Diagram

[sethgrant]

I am definitely missing a huge inference in this game

can you please show me how you diagrammed it?

[noah]

This is a tough game in that it's tough to organize the information around the days of the week. I found it much easier to treat this as an open assignment game and organize around the number of garages of each value. It ends up leading to three frames. Check it out and try playing the game that way and see what you think.



I'd love to know how others solved it . . .

After a bit of review, I'd like to suggest that folks check out Angela's solutions below first - they're quite good and probably what most folks would want to do. When you're done with those, geek out with mine.

- Noah

[gregory.mortenson]

I'm curious to learn more of your thought process for 1) deciding that Open Assigment was a better fit than Numbered Ordering; and 2) decoding the inferences that led you to set up the frames. Was there something in particular about this question that tipped you off to Open Assignment? At one point in the setup did you realize that this game was a good candidate for frames?

I tried this game as Numbered Ordering and stumbled from the beginning, especially notating the constraints. And even after a time-consuming setup process I still missed the key inference that X was limited to 1 of the 5 days -- at what point in your setup did this inference become apparent?

[noah]

Good question.

For the first question, I didn't think numbered ordering would work because none of the questions ask about order -- except for the first orientation question, they're all about some parking lot costing this or that amount. The take-away: if a game seems difficult to classify, scan the questions to see what the focus of the game is. The constraints also had an open assignment smell to them; there's discussion of elements being more prevalent than others. However, I struggled with whether to make the bottom of my open board the dollar amounts or the parking lot types. I just re-played the game and built the board with X, Y and Z on the bottom, and it's possible to do. Go ahead and try it and see what sort of frames you can do -- and post what you figure out.

In terms of how I knew to try making frames, there's a limited number of elements and options, and the fact that X was limited in two ways made it seem like there wouldn't be much "wiggle room" for that element. You don't want to make frames when there's nothing splitting up the options that has further repercussions (so you look for inter-related constraints), and there are lots of possibilities. I recall that the first time I played the game, I got to question 20 (the third one) and realized I didn't have a real handle on the game because I didn't have a clue on how to answer that question other than try everything out, so I looked back to see what I could figure out.

I find that frames come up a lot more than people think. Sometimes the frames are extremely vague, to the point of being relatively useless, and once in a blue moon I see that I can exhaust the possibilities.

Tell me how it goes playing the game using X, Y and Z...

[angela]

I have diagrammed this as well - and explained the questions - and thought it might be helpful. I had a bit of a different take than Noah.

[sgorginian]

Angela - I had the same diagram you did. Except I had difficulty with some questions, especially # 20. I have printed your and Noah's solutions and will look at it - although again mine looks similar to yours.

Thank you for posting and sharing.

Sevan

[johnwonlee]

Hi Noah,

I just have some questions regarding the diagram you posted.

1) The document states "From the last two constraints we can infer that X cannot be $10, Z cannot be $12, Z must have at least two garages, and X can only have one." But couldn't Z be $12?

2) You provide three frames -- one frame with 3 Z's and two frames with 2 Z's -- but isn't there one additional frame with 3 Z's, in which Z is $12, X is $15, and Y is $10?

[noah]

You're the man! I totally goofed. The first one was just a typo - I meant to say that Z cannot be $15, but for the other frame, you're totally right. I'm updating the diagram right now.

I think I'm becoming a fan of Angela's approach!

- Noah

[cardsfan04]

Thanks for the diagrams. I had the same diagram as Angela, but didn't make the inference that X could only appear once. That makes perfect sense in hindsight, but I was going nuts trying to figure out what I was doing wrong.

[contropositive]

I am SOOO confused about question #20. Why can't we place Y for $15 on Wend. and Z for $12 on Friday and obviously X can go on Thursday.
This would satisfy the fact that Wend. is more than Friday. After all, Wend can be $12 or $15 and Friday can be $10 or $12.

[rinagoldfield]

Thanks, Royaimani. If Z is the $12 lot, then X must be the $15 lot, since X costs more than Z (second rule).

