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 Post subject: SSS ORC 2 & 2H
PostPosted: Thu Jun 15, 2017 8:09 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 539
SkyScraper Sum with Odd Row Column

This is a combination of the two puzzle types that I promised. They work well together and give some slightly different solution techniques.

SSS: the clues look into the row or column and are the sum of the numbers seen with large numbers hiding smaller and equal ones.

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-concecutive - diagonally adjacent are not consecutive
NC: adjacent cells are not consecutive

The solution contains 2 three-in-a-rows and 2 four-in-a-rows (either row or column {cannot be diagonal as Anti-King}), which are needed to solve the final slums.

Below the puzzles we give the set of valid combinations. Note that as there are no valid puzzle solutions with a row of six maximum, hence all combinations must contain a seven, eight or nine (contrariwise all even rows and columns must contain a one, two or three).

The number of solutions is small: 46 primary solutions allowing for reflections etc. however about half have bad slums (i.e. cannot be solved by a three-in-a-row) and some have multiple solutions. If I do a lot of these puzzles I'll have to remove the anti-king property.

SSS ORC 2 - an easy starter version:


Image


SSS ORC 2H - a hard one:

Image

Solution:
168524793
841584257
247316859
799993151
426157938
994931516
572961384
395949682
613972485


Possible Number Combinations
7 7 There are no valid puzzles with a row of six max.
8 1 7
8 8
9 1 8
9 2 7
9 9
10 1 9
10 2 8
10 3 7
11 1 3 7
11 2 9
11 3 8
11 4 7
12 1 3 8
12 1 4 7
12 3 9
12 4 8
12 5 7
13 1 3 9
13 1 4 8
13 1 5 7
13 2 4 7
13 4 9
13 5 8
13 6 7
14 1 4 9
14 1 5 8
14 1 6 7
14 2 4 8
14 2 5 7
14 3 4 7
14 5 9
14 6 8
15 1 5 9
15 1 6 8
15 2 4 9
15 2 5 8
15 2 6 7
15 3 4 8
15 3 5 7
15 6 9
15 7 8
16 1 3 5 7
16 1 6 9
16 1 7 8
16 2 5 9
16 2 6 8
16 3 4 9
16 3 5 8
16 3 6 7
16 4 5 7
16 7 9
17 1 3 5 8
17 1 3 6 7
17 1 4 5 7
17 1 7 9
17 2 6 9
17 2 7 8
17 3 5 9
17 3 6 8
17 4 5 8
17 4 6 7
17 8 9
18 1 3 5 9
18 1 3 6 8
18 1 4 5 8
18 1 4 6 7
18 1 8 9
18 2 4 5 7
18 2 7 9
18 3 6 9
18 3 7 8
18 4 5 9
18 4 6 8
18 5 6 7
19 1 3 6 9
19 1 3 7 8
19 1 4 5 9
19 1 4 6 8
19 1 5 6 7
19 2 4 5 8
19 2 4 6 7
19 2 8 9
19 3 4 5 7
19 3 7 9
19 4 6 9
19 4 7 8 There are no valid puzzles with a row of six max.
19 5 6 8 Seven and Eight Combos only done up to 19
20 1 3 7 9
20 1 4 6 9
20 2 4 5 9
20 3 8 9
20 4 7 9
20 5 6 9
21 1 3 8 9
21 1 4 7 9
21 1 5 6 9
21 2 4 6 9
21 3 4 5 9
21 4 8 9
21 5 7 9
22 1 4 8 9
22 1 5 7 9
22 2 4 7 9
22 2 5 6 9
22 3 4 6 9
22 5 8 9
22 6 7 9
23 1 5 8 9
23 1 6 7 9
23 2 4 8 9
23 2 5 7 9
23 3 4 7 9
23 3 5 6 9
23 6 8 9
24 1 3 4 7 9
24 1 3 5 6 9
24 1 6 8 9
24 2 5 8 9
24 2 6 7 9
24 3 4 8 9
24 3 5 7 9
24 4 5 6 9
24 7 8 9
25 1 3 4 8 9
25 1 3 5 7 9
25 1 4 5 6 9
25 1 7 8 9
25 2 6 8 9
25 3 5 8 9
25 3 6 7 9
25 4 5 7 9
26 1 3 5 8 9
26 1 3 6 7 9
26 1 4 5 7 9
26 2 4 5 6 9
26 2 7 8 9
26 3 6 8 9
26 4 5 8 9
26 4 6 7 9
27 1 3 6 8 9
27 1 4 2 5 8 9
27 1 4 6 7 9
27 2 4 1 5 7 9
27 3 7 8 9
27 4 6 8 9
27 5 1 2 3 6 7 9
28 1 3 4 5 6 9
28 1 3 7 8 9
28 1 4 6 8 9
28 1 5 6 7 9
28 2 4 5 8 9
28 2 4 6 7 9
28 3 4 5 7 9
28 4 7 8 9
28 5 6 8 9
29 1 3 4 5 7 9
29 1 4 7 8 9
29 1 5 6 8 9
29 2 4 6 8 9
29 2 5 6 7 9
29 3 4 5 8 9
29 3 4 6 7 9
29 5 7 8 9
30 1 3 4 6 7 9
30 1 5 7 8 9
30 2 4 7 8 9
30 2 5 6 8 9
30 3 4 6 8 9
30 3 5 6 7 9
30 6 7 8 9
31 1 3 4 6 8 9
31 1 3 5 6 7 9
31 1 6 7 8 9
31 2 5 7 8 9
31 3 4 7 8 9
31 3 5 6 8 9
31 4 5 6 7 9
32 1 3 4 7 8 9
32 1 3 5 6 8 9
32 1 4 5 6 7 9
32 2 6 7 8 9
32 3 5 7 8 9
32 4 5 6 8 9
33 1 3 5 7 8 9
33 1 4 5 6 8 9
33 2 4 5 6 7 9
33 3 6 7 8 9
33 4 5 7 8 9
34 1 3 6 7 8 9
34 1 4 5 7 8 9
34 2 4 5 6 8 9
34 4 6 7 8 9
35 1 4 6 7 8 9
35 2 4 5 7 8 9
35 5 6 7 8 9
36 2 4 6 7 8 9


