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Maverick 4

 
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Nasenbaer
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PostPosted: Mon Mar 03, 2008 7:14 pm    Post subject: Maverick 4 Reply with quote

We're all still waiting to hear from Ruud. But, as I said before in Maverick 3, to prevent deprivation I give you a new one. Hopefully there is no shortcut this time. Twisted Evil

Maverick 4
(RS-rating: 1.9 - but I think that's too high, maybe a hard 1.5)



3x3:d:k:3072:3072:3072:6403:6403:6403:4102:4102:4102:5385:2058:5643:5643:6403:5902:5902:1296:3089:5385:2058:5643:5643:6403:5902:5902:1296:3089:5385:4636:4636:5150:5150:5150:4129:4129:3089:3620:3620:4636:5150:6440:5150:4129:5163:5163:4397:3620:5679:5679:6440:4914:4914:5163:5685:4397:4397:5679:5679:6440:4914:4914:5685:5685:4397:1856:4929:3906:6440:3906:3141:3910:5685:1856:4929:4929:3906:3906:3906:3141:3141:3910:

Have fun! Twisted Evil

Cheers,
Nasenbaer
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mhparker
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PostPosted: Mon Mar 03, 2008 7:45 pm    Post subject: Reply with quote

Nasenbaer wrote:
Hopefully there is no shortcut this time.

What, you mean there was one available last time?! Shocked

I guess that (as Einstein would have said) it's all relative...
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Afmob
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PostPosted: Wed Mar 05, 2008 5:56 pm    Post subject: Reply with quote

That was a demanding Killer. The moves might be of rating 1.5 but it took me really long to find them so I rate it higher than Nasenbaer suggested.

By the way, I think we can rest the Mavericks for a while because 2 in one week is more than enough since the gap between M1 and M2 was one month even though there is no Assassin this week.

M4 Walkthrough:

1. R789
a) 15(5) = {12345} locked for N8
b) Innies R89 = 22(3) = 9{58/67} -> 9 locked for R8
c) 15(2): R9C9 <> 6
d) Innies+Outies N7: -2 = R6C1 - R7C3 -> R6C1 <> 8,9; R7C3 <> 1,2
e) Innies+Outies N9: 4 = R6C9 - R7C7 -> R6C9 = (56789), R7C7 = (12345)

2. R456+N569
a) Killer quad (1234) locked in 20(5) + 25(4) for N5 since 20(5) has 3 of (1234)
b) Innies N69 = 5(2+1) = [212/131] -> R4C9 = (12), R6C7 = (13), R7C7 = (12); 1 locked for C7
c) Innies+Outies N9: 4 = R6C9 - R7C7 -> R6C9 = (56)
d) Innies N6 = 9(3) = 1{26/35} -> 1 locked for N6
e) 20(3) <> 6 because {569} blocked by R6C9 = (56)
f) 1 locked in 19(4) = 1{279/369/378} <> 5 since R67C7 <> 5,6,7
g) 12(3) @ N9: R9C8 <> 8,9 because R89C7 <> 1
h) 12(3) @ N6 <> 5 because R4C9 = (12) and R6C9 = (56) blocks {156}
i) 12(3) @ N9 <> 9 because {129} blocked by R7C7 = (12)

3. R123
a) Outies R1 = 8(2) <> 4,8,9
b) Outies = 9(2) = [72/81]

4. C123
a) Innies N4 = 13(3): R6C13 <> 6,7,8,9 because R4C1 >= 7
b) Innies+Outies N7 : -2 = R6C1 - R7C3 -> R7C3 <> 8,9

5. N146 !
a) ! Using Outies R123 = 9(2): Innies N46 = R6C1379 = 13(4) = 14{26/35} because R6C9 = (56)
-> 1,4 locked for R6; 4 locked for N4
-> R6C13 <> 5
b) 4 locked in Innies N4 = 4{18/27} <> 3

6. N347
a) Innies N47 = 15(2+1): R7C3 <> 4 since R4C1+R6C3 would be <= 10
b) Innies+Outies N7: -2 = R6C1 - R7C3 -> R6C1 <> 2; R7C3 <> 5,7
c) 12(3) <> 6 because 2 in R4C9 forces 6 in R6C9 (Innies N6 = [135/216])

7. N7+D/ !
a) Hidden Killer triple (789) in 17(4) since 19(3) can't have more than two of (789)
-> 17(4) <> {2456}
b) ! 19(3) <> 5,6 because 19(3) must have two of (789) since 17(4) can't have more than one of (789)
c) 17(4) <> {1367/2357} because R6C1 = (14) and R7C3 = (36) blocks {1367}
d) 17(4) = 1{259/268/349/457} since Killer pair (38) of 19(3) blocks <> 38{15/24}
-> 1 locked between C1+N7 -> R9C1 <> 1
e) 7(2): R8C2 <> 6
f) Consider canidates of R7C3 -> 17(4) <> 6,8:
- i) R7C3 = 3 -> 19(3) = 8{29/47} -> 17(4) <> {1268}
- ii) R7C3 = 6 -> 17(4) <> {1268}
g) 8 locked in 19(3) = 8{29/47} <> 3
h) 6 locked in R7C3+R9C1 for D/

