SudokuSolver Forum

A few comments on earlier posts in this thread. I'll post my walkthrough later.

When I went through udosuk's walkthrough for A163 I was confused at first by the hidden singles in the second line. Then I realised that Prelims had been assumed. IIRC all previous udosuk walkthroughs that I've gone through started from the basic grid with all nine candidates in each cell.

I've no problem with walkthroughs starting after Prelims but that really ought to be stated. Afmob also omits that statement. IIRC when I first went through one of Afmob's walkthroughs I sent a feedback comment about that but it didn't result in any change.


BTW Thanks udosuk for putting your rating comments in hidden text, as requested by Ed in a different thread.

Thanks Afmob for a challenging Assassin. The first challenge was to colour the cages in my worksheet but I still managed to do it with only four colours. It's amazing how often cage patterns with crossovers can still be done with four colours.

As you will see from my walkthrough and the comments in it and at the end, I made harder work of it than others have. Still I hope you will find some of my steps interesting.


Here is my walkthrough for A163. I had to re-work some early stages because I'd missed some hidden singles and more importantly I hadn't included the 33(5) cage in the Prelims . Those of you who start the lazy way by using a software solver to do the Prelims won't have had that problem. Seriously though, I like using my Excel worksheet with candidates for each cage in a horizontal row of 9 numbers. I don't like the idea of candidates in a 3x3 grid of small numbers which is what software solvers appear to use; well all diagrams from software solvers show them that way. My older eyes prefer normal size numbers.

Thanks Afmob for pointing out errors in steps 15b and 15d and that two of my earlier CPEs had been incomplete. I've re-worked my walkthrough from step 16 onward. Thanks also to Ed for suggesting that some of my combination analysis was unnecessarily complicated. I've deleted step 13 and simplified step 19.

Prelims

a) 10(3) cage in N4 = {127/136/145/235}, no 8,9
b) 19(3) cage in N4 = {289/379/469/478/568}, no 1
c) 11(3) cage in N5 = {128/137/146/236/245}, no 9
d) 9(3) cage in N6 = {126/135/234}, no 7,8,9
e) 22(3) cage in N6 = {589/579}, 9 locked for N6
f) 8(3) cage in N7 = {125/134}, 1 locked for N7
g) 10(3) cage in N9 = {127/136/145/235}, no 8,9
h) 14(4) cage at R1C5 = {1238/1247/1256/1346/2345}, no 9
i) 19(5) cage at R1C3 must contain 1, CPE no 1 in R1C56
j) 32(5) cage at R1C8 = {26789/35789/45689}, no 1
k) 33(5) cage at R7C5 = {36789/45789}, no 1,2, CPE no 7,8,9 in R9C45

1. 45 rule on N7 3 innies R8C3 + R9C23 = 21 = {489/579/678}, no 2,3

2. 45 rule on N9 3 innies R7C89 + R9C7 = 21 = {489/579/678}, no 1,2,3

3. 45 rule on N1 2 outies R45C3 = 1 innie R1C3 + 10
3a. Min R45C3 = 11, no 1
3b. Max R45C3 = 17 -> max R1C3 = 7

4. 45 rule on N6 2 outies R7C89 = 1 innie R4C7 + 4
4a. Min R7C89 = 12 (from step 2) -> R4C7 = 8, clean-up: no 5 in 22(3) in N6
4b. R4C7 = 8 -> R7C89 = 12 -> R9C7 = 9 (step 2), clean-up: no 6 in R7C89 (step 2)
4c. R7C89 = 12 = {48/57} -> R56C7 = 6 = {15/24}, CPE no 4,5 in R78C7
4d. 10(3) cage in N9 = {127/136/235} (cannot be {145} which clashes with R7C89), no 4
4e. 7,8 in 33(5) cage at R7C5 locked in R7C5 + R8C56 + R9C6, locked for N8

5. Naked triple {679} in 22(3) cage in N6, locked for N6

6. R4C7 = 8 -> R45C6 = 9 = {27/36/45}, no 1,9

7. 45 rule on N78 1 remaining outie R6C5 = 9
7a. R7C3 = 9 (hidden single in R7), placed for D/
7b. R8C2 + R9C1 = 7 = {25/34}
7c. R8C4 = 9 (hidden single in R8)
7d. R3C6 = 9 (hidden single in C6)
7e. R2C9 = 9 (hidden single in N3), R3C89 = 7 = {16/25/34}, no 7,8
7f. R5C8 = 9 (hidden single in N6)
7g. R4C2 = 9 (hidden single in R4), R56C1 = 10 = {28/37/46}, no 5
7h. R1C1 = 9 (hidden single in R1), R1C2 + R2C3 = 8 = {17/26/35}, no 4,8