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 Post subject: Re: SSS ORC 2 & 2H
PostPosted: Mon Jul 03, 2017 3:02 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1583
Location: Lethbridge, Alberta, Canada
Thanks HATMAN. A nice combination of SkyScraper Sum and ORC, with other restraints.

I'm glad that you also posted the easier one. I had done two previous ORC puzzles, in this forum in November 2016, but they were simpler ones; a plain vanilla and a repeat killer. It's so easy to make mistakes when not used to a new type of Sudoku; I had several restarts, just like I did with the first SkyScraper Sum puzzle. Remember that cells in the even rows and columns behave differently.

Here's my walkthrough for the easier version:
Sums are for heights of skyscrapers visible from the edges. Only those higher than the previous one(s) count toward the sums.
On the odd-numbered rows and columns, which are normal ones containing 1-9, each sum must include 9, the height of the highest skyscraper, which can be seen from each edge. However this doesn’t necessarily apply for the even-numbered rows and columns where repeated numbers are allowed; for these rows and columns the height of the highest skyscraper may be less than 9.
Since this an ORC puzzle, I’ve stated placements in odd rows/columns.

1a. Since no totals are given as 8, 9 or 17, rows/columns starting/finishing with 8,9 must be ones with unspecified totals.
1b. Unspecified upper totals only in C3 and C8, left-hand total 24 in R1 cannot be R1C123 = [789] because of NC -> R1C8 = 9, R1C3 = 8, both placed for R1, 8 placed for C3, no 7 in R1C24, no 7,8,9 in R2C24, no 7,9 in R2C3, no 8,9 in R2C79, no 8 in R2C8 (AK, FNC, NC)
1c. Right-hand total in R1 = 12, R1C8 = 9 -> R1C9 = 3, placed for R1 and C9, no 2,3,4 in R2C8, no 2,4 in R2C9 (AK, FNC, NC)
1d. Left-hand total in R1 = 24, R1C38 = [89] =17 -> R1C12 must total 7 = [16/25]
1e. R1C1 = {12} -> no 1,2 in R2C12 (AK, FNC, NC)
1f. R1C2 = {56} -> no 5,6 in R2C13 (AK, FNC)
1g. 7 in R1 only in R1C567 -> no 6 in R1C6 (NC)