8. R789
a) 6 locked in R678C9 for C9 -> 22(4) @ C9 = 6{178/259/349/358/457}; R7C8 <> 6
b) R8C37 <> 8 since together with R8C8 = (678) they would leave no combo for Innies R89 = 22(3)
c) 8 locked in 19(3) for R9
d) 15(2): R8C8 <> 7
e) Killer pair (68) locked in Innies R89 + R8C8 for R8
f) 19(3): R9C23 <> 2 because R8C3 <> 8,9
g) 12(3) <> {237} since it's a Killer triple of 22(4) @ N9

9. C123
a) 21(3) <> 5 because R8C1 = (579) blocks {579}
b) 21(3) = 8{49/67} -> 8 locked for C1
c) 12(3) <> {156/237} since they are Killer triples of 8(2)
d) 12(3) <> 6 because {246} blocked by Killer pair (46) of 21(3)

10. R789 !
a) Innies N9 = 18(4) <> {1359} since there is no combo with {359} for 22(4) @ N9
b) ! 12(3) <> 2 since {246} is a Killer triple of Innies N9
c) 2 locked in R7C789 for R7
d) 17(4) = 14{39/57} -> 4 locked between C1+N7 -> R9C1 <> 4
e) 7(2): R8C2 <> 3
f) Hidden Killer pair (89) in 22(4) @ N9 -> 22(4) <> {4567}
g) Innies N7 = 19(4) = {1369/1567/3457} because R7C3 = (36)
h) ! Hidden Killer triple (123) in R7C123 for R7 since R7C7 and R7C89 @ 22(4) must each have one of (123)
-> Innies N7 = 19(4) = 57{16/34} -> 5,7 locked for N7

11. N47+D/
a) 19(3) = {289} -> R8C3 = 2; 9 locked for R9
b) Innies N4 = 13(3) = {148} -> R4C1 = 8; 1 locked for R6+N4
c) Naked pair (36) locked in R7C3+R9C1 for N7+D/

12. C789
a) Innies N6 = 9(3) = [135] -> R4C9 = 1, R6C7 = 3, R6C9 = 5
b) 15(2) = [87] -> R9C9 = 7, R8C8 = 8
c) 22(4) = {2569} -> {269} locked for N9
d) 12(3) @ N9 = {345} -> R9C8 = 3; {45} locked for C7
e) 5(2) = {14} locked for C8+N3
f) 19(4) = {1369} -> R7C7 = 1; {69} locked for C6

13. C123+D/
a) 7(2) = [16] -> R9C1 = 6, R8C2 = 1
b) 21(3) = {489} -> {49} locked for C1+N1
c) 12(3) = {138} -> R1C1 = 3, R1C2 = 8, R1C3 = 1
d) 8(2) = {26} locked for C2+N1
e) 22(4) @ N1 = 57{19/28/46} -> R3C3 = 5, R2C3 = 7

14. Rest is singles without considering diagonals.

Rating: 1.75. One forcing chain, Hidden Killer Triples and some combo analysis were the most difficult moves I used which where hard to spot.


Last edited by Afmob on Wed Mar 12, 2008 10:25 am; edited 4 times in total
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mhparker
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PostPosted: Wed Mar 05, 2008 10:38 pm    Post subject: Reply with quote

Afmob wrote:
That was a demanding Killer. The moves might be of rating 1.5 but it took me really long to find them...

You're telling me! The first 11 moves in my WT (out of the 13 I've managed to find up to now) are all "45" tests(!), which is very unusual for me...

Afmob wrote:
By the way, I think we can rest the Mavericks for a while because 2 in one week is more than enough since the gap between M1 and M2 was one month even though there is no Assassin this week.

Yes, the gap was certainly a bit on the short side! It will be interesting to see whether we get another Assassin on Friday. Things certainly don't look promising at the moment, with no sign of Ruud on the horizon. I'm wondering, if it's just a case of being particularly busy with the day job, why he didn't post any announcement to that effect. Very strange.

As to the difficulty level, Mavericks don't need to always be so hard. Indeed, if no original Assassins appear for a while, we're going to need some in the 1.0 - 1.25 rating range.

In the meantime, I've still got the M4 to enjoy... Rolling Eyes
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Andrew
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PostPosted: Thu Mar 06, 2008 12:21 am    Post subject: Reply with quote

Afmob wrote:
That was a demanding Killer. The moves might be of rating 1.5 but it took me really long to find them...

mhparker wrote:
You're telling me! The first 11 moves in my WT (out of the 13 I've managed to find up to now) are all "45" tests(!), which is very unusual for me...