Anastasia can only park in X once since she parks in Z more times. (If she parked in X 2x, then she’d park in Z 3x and there’d be no room for Y).

Thus, Anastasia must park in X on only Thursday in this scenario. Z must be Wednesday, and Y, the $10 lot, must be Friday.

--Rina

[stacksdoe]

Royaimani-
[color=#BF0080]To clear up some vocabulary: I use slot, space, goes,position all to mean the space on the line diagram that a variable can be placed.

Rina has it, but I'll piggy back off her: There are a few conditional inferences that can be produced during the initial set up and they completely make this game much easier then it seems. Rules two and four produce the following 4 inferences: (1) If X is $15 then Z is either $10 or $15, (2) If X is $12 then Z must be $10 (because as the rule states "lot X cost more than lot Z", (3) If Wednesday is $15 then Friday is either $12 or $15, (4) And lastly, if Wednesday is $12 then Friday must be $10.
As a side note: the numerical relationship/distribution (i.e. the amount of slots to variables ratio is unfixed, i.e. the variables can be placed in more then one space, except for X; with a little bit of analysis, it becomes apparent that X can only be placed once in all cases. Here is how this works, lets say parks in lot X two days, this would mean that she must park in lot Z three days, since she parks in lot Z on more days then in lot X, but this would leave no empty space for lot Y, consequently violating what is stated in the scenario, that "she parks in each of the three lots at least once".
The above four inferences do make # 20 easier, and it also helps gain a better understanding of the whole game and with other questions, especially # 22.

Back to # 20. Instead of going into depth in how to arrive at the correct answer, because Rina has done that, I'll just answer your question. The reason why Y can not be slotted for $15 is because that would force X to be $12 since Z is $10, but the question stem dictates that Z be $12, that's one reason. Additionally, if you attempt to slot Wednesday and Thursday with lot X for $15, you would already violate the last rule( that Z be slotted more days than X) because if X is slotted twice then Z would have to take up three days, thus leaving no empty slot for Y; and as we know, each lot has to be used " at least once".
While I was proof reading this I also noticed an error that is giving you a headache. If you slot Y for $15, then you must place Y on Thursday, because rule 1 states that "Thursday she parks in the $15 lot". And if you did this, you would still force X to be $12, which contradicts the question stem, "that lot Z be $12).

I hope this clears up a lot.
I have found the more inferences you can produce at the onset the better of you will be when attacking the questions. [/color]

[stacksdoe]

I can't imagine how frames would be helpful in this game...
The crucial inference in this game is the numerical relationship. The number of variables compared to the number of available slots is a concept that is crucial to nearly all linear games. Furthermore, more times then not, there will be at least one question that attacks this component.
Here is how to asses the numerical relationship in this game:
After reading the scenario and the first three rules, all we can determined about the number of days compared to the empty slots (the days of the week) is that she will "park in each of the three lots at least once during he workweek". In other words, each lot will be slotted for at least one day, and since they are five empty slots (days) she will park in some lots more then others (which and how many, we find out in the last rule).
The last rule is the key to the numerical relationship: If she parks in lot Z on more days than in lot X, what does this tells us?
Well lets she say she parks in lot X two days, by this rule (rule 4) she would have to park in lot Z three days, but this can not be, because that would take up five days, leaving none for lot Y. Consequently, no matter what, she can only park in lot X on one of the days and only one. Since she will always park once in lot X, we know that she will always park in lot Z at least twice, to satisfy the last rule. But can she park in lot Z three times, damn right: she can park in lot Z three days, lot X one day, and lot Y one day. What about lot two days for lot Z, yup: she could park in lot Z two days, lot X one day, and lot Y two days, totaling five days.

[FaridL797]

Nowhere in the Rules and description does it say that the numbers are kinda fixed…, So how are we supposed to know that if X is 15 Y cannot be 15? It just says that they cost for the day.. Can anyone explain this?
Thank you!