8. R8C3 + R9C23 (step 1) = {678} (only remaining combination), no 4,5
[Alternatively hidden triple {678} in N7.]

9. Min R6C5 + R8C3 + R9C2 = 22 -> max R7C46 = 6, no 6
[Could now use either Killer Quint 1,2,3,4,5 in R7C12, R7C46 and R7C89, locked for R7 or Hidden Killer Triple 6,7,8 in R7C5, R7C7 and R7C89 for R7 -> R7C57 = {678} but either of these would put the rating above the SSscore and Afmob’s estimate.]

10. 45 rule on N4 3 innies R456C3 = 16 = {358/457} (cannot be {178/268/367} which clash with R89C3, ALS block), no 1,2,6, 5 locked for C3 and N4, clean-up: no 3 in R1C2 (step 7h)
[Having avoided using a higher rated move after step 9, I immediately use one here because it is much more effective.]
10a. Killer triple 6,7,8 in R456C3 and R89C3, locked for C3, clean-up: no 1,2 in R1C2 (step 7h)
10b. 1,2 in C3 locked in R123C3, locked for N1
10c. 6 in C3 locked in R89C3, locked for N7
10d. 10(3) cage in N4 = {127/136}, no 4
10e. R56C1 (step 7g) = {28/46} (cannot be {37} which clashes with R456C3), no 3,7

11. 45 rule on N5 1 remaining outie R6C3 = 1 remaining innie R4C5, no 2,6 in R4C5, no 8 in R6C3
11a. R456C3 (step 10) = {358/457}
11b. 8 of {358} must be in R5C3 -> no 3 in R5C3

12. 16(3) cage at R5C4 = {178/358/367/457} (cannot be {268} because no 2,6,8 in R6C3), no 2

13. Deleted.
[In my original walkthrough I had 45 rule on D/ 3 innies R4C6 + R5C5 + R6C4 = 14 and then did some combination analysis, one sub-step including CCC and the other using a clash with the 11(3) which was also a CCC because of a common cell. It’s simpler to wait until step 19 when there are fewer possible combinations.]

14. 6 in R7 locked in R7C57, CPE no 6 in R5C5 using D\

15. Max R6C5 + R8C3 + R9C2 = 24 -> min R7C46 = 4
15a. Hidden killer pair 1,2 in R7C46 and R9C45 for N8, R7C46 must contain one of 1,2 (step 9 and step 15) -> R9C45 must contain one of 1,2
15b. 23(4) cage at R8C4 = {1589/1679/2489/2579} (cannot be {3479/3569} which don’t contain 1 or 2), no 3
15c. 7,8 only in R9C3 -> R9C3 = {78}
15d. Naked pair {78} in R9C23, locked for R9 and N7 -> R8C3 = 6

[Since I’ve had to re-work from here, I’ve used a step that I missed first time.]

16. 45 rule on N14 2 innies R16C3 = 6 = [15/24], clean-up: R4C5 = {45} (step 11)

17. 16(3) cage at R5C4 (step 12) = {358/457} (cannot be {178/367} because R6C3 only contains 4,5), no 1,6
17a. 45 rule on N5 3 remaining innies R4C5 + R56C4 = 16 = {358/457}, 5 locked for N5, clean-up: no 4 in R45C6 (step 6)

18. 1 in N5 locked in 11(3) cage, locked for D\
18a. 11(3) cage = {128/146} (cannot be {137} which clashes with R4C5 + R56C4), no 3,7

19. 45 rule on D/ 3 innies R4C6 + R5C5 + R6C4 = 14 = {167/257/347} (cannot be {158} because no 1,5,8 in R4C6, cannot be {248/356} which clash with R8C2 + R9C1), no 8, 7 locked in R4C6 + R6C4, locked for D/ and N5, clean-up: no 2 in R4C6 (step 6)
[When I first did my re-work I also eliminated {257} from R4C6 + R5C5 + R6C4 because of CCC with R45C6; this is a CCC between hidden 3-cell and 2-cell cages.
Alternatively 45 rule on D/ 2 innies R5C5 + R6C4 = 1 outie R5C6 + 5, IOU no 5 in R6C4, using a hidden 2-cell cage for the I-O.
However the combination analysis for R4C5 + R56C4 below is technically much simpler.]
19a. R5C5 = {124} -> no 4 in R6C4
19b. R4C5 + R56C4 (step 17a) = {358/457}
19c. 8 of {358} must be in R5C4 -> no 3 in R5C4
19d. 5 of {358} must be in R4C5, 7 of {457} must be in R6C4 -> no 5 in R6C4, clean-up: no 2 in R5C5 (step 19)