2a. Unspecified left-hand totals only in R2 and R6, cannot make upper total in C1 = 18 with 9 in R2C1 -> R6C1 = 9, R2C1 = 8, both placed for C1, no 7 in R3C1, no 7,8,9 in R3C2, no 8,9 in R57C2, no 8 in R6C2 (AK, FNC, NC)
2b. Upper total in C1 = 18, R26C1 = [89] = 17 -> R1C1 = 1 (cage sum), placed for R1 and C1
2c. Left-hand total in R1 = 24, R1C138 = [189] =18 -> R1C2 = 6 (cage sum), placed for R1, no 5 in R2C2 (FNC, NC)
2d. Unspecified right-hand totals only in R3 and R5, cannot make upper total in C9 = 19 with both of 8,9 (because R1C9 = 3) -> R3C9 = 9, R5C9 = 8, both placed for C9, 9 placed for R3, 8 placed for R5, no 9 in R2C8, no 8 in R3C8, no 7,8,9 in R46C8, no 7 in R46C9, no 7,9 in R5C8 (AK, FNC, NC)
2e. Upper total in C9 = 19, R13C9 = [39] -> R2C9 = 7 (cage sum), placed for C9, no 6 in R2C8, no 6,7 in R3C8 (AK, FNC, NC)
2f. Unspecified lower totals only in C4 and C8, cannot make right-hand total in R9 = 22 with 9 in R9C8 -> R9C4 = 9, R9C8 = 8, both placed for R9, no 9 in R8C3, no 8 in R8C4, no 8,9 in R8C5, no 7,8,9 in R8C7, no 7,9 in R8C8, no 7 in R9C7 (AK, FNC, NC)
2g. Right-hand total in R9 = 22, R9C45 = [98] = 17 -> R9C9 = 5 (cage sum), placed for R9 and C9, no 4,5,6 in R8C8, no 4,6 in R8C9 (AK, FNC, NC)
2h. R8C9 = {12} -> no 1,2 in R7C89 (AK, FNC, NC)
2i. R7C9 = {46} -> no 5 in R67C8 (FNC, NC)
2j. Upper total in C2 = 15 with R1C2 = 6 can only be [69] -> no 7,8 in R4C2
2k. Lower total in C2 = 10 must contain 9 = [19] -> R9C2 = 1, placed for R9, no 2 in R8C12 + R9C13, no 1,2 in R8C3 (AK, FNC, NC), also no 3,4,5,6,7,8 in R8C2
2l. Right-hand total in R2 = 15 with R2C9 = 7 can only be [87] -> no 9 in R2C56

3a. Right-hand total in R7 = 21 must contain 9 so cannot start with 6 -> R7C9 = 4, placed for R7 and C9, no 3,4 in R6C8, no 3 in R78C8 (AK, FNC, NC)
3b. Right-hand total in R7 must be made up from [984] -> R7C8 = 8 (otherwise one of 6,7 would be visible), placed for R7, no 7,8,9 in R6C7, no 7,9 in R7C7 (AK, FNC, NC)
3c. 9 in R7 must be to the right of 7 -> no 7 in R7C6
3d. 6 in C9 only in R46C9 -> no 5,6 in R5C8 (AK, FNC)
3e. Left-hand total in R7 = 21 must contain 7,9 = [579] (cannot be [21379/2379] because of NC) -> R7C1 = 5, placed for R7 and C1, no 4,5,6 in R6C2, no 6 in R7C2, no 4,6 in R8C1 (AK, FNC, NC)
3f. Left-hand total in R8 = 12 = [39] -> R8C1 = 3, placed for C1, no 2,3 in R7C2, no 4 in R9C1 (AK, FNC, NC)
3g. 2,4 in C1 only in R345C1 -> no 1,3,4,5 in R4C2 (AK, FNC, NC)

4a. Upper total in C4 = 14 must contain 9 = [59] (cannot be [2139/239] because of NC) -> R1C4 = 5, placed for R1, no 4 in R1C5 + R2C3, no 4,6 in R2C4, no 4,5,6 in R2C5 (AK, FNC, NC)
4b. 4 in R1 only in R1C67 -> no 3,5 in R2C6, no 3,4,5 in R2C7 (AK, FNC, NC)
4c. Upper total in C5 = 19 must contain 9 -> no 7 in R1C5
4d. R1C5 = 2, placed for R1 and C5, no 1,2,3 in R2C4, no 1,3 in R2C5, no 1,2 in R2C6 (AK, FNC, NC)
4e. Upper total in C5 = 19 contains 2,9 = [289] -> R2C5 = 8, placed for C5, no 7 in R12C6 + R3C5, no 7,8 in R3C46 (AK, FNC, NC)
4f. R1C6 = 4, R1C7 = 7, placed for C7, no 6,8 in R2C6, no 6 in R2C7, no 7 in R2C8 (AK, FNC, NC)
4g. R2C46 = [54] -> no 4,5,6 in R3C3, no 4,6 in R3C4, no 3,4,5,6 in R3C5, no 3,5 in R3C6, no 3,4,5 in R3C7 (AK, FNC, NC)
4h. R3C5 = 1, placed for R3 and C5 -> no 2 in R3C46, no 1,2 in R4C46 (AK, FNC, NC)
4i. R3C4 = {35} -> no 4 in R4C345 (FNC, NC)
4j. R3C6 = {46} -> no 5 in R4C567 (FNC, NC)
4k. R2C7 = {12} -> no 2 in R3C8 (AK, FNC)
4l. R3C8 = {345} -> no 4 in R4C7 (AK, FNC)

