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HATMAN
PostPosted: Tue Mar 03, 2015 2:19 pm 
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 694
Location: Saudi Arabia
Snakey X Ordered Digital Pairs 5

The cages are ordered increasing top left to bottom right (row first).
The cages are digital - the last digit of the sum is in one of the cells.
The two cells are consecutive - the third one is non-consecutive so obviously NC C C or C C NC
R1c9 and r4c8 are both in two cages.

Standard snakes conflict with X - so I had to bend them a bit.

This one is harder - I would like your views.



Image

Solution:
381479652
759621483
624583917
816754239
235968174
497132568
963217845
172845396
548396721
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wellbeback
PostPosted: Wed Mar 04, 2015 5:40 am 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 282
Location: California, out of London
Thanks HATMAN. You are getting more ingenious all the time! This was harder than the previous one. Although not for the moves being particularly difficult in themselves. Since I do all these puzzles using paper and pencil I find it can take quite some time to actually spot the moves. Especially for the blank part of the puzzle after the cages were solved.

BTW - thanks for all the previous puzzles as well. (And to Ed too!). I do work all the puzzles, but continue to be quite busy and have to be selective in writing WTs.

(Corrections thanks to Andrew)
Hidden Text:
Given the rules the possible cage sequences are:

128 129
237 238
346 347

189 289
278 378
367 467

1. Overlapping cages in n3 and n6 can only be:
237 & 289, or
346 & 378

Since r1c9 and r4c8 are both in Yellow - they must be (23)
So the possible layouts for the cages in n3 and n6 are: (Ordered by rows) (r1, etc.)

a)
__2
_83
9_7
_3_
_74
_68

and

b)
__3
_74
8_6
_2_
_38
_97

The second of those requires for the remaining Yellow cells: 4 in r8c8 and {15} in r79c9.
But this leaves no value possible for r4c9.

An alternative and much simpler way I realized later is that the values in Yellow cells r46c8 must go in c9 in r28c9.
And the latter set of values does not have that.

-> The first set of values above must be the cages in n36

-> HS 9 in n4 -> r4c9 = 9
-> HS 9 in Yellow -> r8c8 = 9
-> NP (15) in Yellow -> r79c9 = {15}
-> NS 6 in c9 -> r8c9 = 6

Also NT r456c7 = {125}
-> HP r13c8 = {15}
-> NP r79c8 = {24}
-> NP r12c7 = {46}
-> NT r789c7 = {378}

2. NS r5c5 = 6
-> r3c6 = 3 and r4c6 = 4
Also HS 3 in Red -> r1c1 = 3

3. Number 6
6 in Green in r9c1 or r9c6
6 in Red in r2c46
-> r12c7 = [64]
-> 6 in r3 in r3c12
-> 6 in r4 in r4c13
-> HS 6 in Blue -> r3c1 = 6
-> HS 6 in r4 -> r4c3 = 6
-> HS 6 in c2 -> r7c2 = 6
-> HS 6 in Green -> r9c6 = 6
-> HS 6 in Red -> r2c4 = 6

4. Number 4
HS 4 in D\ -> r3c3 = 4
-> HS 4 in r1 -> r1c4 = 4
-> HS 4 in n4 -> r6c1 = 4
-> HS 4 in n7 -> r9c2 = 4
-> r79c8 = [42]
-> r8c5 = 4

5. Number 8
Unknowns in D/ = {1357} with r6c4 from (13)
-> Only possibilities for cage r678c4 are [128] or [378]
-> r8c4 = 8
-> HS 8 in D\ -> r7c7 = 8
-> HS 8 in n7/r9 -> r9c3 = 8
-> 8 in n1 in r13c2
-> HS 8 in c1/n4 -> r4c1 = 8
-> HS 8 in r5/n5 -> r5c6 = 8
-> HS 8 in c5/n2 -> r3c5 = 8
-> HS 8 in r1/n1 -> r1c2 = 8

6. Number 5
Whichever of (37) is in r9c7 must go in Green in r8c2
-> 5 in D/ in r7c3 or r9c1
-> HS 5 in r8 -> r8c6 = 5

7. Number 1
-> HS 1 in r8 -> r8c1 = 1
-> HS 1 in D/ -> r6c4 = 1

8. Number 2
-> r7c4 = 2
-> HS 2 in D\ -> r6c6 = 2
-> HS 2 in c5/n2 -> r2c5 = 2
-> HS 2 in r3/n1 -> r3c2 = 2
-> HS 2 in Blue/n4/c1 -> r5c1 = 2
-> HS 2 in r4/n6 -> r4c7 = 2

