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 Post subject: Mean ORC 35
PostPosted: Mon Aug 19, 2019 9:26 pm 
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
Mean ORC 35

Meandoku the following colour clues apply:
Green: the sum of the two adjacent cells is 8 or 9 only
Blue: the sum of the two adjacent cells is ten
Red: the sum of the two adjacent cells is 11 or 12 only
Yellow: the sum of the two cells is below eight
Black: the sum of the two cells is above twelve

Odd Row and Column - ORC - so it is:
ORC: odd rows and columns are 1-9 no repeat; even ones are not (i.e. they can repeat).
NN: no nonets

For all cells:
AK: Anti-King - diagonally adjacent are not equal
FNC: Ferz Non-consecutive - diagonally adjacent are not consecutive
NC: adjacent cells are not consecutive


Diagonal / is non-repeat
Diagonal \ is three ordered Triplets in descending order i.e. zyxzyxzyx or xzyxzyxzy or yxzyxzyxy
To clarify if the three numbers are 4, 7 and 9 they would go in descending order so 974 which means that the nine numbers on the diagonal could be 974974974 or 497497497 or 749749749. thanks Andrew


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 Post subject: Re: Mean ORC 35
PostPosted: Fri Oct 11, 2019 4:53 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks HATMAN for another fun puzzle! I was a bit slow in realising how to use the descending order of the triplets; in hindsight, while it's good to use the triplets, it's probably best to get as far as possible before using the descending order of them.

Here is my walkthrough for Mean ORC 35:
Cells adjacent to yellow lines must total less than 8, blue must total 10.
AK so diagonally adjacent cells cannot be equal, also FNC and NC so horizontally/ vertically/diagonally adjacent cells cannot be {12}, {23}, … {78}, {89}, therefore at least one of the cells adjacent to each yellow mark must contain one of 1,2,3 in even rows/columns or just one of 1,2 in odd rows/columns.
Odd numbered rows and columns are normal; repeats are allowed on even numbered rows and columns.
D/ is a normal non-repeat diagonal, D \ contains three ordered triplets descending down this diagonal, repeated starting at R1C1, R4C4 and R7C7.

Prelims
Delete 5 from cells either side of blue marks in R1, R5 and C7, and from the diagonal blue marks because of AK.
Delete 7,8,9 from cells either side of yellow marks.
Clean-ups, AK, FNC and NC, separately or together, only when stated.

1. R8C89 = [46/55/64] (blue)
1a. R8C8 = {456} -> R8C7 = {12} (yellow, cannot be [34], NC)
1b. R8C9 = {456} -> R9C9 = {12} (yellow, cannot be [43], NC)
Clean-ups:
R8C7 = {12} -> no 1,2 in R79C678
R8C8 = {456} -> no 5 in R7C9 + R9C7
R8C9 = {456} -> no 5 in R79C8
No 1,2 in R7C7 -> no 8,9 in R6C7 (blue)

2. Because of the triplets on D\ R2C2, R5C5 and R8C8 must be equal
2a. R8C8 = {456}, no 5 in R5C5 -> R2C2 = R5C5 = R8C8 = {46}
2b. R2C23 = {46} (blue)
2c. R5C56 = {46} (blue), locked for R5
2d. R8C89 = {46} (blue)
Clean-ups:
R2C23 = {46} -> no 3,7 in R13C23, no 4,6 in R13C3, no 5 in R13C1234 + R2C14
R5C56 = {46} -> no 3,7 in R46C56, no 4,6 in R46C5, no 5 in R4C457 + R5C4 + R6C456
R8C89 = {46} -> no 3,7 in R7C89 + R9C8, no 4,6 in R7C9
R4C6 + R5C7 = {19/28} (blue), no 4,6 in R4C6, no 3,7 in R5C7

3. Because of the triplets on D\ R3C3, R6C6 and R9C9 must be equal
3a. R9C9 = {12} -> R3C3 = R6C6 = {12}
Clean-ups:
R3C3 = {12} -> no 1,2 in R2C4 + R3C24 + R4C234
R34C4 = {37/46} (blue), no 8,9
R6C6 = {12} -> no 1,2 in R5C7 + R7C5
R5C7 = {89} -> R4C6 = {12} (blue)
R4C6 = {12} -> no 1,2 in R3C57
R5C7 = {89} -> no 8,9 in R4C78 + R56C8