5a. R3C37 = [78] (hidden pair in R3), 7 placed for C3, 8 placed for C7, no 6 in R23C2 + R4C23, no 6,7,8 in R4C4, no 7,8,9 in R4C6, no 9 in R4C7 (AK, FNC, NC)
5b. R5C7 = 9 (hidden single in C7), placed for R5, no 8,9 in R6C6 (AK, FNC)
5c. R2C2 = {34} -> no 4 in R3C1 (AK, FNC)
5d. 2 in R3 only in R3C12 -> no 2 in R4C1 (AK)
5e. Left-hand total in R3 = 30, R3C379 = [789] = 24 -> remaining visible cells must total 6 -> R3C12 = [24/6x], no 5 in R3C2 (NC)
5f. R3C2 = {234} -> no 3 in R4C3 (AK, FNC)

6a. Lower total in C3 = 23 must contain 9 but not 7,8 -> remaining visible cells in C3 =14 can only be [356] -> R89C3 = [53], placed for C3, 3 also placed for R9, no 6,9 in R7C3, no 6 in R7C4, no 4,6 in R8C4 (FNC, NC)
6b. Naked pair {12} in R27C3, locked for C3 -> R4C3 = 9, placed for C3
6c. Naked pair {46} in R45C3 -> no 4,5,6 in R5C24, no 5 in R6C4 (AK, FNC, NC)
6d. Left-hand total in R4 = 16, R4C3 = 9 (R4C2 may possibly be 9) -> R4C1 = 7 (cage sum), placed for C1, no 7 in R5C2 (AK)
6e. R359C1 = [246], placed for R3, R5 and R9 respectively -> R56C3 = [64], 6 placed for R5, no 3 in R23C2, no 2 in R4C2, no 5 in R4C4, no 3 in R5C2, no 3,7 in R5C4 + R6C24, no 3 in R7C4 (AK, FNC, NC)
6f. R3C2 = 4, placed for R3 -> R3C6 = 6, no 6,7 in R4C5, no 6 in R4C7 (AK, FNC)
6g. Naked pair {12} in R5C24, locked for R5 -> R5C8 = 3, placed for R5, no 2,3 in R4C7, no 2,4 in R4C8, no 2 in R46C9, no 2,3,4 in R6C7, no 2 in R6C8 (AK, FNC, NC)
6h. R8C9 = 2 (hidden single in C9) -> no 1 in R8C8 (NC)

7a. R4C7 = 1, placed for C7, R2C7 = 2, placed for C7, R9C7 = 4, placed for R9 and C7, R9C5 = 7, placed for R9 and C5 -> R5C5 = 5, placed for R5 and C5, R9C6 = 2, no 1 in R2C8, no 3 in R3C8, no 4,6 in R4C6 + R6C45, no 4,5,6 in R6C6, no 7 in R8C4, no 3,6 in R8C5, no 1,3,4,5,6,7,8 in R8C6, no 3,5 in R8C7 (AK, FNC, NC)
7b. R678C7 = [536], 3 placed for R7, no 2,3 in R6C6, no 6 in R6C8, no 2,6 in R7C6, no 2 in R8C6, no 2 in R8C8 (AK, FNC, NC)
7c. R7C5 = 6 (hidden single in R7) -> no 7 in R6C6 + R7C4, no 5 in R8C4 (FNC, NC)
7d. R8C5 = 4 -> no 3 in R8C4 (NC)
7e. R7C2 = 7 (hidden single in R7)
7f. Naked pair {39} in R46C5 -> no 2 in R5C4 (NC)
7g. R5C24 = [21] -> no 1 in R6C2, no 2 in R6C4 (NC)
7h. 2 in R7 only in R7C34 -> no 1 in R7C34 (NC)
7i. R7C346 = [291], 2 placed for C3, no 2 in R6C2, no 1,8 in R6C4, no 9 in R6C5, no 1 in R8C2, no 1,2 in R8C4 (AK, FNC, NC)

8a. Lower total in C9 = 28, R359C9 = [985] = 22 -> there must also be 6 in a visible cell -> R6C9 = 6, placed for C9
8b. R46C5 = [93]
8c. Right-hand total in R4 = 15, R4C59 = [91] -> R4C8 = 5 (cage sum)

[Now to apply the extra conditions of two three-in-a-row and two four-in-a-row.]
9a. Four 9s in a row in C4 -> need another four-in-a row can only be R4C2345 -> R4C4 = 9
9b. There are already three 5s and three 8s in a row in C8 -> there cannot be another three-in-a-row -> R3C4 = 3
Now to try the harder version, when I've got time.

