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Puzzle Corner 32

Puzzle Corner 32. Hotel Morrison Manzarek Suite [From an original puzzle by Barry Clarke]. What do we know?. Closing door … A unlocks B&E B unlocks C&E C unlocks D&F D unlocks A&F E unlocks B&D F unlocks C&D There are 6 corresponding bolts on door G to be unlocked.

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Puzzle Corner 32

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  1. Puzzle Corner 32 Hotel Morrison Manzarek Suite [From an original puzzle by Barry Clarke]

  2. What do we know? • Closing door … • A unlocks B&E • B unlocks C&E • C unlocks D&F • D unlocks A&F • E unlocks B&D • F unlocks C&D • There are 6 corresponding bolts on door G to be unlocked

  3. A → B&E; B → C&E; C → D&F; D → A&F; E → B&D; F → C&D What do we know? • The doors A and F are entrances to the bedroom and door B is an exit. Lisa must alternate exits and entrances. • Clearly, the minimum number of doors that need to be passed through is 6 (one for each bolt on door G). To achieve this requires each door (A-F) to be used only once. This is possible, and the required order can be deduced before going through any of the doors!

  4. A → B&E; B → C&E; C → D&F; D → A&F; E → B&D; F → C&D Labelling potential solutions • Let the unknown entrance be X. With a "_" denoting an exit door the possible orders are: • (a) _A_F_X ; (b) _F_A_X ; (c) _X_A_F ; (d) _X_F_A ; (e) _A_X_F ; (f) _F_X_A

  5. A → B&E; B → C&E; C → D&F; D → A&F; E → B&D; F → C&D Eliminating impossible solutions • (a) _A_F_X ; (b) _F_A_X ; (c) _X_A_F ; (d) _X_F_A ; (e) _A_X_F ; (f) _F_X_A • Neither (a) nor (c) are possible since A must be followed by B or E and neither opens F.

  6. A → B&E; B → C&E; C → D&F; D → A&F; E → B&D; F → C&D Eliminating impossible solutions • (a) _A_F_X ; (b) _F_A_X ; (c) _X_A_F ; (d) _X_F_A ; (e) _A_X_F ; (f) _F_X_A • Since A can only be preceded by D, (e) and (f) are possible as DA_X_F and _F_XDA, respectively. With D used, door F can only be preceded by C to give DA_XCF and CF_XDA, respectively. Door B must occupy the remaining space in each possibility (third door is exit) so X=E, but C cannot follow E in the former and for the latter, B cannot follow F. This eliminates (e) and (f).

  7. A → B&E; B → C&E; C → D&F; D → A&F; E → B&D; F → C&D Eliminating impossible solutions • (a) _A_F_X ; (b) _F_A_X ; (c) _X_A_F ; (d) _X_F_A ; (e) _A_X_F ; (f) _F_X_A • Since only D precedes A, (b) is possible as _FDA_X and (d) as _X_FDA. For (d), the exit B must occupy a "_" but cannot be followed by F, and if it is the first door, neither the C or E that must follow as X can be followed by the remaining E or C, respectively. This rules out (b).

  8. A → B&E; B → C&E; C → D&F; D → A&F; E → B&D; F → C&D Remaining viable solution • (a) _A_F_X ; (b) _F_A_X ; (c) _X_A_F ; (d) _X_F_A ; (e) _A_X_F ; (f) _F_X_A • This only leaves (b). Since D must precede A, only C (of the possible B,C,E) can precede F as CFDA_X and B must occupy the final "_" as an exit leaving X=E. • Hence, the final sequence must be CFDABE

  9. The End

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