SudokuSolver Forum

A forum for Sudoku enthusiasts to share puzzles, techniques and software
It is currently Thu Mar 28, 2024 9:48 am

All times are UTC




Post new topic Reply to topic  [ 5 posts ] 
Author Message
 Post subject: SkyScraperSum
PostPosted: Fri Mar 17, 2017 10:42 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
Sky Scraper Sum 0

A new variant for me, see:
http://forum.enjoysudoku.com/post256219.html#p256219

At Andrew's suggestion:

Image


Last edited by HATMAN on Mon Mar 20, 2017 4:33 pm, edited 1 time in total.

Top
 Profile  
Reply with quote  
 Post subject: Re: SkyScraperSum
PostPosted: Sun Mar 19, 2017 8:04 pm 
Offline
Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
A new variant for me too!

Suggest that you copy the diagram to this thread, since it's your modified version of an earlier puzzle, in case that other site goes down.

Are we to assume that each of the skyscraper cages must have a foundation, so starts from one of the edges? And that there are two skyscrapers in rows/columns where two totals are given, but not necessarily two (although still probably) when only one total is given?

Not sure what you mean by "ordered" in the phrase "ordered uncaged killer" in your comments.

Possible Hint:
In some rows/columns the totals add up to less than 45, thus giving the total for one of possibly more slum cells. On other rows/columns the totals add up to more than 45, apparently indicating overlapping skyscrapers and giving the total for overlapping cell(s).


Top
 Profile  
Reply with quote  
 Post subject: Re: SkyScraperSum
PostPosted: Mon Mar 20, 2017 5:05 pm 
Offline
Grand Master
Grand Master

Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
Some points:
Each set of clues (left,right, bottom and top) must contain a 9 and a 17.
Each clue sees the highest skyscraper so includes a 9.
The sum of the two clues minus 9 is the sum of the visible skyscrapers.
45 minus this number is the sum of the slums from those two directions.
In looking at combinations the numbers are increasing from the edge, hence ordered.

The two clues on a row give a relatively low number of combinations hence if all 36 clues are given the solution should be pretty easy.
Consider 17 with 18:
89 and 972/963/954/9621/9531/9432
179 and 981 fail
269 and 981
359 and 981
1xx9 and 981 fail.

If you add NonConsecutive into the mix the combinations drop drastically


Top
 Profile  
Reply with quote  
 Post subject: Re: SkyScraperSum
PostPosted: Wed Jun 14, 2017 12:38 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks HATMAN for this new variant and for your helpful post above, after I'd been confused about how to approach it. This puzzle isn't particularly difficult, but it's very easy to make mistakes because of the hidden skyscrapers which don't count towards the sums; it took me at least three partial restarts because of silly mistakes or overlooking possibly valid ordered combinations.

While I wouldn't recommend trying the easier version of this puzzle https://s3.amazonaws.com/gs-geo-images/ ... 2531_l.jpg, which gives more immediate placements but is really a bit too easy for users of this forum, the diagram is useful for checking some of the placements in the border rows and columns.

Here's my walkthrough for SkyScraperSum:
Sums are for heights of cells (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.

1. Upper sum in C7 = 10 -> R12C7 = [19]
1a. Right-hand sum in R2 = 22 -> R2C89 = 13 = [76/85]
1b. Upper total for C3 = 13 must be 4+9 (because no 1 in R1C3) -> R1C3 = 4, 9 in C3 cannot be below R5C3 because a maximum of three hidden skyscrapers, no 9 in R2C3 -> R2C3 must be hidden -> R2C3 = {123}
1c. 4 in N3 only in R3C789, locked for R3

2a. C1 and C5 are the only columns without upper totals specified -> 9 in R1 must be in R1C1 or R1C5 but cannot make right-hand sum in R1 = 18 with R1C1 = 9 (because then sum must see at least 2 in R1C9 and 8 in R1) -> R1C5 = 9
2b. R1 and R7 are the only rows without left-hand totals specified, no 9 in R1C1 -> R7C1 = 9
2c. C2 and C6 are the only columns without lower totals specified, no 9 in R9C2 -> R9C6 = 9
2d. R4 and R8 are the only rows without right-hand totals specified -> 9 in C9 must be in R4C9 or R8C9 but cannot make lower total in C9 = 21 with 9 in R8C9 -> R4C9 = 9
2e. Upper total in C9 = 18, R4C9 = 9 -> visible cells in R123C9 must total 9 -> R12C9 = [36] -> R2C8 = 7 (step 1a), R3C9 is hidden by R2C9 -> no 8 in R3C9
2f. R8C8 = 9 (hidden single in C8), lower total in C8 = 15 -> R9C8 = 6
2g. Lower total in C9 = 21, R4C9 = 9 -> visible cells must total 12 including 8 = [84] (because 3 already placed in C9) -> R9C9 = 4
2h. Right-hand total in R9 = 19, R9C689 = [964] = 19 -> R9C7 must be hidden -> no 7,8 in R9C7
2i. Right-hand total in R1 = 18, R1C5 = 9, R1C7 = 1 (hidden), R1C9 = 3 -> R1C6 = 6, R1C8 must be hidden -> R1C8 = 2, R3C9 = 5
2j. Naked pair {48} in R3C78, locked for R3

