Concentric Squares Killer

[mhparker]

Hi folks,

If there aren't any really fiendish variants I might get a chance to do something else this weekend!

We wouldn't want that now, would we?!


Concentric Squares (CS) Killer (Est. rating: 1.25)



3x3::k:4608:4608:460817953077:4359:4359:4873:46082571:4109:41092575:4359:48732571:4373:4373:4375:43751562:48732333:4373:5407:5407:43751562:41322333:4391:5407:5407233841322863:4391:43912858:46604132:543128633386:4660:4660:59505431:543131382628:5950:59502623365839173917:5950:

Happy weekend!

[gary w]

Gee thanks Mike!!

Gary

[Andrew]

Nice cage pattern Mike!

An easy one to colour, it only requires three colours, but it looks better in black and white. Still I'll stick with the coloured version when I work on it; the colours make it easier to see the cage boundaries.

[gary w]

Hi Mike,

I take it back..this turned out to be a rather lovely puzzle.I've outlined my solution.



Prelims

1.r18c3=11 -> r3c1<>9/8/1 cannot have a 2/3 at r8c3).In c3 9 must be at r89.r2c56={79}
2.r7c29=15-> 13/2 cage N8 <> {67}
3.Outies on r123-> r279c4+r9c5=10 r2c4 max 4 r23c3 min 6
4.In N8 one of 1/2 must go in r7c4 and r9c45.Thus from 4a below r7c4=1/2......4a Outies N9 -> r6c8+r89c6=21


5.From 2. 13/2 cage N8 = {58} or (49}-> 4/8 so 12/2 cage N8 <>{48}
6.So 12/2 cage N8 = {57} or {39} -> 3/7 so 10/2 cage N89 <> {37}
7.Consider placement of 3 in N8.In all cases,whether it is at r8c45 or r9c456 it must enforce r9c3=9.Also now 21/3 cage N7= {678} and r7c13=5
8.Now in N8 6 is at r89c6 ..only places
9Thus from 4a r6c8<>9 so in N6 HS 9 at r4c7
10.N4 9 must be at r56c1
11.N3 9 at r1c89
12.HS 9 at r3c2 -> r4c2=3 -> r56c2={14} 9/2 cage N4= {27} r7c1<>4 and thus from 7. r7c3<>1So r7c3 <> 1 or 2 so r6c3<>8
so HS c3 r8c3=8
13 c3 1 at r23and N1 8 at r12c2 so 18/4 cage N1 ={2358} thus other number in r23c3=6 ..cannot be 4 because of constraints on 10/3 cage
14.R2C4=3 R7C4=2 (R2C4+ R7C134 =10 outies on 1/4)
15. r7c1=1 r7c3=4

Mop up now.
The effect of the placements of the 9s was very attractive.
And,of course,the last placements were the "centre" square.


Now I can enjoy the rest of my weekend!!Seriously though Mike..thanks for a very nice puzzle.

Regards

Gary

[Afmob]

Cool cage pattern. It required lots of litte steps but nothing to complicated.

CS Walkthrough:

1. C1234
a) Outies C12 = 11(2): R1C3 <> 1,8,9
b) Outies C1234 = 18(5) = 12{348/357/456} -> 1,2 locked for C5

2. R12
a) 16(2) = {79} locked for C2+N2
b) Outies R1 = 9(2) = {18/36/45}
c) 12(2) @ R1: R1C7 <> 3,5

3. C9
a) Outies C9 = 10(2) -> no 5
b) Killer pair (45) of 6(2) blocks {45} of 9(2) @ C9

4. R6789
a) Innies+Outies R9: -6 = R8C1 - R9C9 -> R8C1 = (123); R9C9 = (789)
b) 12(2): R8C4 <> 3
c) Outies R89 = 15(2) = {69/78}
d) Killer pair (67) of Outies R89 blocks {67} of 13(2)
e) Killer pair (89) locked in 13(2) + Outies R89 for R7
f) Innies+Outies R789: 12 = R6C38 - R7C1
-> R7C1 = (12345)
-> R6C38 = 13/14/15/16/17 -> no 1,2,3; R6C8 <> 4 since R6C3 <= 8
g) 9 locked in R89C3 for N7
h) 21(3): R8C3 <> 4,5 because R78C2 <> 9

