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SSS ORC UK D 27 28 SSS LS D 28 SSS ORC 29 http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=13&t=1494 |
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Author: | Andrew [ Sun Jun 16, 2019 8:07 pm ] |
Post subject: | Re: SSS ORC UK D 27 28 SSS LS D 28 SSS ORC 29 |
Skyscraper Sum LS D 28 is fairly easy; a good starter puzzle for anyone who hasn't yet tried Skyscraper Sum puzzles. Rather than posting a full walkthrough, I've posted a solving outline which will be of help to anyone who can't see how to get started. Solving Outline: Sums are for heights of skyscrapers visible from the edges. Only those higher than the previous one(s) count toward the sums. x not 0, y not 7, w is unknown. AK, FNC and NC. 1. 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 or 2w cannot be 8, 9 or 17, rows/columns starting/finishing with 8,9 must be ones with unspecified totals. 1a. Unspecified upper totals in C15 -> R1C15 = {89}, locked for R1 1b. Unspecified left-hand totals in R18 -> R18C1 = {89}, locked for C1 1c. Unspecified right-hand totals in R35 -> R35C9 = {89}, locked for C9 1d. Unspecified lower totals in C389 but no 8,9 in R9C9 -> R9C38 = {89}, locked for R9 Clean-ups: R1C1 = {89} -> no 8,9 in R2C2 R1C5 = {89} -> no 8,9 in R2C456 R3C9 = {89} -> no 8,9 in R234C9 R5C9 = {89} -> no 8,9 in R56C9 R8C1 = {89} -> no 8,9 in R78C2 R9C3 = {89} -> no 8,9 in R8C34 R9C8 = {89} -> no 8,9 in R8C78 Naked pair {89} in R35C9 -> no 7 in R4C89 Continue, using hidden and naked pairs, until all the {89} pairs have been placed. Once R4C7 has been limited to {89} use the right-hand total 13 in C4 to make R4C7 = 9, from which all the 8s and 9s will be placed. Then after clean-ups, place the 7s, then the 6s and so on until the 3s have been placed, when one clean-up removing a 2 will leave the puzzle as naked singles. Solution: 9 2 4 6 8 1 3 5 7 5 7 9 2 4 6 8 1 3 1 3 5 7 9 2 4 6 8 6 8 1 3 5 7 9 2 4 2 4 6 8 1 3 5 7 9 7 9 2 4 6 8 1 3 5 3 5 7 9 2 4 6 8 1 8 1 3 5 7 9 2 4 6 4 6 8 1 3 5 7 9 2 |
Author: | Andrew [ Sat Jun 22, 2019 11:01 pm ] |
Post subject: | Re: SSS ORC UK D 27 28 SSS LS D 28 SSS ORC 29 |
SSS ORC UK D 28 wasn't particularly difficult, comparable with some of the earlier SSS ORC UK puzzles in this forum. Hints, which aren't spoilers: Don't try using the killer cages too early; it's good to make as many placements and eliminations as possible first. If you are having problems working out numbers which make up totals, my post Basic Techniques and Cage Combinations in the Killer Techniques forum may be helpful. Here is my walkthrough for SSS ORC UK D 28: 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. x not 0, y not 7, w is unknown. AK, FNC and NC. The four killer cages have the same total, repeats may be possible in the L-shaped cages. There are two totals on the non-repeating diagonal, one total on the repeating diagonal. 1. Skyscrapers starting with 8 or 9 must total 8 (possible for even rows/columns), 9 or 17. Totals given as xy or 2w cannot be 8, 9 or 17 so rows/columns starting/finishing with 8,9 must be ones with unspecified totals or total 17. 