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Subject: cityscape puzzle #1 rss

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Dave Dyer
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Playa Del Rey
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Cityscape ought to be a rich source of puzzles. I'm officially
offering 5 GG for improvement in the solution to the following
puzzle (or for a proof that the current solution is optimum)

Highest Total score (sum of all player scores) in a 4-player game.
Baseline: 600 points for the referenced picture.

To claim your gold, simply post a picture with your claim as a caption.



 
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mrbass
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this game is 50% off for $16.50 at amazon now. Just saying.
 
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I would do all the things I have ever dreamed of doing. I would love to become a professional whistler.I'm pretty amazing at it now, but I wanna get, like, even better. Make my living out of it.
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I'll have to double check, but I think it may be possible to score 640 points. (Although, I'm probably wrong.) However, if this is possible, I'm positive that 640 is the absolute highest.

*Update*
Nope, I'm in agreement, 600 is the best score possible.
 
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Alex Churchill
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600 is the highest possible score, with the arrangement you show.

Proof:
The maximum score for any player is 160. If each player scores 150, you get the board shown above, scoring 600. So for a score to be higher than 600, at least one player must score 160: i.e., must score 40 for each of their 4 rows.
Note that the theoretical maximum score for all players would be 640 (4 x 160). So if a total score is to be higher than 600, not only must at least one player score 160, but all players must score at least 130 (as 120+160+160+160 = 600).

How can a player score 40 for each of their 4 rows? The only plays that can score 40 are 4, 5 or 6. Consider each of these in turn.

If a player were scoring a 4, then the opposite player could score a maximum of 10 for that row (a 1 or a 6 could score 10, while nothing would be scored for a 2, 3, 4 or 5). So this opposite player's score would be at most 130 (40+40+40+10). Thus, there can be at most one 4 scored anywhere on the board. 4s are not the way.

If a player is using a 6 to score 40, a 5 would also have scored 40 for the same row. Thus, we can assume nobody is using 6s (by replacing them all with 5s).

In other words, we can assume at least 14 of the 16 dice around the board are 5s, with at most one 4, opposed by a 1. But if there were such a 4, then at least one of the 5s on the board would not be maximised. (This can be proved - it's fairly obvious, but fiddly in the detail. I can provide a proof if anyone wants.) Thus a score containing a 4 can't be greater than the 600-point configuration shown above.

So there are no 4s or 6s: everyone's chosen 5. Now it just remains to prove that the arrangement in the picture above is optimal.

This isn't quite as obvious as it looks, because there could be a set of heights a, b, c and d (some of which might be the same) which could be distributed as follows, to let everyone score fully for their 5s:
a a b b
a a b b
c c d d
c c d d

However, here's where parity comes to our rescue. The available blocks are 5 copies of each of 1, 2, 3, 4 and 5. So the total height is 75. This is an odd number, so it can't be divided into matching pairs or fours. So there must be an odd number of towers which have a different height to all the others. Thus, at least one vertical row will have either 1 or 3 of this height in it, as will at least one horizontal row. Thus, for each of the players, at least one of their 5s will score 30 or less, making the maximum score 600.

This proves that the board above is optimal.

Note that if we were on a board of size 5x5 or larger, parity wouldn't help, because we could get rid of an odd number of heights in a 3x3 block like this:
a a a b b
a a a b b
a a a b b
b b b b b
b b b b b

Note also that if you don't require all the blocks to be played to the board, the maximum score of 640 is easily possible:
5 5 4 4
5 5 4 4
3 3 2 2
3 3 2 2
 
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Danny Leenders
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The last example cant be 640 points.

5 5 4 4
5 5 4 4
3 3 2 2
3 3 2 2

This is because the player to the left can only see 2 buildings of the same height on the first row for him, the 4 is hiding behind it.


Left player eye > ---- 5 5 4 4

The 4 height is hiding behind the 5.

The only way to get 640 is when you could make all the buildings the same height.

 
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Alex Churchill
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No, the idea is that each player plays 5 for each of their dice. A 5 scores for the whole row even if you can't see part of it.
 
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