I'm still working at it, making some progress each session. So far 20(!) different "45" tests in 31 steps. Unusually for a Killer-X I haven't yet managed to use either diagonal.
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Nasenbaer
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PostPosted: Thu Mar 06, 2008 11:54 am    Post subject: Reply with quote

Andrew wrote:
I'm still working at it, making some progress each session. So far 20(!) different "45" tests in 31 steps. Unusually for a Killer-X I haven't yet managed to use either diagonal.

Sadly the diagonals are only there because of uniqueness this time, sorry for that.
Afmob wrote:
By the way, I think we can rest the Mavericks for a while because 2 in one week is more than enough since the gap between M1 and M2 was one month even though there is no Assassin this week.

As you wish. I only gave Maverick 4 to you because everybody seemed to hunger for a new challenge indicated by the amount of walkthroughs for Maverick 3. Wink

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Andrew
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PostPosted: Thu Mar 06, 2008 6:56 pm    Post subject: Reply with quote

Andrew wrote:
I'm still working at it, making some progress each session. So far 20(!) different "45" tests in 31 steps. Unusually for a Killer-X I haven't yet managed to use either diagonal.

Nasenbaer wrote:
Sadly the diagonals are only there because of uniqueness this time, sorry for that.

No need to apologise. Uniqueness is a good reason to make a Killer into a Killer-X.

It's good to know that I hadn't missed anything important on the diagonals. Smile

I've actually got 23 different "45" tests so far but the other 3 are just in my notes, at the end of my partial walkthrough, of things that I haven't yet been able to use for candidate eliminations.
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Andrew
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PostPosted: Fri Mar 07, 2008 7:42 am    Post subject: Reply with quote

Many thanks to Nasenbaer for the latest challenging puzzle. Very Happy

As Afmob said, it was a very demanding killer and the moves were very hard to find. On Tuesday and Wednesday I think I only found 6 or 7 moves each day but I never ground to a complete halt. Then today I found steps 33 and 34, after which it was fairly straightforward even though the first placement didn't come until step 43.

I didn't find this one quite as difficult as Maverick 1, which most people rated as a Hard 1.5, so I'll also rate Maverick 4 as a Hard 1.5. I didn't feel that my combination analysis justified a higher rating; it was fairly routine after the heavy analysis used for some recent puzzles.

I'll be interested to see how Afmob solved it. My walkthrough looks a lot longer.

Here is my walkthrough. Some of the key moves were steps 11a, which I added today, 26, 33, 34 and the simpler 36.

Nasenbaer wrote:
Sadly the diagonals are only there because of uniqueness this time, sorry for that.

Now that I've finished the puzzle I must disagree with that. The diagonals were very helpful from step 42 onward.

Thanks Afmob for the comments. I've corrected steps 17 and 26b and simplified step 41. I've also re-phrased step 10b and added a couple of clean-ups for step 43 which allowed me to simplify the remaining steps.

This is a Killer-X. I've included eliminations on the diagonals because it's easy to overlook them.

Prelims

a) R23C2 = {17/26/35}, no 4,8,9
b) R23C8 = {14/23}
c) R8C2 + R9C1 = {16/25/34}, no 7,8,9
d) R8C8 + R9C9 = {69/78}
e) R234C1 = {489/579/678}, no 1,2,3
f) 20(3) cage in N6 = {389/479/569/578}, no 1,2
g) 19(3) cage in N7 = {289/379/469/478/568}, no 1
h) 15(5) cage in N8 = {12345}, locked for N8

1. 45 rule on R1 2 outies R23C5 = 8 = {17/26/35}, no 4,8,9

2. 45 rule on R123 2 outies R4C19 = 9 = {45}/[63/72/81], no 9, no 6,7,8 in R4C9

3. 45 rule on R89 3 innies R8C159 = 22 = {589/679}, 9 locked for R8, clean-up: no 6 in R9C9

4. 45 rule on N3 2 innies R23C7 = 1 outie R4C9 + 12, min R23C7 = 13, no 1,2,3

5. 45 rule on N7 1 innie R7C3 = 1 outie R6C1 + 2, no 8,9 in R6C1, no 1,2 in R7C3

6. 45 rule on N9 1 outie R6C9 = 1 innie R7C7 + 4, no 1,2,3,4 in R6C9, no 6,7,8,9 in R7C7

7. 45 rule on C12 2 innies R49C2 = 1 outie R1C3 + 11
7a. Min R49C2 = 12, no 1,2
7b. Max R49C2 = 17 -> max R1C3 = 6

8. 45 rule on C89 2 innies R49C8 = 1 outie R1C7
8a. Max R49C8 = 9, no 9
8b. Min R49C8 = 3 -> min R1C7 = 3

9. 45 rule on N5 2 outies R78C5 = 2 innies R6C46
9a. Min R78C5 = 13 -> min R6C46 = 13, no 1,2,3

