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| Thread: JJ's logic puzzles: Sleuth, Mastermind, Egghead |  This thread is pages long: 1 2 · «PREV |
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Asheera
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posted November 16, 2008 06:33 PM |
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D1 D2 D3 P1 P2 P3 O1 O2 O3
R 1
B 1 1
G 1 1
Y 1 1
Blue Opal 2 is player C's. Here's why:
Player C has 3 Blues, 0 Blue Triples. Player D has 1 Blue Twin and 2 Blue Pearls, and Player E has 1 Blue Solitaire.
Because of Player D, a maximum of 1 Blue Pearl can be Player C's, therefore the only possibility of Player D not having Blue Opal 2 is if he has both Blue Diamonds. However, in that case, Player E (with 1 Blue Solitaire) won't fit the description, since all Diamonds and Pearls have been taken by others, and the Blue Opal Solitaire is Player A's.
Moreover, Player D's Blue Twin must be a Pearl, otherwise we have the same situation: Player D has 2 Blue Pearls (1 and 3, in this case), AND a Blue Dia 2 as well. Then, the other 2 Blues of Player C must be Blue Dia 1 and Blue Pearl 2. Then Player E's 1 Blue Solitaire won't fit anywhere. So, Player D must have Blue Pearl 2.
Player D also must have Blue Pearl 3, because in the case of Blue Pearl 1, Player C's 2 remaining Blues must be the two Diamonds (since he has no Blue Triples), and then Player E's 1 Blue Solitaire won't fit again.
Also, for the same reason, one of the two Blues of Player C must be Blue Dia 2.
Here's what we have so far:
D1 D2 D3 P1 P2 P3 O1 O2 O3
R 1
B 3 1 4 4 1 3
G 1 1
Y 1 1
Player C's Green Diamond can't be Green Dia 3, because if it is, then the 2 Yellow Pearls must be 1 and 2 since he has only one Triple. Then, Player D's 1 Yellow Twin must be Yellow Dia 2, also resulting in Player B's 2 Diamond Twins being Green and Red. In this case, Player D's 2 Red Diamonds will be 1 and 3, and Player E's 2 Diamond Triples won't fit (he'll have space for only Yellow Diamond 3)
So, Player C's Green Diamond is Green Dia 2, and one of his Yellow Pearls must be a Triple as well (otherwise he'll have 8 cards, because he has 2 known, one more Blue non-Triple, one Green Diamond Twin, one Opal Solitaire, one Triple and two Yellow Pearls non-triple (in our case))
Moreover, Player B's two Diamond Twins must be Red Dia 2 and Yellow Dia 2, Player D's Yellow Twin must be Yellow Pearl 2, Player C's second Yellow Pearl must be a Solitaire in this case, Player E's 2 Pearl Twins are now known as Red and Green, Player D's 2 Red Diamonds have no other possibility than 1 and 3, and finally Player E's 2 Diamond Triples must be Green and Yellow in this case.
Now, since Blue Dia 1 and Blue Pearl 1 are taken (by Player C's 1 more Blue non-triple and Player E's 1 Blue Solitaire), Player B's 2 Pearls are Red Pearl 1 and Green Pearl 3.
So far, we have:
D1 D2 D3 P1 P2 P3 O1 O2 O3
R 4 2 4 2 5 1
B 3 1 4 4 1 3
G 1 3 5 1 5 2
Y 1 2 5 3 4 3 1
It seems the missing one is an Opal.
Player D has 1 more Opal Solitaire (because the Diamond and Pearl Solitaires will be taken by C and E) and 1 more Opal Triple, Player E has 2 more Red Opals, Player C has 1 more Opal Solitaire, and Player B has 1 more Red Opal and an Opal Twin (could be this Red one) and some other card(s).
The Red Opals are taken by Player E and B, while the Opal Solitaires are completed by Player C and D.
So the missing one is either a Blue, Green or Yellow Opal Triple, or Green Opal 2.
I actually got stuck now. I only know that Player D has 1 Opal Triple (not Red since all of those were taken), but I don't have enough information about Player B's cards...
btw, I'm getting pretty busy these days, I don't think I'll be as active as before to solve them daily...
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JollyJoker
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posted November 16, 2008 09:02 PM |
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Take your time. I'm not giving one daily; I give one when the last one is solved.
Which means, I give a new one after you have solved it. 
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JollyJoker
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posted November 17, 2008 12:08 PM |
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I've read your solution so far, and it's basically solved - you just need to look at the info again.
For E you can make a pretty definite statement concerning his Red Opals and all 7 of his Gems. You can make a pretty definite statement about B's red Opal as well then and while you cannot pinpoint B's cards, you can pinpoint something B cannot have.
For C and D you have already pinpointed them all or can make definite statements as well what they are and therefore cannot be.
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Asheera
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Elite Assassin
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posted November 17, 2008 02:08 PM |
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Aha, I get it. Player E can't have another twin, so he has Red Opal 1 and 3, therefore Player B has Red Opal 2, which happens to be his only Opal Twin as well. Player C has 1 more Opal Solitaire, and Player D has 1 Opal Solitaire and 1 Opal Triple.
So the Green Opal 2 is missing.
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JollyJoker
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posted November 17, 2008 03:02 PM |
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Exactly.
Do you want another one or would you like to try Mastermind or Egghead. Or do you have enough of it?