9...
-> r56c7 = [15]
-> HS 1 in r4/n4 -> r4c2 = 1
Also HS 9 in Blue -> r6c2 = 9
-> HS 9 in r5/n5 -> r5c4 = 9
-> HS 3 in c4 -> r9c4 = 3
-> r89c7 = [37]
-> r7c3 = 3, r8c2 = 7, r9c1 = 5
etc.
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Andrew
PostPosted: Sun Mar 08, 2015 12:13 am 
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Joined: Wed Apr 23, 2008 6:04 pm
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Thanks HATMAN for your latest ingenious killer! Definitely harder than your previous snake puzzle.

wellbeback commented on the difficulty in solving this puzzle using paper and pencil, but maybe it's the best way to do it when there are different types of constraints which overlap. He made better use of the snakes in the early stages than I did. I have two diagrams, side by side, one for the cages and the other for the snakes; that makes it harder to focus on what's in both cages and snakes.

Here's my walkthrough:
Cages are ordered increasing from top to bottom and left to right.
The cages are digital, the last digit of the sum is in one of the cells.
Two cells have consecutive values, with the other non-consecutive so either C C NC or NC C C
R1C1 and R1C9 are each in two cages.

1. Each cage must contain the last digit of the cage sum, with the other two digits totalling 10, so cage sums must be in the range 11 to 19.

11 = [128]
12 = [129/237]
13 = [238/346]
14 = [347]
No possible permutations for 15, since [456] would have three consecutive values
16 = [367]
17 = [278/467]
18 = [189/378]
19 = [289]

2. Since there are fewer cages these days, I’ll start by limiting values in each of the cages
2a. All cages must start with 1,2,3 or 4 -> R1C9, R3C6, R4C8 and R6C4 must all be {1234}
2b. Second cells of cages cannot be 1,5 or 9 -> no 1,5,9 in R2C89, R4C6, R5C89 and R7C4
2c. Last cells of cages can only be one of 6,7,8 or 9 -> R3C79, R5C5, R6C89 and R8C4 must all be {6789}
2d. 5 on D/ only in R7C3 + R8C2 + R9C1, locked for N7

3a. The two cages at R1C9 share the same cell and are in the same nonet so must contain five different values. Only possibilities are [237/289] and [346/378] -> R1C9 = {23}, no 2,6 in R2C89, 7,8 locked in the two cages for N3
3b. Similarly for the two cages at R4C9; they can only be [237/289] and [346/378] -> R4C8 = {23}, no 2,6 in R5C89, 7,8 locked in the two cages for N6
3c. Naked pair {23} in R1C9 and R4C8, locked for right-hand snake
3d. Naked pair {23} in R1C9 and R4C8, CPE no 2,3 in R13C8 + R4C1, also no 2,3 in R4C6 using D/
3e. Diagonal cage at R1C9 cannot be [237] -> no 7 in R3C7
3f. Vertical cage at R1C9 cannot be [289] -> no 8 in R2C9, no 9 in R3C9
3g. 1,5 in right-hand snake only in R79C9 + R8C8, locked for N9

4a. Vertical cage at R1C9 cannot be [378] because the cage R4C8 + R5C9 + R6C8 cannot contain 2 in R4C8 and 4 in R5C9
4b. Diagonal cage at R1C9 cannot be [346]
4b. Vertical cage at R1C9 = [237/346], 3 locked for C9 and N3, R2C9 = {34}, R3C9 = {67}
4c. Diagonal cage at R1C9 = [289/378], 8 locked for D/ and top snake, R2C8 = {78}, R3C7 = {89}