5. N5
a) Innies N5 = 7(2) -> no 7,8,9
b) 17(3) @ R3C4 = 8{36/45} -> 8 locked for C3+N2
c) Innies N5 = 7(2): R6C6 <> 5,6

6. N8
a) Killer pair (48) of 13(2) blocks {48} of 12(2)
b) Killer pair (59) locked in 13(2) + 12(2)
c) 11(3): R7C34 <> 7 because R6C3 is at least 4
d) 10(2): R8C7 <> 1

7. N9
a) Outies = 21(2+1) <> 1,2
b) 10(2): R8C7 <> 8,9

8. N3
a) 12(2) = [39/48/57]

9. C1234
a) Outies C12 = 11(2) = [29/38/47/56]
b) 12(2) @ C2: R4C2 <> 4
c) Innies+Outies C123: -4 = R27C4 - R9C3
-> R9C3 = (789) since R27C4 is at least 3
-> R27C4 = 3/4/5 -> no 5,6
d) 21(3) = {489/579/678} -> has 2 of (789) + R9C3 = (789) -> locked for N7

10. R789
a) Killer pair (37) of 12(2) blocks {37} of 10(2) -> 10(2) = [46/64/82]
b) 21(3): R7C2 <> 7 since R8C23 would be {59/68}
-> blocked by Killer pairs (59,68) of 12(2) and 10(2)
c) 21(3) <> 5 because R7C2 = (68) -> 21(3) = 8{49/67}, 8 locked for N7
d) 7 locked in R7C789 for N9
e) 14(3): R9C45 <> 7,8 since R9C3 = (79)
f) Outies R89 = 15(2) = [69/87]
g) Innies+Outies R9: -6 = R8C1 - R9C9; R9C9 = (89) -> R8C1 = (23)
h) 10(3) <> 4 because R8C1 = (23)
i) 10(3) = 3{16/25} -> 3 locked for N7
j) 1 locked in 23(4) for N9 -> 23(4) = 19{58/67} -> 9 locked for N9

11. N8 !
a) ! 15(3) <> {456} since it must have 7 xor 8 because 15(3) + R9C39 are
the only positions in R9 where 7,8,9 are possible
-> 15(3) = {258/267/348/357}
b) 14(3) <> 2 because {239} is blocked by Killer pair (23) of 15(3)
-> 14(3) = {149/167/347}
c) Hidden Single: R7C4 = 2
d) 1 locked in 14(3) = 1{49/67} for R9

12. N7
a) 10(3) = {235} locked; 5 locked for R9
b) 11(3) = 2[54/81]
c) 6 locked in 21(3) -> 21(3) = {678} locked; 7 locked for R8
d) R9C3 = 9, R9C9 = 8
e) Hidden Single: R9C6 = 7 @ R9, R8C5 = 3 @ N8
f) R2C6 = 9, R2C5 = 7, R8C4 = 9, R8C1 = 2
g) Hidden Single: R7C9 = 9 @ N9

13. N9
a) 23(4) = {1589} -> R8C9 = 5, R8C8 = 1
b) 10(2) = {46} locked for R8
c) 2 locked in R9C78 -> 15(3) = {267}; 2,6 locked for R9+N9
d) R8C7 = 4, R8C6 = 6
e) Hidden Single: R7C2 = 6 @ R7
f) 18(3) = {378} -> R6C8 = 8
g) R6C3 = 5, R7C3 = 4, R7C1 = 1

14. N3
a) Hidden Single: R1C7 = 8 -> R1C6 = 4
b) Hidden Single: R1C8 = 9 @ C8, R4C7 = 9 @ C7, R5C5 = 9 @ C5
c) Hidden Single: R6C1 = 9 @ R6, R3C2 = 9 @ C2
d) R5C1 = 6, R4C2 = 3, R9C2 = 5, R9C1 = 3
e) 6(2) = {24} locked for C9

15. N6
a) 9(2) @ C9 = {36} -> R6C9 = 6, R5C9 = 3

16. Rest is clean-up and singles.

[CathyW]

Nice one Mike. I didn't keep a WT but the outies of c123 and innies of N2356 proved quite helpful.

