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 Post subject: SkyScraper Sum NC 1
PostPosted: Thu Mar 23, 2017 11:26 am 
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Grand Master
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SkyScraper Sum NC 1 and 1 KiMo

SSS the clues are the sum of the numbers visible (with larger numbers hiding smaller ones)

Non-Consecutive (horizontally and vertically)

SSS NC 1 (corrected)


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SSS NC 1 KiMO

Same puzzle as a KiMo - slightly harder

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 Post subject: Re: SkyScraper Sum NC 1
PostPosted: Sat Jun 17, 2017 4:50 am 
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Location: Lethbridge, Alberta, Canada
Thanks HATMAN for another nice puzzle. Having just done the first SkyScraper Sum puzzle, I found this one quite a lot easier, even though some of the totals were in KiMo form. I must be getting more used to the SkyScraper form, not restarts this time. :)

Here is my walkthrough:
Sums are for heights of skyscrapers visible from the edges. Only those higher than the previous one(s) count toward the sums.
Therefore each sum must include 9, the height of the highest skyscraper, which can be seen from each edge. Since there are no totals given as x9, each sum must include the total of at least two skyscrapers so x cannot be zero.
Non-consecutive horizontally and vertically

Edited. First time through I'd missed a couple of NCs, including one important one; step 3g now deleted.

1a. Left-hand sum in R6 = 11 -> R6C12 = [29] (cannot have R6C123 = [219] because of NC), no 1,3 in R57C1, no 8 in R57C2 + R7C3 (NC)
1b. R1 and R5 only rows with total unspecified but 9 already locked for N4 -> R1C1 = 9, no 8 in R1C2 + R2C1 (NC)
1c. Left-hand sum in R7 = 13 -> visible cells must be [49] -> R7C1 = 4, 9 must be R7C345, locked for R7, no 3,5 in R7C2 + R8C1 (NC)
1d. R7C2 must be a hidden cell in R7 -> R7C2 = {12} -> no 1,2 in R7C3 + R8C2 (NC)
1e. Lower sum in C2 = x2 = 22 (cannot be 12 because min R789C2 = 6) -> visible cells total 13 -> R89C2 = [85] (cannot be [76] because of NC), no 7 in R8C13, no 6 in R9C13 (NC)
1f. 9 in N7 only in R789C3, lower sum in C3 = x8 must be 18 -> R789C3 = [963], R89C1 = [17], R7C2 = 2, no 8 in R7C4, no 5,7 in R8C4, no 2,4 in R9C4 (NC)

2a. C4 and C9 only columns with lower total unspecified but cannot make lower-left total in C9 = x4 with 9 in R9C4 -> R9C9 = 9, no 8 in R9C8 (NC)
2b. Only possible position in C1 for 8 is R5, because other left-hand totals cannot be 17 -> R5C1 = 8, no 7 in R5C2 (NC)
2c. Only possible position in C9 for 8 is R2, because other right-hand totals cannot be 17 -> R2C9 = 8, no 7 in R13C9, no 7,9 in R2C8 (NC)
2d. Upper total in C9 = 19 -> R1C9 = 2, no 1,3 in R1C8 (NC)
2e. Only possible position in R1 for 8 is C5, because other upper totals cannot be 17 -> R1C5 = 8, no 7 in R1C46, no 7,9 in R2C5 (NC)
2f. Only possible position in R9 for 8 is C4, because other lower totals cannot be 17 -> R9C4 = 8, no 9 in R8C4 (NC)
2g. Right-hand total in R1 = 29, R1C1 = 9, R1C5 = 8, R1C9 = 2 -> sum of remaining visible cells = 10 = [64] (cannot be [73] because no 1,2,3 in R1C8) -> R1C8 = 4, 6 in one of R1C67, locked for R1, no 5,7 in R1C67, no 3 in R1C7, no 3,5 in R2C8 (NC)
2h. Sum of visible cells starting with 7 must be either 16 or 24, upper totals for C2 and C3 are x6 and x2 -> R16C2 = [79], no 6 in R2C2 (NC)
2i. Upper total in C3 must include 8 in one of R23C3 and 9 in R7C3, upper total x2 must be 22 -> R1C3 = 5
2j. R4C1 = 5 (hidden single in C1) -> no 6 in R3C1, no 4,6 in R4C2 (NC)
2k. R23C1 = [63] -> no 4 in R3C2 (NC)
2l. R23C2 = [41] -> no 2 in R3C3 (NC)
2m. R23C3 = [28] -> no 1,3 in R2C4, no 7,9 in R3C4, no 7 in R4C3 (NC)
2n. R45C2 = [36] -> no 4 in R4C3, no 7 in R5C3 (NC)
2o. R456C3 = [147] -> no 2 in R4C4, no 3,5 in R5C4, no 6 in R6C4 (NC)
2p. R2C8 = 1, R1C7 = 6, R3C9 = 5, R2C7 = 3 (hidden single in N3), R2C5 = 5, no 4,6 in R3C5 + R4C9 (NC)
2q. R4C9 = 7 -> no 6,8 in R4C8 (NC)
2r. Right-hand total in R3 = 21, R3C9 = 5 -> sum of visible cells = 16 -> R3C78 = [97], no 8 in R4C7 (NC)
2s. R4C6 = 8 (hidden single in R4) -> no 9 in R4C5, no 7,9 in R5C6 (NC)
2t. R4C9 = 7, right-hand total in R4 = x6 must be 16 -> R4C8 = 9

