# Help with puzzle (May's Mystery 103)

**URL:** <https://boards.straightdope.com/t/help-with-puzzle-mays-mystery-103/610931>\
**Category:** The Game Room\
**Created:** [January 28, 2012, 12:48am UTC](https://boards.straightdope.com/t/help-with-puzzle-mays-mystery-103/610931 "2012-01-28T00:48:45Z")\
**Posts on this page:** 7\
**Page:** 1

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**Author:** ![Mr\_Shine](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/mr_shine/32/464_2.png) [@Mr\_Shine](https://boards.straightdope.com/u/Mr_Shine)\
**Post date:** [January 28, 2012, 12:48am UTC](https://boards.straightdope.com/t/help-with-puzzle-mays-mystery-103/610931/1 "2012-01-28T00:48:45Z")

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I just rented May’s Mystery’s for the DS. Some of the puzzles are good, some are pretty poor, some I just legitimately can’t solve. Puzzle 103 though just seems to me to be trivially unsolvable. Can anyone help?

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All blank spaces have roads, All roads go both horizontally and vertically and are 1km long.  
You need to place a school, church, bus station, hospital, restaurant and supermarket on letters A-F

The road from church to the bus station is 6km longer than church to supermarket

All buildings are different distances from the school

The hospital is the same distance from school and restaurant.

My reasoning: the Church has to go in spot A and the bus station in spot C as that is the only way to get a difference of 6km by road, and this assumes that you cannot go though the buildings. This means the supermarket must go in E or B.

The only places the Hospital, (and the school and restaurant) can go are  
D (EB), A(EB), C(AF) or F(BC) all 4 of which contradict the church, bus station and supermarket logic.

Anybody see what I’m missing?

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**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\
**Post date:** [January 28, 2012, 4:21am UTC](https://boards.straightdope.com/t/help-with-puzzle-mays-mystery-103/610931/2 "2012-01-28T04:21:59Z")

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> [@Mr\_Shine](#):
>
> My reasoning: the Church has to go in spot A and the bus station in spot C as that is the only way to get a difference of 6km by road, and this assumes that you cannot go though the buildings.

What about Church in A, and Bus Station in D, taking the path from A to D that goes through B?

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**Author:** ![Omphaloskeptic](https://avatars.discourse-cdn.com/v4/letter/o/bcef8e/32.png) [@Omphaloskeptic](https://boards.straightdope.com/u/Omphaloskeptic)\
**Post date:** [January 28, 2012, 4:27am UTC](https://boards.straightdope.com/t/help-with-puzzle-mays-mystery-103/610931/3 "2012-01-28T04:27:26Z")

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> [@Mr\_Shine](#):
>
> The only places the Hospital, (and the school and restaurant) can go are  
> D (EB), A(EB), C(AF) or F(BC) all 4 of which contradict the church, bus station and supermarket logic.

The shortest path from C to A is longer than the shortest path from C to F, so C(AF) probably isn’t right (unless we consider the scenic route from C). But as I count them there are two more possibilities you haven’t listed here, one with the hospital at C and one with it at F.

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**Author:** ![Mr\_Shine](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/mr_shine/32/464_2.png) [@Mr\_Shine](https://boards.straightdope.com/u/Mr_Shine)\
**Post date:** [January 28, 2012, 10:27am UTC](https://boards.straightdope.com/t/help-with-puzzle-mays-mystery-103/610931/4 "2012-01-28T10:27:30Z")

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> [@ZenBeam](#):
>
> What about Church in A, and Bus Station in D, taking the path from A to D that goes through B?

But the other route only takes 6 km

> [@](#):
>
> The shortest path from C to A is longer than the shortest path from C to F, so C(AF) probably isn’t right (unless we consider the scenic route from C). But as I count them there are two more possibilities you haven’t listed here, one with the hospital at C and one with it at F.

I’m assuming that when unqualified “distance” means “as the crow flies” rather than by road, otherwise the other condition about the school would not make any sense.

Thanks for looking, anyway.

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**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\
**Post date:** [January 28, 2012, 1:19pm UTC](https://boards.straightdope.com/t/help-with-puzzle-mays-mystery-103/610931/5 "2012-01-28T13:19:20Z")

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> [@Mr Shine](#):
>
> The only places the Hospital, (and the school and restaurant) can go are  
> D (EB), A(EB), C(AF) or F(BC) all 4 of which contradict the church, bus station and supermarket logic.

I don’t think C(AF) works, and what about C(BD) and F(DE)?  
“All buildings are different distances from the school”

Therefore, the school must be E. And therefore the supermarket is B, the hospital is F, and the restaurant D.

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**Author:** ![Mr\_Shine](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/mr_shine/32/464_2.png) [@Mr\_Shine](https://boards.straightdope.com/u/Mr_Shine)\
**Post date:** [January 28, 2012, 1:28pm UTC](https://boards.straightdope.com/t/help-with-puzzle-mays-mystery-103/610931/6 "2012-01-28T13:28:29Z")

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Cheers, just worked that out myself. :smack:

Somehow I’d got into the mindset that E had an identical road length somewhere so got too focused on the linear distances.

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**Author:** ![Mrs\_Smiley](https://avatars.discourse-cdn.com/v4/letter/m/7ba0ec/32.png) [@Mrs\_Smiley](https://boards.straightdope.com/u/Mrs_Smiley)\
**Post date:** [July 15, 2012, 2:42pm UTC](https://boards.straightdope.com/t/help-with-puzzle-mays-mystery-103/610931/7 "2012-07-15T14:42:26Z")

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A = Church  
B = Supermarket  
C = Bus Station  
D = Restaurant  
E = School  
F = Hospital