20. 8 on D/ locked in R1C9 + R2C8, locked for N3
20a. 15(3) cage = {168/258/348}

21. 32(5) cage at R1C8 = {35789/45689} (cannot be {26789} because R4C5 only contains 4,5), no 2
21a. R3C4 = 8 (hidden single in 32(5) cage)

22. Naked pair {45} in R4C5 + R5C4, locked for N5 -> R5C5 = 1, placed for D/, R6C4 = 7 (step 17a), R6C8 = 6, R4C9 = 7, R6C6 = 8 (hidden single in N5), R4C4 = 2 (step 18a), both placed for D\, R4C6 = 6 (step 19), placed for D/, R5C6 = 3, clean-up: no 1 in R3C9 (step 7e), no 2,4 in R5C1 (step 7g), no 5 in R6C7, no 5 in R7C8 (both step 4c)

23. 45 rule on N2 2 remaining outies R1C37 = 5 = [14/23]
23a. R3C89 (step 7e) = [16/25/52] (cannot be {34} which clashes with R1C7), no 3,4
23b. 15(3) cage in N3 (step 20a) = {258} (only remaining combination, cannot be {348} which clashes with R1C7), locked for N3 and D/ -> R3C89 = [16]

24. 32(5) cage at R1C8 (step 21) = {35789} (only remaining combination) -> R4C5 = 5, R1C8 + R2C7 = {37}, locked for N3 -> R1C7 = 4, R1C3 = 1 (step 23), R5C4 = 4, R6C3 = 5, clean-up: no 2 in R56C7 (step 4c)
24a. R56C7 = [51], R3C7 = 2, R5C9 = 2, clean-up: no 7 in R7C8 (step 4c)
24b. Naked pair {48} in R7C89, locked for R7 and N9

25. 10(3) cage in N9 (step 4d) = {235} (only remaining combination) -> R8C7 = 3, R9C9 = 5, R8C8 = 7, R7C7 = 6, all three placed for D\

and the rest is naked singles.

I made hard work of this one by missing
45 rule on N78 2 outies R6C5 + R9C7 = 18 = [99] although I used it later to get R6C5
45 rule on N69 2 innies R49C7 = 17 = {89}
and 45 rule on N14 2 innies R16C3 = 6 = {15/24}; I used this in my re-work.
Also 45 rule on N8 4 remaining innies R7C46 + R9C45 = 12 would have been simpler than steps 15 and 15a.

First, thankyou very much udosuk for hiding your ratings comments. Much appreciated.I take this approach too since I find it difficult to do (and follow in a WT). I must have missed some important things. I took a very different path to udosuk and Andrew and had to use a really nice (but hard-to-see) cage-block to get anywhere (see below step 7a). Then it was a really long solution so won't worry about a full WT. Perhaps someone can finish off this start in a quick way.

Finally, thanks for the puzzle Afmob!

Start of A163 solution.
To show I'm not just being lazy with only a partial WT, I included the prelims.
Prelims
i. 14(4)n2: no 9
ii. 32(5)n3: no 1
iii. 10(3)n4: no 8,9
iv. 19(3)n4: no 1
v. 11(3)n5: no 9
vi. 9(3)n6: no 7,8,9
vii. 22(3)n6: no 1,2,3,4
viii. 8(3)n7: no 6..9
ix. 33(5)n8: no 1,2
x. 10(3)n9: no 8,9

1. 22(3)n6 = {589/679}
1a. 9 locked for n6

2. "45" n69: 2 innies r49c7 = 17 = [89]

3. "45" n78: 1 remaining outie r6c5 = 9
3a. hidden single 9 at r8c4, r7c3 (placed for D/), r3c6, r2c9, r5c8, r4c2, r1c1

4. "45" n9: 2 remaining outies r56c7 = 6 = {15/24}(no 3,6,7)

5. "45" n9: 2 remaining innies r7c89 = 12 = {48/57}(no 1,2,3,6} = [4/7,5/8..]