10. 45 rule on C789 4 innies R2367C7 = 17
10a. Min R23C7 = 13 (step 4) -> max R67C7 = 4 -> R67C7 = {12/13}, 1 locked for C7, clean-up: no 8,9 in R6C9 (step 6)
10b. Min R67C7 = 3 -> max R23C7 = 14 -> no 3,4,5 in R4C9 (step 4), clean-up: no 4,5,6 in R4C1 (step 2)
10c. Max R67C7 = 4 -> min R67C6 = 15, no 4,5

11. 45 rule on N6 3 innies R4C9 + R6C79 = 9 = {126/135} (cannot be {234} because no 2,3,4 in R6C9), no 7, 1 locked in R4C9 + R6C7 for N6, clean-up: no 3 in R7C7 (step 6)
11a. 2 of {126} must be in R4C9 (cannot be [126] which would give R7C7 = 2 (step 6) when R6C7 clashes with R7C7) -> no 2 in R6C7

12. 12(3) cage in N9 = {138/147/156/237/246/345} (cannot be {129} which clashes with R7C7), no 9
12a. 1 of {138} must be in R9C8 -> no 8 in R9C8
12b. 7 of {237} must be in R89C7 (cannot be {23}7 which clashes with R67C7) -> no 7 in R9C8

13. 45 rule on N58 4 outies R67C37 = 11
13a. R67C7 = 3,4 (steps 10b and 10c) -> R67C3 = 7,8, no 8,9, no 6,7 in R6C3, clean-up: no 6,7 in R6C1 (step 5)

14. 45 rule on N1 2 innies R23C3 = 1 outie R4C1 + 4, min R4C1 = 7 -> min R23C3 = 11, no 1

15. R234C9 = {129/138/147/237/246} (cannot be {156} which clashes with R6C9, cannot be {345} because R4C9 only contains 1,2), no 5

16. 20(3) cage in N6 = {389/479/578} (cannot be {569} which clashes with R6C9), no 6

17. 16(3) cage in N6 = {268/349/358/457} (cannot be {259/367} which clash with 20(3) cage and with R4C9 + R6C79)
17a. 6 of {268} must be in R4C78 (cannot be {28}6 which clashes with R4C19) -> no 6 in R5C7

18. R67C3 = 7,8 -> R67C4 = 14,15, no 4 in R6C4

19. R1C123 = {129/138/147/246/345} (cannot be {156/237} which clash with R23C2), if {246/345} R23C2 = {17} -> R1C123 + R23C2 must contain one of 7,8,9
19a. R234C1 = {489/579/678} -> R23C1 must contain one of 7,8,9
19b. Hidden killer triple 7,8,9 in R1C123 + R23C2, R23C1 and R23C3 for N1 -> R23C3 must contain one of 7,8,9
19c. R4C1 = {78} -> R23C3 = 11,12 (step 14), no 6 in R23C3

20. 45 rule on N9 4 innies R7C789 + R8C9 = 18 = {1269/1278/2349/2358/2457} (cannot be {1368/1467/2367} which clash with 15(2) cage, cannot be {1359/1458} which clash with 22(4) cage at R6C9 which would be 5{359}/5{458}, cannot be {3456} because R7C7 only contains 1,2), 2 locked for R7 and N9

21. 12(3) cage in N9 (step 12) = {138/147/156/345}
21a. 1 of {156} must be in R9C8 -> no 6 in R9C8

22. 45 rule on N7 4 innies R7C123 + R8C1 = 19 = {1369/1378/1468/1567/3457} (cannot be {1459} which clashes with 7(2) cage)
22a. 3 of {1369} must be in R7C12 (3 cannot be in R7C3 which would make 17(4) cage at R6C1 1{169} -> 6 of {1369} must be at R7C3 and 9 at R8C1 -> no 9 in R7C12

23. 19(3) cage in N7 = {289/379/469/478} (cannot be {568} which clashes with R7C123 + R8C1), no 5

24. R23C3 = 11,12 (step 14) -> R23C4 = 10,11
24a. 45 rule on C123 4 outies R2367C4 = 25 = {1789/2689/3589/3679/4579/4678}
24b. 5 of {3589/4579} must be in R6C4 (cannot make R23C4 = 10,11 including 5) -> no 5 in R23C4

25. 45 rule on N3 4 innies R23C79 = 24 = {1689/2589/2679/3579/3678/4569/4578} (cannot be {3489} which clashes with R23C8)
25a. 4 of {4569} must be in R23C9 because no {156} in R234C9 (step 15)
25b. 4 of {4578} must be in R23C9 (only way to make R234C9 total 12)
25c. -> no 4 in R23C7

26. R4C19 (step 2) = [72/81]
26a. 45 rule on N4 3 innies R4C1 + R6C13 = 13 = {148/157/238/247}
26b. Cannot be 7{15} which clashes with R4C9 + R6C79 = [216]
26c. Cannot be 8{23} which clashes with R4C9 + R6C79 = [135]
26d. -> R4C1 + R6C13 = {148/247}, no 3,5, 4 locked for R6 and N4, clean-up: no 5,7 in R7C3 (step 5)