HATMAN wrote:
The number of solutions is small: 46 primary solutions allowing for reflections etc. however about half have bad slums (i.e. cannot be solved by a three-in-a-row) and some have multiple solutions. If I do a lot of these puzzles I'll have to remove the anti-king property.
I hope not. A lot of my candidate eliminations made use of the anti-king property; removing that would have for a much longer solving path.


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 Post subject: Re: SSS ORC 2 & 2H
PostPosted: Mon Jul 03, 2017 3:07 am 
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Joined: Wed Apr 23, 2008 6:04 pm
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Location: Lethbridge, Alberta, Canada
It was stated that there are no valid combinations for repeat columns with a highest number 6. Maybe I'm missing something but I don't see why. I'm assuming given SkyScraper Sums are at least 10, so that one can make the same starting approach as in the SSS puzzles posted so far.

Consider an even-numbered row or column with totals of 10 and 12 at the two ends. There seem to be several possibilities including [11336664422] What's wrong with this? I don't think it prevents 1-9 being placed in the adjacent odd-numbered rows/columns, assuming that AK and FNC apply and that the adjacent rows are NC.


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 Post subject: Re: SSS ORC 2 & 2H
PostPosted: Wed Jul 05, 2017 6:38 pm 
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 539
Andrew

I removed 789 from row 2 and ran JSudoku "recursively solve" for hours giving no solution. I then reset is and removed 789 from row 4 - again JSudoku could not solve it.
I've added this as a given for the puzzles to avoid wasting time.

Maurice


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 Post subject: Re: SSS ORC 2 & 2H
PostPosted: Wed Jan 24, 2018 12:43 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1583
Location: Lethbridge, Alberta, Canada
Yes, 2H was quite a hard one. I originally started it without the 7,8,9 restriction on even rows and columns and didn't get very far. When I came back to it, using that restriction, it gave another early placement but then I had to start working. I used several fairly short forcing chains, so I expect HATMAN and JSudoku needed to use fishes.

Here is my walkthrough for SSS ORC 2H:
Sums are for heights of skyscrapers visible from the edges. Only those higher than the previous one(s) count toward the sums.
On the odd-numbered rows and columns, which are normal ones containing 1-9, each sum must include 9, the height of the highest skyscraper, which can be seen from each edge. However this doesn’t necessarily apply for the even-numbered rows and columns where repeated numbers are allowed; for these rows and columns the height of the highest skyscraper may be less than 9.
Since this an ORC puzzle, I’ve stated placements in odd rows/columns.
This is a KiMo, with the totals being xy where it has been specified that x is never 0, y never 7.
There are two three-in-a-rows and two four-in-a-rows, which are required to give a unique solution. Since it is not specified otherwise, these repetitions may possibly be in the same row/column.
It has been specified that all totals in the even-numbered rows and columns must contain at least one of 7,8,9; I originally started without that specification, but have re-worked after step 1 to take account of it.

