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PostPosted: Sat Nov 17, 2012 1:10 pm 
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Grand Master
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Location: Saudi Arabia
Nice Groups 1 X Single Repeat

Each twin killer cage (solid line) contains a single repeat i.e. AABC. Note JSudoku solves it without this assumption. The 30(4) and 26(4) cages are normal.

It is X.

There are Nonets plus the set of extra groups.

The set of extra groups give some interesting Law of Leftovers. I'll do some more with this format so if you use JSudoku save it. Note when you are inserting extra groups the seventh one is grey and it is visually easier if this is the centre cross.

I'd guess it is about 1.0 - but then I'm used to repeat cages.

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PostPosted: Sun Jan 08, 2017 3:24 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
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Location: Lethbridge, Alberta, Canada
Another one from a long time ago, actually just over 4 years rather than almost 5. ;)

An enjoyable puzzle. Thanks HATMAN for clarifying the rules and for pointing out that I'd missed out a possible repeat combination when I sent you my Prelims.

I took a few tries at this but that was due to my carelessness, not because of any difficulty.

Here is my walkthrough:
This is a jigsaw+regular killer, single repeat in cages except for the 30(4) and 26(4) cages. I’ve identified the jigsaw groups by their upper left cells, for example JR1C1.

Prelims

a) 10(4) repeat cage at R1C6 = {1126/1135/2215}, no 4,7,8,9
b) 30(4) cage at R1C8 = {6789}
c) 11(4) repeat cage at R3C8 = {1127/1136/1145/2234/3314/4412}, no 8,9
d) 11(4) repeat cage at R6C8 = {1127/1136/1145/2234/3314/4412}, no 8,9
e) 26(4) cage at R8C8 = {2789/3689/4589/4679/5678}, no 1

1. 30(4) cage at R1C8 = {6789}, locked for N3
1a. 4 in N3 only in R3C789, locked for R3

2. The repeated numbers in cages must be diagonally opposite and in different jigsaw groups; this immediately identified the repeated number cells for 10(4) repeat cage at R1C6 as R1C6 & R2C7, 15(4) repeat cage at R3C1 as R3C2 & R4C1, 11(4) repeat cage at R6C8 as R6C9 & R7C8 and 16(4) repeat cage at R8C3 as R8C3 & R9C4.
2a. 10(4) repeat cage at R1C6 = {1126/1135/2215}, repeated numbers in R1C6 & R2C7 -> R1C6 & R2C7 = {12}
2b. 10(4) cage must contain 1 in R1C67 + R2C6, locked for JR1C6
2c. 11(4) repeat cage at R3C8 cannot contain more than one 1 = {2234/3314/4412}, no 5,6,7
2d. 1 in C9 only in R67C9 -> 11(4) repeat cage at R6C8 = {1127/1136/1145/3314/4412}, repeated numbers in R6C9 & R7C8 -> R6C9 & R7C8 = {134}
2e. 45 rule on C89 2 innies R5C89 = 12 = {39/48/57}, no 1,2,6
2f. 6 in JR1C6 only in R1C89 + R2C69, CPE no 6 in R2C8
2g. 30(4) cage at R1C8 = {6789}, 6 locked for JR1C6
2h. 10(4) repeat cage at R1C6 = {1135/2215}, 5 locked for JR1C6, clean-up: no 7 in R5C8
2i. Law of Leftovers (LoL) for R1234 2 innies R34C5 must exactly equal two outies R5C89 -> R34C5 = {39/48/57}, no 1,2,6, no 8 in R4C5
2j. Double hidden killer triple 7,8,9 in 30(4) cage at R1C8, R5C89 and 26(4) cage at R8C8 for C89, 30(4) cage contains all of 7,8,9, R5C89 contain one of 7,8,9 -> 26(4) cage can only contain two of 7,8,9 = {3689/4589/4679/5678}, no 2
2k. Double killer triple 7,8,9 in 30(4) cage at R1C8, R5C89 and 26(4) cage at R8C8 for C89, locked for C89
2l. Hidden killer pair 1,2 in 11(4) repeat cage at R3C8 and 11(4) repeat cage at R6C8 for C8, 11(4) repeat cage at R3C8 only contains one of 1,2 in C8 -> 11(4) repeat cage at R6C8 must contain one of 1,2 in C8 plus 1 in C9 = {1136/1145} (cannot be {4412} which clashes with 11(4) cage at R3C8)
2m. 11(4) repeat cage at R6C8 = {1136/1145}, no 2 -> R6C9 = 1, placed for JR6C8, R7C8 = 1, placed for JR6C6
2n. 11(4) repeat cage at R3C8 = {2234}, 2 locked for R34, JR1C6 and N6
2o. R1C6 = 1, R2C7 = 1, R1C7 + R2C6 = {35}, locked for JR1C6, clean-up: no 9 in R5C8 (step 2e)
2p. R34C9 = {24}, locked for C9, R34C8 = {23}, locked for C8 and JR2C7, clean-up: no 8 in R5C8, no 9 in R5C9 (both step 2e), no 5 in R6C8, no 6 in R7C9 (both step 2m), no 3,9 in R34C5 (step 2i)

