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| Thread: JJ's logic puzzles: Sleuth, Mastermind, Egghead |  This thread is pages long: 1 2 · NEXT» |
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JollyJoker
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posted November 13, 2008 08:29 AM |
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JJ's logic puzzles: Sleuth, Mastermind, Egghead
*Pulled from the Math thread.*
Half a lifetime ago I used to make logic puzzles for a German gaming magazine. They were pretty popular then, and maybe you'll enjoy them as well.
I'll begin with an introductory riddle for Sid Sackson's well-known boardgame SLEUTH, a card game where up to 7 players try to find one or more missing gems. The game consists if 36 Gem cards and a plethora of asking cards. Gems have three defining qualities:
1) color (red, blue, green, yellow)
2) kind (diamonds, pearls, opals)
3) number (solitaire, twin, triple)
Asking cards either ask for numbers of one or two of those qualities.
Gem cards are shuffled and one or more of them will be pulled out - the missing ones. The rest is dealt face down to the players in equal parts. An eventual rest will be turned open, so everyone sees them. Game consists of getting informations about the other players' cards and concluding what the missing gem(s) are.
What you need for solving this, is a piece of paper and a pen that you will divide in FOUR rows and 9 columns (3x3), so you get a roster. You'll name the rows from top to bottom:
Red
Blue
Green
Yellow
and the columns from left to right:
Dia 1, Dia 2, Dia 3, Pearl 1, Pearl 2, Pearl 3, Opal 1, Opal 2, Opal 3
In this introductory difficulty beginner example ONE card is missing, and FOUR (4) players are participating, which means that each player is dealt 8 cards face-down and the remaining three cards are turned face-up.
The 3 open cards are: red Dia 1, blue Dia 3, green Pearl 1
You are Player A and you have: blue Dia 1, green Dia 1, green Dia 2, yellow Dia 2, yellow Pearl 1, blue Pearl 3, blue Opal 1, red Opal 3.
You have the following informations about the other players cards:
Player B: 4 Pearls, 2 Triples, 2 Green, 0 Pearl triples, 0 yellow Pearls, 1 yellow twin.
Player C: 2 Diamonds, 3 Reds, 2 Blues, 3 Opal triples, 0 red Opals, 1 red solitaire, 2 green triples
Player D: 3 Opals, 4 Twins, 4 Yellows, 1 blue Diamond, 0 red triples
Find the missing one.
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Asheera
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Elite Assassin
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posted November 13, 2008 03:47 PM |
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This is what we have for a start:
D1 D2 D3 P1 P2 P3 O1 O2 O3
R 0 1
B 1 0 1 1
G 1 1 0
Y 1 1
You should notice the notation: a 0 are those that are open cards, a 1 are player A's, a 2 player B's, etc.
Player B: 1 Yellow twin and no Yellow Pearls gives the only possibility of Yellow Opal 2.
Player C: 3 Opal Triples gives the only possibility of Blue + Green + Yellow Opal 3.
Player D: 1 Blue Diamond gives the only possibility of Blue Diamond 2.
So far, we have:
D1 D2 D3 P1 P2 P3 O1 O2 O3
R 0 1
B 1 4 0 1 1 3
G 1 1 0 3
Y 1 1 2 3
Player C: 1 Red Solitaire & 0 Red Opals  Red Pearl 1.
Player B: 4 Pearls & 0 Pearl Triples & 0 Yellow Pearls Red + Blue + Green Pearl 2, Blue Pearl 1
Player C: 2 Blues Blue Opal 3 (which we had checked) and Blue Opal 2.
So far, we have:
D1 D2 D3 P1 P2 P3 O1 O2 O3
R 0 3 2 1
B 1 4 0 2 2 1 1 3 3
G 1 1 0 2 3
Y 1 1 2 3
Now, Player B has two triples and no Pearl triples, which means that he has 2 Diamond Triples. Then, Player C also has 2 Diamonds, leaving only one more diamond left. And, since Player D has 4 Yellows, it means he has Yellow Opal 1, Yellow Pearl 3, Yellow Pearl 2 AND one of the remaining yellow diamonds (Yellow Diamond 3 or Yellow Diamond 1). We don't know which one yet, but at least we know the other 3 yellows.
Player D also has 4 Twins. He has Yellow Pearl 2 (already checked), Blue Diamond 2 (already checked), Red Diamond 2 can't be as all the diamonds have been occupied by the other players, leaving us with with Red Opal 2 + Green Opal 2 to check for player 4.