1a. Skyscrapers starting with 8 or 9 must total 8 (possible for even rows/columns), 9 or 17. Since no totals are given as 17 and xy cannot be 8, 9 or 17, rows/columns starting/finishing with 8,9 must be ones with unspecified totals.
1b. Unspecified upper totals only in C3 and C8, right-hand total 12 in R1 must contain 9 = [93] -> R1C89 = [93], R1C3 = 8, all placed for R1, 8 placed for C3, 3 placed for C9, no 7 in R1C24, no 7,8,9 in R2C24, no 7,9 in R2C3, no 8,9 in R2C7, no 2,3,4,8 in R2C8, no 2,4,8,9 in R2C9 (AK, FNC, NC)
1c. Unspecified right-hand totals only in R3 and R5, upper total 19 in C9 with R1C9 = 3 cannot contain 8 -> R3C9 = 9, R5C9 = 8, both placed for C9, 9 placed for R3, 8 placed for R5, no 9 in R2C8, no 8 in R3C8, no 7,8,9 in R46C8, no 7 in R46C9, no 7,9 in R5C8 (AK, FNC, NC)
1d. Upper total in C9 = 19, R13C9 = [39] = 12 -> R2C9 = 7 (cage sum), placed for C9, no 6 in R2C8, no 6,7 in R3C8 (AK, FNC, NC)
1e. Unspecified left-hand totals only in R2 and R6, cannot make upper total in C1 = 18 with 9 in R2C1 -> R2C1 = 8, R6C1 = 9, both placed for C1, no 7 in R13C1, no 7,8 in R3C2, no 9 in R5C2, no 8 in R6C2, no 8,9 in R7C2 (AK, FNC, NC)
1f. Upper total in C1 = 18, R26C1 = [89] = 17 -> R1C1 = 1 (cage sum), placed for R1 and C1, no 2 in R1C2, no 1,2 in R2C2 (AK, FNC, NC)
1g. R1C2 = {456} -> no 5 in R2C3 (AK, FNC)
1h. Unspecified lower totals only in C4 and C8, left-hand total in R9 = 15 must contain 9 = [249/69] -> R9C1 = {26}, R9C4 = 9, R9C8 = 8, both placed for R9, no 9 in R8C3, no 8 in R8C4, no 8,9 in R8C5, no 7,8,9 in R8C7, no 7,9 in R8C8, no 7 in R9C7 (AK, FNC, NC)
1i. Left-hand total in R9 = 15 = [249/69] cannot contain 7 -> no 7 in R9C23, also no 5,6 in R9C2, no 6 in R9C3 (NC)
1j. 7 in R1 only in R1C567 -> no 6 in R1C6, no 6,8 in R2C6 (FNC, NC)
1k. 7 in R9 only in R9C56 -> no 6,7 in R8C5, no 6,8 in R8C6, no 6 in R9C56 (AK, FNC, NC)
1l. Right-hand total in R9 = xy must be at least 10, R9C8 = 8 -> no 1 in R9C9

2a. C2 must contain at least one of 7,8,9, lower total in C2 = 10 = [713/73/82/91] with possible repeats (cannot be [721] because of NC) -> R9C2 = {123}, no 2 in R89C13 (AK, FNC, NC)
2b. R9C1 = 6, placed for R9 and C1, no 5,7 in R8C1, no 5,6,7 in R8C2 (AK, FNC, NC)
2c. R8C1 = {34} -> no 3,4 in R7C12, no 3 in R9C2 (AK, FNC, NC)
2d. R9C2 = {12} -> no 1 in R89C3 (AK, FNC, NC)
2e. R9C3 = {345} -> no 4 in R8C234 (AK, FNC)

3a. Upper total in C5 = 19 must contain 9 -> total of remaining visible cells = 10 = [28/46] (cannot be [235] or [2135] because of NC) -> R1C5 = {24}
3b. R1C5 = {24} -> no 3 in R2C456 (FNC, NC)
3c. Remaining visible cells = [28/46] -> R2C5 = {1268}
3d. Either R2C5 is an invisible cell or R2C5 = {68} -> no 7 in R3C5 (NC)
3e. 7 in R1 only in R1C67 -> no 6 in R1C7, no 6,7 in R2C7 (AK, NC)
3f. 6 in R1 only in R1C24 -> no 6 in R2C3 (AK)

4a. Left-hand total in R3 = 30 must contain both of 8,9 since 9 is in R3C9 -> remaining visible cells must total 13 = [247/256/346] -> R3C1 = {23}, no 3 in R2C2, no 2,3 in R3C2 + R4C12 (AK, FNC, NC)
4b. The second visible digit must be 4 or 5 -> no 6 in R3C2, also no 6 in R3C3 because [247] can only have 6 after 7 while [256/346] must include one hidden cell because of NC
4c. 6 of [256/346] cannot be further left that R3C4 while 7 of [247] cannot be further left than R3C3 -> no 8 in R3C4 (NC)
4d. 2 of [247/256] must be in R3C1, [346] must start [31] -> no 2 in R3C3 (NC)
4e. 8 in R3 only in R3C567 -> no 7,9 in R24C6, no 7 in R3C6 (AK, FNC, NC)
4f. R2C2 = {456} -> no 5 in R3C3 (AK, FNC)
4g. R45C1 cannot both be {45} (NC) -> R78C1 must contain one of 4,5 -> no 5 in R7C2 (NC)

5a. Left-hand total in R4 = 16, R4C1 = {457} -> only possible remaining visible permutations are [457/79] -> R4C1 = {47}
5b. [457/79] -> no 6,8 in R4C2 (NC)
5c. 5 in C1 only in R57C1 -> no 4,5,6 in R6C2 (AK,FNC)
5d. Lower total in C2 (step 2a) = [82/91], may possibly contain 1 in hidden cells or R89C2 = [82] (only remaining 8 in C2) -> no 2,3 in R8C2