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Asheera
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Elite Assassin
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posted November 17, 2008 04:29 PM |
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You can post another one. Although I'll not be very active I'll try to solve it at some time.
At worst I'll get very busy and forget about this thread... I hope not though.
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Jiriki9
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Altar Dweller
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posted November 17, 2008 05:05 PM |
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Give another one, I'll take a try! I like this very much allthough I was not very succesfull up to now...
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JollyJoker
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posted November 17, 2008 05:30 PM |
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This next one then sees 6 players with TWO Gems missing. That means, evey player has FIVE cards, plus there are FOUR open ones:
Open Cards: Blue Diamond Twin, Red Pearl Triple, Yellow Pearl Triple, Yellow Opal Twin
Player A: Red Diamond Solitaire, Green Diamond Triple, Yellow Diamond Solitaire, Green Pearl Twin, Green Opal Solitaire
Player B: 2 Diamonds, 2 Solitaires, 2 Greens, 2 Pearl Twins, 1 Red Triple,
Player C: 1 Diamond TRiple, 1 Yellow Opal, 1 Blue Solitaire, 1 Green Twin, 1 Green Triple
Player D: 3 Pearls, 0 Reds, 1 Pearl Solitaire, 2 Opal Twins
Player E: 0 Opals, 2 Blues, 2 Triples, 1 Diamond Twin, 1 Red Pearl, 1 Yellow Solitaire,
Player F: 2 Yellows, 1 Diamond Solitaire, 2 Opal Triples, 1 Red Opal, 1 Red Twin
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Asheera
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Elite Assassin
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posted November 18, 2008 05:06 PM |
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D1 D2 D3 P1 P2 P3 O1 O2 O3
R 1 0
B 0
G 1 1 1
Y 1 0 0
First of all Player E's 1 Yellow Solitaire can be only Yellow Pearl 1 (because he has 0 Opals)
Player F must have Yellow Opal 3. Otherwise he'll have more than 5 cards, because he'll have 2 Yellows + 1 Diamond Solitaire + 2 Opal Triples + 1 Red Twin = 6 cards.
Now, Player C's 1 Yellow Opal = Yellow Opal 1.
Player E has 1 Diamond Twin, 1 Red Pearl, 1 Yellow Solitaire and 2 Blues. Those can't 'intersect' (because Blue Dia 2 is an open card), so that makes 5 different cards. Therefore, the 2 Triples 'intersect' with those. With the 1 Yellow Solitaire, 1 Diamond Twin and 1 Red Pearl they can't (Red Pearl 3 is an open card), so that leaves the 2 Blues. Which means, Player E has 2 Blue Triples, and because of 0 Opals, this means Blue Dia 3 and Blue Pearl 3.
Because of Player B, two Pearl Twins are filled so there's only one space left. Now, moving to Player D, which has 3 Pearls and only 1 Pearl Solitaire, this means he must have 1 Pearl Solitaire, 1 Pearl Twin and Green Pearl 3. Moreover, Player D has 0 Reds so Red Pearl 2 can only be Player B's.
This way we also found out Player E's Red Pearl  Red Pearl 1.
This is what we have so far:
D1 D2 D3 P1 P2 P3 O1 O2 O3
R 1 5 2 0
B 0 5 5
G 1 1 4 1
Y 1 5 0 3 0 6
Player C's 1 Green Triple Green Opal 3.
Player D has 0 Reds and 2 Opal Twins Blue & Green Opal 2.
Now, Player C's 1 Green Twin Green Dia 2.
Player B has 2 Greens, so he takes the two more which are available: Green Dia 1 and Green Pearl 1.
Now, Player F's 1 Diamond Solitaire Blue Dia 1.
Player B has 3 cards displayed now, 1 more Diamond, 1 more Pearl Twin and 1 Red Triple. Therefore, the Red Triple must be a Diamond (can't be a Pearl Twin), so Player B has Red Dia 3.
And now Player C's 1 Diamond Triple is Yellow Dia 3, and Player F must have a Yellow Dia 2 in this case (the Yellow Pearl Twin is taken by either B or D)
We also know Player D's Pearl Solitaire, which is the remaining Blue Pearl 1.
So far, we have:
D1 D2 D3 P1 P2 P3 O1 O2 O3
R 1 2 5 2 0
B 6 0 5 4 5 4
G 2 3 1 2 1 4 1 4 3
Y 1 6 3 5 0 3 0 6
Player C's 1 Blue Solitaire Blue Opal 1. Now we know all C's cards.
Player E's 1 Diamond Twin Red Dia 2. Now we know all of E's cards as well. And it seems the missing ones are Opals.
Player F's 1 Red Twin Red Opal 2. And, because he has only 1 Red Opal and one more Opal Triple, he must have Blue Opal 3. We know all of F's cards as well.
So, we have:
D1 D2 D3 P1 P2 P3 O1 O2 O3
R 1 5 2 5 2 0 6
B 6 0 5 4 5 3 4 6
G 2 3 1 2 1 4 1 4 3
Y 1 6 3 5 0 3 0 6
Since the Pearl Twins are taken by B and D, the missing ones are Red Opal 1 and Red Opal 3.
@Jiriki: Do you want me to let you solve these as well? (that means I won't post them for a while)
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JollyJoker
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posted November 18, 2008 06:34 PM |
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Very nice. I'll try to post a new one today.
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