).

[chewdak]

This is the only game among hundreds I did completely in my head.

Calculating that each board value adds to 9 and the processors it holds is a very simple version of Gauss summation: 1 + 8, 2 + 7, 3 + 6, 4 + 5.

From the rules,
T cannot hold 3-6 and 1-8, then it can hold 4-5 and 2-7.
U cannot hold 4-5.

The Gauss summation is well-known. If you are not familiar with it you might lose time.

http://www.newton.dep.anl.gov/askasci/m ... h99155.htm

Edit: I just read the Q7 discussion, I did find it more difficult than others
but it's easy once you remember that
T must hold 4-5 or 2-7
if it holds 2-7, U must hold 1-8, and T holds higher value processor (7) than any in R,S
if T holds 4-5, every other board will have higher value processor than those two.