6. combining steps 4 & 5: 18(4)n6 = {1458/2457} = [2/8..]
6a. must have 4 & 5 -> no 4 or 5 in r78c7 (CPE)

7. 14(3)n9: {158/347} blocked by h12(2)n9 (step 5)
7a. {248} must be [2]{48} but this clashes with the 18(4)n6 (step 6)
7b. 14(3) = {167/257/356}(no 4,8)

8. 8 in n9 only in r7c89 in h12(2) = {48}: both locked for n9, r7 & r56c7
8a. r56c7 = h6(2) = {15}: both locked for n6 & c7

It was a long struggle from there on for me but this allowed progress.

Cheers
Ed

Thanks for the comments Andrew, giving me a better understanding about the rating system.

And also thanks for letting me know how much you guys (Ed & Andrew) want to see the rating comments hidden. I'll try my best to hide them from now on.

And after reading manu's walkthrough (for V2) I revised mine a little bit (just reordering my most critical move earlier). It doesn't change the difficulty of the whole walkthrough (because the most critical move is the same) but it does make other moves a tad bit more elegant.

_________________
ADYFNC HJPLI BVSM GgK Oa m

Actually I found the V2 easier, at a human solving level, than A163. That was because I used the naked killer quint, which I'd spotted for A163 but decided wasn't appropriate to use for it.

I felt that the "unusual combined cages" in udosuk's first walkthrough and the replacement 3+3 innies-outies in his second walkthrough were a bit harder, at least at a human level, but not enough to affect the rating.

All the posted walkthroughs, including mine, use the killer quint / hidden killer triple. Apart from that my other steps are IMHO simpler, even so my walkthrough isn't particularly long. No fishes!


Here is my walkthrough for A163 V2.

Prelims

a) 11(3) cage in N5 = {128/137/146/236/245}, no 9
b) 8(3) cage in N7 = {125/134}, 1 locked for N7
c) 10(3) cage in N9 = {127/136/145/235}, no 8,9
d) 14(4) cage at R1C5 = {1238/1247/1256/1346/2345}, no 9
e) 19(5) cage at R1C3 must contain 1, CPE no 1 in R1C56
f) 32(5) cage at R1C8 = {26789/35789/45689}, no 1
g) 33(5) cage at R7C5 = {36789/45789}, no 1,2, CPE no 7,8,9 in R9C45

[I’ve used some of the steps that I missed in A163. No point in making things too difficult for myself.]

1. 45 rule on N78 2 outies R6C5 + R9C7 = 18 = [99]
1a. 7,8 of 33(5) cage at R7C5 locked in R78C5 + R89C6, locked for N8

2. 45 rule on N69 1 remaining innie R4C7 = 8
2a. R45C6 = 9 = {27/36/45}, no 1

3. R8C4 = 9 (hidden single in N8)
3a. R3C6 = 9 (hidden single in C6)
3b. R7C3 = 9 (hidden single in R7), placed for D/
3c. R8C2 + R9C1 = 7 = {25/34}
3d. R2C9 = 9 (hidden single in N3), R3C89 = 7 = {16/25/34}, no 7,8
3e. R1C1 = 9 (hidden single on D\), R1C2 + R2C3 = 8 = {17/26/35}, no 4,8

4. 45 rule on N7 3 innies R8C3 + R9C23 = 21 = {678} (only remaining combination)
[Alternatively hidden triple in N9]

5. 45 rule on N9 2 innies R7C89 = 12 = {48/57}
5a. 45 rule on N9 2 outies R6C89 = 9 = {27/36} (cannot be {45} which clashes with R7C89)
5b. 10(3) cage in N9 = {127/136/235} (cannot be {145} which clashes with R7C89), no 4

6. Min R6C5 + R8C3 + R9C2 = 22 -> max R7C46 = 6, no 6

7. Killer quint 1,2,3,4,5 in R7C12, R7C46 and R7C89, locked for R7
[I’d spotted this step for A163 but decided not to use it for that Assassin.
Alternatively hidden killer triple 6,7,8 in R7C5, R7C7 and R7C89 -> R7C5 = {678}, R7C7 = {67}]

8. Min R7C7 = 6 -> max R8C89 = 8, no 8
8a. 8 in N7 locked in R7C89 (step 5) = {48}, locked for R7 and N9
8b. Naked pair {67} in R7C57, CPE no 6,7 in R5C5 using D\

