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HATMAN
 Post subject: Old Lace Test
PostPosted: Mon Oct 11, 2010 5:37 pm 
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Grand Master
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Old Lace Test: Windoku X OL

For those of you who have completed one of Børge's Assassin 201s here is a test to see if you fully understood the implications of the Old Lace group.

If you have this is reasonably easy.

Image

Note r5c5 is seen by the other 8 cells so is automatically in the old lace group.
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simon_blow_snow
 Post subject: Re: Old Lace Test
PostPosted: Wed Oct 13, 2010 9:53 am 
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Spoiler Warning: Don't read this if you really want to train your skills on the old lace pattern. Only read this after you find it too hard.

8 steps walkthrough:
1:
6(3)={123} [W:R2C6]
7(3)={124} [N4]
8(3)={125/134} must include one of {35}, two of {124}
--> R5C1289 must include all of {124} [R5], R4C9<>{35}

2:
Old lace & N5 interaction:
R37C5 clone R5C46, R46C5 clone R5C37
--> together R34567C5 clone R5C34567

3:
Since R5C34567 must include all of {6789} of R5
--> R34567C5 include all of {6789} [C5]
18(3): R12C5={45} [C5,N2], R1C4=18-4-5=9

4:
Since R34567C5<>{5}, R5C34567<>{5}
--> R5C89 must include {5} of R5 --> 8(3)={125} [N6]
--> R5C34567={36789} --> R34567C5={36789}

5:
Hidden pair C5: R89C5={12} --> 19(3)=[829], R8C5=1
Hidden pair R6: R6C46={12} [N5]
Hidden pair N5: R4C46={45} [R4]

6:
Naked triple D/: D/236={123} [D/]
--> 9(3)=[432] [W:R6C2] [{4} D/]
--> R46C46=[4512] [{15} D/] [{1} W:R6C2] [{2} W:R6C6]

7:
Hidden single R7: R7C9=2 --> 8(3)=[125] --> 6(3)=[132]
Hidden singles N8: R8C46=[54] [{5} W:R6C2]
Hidden single R6: R6C1=5

8:
10(3): R7C7=1, R68C7={36} [C7,W:R6C6]
--> R7C56=[67] --> R7C12={89} [R7,N7]
--> R7C8=5, D/89={67} [D/,N7]
--> R8C1=3, D/15=[89] --> 10(3)=[316] --> D/89=[76]
--> R8C89=[89] --> R6C8=9 (W:R6C6) --> R45C7={78} [C7,N6]
--> W:R2C6=[193824576] --> R1C7=5 --> 18(3)=[945]
--> D\=[765492183]

Finished

723946518
468251937
915738246
239485761
147693825
586172394
894367152
372514689
651829473


PS: I am glad I did not even need to use the 5 hidden Windoku groups at all.
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Andrew
 Post subject: Re: Old Lace Test
PostPosted: Thu Oct 14, 2010 6:26 am 
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Joined: Wed Apr 23, 2008 6:04 pm
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Thanks HATMAN for a fun puzzle. I took a break from the Assassin 201 puzzles, I've done two of them so far but haven't yet posted any walkthroughs, after seeing your message there about the Old Lace test.

This is the first time that I've tried a Windoku so I thought I might not make most efficient use of that extra feature but, after going through Simon's walkthrough I got the impression that I didn't miss much relating to them.

Nice walkthrough Simon!:
You got to the key breakthrough immediately; I took some time to find it. It's possible that you wrote an optimised walkthrough.

I was a bit surprised, when going through Simon's walkthrough to see the word "clone". I effectively used them except that I saw those steps as Law of Leftovers and didn't think about them as being "clones", even though I've done several puzzles on the Killer Forum using "clones".

Here is my walkthrough:
This is a Windoku Killer-X. R234C234, R234C678, R678C234 and R678C678 are also windows/nonets.

I’ve treated the Old Lace pattern as a 45(9) cage, as in the A201 puzzles.

Placements on diagonals are given in red, placements for the windows are given in blue; any placements for both are stated.