27. R67C3 (step 13a) = 7,8 = [16/26/43], no 4 in R7C3, clean-up: no 2 in R6C1 (step 5)

28. R7C123 + R8C1 (step 22) = {1369/1567/3457} (cannot be {1378/1468} which would make 17(4) cage at R6C1 1{178}/4{148}), no 8
28a. 6 of {1567} must be in R7C3 -> no 6 in R7C12 + R8C1

29. 8 in N7 locked in 19(3) cage = {289/478}, no 3,6

30. R23C3 must contain one of 7,8,9 (step 19b)
30a. 45 rule on C123 4 innies R2367C3 = 19 = {1369/1468/1567/2368/2467/3457} (cannot be {1279/1378} which contain two of 7,8,9, cannot be {2458} because R7C3 only contains 3,6, cannot be {1459/2359} which arenít consistent with R67C3 = [16/26/43])
30b. 2 of {2368/2467} must be in R6C3 -> no 2 in R23C3

31. 45 rule on N1 4 innies R23C13 = 25 = {3589/4579/4678} (cannot be {3679} which clashes with R23C2)
31a. 9 of {3589/4579} must be in R23C1 -> no 9 in R23C3

32. R1C123 (step 19) = {129/138/147/246} (cannot be {345} which clashes with R23C13), no 5

33. R67C37 = 11 (step 13)
33a. R67C3 (step 27) = [16/26/43]
33b. If R67C3 = [26] => R67C7 = 3 = [12]
33c. If R67C3 = [43] => R6C1 = 1 (step 26d)
33d. -> 1 in R6 locked in R6C137, no 1 in R6C5

34. R67C3 (step 27) = [16/26/43]
34a. R7C123 + R8C1 (step 28) = {1369/1567/3457}
34b. 6 of {1369/1567} must be in R7C3 -> R7C37 cannot be [61] => R67C7 = [12]
34c. -> no 1 in R6C3

35. R2367C3 (step 30a) = {2368/2467/3457}
35a. 19(3) cage (step 29) = {289/478}
35b. 4 of {478} must be in R89C3 (because R89C3 = {78} would clash with R23C3) -> no 4 in R9C2

36. Killer pair {24} in R6C3 and R89C3, locked for C3

37. R2367C3 (step 35) = {2368/3457} (cannot be {2467} because 2,4,6 only in R67C3), 3 locked for C3
37a. R23C3 = {38/57}

38. R1C123 (step 32) = {129/138/147/246}
38a. R1C3 = {16} -> no 1,6 in R1C12

39. R23C3 = {38/57} (step 37a)
39a. R23C13 (step 31) = {3589/4579} (cannot be {4678} which clashes with R23C3), 9 locked in R23C1, locked for C1 and N1, 5 locked for N1, clean-up: no 3 in R23C2
39b. R23C1 = {49/59}, no 6,7,8

40. Hidden killer quad 1,3,5,6 in R1C3, R23C3, R45C3 and R7C3 for C3 -> R45C3 must contain one of 1,5,6
40a. 18(3) cage in N4 = {189/369/567} (cannot be {378} which clashes with R4C1)
40b. R45C3 cannot be {56} -> no 7 in R4C2

41. 17(4) cage at R6C1 = {1457} (only remaining combination), no 3, 5,7 locked for N7, clean-up: no 2 in 7(2) cage, no 4 in R89C3 (step 29)
[Step 41 simplified at Afmobís suggestion. I originally used the remaining combinations for R7C123 + R8C3.]

42. Killer pair 3,6 in R7C3 and 7(2) cage, locked for D/, clean-up: no 2 in R3C8

43. R6C3 = 4 (hidden single in C3), R7C3 = 3 (step 34), R6C1 = 1, R6C7 = 3, R7C7 = 1 (hidden single in C7, locked for D\), clean-up: no 7 in R3C2, no 8 in R23C3 (step 37a), no 4 in 7(2) cage in N7
43a. R9C1 = 6, R8C2 = 1, locked for D/, clean-up: no 7 in R2C2, no 4 in R3C8

44. R1C3 = 1 (hidden single in C3)
44a. Naked pair {26} in R23C2, locked for C2 and N1
44b. Naked pair {57} in R23C3, locked for C3, N1 and 22(4) cage
44c. Naked pair {49} in R23C1, locked for C1 and N1
44d. Naked pair {38} in R1C12, locked for R1
44e. Naked pair {57} in R78C1, locked for C1 and N7 -> R7C2 = 4, R4C1 = 8, R1C12 = [38], 3 locked for D\, R5C1 = 2, R9C2 = 9, R4C9 = 1 (step 2), R6C9 = 5 (step 11), R6C2 = 7, clean-up: no 6 in R8C8