1a. Unspecified upper totals only in C15 -> R1C15 = {89}, placed for R1 1b. Unspecified left-hand totals only in R18 -> R18C1 = {89}, locked for C1 1c. Unspecified right-hand totals only in R35, R35C9 = {89}, placed for C9 1d. Unspecified lower totals only in C38 with no 8,9 in R9C9 -> R9C38 = {89}, locked for R9 1e. Lower total in C1 = 21 with R18C1 = {89} can only be R189C1 = [984], 9 placed for R1 and D\, 4 placed for R9 and C1 -> R1C5 = 8, placed for C5 Clean-ups: R1C1 = 9 -> no 8 in R2C2 R1C5 = 8 -> no 7 in R1C46, no 7,8,9 in R2C46, no 7,9 in R2C5 R3C9 = {89} -> no 8,9 in R234C8 R5C9 = {89} -> no 8,9 in R56C8 R8C1 = 8 -> no 7 in R7C1 + R9C2, no 7,8,9 in R7C2, no 7,9 in R8C2 R9C1 = 4 -> no 3,4,5 in R8C2, no 3,5 in R9C2 R9C3 = {89} -> no 8 in R8C2, no 8,9 in R8C34 R9C8 = {89} -> no 8,9 in R8C7 Naked pair {89} in R35C9 -> no 7 in R4C89 2a. Right-hand total in R9 = 11 -> R9C89 = [83/92] 2b. Lower total in C9 = 17 with R9C9 = {23} cannot include both of 8,9 -> R35C9 = [89], 8 placed for R3, 9 placed for R5, no 7 in R2C9 2c. Total of 17 with R9C9 = {23} must be [962/953] with some lower numbers between R59C9 -> no 7 in R678C9 2d. R1C9 = 7, placed for R1 2e. Upper-right diagonal total = 7 -> no 8,9 on D/ 2f. Lower-left diagonal total = 11 = [47] -> no 6 in R8C2 Clean-ups: R1C9 = 7 -> no 6 in R1C8 + R2C9, no 6,7 in R2C8 R3C9 = 8 -> no 7 in R3C8 R8C2 = {12} -> no 1,2 in R7C13 R9C9 = {23} -> no 2,3 in R8C89 3. Left-hand total in R3 = 25 must contain four or five digits ending in 9, because 8 is in R3C9, starting with one of 1,2,3 (since 4 already placed for C1) -> R3C1 = {123}, no 7,9 in R3C2 Clean-up: R3C1 = {123} -> no 2 in R24C12 + R3C2 4a. Left-hand total in R9 = 2w, at least 20, must include both of 8,9, since max R9C123 = [469] = 19 -> R9C3 = 8, placed for C3, R9C8 = 9 -> R9C9 = 2 (step 2a), placed for R9, C9 and D\, no 1 in R8C9 4b. Lower total in C9 = 17 must be [962] (step 2c) -> R8C9 = 6, placed for C9 Clean-ups: R8C9 = 6 -> no 5,6,7 in R7C8, no 5 in R7C9, no 5,7 in R8C8 R9C3 = 8 -> no 7 in R8C34 + R9C4 R9C8 = 9 -> no 8 in R8C8 R9C9 = 2 -> no 1 in R8C8 5a. Lower-right diagonal total = 29 includes 2,8,9 -> the remaining digits total 10 must be [64] -> R8C8 = 4, placed for D\, no 3,5,7,8 in R7C7, no 3,4 in R7C9 5b. R7C9 = 1, placed for R7 and C9 -> R7C7 = 6, placed for R7, C7 and D\ Clean-ups: R7C7 = 6 -> no 5,7 in R6C67 + R7C6 + R8C7, no 5,6,7 in R6C8 + R8C6 R7C9 = 1 -> no 1,2 in R6C8, no 2 in R7C8 R8C8 = 4 -> no 3 in R7C8 + R8C7 R6C8 = {34} -> no 3,4 in R5C7 R7C1 = {35} -> no 4 in R67C2 8 on D\ only in R4C4 + R6C6 -> no 7 in R5C5 7 in R7 only in R7C345 -> no 6 in R68C4, no 8 in R7C4 8 in R7 only in R7C68 -> no 8,9 in R6C7 6. 8,9 in C7 only in R2C7 and one of R45C7 (R45C7 cannot be 8,9 NC) -> R2C7 = {89} Clean-ups: R2C7 = {89} -> no 9 in R3C6 One of 8,9 in R45C7 -> no 8 in R5C6 9 in R3 only in R3C45 -> no 8 in R4C4, no 9 in R4C5 7a. R6C6 = 8 (hidden single on D\), no 7,8 in R5C7 7b. R24C7 = {89} (hidden pair in C7), no 7 in R3C7 7c. R9C7 = 7 (hidden single in C7), placed for R9 Clean-ups: R6C6 = 8 -> no 7 in R5C6, no 7,9 in R67C5, no 9 in R7C6 R9C7 = 7 -> no 8 in R8C6, no 6 in R9C6 8a. Left-hand total in R3 = 25 must contain four or five digits ending in 9 (step 3) = {13579/36.79} (cannot be {12679/14569/23479/23569}, NC because 9 only in R3C45) with a lower