[Andrew]

Nice puzzle Mike. I'll agree with your rating of 1.25.

I didn't keep a WT but the outies of c123 and innies of N2356 proved quite helpful.

I used innies/outies for C123 (see comment about the difference between outies and innies/outies after step 28) but missed innies of N2356. At one stage I looked at multiple nonets but couldn't see any useful ones; don't know how I missed those ones which are probably the only useful multiple nonets. As it happens I had no problems with eliminations from R6C8 so it probably didn't have much affect on my solution path but it's annoying that I missed it; in a different puzzle it might have been important.

At one stage I wondered whether it would be possible to use "concentric" as a solving step. However there were also two outies from the outer ring so that didn't help. Still some puzzle creator might be able to use the idea of an outer concentric ring so that overlaps give the total for R19C9.

Here is my walkthrough. It took me until step 51, including prelims, to make the first placements but then it fell fairly quickly.

Didn't have time to check my walkthrough properly last night. I've now done that and made the necessary changes.

Thanks Mike for pointing out some more corrections to typos and extensions to a few of my steps.

1. R1C45 = {16/25/34}, no 7,8,9

2. R1C67 = {39/48/57}, no 1,2,6

3. R2C56 = {79}, locked for R2 and N2, clean-up: no 3,5 in R1C7

4. R34C2 = {39/48/57}, no 1,2,6

5. R34C9 = {15/24}

6. R45C3 = {18/27/36/45}, no 9

7. R45C8 = {18/27/36/45}, no 9

8. R56C2 = {14/23}

9. R56C9 = {18/27/36} (cannot be {45} which clashes with R34C9), no 4,5,9

10. R7C56 = {49/58/67}, no 1,2,3

11. R8C45 = {39/48/57}, no 1,2,6

12. R8C67 = {19/28/37/46}, no 5

13. R234C1 = {289/379/469/478/568}, no 1

14. 10(3) cage at R2C3 = {127/136/145/235}, no 8,9

15. 10(3) cage in N3 = {127/136/145/235}, no 8,9

16. 11(3) cage at R5C7 = {128/137/146/236/245}, no 9

17. 11(3) cage at R6C3 = {128/137/146/236/245}, no 9

18. 21(3) cage in N7 = {489/579/678}, no 1,2,3

19. 10(3) cage in N7 = {127/136/145/235}, no 8,9

[Mike commented "You could have gone one step further (with step 13) and eliminated the 3 from R34C1 due to {79} being unavailable in R2C1." I'm putting this comment here because it's not really part of a preliminary step.]

20. 45 rule on R1 2 outies R2C29 = 9 = {18/36/45}, no 2

21. 45 rule on R9 1 innie R9C9 – 6 = 1 outie R8C1, R8C1 = {123}, R9C9 = {789}

22. 45 rule on C1 1 innie R1C1 = 1 outie R9C2, no 8,9 in R1C1

23. 45 rule on C9 2 outies R18C8 = 10 = {19/28/37/46}, no 5

24. 45 rule on C12 2 outies R18C3 = 11 = [29/38]/{47/56}, no 1, no 8,9 in R1C3

25. 45 rule on R89 2 outies R7C29 = 15 = {69/78}
25a. R7C56 (step 10) = {49/58} (cannot be {67} which clashes with R7C29)
25b. Combined cage R7C2569 = 28 = 89{47/56}

26. Killer pair 8,9 in R7C29 and R7C56, locked for R7
26a. 18(3) cage at R6C8, 8,9 only in R6C8 -> no 1
[Mike added "Similarly, no {89} in R7C78 -> no 2,3,4 in R6C8".]

27. R8C45 (step 11) = {39/57} (cannot be {48} which clashes with R7C56), no 4,8
27a. R8C67 (step 12) = {19/28/46} (cannot be {37} which clashes with R8C45), no 3,7
27b. Combined cage R7C56 + R8C45 = {3589/4579} = 59{38/47}, 5,9 locked for N8, clean-up: no 1 in R8C7

28. 45 rule on C123 1 innie R9C3 – 4 = 2 outies R27C4, min R27C4 = 3 -> min R9C3 = 7, max R27C4 = 5, no 5,6,7
28a. Min R9C3 = 7 -> max R9C45 = 7, no 7,8
[I missed 45 rule on C123 4 outies R279C4 + R9C5 = 10. It's interesting to note the difference between using the two 45s. Innies/outies limits R27C4 to {1234} leaving R9C45 as {1..6}. Outies limits R2C4 and R9C5 to {1234} leaving R79C4 as {1..6}. If both are used then R27C4 and R9C5 are limited to {1234} leaving just R9C4 as {1..6}.]

