SudoCue Users

i would guess it's about 1.25 on the basis it took me 1 to 2 hours

walkthrough principles are


1 row6 innies are 689
2 row 6 column9 is 5 or 7 (row 9 column 7 is 1 or 3)
3 rows1,3,4 and5 column7 together with row 6 column 9 equals 19
4 given row 9 column 7 is 1 or 3 the only possibilities are 1245 and 7 or 2345 with 5 however 2345 with 5 produces impossibility in row 3 columns 8 and 9 being 6 yet there being both 4 and 5 (presumed) in nonet 3
5 so we have 1245 and 7 and 3 in row9 column7
6 because row 7 column 8 can only be 1 2 or4 it follows rows 7 and 8 of column 9 cant be 2 and 9 (because of 15 innie in row 8) so they are 5 and 6
7 fairly straightforward now nonets 7 8 and 9 fall out

apologies i havent done the tiny text correctly

Gooders, there is nothing wrong with making a mistake but you should correct it by using the edit function.

I tried to avoid those little eliminations that led to nothing as best as I could since there were lots of them in this Assassin.

A83 Walkthrough:

1. R6789
a) Innies = 23(3) = {689} locked for R6
b) 16(3): R7C6 <> 1,2,3 because R6C56 <= 12
c) Innies+Outies N9: 4 = R6C9 - R9C7
-> R9C7 = (57), R6C9 = (13)
d) 18(3) <> 1 because R6C9 <= 7
e) 13(3) <> 9 because R9C7 = (13) blocks {139}
f) 16(3): R7C6 <> 4 because R6C9 = (57) blocks {457}
g) Innies+Outies R9: -5 = R8C8 - R9C67
-> R9C6 <> 1,2,3 and R8C8 <> 8,9 because R9C7 = (13)

2. R123
a) Outies R1 = 17(3): R2C2 <> 1,2,3,4 because R2C45 <= 12 since R2C45 @ 15(4)
b) Innies+Outies N1: 5 = R4C1 - R1C3
-> R4C1 = (6789), R1C3 = (1234)

3. N7 !
a) Innies+Outies: -11 = R6C2 - R79C3; R6C2 = (124)
-> R79C3 = 12/13/15(2) <> 1,2 and R7C3 <> 3
b) ! Innies+Outies: -11 = R6C2 - R79C3
-> R9C3 <> 3 since it would force R7C3 = 9 and R6C2 = 1 -> no combo for 12(3)
c) 3 locked in 14(3) = 3{29/47/56} -> 3 locked for R8
d) 1 locked in 7(3) for R7, R6C2 <> 1

4. C789
a) Outies C89 = 30(4) = {6789} locked for C7
b) 17(3) must have 2,3 xor 4 and it's only possible @ R7C8 -> R7C8 = (234)
c) Innies+Outies N3: -2 = R4C8 - R13C7
-> R4C8 <> 2,8,9 because R13C7 <= 9 and R13C7 = {13} blocked by R9C7 = (13)
d) Innies N9 = 14(3): R7C9 <> 3 because R9C7 = (13)
e) 18(3) = {279/459/567} because R6C9 = (57)
f) 17(3) <> 9 because {269} blocked by Killer pair (69) of 18(3)
g) Innies+Outies N3: -2 = R4C8 - R13C7
-> R4C8 <> 6 because R13C7 <> 6,7 and R3C7 <> 3,5 (8(2) impossible)
h) 6,8,9 locked in 28(5) = 689{14/23}

5. N24
a) Innies+Outies N2: 4 = R1C37 - R3C4
-> R3C4 <> 6,7,8,9 because R1C37 <= 9
b) 6,7,8,9 locked in 30(5) + R4C1 for N4

6. C789 !
a) ! Innies+Outies N3: -2 = R4C8 - R13C7
-> R4C8 <> 7 since R13C7 would be [54] which is a Killer pair of 13(3) = 7{15/24}
b) Hidden Single: R6C9 = 7 @ N8
c) 18(3) = 7{29/56}
d) Innies+Outies N9: 4 = R6C9 - R9C7 -> R9C7 = 3
e) 17(3) = 7{28/46} -> 7 locked for C7+N9
f) Killer pair (26) locked in 17(3) + 18(3) for N9

