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 Post subject: SSS ORC 10
PostPosted: Sat Jul 22, 2017 5:10 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
SkyScraper Sum with Odd Row Column 10

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

There are no three-in-a-rows or higher (but I did not use this property).

The two red sums are needed for uniqueness - chose just 3 more sums and solve it.


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 Post subject: Re: SSS ORC 10
PostPosted: Thu Mar 08, 2018 10:20 pm 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks HATMAN for another interesting puzzle.

The challenge of selecting 3 more totals wasn't as difficult a challenge as I'd expected. Don't think I'm giving anything away by saying that I chose totals which led to multiple placements. I wonder whether there's a better group of 3 than the ones which I chose.

Here is my walkthrough for SSS OCR 10:
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.
There are no three-in-a rows or more in any columns/rows; although HATMAN says that he didn’t use that property to solve this puzzle.
Since this an ORC puzzle, I’ve stated placements in odd rows/columns.
It has been specified that all totals in the even-numbered rows and columns must contain at least one of 7,8,9.
This puzzle’s post states that lower totals in C4 and C8 are required for uniqueness; the challenge is to solve the puzzle using only three of the other totals.
I checked with HATMAN who confirmed that blank totals may also be used. They are probably necessary to solve the puzzle with only using five of the given totals, of which two have been specified.

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, C6 and C7, 8,9 cannot both be in R1C67 because of NC -> R1C3 = {89}, one of R1C67 must contain one of 8,9, no 8,9 in R1C124589, no 8,9 in R2C2347 (AK, FNC, NC)
1c. Unspecified left-hand totals only in R7 and R9 -> R79C1 = {89}, locked for C1, no 8,9 in R679C2, no 7 in R8C1, no 7,8,9 in R8C2 (AK,FNC, NC)
1d. Unspecified right-hand totals only in R4 and R8 -> R48C9 = {89}, locked for C9, no 8,9 in R3579C2 (AK,FNC)
1e. Unspecified lower totals only in R1 and R5 -> R9C15 = {89}, locked for R9, no 8,9 in R8C456 (AK,FNC, NC)

2. Lower total in C8 = 12 must include one of 7,8,9 = {138/147/39/48/57} (cannot be {129/237} NC) -> R9C8 = {1345}

3a. Lower total in C4 = 26 must contain at least four digits, and cannot be made from four of 6,7,8,9 -> no 6,7 in R9C4
3b. 26 must either contain 9 or be {5678} with hidden digits -> no 9 in R7C4

[Now I need to decide which of the other totals to use. Left-hand total in R3 = 11 looks an obvious one to start with, but I’ll try at least one other first.]

4a. Lower total in C7 = 35 must contain 7,8,9 with two or three other digits (cannot be 6,8,9 because there isn’t enough space to hide 7 and make up the remaining 12 without NC) = {146789/245789/56789} -> R135C7 = [987], placed for C7, 9 placed for R1, 8 placed for R3 and 7 placed for R5, R1C3 = 8, placed for R1 and C3, no 7 in R1C24 + R2C234 + R3C8, no 7,8,9 in R2C68, no 7,9 in R3C6, no 6,7,8,9 in R4C68, no 6 in R46C7 + R5C8, no 6,8 in R5C6, no 6,7,8 in R6C68 (AK,FNC, NC)
4b. R9C7 = {125}, second digit must be either 4 or hidden -> no 5,6 in R8C7
4c. R8C7 = 4 or R9C7 = 5 -> no 5 in R8C68, no 4,5 in R9C68 (AK,FNC,NC)
4d. R9C8 = {13} -> no 2 in R8C78 + R9C79 (FNC,NC)
4e. Total 35 = {56789} (cannot be {146789} which clashes with R9C8 = {13} FNC) -> R79C7 = [65], placed for C7, 6 placed for R7 and 5 placed for R9, no 5 in R6C68, no 5,7 in R7C68, no 4,6,7 in R8C68, no 4 in R8C7, no 6 in R9C6 (AK,FNC,NC)
4f. R8C7 = {13} -> no 2 in R7C68 + R89C6 (FNC,NC)
4g. R8C6 = {13} -> no 2 in R78C5 (FNC,NC)
4h. 2 in R9 only in R9C234 -> no 1,2,3 in R8C3, no 1,3 in R9C3 (AK,FNC,NC)

[And now I need to select another total which gives multiple placements.]

