Save Dr. Lucky is a new Cheapass game which is a wonderfully clever meta sequal to Kill Dr. Lucky. The mechanics are essentially identical, except you're on the sinking Titanic and you want to save instead of kill Dr. Lucky. To save him you must be seen by another player (as opposed to Kill Dr. Lucky where to kill him you must not be seen by another player). The game's pace is faster than Kill Dr. Lucky since the ship is sinking, which actually destroys successive levels of the board, and so gameplay can't go on indefinitely!

The 10 F-2 failure cards each contain part of a logic puzzle. Here are the clues:

- Aboard the Ship were four couples: the
**Andersons**, the**Browns**, the**Carvers**, and the**Dawsons**. - The husbands' names, in no particular order, were
**Frank**,**Gary**,**Henry**, and**Isaac**. - The wives' names, in no particular order, were
**Selma**,**Thelma**,**Ursula**, and**Victoria**. - At 1:15 AM on the fateful day, Frank was jolted awake by the Browns, arriving home drunk to their stateroom.
- At 4:15 AM, Thelma left her stateroom and descended three decks for a secret rendezvous with Mr. Dawson, whom her husband never met.
- At 8:17 AM, over breakfast in the dining room, Isaac informed his wife that he was not on speaking terms with the Carvers.
- At 9:25 PM, Henry and Frank joined the Andersons in the Observation Lounge for a friendly game of Bridge.
- At 11:40 PM, three women heard a terrible scraping sound from their neighboring staterooms. They were Selma, Isaac's wife, and Mrs. Carver.
- At 12:15 AM on the morning after the incident, Ursula and Mrs. Dawson were the last to board Lifeboat number five.
- At 2:24 AM, as the stern slipped effortlessly in to the icy sea, Victoria's husband remarked aloud that he had no patience for games of chance.

Here is my solution:

We have 8 entities: F, G, H, I and S, T, U, V. There are 4 properties A, B, C, D. I write A(F) to mean Frank is an Anderson. I write m(F,S) to mean Frank and Selma are married.

Here's what we know from each rule in isolation:

4: not B(F)

5: not D(T)

6: not C(I)

7: not A(H) and not A(F)

8: not m(S,I) and not C(S) and not C(I)

9: not D(U)

In isolation, clue 6 tells us nothing clue 8 doesn't also tell us. But clue 6 is important since it interacts with clue 7: Isaac wouldn't play bridge with Mr. Carver if he's not on speaking terms with the Carvers, so if Isaac played bridge, Henry and Frank are not Carvers:

6&7: A(I) implies not C(H) and not C(F)

Equivalently, if Henry or Frank is a Carver, then Isaac did not play bridge:

6&7: C(H) or C(F) implies not A(I)

We deduce from clue 10 that Victoria's husband would not play bridge, so

7&10: not m(V,F) and not m(V,H) and not A(V)

We also know from clue 5 that Thelma's husband never met Mr. Dawson, therefore it's not possible that the 2 men played bridge (clue 7). So if Thelma's husband was a bridge player, Dawson was not:

5&7: m(T,H) implies not D(F)

5&7: m(T,F) implies not D(H)

5&7: A(T) implies not D(H) and not D(F)

Equivalently, If Dawson was a bridge player, Thelma's husband was not:

5&7: D(F) implies not m(T,H) and not A(T)

5&7: D(H) implies not m(T,F) and not A(T)

These puzzles get a little annoying with things like clues 5 & 6; can we assume that Isaac wouldn't talk about whether he's on speaking terms with the Carvers if he hadn't met Mr. Carver? I'll assume so:

5&6: D(I) implies not C(T)
Equivalently,

5&6: C(T) implies not D(I)

Here's a table showing what we know (X means an impossible combination; the subscript tells the rule(s) implying it). (The table can't show the implications like 6&7.)

Frank | Gary | Henry | Isaac | Anderson | Brown | Carver | Dawson | |

Selma | X_{8} |
X_{8} |
||||||

Thelma | X_{5} |
|||||||

Ursula | X_{9} |
|||||||

Victoria | X_{7&10} |
X_{7&10} |
X_{7&10} |
|||||

Anderson | X_{7} |
X_{7} |
||||||

Brown | X_{4} |
|||||||

Carver | X_{6,8} |
|||||||

Dawson |

Let's assume A(I) and see what happens. This tells us not A(H) and not B(I) and not D(I). Using 6&7 we know not C(F) and not C(H). That tells us D(F), which tells us not D(G) and not D(H), which tells us B(H), which tells us not B(G), which tells us C(G). This completes the lower left portion, telling us all the mens' surnames:

Frank Gary Henry Isaac Anderson Brown Carver Dawson Selma X _{8}X _{8}Thelma X _{5&7}X _{5&7}X _{5}Ursula X _{9}Victoria X _{7&10}X _{7&10}X _{7&10}Anderson X _{7}X _{7}X _{A(I)}YES _{assume}Brown X _{4}X _{B(H)}YES X _{A(I)}Carver X _{6&7}YES X _{6&7}X _{6,8}Dawson YES X _{D(F)}X _{D(F)}X _{A(I)}Using 5&7 and D(F), we know not m(T,H) and not A(T). A(I), therefore not m(I,T) and not m(T,V) (since in these puzzles married couples have the same last name). Thus m(I,U) and m(G,V), so not m(G,S) and not m(G,T) and not m(G,U). Also by the marriage last name rule, Frank Dawson can't be married to Thelma or Ursula, so m(F,S). But now we reach a contradiction: Henry can only be married to Selma or Ursula, but they're already married to Frank and Issac. And poor Thelma isn't married to anyone. Therefore our initial assumption A(I) has led to a contradiction, and so it must be that not A(I). Here's the table leading to the contradiction:

Frank Gary Henry Isaac Anderson Brown Carver Dawson Selma YES X X _{8}X _{8}Thelma X X X _{5&7}X X _{5&7}X _{5}Ursula X X YES X _{9}Victoria X _{7&10}YES X _{7&10}X X _{7&10}Anderson X _{7}X _{7}X _{A(I)}YES _{assume}Brown X _{4}X _{B(H)}YES X _{A(I)}Carver X _{6&7}YES X _{6&7}X _{6,8}Dawson YES X _{D(F)}X _{D(F)}X _{A(I)}

So let's backtrack now that we know not A(I). That implies A(H), which implies not B(H) and not C(H) and not D(H). So we have:

Frank | Gary | Henry | Isaac | Anderson | Brown | Carver | Dawson | |

Selma | X_{8} |
X_{8} |
||||||

Thelma | X_{5} |
|||||||

Ursula | X_{9} |
|||||||

Victoria | X_{7&10} |
X_{7&10} |
X_{7&10} |
|||||

Anderson | X_{7} |
X_{7} |
YES | X_{proved} |
||||

Brown | X_{4} |
X_{A(H)} |
||||||

Carver | X_{A(H)} |
X_{6,8} |
||||||

Dawson | X_{A(H)} |

Now let's assume m(T,H) since we'll be able to use 5&7 and this gives us lots of Xs in the upper left (not m(H,S) and not m(H,U) and not m(F,T) and not m(G,T) and not m(I,T)). 5&7; tells us not D(F). That gives us C(F), which gives us not C(G). Thus we have:

Frank | Gary | Henry | Isaac | Anderson | Brown | Carver | Dawson | |

Selma | X_{m(T,H)} |
X_{8} |
X_{8} |
|||||

Thelma | X_{m(T,H)} |
X_{m(T,H)} |
YES_{assume} |
X_{m(T,H)} |
X_{5} |
|||

Ursula | X_{m(T,H)} |
X_{9} |
||||||

Victoria | X_{7&10} |
X_{7&10} |
X_{7&10} |
|||||

Anderson | X_{7} |
X_{7} |
YES | X_{proved} |
||||

Brown | X_{4} |
X_{A(H)} |
||||||

Carver | YES | X_{C(F)} |
X_{A(H)} |
X_{6,8} |
||||

Dawson | X_{5&7} |
X_{A(H)} |

Now again use that married people have the same last name. Frank Carver can't be married to Selma, so m(F,U), which tells us not m(G,U) and not m(I,U). Now Isaac can only be married to Victoria, so not m(G,V), so m(G,S).

Frank | Gary | Henry | Isaac | Anderson | Brown | Carver | Dawson | |

Selma | X | YES | X_{m(T,H)} |
X_{8} |
X_{8} |
|||

Thelma | X_{m(T,H)} |
X_{m(T,H)} |
YES_{assume} |
X_{m(T,H)} |
X_{5} |
|||

Ursula | YES | X_{m(F,U)} |
X_{m(T,H)} |
X_{m(F,U)} |
X_{9} |
|||

Victoria | X_{7&10} |
X_{m(I,V)} |
X_{7&10} |
YES | X_{7&10} |
|||

Anderson | X_{7} |
X_{7} |
YES | X_{proved} |
||||

Brown | X_{4} |
X_{A(H)} |
||||||

Carver | YES | X_{C(F)} |
X_{A(H)} |
X_{6,8} |
||||

Dawson | X_{5&7} |
X_{A(H)} |

Now we know who's married to whom. Let's carry over the known last names of husbands to wives. m(F,U) and C(F) imply C(U), and thus not A(U) and not B(U) and not C(T) and not C(V). m(H,T) and A(H) imply A(T) and thus not B(T) and not A(S).

Frank | Gary | Henry | Isaac | Anderson | Brown | Carver | Dawson | |

Selma | X | YES | X_{m(T,H)} |
X_{8} |
X_{A(T)} |
X_{8} |
||

Thelma | X_{m(T,H)} |
X_{m(T,H)} |
YES_{assume} |
X_{m(T,H)} |
YES_{m(T,H)} |
X_{A(T)} |
X_{C(U)} |
X_{5} |

Ursula | YES | X_{m(F,U)} |
X_{m(T,H)} |
X_{m(F,U)} |
X_{C(U)} |
X_{C(U)} |
YES_{m(U,F)} |
X_{9} |

Victoria | X_{7&10} |
X_{m(I,V)} |
X_{7&10} |
YES | X_{7&10} |
X_{C(U)} |
||

Anderson | X_{7} |
X_{7} |
YES | X_{proved} |
||||

Brown | X_{4} |
X_{A(H)} |
||||||

Carver | YES | X_{C(F)} |
X_{A(H)} |
X_{6,8} |
||||

Dawson | X_{5&7} |
X_{A(H)} |