7. R6789
a) 2 locked in 13(3) = 2{47/56} for N8
b) Killer pair (67) of 13(3) blocks {67} of 13(2)
c) Killer pair (45) locked in 13(2) + 13(3) for R9
d) 3 locked in 12(3) @ N8 = 3{18/45}
e) Killer pair (58) locked in 12(3) + 21(4) for N8
f) 4 locked in R78C8 for C8
g) 1 locked in R9C89 for N9
h) 16(3) <> 1 because R6C56 <> 6,9
i) 16(3) <> 6 because R6C56 <= 9
j) Killer pair (24) locked in R6C2 + 16(3) for R6

8. R6789
a) 1 locked in 12(3) @ R6 = 1{38/56} -> R7C3 = (68)
b) Innies R9 = 16(3) = {169/178} -> R9C6 = (67)
c) 21(4) <> 1,9 because R9C6 = (67)
d) Hidden Single: R8C4 = 1, R7C6 = 9
e) 12(3) @ N8 = {138} -> {38} locked for R7+N8
f) 12(3) @ N7 = {156} -> R7C3 = 6, R6C3 = 1, R6C4 = 5
g) 16(3) = {349} -> {34} locked for R6+N8
h) Hidden Single: R7C9 = 5 @ R7 -> R8C9 = 6

9. C789
a) 15(2) = [69/78]
b) 17(3) = {278} -> R7C7 = 7, R7C8 = 2, R8C7 = 8
c) 15(3) <> 3 because R2C7 = (69)
d) 3 locked in 13(3) for R3 = 3{19/28/46}; R4C8 <> 3
e) 13(3) = 1{39/48} -> R4C8 = 1; R3C9 <> 8
f) 7(3) = {124} -> R4C6 = 2, R4C7 = 4, R3C7 = 1
g) Killer pair (89) locked in 15(2) + 13(3) for N3
h) R2C7 = 6, R1C8 = 7 -> R1C9 = 8, R2C8 = 5
i) 16(3) = {259} -> R1C7 = 2, R1C6 = 5, R1C5 = 9

10. N2
a) 18(3) = 8{37/46} -> 8 locked
b) 15(3) = {267} because (24) only possible @ R3C4
-> R3C4 = 2, {67} locked for R4+N5
c) 15(4) = {1347} -> R2C5 = 1, R2C4 = 7, {34} locked for R1

11. N1
a) 16(3) = {169} -> R2C2 = 9, {16} locked
b) 17(3) @ R2C3 = 8{27/45} -> 8 locked

12. Rest is singles.

Rating: Hard 1.25. It needed some detailed Innies+Outies analysis.

Yes,I'ld agree with a 1.25 rating.Once you see the key areas of the puzzle to work on r6c9,r9c7 plus I-O of r789/N7 it comes out quite readily.

Regards

Gary

Hi folks,

Ruud wrote:A tough Assassin with a few large cages. Also a test to see if a high rating by a solver program really represents a tough killer. I think the human brain outsmarts every computer program, but it needs a little more time.

Well, I'd love to be able to say that I found it easy, thus going 1 up on the automated solvers, but unfortunately I can't. Indeed, it took me a long time to even find a way into the puzzle without using some sort of contradiction move. Therefore, I have no hesitation in giving this one a rating of 1.5.

Ooops - I see I'm out of sync with everyone else again with my rating. At least SudokuSolver is keeping me company. I've namely just seen that it's scoring a 1.63 for this one.

Edit 2: minor amendments to WT
Edit 3: More corrections and comments (thanks, Andrew!)