So far, we have:
D1 D2 D3 P1 P2 P3 O1 O2 O3
R 0 3 2 4 1
B 1 4 0 2 2 1 1 3 3
G 1 1 0 2 4 3
Y 1 1 4 4 4 2 3
Knowing that player D has 8 cards, and:
- we have 6 displayed
- one of them is a Yellow Diamond, and all other diamonds are filled by other players.
- the other is not an Opal, since we already have 3 displayed and he has only 3 Opals.
- has no Red Triples
Therefore, the only possibility for his other card is Green Pearl 3
Now, for Player C: 2 Green Triples, leads to the only possibility of Green Opal 3 (checked) and Green Diamond 3 (which we check now)
So far, we have:
D1 D2 D3 P1 P2 P3 O1 O2 O3
R 0 3 2 4 1
B 1 4 0 2 2 1 1 3 3
G 1 1 3 0 2 4 4 3
Y 1 1 4 4 4 2 3
Player B: 2 Triples & No Pearl Triples Red Diamond 3 and Yellow Diamond 3.
Player D: 4 Yellows Yellow Diamond 1
Player C: 2 Diamonds Red Diamond 2
So far, we have:
D1 D2 D3 P1 P2 P3 O1 O2 O3
R 0 3 2 3 2 4 1
B 1 4 0 2 2 1 1 3 3
G 1 1 3 0 2 4 4 3
Y 4 1 2 1 4 4 4 2 3
Player C: 3 Reds & No Red Opals Red Pearl 3
Player B: 2 Green Green Opal 1.
So the result:
D1 D2 D3 P1 P2 P3 O1 O2 O3
R 0 3 2 3 2 3 4 1
B 1 4 0 2 2 1 1 3 3
G 1 1 3 0 2 4 2 4 3
Y 4 1 2 1 4 4 4 2 3
It seems Red Opal 1 is missing 
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JollyJoker
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Undefeatable Hero
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posted November 13, 2008 04:49 PM |
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Right. Nice work, Ash.
But of course it's only introductory level. 
So I'd like some feedback, whether
a) I should continue this
and b) how to do it to maximize the fun concerning posting of solutions.
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Asheera
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Elite Assassin
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posted November 13, 2008 05:18 PM |
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It was only the introductory level? It was pretty hard
Maybe you should continue I'm not sure I'll be able to solve all of them if this was the easiest, but I hope some others join in as well 
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JollyJoker
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Undefeatable Hero
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posted November 14, 2008 08:49 AM |
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Ok, Ash, it looks like no one would be interested in even writing a comment, so I'll give you another introductory one, just for you then.
The setting is the same as above.
Open Gems: Red Dia 1, Yellow Pearl 2, Blue Opal 2
Player A: Red Dia 2, Green Dia 2, Red Pearl 2, Blue Pearl 1, Blue Pearl 2, Blue Opal 3, Green Opal 2, Yellow Opal 1.
Player B: 4 Pearls, 2 Greens, 2 Red Opals, 2 Blue Diamonds, 1 Yellow Solitaire
Player C: 3 Greens, 1 Opal Solitaire, 2 Opal Triples, 1 Green Pearl, 0 Blue Solitaires
Player D: 3 Yellows, 2 Diamond Triples, 1 Red Diamond, 1 Red Opal, 0 Yellow Twins
This one should be easy indeed.
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Asheera
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Elite Assassin
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posted November 14, 2008 02:39 PM |
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We start with:
D1 D2 D3 P1 P2 P3 O1 O2 O3
R 0 1 1
B 1 1 0 1
G 1 1
Y 0 1
Played D has 1 Red Diamond Red Diamond 3
Now, since Player B has 2 Red Opals, and Player D has 1 Red Opal, then Player C can't have any Red Opals. And we also know that Player C has 1 Opal Solitaire and no Blue Solitaires, so the only possibility is the Green Opal 1.
Player C also has 2 Opal Triples, and according to the fact that he can't have any Red Opals, the only possibility is to have Green Opal 3 + Yellow Opal 3.