5a. Lower total in C8 = 12 (step 2) = {147/39} (cannot be {138} because 8 in C8 only in R8C8) -> no 8 in R8C8
5b. Right-hand total in R7 = 29 must contain 9 and 6 in R7C7 = {24689} (cannot be {34679} NC) -> R7C1 = 9, placed for R7 and C1, R7C89 = [42], placed for R7, 2 placed for C9, R9C1 = 8, placed for R9, R9C5 = 9, placed for C5, no 3,4 in R6C7, no 1,2,3 in R6C8, no 1,3,4,5 in R6C9, no 3 in R8C7, no 1,3 in R8C8, no 7 in R9C2 (AK,FNC,NC)
5c. R8C8 = 9 -> no 8 in R8C9 (NC)
5d. R8C9 = 9, placed for C9, R4C9 = 8, no 7 in R3C9 (NC)
5e. R68C7 = [21], placed for C7, no 1,2,3 in R5C68, no 1,3 in R67C6, no 1 in R9C68 (AK,FNC,NC)
5f. R9C68 = [73], placed for R9, no 6,7 in R8C5, no 4 in R9C9 (AK,FNC,NC)
5g. R7C6 = 8, placed for R7, no 7,8 in R6C5, no 9 in R6C6, no 7 in R7C5 (AK,FNC,NC)
5h. R5C8 = {45} -> no 4 in R4C7 (AK,FNC)
5i. R24C7 = [43] -> no 3,4,5 in R1C68, no 3,5 in R2C68, no 2,3,4,5 in R3C68, no 2,4 in R4C68, no 4 in R5C68 (AK,FNC,NC)
5j. R5C68 = [95], placed for R5, no 8 in R45C5, no 4,6 in R5C9, no 4 in R6C8 (FNC,NC)
5k. Naked pair {16} in R3C68, locked for R3
5l. 4,5 in C9 only in R123C9 -> no 4,5 in R2C9 (NC)
5t. R13C9 = {45} (hidden pair in C9), no 4,6 in R2C8, no 3,6 in R2C9 (AK,NC)
5u. R5C9 = 3 (hidden single in C9), placed for R5, no 3 in R4C8 (AK)
5v. R6C6 = {24} -> no 3 in R67C5 (FNC,NC)
5w. 8 in R5 only in R5C24, no 7,9 in R46C3 (FNC)
5x. 3,7 in R7 only in R7C234 -> no 2,3,4,6 in R6C3, no 4,6 in R8C3 (AK,FNC)
5y. 4 in R9 only in R9C234 -> no 5 in R8C3 (FNC)

[Time to select a third total.]