4. Because of the triplets on D\ R1C1, R4C4 and R7C7 must be equal
4a. R4C4 = {3467} -> R1C1 = R7C7 = {3467}
Clean-up:
R34C4 = {37/46} -> no 3,4,6,7 in R3C5
No 8,9 in R7C7 -> no 1,2 in R6C7 (blue)
R67C7 = {37/46} -> no 3,4,6,7 in R7C6, no 4,6 in R7C8
R7C8 = {89} -> no 8,9 in R67C9
4b. Naked pair {12} in R79C9, locked for C9
Clean-up:
R7C9 = {12} -> no 1,2 in R6C8

5. 5 in C7 only in R23C7 -> no 4,6 in R2C678 + R3C7, no 4,5,6 in R3C68

6. R8C89 = {46}, R5C5 = R8C8 from triplets -> R5C5 + R8C9 = naked pair {46}, CPE no 4,6 in R1C9 using D/

[Now to apply descending order of triplets on D\, if I’ve understood the requirement correctly the three numbers have to be in descending order but can start at R1C1, R2C2 or R3C3.]
7. R7C7 + R8C8 + R9C9 = {67}[4]{12} (cannot be {34}[6]{12} which aren’t in descending order) -> R7C7 = {67}, R8C8 = 4, R8C9 = 6 (blue) -> R9C9 = 1 (yellow), 1,6 placed for C9, 1 placed for R9
7a. R7C9 = 2, placed for R7
7b. R8C8 = 4 -> R2C2 = 4, R2C3 = 6 (blue), placed for C3, R5C5 = 4, placed for C5 and D/, R5C6 = 6 (blue)
7c. R9C9 = 1 -> R3C3 = 1, placed for R3 and C3, R6C6 = 1
7d. R7C7 = {67} -> R1C1 = R4C4 = {67}
Clean-ups:
R4C4 = {67} -> R3C4 = {34} (blue)
R7C7 = {67} -> R6C7 = {34} (blue)
R7C3 = {3789} -> R6C2 = {1237} (blue)
R2C2 = 4 -> no 3 in R2C1, no 3,4 in R3C1
R2C3 = 6 -> no 6 in R13C2, no 6,7 in R1C4, no 7 in R2C4
R5C5 = 4 -> no 3 in R56C4
R5C6 = 6 -> no 6,7 in R4C7
R6C6 = 1 -> no 2 in R6C5
R7C9 = 2 -> no 3 in R6C89
R8C8 = 4 -> no 3,4 in R9C7
R8C9 = 6 -> no 6 in R9C8
R1C1 = {67} -> no 6,7 in R2C1
R3C4 = {34} -> no 3 in R2C5, no 3,4 in R4C3
R4C4 = {67} -> no 7 in R5C3
R6C7 = {34} -> no 3 in R5C8
R7C7 = {67} -> no 6,7 in R6C8 + R8C6
R6C8 = {45} -> no 5 in R5C9

8a. R3C1 = 6 (hidden single in R3), placed for C1
8b. R1C1 = 7, placed for R1 and C1
8c. R1C1 = 7 -> R4C4 = 7 -> R3C4 = 3 (blue), placed for R3
8d. R1C1 = 7 -> R7C7 = 7, placed for R7 and C7, R6C7 = 3 (blue), placed for C7
Clean-up:
R7C7 = 7 -> no 8 in R7C68 + R8C6
8e. R7C68 = [59], placed for R7
Clean-ups:
No 7 in R1C7 -> no 3 in R1C6 (blue)
No 7,9 in R7C3 -> no 1,3 in R6C2 (blue)
R1C1 = 7 -> no 8 in R1C2 + R2C1
R3C1 = 6 -> no 5 in R4C1, no 5,6,7 in R4C2
R3C4 = 3 -> no 4 in R2C4, no 2 in R24C5
R4C4 = 7 -> no 8 in R3C5 + R4C35 + R5C34
R6C7 = 3 -> no 2 in R5C8, no 4 in R6C8
R6C8 = 5 -> no 4 in R6C9
R7C6 = 5 -> no 6 in R7C5, no 5,6 in R8C5, no 4 in R8C6
8f. Naked pair {38} in R7C35, locked for R7
Clean-up:
Naked pair {38} in R7C35 -> no 2,7,8,9 in R6C4, no 4 in R7C4, no 2,3,4,7,8,9 in R8C4
4 in R7 only in R7C12 -> no 3,4,5 in R68C1, no 3,5 in R8C2