number between 6,7 of {3679} -> R3C45 = [79], placed for R3, 9 also placed for C5 8b. R3C123 = [135/361], 1,3 locked for R3 Clean-ups: R3C4 = 7 -> no 6,7 in R2C3 + R4C35, no 6 in R2C45 2 in C1 only in R56C1 -> no 1 in R56C1, no 1,2 in R5C2 8 in R5 only in R5C24 -> no 9 in R4C3, no 7 in R5C3, no 7,9 in R6C3 7 in R7 only in R7C34 -> no 6 in R68C3 R3C123 = [135/361] -> no 3 in R2C1, no 1 in R2C2 9a. R27C3 = [97] (hidden pair in C3), 7 placed for R7 9b. R8C5 = 7 (hidden single in C5), no 6 in R9C45 9c. R6C5 = 6 (hidden single in C5) 9d. R9C2 = 6 (hidden single in R9) Clean-ups: R6C5 = 6 -> no 5,6,7 in R5C4, no 5 in R5C5 + R7C45, no 5,6 in R5C6, no 5,7 in R6C4 R7C3 = 7 -> no 6,7,8 in R6C2 R8C5 = 7 -> no 8 in R7C6 R9C2 = 6 -> no 5 in R8C3 R5C5 = {13} -> no 2 in R4C56 + R5C46 + R6C4 10a. R7C48 = [98] (hidden pair in R7) 10b. R7C56 = {24} (cannot be {23/34}, NC), locked for R7, no 3 in R6C7 10c. R1C7 = 3 (hidden single in C7), placed for R1 Clean-ups: R1C7 = 3 -> no 2,4 in R1C68, no 2,3,4 in R2C68 R7C2 = {35} -> no 4 in R6C3 Naked pair {35} in R7C12 -> no 3,5 in R6C1 5 in C7 only in R35C7 -> no 4,5,6 in R4C68 [Time to use the killer cages.] 11. Cage at R1C5 = [813/853/863] = 12,16,17, max. total for cage at R1C4 = [555] = 15 (cannot be [655], NC), all cages must have the same total = 12 -> R1C6 = 1, placed for R1, R1C8 = 5, placed for R1, no 4,5 in R2C9 11a. R2C9 = 3, placed for C9, no 2,4 in R3C8 11b. R3C8 + R4C9 = [64] (cannot be [54/55/65], AK, FNC) -> R4C8 = 2 (cage total), 6 placed for R3, 4 placed for C9, no 2 in R3C7 11c. R3C123 = [135], 1 placed for C1, 5 placed for R3 and D\ 11d. R3C67 = [24], 4 placed for C7 11e. Naked pair {12} in R68C7, locked for C7, no 2 in R7C6 11f. R5C7 = 5, placed for R5 11g. R7C56 = [24], 2 placed for C5 Clean-ups: R1C6 = 1 -> no 1 in R2C5 R3C1 = 1 -> no 1 in R4C2 R3C2 = 3 -> no 3 in R4C1, no 4 in R4C2, no 2,3,4 in R4C3 R3C6 = 2 -> no 3 in R2C5, no 1 in R2C6, no 1,3 in R4C56 R3C7 = 4 -> no 5 in R2C68 R4C6 = 7 -> no 8 in R4C7 R4C8 = 2 -> no 1,3 in R5C8 R4C9 = 4 -> no 4 in R5C8 R7C5 = 2 -> no 1,3 in R6C4, no 1,2,3 in R8C46 R6C4 = 4 -> no 3,4 in R5C34, no 3 in R5C5 + R6C3 R6C9 = 5 -> no 6 in R5C8, no 4 in R6C8 R6C8 = 3 -> no 2 in R5C8 + R6C7 12a. R5C5 = 1, placed for R5, C5 and D\ 12b. R14568C3 = [41623], 4 placed for R1, 6 placed for R5, no 3 in R2C2 12c. R2C2 = 7, placed for D\ -> R4C4 = 3, no 6 in R1C2, no 4 in R4C5 12d. R1C24 = [26] 12e. R249C5 = [453], 3 placed for R9, no 4 in R5C6 12f. R68C7 = [12], no 1 in R9C6 12g. R9C46 = [15] 12h. R1C4 + R2C5 = [64] -> R2C4 = 2 (cage total) 12i. R3C2 + R4C3 = [31] -> R4C2 = 8 (cage total) 12j. R5C12 = [24] (hidden pair in R5), 2 placed for C1 12k. R7C1 = 3 (hidden single in C1) -> R7C2 = 5 Clean-ups: R2C2 = 7 -> no 6 in R2C1 R5C1 = 2 -> no 1,2,3 in R6C2 R5C3 = 6 -> no 5 in R6C2 R7C1 = 3 -> no 2 in R8C2 R7C2 = 5 -> no 6 in R6C1 R9C5 = 3 -> no 4 in R8C46 The rest is naked singles, without using AK, FNC, NC or the repeating diagonal. Solution: 9 2 4 6 8 1 3 5 7 5 7 9 2 4 6 8 1 3 1 3 5 7 9 2 4 6 8 6 8 1 3 5 7 9 2 4 2 4 6 8 1 3 5 7 9 7 9 2 4 6 8 1 3 5 3 5 7 9 2 4 6 8 1 8 1 3 5 7 9 2 4 6 4 6 8 1 3 5 7 9 2 Note that this is the same solution as for SSS LS D 28, but a different puzzle. The fact that both are number 28 might have given a hint to some people that they could have the same solution. |