45. Naked pair {89} in R6C48, locked for R6 -> R6C56 = [26]
, 6 locked for D\, R7C6 = 9 (cage sum), R23C2 = [26], R23C8 = [41], 4 locked for D/, R23C1 = [94], clean-up: no 6,7 in R2C5

46. Naked pair {78} in R8C8 + R9C9, locked for N9 and D\ -> R3C3 = 5, locked for D\, R5C5 = 9, locked for D/, R2C3 = 7, R45C3 = [96], R4C2 = 3 (step 40a), R5C2 = 5, R4C4 = 4, R6C4 = 8, locked for D/, R3C7 = 7, R1C9 = 2, R3C5 = 3, R2C5 = 5 (step 1)

and the rest is naked singles

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

In my previous message, I mentioned that I'd used 20 different "45" tests which are all included above. In addition I eventually looked at another 5 of them which never led to any candidate eliminations, although some produced combination eliminations.

This walkthrough was posted before I went through Afmob's walkthrough. Some comments on that and Mike's walkthrough are given in my later message including pointing out a couple of 45s that I missed, one of which I ought to have seen and the other was a less obvious one.

So, if my counting is correct, that's at least 27 45s in this puzzle! That must be some sort of record.


Last edited by Andrew on Fri Mar 21, 2008 2:09 am; edited 1 time in total
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mhparker
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PostPosted: Sun Mar 09, 2008 5:54 pm    Post subject: Reply with quote

Hi Afmob + Andrew,

I'm still doing this puzzle, so please don't convert your walkthroughs to normal text for the moment. I look forward to taking a look at them when I've finished (if I finish... Rolling Eyes ).
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Mike
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mhparker
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PostPosted: Tue Mar 11, 2008 9:39 pm    Post subject: Reply with quote

Hi folks,

Well, I finally did it, even if some of my moves were maybe a bit "OTT" (i.e., "over the top")! Very Happy

Thanks to Nasenbaer for an (ultimately!) rewarding challenge! Contrary to his statement above, I even managed to make some use of the diagonals!

Afmob and Andrew will probably have a good laugh at this one, when they see what techniques I was resorting to! I suspect I've missed something important, but at least my approach should make for some interesting reading if nothing else! I hope so.

As to the rating, it definitely deserves a 1.75 from the point-of-view of the techniques I was using. However, if they were indeed OTT, Andrew's rating of a "hard 1.5" may prove to be more accurate.

In the meantime, happy reading!

Edit: Many thanks to Afmob for pointing out two things I missed!

Quote:
M4 Walkthrough

Prelims:

a) 21(3) at R2C1 = {489/579/678} (no 1..3)
b) 8(2) at R2C2 = {17/26/35} (no 4,8,9)
c) 5(2) at R2C8 = {14/23}
d) 20(3) at R5C8 = {389/479/569/578} (no 1,2)
e) 7(2) at R8C2 = {16/25/34} (no 7..9)
f) 19(3) at R8C3 = {289/379/469/478/568} (no 1)
g) 15(5) at R8C4 = {12345}, locked for N8
h) 15(2) at R8C8 = {69/78}

1. Innies R89: R8C159 = 22(3) = {589/679} (no 1..4)
1a. 9 locked for R8
1b. cleanup: no 6 in R9C9

2. Outies R1: R23C5 = 8(2) = {17/26/35} (no 4,8,9)

3. Outies R123: R4C19 = 9(2) = [45/54/63/72/81]
3a. -> no 9; no 6..8 in R4C9

4. I/O diff. N7: R7C3 = R6C1 + 2
4a. -> no 1,2 in R7C3; no 8,9 in R6C1

5. I/O diff. N9: R6C9 = R7C7 + 4
5a. -> no 6..9 in R7C7; no 1..4 in R6C9

6. I/O diff. N3: R23C7 = R4C9 + 12
6a. -> min R23C7 = 13
6b. -> no 1..3 in R23C7

7. I/O diff. C12: R49C2 = R1C3 + 11
7a. min. R1C3 = 1 -> min. R49C2 = 12
7b. -> no 1,2 in R49C2
7c. max. R49C2 = 17 -> max. R1C3 = 6
7d. -> no 7..9 in R1C3

8. I/O diff. C89: R49C8 = R1C7
8a. min. R49C8 = 3 -> min. R1C7 = 3
8b. -> no 1,2 in R1C7
8c. max. R1C7 = 9 -> max. R49C8 = 9
8d. -> no 9 in R49C8

9. Innies N6: R4C9+R6C79 = 9(3) = {126/135} (no 4,7..9)
(Note: not {234}, because none of these digits in R6C9)
9a. 1 locked for N6
9b. {5/6} must go in R6C9
9c. -> no 5,6 in R4C9+R6C7
9d. cleanup: R7C7 = {12} (step 5); no 4,5 in R4C1 (step 3)

10. 12(3) at R8C7 = {138/147/156/237/246/345} (no 9)
(Note: {129} blocked by R7C7)

11. 20(3) at R5C8 = {389/479/578} (no 6)
(Note: {569} blocked by R6C9)

12. Innies N69: R4C9+R67C7 = 5(1+2) = [131/212]
12a. -> no 3 in R4C9; no 2 in R6C7
12b. R4C9+R6C79 (step 9) = [216/135]
12c. cleanup: no 6 in R4C1 (step 3)
Afmob pointed out that I could have added that 1 is locked in R67C7 for C7 here.