Assassin 83 Walkthrough

Prelims:

a) 15(2) at R1C8 = {69/78}
b) 7(3) at R3C7 and R6C2 = {124}; no 1,2,4 in R78C2 (CPE)
c) 13(2) at R9C1 = {49/58/67}; no 9 in R9C1

1. Innies R6789: R6C178 = 23(3) = {689}, locked for R6

2. Outies C89: R2678C7 = 30(4) = {6789}, locked for C7

3. 17(3) at R7C7: min. R78C7 = 13
3a. -> R7C8 = {1..4}

4. Innie/Outie (I/O) diff. N9: R6C9 = R9C7 + 4
4a. -> R6C9 = {57}, R9C7 = {13}

5. 15(3) at R2C7 = {159/249/258/348/357/456}
(Note: {168/267} both blocked by 15(2) (prelim a))
5a. -> must contain exactly 1 of {6789}, which must go in R2C7
5b. -> no 6..9 in R2C89

6. 15(2) at R1C8, 15(3) at R2C7 and 13(3) at R3C8 form hidden killer quad (type 2:1:1) on {6789} within N3
6a. 13(3) at R3C8 can only contain 1 of {6..9}
6b. -> no 6..9 in R4C8

7. All 3 of {689} in N6 locked in 28(5) at R3C9 = {(14/23)689} (no 5,7)

8. Hidden single (HS) in N6 at R6C9 = 7
8a. -> R9C7 = 3 (step 4)
8b. cleanup: no 8 in R1C8
8c. split 11(2) at R78C9 = {29/56} (no 1,4,8)

9. Naked triple at R7C128 = {124}, locked for R7
9a. cleanup: no 9 in R8C9 (step 8c)

10. 16(3) at R6C5: max. R6C56 = 9 -> min. R7C6 = 7
10a. -> R7C6 = {789}
10b. max. R7C6 = 9 -> min. R6C56 = 7
10c. -> no 1 in R6C56

11. 12(3) at R7C4 must contain 1 of {124}
(otherwise lowest possible 3-cell cage sum would be {356} = 14(3))
11a. only place for {124} is R8C4
11b. -> R8C4 = {124}

12. 3 in R8 locked in 14(3) at R8C1
12a. -> 14(3) at R8C1 = {239/347/356} (no 1,8)
12b. 3 locked for N7

13. I/O diff. R89: R8C47 = R7C9 + 4
13a. -> no 4 in R8C4 (IOU)
13b. max. R8C47 = 11 -> max. R7C9 = 7
13c. -> no 9 in R7C9
13d. cleanup: no 2 in R8C9 (step 8c)

14. Naked pair (NP) on {56} at R78C9, locked for C9 and N9
14a. cleanup: no 9 in R1C8

15. 17(3) at R7C7 = {179/278} (no 4)
15a. 7 locked in R78C7 for C7 and N9

16. 4 in R7 locked in R7C12 -> not elsewhere in N7
16a. no 4 in R6C2
16b. cleanup: no 9 in R9C2; no 7 in R8C123 (step 12a)

17. 3 in R7 locked in 12(3) at R7C4 = {138/237} (no 5,6,9)

18. R7C6, 12(3) at R7C4 and split 18(3) at R8C56+R9C6 form hidden killer triple on {789}
18a. -> {189/279} blocked for split 18(3) = {459/468/567} (no 1,2)

Note: Andrew correctly pointed out that I could have eliminated the 7,8,9 from R9C45 at this stage, rather than in the next step.

19. R8C4 and R9C45 form hidden killer pair on {12} within N8
19a. -> 13(3) at R9C3 = {(1/2)..} = {148/247} (no 5,6,9)
(Note: {157/256} both blocked by 13(2) at R9C1)
19b. 13(3) cannot have both of {12}
19c. -> no 1,2 in R9C3
19d. 13(3) must have exactly 1 of {78}, which must go in R9C3
19e. -> no 7,8 in R9C45

20. 1 in N7 locked in R7C12 -> not elsewhere in R7
20a. no 1 in R6C2

21. R6C2 = 2, R7C8 = 2 (naked singles)
21a. split 15(2) at R78C7 = {78} (last combo), 8 locked for C7 and N9

22. R2C7 blocks {69} combo for 15(2) at R1C8
22a. -> R1C89 = [78]

23. 2 in R9 locked in N8 -> not elsewhere in N8

24. Naked single (NS) at R8C4 = 1
24a. -> R7C9+R8C7 = [58] (step 13)
24b. -> R7C7 = 7, R8C9 = 6

25. Hidden single (HS) in R6C3 = 1
25a. -> split 11(2) at R6C4+R7C3 = [56]
(Note: [38] blocked by R7C4)
25b. cleanup: no 7 in R9C12