Author: | Andrew [ Tue Jun 25, 2019 4:01 am ] |
Post subject: | Re: SSS ORC UK D 27 28 SSS LS D 28 SSS ORC 29 |
SSS ORC UK D 27 was harder at first but got easier after the killer cage total had been found. Here is my walkthrough for SSS ORC UK D 27: 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. x not 0, y not 7, w is unknown. AK, FNC and NC. The six killer cages have the same total. 35(7) = {1235789/1245689} 1. Lower total in C6 = 35 can be only made up as [9.8.7.6.5/9.8.7.641/9.8.75.42] (cannot be [9.8.7.63.2/9.8.753.2.1/9.86.5.4.3/9.86.5.42.1/97.6.5.4.31/8.7.6.5.4.3.2] which require more than 9 cells to provide gaps between consecutive digits for NC) -> R1C6 = 9, placed for R1, R3C6 = 8, placed for R3, R5C6 = 7, placed for R5, R7C6 = {126}, R9C6 = {125}, max R6C6 = 5, max R8C6 = 4 [Note. Since these are long permutations, I worked them out by looking at groups of the 2, 3 or 4 missing digits which total 10.] Clean-ups: R1C6 = 9 -> no 8 in R1C57 + R2C6, no 8,9 in R2C57 R3C6 = 8 -> no 7 in R2C57, no 7,9 in R24C6 + R3C57, no 7,8,9 in R4C57 R5C6 = 7 -> no 6 in R4C57, no 6,8 in R4C6 + R5C57, no 6,7,8 in R6C57 2. 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 or 2w cannot be 8, 9 or 17, rows/columns starting/finishing with 8,9 must be ones with unspecified totals. 2a. Unspecified upper totals only in C167, R1C6 = 9, no 8,9 in R1C7 -> R1C1 = 8, placed for R1 and C1 2b. Unspecified left-hand totals only in R18C1, R1C1 = 8 -> R8C1 = 9, placed for C1 2c. Unspecified right-hand totals only in R46 -> R46C9 = {89}, locked for C9 2d. Unspecified lower totals only in C38 -> R9C38 = {89}, locked for R9 Clean-ups: R1C1 = 8 -> no 7 in R1C2 + R2C1, no 7,8,9 in R2C2 R8C1 = 9 -> no 8,9 in R7C2, no 8 in R8C2 R4C9 = {89} -> no 9 in R3C8, no 8,9 in R5C8 R6C9 = {89} -> 8,9 in R7C8 R9C3 = {89} -> no 9 in R8C2, no 8,9 in R8C34 R9C8 = {89} -> no 8,9 in R8C7 3a. Right-hand total in R2 = 18 -> no 9 in R2C8, max R2C9 = 5 3b. Left-hand total in R2 = 24, must contain at least four digits -> no 9 in R2C3, max R2C1 = 4 (since no 7 in R2C1) 3c. 9 in R2 only possible in R2C4, 8 in R2 only possible in R2C38 but cannot have R2C34 = {89}, NC so, before making further eliminations, left-hand total = 24 can only be {2679/3579/3678/4569/4578/12678/13578/14568/23478/23568/24567/123468/123567} 3d. Right-hand total = 18 can only contain 8 in R2C8 if it also contains 9 -> left-hand total = 24 cannot contain 8 since it cannot contain both of 8,9 and cannot contain 8 in R2C34 because of NC = {2679/3579/4569/24567/123567} 3e. Left-hand total = 24 = {2679/3579/4569} (cannot be {24567/123567} which prevent R2C8 = 8 and clash with 18 total = {567} 3f. R2C1234 = 24 = {3579} = [3579] (cannot be 2679/4569}, NC), 3 placed for C1, 7 placed for C3 Clean-ups: R2C1 = 3 -> no 2,3,4 in R13C2, no 2,4 in R3C1 R2C2 = 5 -> no 6 in R13C2, no 4,5,6 in R13C3, no 5,6 in R3C1 R2C3 = 7 -> 6,7 in R13C4, no 7 in R3C2 R2C4 = 9 -> no 9 in R3C3 R1C3 = {123} -> no 2 in R1C4 R3C3 = {123} -> no 2 in R3C4 + R4C234 4a. Killer cage R2C123 = [357] = 15 -> all six killer cages must total 15, some possibly with repeats 4b. Right-hand total in R2 = 18 contains 9 in