SudokuSolver Forum :: View topic - Semi-Symmetric Killer 1

Author:

Andrew [ Thu Dec 05, 2019 12:05 am ]

Post subject:

Re: Semi-Symmetric Killer 1

Thanks HATMAN for an interesting puzzle. I think I'm now getting used to semi-symmetry having so far done a couple of the Triankle puzzles in the Other Variants forum; killer fans should also try them.

My solving path was a bit longer as I didn't spot wellbeback's very powerful step 1. After I'd reached the results of that, we followed fairly similar paths but saw the 'critical step' differently.

Here is my walkthrough for Semi-Symmetric Killer 1:

Solution:

HATMAN wrote Using the semi-symmetry I would guess it is at or below 1.0:

Author:	HATMAN [ Tue Sep 17, 2019 11:21 am ]
Post subject:	Semi-Symmetric Killer 1
Semi-Symmetric Killer 1 I'm pushing the boundaries again, we have allowed symmetry as part of the solution method in the past - this is a slight extension. I've come up with a new sudoku layout based on 10 givens which I will publish shortly. In creating an initial solution for simplicity I used symmetry however I made a mistake as in some cases the number was symmetrical with its partner and in some cases symmetrical with itself. As it solved I thought why not try with normal killer so here is the first one: This is a simple killer however the solution is semi-symmetrical. There are four unknown number pairs and an odd number (which of course goes in r5c5). Each number pair is semi-symmetrical, consider for example that the number pair is 4,7, so if r3c1 contains a 7 then r7c9 must contain a 4 or a 7. So deduce the number pairs and solve the sudoku. Both solvers actually solve this directly SS gives it 3.35 and JS uses 41 fishes. Using the semi-symmetry I would guess it is at or below 1.0. Comment welcome as usual. On a style point as this is about symmetry so the cages should be symmetrical too, as below. On the other hand I can imagine opposing cages that partially overlap giving different information, so I will not rule these out.

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Author:	wellbeback [ Sat Sep 21, 2019 11:18 pm ]
Post subject:	Re: Semi-Symmetric Killer 1
Very cool! I imagine not having symmetrical cages could make it much harder. Here's how I started... Semi-Symmeric Killer 1 WT: 1. Innies n12 = +45(6) Can only be r1c56 = {89}, r3c123 = {789}, r3c4 = 4 -> r1c7 = 4 -> Innies n3 -> r3c9 = 3 Also r4c45 = {12} Also r12c4 = +8(2) 2. 4 in n1 in r2c123 17(5)r1c3 = {123(47\|56)} 4 cannot go in r2c1 since that would put whichever of {35} or {26} is in r1c12 also in r12c4 -> 4 in r2c23 -> 12(3)n1 = {156} -> 27(5)r1c3 = {234} in n1 and {17} in r12c4 3. -> 1 in r3 in n3 in r3c78 -> 2 in n3 in 16(3) -> 7 in n3 in r2c78 -> r12c4 = [71] Also 3 in n2 in r2c56 -> r1c3 = 3 and r2c23 = {24} -> 2 in n3 in r1c89 -> 2 in n2 in r3c56 Also r4c45 = [21] -> 1 in c6 in n8 in r89c6 -> since 1 also in 8(3)n9 and since 1 in c3 only in n7 -> r7c3=1 4! Critical Step 11(4)n6 = {1235} 28(4)n4 = {(47\|56)89} Can that be {4789}? That puts r789c3 = [165] (6 cannot go in r9c3 since innies n7 = r7c1,r9c3 = +12(2)) But that leaves no solution for 13(4)n7 since r2c23 = {24} -> 28(4)n4 = {5689} 5! Breaking step -> Since 11(4)n6 and 28(4)n4 must be semi-symmetrical -> {123} must pair with {689} (It can't be that one of (123) in n6 pairs with the 5 in n4 since that would require the 5 in n6 pairing with a (123) in the 28(4)n4). -> 4 must pair with one of (57) or be unpaired -> 20(3)r6c5 must have two of (689) in r6c56 and one of (457) in r7c6 Only solution is [{69}5] -> (12) pairs with (69) -> (38) is a pair Also (45) is a pair -> 7 is unpaired (and goes in r5c5). 6. If 3 were in 8(3)n9 one of (38) would be in 12(3)n1 Since that's not the case -> 8(3)n9 = {125} -> (26) are a pair and (19) are a pair. Straightforward from here

Author:	HATMAN [ Sun Sep 29, 2019 6:35 pm ]
Post subject:	Re: Semi-Symmetric Killer 1
Excellent solution On symmetric cages: they help identify the pairs but give less on the overall solution.