26. R9C36 = [76] (hidden singles, R9)
26a. -> split 12(2) at R8C56 = {57}, 5 locked for R8

27. R7C6 = 9, R8C8 = 4 (hidden singles, R7/R8)

28. Naked pair (NP) at R6C56 = {34}, locked for N5

29. 4 in 7(2) at R3C7 locked in C7 -> not elsewhere in C7

30. 16(3) at R1C5 = {259} (last combo)
30a. -> R1C5 = 9; R1C67 = {25}, locked for R1

31. 1 in R1 locked in 16(3) at R1C1 = [169/619]
31a. -> R2C2 = 9; R1C12 = {16}, locked for R1 and N1

32. 15(3) at R2C7 = [654] (last combo/permutation)

33. R2C45 = 8(2) (15(4) cage split)
33a. -> R2C45 = [71] (last combo/permutation)

34. 18(3) at R2C6 = [864] (last combo/permutation)

35. 17(3) at R2C3 = [278] (last combo/permutation)

36. Hidden pair in N3 at R3C89 = {39}, 3 locked for R3
36a. -> R4C8 = 1 (cage split)

Now all naked singles to end.

Interesting to see the different ways to solve A83.

I must admit that I couldn't follow goooders step 4. From my understanding of the first four steps I still had R34C7 as {24} rather than [42] but even if they were [42] I can't see why 13(3) at R3C8 can't be {39}1.

Gary said the key moves were to work on r6c9,r9c7 plus I-O of r789/N7. I used the three of those but, although I tried it at several times, I never managed to use I-O of r789 and I don't think the other posted walkthroughs did either.

Mike wrote:Indeed, it took me a long time to even find a way into the puzzle without using some sort of contradiction move.

I was the same with this puzzle. Although it may not look it to go through my walkthrough, I found it difficult to make progress in the early stages. My key breakthrough, step 22, could have been made earlier but I didn't see it until then. I think there are some interesting moves between when it might first have appeared, probably after step 12, and when I spotted it.

Because of the difficulty of getting into this puzzle, I rate it a moderate 1.5.

Here is my walkthrough, with step 15 edited for clarity.

Prelims

a) R1C89 = {69/78}
b) R9C12 = {49/58/67}, no 1,2,3
c) 7(3) cage at R3C7 = {124}
d) 7(3) cage at R6C2 = {124}, CPE no 1,2,4 in R89C2, clean-up: no 9 in R9C1

1. 45 rule on N1 1 outie R4C1 = 1 innie R1C3 + 5, R1C3 = {1234}, R4C1 = {6789}

2. 45 rule on N9 1 outie R6C9 = 1 innie R9C7 + 4, R6C9 = {56789}, R9C7 = {12345}

3. 45 rule on R6789 3 innies R6C178 = 23 = {689}, locked for R6, clean-up: no 2,4,5 in R9C7 (step 2)
3a. Max R6C56 = 11 (cannot be {57} which clashes with R6C9) -> min R7C6 = 5

4. R678C9 = {279/378/459/567} (cannot be {189/369/468} because R6C9 only contains 5,7), no 1

5. 45 rule on N7 2 innies R79C3 = 1 outie R6C2 + 11, min R79C3 = 12, no 1,2

6. 45 rule on C89 2 outies R26C7 = 1 innie R7C8 + 13
6a. Max R26C7 = 17 -> max R7C8 = 4
6b. Min R26C7 = 14, no 1,2,3,4

7. 17(3) cage in N9 = {179/269/278/359/368/458/467}
7a. R7C8 = {1234} -> no 1,2,3,4 in R78C7

8. 45 rule on R1 4 innies R1C1234 = 14 = {1238/1247/1256/1346/2345}, no 9

9. 45 rule on N4 4 innies R4C1 + R5C3 + R6C23 = 15 = {1239/1248/1257/1347/1356/2346}
9a. R4C1 = {6789} -> no 6,7,8,9 in R56C3

10. 45 rule on C89 4 outies R2678C7 = 30 = {6789}, locked for C7
10a. Min R2C7 = 6 -> max R2C89 = 9, no 9

11. 45 rule on N3 2 innies R13C7 = 1 outie R4C8 + 2
11a. Max R13C7 = 9 -> max R4C8 = 7