Prelims

a) 6(3) cage at R2C6 = {123}
b) 7(3) cage in N4 = {124}
c) 8(3) cage in N6 = {125/134}
d) 10(3) cage at R6C7 = {127/136/145/235}, no 8,9
e) 9(3) cage at R7C3 = {126/135/234}, no 7,8,9
f) 19(3) cage in N8 = {289/379/469/478/568}, no 1

Steps resulting from Prelims
1a. Naked triple {123} in 6(3) cage at R2C6, locked for window R234C678
1b. Naked triple {124} in 7(3) cage, locked for N4
1c. 8(3) cage in N6 = {125/134}, 1 locked for N6

2. 1 in R6 only in R6C456, locked for N5, CPE no 1 in R37C5
2a. 1 in 45(9) cage only in R6C46, locked for R6

3. 8(3) cage in N6 = {125/134}
3a. 3,5 must be in R5C89 (R5C89 cannot be {12/14} which clash with 7(3) cage in N4, ALS block) -> no 3,5 in R4C9

4. 7(3) cage in N4 = {124} and 8(3) cage in N6 = {125/134} must contain two 1s and either two 2s and one 4 or two 4s and one 2
4a. R4C19 must contain one and one of 2,4 -> R5C1289 must contain both of 2,4, killer pair for R5

5. 2,4 in 45(9) cage only in R37C5 + R46C46, CPE no 2,4 in R46C5
5a. 2,4 in N5 only in R46C46, locked for 45(9) cage, no 2,4 in R37C5
[Alternatively step 5a can be considered Law of Leftovers.]

6. Max R3C7 + 10(3) cage at R6C7 = 13 must contain 1, locked for C7

7. Consider placement of 1 in R6
R6C4 = 1, placed for D/ => R2C6 = 1 (only remaining 1 in 6(3) cage at R2C6)
or R6C6 = 1
-> 1 locked in R26C6, locked for C6, CPE no 1 in R2C2 using D\

8. Similarly consider placement of 2 in N5
R6C4 = 2, placed for D/ => R2C6 = 2 (only remaining 2 in 6(3) cage at R2C6)
or 2 in R4C4 + R6C6
-> 2 locked in R2C6 + R4C4 + R6C6, CPE no 2 in R2C2 using D\

9. 1 in window R234C234 only in R2C34 + R3C234
9a. Grouped X-Wing for 1 in R2C34 + R3C234 and 6(3) cage at R2C6, no other 1 in R23

10. Law of Leftovers R37C5 = R5C46 (each pair must have exactly the same two candidates) and R46C5 = R5C37 (same applies)

11. 6,7,8,9 in R5 only in R5C34567, R5C46 = R37C5 (same pair of candidates) -> 6,7,8,9 locked in R37C5 + R5C357, locked for 45(9) cage, no 6,7,8,9 in R46C46

12. Hidden killer pair 2,4 in R4C19 and R4C46 for R4, R4C19 contains one of 2,4 -> R4C46 cannot contain more than one of 2,4
12a. Hidden killer pair 2,4 in R4C46 and R6C46 for N5, R4C46 cannot contain more than one of 2,4 -> R6C46 must contain at least one of 2,4 -> R6C46 = {124}, no 3,5
12b. Hidden killer pair 2,4 in R4C46 and R6C46 for N5, R6C46 contains one of 2,4 -> R4C46 must contain one of 2,4
12c. Killer pair 2,4 in R4C19 and R4C46, locked for R4

13. Not sure whether I can extend the logic of step 11 further but I’ll try and hope it’s correct.
6,7,8,9 locked in R37C5 + R5C357, R5C37 = R46C5 (same pair of candidates) -> 6,7,8,9 locked in R34567C5, locked for C5

14. 18(3) cage in N2 = {459} (only remaining combination) -> R1C4 = 9, R12C5 = {45}, locked for C5 and N2

15. R8C5 = 1 (hidden single in C5)

16. R9C5 = 2 (hidden single in C5) -> R9C46 = 17 = [89]

17. 3 in C5 only in R34567C5 -> 3 locked in R357C5 + R5C37, locked for 45(9) cage, no 3 in R4C4
17a. Naked quad {1245} in R4C1469, 5 locked for R4
17b. Naked quad {1245} in R46C46, 5 locked for N5
17c. 3 locked in R357C5 + R5C37 -> 3 locked in R5C34567, locked for R5

18. 8(3) cage in N6 = {125} (only remaining combination), locked for N6, 5 also locked for R5
18a. 2 in R6 only in R6C46, locked for N5
18b. Naked pair {45} in R4C46, locked for R4 and N5

