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 Post subject: CNC 46 Argyle Old Lace
PostPosted: Wed Oct 02, 2013 1:25 pm 
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Grand Master
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CNC 46 Argyle Old Lace

There is one number missing from all the short diagonals i.e. a common number is absent from the four diagonals - this is needed for solution.

Killer cages are non-consecutive i.e. no consecutive entries within the cage (called nabmer from now on).

Twin killer cages are remban groups so fully consecutive entries in any order.

Note r1c3 is in three cages.


Clue: The missing number must be in r5c5 from old lace properties. Note even without old lace this has to be true due to symmetry.

Thanks for pointing out the error Andrew.

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Solution:
634279158
634279158
275618394
918534672
581927436
492356817
367481925
743165289
826793541
159842763


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PostPosted: Thu Oct 03, 2013 9:11 pm 
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Location: Lethbridge, Alberta, Canada
Thanks HATMAN for another enjoyable puzzle.

An easy start, to draw one in; then it continued to be fairly easy until the final breakthrough step.

Here is my walkthrough for CNC 46 Argyle Old Lace:
Red border cages contain consecutive numbers, not necessarily in order; black border cages cannot contain consecutive numbers.

1. 20(3) cage at R1C3 = {479} (only combination without consecutive numbers), locked for N1
1a. 11(3) cage at R1C5 = {137/146} (other combinations contain consecutive numbers), 1 locked for C5 and N2
1b. 12(3) cage at R1C3 = {246} (only remaining combination without consecutive numbers) -> R1C3 = 4, placed for D\ at R1C3, R12C4 = {26}, locked for C4 and N2
1c. 11(3) cage at R1C5 = {137}, locked for C5 and N2

2. Remban group at R1C1 = {567/678} (cannot be {789} which clashes with 20(3) cage at R1C23) -> R2C2 = 7, placed for D\, R1C1 + R3C3 = {56/68}, 6 locked for N1 and D\ at R1C1, R3C1 = 9, placed for D\ at R3C1
2a. 9 in N2 only in R12C6, locked for C6

3. Remban group at R7C7 = {123/234/345}, no 8,9, 3 locked for D\ and N9

4. 10(3) cage at R6C6 = {136} (only combination without consecutive numbers) -> R7C5 = 6, placed for D\ at R3C1, D/ at R3C9 and Old Lace, R8C6 = 3, placed for D\ at R3C1, R6C6 = 1, placed for D\ at R1C1, D/ at R3C9 and Old Lace
4a. R5C6 = 6 (hidden single in C6)

5. Remban group at R7C7 (step 3) = {234/345}, 4 locked for D\ and N9

6. 9 on D\ at R1C1 only in R4C4 + R5C5, locked for N5 and Old Lace
6a. Killer cage at R4C4 cannot contain both of 1,2 and both of 8,9 -> R4C4 + R5C5 = {59}, locked for N5, D\ at R1C1 and Old Lace
6b. Naked pair {68} in R1C1 + R3C3, locked for N1
6c. Naked triple {234} in Remban group at R7C7, locked for N9

7. R5C4 = 3 (hidden single in N5)
7a. R3C5 = 3 (hidden single in Old Lace), placed for D\ at R1C3 and D/ at R1C7, R12C5 = [71]
7b. R6C4 = 4 (hidden single in Old Lace), placed for D/ at R1C9 and D\ at R3C1
7c. Naked pair {28} in R46C5, locked for C5 and N5 -> R4C6 = 7, placed for D\ at R1C3, D/ at R1C9 and Old Lace
7d. Naked pair {28} in R5C37, locked for R5
7e. Naked pair {28} in R5C37, CPE no 2,8 in R1C7 using D/ at R1C7, no 2,8 in R9C3 using D/ at R3C9
7e. 1 in N1 only in R13C2, locked for C2
7f. 4 in C5 only in R89C5, locked for N8

8. The only numbers missing from all the short diagonals are 2,5,8
8a. D\ at R3C1 contains two of 2,5,8 in R4C2 and R5C3 -> no 5,8 in R9C7

[I was forgetting about this cage …]
9. R1C3 = 4 -> Remban group at R1C2 = {234/345}, no 1, 3 locked for N1
9a. R3C2 = 1 (hidden single in N1)
9b. 2 in R3 only in R3C789, locked for N3

[Then I was a bit slow to spot …]
10. R5C37 = {28} are both on short diagonals so the only number which can be missing from all the short diagonals is 5 -> no 5 in R1C7, R2C6, R3C9, R4C248, R6C28, R7C19, R8C4 and R9C3
10a. Naked pair {28} in R4C2 + R5C3, locked for N4
10b. Naked pair {28} in R4C25, locked for R4

11. R4C4 = 9, placed for D/ at R1C7, R5C5 = 5, placed for D/ at R1C9, R6C2 = 6, placed for D/ at R1C7, R1C7 = 1, placed for D/ at R1C7, R9C7 = 7
11a. R1C6 = 9 (hidden single in C6)
11b. 5 in N2 only in R3C46, locked for R3

12. 1 on D/ at R1C9 only in R7C3 + R9C1, locked for N7
12a. R8C9 = 1 (hidden single in R8)
12b. R5C8 = 1 (hidden single in N6)

13. Naked triple {289} in R5C7 + R6C8 + R7C9, locked for D\ at R1C3, 2 also locked for N6 -> R2C4 = 6, R1C4 = 2
13a. Naked triple {289} in R5C7 + R6C8 + R7C9, CPE no 8,9 in R56C9

14. R5C2 = 9 (hidden single in R5)
14a. R5C1 = 4 (hidden single in N4), R5C9 = 7
14b. R3C8 = 7 (hidden single in N3)
14c. Naked pair {28} in R48C2, locked for C2
14d. Naked triple {357} in R6C139, locked for R6

15. 9 on D/ at R1C9 only in R2C8 + R7C3, CPE no 9 in R7C8

At this stage I looked at placements for R9C3 but I’ve optimised slightly, omitting those steps, since the next step results in a placement for R9C3

16. Consider placements for R5C3
R5C3 = 2 => R4C2 = 8, R8C2 = 2, placed for D/ at R1C9
or R5C3 = 8 => R5C7 = 2
-> no 2 in R3C7
[Maybe this step is a finned X-Wing?]
16a. Naked pair {68} in R3C37, locked for R3 -> R3C46 = [54], R3C9 = 2, R2C6 = 8, placed for D/ at R1C7, R5C3 = 2, placed for D/ at R1C7, R4C2 = 8, R7C1 = 7, R8C2 = 2, R8C8 = 4, R4C8 = 3, placed for D/ at R3C9, R9C3 = 9

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough for CNC 46 at Easy 1.5. I used one short forcing chain.


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PostPosted: Fri Oct 04, 2013 12:50 pm 
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Grand Master
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Andrew I started: D\ 2*Remban(3) and NC(3) means NC(3) = {159} 11(3) must have 1 implies r5c5= 5/9
Old Lace properties means that in N5 the missing number must be at r5c5 but 20(3) means 9 at r1c3 or r3c1 hence not missing hence r5c5 = 5.
Actually on pure symmetry considerations r5c5 must contain the missing number.


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PostPosted: Fri Oct 04, 2013 8:33 pm 
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Thanks HATMAN for showing the more direct way to get started.

Andrew


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