12. 13(3) cage at R3C8 = {139/148/157/238/247/256/346}
12a. 7 of {157/247} must be in R3C89 (7 cannot be in R4C8 because R3C89 = {15/24} clash with R13C7 = [54], step 11) -> no 7 in R4C8

13. 45 rule on N12 2 outies R1C7 + R4C1 = 1 innie R3C4 + 9
13a. Max R1C7 + R4C1 = 14 -> max R3C4 = 5

14. 45 rule on R89 3 innies R8C479 = 15, min R8C7 = 6 -> max R8C49 = 9, no 9, clean-up: no 2,4 in R7C9 (step 4)

15. Hidden killer quad 1,2,3,4 in R7C12, R7C45 and R7C8 for R7 -> R7C45 cannot contain more than one of 1,2,3,4 -> min R7C45 = {15} = 6 -> max R8C4 = 6

16. 45 rule on R9 2 innies R9C67 = 1 outie R8C8 + 5
16a. Max R9C67 = 12 -> max R8C8 = 7
16b. Min R9C67 = 6 -> min R9C6 = 4 (R9C67 cannot be [33])

17. 45 rule on N8 2 outies R9C37 = 1 innie R7C6 + 1
17a. Min R7C6 = 5 -> min R9C37 = 6, min R9C3 = 4 (R9C37 cannot be [33])
17b. Min R9C3 = 4 -> max R9C45 = 9, no 9
17c. R9C45 cannot be {13} which clashes with R9C7 -> no 9 in R9C3, clean-up: no 3 in R7C3 (step 5)

18. R678C9 = {279/378/459/567}
18a. 14(3) cage in N9 = {149/158/167/239/248/257/347/356}
18b. Hidden killer quad 6,7,8,9 in R78C7, R78C9 and 14(3) cage -> R78C9 and 14(3) cage must each contain one of 6,7,8,9
18c. 7 of {279/378/567} must be in R6C9 (R678C9 cannot be 5{67} because of step 18b) -> no 7 in R78C9
18d. 14(3) cage = {149/158/239/248/257/347/356} (cannot be {167} because of step 18b)

19. 3 in N7 locked in R8C123, locked for R8, clean-up: no 8 in R7C9 (steps 18 and 18c)
19a. R8C123 = 3{29/47/56}, no 1,8

20. 1 in N7 locked in R7C12, locked for R7 and 7(3) cage -> no 1 in R6C2, clean-up: no 4 in R7C3 (step 5)
20a. Min R7C3 = 5 -> max R6C34 = 7, no 7

21. Hidden killer quad 1,2,3,4 in R7C12, R7C8 and R7C459 for R7 -> R7C459 can only contain one of 2,3,4
21a. 12(3) cage in N8 = {129/138/147/156/237/246/345}
21b. 5,6 of {246/345} must be in R7C45 (step 21) -> no 5,6 in R8C4

22. 45 rule on N6 4 innies R4C78 + R5C7 + R6C9 = 17 = {1367/1457/2357} (cannot be {2456} because R345C7 = 1{24} => R9C7 = 3 => R6C9 = 7) -> R6C9 = 7, R9C7 = 3 (step 2); clean-up: no 8 in R1C8
22a. R78C9 = 11 = {56}/[92], no 4,8
22b. 1 in N9 locked in 14(3) cage (step 18d) = {149/158}, no 2,6,7
22c. R4C78 + R5C7 + R6C9 = {1457/2357} (cannot be {1367} because 3,6 only in R4C8), no 6, 5 locked for N6

23. 7 in N9 locked in 17(3) cage = {278/467}, no 9, 7 locked for C7

24. Naked triple {124} in R7C128, locked for R7
24a. 3 in R7 locked in R7C45, 12(3) cage in N8 = 3{18/27/45}, no 6,9

25. 16(3) cage at R6C5 = {259/268/349/358/367/457} (cannot be {169/178} because no 6,7,8,9 in R6C56), no 1
25a. 7,8,9 only in R7C6 -> R7C6 = {789}

26. 1 in R6 locked in R6C34
26a. 12(3) cage at R6C3 = 1{29/38/47/56}
26b. 5 of {156} must be in R6C34 -> no 5 in R7C3