So, we have:
D1 D2 D3 P1 P2 P3 O1 O2 O3
R 0 1 4 1
B 1 1 0 1
G 1 3 1 3
Y 0 1 3
Player B has 2 Blue Diamonds, 2 Red Opals, 2 Greens, 1 Yellow Solitaire and 4 Pearls: since 2 Blue Diamonds, 2 Red Opals and 1 Yellow Solitaire can't be Green, this means 2 of those 4 Pearls are the 2 Greens, and that the Yellow Solitaire is also a Pearl, leaving the only possibility of Yellow Pearl 1 (because we know there are 8 cards for a player)
Now Player C also has 1 Green Pearl, which means Player D has 2 Green Diamonds (or 1, if one of them is the one missing)
Also, Player D has 3 Yellows, but no Yellow Twins, which now gives the only possibility of Yellow Dia 1, Yellow Dia 3 and Yellow Pearl 3.
So, we have:
D1 D2 D3 P1 P2 P3 O1 O2 O3
R 0 1 4 1
B 1 1 0 1
G 1 3 1 3
Y 4 4 2 0 4 1 3
Well I just realized now that Green Diamond 3 can't be present to any player.
Player B has 2 Blue Diamonds, 2 Red Opals and 4 Pearls, and can't have more cards.
Player C has three Greens, one of which is a Green Pearl and the others the two Green Opals (1 and 3)
Player D has only two Diamond Triples, and these are Red Dia 3 and Yellow Dia 3. So no Green Dia 3 here either.
So the missing one is Green Diamond 3, and I didn't even need to complete the whole table
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JollyJoker
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posted November 14, 2008 03:36 PM |
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That's right. You didn't need to complete the whole table.
So, since you seem to enjoy this, I'll give you another one which should give you a bit more to think about.
Now, this one involves FIVE players, one card missing, so everyone has 7 cards, no open ones.
Player A: Green Dia 2, Green Dia 3, Red Pearl 3, Yellow Pearl 2, Red Opal 3, Blue Opal 2, Yellow Opal 3.
Player B: 1 Blue, 4 Greens, 2 Diamond Solitaires, 1 Diamond Twin, 1 Red Pearl
Player C: 2 Opals, 2 Pearl Triples, 1 Yellow Diamond, 1 Green Opal, 1 Red Twin, 1 Blue Twin
Player D:: 4 Yellows, 2 Blue Diamonds, 1 Red Solitaire
Player E:: 1 Diamond, 3 Reds, 2 Opal Twins, 1 Green Pearl, 1 Blue Opal, 1 Yellow Triple
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Asheera
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Elite Assassin
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posted November 14, 2008 04:45 PM |
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Edited by Asheera at 20:17, 14 Nov 2008.
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Ok I'm pretty busy so I'll try to solve it later or tomorrow.
EDIT: Just a quick look up and I see that Player C has only 6 cards (since 2 Opals includes 1 Green Opal) - is the last card intended to be not described?
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JollyJoker
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Undefeatable Hero
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posted November 14, 2008 08:29 PM |
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Well, we have a lot of players in the examples until now that don't have all cards described, haven't we? We have an example where we could pinpoint the missing card of a player and we have an example where we didn't need all cards of a player...
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Asheera
Honorable
Undefeatable Hero
Elite Assassin
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posted November 14, 2008 09:32 PM |
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Ok I started this for a bit and here's what I got so far. I'll continue tomorrow.
The start:
D1 D2 D3 P1 P2 P3 O1 O2 O3
R 1 1
B 1
G 1 1
Y 1 1
Player B has 7 cards, so the fact that he has 2 Diamond Solitaires, 1 Diamond Twin, 1 Red Pearl and 4 Greens means that one of the first three is Green. The Red Pearl can't obviously and neither can the Diamond Twin (Green Diamond 2 is Player A's). So this means that player B has a Green Diamond 1.
And because he has 1 Blue, it also means that he has a Blue Dia 1 or Blue Dia 2 as well. Also, Player D has 2 Blue Diamonds, which means that Player C, who has 1 Blue Twin, must have the Blue Pearl 2 (Blue Opal 2 has been taken by Player A). Moreover, this also means that Player D has Blue Dia 3.
Player E has 7 cards: 3 Reds, 2 Opal Twins, 1 Green Pearl, 1 Blue Opal and 1 Yellow Triple. The previous list points to 8 cards, unless some of them 'intersect'. The only possibility is the 3 Reds to intersect with the 2 Opal Twins. Therefore, Player E has Red Opal 2.
So, we have:
D1 D2 D3 P1 P2 P3 O1 O2 O3
R 1 5 1
B 4 3 1
G 2 1 1
Y 1 1
btw, I just analyzed the whole thing a lot and figured out that nobody can have Blue Pearl 1 except Player C since he has one card not described. Every description either has a different color, kind or number, except Player B's: 1 Blue, which we found out above that it's a Blue Diamond, so that one gets ruled out as well.