6a. Remaining visible cells at left-hand end of R3 (step 4a) = [247/256/346] -> R3C12 = [24/25/31] -> no 4 in R4C1 (AK, FNC, NC)
6b. R4C1 = 7, placed for C1, no 6,7 in R5C2 (AK, FNC)
6c. R3578C1 = [2453/3524] -> R5C1 = {45}, no 4,5 in R45C2 (NC)
6d. R5C2 = {123} -> no 2 in R456C3 (AK, FNC, NC)
6e. Consider placements for R3C1 = {23}
R3C1 = 2, placed for C1
or R3C1 = 3, R3C2 = 1 => no 2 in R2C3 (FNC) => R7C3 = 2 (hidden single in C3)
-> R7C1 = 5, R358C1 = [243], 2 placed for R3, 4 placed for R5, 5 placed for R7, no 1 in R34C2, no 3 in R56C2, no 2,6 in R7C2, no 2 in R9C2 (FNC, NC)
6f. R9C2 = 1, placed for R9, no 8 in R8C2 (step 5d)
6g. R3C2 = {45} -> no 4 in R23C3, no 4,5 in R4C3 (AK, FNC, NC)
6h. R2C3 = {123} -> no 2 in R1C4 (AK, FNC)
6i. 2 in R1 only in R1C567 -> no 1 in R2C6 (FNC, NC)
6j. R5C2 = {12} -> no 1 in R456C3 (AK, FNC, NC)

7a. Consider placements for 2 in C3
R2C3 = 2 => no 1,3 in R3C3 (NC) => R3C3 = 7
or R7C3 = 2 => no 1 in R7C2, no 3 in R68C3 (NC), R7C2 = 7 => no 6,7 in R8C3 (AK, FNC), R8C3 = 5 => no 4 in R9C3 (NC), R9C3 = 3, R23C3 = [17]
-> R3C3 = 7, placed for R3 and C3
7b. R3C3 = 7 -> no 6 in R2C24 + R3C4 + R4C3, no 7 in R4C2, no 6,7,8 in R4C4 (AK, FNC, NC)
7c. R4C2 = 9 -> no 9 in R5C3 (AK)
7d. Right-hand total in R4 = 15 must contain 9 so no 7,8 visible, 9 repeated can be as far right at R4C7 -> no 7,8 in R4C7, no 8 in R4C6 (NC)
7e. R27C3 = {12} (hidden pair in C3) -> no 1 in R3C4 + R78C2, no 1,2 in R6C2 + R678C4 (AK, FNC, NC)
7f. R7C2 = 7, placed for R7, no 6 in R68C3 (FNC)
7g. R8C3 = {35} -> no 4 in R9C3 (NC)
7h. Naked pair {35} in R89C3, locked for C3, R456C3 = [964], 6 placed for R5, no 5 in R4C4, no 3,5,7,9 in R5C4, no 7 in R6C2, no 3,5,6,7 in R6C4, no 3,4 in R7C4 (AK, FNC, NC)
7i. Naked pair {12} in R5C24, locked for R5
7j. R5C4 = {12} -> no 1,2 in R46C5 (AK, FNC)
7k. R5C8 = {35} -> no 4 in R46C789 (AK, FNC, NC)
7l. R3C13 = [27] -> R3C2 (step 6a) = 4, placed for R3, no 5 in R2C2 (NC)
7m. R2C2 = 4 -> no 5 in R1C2 (NC)
7n. R3C4 = {35} -> no 4 in R2C4 + R4C45 (FNC, NC)
7o. 7 in R5 only in R5C567 -> no 6 in R4C6, no 6,8 in R6C6 (AK, FNC, NC)

8a. Right-hand total in R4 = 15 must contain 9 -> visible cells = [942/951/96], no 5 in R4C9
8b. 4 in C9 only in R789C9 -> no 3,5 in R8C8, no 5 in R8C9 (FNC, NC)
8c. Consider placements for 5 in C9
R6C9 = 5 => no 4,6 in R7C9
or R9C9 = 5 => no 4,6 in R8C9
-> at least one of R78C9 must be {12} but cannot both be {12} (NC) -> R4C9 = {12}
8d. Right-hand total in R4 = 15, no 4 in R4C8 -> visible cells = [951], 1 placed for C9, no 1 in R3C8, no 6 in R4C7, no 2,3,6 in R4C8 (AK, FNC, NC)
8e. R3C8 = {35} -> no 4 in R2C7 (FNC)
8f. Lower total in C9 = 28 must contain 8,9 -> remaining total = 11 = [56] (cannot be [245] because cannot then place 6 in C9, NC) -> R9C9 = 5, placed for R9 and C9, R89C3 = [53], 3 placed for R9, no 6 in R7C4, no 3,6 in R8C4, no 4,6 in R8C9 (FNC, NC)
8g. R8C9 = 2, placed for C9 -> R67C9 = [64], 4 placed for R7, no 5 in R5C8, no 3,5 in R6C8, no 1,2,3,6 in R7C8, no 1,4,6 in R8C8 (AK, FNC, NC)
8h. R5C8 = 3, placed for R5, no 2,3 in R46C7, no 2 in R6C8 (AK, FNC, NC)
8i. R9C7 = {24} -> no 3 in R8C67 (NC)
8j. R9C567 = {247} -> no 1,5 in R8C6 (AK, FNC, NC)