3a. Upper totals in C2 and C4 are greater than 17 -> no 8 in R1C24
3b. R1C1 = 8 (hidden single in R1)
3c. Lower totals in C3, C4 and C5 are greater than 17 -> no 8 in R9C345
3d. R9C2 = 8 (hidden single in R9)
3e. Left-hand total in R9 = 20, R9C2 = 8, R9C6 = 9 -> R9C1 = 3 (all other cells in R9 hidden by R9C2 = 8)
3f. Lower total in C1 = 18, R7C1 = 9, R9C1 = 3 -> R8C1 = 6
3g. Left-hand total in R8 = 22, R8C1 = 6, R8C8 = 9 -> remaining visible cell must be 7 -> no 8 in R9C4567
3h. R8C9 = 8 (hidden single in R8)
3i. 4 in N7 only in R8C2, locked for C2
3j. 1 in N9 only in R7C89, locked for R7

4a. Upper total in C2 = 20 -> visible cells must be [569] -> R1C2 = 5, R1C4 = 7
4b. Naked triple {123} in R2C123, locked for R2 and N1 -> R3C1 = 7, R3C23 = [69]
4c. From step 4a, 7 cannot be visible above the 9 -> no 7 in R45C2
4d. Left-hand total in R2 = 22, R2C7 = 9, visible cells must total 13 including 8 -> R2C123 = [213/231] -> R2C1 = 2
4e. Upper total in C6 = 30, R19C6 = [69] -> visible cells must be [78] -> no 8 in R24C6, no 7 in R78C6
4f. Upper total in C4 = 24, R1C4 = 7, visible cells must total 17 = [89] -> no 8 in R67C4
4g. Lower total in C4 = 24 cannot be [9654] because no 4 in R9C4 -> R5C4 = 9
4h. R6C2 = 9 (hidden single in N4), left-hand total in R6 = 14 -> R6C1 = 5
4i. Naked pair {14} in R45C1, locked for N4
4j. Naked pair {23} in R45C2, locked for C2 and N4 -> R2C23 = [13]
4k. Naked pair {47} in R78C2, locked for N7
4m. R9C5 = 7 (hidden single in R9)

5a. Left-hand total in R5 = 21, R5C4 = 9 -> visible cells must total 12 = [138/48] -> R5C3 = 8
5b. Right-hand total in R5 = 28, visible cells must total 19 and contain 7 which cannot be in R5C79 -> R5C6 = 7
5c. Lower total in C3 = 29, R35C3 = [98] -> visible cells must total 12 = [651/75] -> 5 in R89C3, locked for C3 -> R7C3 = 2
5d. Left-hand total in R4 = 33 cannot have 7 in R4C3 because R4C12 cannot total 9 -> R4C3 = 6, R6C3 = 7, R9C3 = 5 (step 5c), R9C47 = [12]
5e. R7C9 = 7 (hidden single in C9), naked pair {35} in R78C7, locked for C7 and N9
5f. R4C7 = 7 (hidden single in C7)
5g. Left-hand total in R4 = 33 must contain all of 6,7,8,9 (cannot just contain 6,8,9 because max R4C12 = 7) -> R4C8 = 8, remaining visible cells must total 3 -> R4C12 = [12], R5C12 = [43], R3C78 = [84], R567C8 = [531]
5h. Lower total in C7 = 37, R2345C7 = [9876], R9C7 = 2 -> remaining visible cells must be 5 -> R8C7 = 5
5i. Lower total in C4 = 24, R59C4 = [91] -> visible cells must total 14 -> R678C4 = [653]
5j. Right-hand total in R5 = 28, R5C4678 = [9765] -> R5C9 = 1

and the rest is naked singles.

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


Top
 Profile  
Reply with quote  
 Post subject: Re: SkyScraperSum
PostPosted: Wed Jun 14, 2017 7:15 pm 
Offline
Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
In the first part of the post quoted by HATMAN http://forum.enjoysudoku.com/skyscraper ... ml#p256219 gabor listed two SkyScraper puzzles and suggested that the one with sums as clues was harder than the one with number of visible skyscrapers as clues. I wonder whether that's really true. Working from the number of visible skyscrapers would require a very different approach. I've been doing killer sudokus for more than ten years and they quickly became my favourite type of sudokus, along with various killer variants to extend the challenge.


Top
 Profile  
Reply with quote  
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 5 posts ] 

All times are UTC


Who is online

Users browsing this forum: No registered users and 5 guests


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
cron
Powered by phpBB® Forum Software © phpBB Group