by Mike Selinker

Solution: DOING TIME

Nomenclature for this answer: The ten characters are SEJ, COC, SHE, BIL, REV, MAD, DOC, KID, INS, and JEN. For each character, their first and second clues are XXX1 and XXX2. Upper cabins are #U and lower cabins are #D.

From JEN1, KID rented either 7U&1D or 14U&2D. From REV2, COC either rented 7D or 14D. From MAD1, COC only rented 2 cabins on the bottom floor, where one had a number twice the other. If COC rented 7D, he also rented 14D and vice versa. So COC rented 7D and 14D.

From SHE2, JEN only rented 1 cabin, one lower than COC's lowest bottom cabin. This is 6D. From KID2, REV rented a cabin 1 higher than one of COC's cabin. This can only be 8D. From COC2, 8U is empty.

From KID1 4U is empty. From KID1, 10U is empty. From MAD2, all bottom cabins are rented, and among the top cabins, there are two stretches of at least three cabins, and no other empty cabins. From REV1, he rented either 3D, 6D,

9D, and/or 12D. He did not rent 6D (above) or 9D (INS1). He rented either 3D or 12D, and either 3U or 12U is empty.

The following individuals had top cabins: MAD (BIL1), KID (JEN1), INS (COC1), SEJ (INS2), and DOC*2 (DOC2 and INS2). This accounts for 6 cabins of the 14. MAD2 accounts for 6 others, leaving another 2 that may or may not be empty. The most empty cabins there can be is 8.

If 12U is empty, then the fewest empty cabins there can be that connect the existing empty cabins is 8 (either 2-4/3-5/4-6 and 8-12 or 4-8/10-12).

* Suppose the empty cabins are 4-8/10-12. INS is between two occupied cabins (COC1) must be in 2U. KID must be 14U. This leaves 1, 3, 9, and 13. DOC is in two cabins that add up to SEJ's. There are no two numbers that make this possible.

* Suppose 8-12 are empty. If 3-5 is empty, there is no place for INS.

** If 2-4 is empty, INS is in 6, leaving 1,5,7,13,14.

** If 4-6 is empty, INS is in 2, leaving 1,3,7,13,14.

In neither case do two numbers add up to another.

12U cannot be empty, therefore REV rented 3D, and 3U is empty. The cabin at 10U is part of three consecutive empty cabins, which cannot include 12U. Therefore 9U is also empty.

SHE's only room is 1/4 the number of SEJ's lowest room (DOC1). SHE did not rent either 3U (empty) or 3D (REV). SEJ didn't rent either 8U (empty) or 8D (REV). Therefore, SHE rented either 1U or 1D, and SEJ rented 4U or 4D. If SEJ rented 4U, DOC's lowest top cabin must be 1U or 3U (INS2), but 3U is empty. SEJ rented 4D. Any bottom cabin rented by SEJ is next to another of his cabins. Since 3D is REV, SEJ rented 5D. This gives us:

???|???|   |   |???|???|???|   |   |   |???|???|???|???

MAD rented 2 cabins (BIL1), NU and N+3D. He can only have rented 6U&7D or 7U&10D. DOC's highest number cabins on the top deck must differ by 6 or 7.

If MAD rented 6, DOC rented 7&1, 11&5, or 13&7.

**If DOC rented 7&1, these must be the same 2 cabins in INS2, which puts SEJ in 8U, which cannot be true.

**If DOC rented 11&5, they cannot be the 2 cabins in INS2. 5U is rented, so 2U is empty. DOC must have also rented 1U, making SEJ 6U, which is already MAD's. This cannot be true.

**If DOC rented 13&7, they cannot be the 2 cabins in INS2. DOC rented 1U, 2U, and/or 5U. If DOC rented 1U and 5U, SEJ is 6U, which is already MAD's. If DOC rented either 1U or 2U only, SEJ would conflict with an empty cabin. DOC rented 5U and SEJ 12U. But now there is no place for INS. So this is not the case.

MAD rented 7U and 10D. KID must have rented 14U and 2D. DOC rented either 13&6 or 12&5.

* If it's 13&6, then these are not the 2 cabins in INS2. If DOC rented either 1U or 2U, SEJ would conflict with an occupied or empty cabin.

DOC rented 12&5. These are not the 2 cabins in INS2. If he rented 2U, SEJ would conflict with MAD. DOC rented 1U, leaving SHE with 1D. SEJ rented 6U, and 2U is empty. BIL rented a bottom cabin less than 10; this can only be 9D. INS can only be 13U.

DOC|   |   |   |DOC|SEJ|MAD|   |   |   |???|DOC|INS|KID

Now the question is who, if anyone, is in 11U. Of the 10, BIL, SHE, JEN, and REV are bottom only, KID, MAD, SEJ and INS only have 1 room, DOC has only 3, leaving COC or emptiness. Neither can be eliminated by the clues.

Of 11D-13D, BIL, SHE, JEN, KID, MAD, and INS have had all their rooms accounted for. COC is accounted for on the bottom level. This leaves SEJ, DOC, and REV (as we know they were rented, MAD2). DOC's 12U is adjacent to all 3, eliminating DOC (INS). REV cannot be in 13D (13U is not empty), so 13D is SEJ. This leaves 11D, which could be either SEJ or REV.

So we have these three possible solutions:

DOC|   |   |   |DOC|SEJ|MAD|   |   |   |   |DOC|INS|KID
DOC|   |   |   |DOC|SEJ|MAD|   |   |   |COC|DOC|INS|KID
DOC|   |   |   |DOC|SEJ|MAD|   |   |   |   |DOC|INS|KID

From SHE1, however, only the last of these possibilities is allowed, as the COC and REV have an even number of cabins.

Mapping the football plays onto the appropriate cabins produces this result:

                        __                          __
|\              |\   | |                   |\   /| |
| \  __ __      | \  | |   __  __   __     | \ / | |__
| / |     |  |  |  \ | |     |   | |    |  |     | |
|/  |__ __|  |  |   \| |__ __|   | |    |  |     | |__


Page title fixed in 2010 by the archivists of Beginner's Luck. Was "For each character, their first and second clues are XXX1 and XXX2"