Logic bash..
So we can ignore left + right and just consider adjacency, and thus we know one of each of these pairs are adjacent:
bd or cd
pq or cq
bk or ck
cp or dp
cg or ck
dg or cg
bp or bg
Case 1: If ck:
Case 1a: if cd, then we have dck =>
bdck =>
impossible
Case 1b: if cq, then we have qck =>
qck,bd =>
qck,bdp =>
qck,gbdp =>
impossible
Case 1c: otherwise, we have ck,bd,pq =>
either kcpq,bd=>
kcpq,bdg =>
impossible, or ck,bdpq =>
gck,bdpq =>
kcgbdpq.
Case 2: Otherwise, must have bk,gc. b is next to p or g, therefore not d, so c is next to d. So we have bk,gcd =>
bk,gcdp =>
kb,gcdpq =>
kbgcdpq.
So, they both work; the former had both of Crang's statements true, which is still logically valid but I guess the second one was intended.
I heard the following cited from a book called Puzzles in Math and Logic. I don't know the answer, so the aim is to convince me your answer is correct
If we assume all planet's orbits are perfect circles centered around the sun (and we assume they all move at different speeds so that you could pick a spot for each planet in its orbit and at some point they will all be at the positions you choose), what planet is closest to Pluto most of the time?
[Last edited by TripleM at 07-17-2008 05:13 AM]