However, the problem is that the Blue Pearl 1 can also be the one missing so I can't write it as Player C's.
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jiriki9
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Altar Dweller
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posted November 15, 2008 10:42 AM |
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Hmmm, I tried a bit here, too, but I do not see, why Player B necessarily should have Blue Diamond 1 or Blue Diamond two I mean, why shouldn't it be a Blue Pearl or Opal?!?
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JollyJoker
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Undefeatable Hero
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posted November 15, 2008 11:22 AM |
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Ash has established that B must have Green Diamond Solitaire. This finishes the green Diamonds completely, do B cannot have more than 1 green Diamond. If B has NO blue Diamond, the remaining 2 of his diamonds must be either Red or Yellow, which would give him 4 Greens, 1 Blue, 1 Red Pearl plus another 2 red/yellow diamonds, giving him, 8 cards.
Which means that his 1 Blue must be Dia 1 or Dia 2.
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JollyJoker
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Undefeatable Hero
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posted November 15, 2008 11:36 AM |
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Err,
SORRY!
I mistyped something. 
C has TWO pearl triples instead of ONE.
I corrected this in listing.
Sorry, again, this shouldn't happen. 
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jiriki9
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Altar Dweller
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posted November 15, 2008 12:35 PM |
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That could help a bit...anyway, I'm away for a short while now (not out of the houee, just downstairs^^), but I'll give it a try later - AND: All the rest of the blue cards, that is at least 2 (if a blue card is missing), are in the hand of player C (Tere have to be at least two mroe blue cards, and no other player oculd have them)! I feel that could be of importance further on^^
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Asheera
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Elite Assassin
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posted November 15, 2008 01:51 PM |
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Oh, so now that we have all Player C's cards described in some way, I can conclude that nobody can have Blue Pearl 1, so that's the missing one
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JollyJoker
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Undefeatable Hero
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posted November 15, 2008 02:15 PM |
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While this right, you can pinpoint things a lot better.
E has a Yellow Triple which may be Pearl or Diamond. If it was a Diamond, E had no red Diamonds (since he has only one). In that case E had one Red Pearl (A and B have one each) and the Red Opal Solitaire to his Red Opal Twin. D's Red Solitaire and C's Red Twin were Diamonds then. We have one unaccounted-for Diamond Solitaire/Pair for B (1 of his Diamonds is Green, the other Blue) which had to be yellow then. C has a yellow Diamond as well (and we started with E having the yellow triple). We could fill in now D's 4 yellows. E's second opal twin would be green. C has a green Opal and E a green Pearl, and B's remaining 3 Greens would fill the greens up, but in that case C couldn't have 2 Pearl Triples anymore.
Which means that E's Yellow Triple must be the Pearl.
C's two Pearl Triples are then blue and green. We can fill the Green row up now with B's remaining Greens, C's Green Opal and E's Green Pearl. Which means, that E's second Opal Twin is Yellow. D's 4 Yellows plus C's Yellow Diamond fill up the Yellow row completely, so B's last missing Diamond must be Red. We can fill now the Red row completely with B's 2 Reds, (A's two Reds), E's 3 Reds plus C's Red Twin and D's Red Solitaire.
C's Blue Twin has to be a Pearl. Moreover C has two Opals, one of which is Green, since the yellow and red ones are filled, C must have a blue Opal as his second one. E has a Blue Opal as well.
Which leaves the Blue Pearl Solitaire.
I'll give you a weekend puzzle in a few.
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JollyJoker
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posted November 15, 2008 03:03 PM |
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Okay, this may well be the last puzzle I'm doing, since this doesn't seem to find too many friends.
Anyway, we are back to 4 players, but this time TWO Gems are missing.
Open Cards: Green Dia 2, Green Opal 1
Player A: Red Dia 3, Yellow Dia 1, Blue Pearl 2, Green Pearl 3, Yellow Pearl 3, Red Opal 3, Blue Opal 3, Yellow Opal 2
Player B: 2 Greens, 0 Diamond Solitaires, 2 Opal Twins, 1 Blue Diamond, 1 Green Solitaire, 1 Blue Twin, 1 Yellow Triple
Player C: 3 Opals, 3 Reds, 2 Pearl Twins, 2 Blue Solitaires, 1 Yellow Solitaire
Player D: 2 Triples, 3 Yellows, 0 Diamond Twins, 1 Diamond Triple, 1 Yellow Opal, 1 Red Twin, 1 Blue Triple.