29. Killer triple 7,8,9 in R9C3 and 21(3) cage, locked for N7, clean-up: no 7 in R1C1 (step 22)

30. 9 in C3 locked in R89C3, locked for N7, clean-up: no 6 in R7C9 (step 25)

31. 21(3) cage in N7 (step 18) = {489/579/678}
31a. 9 of {489/579} must be in R8C3 -> no 4,5 in R8C3, clean-up: no 6,7 in R1C3 (step 24)

32. 45 rule on R789 2 outies R6C38 – 12 = 1 innie R7C1, min R6C38 = 13, no 1,2,3, no 4 in R6C8, max R6C38 = 17 -> max R7C1 = 5
32a. Max R7C1 = 5 -> min R56C1 = 11, no 1

33. 18(3) cage at R6C8 = {279/369/378/567} (cannot be {459} which clashes with R7C56, cannot be {468} which clashes with R7C2569), no 4
33a. R7C78 cannot be {67} (clashes with R7C29) -> no 5 in R6C8
33b. R7C78 cannot be {57} (clashes with R7C2569) -> no 6 in R6C8
33c. R7C78 = {27/36/37/56}

34. Hidden killer quad 6,7,8,9 in R7C2, R7C56, R7C78 and R7C9 -> no 6 in R7C3

35. 45 rule on N3 3 innies R1C7 + R3C79 = 18, max R3C9 = 5 -> min R13C7 = 13, no 1,2,3
35a. 17(3) cage in N3 cannot contain both 8,9 -> R13C7 must contain at least one of 8,9

36. 45 rule on N9 3 outies R6C8 + R89C6 = 21, max R6C8 = 9 -> min R89C6 = 12, no 1,2,3, clean-up: no 8,9 in R8C7
[I could have done step 44 here but didn’t see it until later.]

37. 1,2 in N8 locked in R7C4 and R9C45
37a. R9C45 cannot contain both 1 and 2 (can’t make cage sum) -> R7C4 = {12}
37b. R9C345 = 7{16}/8{24}/9{14}/9{23} (cannot be 7{34} which doesn’t contain 1/2)

38. 45 rule on N5 2 innies R4C4 + R6C6 = 7 = {16/25/34}, no 7,8,9

39. 17(3) cage at R3C4 = {368/458} (only remaining combinations) = 8{36/45}, no 1,2, 8 locked in R3C45 for R3 and N2, clean-up: no 4 in R1C7, no 4 in R4C2, no 5,6, in R6C6

40. 45 rule on N6 3 innies R4C79 + R6C8 – 16 = 1 outie R6C6, min R4C79 + R6C8 = 17, max R4C9 + R6C8 = 14 -> min R4C7 = 3

41. 45 rule on C6789 4 outies R2457C5 = 27 = {3789/4689/5679}, no 1,2, 9 locked for C5, clean-up: no 3 in R8C4

42. 45 rule on R12 2 outies R3C38 = 1 innie R2C1, min R3C38 = 3 -> min R2C1 = 3

43. 23(4) cage in N9 = {1589/1679/2489/2579/2678/3479/3578} (cannot be {3569/4568} which only contain one of 7,8,9) -> R8C89 = {15/16/24/25/26/34/35}, no 7,8,9; clean-up: no 1,2,3 in R1C8 (step 23)

44. R6C8 + R89C6 = 21 (step 36) = [768/786/867] (cannot be [948/984] which clash with R7C56), no 9 in R6C8, no 4 in R89C6, clean-up: no 6 in R8C7
44a. 6 locked in R89C6, locked for C6 and N8

45. 18(3) cage at R6C8 (step 33) = {378/567}, no 2
45a. R7C78 = {37/56}
[Mike added "You could have also noted here that 7 is locked in the 18(3) cage -> no 7 in R9C8" (IOU).]

