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 Post subject: MeanDoku NC X 16 Extra
PostPosted: Sun Oct 08, 2017 9:42 pm 
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Grand Master
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MeanDoku NC X 16 Extra


I find it easier to think in terms of sums not averages so:
The following colour clues apply:
Green: the sum of the two adjacent cells is below ten
Blue: the sum of the two adjacent cells is ten
Red: the sum of the two adjacent cells is above ten
Yellow: the sum of the two cells is below eight
Grey: the sum of the two cells is above twelve

I have marked the centre nonet strictly so if the total is below eight I have marked yellow not green and over twelve is grey not red. This means that in the centre nonet green is 8 or 9 and red is 11 or 12. this is not true elsewhere.

NC: horizontally and vertically adjacent cells are non-consecutive.

X: diagonals do not repeat.

I used the following two facts in my solution:

1 all the blue clues are marked
2 all the blue clues have the same pair


Image


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PostPosted: Wed Oct 11, 2017 11:30 pm 
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Very clever! One question before I go any further, though: I am assuming that your caveat that all blue clues are marked refers to adjacent cells ONLY within the same nonet. Is this correct?

Thanks, HATMAN!


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PostPosted: Tue Oct 17, 2017 4:57 pm 
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Thanks again, Hatman!

My rating: Fun-but-not-too-difficult. Which is good, because I like a mix of puzzles I struggle with and puzzles I can complete in spare time.

BTW, how're you numbering these? Did I miss a bunch between 5 & 15?


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PostPosted: Wed Oct 25, 2017 12:12 pm 
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azpauli

Numbering is based on my records. For various reasons I don't post all the puzzles I create

Maurice


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PostPosted: Thu Oct 26, 2017 4:51 pm 
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Thanks for that clarification, Maurice!

Looking forward to the next one that you decide to post!

--Paul


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PostPosted: Tue Feb 26, 2019 1:19 am 
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After MeanDocku NC X 15 I immediately moved on to this one. As azpaull said "fun-but-not-too-difficult". I used one step which was harder than any I used in no.15.

Here is my walkthrough for MeanDocku NC X 16 Extra:
Cells adjacent to green marks must total less than 10, blue must total 10, red must total more than 10. However in N5 green means total 8 or 9, red means total 11 or 12, yellow total less than 8, grey total more than 12.
Also NC so horizontally/vertically adjacent cells cannot be {12}, {23}, … {78}, {89}, therefore at least one of the cells adjacent to each green or yellow mark must contain one of 1,2,3 as green cannot be {45}; similarly at least one of the cells adjacent to each red or grey mark must contain one of 7,8,9 as red cannot be {56}.

Prelims.
Delete 9 from cells either side of green marks.
Delete 7,8,9 from cells either side of yellow marks.
Delete 5 from cells either side of blue marks.
Delete 1 from cells either side of red marks.
Delete 1,2,3 from cells either side of grey marks.
Also delete 2 from R1C89, R5C6 and R8C89 (two red marks)
Also delete 8 from R2C79, R3C3, R7C78 and R8C1378 (two green marks)
Also delete 7,8 from R2C8 (three green marks)
R1C2 = {789} (only way to satisfy four red marks)
R5C5 = {789} (only way to satisfy two red and two grey marks)
R8C2 = {123} (only way to satisfy four green marks)
Clean-ups and NCs only as stated.

1. 9 in N5 only in R4C6 + R5C5, R45C6 and R5C56 must both total 11 or 12 (red) -> R5C6 = 3, R4C6 + R5C5 = {89}, locked for N5 and D/
1a. R56C5 must total 11 or 12 (red) = [84/92]
1b. R56C6 must total 8 or 9 (green) = [35/36]
1c. R45C4 = [17] (hidden pair in N5), 1 placed for D\, no 8 in R5C5, no 6 in R6C4 (NC)
1d. R5C5 = 9, placed for D\, R6C5 = 2 (red), R4C6 = 8
1e. R6C5 = 2 -> R6C6 = 5 (yellow), placed for D\, R6C4 = 4, placed for D/, R4C5 = 6
NCs: no 2 in R3C4 + R4C3, no 5,7 in R3C5, no 7,9 in R3C6 + R4C7, no 6,8 in R5C3, no 2,4 in R5C7, no 3 in R6C3, no 6 in R6C7, no 3,5 in R7C4, no 1,3 in R7C5 and no 4,6 in R7C6

2. R2C2 = {78} -> no 7,8 in R13C2 + R2C13 (NC)
2a. No 9 in R2C2 -> no 2 in R13C2 + R2C13 (red)
Clean-ups: no 2,3,8 in R1C1, no 9 in R1C2 (blue), no 7 in R3C3 (green)
2b. R2C3 + R3C23 = {23456} must contain at least one of 4,6 (cannot be [352/532], NC) -> R1C12 = [73] (cannot be {46} which clashes with R2C3 + R3C23), R2C2 = 8, 7,8 placed for D\, no 6,9 in R2C1 (NC)
2c. Naked triple {456} in R2C13 + R3C2, locked for N1 -> R3C3 = 2, placed for D\

3. Naked triple {346} in R7C7 + R8C8 + R9C9, locked for N9
3a. R8C8 + R9C9 = {346} -> no 5 in R8C9 (red)
3b. R8C9 = {789} -> no 8 in R7C9 (NC)
3c. No 2,9 in R9C8, no 6 in R9C9 (blue)
3d. R9C8 = {78} -> no 7,8 in R9C7 (NC)
3e. 6 in N9 only in R7C7 + R8C8 -> no 5,7 in R7C8 + R8C7 (NC)
3f. Naked pair {12} in R7C8 + R8C7, locked for N9, no 1 in R6C8 (NC)
3g. Naked pair {12} in R8C27, locked for R8
Clean-up: no 8,9 in R8C4, no 8 in R8C5 (blue)
3h. R8C9 = 8 (hidden single in R8), R9C89 = [73] (blue), no 9 in R7C9 (NC)
3j. R7C9 = 5, R9C7 = 9, no 6 in R6C9 (NC)

4. R7C1 + R9C3 = [98] (hidden pair in N7) -> R1C3 + R3C1 = [91], no 8 in R6C1 (NC)
4a. R8C6 = 9 (hidden single in N8)
4b. R8C45 = [37] (blue, cannot be [64] which clashes with R8C8), no 2 in R79C4, no 8 in R7C5, no 4 in R8C3 (NC)
4c. R7C5 = 4, R7C7 + R8C8 = [64], R7C4 = 8
4d. R7C23 = [73] (blue), 3 placed for D/
4e. Naked pair {56} in R8C13, locked for N7 -> R9C1 = 2, placed for D/, R89C2 = [14], R8C7 = 2, R7C68 = [21]

5. R1C9 + R2C8 + R3C7 = [657], R13C8 = [83], no 6 in R3C6, no 2 in R4C8 (NC)
5a. R456C9 = [926], no 1 in R5C79 (NC)

The rest is naked singles, without using NC, diagonals or coloured marks.

Solution:
7 3 9 2 5 1 4 8 6
4 8 6 9 3 7 1 5 2
1 5 2 6 8 4 7 3 9
5 2 4 1 6 8 3 9 7
8 6 1 7 9 3 5 2 4
3 9 7 4 2 5 8 6 1
9 7 3 8 4 2 6 1 5
6 1 5 3 7 9 2 4 8
2 4 8 5 1 6 9 7 3

I was intrigued by HATMAN's small print solving method:
At what stage does it follow that all the blue pairs must be the same and why?
In the initial position why can't the values in the blue pair in R1C12 be in R2C789 and R3C456?


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