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Rectangles Puzzle #1778 by Maarten
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mathgrant


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Joined: 19/08/2008 20:52:44
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Number: Puzzle #1778
Genre: Rectangles
Author: Maarten
Appeared at: April 17, 2011

It's interesting to see this variation on Shikaku (which I first saw on motris's blog) implemented here. I was considering trying to make some myself, but now I definitely must.

That being said, I didn't see how to tackle this puzzle without branching. Anyone willing to post or e-mail a walkthrough? We could compare notes or something.

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anurag.sahay



Joined: 25/06/2009 11:29:52
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As you might have seen,it starts off well, you can complete the 10 and the 5 on its left.From here,start filling the 6 on the right.It can be completed in 3 ways- vertical 1x6 or horizontal 2x3 or vertical 2x4.I found that the 4 and 3 on R4C7 and R6C8 cant complete if you choose either of the first 2 ways for the 6.From here,it was quick.
mathgrant


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anurag.sahay wrote:
As you might have seen,it starts off well, you can complete the 10 and the 5 on its left.From here,start filling the 6 on the right.It can be completed in 3 ways- vertical 1x6 or horizontal 2x3 or vertical 2x4.I found that the 4 and 3 on R4C7 and R6C8 cant complete if you choose either of the first 2 ways for the 6.From here,it was quick. 


As far as I can tell, the 10 can be contained in R5-10C9-110 or R7-10C8-10. The easiest way I can see to solve the 10 is to observe that the former forces R10C8 to be in R10C4-8, and R9C7 and R10C3 to both require the 7 at R9C3, a contradiction. With the 10 in a 3x4 rectangle, R9C7 can only be contained in R9-10C5-7, solving the 5 as you mentioned.

I don't see how the 4 and the 3 are constrained by the first two choices for the 6 unless you're looking further ahead and not mentioning it.

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Bram



Joined: 04/03/2008 13:59:34
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I'll have a look at a walkthrough. I did it logically for verification.
Bram



Joined: 04/03/2008 13:59:34
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Here's the walkthrough. It's not my best genre, and found writing it out a bit messy butI think it should be clear.

Walkthrough (in white as to not spoil)

Start

R710C910 have to belong to the 10 in R8C9.
R9C7 has to belong with the 5 below it. The only other valid option could be the 7 in R9C3, except then the bottom row wouldn't work.
R10C4 has to go with the 7 in R9C3.
R8C1 has to go with the 5 above it.
R3C1 has to go with the 7 in R3C2.
R910C2 have to go with the 7 in R9C3.
R10C6 now has to go with the 5 in R10C7.
Now th only valid rectangle with that 5 is R910C57. The 10 has to be R710C810.

The 7 in R7C4 has to use R6C4 as all valid rectangles from this one all use this square.
R7C6 has to use R7C5 as all rectangles that use this square also use R7C5.
Now R6C5 can only be used by the 7 in R7C4, which makes this rectangle R57C46.

The 6 in R5C3 has to use R67C3 as all valid rectangles use these squares.
The 5 in R7C1 now can't reach any lakes, so it has to be a vertical 1 wide rectangle and always uses R6C1.
R567C2 now can only be used by the 6 in R5C3, which makes this rectangle R58C23.

The rest is just filler work. Start with the 7 in R9C3.


End

Hope this was clear and that it helped. Think it is all logically sound.
mathgrant


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Joined: 19/08/2008 20:52:44
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I find your logic to be sound, Bram. Phew.

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