27. 2 in R9 locked in R9C45, locked for N8, clean-up: no 7 in R7C45 (step 24a)
27a. R9C345 = 2{47/56}, no 1,8
27b. R9C12 = [49]/{58} (cannot be {67} which clashes with R9C345), no 6,7

28. 1 in R9 locked in R9C89, locked for N9
28a. 14(3) cage (step 22b) = {149/158}
28b. 4,5 must be in R8C8 -> no 4,5 in R9C89

29. Killer pair 8,9 in R9C12 and R9C89, locked for R9
29a. Killer pair 4,5 in R9C12 and R9C345, locked for R9

30. 21(4) cage at R8C5 = {3468/3567} (cannot be {1389/3459} because R9C6 only contains 6,7), no 1,9

31. R8C4 = 1 (hidden single in R8), R7C45 = {38}, locked for R7 and N8

32. 21(4) cage at R8C5 (step 30) = {3567} (only remaining combination), 5,6,7 locked for N8 -> R7C6 = 9
32a. 5 locked in R8C56, locked for R8 -> R8C8 = 4, R7C8 = 2, R78C9 = [56], R78C7 = [78], R7C3 = 6; clean-up: no 9 in R1C8
[While checking the walkthrough I saw that I could now have fixed R1C89 because of the clash with R2C7. Missing this probably didn’t make much difference since the remaining steps are straightforward.]

33. Naked pair {57} in R8C56, locked for R8 and N8 -> R9C6 = 6

34. Naked pair {14} in R7C12, locked for N7 and 7(3) cage -> R6C2 = 2, clean-up: no 9 in R9C2

35. R7C6 = 9 -> R6C56 = 7 = {34} (only remaining combination), locked for R6 and N5, R7C23 = [15], clean-up: no 6 in R4C1 (step 1)

36. 4 in 7(3) cage locked in R34C7, locked for C7

37. R9C3 = 7 (hidden single in R9)

38. R1C7 + R4C1 = R3C4 + 9 (step 13)
38a. R4C1 cannot be 8 more than R3C4 -> no 1 in R1C7

38. R1C567 = {259/358/457} (cannot be {169/178/349/367} because R1C7 only contains 2,5, cannot be {268} which clashes with R1C89), no 1,6
38a. 5 of {358/457} must be in R1C7
38b. 9 of {259} must be in R1C5
38c. -> no 2,5 in R1C5

39. 15(3) cage at R3C4 = {249/267} (cannot be {168} because R3C4 only contains 2,3,4, cannot be {348} because 3,4 only in R3C4), no 1,3,8
39a. CPE no 2 in R5C4

40. 45 rule on N2 2 outies R1C37 = 1 innie R3C4 + 4, R3C4 = {24} -> R1C37 = 6,8 = [35/42], no 2 in R1C3

41. R2C789 = {159/168/249/267/456} (cannot be {258/348/357} because R2C7 only contains 6,9), no 3
41a. 6 of {168/267/456} must be in R2C7 -> no 6 in R2C8

42. 3 in N3 locked in R3C89, locked for R3 and 13(3) cage -> no 3 in R4C8
42a. 13(3) cage at R3C8 = {139} (only remaining cage) -> R4C8 = 1, R4C67 = [24], R3C7 = 1, R3C89 = {39}, locked for R3 and N3, R1C89 = [78], R2C789 = [654], R1C7 = 2, R5C7 = 5, R6C7 = 9, R9C89 = [91], R3C89 = [39], R45C9 = [32], clean-up: no 7 in R4C1 (step 9)

43. R1C7 = 2 -> R1C56 = 14 = [95], R8C56 = [57]

44. 1 in R1 locked in R1C12, locked for N1
44a. 16(3) cage = {169} (only remaining combination), locked for N1 -> R2C2 = 9, R1C12 = {16}, locked for R1, R8C2 = 3

45. R1C34 = {34}, locked for 15(4) cage at R1C3 -> R2C45 = 8 = [71]

46. 8 in N2 locked in 18(3) cage = [864] (only remaining permutation)

and the rest is naked singles, a hidden single and a cage sum