Author:	HATMAN [ Thu Dec 05, 2019 8:16 pm ]
Post subject:	Re: Semi-Symmetric Killer 1
I also missed that step.

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Author:	Andrew [ Thu Dec 05, 2019 12:05 am ]
Post subject:	Re: Semi-Symmetric Killer 1
Thanks HATMAN for an interesting puzzle. I think I'm now getting used to semi-symmetry having so far done a couple of the Triankle puzzles in the Other Variants forum; killer fans should also try them. My solving path was a bit longer as I didn't spot wellbeback's very powerful step 1. After I'd reached the results of that, we followed fairly similar paths but saw the 'critical step' differently. Here is my walkthrough for Semi-Symmetric Killer 1: “This is a simple killer however the solution is semi-symmetrical. There are four unknown number pairs and an odd number (which of course goes in r5c5). Each number pair is semi-symmetrical, consider for example that the number pair is 4,7, so if R3C1 contains a 7 then R7C9 must contain a 4 or a 7. So deduce the number pairs and solve the sudoku.” Prelims a) 21(3) cage at R1C5 = {489/579/678}, no 1,2,3 b) 7(3) cage at R3C4 = {124} c) 19(3) cage at R5C9 = {289/379/469/478/568}, no 1 d) 20(3) cage at R6C5 = {389/479/569/578}, no 1,2 e) 20(3) cage at R8C1 = {389/479/569/578}, no 1,2 f) 8(3) cage at R8C9 = {125/134} g) 28(4) cage at R4C3 = {4789/5689}, no 1,2,3 h) 11(4) cage at R4C7 = {1235} i) 13(4) cage at R7C2 = {1237/1246/1345}, no 8,9 j) 17(5) cage at R1C3 = {12347/12356}, no 8,9 Steps resulting from Prelims 1a. 8(3) cage at R8C9 = {125/134}, 1 locked for N9 1b. 28(4) cage at R4C3 = {4789/5689}, 8,9 locked for N4 1c. 11(4) cage at R4C7 = {1235}, locked for N6 1d. 13(4) cage at R7C2 = {1237/1246/1345}, 1 locked for N7 1e. 7(3) cage at R3C4 = {124}, CPE no 1,2,4 in R56C4 2. 45 rule on N3 2 innies R1C7 + R3C9 = 7 = [43/52/61] 2a. 21(3) cage at R1C5 = {489/579/678}, R1C7 = {456} -> no 4,5,6 in R1C56 3. 45 rule on N7 2 innies R7C1 + R9C3 = 12 = {39/48/57}, no 1,2,6 3a. 20(3) cage at R8C1 = {389/569/578} (cannot be {479} which clashes with R7C1 + R9C3), no 4 3b. 13(4) cage at R7C2 = {1237/1246} (cannot be {1345} which clashes with R7C1 + R9C3), no 5 3c. 45 rule on R89 4 outies R7C2345 = 13 = {1237/1246/1345}, no 8,9 4a. 45 rule on R12 4 outies R3C5678 = 14 = {1256/1346/2345} (cannot be {1238} which clashes with R3C9, cannot be {1247} which clashes with R3C4), no 7,8,9 4b. R3C123 = {789} (hidden triple in R3), locked for N1 4c. 12(3) cage at R1C1 = {156/246/345} 4d. 6 in R3 only in R3C5678 = {1256/1346}, 1 locked for R3, clean-up: no 6 in R1C7 (step 2) 4e. 21(3) cage at R1C5 = {489/579}, 9 locked for R1 and N2 4f. 7(3) cage at R3C4 = {124}, 1 locked for R4 and N5 4g. Min R3C1 = 7 -> max R45C1 = 7, no 7 in R4C1, no 6,7 in R5C1 5. 45 rule on N1 using R3C123 = {789} = 24, 2 outies R12C4 = 8 = {17/26/35}, no 4 5a. 45 rule on N2 using R12C4 = 8, 1 outie R1C7 = 1 innie R3C4 -> R1C7 = 4, R3C4 = 4, R3C6 = 3 (step 2), R4C45 = {12}, 2 locked for R4 and N5 5b. R1C7 = 4 -> R1C56 = 17 = {89}, 8 locked for R1 and N2 [wellbeback got some of these results more directly from 45 rule on N12 6 innies R1C56 + R3C1234 = 45.] 