13. 12(3) at R2C9 = {129/138/147/237} (no 5,6)
(Note: not {345}, because none of these digits in R4C9;
{156} blocked by R6C9; {246} blocked by h9(3)n6 (step 12b))

14. I/O diff. N5: R78C5 = R6C46
14a. min. R78C5 = 13 -> min. R6C46 = 13
14b. -> no 1..3 in R6C46
(Note: could stretch the logic here to also eliminate 4 from R6C46, but that's done in next step)

15. Outies N5789: R6C1379 = 13(4) = {1246/1345} (no 7..9)
(Note: not {1237}, because none of these digits in R6C9)
15a. 1,4 locked for R6
15b. 4 locked in R6C13 for N4
15b. possible permutations h13(4): [4216/1435/4135]
(Note: [2416] blocked by R7C3 (step 4); [3415/4315] blocked by h9(3)n6 (step 12b))
15b. -> R6C1 = {14}; R6C3 = {124}
15c. cleanup: R7C3 = {36} (step 4)

16. 17(4) at R6C1 = {1259/1349/1457} (no 6,8) = {(7/9)..}
(Note: not {2357}, because none of these digits in R6C1;
{1358/1367/2456} blocked by R7C3 (step 4);
{1268} blocked because it would force R7C3 to 3 and thus leave no combos for 7(2)n7;
{2348} blocked because it would force R7C3 to 6 and thus leave no combos for 7(2)n7)
16a. no 1 in R9C1 (CPE)
16b. cleanup: no 6 in R8C2

17.Hidden killer pair on {79} within N7 as follows:
17a. 17(4) at R6C1 must have exactly 1 of {79} within N7 (step 16)
17b. -> 19(3) at R8C3 (only other place for {79} in N7) must also have exactly 1 of {79}
17c. -> {379/568} combos both blocked
17d. Furthermore, 8 of N7 locked in 19(3) = {8..}
17e. -> 19(3) at R8C3 = {289/478} (no 3,5,6)

18. 6 in N7 locked in D/ -> not elsewhere in D/
18a. 6 in C9 locked in R678C9
18b. -> no 6 in R7C8
18c. 22(4) at R6C9 = {6..} = {1678/2569/3469/3568/4567} (no eliminations yet)

19. 21(3) at R2C1 = {489/678} (no 5)
(Note: {579} blocked by R8C1)
19a. 8 locked for C1

20. Grouped AIC removes combos {138/147} from 12(3) at R8C7:
(6)r9c78=(6)r9c1,(1)r8c2-(1)r7c12=(1)r7c789
This can be explained in verbose form as follows:
20a. Either R9C78 contains a 6 or...
20b. ...R9C78 does not contain a 6
20c. => R9C1 = 6 (strong link, R9); R8C2 = 1 (neutral link, 7(2))
20d. -> R7C12 does not contain a 1 (weak link, N7)
20e. => R7C789 contains a 1 (strong link, R7)
20f. -> no 1 in 12(3) at R8C7
20g. Thus, if 12(3) at R8C7 does not contain a 6, it cannot contain a 1
20h. -> {138/147} combos blocked
20i. -> 12(3) at R8C7 (step 10) = {156/237/246/345} (no 8)

21. Distribution of {13} across N79:
21a. 7(2) at R8C2 cannot contain both of {13}
21b. -> R7C123 (only other place for {13} in N7) must contain at least 1 of {13}
21c. 12(3) at R8C7 cannot contain both of {13}
21d. -> R7C789 (only other place for {13} in N9) must contain at least 1 of {13}
21e. from steps 21b and 21d, R7C123 and R7C789 must each contain exactly 1 of {13}
21f. -> 7(2) at R8C2 and 12(3) at R8C7 must also each contain exactly 1 of {13}
21g. -> 7(2) at R8C2 = {16/34} (no 2,5) = {(3/6)..}
21h. 12(3) at R8C7 (step 20i) = {(1/3)..} = {156/237/345} (no eliminations)

22. 7(2) at R8C2 (step 21g) and R7C3 form killer pair on {36}
22a. -> no 3 elsewhere in N7 and D/
22b. cleanup: no 2 in R3C8

