Which one of the following could be the toys included in the display?

Cal on May 29, 2019


I am missing something mentally on this, because going through I didn't have enough to set up. It doesn't say anything in the book that each color has to be used, whereas it will in the conditions or prompt in all the others.

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Ravi on May 29, 2019


Could you explain more what your question is? You're right, the rules
never say that each color has to be used. We just know that there are
two mauve dinosaurs and that the stegosaur is red.

Let's take a look at this game's setup.

We know that there are 7 dinosaurs—I, L, P, S, T, U, and V. We also
know that the dinosaurs on display are 1 of 4 colors—G, M, R, or Y.
The display has 5 toys, so 2 are out.

Rule 1 says 2 mauve toys are in.

m m _ _ _ | _ _
_ _ _ _ _ | _ _

Rule 2 says that S is red and is in the display

m m r _ _ | _ _
_ _ S _ _ | _ _

Rule 3 says that I is in only if it's green

I - >gI

Rule 4 says P is in only if it's yellow

P - >yP

Rule 5 says that V is in only if U is out

V - >/U
U - >/V

m m r _ _ |
_ _ S _ _ | U/V _

Note that for the out pieces, we don't need to keep track of their
colors, so we've removed the row above U/V

We also know that if at least 1 of U or V has to always be out, then I
and P can't ever both be out at the same time. This means that I or P
has to go in.

Rule 6 says that if both L and U and in, then at least one of them isn't mauve

L and U - >at least 1 not m

m m r _ _ |
_ _ S _ _ | U/V _

In looking at the board, we know that if L or U are both in, at least
1 can't be mauve. However, we know that I or P have to be in, so this
means that T has to be in as a mauve dinosaur.

L and U - >at least 1 not m and mT

With this setup, you should be well-equipped to tackle the questions.

Does this make sense? Let us know if you need any more clarification!

Cal on May 29, 2019

It does make more sense and I can follow how to get through the problems now, just was thrown off, had not seen a similar problem to this quite yet. Thank you.

Ravi on May 30, 2019

@Cal, you're welcome! Happy to see you know how to solve it now.