19. Naked triple {123} in R2C8 + R3C7 + R6C4, locked for D/, 3 also locked for N3 and window R234C678

20. 9(3) cage at R7C3 = {234} (only remaining combination) -> R8C3 = 2, R7C4 = 3, R7C3 = 4, placed for D/ and window R678C234, R4C6 = 5, placed for D/ and window R234C678, R4C4 = 4, placed for D\ and window R234C234, R6C4 = 1, placed for D/ and window R678C234, R6C6 = 2, placed for D\ and window R678C678, R2C6 = 1
20a. Naked pair {23} in R2C8 + R3C7, 2 locked for N3
20b. R1C8 = 1 (hidden single in N3)

21. 10(3) cage at R6C7 = {136/145} -> R7C7 = 1, R68C7 = [36/45/63], no 7, no 4 in R8C7
21a. R3C2 = 1 (hidden single in N1)
21b. R9C3 = 1 (hidden single in N7)

22. Naked pair {67} in R7C56, locked for R7 and N8 -> R8C4 = 5, R8C6 = 4
22a. Naked pair {36} in R68C7, locked for C7 and window R678C678 -> R3C7 = 2, R2C8 = 3, R7C6 = 7, R7C6 = 6

23. R7C9 = 2 (hidden single in N9), R4C9 = 1, R5C9 = 5, R5C8 = 2, R4C1 = 2, R5C2 = 4, R5C1 = 1

24. R1C2 = 2 (hidden single in R1)
24a. R6C1 = 5, R6C9 = 4 (hidden singles in R6)

25. Naked pair {89} in R7C12, locked for R7 and N7 -> R7C8 = 5
25a. Naked pair {67} in R8C2 + R9C1, locked for N7 and D/ -> R8C1 = 3, R9C2 = 5, R8C7 = 6, R6C7 = 3, R8C2 = 7, R9C1 = 6, R1C9 = 8, R5C5 = 9, R8C8 = 8, R8C9 = 9, R2C2 = 6, R1C1 = 7, R9C9 = 3, R3C3 = 5

and the rest is naked singles.

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

Simon wrote:
PS: I am glad I did not even need to use the 5 hidden Windoku groups at all.
Excuse my ignorance but what does this mean? As I said above, this is the first time I've done a puzzle involving Windoku so I've no idea what the hidden Windoku groups are.
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simon_blow_snow
 Post subject: Re: Old Lace Test
PostPosted: Fri Oct 15, 2010 12:47 am 
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Joined: Thu Oct 07, 2010 3:21 pm
Posts: 170
Andrew


I do not know (or care) much about jargons. When I write a walkthrough I just write whatever comes to my mind.

In hindsight LoL is probably a more precise terminology for that move, and there is probably another move called "cloning", but for me whenever I see two group of cells must contain the exact same set of values, I will say they "clone" each other. Hope it doesn't cause too much discomfort for you guys who are familiar with all the jargons.

Regarding the 5 hidden windoku groups, it can be found on various resources (including sudopedia) that there are 5 hidden groups of 9 cells which must contain no repeating values, deduced from the 4 given 3x3 "windows":

Code:
EAAAEBBBE
CWWWCXXXC
CWWWCXXXC
CWWWCXXXC
EAAAEBBBE
DYYYDZZZD
DYYYDZZZD
DYYYDZZZD
EAAAEBBBE


I think some players can spot this logic straight away, while it takes others a bit longer to understand.
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Andrew
 Post subject: Re: Old Lace Test
PostPosted: Fri Oct 15, 2010 6:21 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Simon

Thanks for the reply. You were right to call them clones. Maybe I didn't express myself clearly but all I was saying was that I'd used Law of Leftovers and I'd never realised that they were clones until I went through your walkthrough. Law of Leftovers must generate a group of clones. Maybe I tend to think of clones as individual cells, that's how I've used them in the past, but in the case of the Old Lace Test using them as a group was very helpful.

Thanks also for your diagram explaining the 5 hidden windoku groups. As I said in my post, it was the first time I'd tried a puzzle involving windoku. I expect that the 5 hidden windoku groups can probably be worked out using Law of Leftovers although I haven't (yet) tried doing that. I later found, while doing the first puzzle in the More Lace Practice thread, that 4 of these groups can be seen more easily by looking at groups of rows/columns; then the 5th group is what is left.

Another aspect of the Old Lace pattern which neither of us used is that the Innies and Outies for N5 are also a set of clones.

Andrew
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