6. Hidden killer pair 8,9 in R2C78 and R2C9 for N2, 22(4) cage at R2C7 cannot contain both of 8,9 (because R3C78 cannot total 5) -> R2C78 must contain one of 8,9 and R2C9 = {89} 6a. 19(3) cage at R5C9 = {469/478/568} (cannot be {289} which clashes with R2C9), no 2 6b. Killer pair 8,9 in R2C9 and 19(3) cage, locked for C9 7. 17(5) cage at R1C3 = {12347/12356}, R12C4 = 8 (step 5) 7a. 12(3) cage at R1C1 (step 4c) = {156} (cannot be {246/345} = {26}4/{35}4 which clash with R1C3 + R2C23 + R12C4 = {135}{26}/{126}{35}), locked for N1 [I spent a lot of time trying to find a way to express this step as a forcing chain without contradictions.] 7b. R1C3 + R2C23 = {234} -> R12C4 = {17}, locked for C4 and N2 -> R4C45 = [21] 7c. 1 in R3 only in R3C78, locked for N3 7d. 22(4) cage at R2C7 contains 1 = {1579/1678}, no 2, 7 locked for N3 7e. R1C4 = 7 (hidden single in R1) -> R2C4 = 1 7f. R1C3 = 3 (hidden single in R1), clean-up: no 9 in R7C1 (step 3) 7g. Naked pair {24} in R2C23, 2 locked for R2 7h. R2C789 = {789} (hidden triple in R2) 8. 28(4) cage at R4C3 = {4789/5689} 8a. Consider combinations for 13(4) cage at R7C2 (step 3b) = {1237/1246} 13(4) cage = {1237} => 6 in C3 only in 28(4) cage or 13(4) cage = {1246}, 4 locked for N7 => 4 in N4 only in R456C1 -> 28(4) cage = {5689}, locked for N4 9. 28(4) cage at R4C3 = {5689} and 11(4) cage at R4C7 = {1235} correspond to each other -> 1,2,3 must correspond with 6,8,9 9a. R4C45 = [21] corresponds with R6C56 -> R6C56 = {689} 9b. 20(3) cage at R6C5 = {69}5/{89}3 -> R7C6 = {35} 9c. R7C6 = {35} corresponds with R3C4 = 4 but 3 corresponds with one of 6,8,9 -> R7C6 = 5, R6C56 = {69}, locked for R6 and N5 9d. 4 corresponds with 5, 1,2 correspond with 6,9 -> 3 corresponds with 8 leaving 7 as the remaining number which must be in the central cell -> R5C5 = 7 9e. 19(3) cage at R5C9 (step 6a) = {469/478}, 4 locked for C9 9f. 8(3) cage at R8C9 = {125} (only remaining combination), locked for N9 9g. 1 in C9 only in R89C9, locked for N9 9h. 4 in N5 only in R45C6, locked for C6 9i. 7 in C6 only in R89C6, locked for 26(5) cage at R8C6 9j. 7 in N9 only in R7C789, locked for R7 [Now a lot more use of corresponding cells.] 10. 12(3) cage at R1C1 = {156} corresponds with 8(3) cage at R8C9 = {125} -> 2 corresponds with 6 and 1 with 9 10a. R2C23 = {24} correspond with R8C78, no 2,5 in R8C78 -> R8C78 = [64] 10b. R89C6 = {17} (hidden pair in C6), R9C7 = 8 (cage sum) 10c. Naked triple {379} in R7C789, 3 locked for R7 -> R7C4 = 6 10d. R7C1 + R9C3 (step 3) = [84] (only remaining permutation), R2C23 = [42], R7C23 = [21], R8C23 = [37], R89C6 = [17], R46C2 = [71] 10e. R7C45 = [64] -> R8C45 = 10 = [82] 10f. R3C1 = 7 (hidden single in N1) -> R45C1 = 7 = {34}, locked for C1 10g. R6C1 = 2 corresponds to R4C9, no 2 in R4C9 -> R4C9 = 6 10h. 19(3) cage at R5C9 (step 9e) = {478} = {48}7, 8 locked for C9 and N6 10i. Naked pair {35} in R6C47, 5 locked for R6, R6C3 = 8 -> R56C9 = [84] 10j. R6C9 = 4 corresponds to R4C1, no 5 in R4C1 -> R4C1 = 4 10k. R8C9 = 5, R12C9 = [29] -> R1C8 = 5 (cage sum) and the rest is naked singles, without using corresponding pairs. Solution: 1 6 3 7 8 9 4 5 2 5 4 2 1 6 3 7 8 9 7 8 9 4 5 2 1 6 3 4 7 5 2 1 8 3 9 6 3 9 6 5 7 4 2 1 8 2 1 8 3 9 6 5 7 4 8 2 1 6 4 5 9 3 7 9 3 7 8 2 1 6 4 5 6 5 4 9 3 7 8 2 1 HATMAN wrote Using the semi-symmetry I would guess it is at or below 1.0: I would put it somewhat higher. I used a short forcing chain; also step 7a feels like it's in the 1.25 range.