3. 9 in JR1C6 only in R1C89 + R2C9, locked for N3
3a. 3,5 in C9 only in R789C9, locked for N9
3b. 2 in JR6C8 only in R9C567, locked for R9
3c. R5C8 = 5 (hidden single in C8), placed for JR2C7, R5C9 = 7 (step 2e), placed for JR1C6, R3C7 = 4, placed for D/, R34C9 = [24], R34C8 = [32], clean-up: no 8 in R3C5 (step 2i)
3d. R6C8 = 6, R7C9 = 3 (cage sum), both placed for JR6C8
3e. Naked pair {57} in R34C5, locked for C5
3f. R1C7 = 5, R2C6 = 3
3g. R2C8 = 7 (hidden single in N3), placed for JR2C7 and D/
3h. Naked triple {389} in R456C7, locked for C7
3j. Naked triple {689} in R34C6 + R4C7 for JR2C7, 6 locked for C6
3l. R5C5 = 1 (hidden single in C5), placed for both diagonals
3m. LoL for C1234 2 outies R12C5 must exactly equal 2 innies R5C34, no 3 in R12C5 -> no 3 in R5C34

4. 17(4) repeat cage at R8C6 = {2267} (only possible combination, cannot be {2249/2258} because 4,5,8,9 only in R89C6) -> R8C7 = 6, R8C6 & R9C7 = 2, R9C6 = 7, R7C7 = 7, placed for D\
4a. 4 in JR6C8 only in R9C58, locked for R9
4b. R4C5 = 7 (hidden single in N5), R3C5 = 5

5. 2 on D\ only in R1C1 + R2C2, locked for N1
5a. 2 on D/ only in R6C6 + R7C7, locked for JR5C2

6. R8C3 = 1 (hidden single in JR5C2), R9C4 = 1 (hidden single in C4), R8C4 + R9C3 = 14 = [59/86/95]

7. 2,4,7 in JR5C1 only in R5678C1, locked for C1
7a. R2C2 = 2 (hidden single in N1)
7b. R3C2 = 1 (hidden single in JR2C2), R4C1 = 1 (hidden single in C1) -> R3C1 + R4C2 = 13 = [85], 8 placed for JR1C1
7c. R3C4 = 7 (hidden single in R3)

8. R2C3 = 5 (hidden single in JR2C2), clean-up: no 9 in R8C4 (step 6)
8a. Naked pair {69} in R39C3, locked for C3
8b. R2C3 = 5 -> 22(4) repeat cage at R1C3 = {4459} -> R1C3 & R2C4 = 4, R1C4 = 9, placed for JR1C1
8c. R2C1 = 6, R3C3 = 9, placed for D\
8d. R4C4 = 6 (hidden single on D\)
8e. R9C3 = 6 -> R8C4 = 8 (cage sum), placed for JR5C1, R8C8 = 4, placed for D\
8f. R57C4 = [25], R6C4 = 3, placed for JR5C2 and D/

and the rest is naked singles, without using the diagonals or remaining repeat cage property.

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

It looks like there are some more killer variants currently on page 3 of this forum, so I'll probably try them.


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PostPosted: Sun Jan 15, 2017 7:26 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
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Location: Saudi Arabia
Nice Solution Andrew. I mostly used combination elimination between cages which was not very satisfactory in this one.

On notation: in your solution for the repeat cells you write r8c6 + r9c7 = 2, I slightly prefer r8c6 & r9c7 = 2.

Does anyone have any views on this?

Maurice


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PostPosted: Tue Jan 17, 2017 5:51 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
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Location: Lethbridge, Alberta, Canada
Thanks Maurice. :D

I've changed my WT as you've suggested. I'll try to remember to use this notation for future puzzles with cage repeats, including the Decidoku which you posted earlier this week.

Andrew


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