23. Grouped Turbot Fish removes 1 from R7C89:
(1)r789c7=(1)r6c7-(1)r6c1=(1)r7c12
This can be explained in verbose form as follows:
23a. Either R789C7 contains a 1 or...
23b. ...R789C7 does not contain a 1
23c. => R6C7 = 1 (strong link, C7)
23d. -> R6C1 <> 1 (weak link, R6)
23e. => R7C12 = 1 (strong link, 17(4))
23f. Thus, either R789C7 or R7C12 (or both) must contain a 1
23g. -> no 1 in R7C89 (common peers)

24. 1 now unavailable to 22(4) at R6C9 (step 18c) = {2569/3469/3568/4567}
24a. Hidden killer pair on {89} blocks {4567} combo, as follows:
24b. Only places for {89} in N9 are 15(2) and 22(4), neither of which can contain both
24c. -> each of 15(2) at R8C8 and 22(4) at R6C9 must contain exactly 1 of {89}
24d. -> {4567} combo blocked
24e. -> 22(4) at R6C9 = {2569/3469/3568} (no 7) = {(2/3}..}
Afmob pointed out here that step 23 was not really necessary, because even without it, 22(4) at R6C9 = {1678/2569/3469/3568} = {(2/3/7)..} (within N9), thus also blocking the {237} combo for 12(3) in the next step and thus likewise breaking the puzzle.

25. 12(3) at R8C7 (step 21h) = {156/345} (no 2,7)
(Note: {237} blocked by 22(4) at R6C9 (step 24e))
25a. 5 locked for N9

26. 7 in N9 locked in 15(2) at R8C8 = {78} (no 6,9), locked for N9 and D\
26a. cleanup: no 1 in R3C2

27. 9 in R9 locked in R9C23 for N7
27a. 19(3) at R8C3 = {9..} = {289} (no 4,7) (last combo)
27b. 2 locked in R89C3 for C3 and N7

28. Hidden single (HS) in R9 at R9C9 = 7
28a. -> R8C38 = [28]

29. Naked pair (NP) at R6C13 = {14}, locked for R6 and N4
29a. -> R6C7 = 3
29b. -> R46C9 (step 12b) = [15]
29c. -> R7C7 = 1 (step 5); R4C1 = 8 (step 3, but also obtainable via cage split of N4 innies)

30. 6 in C9 locked in N9 -> not elsewhere in N9
30a. 12(3) at R8C7 = {345} (last combo)
30b. -> R9C8 = 3; R89C7 = {45}, locked for C7 and N9
30c. cleanup: no 2 in R2C8

31. HS in R9 at R9C1 = 6
31a. -> R8C2 = 1; R7C3 = 3
31b. -> R6C1 = 1 (step 4)
31c. -> R6C3 = 4
31d. cleanup: no 7 in R23C1 (step 19)

32. R23C8 = [41]
32a. -> R23C1 = [94]

33. 12(3) at R1C1 = [381/561]
33a. -> R1C1 = {35}; R1C2 = {68}; R1C3 = 1
33b. cleanup: no 7 in R3C2

Only singles and simple cage sums left now.

_________________
Cheers,
Mike


Last edited by mhparker on Wed Mar 12, 2008 9:36 pm; edited 1 time in total
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Afmob
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PostPosted: Wed Mar 12, 2008 3:11 pm    Post subject: Reply with quote

There is nothing wrong with different solving approaches. Better than reading the same wt thrice. It just shows that this Killer is well crafted, so thanks again Nasenbaer!

Though we all used different moves it seems that the key areas to crack this puzzle are N7 and N9 by either combo analyzing their Innies or using chains in that region.
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Andrew
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PostPosted: Fri Mar 21, 2008 2:03 am    Post subject: Reply with quote

Mike wrote:
Afmob and Andrew will probably have a good laugh at this one, when they see what techniques I was resorting to! I suspect I've missed something important, but at least my approach should make for some interesting reading if nothing else! I hope so.

Good to see that you posted your walkthrough, Mike. Smile As Afmob said we had three different solving paths.

Some interesting steps although it did at times seem to have been hard work. Step 21 looks as if it ought to contain the term "hidden killer pair" in some sub-steps.

Having gone through all three walkthroughs again today, I wondered whether I was right with my rating but then remembered the reason I gave with my walkthrough; I found Maverick 1 harder to solve than this one, particularly in how long they took.

Afmob and Mike both found a couple of important 45s that I missed. They both had innies for N69 which was more direct than how I got that result; that's a 45 that I shouldn't have missed. They also both got the less obvious R6C1379. I particularly liked the way Afmob used N46 to get that. Another good move that only Afmob had was the killer quad in N5 (step 2a).

My walkthrough seems to have made more use of sets of 4 innies and I think I was the only one of us three to use 4 outies for N58.

Thanks again Nasenbaer! A challenging puzzle but at the same time allowing plenty of scope for different solving paths.

Afmob wrote:
There is nothing wrong with different solving approaches. Better than reading the same wt thrice.

We'll all agree with that. I certainly wouldn't want the same walkthrough thrice. However I think it's acceptable to have two very similar walkthroughs if they show different logic or thought patterns to achieve the similar results.
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