3a. Upper total in C6 = x5 must be at least 15, R1C6 {13} -> no 9 in R2C6 -> R2C46 = [97], no 6 in R2C6 (NC)
3b. R8C6 = 9 (hidden single in C6)
3c. Upper total in C6 contains 7,8,9 must be 25 -> R1C6 = 1, R1C4 = 3
3d. R8C6 = 9, lower total in C6 = x5 must be 15 -> R9C6 = 6
3e. R8C4 = {24} -> no 3 in R8C5 (NC)
3f. 3 in R8 only in R8C89, R9C6 = 9, right-hand total in R8 = x4 must contain 3,9 -> sum of remaining visible cells must be 12 -> R8C789 = [753], no 8 in R7C7, no 6 in R7C8 (NC)

and the rest is naked singles.

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


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 Post subject: Re: SkyScraper Sum NC 1
PostPosted: Sat Jun 17, 2017 8:43 am 
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
SSS NC 1 EH

I've tightened the puzzle up (the only problem with this approach is that it makes the puzzle path a bit obvious).

Image

puzzle corrected


Last edited by HATMAN on Wed Jun 21, 2017 3:51 pm, edited 1 time in total.

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 Post subject: Re: SkyScraper Sum NC 1
PostPosted: Sun Jun 18, 2017 10:49 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
And then I tried the KiMo version. HATMAN said "slightly harder"; a bit of an underestimate, unless it's because I haven't done many KiMo puzzles, I thought it was about as hard as the original Sky Scraper Sum 0.

Here is my walkthrough for the KiMo version:
Sums are for heights of skyscrapers visible from the edges. Only those higher than the previous one(s) count toward the sums.
Therefore each sum must include 9, the height of the highest skyscraper, which can be seen from each edge. This is a KiMo, with the totals being xn where it has been specified that x is never 0.

1a. R2 and R9 are the only rows with right-hand total unspecified, cannot make upper total x9 in C9 with 9 in R2C9 -> R9C9 = 9, no 8 in R8C9 + R9C8 (NC)
1b. C4 and C9 are the only columns with lower total unspecified, and none of the lower totals are x7, R9C9 = 9 -> R9C4 = 8, no 7,9 in R8C4, no 7 in R9C35 (NC)
1c. R1 and R5 are the only rows with left-hand total unspecified, and none of the left-hand totals are x7 -> R15C1 = {89}, locked for C1
1d. C1 and C5 are the only columns with upper total unspecified, and none of the left-hand totals are x7 -> R1C15 = {89}, locked for R1
1e. R2 and R9 are only rows with right-hand total unspecified, R6 has right-hand total x7, R9C9 = 9 -> 8 in C9 must be in one of R26C9, locked for C9
1f. No 8,9 in R2C5 and R5C2 (NC)