9. 45 rule on N14 2 innies R16C3 = 6 = {15/24}

10. 45 rule on N5 1 outie R6C3 = 1 remaining innie R4C5, no 3,6,7 in R4C5, no 1 in R6C3, clean-up: no 5 in R1C3 (step 9)

11. 45 rule on N5 3 innies R4C5 + R56C4 = 16 = {268/358/457} (cannot be {178/367} because R4C5 only contains 2,4,5), no 1
11a. 2 of {268} must be in R4C5 -> no 2 in R56C4

12. 1 in N5 locked in 11(3) cage, locked for D\
12a. 11(3) cage = {128/137/146}, no 5

13. 45 rule on D/ 3 innies R4C6 + R5C5 + R6C4 = 14 = {158/167/257/347} (cannot be {248/356} which clash with R8C2 + R9C1)
13a. 1,2 of {158/257} must be in R5C5 -> no 2 in R4C6, no 8 in R5C5, clean-up: no 7 in R5C6 (step 2a)
[I originally also did combination analysis on {347}, using a clash with the 11(3) cage, but this proved unnecessary.
At this stage I could have done 45 rule on D/ R5C5 + R6C4 = 1 outie R5C6 + 5, using hidden 2-cell cage R45C6, IOU no 5 in R6C4. This doesn’t achieve much here so I didn’t use it.]

14. 45 rule on N8 4 remaining innies R7C46 + R9C45 = 12 = {1236/1245}
14a. R8C4 = 9 -> R9C345 = 14 = {158/167/248/257} (cannot be {347/356} which clash with combinations for R7C46 + R9C45), no 3
14b. 7,8 only in R9C3 -> R9C3 = {78}

15. 6 in N7 locked in 28(5) cage at R6C5 = {15679/23689} -> R7C46 = {15/23}

16. 8(3) cage in N7 = {125} (cannot be {134} because {13}4 clashes with R7C46), locked for N7
16a. {125} can only be {15}2 (R7C12 cannot be {12/25} which clash with R7C46) -> R8C1 = 2, R7C12 = {15}, locked for R7 and N7
16b. Naked pair {23} in R7C46, locked for N8
16c. 28(5) cage at R6C5 (step 15) = {23689} (only remaining combination), no 7

17. R9C3 = 7 (hidden single in N7), R9C45 (step 14a) = {16} (only remaining combination), locked for R9 and N8 -> R9C2 = 8, R8C3 = 6, R7C5 = 7, R7C7 = 6, placed for D\, clean-up: no 1,2 in R1C2 (step 3e), no 4 in 11(3) cage in N5 (step 12a)

18. 4 on D\ locked in R2C2 + R3C3, locked for N1, clean-up: no 2 in R6C3 (step 9), no 2 in R4C5 (step 10)
18a. R4C5 + R56C4 (step 11) = {358/457}, no 6, 5 locked for N5, clean-up: no 4 in R5C6 (step 2a)
18b. 6 in N5 locked in R45C6 (step 2a) = {36}, locked for C6 and N5 -> R7C6 = 2, R7C4 = 3, clean-up: no 7 in R4C4 + R6C6 (step 12a), no 8 in R56C4 (step 18a)

19. Naked pair {12} in R4C4 + R5C5, locked for D\ -> R6C6 = 8, placed for D\
19a. Naked pair {45} in R89C6, locked for C6 and N8 -> R12C6 = [71], R8C5 = 8

20. Naked pair {34} in R8C2 + R9C1, locked for D/ -> R4C6 = 6, placed for D/, R5C6 = 3

21. 15(3) cage in N3 = {258} (only remaining combination), locked for N3 and D/ -> R5C5 = 1, R4C4 = 2, R6C4 = 7, R9C45 = [16], clean-up: no 2 in R6C89 (step 5a)
21a. Naked pair {36} in R6C89, locked for R6 and N6

22. 10(3) cage in N9 (step 5b) = {235} (only remaining combination), cannot be {127} because {17} only in R8C7), locked for N9 -> R8C8 = 7, placed for D\, R8C9 = 1, clean-up: no 6 in R3C8 (step 3d)

23. Naked triple {346} in R1C78 + R3C9, locked for N3 -> R2C7 = 7, R3C8 = 1, R3C9 = 6 (step 3d), R6C89 = [63], R9C9 = 5, placed for D\

and the rest is naked singles.