R2C4 so remaining digits must total 9 and be in two cells because of killer cage R2C789 = 15 -> R2C789 = [663/681] (cannot be [672/654], NC) -> R2C7 = 6, placed for C7, R2C8 = {68}, R2C9 = {13} Clean-up: R2C7 = 6 -> no 5,7 in R1C7, no 5,6,7 in R13C8, no 5 in R2C6 + R3C7 5a. R7C7 = 8 (hidden single in C7), placed for R7, no 9 in R6C7, no 7 in R8C7 5b. R5C7 = 9 (hidden single in C7), placed for R5 5c. R9C7 = 7 (hidden single in C7), placed for R9, no 8 in R9C8 5d. R9C8 = 9 -> R9C3 = 8, placed for C3 5e. 9 on D/ only in R6C4 + R7C3 -> no 8 in R6C4 5f. R2C8 = 8 (hidden single on D/) -> R2C9 = 1 (killer cage total), placed for C9 Clean-ups: R2C8 = 8 -> no 7 in R13C9 R2C9 = 1 -> no 1,2 in R13C8, no 2 in R12C9 R5C7 = 9 -> no 8,9 in R46C8 R7C7 = 8 -> no 7 in R67C8, no 7,8,9 in R8C8 R9C7 = 7 -> no 6 in R8C8 R1C8 = {34} -> no 3,4 in R1C79 R3C8 = {34} -> no 3,4 in R3C79 + R4C7 6a. Naked pair {12} in R13C7, locked for C7 -> R4C7 = 5, placed for C7, no 4 in R3C8 6b. R3C8 = 3, placed for R3, no 2 in R3C7 6c. R13C7 = [21], 2 placed for R1, 1 placed for R3 and D/, no 3 in R1C8 6d. R3C1 = 7, placed for C1 6e. R1C5 = 7 (hidden single in R1), placed for C5 6f. R8C5 = 8 (hidden single in C5), no 9 in R7C5 6g. R6C5 = 9 (hidden single in C5) 6h. R1C8 = 4, placed for R1, no 5 in R1C9 6i. R1C9 = 6, placed for C9 and D/ 6j. R3C9 = 5, placed for R3 and C9 6k. R3C2345 = [9246], 2 placed for C3, 6 placed for C5 Clean-ups: R1C7 = 2 -> no 1,2,3 in R2C6 R3C1 = 7 -> no 6 in R4C1, no 6,7,8 in R4C2 R3C2 = 9 -> no 9 in R4C3 R3C3 = 2 -> no 1,3 in R4C234 R3C4 = 4 -> no 4,5 in R4C3, no 5 in R4C4, no 3,4,5 in R4C5 R3C5 = 6 -> no 5 in R2C5 + R4C6, no 6 in R2C6, no 6,7 in R4C4 R3C7 = 1 -> no 2 in R4C6, no 1,2 in R4C8 R3C9 = 5 -> no 4,5,6 in R4C8 R2C6 = 4 -> no 3 in R2C5 R4C7 = 5 -> no 4 in R4C6, no 4,5,6 in R5C8 R4C6 = 3 -> no 2 in R4C5, no 2,3,4 in R5C5 R6C5 = 9 -> no 8 in R5C4, no 9 in R7C4 R8C5 = 8 -> 7 in R78C4 Naked pair {34} in R68C7 -> no 2 in R7C6, no 2,3,4,5 in R7C8 7a. R4C356 = [613], 6 placed for C3, 1 placed for C5, 3 placed for D/, no 5 in R4C2 7b. R5C5 = 5, placed for R5, C5 and D/ 7c. Killer cage total = 15, R4C35 = [61] =7 -> R4C4 = 8 7d. Killer cage total = 15, R6C5 = 9 -> R6C67 = 6 = [24/33] 7e. R5C2 = 8 (hidden single in R5), no 9 in R4C2 7f. R4C2 = 4 -> no 3,4 in R5C3 7g. R5C3 = 1, placed for R5 and C3 7h. R1C3 = 3, placed for R1 and C3 7i. Killer cage total = 15, R8C1 = 9 -> R8C23 = 6 = [24], 2 placed for D/, 4 placed for C3 7j. R67C3 = [59], 9 placed for D/ 7k. R9C1 = 4, placed for R9, C1 and D/ 7l. Naked pair {23} in R9C59, locked for R9 Clean-ups: R4C2 = 4 -> no 5 in R4C1 R4C3 = 6 -> no 6 in R5C4 R4C5 = 1 -> no 2 in R5C4 R5C2 = 8 -> no 7,9 in R6C2 R5C3 = 1 -> no 1,2 in R6C2 R5C5 = 5 -> no 4 in R5C4 R6C3 = 5 -> no 4,6 in R6C2, no 4,5,6 in R7C24 R7C3 = 9 -> no 8 in R6C2 R8C2 = 2 -> no 1,2 in R7C1, no 1,3 in R7C2, no 1 in R9C2 R8C3 = 4 -> no 3 in R7C4, no 3,5 in R8C4, no 5 in R9C24 8a. R5C1489 = [6324], 6 placed for C1, 4 placed for C9 8b. R7C1 = 5, placed for C1, no 5 in R6C2 8c. R6C2 = 3, no 2 in R6C1 + R7C2 8d. R7C2 = 7, placed for R7 8e. R7C5 = 4 (hidden single in R7), placed for C5 -> R29C5 = [23], 3 placed for R9 8f. R789C9 = [372], 3 placed for R7, no 6 in R7C8 8g. R9C246 = [615] 8h. R7C468 = [261] Clean-ups: R2C5 = 2 -> no 1 in R1C4 R5C8 = 2 -> no 3 in R4C8 + R6C7, no 1,3 in R6C8 R5C9 = 4 -> no 4,5 in R6C8 R4C8 = 7 -> no 8 in R4C9 R7C4 = 2 -> no 1 in R8C4 R7C5 = 4 -> no 3 in R6C6, no 4 in R8C4, no 3,4 in R8C6 R7C8 = 1 -> no 2 in R68C8 R7C9 = 3 -> no 3,4 in R8C8 R9C5 = 3 -> no 2 in R8C46 R9C6 = 5 -> no 4 in R8C7 R9C9 = 2 -> no 1 in R8C8 The rest is naked singles, without using AK, FNC, NC or the repeating diagonal. Solution: 8 1 3 5 7 9 2 4 6 3 5 7 9 2 4 6 8 1 7 9 2 4 6 8 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 Interesting that SSS ORC UK D 27 and 28 both have 1-9 in all the even rows and columns. Maybe they were posted as ORC puzzles to make them a bit harder. |
Author: | Andrew [ Wed Jun 26, 2019 10:33 pm ] |
Post subject: | Re: SSS ORC UK D 27 28 SSS LS D 28 SSS ORC 29 |
I had several tries at SSS ORC 29 and kept reaching impossible positions, so I then tried the two puzzles with unknown total killer cages before coming back to this one and realising the flawed assumption that I'd been making (see comment at the relevant step). As with the previous batch, I think HATMAN posted these four in order of difficulty. Here is my walkthrough for SSS ORC 29: 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. x not 0, y not 7, w is unknown. The even-numbered have 5 three in a row, 1 four in a row, horizontally and/or vertically. 1. 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 or 2w cannot be 8, 9 or 17, rows/columns starting/finishing with 8,9 must be ones with unspecified totals. 1a. Unspecified upper totals only in C79 -> R1C79 = {89}, locked for R1 1b. Unspecified left-hand totals only in R36 -> R36C1 = {89}, locked for C1 1c. Unspecified right-hand totals only in R15 -> R15C9 = {89}, locked for C9 1d. Lower total in C9 = 30 must contain both of 8,9 (cannot be [97.6.53] in R56789C9 because of NC while other sums containing 9 but not 8 contain 6 or 7 digits) -> R1C9 = 9, R1C7 = 8, placed for C7, R5C9 = 8, placed for R5 1e. Unspecified lower totals only in C247, no 8 in R9C7 -> 8 must be in one of R9C24, locked for R9, 9 must be in one of R9C247, locked for R9 Clean-ups: R1C7 = 8 -> no 7 in R1C68, no 7,8,9 in R2C68, no 7,9 in R2C7 R5C9 = 8 -> no 7,8,9 in R46C8, no 7 in R46C9, no 7,9 in R5C8 R3C1 = {89} -> no 8,9 in R234C2 R6C1 = {89} -> no 9 in R5C2, no 8,9 in R7C2 8 in one of R9C24 -> no 7,8,9 in R8C3, no 7 in R9C3 2. Right-hand total in R7 = 20 must contain at least three digits with the lowest a maximum of 5 -> no 6,7 in R7C9, no 9 in R7C8 2a. Lower total in C9 = 30 contains 8,9 -> the remaining visible digits must total 13 = {256/346} (cannot be {1237/1246/1345/67} because of NC, cannot be {157/247} because 7 only in R89C9) = [6.52/64.3] -> R6C9 = 6, placed for C9, R9C9 = {23}, no 5 in R7C9, no 2,3,4,7 in R8C9 2b. 7 in C9 only in R23C9 -> no 6 in R2C8, no 6,7,8 in R3C8 Clean-ups: R6C9 = 6 -> no 5,6 in R5C8, no 5 in R6C8, no 5,6,7 in R7C8 R9C9 = {23} -> no 2,3 in R89C8 3. Lower total in C8 = 24 must contain 9 (cannot be {24567/123567} because 7 only in R89C8, also can only be {789} with intermediate lower numbers) -> R3C8 = 9, placed for R3, no 8,9 in R8C8 3a. 7,8 of {789} must be in R79C8 -> no 7 in R8C8 3b. Total of 24 can only start with 