5. Looking at the interactions between the cages at R1C9 and the cages at R4C8
5a. For vertical cage at R1C9 = [237] => cage R45C8 + R6C9 = [378], cage R4C8 + R5C9 + R6C8 = [346]
For vertical cage at R1C9 = [346] => cage R45C8 + R6C9 = [237], cage R4C8 + R5C9 + R6C8 = [289]
-> R45C8 + R6C9 = [237/378], 3 locked for C8 and N6, R5C8 = {37}, R6C9 = {78}
R4C8 + R5C9 + R6C8 = [289/346] -> R5C9 = {48}, R6C8 = {69}
5b. 8 in N6 only in R56C9, locked for C9
5c. 7 in C7 only in R789C7, locked for N9
5d. Vertical cage at R1C9 = [237] => cage R45C8 + R6C9 = [378], cage R4C8 + R5C9 + R6C8 = [346] or vertical cage at R1C9 = [346] -> 4 in R25C9, locked for C9
5e. Vertical cage at R1C9 = [237] => cage R4C8 + R5C9 + R6C8 = [346] or vertical cage at R1C9 = [346] -> 6 in R3C9 + R6C8, locked for right-hand snake
5f. 6 in R3C9 + R6C8, CPE no 6 in R13C8 + R4C1
5g. Vertical cage at R1C9 = [237] => diagonal cage at R1C9 = [289] or vertical cage at R1C9 = [346] => R5C9 = 8 -> 8 in R2C8 + R5C9, locked for right-hand snake

6. Cage at R3C6 = [346/347/467] -> R3C6 = {34}, R4C6 = {46}, R5C5 = {67}, 4 in R34C6, locked for C6

7. Vertical cage at R1C9 = [237] => cage R45C8 + R6C9 = [378] or vertical cage at R1C9 = [346] => diagonal cage at R1C9 = [378], locked for D/ -> R5C5 = 6, placed for both diagonals, R4C6 = 4, placed for D/, R3C6 = 3
7a. 6 in top snake only in R2C46, locked for R2 and N2

[Please feel free to skip step 8 and go directly to step 9, when step 9e will be only place for 3 in top snake.]
8. 3 in top snake only in R1C19 + R2C2
8a. 3 in R1C9 or 3 in R2C9 => 3 in top snake only in R1C1 -> 3 in R1C19, locked for R1
8b. 3 in R1C19, locked for top snake, no 3 in R2C2
8c. 3 in R1C19, CPE no 3 in R9C1 using the diagonals

[I was a bit slow to spot …]
9. Naked triple {159} in R479C9, locked for C9
9a. 9 in C9 only in R4C9 => R6C8 = 6 or 9 in R79C9, locked for right-hand snake => R6C8 = 6 -> R6C8 = 6, placed for right-hand snake, R3C9 = 7, R2C8 = 8, R3C7 = 9, placed for D/ and top snake, R6C9 = 8, R5C9 = 4, placed for right-hand snake, R2C9 = 3, R1C9 = 2, placed for D/, top snake and right-hand snake, R45C8 = [37], R8C9 = 6
9b. Naked triple {125} in R456C7, locked for C7 and N6 -> R12C7 = [64], R4C9 = 9
9c. Naked pair {15} in R79C9, locked for N9 -> R8C8 = 9, placed for D\ and bottom snake
9d. R1C1 = 3 (hidden single in R1), placed for D\
9e. 7 on D/ only in R7C3 + R8C2 + R9C1, locked for N7

10. R3C3 = 4 (hidden single on D\), locked for top snake and left-hand snake
10a. R1C4 = 4 (hidden single in R1)
10b. R6C1 = 4 (hidden single in N4)

11. Cage at R6C4 = [128/378] -> R7C4 = {27}, R8C4 = 8, placed for bottom snake
11a. R8C5 = 4 (hidden single in R8) -> R9C2 = 4 (hidden single in R9), R79C8 = [42]

12. R7C7 = 8 (hidden single on D\) -> R9C3 = 8 (hidden single in R9), placed for left-hand snake -> R4C1 = 8 (hidden single in C1), R5C6 = 8 (hidden single in R5), R3C5 = 8 (hidden single in N2), R1C2 = 8 (hidden single in N1)

13. 6 in left-hand snake only in R3C1 + R4C2, CPE no 6 in R3C2
13a. R3C1 = 6 (hidden single in R3), locked for left-hand snake -> R4C3 = 6 (hidden single in R4), R7C2 = 6 (hidden single in N7)
13b. R7C1 = 9 (hidden single in N7)
13c. R6C2 = 9 (hidden single in left-hand snake)

14. R8C3 = 2 (hidden single in bottom snake), R8C1 = 1

15. R6C4 = 1 (hidden single on D/) -> cage at R6C4 (step 11) = [128] -> R7C4 = 2, R3C4 = 5, R4C4 = 7, placed for D\, R2C4 = 6
15a. Naked pair {17} in R1C5 + R2C6, locked for N2 and top snake

and the rest is naked singles, without using cages, diagonals or snakes.
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