9a. Consider placement for R1C2 = {46}
R1C2 = 4
or R1C2 = 6 => R1C4 = {45}, no 4 in R1C5 (NC)
-> R1C5 = 2, placed for R1 and C5
9b. Upper total in C5 = 19 must contain 9 -> total of remaining visible cells = 10 (step 3a) = [28] -> R2C5 = 8, placed for C5, no 7 in R1C6, no 8 in R3C6 (AK, FNC)
9c. R1C5 = 2 -> no 2 in R2C46 (AK)
9d. R1C7 = 7 (hidden single in R1), placed for C7, no 7 in R2C8 (AK)
9e. R3C7 = 8 (hidden single in R3), placed for C7
9f. R1C6 = {45} -> no 5 in R2C7 (AK, FNC)
9g. 2 in R9 only in R9C67 -> no 1,2 in R8C7 (AK, FNC, NC)
9h. 4 in C7 only in R89C7 -> no 4 in R9C6 (AK)
9i. 6 in C7 only in R678C7 -> no 6 in R7C6 (AK)

10a. R3C56 = {16} (hidden pair in R3) -> no 5 in R2C6 (FNC, NC)
10b. R2C6 = 4 -> no 5 in R1C6, no 3 in R2C7 (NC)
10c. R1C6 = 4, placed for R1 -> R1C24 = [65]
10d. R7C7 = 3 (hidden single in C7), placed for R7, no 2,3,4 in R6C6, no 2 in R7C6, no 2,4 in R8C6, no 4 in R8C7, no 2 in R8C8 (AK, FNC, NC)
10e. R9C7 = 4 (hidden single in C7), placed for R9 -> R9C56 = [72], 7 placed for C5, R5C6 = 7 (hidden single in R5), no 6 in R4C5 + R6C57, no 7 in R8C46, no 1,3 in R8C5, no 5 in R8C7 (AK, FNC, NC)
10f. Naked pair {59} in R5C57 -> no 5 in R4C6, no 5,9 in R6C6 (AK)
10g. R2C7 = 2 (hidden single in C7), no 1 in R2C8 + R3C6, no 3 in R3C8 (FNC, NC)
10h. R3C68 = [65], placed for R3, R3C5 = 1, placed for C5, no 1 in R2C4, no 1,2 in R4C46, no 5 in R4C57 (AK, FNC)
10i. R7C5 = 6 (hidden single in R7 and C5), no 7 in R6C6, no 5 in R8C45 (FNC, NC)
10j. 3 in C5 only in R46C5 -> no 2 in R5C4 (FNC)
10k. R5C4 = 1, placed for R5

[And now to apply the final condition, 4 in a row twice and 3 in a row twice.]
11a. 4 in a row is only possible in R4 and C4 -> R4C2345 = [9999], R6789C4 = [9999], 9 placed for R7 and C5 -> R5C5 = 5, placed for R5 and C5, R5C7 = 9, placed for C7, R68C5 = [34], R7C368 = [218], R2C3 = 1, no 4 in R4C6 (FNC)
11b. R7C6 = 1 -> no 1 in R6C7 (AK)
11c. R6C7 = 5 -> no 6 in R6C8 (NC)
11d. Right-hand total in R4 = 15, visible cells must be [951] -> R4C8 = 5 (alternatively R234C8 are the second 3 in a row)

Now we are down to naked singles.

I've checked through my WT several times making detail changes. I hope I've now got it right; if you spot any errors, particularly in the AK/FNC/NC eliminations, please send me a PM.


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 Post subject: Re: SSS ORC 2 & 2H
PostPosted: Wed Feb 07, 2018 5:31 pm 
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Grand Master
Grand Master

Joined: Wed Apr 30, 2008 9:45 pm
Posts: 539
Excellent walk through Andrew, a few solving ideas I had not seen. I particularly missed using the information from the xy clue to eliminate a number.

My solution was similar but in a different order and when I'm down to two or three possibilities for a clue I like to use interactions to eliminate adjacent column/row numbers (while you did this a couple of times I do it a lot).


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