1, 2, 3, 4 or 7 -> R9C8 = {147} (2,3 have been eliminated by clean-ups) 3c. Total of 24 cannot be {4569} because 5 only in R48C8 -> no 4 in R9C8 3d. Remaining possible ways to make total of 24 = {1689/789} (cannot be {12579/12678/13479} because 7 only in R9C8, cannot be {13569} because 5 only in R48C8, cannot be {12489} because 8 only in R8C8, cannot be {123459} because of NC) -> R7C8 = 8, placed for R7 3e. Right-hand total = 20 must be [9.83] -> R7C9 = 3, R89C9 = [52] (step 2a), 2,3,5 placed for C9, 3 placed for R7, 2 placed for R9, no 1 in R9C8 3f. R9C8 = 7, placed for R9 3g. R3C1 = 8, placed for R3 and C1, R6C1 = 9 3h. Upper total in C8 = 12 contains 9 in R3C8 = [319/339] (cannot be [129], NC) -> R1C8 = 3, placed for R1, R2C8 = {13} Clean-ups: R1C8 = 3 -> no 2,3,4 in R2C7, no 4 in R2C9 R2C8 = {13} -> no 2 in R3C7 R3C1 = 8 -> no 7 in R24C12 + R3C2 R3C8 = 9 -> no 9 in R4C7 R6C1 = 9 -> no 8 in R6C2 R7C8 = 8 -> no 7,9 in R678C7 R7C9 = 3 -> no 2,3,4 in R6C8, no 4 in R8C8 R8C9 = 5 -> no 6 in R8C8 R9C9 = 2 -> no 1 in R8C8 R8C8 = 5 -> no 4,5,6 in R79C7, no 4,6 in R8C7 R7C7 = {12} -> no 1,2 in R68C67 + R7C6, no 1 in R6C8 R6C8 = 6 -> no 5,6,7 in R5C7, no 5 in R6C7 R8C7 = {35} -> no 4 in R789C6 4 in C9 only in R34C9 -> no 3,5 in R4C8 4. Right-hand total in C8 = 14, R8C9 = 5 -> can only be [95] including repeated or lower numbers 4a. Left-hand total in C8 = 11 contains 9 -> R8C1 = 2, placed for C1 4b. Left-hand total = [29], right-hand total = [95], possibly including repeated numbers -> R8C2 = {29} (left-hand total cannot contain 1 because of NC) 4c. 6,7,8 only possible in R8 between two 9s -> no 6,7,8 in R8C6, no 8 in R8C5 [This was where I went wrong on my earlier attempts, I’d mistakenly thought that 6,7,8 could be eliminated from all cells in R8.] Clean-up: R8C1 = 2 -> no 1 in R7C1, no 1,2 in R7C2, no 1,3 in R9C12 5. Left-hand total in R4 = 10 must include at least two numbers with the first not higher than 4 -> R4C1 = {134} 6. Left-hand total in R5 = 25 must use at least four cells and contain 9 (but not 8 which is in R5C9) with first digit not higher than 4 -> R5C1 = {134}, no 7 in R5C2, no 9 in R5C3 6a. R45C1 only contain 1,3,4, cannot be {34} (NC) -> 1 in R45C1, locked for C1 Clean-ups: 1 in R45C1 -> no 2 in R4C2, no 1,2 in R5C2 R45C1 contain one of 3,4 -> no 3,4 in R5C2 6b. Only remaining possible way to make 25 with R5C2 = {56} and containing 9 = {3679} (cannot be {4579} NC) -> R5C1 = 3, R5C2 = 6, both placed for R5, 3 also placed for C1 Clean-ups: R5C1 = 3 -> no 4 in R4C1, no 3,4 in R4C2, no 2,3,4 in R6C2 R5C2 = 6 -> no 5 in R4C2, no 5,6,7 in R46C3, no 5,7 in R5C3 + R6C2 R4C1 = 1 -> no 1,2 in R3C2 7a. Left-hand total in R4 = 10 cannot start [16] -> R4C2 = 1, no 1,2 in R5C3 7b. Total = 10 cannot start [118] -> no 8 in R4C3 7c. R5C3 = 4, placed for R5 and C3 Clean-ups: R4C2 = 1 -> no 1,2 in R3C3, no 2 in R4C3 R5C3 = 4 -> no 3 in R46C3, no 3,4,5 in R46C4, no 5 in R5C4 R5C8 = {12} -> no 1,2 in R45C7, no 1 in R4C9 8a. R4C9 = 4, placed for C1 8b. R5C7 = 9, placed for R5 and C7 8c. R7C7 = 2 (hidden single in C7), placed for R7, no 3 in R8C7 8d. R8C7 = 5, placed for C7 8e. R9C24 = {89} (hidden pair in R9) 8f. Right-hand total in R9 = 18 contains 2,7 = {279} -> R9C24 = [89] Clean-ups: R5C7 = 9 -> no 8,9 in R46C6 R7C7 = 2 -> no 3 in R6C67 + R8C6 R9C2 = 8 -> no 9 in R8C2 R9C4 = 9 -> no 8 in R8C4, no 9 in R8C5 R8C2 = 2 -> no 1 in R7C3, no 1,3 in R89C3 R8C7 = 5 -> no 5,6 in R79C6 R9C3 = {56} -> no 5 in R8C34 9a. Left-hand total in R8 = 11 = [29] (step 4b) -> R8C3 = 2, placed for C3 9b. Naked pair {13} in R9C67, locked for R9 9c. Left-hand total in R8 = 11 = [29] -> R8C4 = {29} Clean-ups: R8C3 = 2 -> no 1 in R7C4 R6C7 = {46} -> no 5 in R56C6 R7C6 = {79} -> no 8 in R6C5 R9C5 = {456} -> no 5 in R8C5 R9C6 = {13} -> no 2 in R8C5 10a. Left-hand total 22 in R7 must contain 9 but 8 is placed in R7C8 so only possible way is {679} (cannot be {1579} because 1 in R7 only in R7C5) with possible hidden lower numbers -> R7C1 = 6, placed for R7 and C1, no 5,7 in R7C2 10b. R7C2 = 4, placed for R7 10c. R1C1 = 7 (hidden single in C1), placed for R1 10d. R7C5 = 1 (hidden single in R7), placed for C5 Clean-ups: R1C1 = 7 -> no 6 in R12C2 R7C1 = 6 -> no 6 in R6C2 R7C2 = 4 -> no 5 in R7C3 R7C5 = 1 -> no 1,2 in R6C4, no 2 in R6C5 + R8C4 R8C4 = 9 -> no 9 in R7C3 R7C3 = 7 -> no 8 in R6C3, no 6,7,8 in R6C4 R2C1 = {45} -> no 4,5 in R13C2 R1C2 = {12} -> no 1 in R12C3 R1C3 = {56} -> no 5,6 in R1C4 + R2C34, no 5 in R2C2 R9C5 = {456} -> no 5 in R8C6 11a. R7C346 = [759], 7 placed for C3 11b. R2C3 = 8 (hidden single in C3) Clean-ups: R2C3 = 8 -> no 7,9 in R2C4, no 7 in R3C4 R7C4 = 5 -> no 4,5,6 in R6C5, no 4,6 in R8C5 R7C6 = 9 -> no 9 in R6C5 Naked pair {19} in R46C3 -> no 1,2 in R5C4 12a. R5C4 = 7, placed for R5, no 7 in R6C5 12b. R68C5 = [37], placed for C5 12c. R5C5 = 5 (hidden single in R5), placed for C5 Clean-ups: R5C4 = 7 -> no 6,8 in R4C45 R5C5 = 5 -> no 4 in R4C5, no 4,5,6 in R4C6, no 4,6 in R6C6 R6C5 = 3 -> no 2 in R5C6 R6C6 = 7 -> no 6 in R6C7 R5C6 = 1 -> no 2 in R4C5 13a. R6C7 = 4, placed for C7 13b. R2C5 = 8 (hidden single in C5) 13c. 7 in C7 only in R34C7 -> no 6 in R34C7, NC 13d. R2C7 = 6 (hidden single in C7), no 7 in R3C7 13e. R4C7 = 7 (hidden single in C7) Clean-ups: R2C7 = 6 -> no 5,6 in R1C6, no 5 in R2C6, no 5,6,7 in R3C6 R4C7 = 7 -> no 6 in R4C8 R3C7 = {13} -> no 2 in R234C6 + R4C8 2 in C5 only in R13C5 -> no 1,2,3 in R2C4, no 1,3 in R2C6 [I ought to have spotted this earlier.] 14a. R2C1 + R3C2 = [46] (cannot be [45/55], AK, FNC), 4 placed for C1, 6 placed for R3 14b. R9C135 = [564], 6 placed for C3, 4 placed for C5 14c. R1C3 = 5, placed for C3, R3C3 = 3, placed for R3 14d. R3C45679 = [52417], 2 placed for C5, 1 placed for C7, 7 placed for C9 Clean-ups: R1C3 = 5 -> no 4 in R1C4 + R2C24 R1C5 = 6 -> no 6 in R2C6 R3C3 = 3 -> no 2,3 in R2C2, no 2 in R4C4 R3C5 = 2 -> no 1 in R4C4, no 1,3 in R4C6 R3C7 = 1 -> no 1 in R24C8 R2C2 = 1 -> no 2 in R1C2 15a. R1C246 = [124] 15b. R5C68 = [12] 15c. R9C67 = [13] [Now to apply five three-in-a-rows and one four-in-a-row] 16a. Four-in-a-row only possible in R6C1234 = [9999], 9 placed for C3 16b. Left-hand total in R4 = 10 = [19] -> R4C4 = 9 This leaves five three-in-a-rows, 8s in R2, 1s in R4, 2s and 5s in R8 and 4s in C6 Solution: 7 1 5 2 6 4 8 3 9 4 1 8 8 8 4 6 3 1 8 6 3 5 2 4 1 9 7 1 1 1 9 9 7 7 4 4 3 6 4 7 5 1 9 2 8 9 9 9 9 3 7 4 6 6 6 4 7 5 1 9 2 8 3 2 2 2 9 7 9 5 5 5 5 8 6 9 4 1 3 7 2 |
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