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TheDeath
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with serious business
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posted November 15, 2008 03:05 PM |
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Hey keep them coming, I may not be very active since I don't really have a lot of time but I like to read Ash's or anyone else's reasoning about it
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The above post is subject to SIRIOUSness.
No jokes were harmed during the making of this signature.
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Asheera
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Elite Assassin
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posted November 15, 2008 09:43 PM |
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The start:
D1 D2 D3 P1 P2 P3 O1 O2 O3
R 1 1
B 1 1
G 0 1 0
Y 1 1 1
Player B has 0 Diamond Solitaires and 1 Green Solitaire, which gives the only possibility of Green Pearl 1 (since Green Opal 1 is an open card)
Player D has 3 Yellows, 0 Diamond Twins and only 1 Yellow Opal - this means that he has at least 1 Yellow Pearl. Player C has 2 Pearl Twins - if one of them is Yellow Pearl 2, then Player D must have Yellow Pearl 1, Yellow Diamond 3, and one Yellow Opal (1 or 3). However, either Player B with a Yellow Triple or Player C with a Yellow Solitaire won't fit their description (no more yellow solitaires or triples, depending on which yellow opal Player D has in this case). Therefore, the two Pearl Twins of Player C are Red and Green.
Player D also has 1 Red Twin but 0 Diamond Twins, and since the Red Pearl 2 belongs to another player (as we found above), the only possibility is Red Opal 2.
Because of this, we know Player B's 2 Opal Twins as well: Blue Opal 2 and Green Opal 2.
So far, we have:
D1 D2 D3 P1 P2 P3 O1 O2 O3
R 1 3 4 1
B 1 2 1
G 0 2 3 1 0 2
Y 1 1 1
Player B has only 1 Blue Twin, and that one is Blue Opal 2 as we found out. He also has 0 Diamond Solitaires, so his Blue Diamond must be a Triple.
Player D can't have Yellow Opal 3 because he has only 2 Triples, and one of them being Yellow Opal 3 (in this case), the other one must be Blue Diamond 3 (since he also has 1 Diamond Triple and 1 Blue Triple in his description). So Player D has Yellow Opal 1.
Player D's 1 Blue Triple also gives the only possibility of Blue Pearl 3.
Player C, with his 1 Yellow Solitaire gives the only possibility of Yellow Pearl 1.
And now, Player D must have Yellow Dia 3 and Yellow Pearl 2 (because he has 0 Diamond Twins, only one Yellow Opal (which is Yellow Opal 1 as we found out), and 3 Yellows)
Player B's 1 Yellow Triple Yellow Opal 3.
The three Opals by Player C are the three remaining ones: Red 1, Blue 1 and Green 3.
So far, we have:
D1 D2 D3 P1 P2 P3 O1 O2 O3
R 1 3 3 4 1
B 2 1 4 3 2 1
G 0 2 3 1 0 2 3
Y 1 4 3 4 1 4 1 2
Green Dia 3 seems to be missing. Player D has no more Diamond Triples, Player B has no more Greens, and Player C has 6 cards displayed, with the other 2 being 1 Blue Solitaire and 1 Red.
Blue Dia 2 seems to be missing as well. Player B has only 1 Blue Diamond (Blue Dia 3), Player C has only two cards more not displayed, and those are 1 Blue Solitaire and 1 Red so no chance for Blue Dia 2 either. Player D has no Diamond Twins.
So the missing ones are Green Dia 3 and Blue Dia 2.
I hope I didn't do any mistakes.
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JollyJoker
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posted November 16, 2008 09:56 AM |
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You mistyped some things, but you did well.
Here comes your Sunday puzzle:
5 Players, 1 missing Gem:
Player A: Blue Dia 3, Green Dia 1, Yellow Dia 1, Red Pearl 3, Green Pearl 1, Blue Opal 1, Yellow Opal 2.
Player B: 2 Pearls, 2 Diamond Twins, 1 Opal Twin, 1 Red Opal
Player C: 1 Triple, 3 Blues, 1 Opal Solitaire, 1 Green Diamond, 2 Yellow Pearls, 0 Blue Triples
Player D: 2 Solitaires, 1 Opal Triple, 2 Red Diamonds, 2 Blue Pearls, 1 Blue Twin, 1 Yellow Twin
Player E: 2 Twins, 3 Reds, 1 Yellow, 2 Diamond Triples, 2 Pearl Twins, 1 Blue Solitaire
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