2a. Row starting 7 must total either 16 or 24, R9 is the only row with left-hand total either x4 or x6 (unspecified totals already used) -> R9C1 = 7, no 6 in R8C1 + R9C2 (NC)
2b. Column starting 7 must total either 16 or 24, C2 is the only column with upper total either x4 or x6 (unspecified totals already used) -> R1C2 = 7, no 8 in R1C1, no 6 in R1C3, no 6,8 in R2C2 (NC)
2c. R1C1 = 9, R1C5 = 8, R5C1 = 8, no 7 in R2C5 (NC)
2d. Upper total in C2 = 16 = [79] -> no 8 in C2 above the 9 -> no 8 in R3C2
2e. 8 in C2 only in R78C2, locked for N7
2f. 8 in R78C2 -> no 9 in R78C2 (NC)
2g. 9 in N7 only in R78C3, locked for C3
2h. Lower total in C3 = x8 contains 9 in R78C3 must be 18 = [963] (cannot be [954] because of NC), no 8 in R7C2, no 5 in R8C24, no 2,4 in R9C2 (NC)
2i. R8C2 = 8 (hidden single in C2)
2j. Lower total in C2 = x2 contains 9 in R46C2 and 8 in R8C2 must be 22 -> R9C2 =5

3a. Row ending 7 must total 16, 24 or have unspecified total -> 7 in C9 must be in one of R24C9 (7 cannot be in R8C9 because 8 in R8 is to the left of the 9), locked for C9
3b. 7 in R24C9 -> no 6 in R3C9 (NC)
3c. 8 in N9 only in R7C78 -> no 7 in R7C78 (NC)
3d. 7 in N9 only in R8C7 (7 cannot be in R8C7 because cannot make lower total x4 in C8), no 6,8 in R7C7 (NC)
3e. R7C8 = 8 (hidden single in N9), no 7,9 in R6C8 (NC)
3f. Lower total x4 in C8 contains 8,9 must be 24 -> remaining cells must total 7 -> R89C8 = [52], no 4 in R8C9, no 1 in R9C7 (NC)
3g. Right-hand total in R8 = x4 contains 5,7,9 must total 24 -> R8C9 = 3, no 4 in R7C9 (NC)
3h. 1 in R9 only in R9C56, locked for N8
3i. R8C1 = 1 (hidden single in R8), no 2 in R7C1 (NC)
3j. R7C12 = [42], R7C79 = [16], R9C7 = 4, no 3,5 in R6C1, no 1,3 in R6C2, no 2 in R6C7, no 5 in R6C9 (NC)

4a. Lower total in C5 = x1 cannot be made from [69] or [679] -> R9C5 = 1, R9C6 = 6, no 2 in R8C5 (NC)
4b. Lower total in C5 = x1 cannot be made with 9 in R8C5 -> R8C5 = 4, R8C4 = [29], no 3 in R7C4, no 3,5 in R7C5 (NC)
4c. R7C456 = [573], no 4,6 in R6C4, no 6 in R6C5, no 2,4 in R6C6 (NC)

5a. Left-hand total in R6 = x1 cannot be made from [69] -> R6C1 = 2
5b. 3 in C1 must be in R23C1 (R23C1 cannot be {56} NC)
5c. R4C1 = {56} -> no 5,6 in R3C1, no 6 in R4C2 {NC)
5d. R3C1 = 3 (hidden single in C1), no 4 in R3C2 (NC)
5e. Left-hand total in R4 = x8 cannot be made from [59/69] -> no 9 in R4C2
5f. R6C2 = 9 (hidden single in C2)