6a. Upper total in C8 = 16 = {79} (cannot be {169} because R13C8 = [16] clashes with R2C9 = {17} AK,FNC, cannot be {259} because R123C8 = [226] (no 1 in R23C8, NC) makes 6 ‘visible’) -> R1C8 = 7, placed for R1, no 7 in R2C9 (AK)
6b. R2C9 = 1, placed for C9, no 2 in R2C8, no 1 in R3C8 (AK,NC)
6c. R9C9 = 6, placed for R9 and C9
6d. R3C8 = 6, placed for R3, no 5 in R3C9 + R4C8 (NC)
6e. R13C9 = [54], 5 placed for R1, 4 placed for R3
6f. R3C6 = 1 -> no 1,2 in R24C5, no 2 in R2C6 + R3C5 (AK,FNC,NC)
6g. 1 in R9 only in R9C24 -> no 2 in R9C3 (NC)
6h. R9C3 = 4, placed for R9 and C3, no 3,4,5 in R8C24 (AK,FNC)
6i. R9C2 = {12} -> no 1,2 in R8C1 (AK,FNC)
6j. R9C4 = {12} -> no 1 in R8C5 (AK,FNC)
6k. R2C5 = 8 (hidden single in C5) -> no 7,9 in R3C4, no 7 in R3C5 (FNC,NC)
6l. R4C5 = 7 (hidden single in C5) -> no 6,8 in R45C4, no 6 in R5C5 (FNC,NC)
6m. R3C5 = {35} -> no 4 in R2C46 + R4C4 (FNC,NC)
6n. R5C2 = 8 (hidden single in R5) -> no 7 in R4C1 + R6C12, no 7,9 in R4C2 (FNC,NC)
6o. 6 in R5 only in R5C13 -> no 5,6 in R46C2 (AK,FNC)
6p. 7 in C1 only in R23C1 -> no 6 in R2C12, no 7 in R3C3 (AK,FNC,NC)
6q. 7 in R3 only in R3C13 -> no 8 in R4C2 (FNC)
6r. R3C45 = [25]/{35} (cannot be {23} NC), 5 locked for R3, no 6 in R2C4 (FNC,NC)
6s. R7C5 = 1 (R78C5 cannot be [53] which clashes with R3C5, cannot be [54] NC), placed for C5 and R7, no 1,2 in R68C4, no 2 in R6C56, no 1 in R8C6 (AK,FNC,NC)
6t. R6C6 = 4 -> no 4 in R5C5, no 5 in R6C5 (AK,NC)
6u. R5C5 = 2, placed for R5 and N5, no 1,2,3 in R4C4, no 1,3 in R4C6, no 1 in R5C4, no 3 in R6C4 (AK,FNC,NC)
6v. R4C6 = 5 -> no 5 in R3C5 (AK)
6w. R3C5 = 3, placed for R3 and C5, no 2,3 in R2C4, no 2 in R3C4 (AK,FNC,NC)
6x. R5C4 = 4, placed for R5, no 3,5 in R4C3, no 5 in R4C4 + R6C34, no 4 in R6C5 (AK,FNC)
6y. R168C5 = [465], 4 placed for R1, no 3 in R1C4, no 5 in R2C4, no 7 in R6C4, no 5,7 in R7C4, no 6 in R8C4 (FNC,NC)
6z. R3C4 = 5 -> no 5,6 in R2C3, no 6 in R2C4 + R4C3
6aa. Naked pair {16} in R56C3, locked for C3 -> R24C3 = [32], no 2,3 in R1C2, no 2 in R1C4 + R3C2, no 2,4 in R2C2, no 1,3 in R4C2, no 1 in R5C3 (AK,FNC,NC)
6ab. R3C123 = [297], 2 placed for C1, 7 placed for C3, R5678C3 = [6159], 6 placed for R5, 5 placed for R7, no 1,3 in R2C12 + R4C1, no 2 in R4C2, no 7 in R4C4, no 2,4 in R6C2, no 4,6 in R6C4 (AK,FNC,NC)
6ac. R7C24 = [73] -> no 6 in R6C1 + R8C12 (FNC,NC)
6ad. R1C6 = 2 (hidden single in R1) -> no 1 in R2C6 (NC)
6ae. R1C1 = 3 (hidden single in R1), placed for C1
6af. R2C1 = 7 (hidden single in C1) -> no 6 in R1C2
6ag. R5C1 = 1, placed for C1, no 1 in R6C2 (AK)
6ah. R6C2 = 3 -> no 4 in R6C1 (NC)

7. Lower total in C4 = 26 with R48C4 = [97] must be {2789} -> R69C4 = [82], R9C2 = 1 -> R8C2 = 1 (NC)

and the rest is naked singles, using odd rows/columns.

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


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