9. 2 in R3 only in R3C68 -> no 1,2 in R24C7
9a. R4C7 = 4, placed for C7
Clean-up: no 5 in R3C7
9b. Naked pair {89} in R35C7, locked for C7 -> R2C7 = 5, R9C7 = 6, placed for R9 and C7
Clean-ups:
R1C7 = {12} -> R1C6 = {89} (blue)
R2C7 = 5 -> no 4,5,6 in R1C8
R4C7 = 4 -> no 3,5 in R4C8, no 5 in R5C8
R9C7 = 6 -> no 5 in R8C6, no 5,7 in R9C6
R1C6 = {89} -> no 8,9 in R12C5
R1C7 = {12} -> no 1,2 in R1C8 + R2C68
R3C7 = {89} -> no 8,9 in R23C68
Naked pair {89} in R35C7 -> no 7 in R4C8
9c. Naked pair {27} in R3C68, 7 locked for R3

10a. R1C5 = 6 (hidden single in R1), placed for C5
Clean-up: R1C5 = 6 -> no 6 in R2C4, no 5,7 in R2C56
10b. R23467C5 = [15983], all placed for C5, 5 placed for R3, 3 placed for R7
Clean-up:
R7C5 = 3 -> no 2 in R8C5
10c. R8C5 = 7 -> R9C5 = 2, placed for R9
10d. R7C3 = 8, placed for C3 and D/ -> R6C2 = 2 (blue)
10e. R3C7 = 9, placed for R3, C7 and D/
10f. R5C7 = 8, placed for R5 -> R4C6 = 2 (blue), placed for D/
Clean-up:
R2C6 = 3 -> no 2 in R1C7 + R3C6
10f. R1C7 = 1 -> R1C6 = 9 (blue), both placed for R1, 1 placed for C7
10g. R1C23 = [42], both placed for R1, 2 placed for C3
10h. R1C9 = 5 (hidden single in R1), placed for C9 and D/
10i. R9C1 = 3, placed for R9, C1 and D/ -> R2C8 = 7, placed for D/
10j. R6C9 = 7, placed for C9
Clean-ups:
R1C2 = 4 -> no 4 in R2C1
R1C3 = 2 -> no 3 in R12C4
R1C4 = 8 -> no 9 in R2C4
R2C8 = 7 -> no 8 in R1C8 + R23C9
R1C8 = 3 -> no 3,4 in R2C9
R4C5 = 9 -> no 9 in R5C4
R6C2 = 2 -> no 1,2 in R5C1, no 1,3 in R5C2, no 3 in R56C3, no 1 in R6C1 + R7C12
R6C9 = 7 -> no 7 in R5C8
R5C8 = 1 -> no 2 in R4C8
R7C3 = 8 -> no 7,9 in R68C3
R8C5 = 7 -> no 6 in R78C4, no 7,8 in R9C4, no 8 in R9C6
R9C1 = 3 -> no 2 in R8C1, no 4 in R9C2
R9C5 = 2 -> no 1 in R8C4, no 1,2,3 in R8C6
R6C3 = {45} -> no 5 in R5C23

11a. R3C2689 = [8724]
Clean-up:
R3C2 = 8 -> no 7,9 in R4C3
11b. R245C9 = [983]
11c. R5C138 = [591], 1,9 placed for R5, 9 placed for C3
11d. R7C12 = [46], 4 placed for C1
11e. R4689C3 = [5437], 7 placed for R9
Clean-up:
R9C3 = 7 -> no 6 in R8C2
11f. R8C2 = 1, placed for D/
Clean-up:
R6C4 = 6 -> no 7 in R5C4
11g. R5C4 = 2 -> R5C2 = 7
Clean-ups:
R3C2 = 8 -> no 9 in R2C1, no 8,9 in R4C1, no 9 in R4C2
R3C8 = 2 -> no 1 in R4C8
R3C9 = 4 -> no 4 in R4C8
R4C3 = 5 -> no 4 in R4C2
R5C2 = 7 -> no 8 in R4C2 + R6C1
R4C2 = 3 -> no 2 in R4C1
R8C3 = 3 -> no 4 in R9C4

12a. R2468C1 = [2198]
Clean-up: no 8,9 in R9C2
12b. R9C2468 = [5948]

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


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