6a. Upper total in C3 = x2 contains 8,9 must be 22 = [589] (cannot be [1489] which clashes with R2C2) -> R1C3 = 5, R24C1 = [65], R23C2 = [41], R45C2 = [36], no 4,6 in R1C4, no 2 in R3C3, no 4 in R4C3, no 7 in R5C3 (NC)
6b. R23C3 = [28], no 1,3 in R2C4, no 7,9 in R3C4, no 7 in R4C3 (NC)
6c. R456C3 = [147], no 3 in R5C4 (NC)
6d. Upper total in C4 = x2 must be 12 (cannot be 22) -> R12C4 = [39], R2C5 = 5, R56C4 = [71], no 6 in R4C4 (NC)
6e. R34C4 = [64], R3456C5 = [2693]
6f. 6 in R6 only in R6C78 -> no 5 in R6C7 (NC)
6g. R6C6 = 5 (hidden single in R6) -> R45C6 = [82], no 7 in R3C6, no 9 in R4C7, no 6 in R6C7 (NC)

Now to try the EH version and find out whether EH means Extra Hard or Extremely Hard.


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 Post subject: Re: SkyScraper Sum NC 1
PostPosted: Sun Jun 25, 2017 2:27 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
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Thanks HATMAN for correcting the EH version. I initially tried it with x not 0, as in the earlier versions, but y unspecified. After getting about 25 placements I was struggling; it probably can't be solved or, if it can, it would involve a lot of T&E.

With x not 0 and y not 7, it's not a difficult puzzle; it felt about the same level as the earlier SkyScraper Sum puzzles that I've done recently.

Here is my walkthrough for the EH version:
Sums are for heights of skyscrapers visible from the edges. Only those higher than the previous one(s) count toward the sums.
Therefore each sum must include 9, the height of the highest skyscraper, which can be seen from each edge. This is a KiMo, with the totals being xy where it has been specified that x is never 0, y never 7.

1a. Since no totals are given as 17 and xy cannot be 17, rows/columns starting/finishing with 8,9 must be ones with unspecified totals.
1b. R2 and R9 are the only rows with right-hand total unspecified, cannot make upper total 19 in C9 with 9 in R2C9 -> R9C9 = 9, R2C9 = 8, no 7 in R13C9, no 7,9 in R2C8, no 8 in R8C9 (NC)
1b. Upper total in C9 = 19 must be [289] -> R1C9 = 2, no 1,3 in R1C8 (NC)
1c. Left-hand total in R6 = 11 -> R6C12 = [29] (cannot be [219] because of NC), no 1,3 in R57C1, no 8 in R57C2 + R6C3 (NC)
1d. R1 and R5 are the only rows with left-hand total unspecified, but 9 already placed for N4 -> R1C1 = 9, R5C1 = 8, no 8 in R1C2, no 7 in R4C1 + R5C2 (NC)
1e. Only remaining unspecified upper total is in C5 -> R1C5 = 8, no 7 in R1C46, no 7,9 in R2C5 (NC)
1f. Only remaining unspecified lower total is in C4 -> R9C4 = 8, no 7,9 in R8C4, no 7 in R9C35 (NC)
1g. 9 in C3 only in R78C3 -> no 8 in R78C3 (NC)
1h. R8C2 = 8 (hidden single in N7) -> no 7 in R79C2 + R8C1, no 7,9 in R8C3 (NC)
1i. R7C3 = 9 (hidden single in N7)
1j. R3C3 = 8 (hidden single in N1) -> no 7 in R24C3 + R3C2, no 7,9 in R3C4 (NC)
1k. 2 in N1 only in R2C23 + R3C2 -> 1,3 in R2C2 (NC)
1l. 7 in N7 only in R79C1, locked for C1
1m. 7 in R79C1 -> no 6 in R8C1 (NC)
1n. 7 in N1 only in R1C23 + R2C2 -> no 6 in R1C2 (NC)
1o. 8 in N9 only in R7C78 -> no 7 in R7C78 (NC)