46. R8C89 (step 43) = {15/16/25/26/34} (cannot be {24} which clashes with R8C7, cannot be {35} which clashes with R7C78 and with R8C45)

47. R9C345 (step 37b) = 8{24}/9{14}/9{23} -> no 7 in R9C3
47a. 7 in N7 locked in 21(3) cage = {579/678}, no 4

48. 10(3) cage in N7 (step 19) = {136/145/235}
48a. 45 rule on N7 3 innies R7C13 + R9C3 = 14 = {149/239/248} (cannot be {158} which clashes with 10(3) cage), no 5
48b. Max R7C34 = 6 -> no 4 in R6C3

49. 21(3) cage in N7 (step 18) = {579/678}
49a. 7 must be in R8C23 (R8C23 = {59} clashes with R8C45, R8C23 = {68} clashes with R8C6) -> no 7 in R7C2

50. 7 in N7 locked in R8C23, locked for R8, clean-up: no 5 in R8C23

51. R8C45 = [93], clean-up: no 4 in R1C4, no 4 in R7C56, no 5 in R8C2 (step 49)

52. R9C3 = 9 (hidden single in C3), R9C45 = 5 = {14} (only remaining combination)
52a. R7C9 = 9 (hidden single in R7)
52b. R1C8 = 9 (hidden single in C8), clean-up: no 3 in R1C6
52b. R9C3 = 9 -> R7C13 = 5 (step 48a) = {14/23}
52c. R1C8 = 9 -> R12C9 = 8 = [26/35/53/71]

53. Naked pair {58} in R7C56, locked for R7 and N8 -> R7C2 = 6, R8C6 = 6, R8C7 = 4, R9C6 = 7, R9C9 = 8, R2C56 = [79], clean-up: no 1 in R56C9

54. Naked pair {14} in R9C45, locked for R9 and N8 -> R7C4 = 2, clean-up: no 5 in R1C5, no 3 in R7C13 (step 52b)
54a. R67C3 = 9 = [54/81]

55. 23(4) cage in N9 R79C9 = [98] -> R8C89 = 6 = [15] -> R8C1 = 2, clean-up: no 3 in R12C9 (step 52c), no 1 in R34C9, no 8 in R45C8

56. Naked pair {24} in R34C9, locked for C9, clean-up: no 7 in R56C9

57. Naked pair {36} in R56C9, locked for C9 and N6 -> R12C9 = [71], R1C67 = [48], R2C4 = 3

58. 10(3) cage in N3 = {235} (only remaining combination), locked for N3 -> R3C7 = 6, R3C8 = 3, R34C9 = [42], R7C78 = [37], R6C8 = 8, R67C3 = [54], R7C1 = 1, clean-up: no 7 in R3C2, no 8,9 in R4C2, no 7 in R5C3

59. Naked pair {58} in R3C45, locked for R3 and N2 -> R34C2 = [93], R3C1 = 7, R9C12 = [35], R9C78 = [26], R2C78 = [52], R2C3 = 6, R3C3 = 1, R1C123 = [523], R2C2 = 8, R2C1 = 4, R8C23 = [78], R4C1 = 8, R45C3 = [72]

60. R3C45 = {58} -> R4C4 = 4

and the rest is naked singles and a cage sum

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

[Andrew]

Step 7 of Gary's outline was a nice shortcut.

I must admit that I didn't understand why 3 in r9c6 forces r9c3 = 9 until Gary explained it. For those who didn't understand it either, step 7 could be expanded to become

7.Consider placement of 3 in N8.In all cases,whether it is at r8c45 or r9c456 it must enforce r9c3=9 (if r9c6=3 then r8c6=9 from 4a). Also now 21/3 cage N7= {678} and r7c13=5

I was interested to note that Mike's breakthrough step 22 for A72V2 was remarkably similar, being what I thought of as the inverse of Gary's step 7. Gary used all remaining positions for 3 in N8 to fix 9 in r9c3. For A72V2, Mike effectively used all remaining positions for 1 in N7 to show that r8c9 cannot be 1 and therefore to fix r8c9.

Mike hadn't noticed the similarity until I pointed it out to him. I wonder if Gary spotted it; he commented on Mike's step 22 in the A72 thread.