2a. Right-hand total in R1 = 29 must include 2,8,9 -> sum of remaining visible cells = 10 = [64] (because min R1C8 = 4) -> R1C8 = 4, no 3,5 in R1C7 + R2C8 (NC)
2b. This total doesn’t include 7 -> 7 in R1 must be to the left of the 8 -> no 7 in R1C7 (NC)
2c. Naked pair {16} in R1C7 + R2C8, locked for N3
2d. Naked pair {16} in R1C7 + R2C8 -> no 5,7 in R2C7 (NC)
2e. 5,7 in N3 only in R3C789, locked for R3
2f. R3C9 = {35} -> no 4 in R4C9 (NC)
2g. 7 in R1 only in R1C23, locked for N1
2h. 7 in R1C23 -> no 6 in R1C3 (NC)

3a. Left-hand total in R2 = 15 must be [159/69] (cannot be [1239] because of NC) -> R2C1 = {16}, no 6 in R2C23
3b. Total doesn’t contain 7 -> 9 in R2 must be to the left of 7 -> no 7 in R2C4, no 9 in R2C67
3c. R2C7 = 3, R3C9 = 5, no 2,4 in R2C6, no 6 in R4C9 (NC)
3d. Naked pair {16} in R2C18, locked for R2
3e. Naked pair {79} in R3C78, locked for R3
3f. R3C78 = {79} -> no 8 in R4C78 (NC)
3g. R2C46 = [97] (hidden pair in R2), no 6 in R13C6 (NC)

[Here I initially looked at positions for 7 in C9, but then realised that it may be better to look at 6 first, so I’ve done that.]
4a. Only possible totals starting with 6 are [69] = 15, [679] = 22, [689] = 23 and [6789] = 30, right-hand totals in R5 and R6 are x8 and 27 -> no 6 in R56C9
4b. 6 in C9 only in R78C9, locked for N9
4c. 6 in R78C9 -> no 7 in R78C9 (NC)
4d. Only possible totals starting with 7 are [79] = 16 and [789] = 24, right-hand totals in R4, R5 and R6 are x6, x8 and 27 -> no 7 in R56C9
4e. R4C9 = 7 (hidden single in C9), no 6 in R4C8 (NC)
4f. Right-hand total 27 in R6 must contain 8,9 and at least two more visible numbers -> no 8 in R6C8
4g. R6C7 = 8 (hidden single in N6) -> no 7,9 in R5C7, no 7 in R6C8 (NC)
4h. Right-hand total 27 in R6 contains 8,9 -> R6C89 = 10 = [64], no 5 in R57C8, no 3 in R57C9 (NC)
4i. R578C9 = [163] -> no 2 in R58C8 (NC)
4j. Right-hand total in R5 = x8 must have at least three visible numbers including 9 -> no 9 in R5C8
4k. R5C8 = 3 -> no 2 in R4C8 + R5C7 (NC)
4l. R5C7 = 5, R4C78 = [29], R3C78 = [97], no 1,3 in R4C6, no 4,6 in R5C6 (NC)
4m. R2C18 = [61], R1C7 = 6 -> no 5 in R1C6 + R2C2 (NC)
4n. R789C8 = [852] -> no 4 in R8C7, no 1 in R9C7 (NC)
4o. R4C6 = 8 (hidden single in R4) -> no 9 in R5C6 (NC)
4q. R5C5 = 9 (hidden single in R5) -> R8C6 = 9 (hidden single in R8)
4r. R9C6 = 6 (hidden single in C6) -> no 5 in R9C5, no 7 in R9C7 (NC)
4s. R789C7 = [174], R5C6 = 2
4t. R9C5 = {13} -> no 2 in R8C5 (NC)

5a. 7 in R5 only in R5C34 -> no 6 in R5C34 (NC)
5b. R5C2 = 6 (hidden single in R5) -> no 5 in R4C2, no 7 in R5C3 (NC)
5c. R5C34 = [47] -> no 3,5 in R46C3, no 6 in R4C4 (NC)
5d. R4C123 = [531] -> no 4 in R3C1, no 2,4 in R3C2 (NC)
5e. R8C3 = 6 (hidden single in C3) -> R8C4 = 2 (hidden single in R8), no 3 in R7C4, no 1 in R8C5 (NC)

and the rest is naked singles.


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