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 Post subject: Max Repeat NC FNC LS 4
PostPosted: Sun Feb 19, 2017 11:59 am 
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Grand Master
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Maximum Repeat Non-Consecutive Fers NC Latin Square 4

Please note the red "2" in r5c5 is a negative clue, R5C5 is not 2, R5C5 = 1/3/4/5/6/7/8/9.
(I've rebuilt the puzzle in Excel - which do you think is better?)

The cages are maximum repeat (i.e. they contain no unrepeated numbers).
It is a Latin Square so no nonets.
It is Non-consecutive and Fers Non-consecutive so adjacent squares horizontally, vertically and diagonally are not consecutive.

Image

Image

Solution:
168359247
936815724
691483572
247168359
724936815
572691483
359247168
815724936
483572691


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PostPosted: Sat Mar 04, 2017 7:45 pm 
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Grand Master
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Location: California, out of London
Thanks HATMAN. Not too difficult. Here is my walk through. (At least the start of it...)

MaxRepeat NC FNC LS 4:
1. 15(6) is Max Repeat. Cannot be {AAAABB} or {AABBCC} since both sum to an even number. -> must be {AAABBB}.
Since also NC -> 15(6) = {111444}.
-> Each 15(6) from [114414] or [441141]
-> r27c5 = {14}

2. 29(5) is Max Repeat -> must be {AAABB} with 'A' odd -> must be from {99911} {77744} {55577}

3. r3c5, r4c5, r6c5 all from (6789)
Trying r5c5 also from (6789)
puts r12789c5 = [241{35}]
which leaves no place for 2 in c4.
-> r5c5 from (35)
-> r4c4 = r6c6 = 1
-> Both 15(6)s are [114414]

4. -> 2 in c5 in r89c5
-> HS 3 in c4 -> r1c4 = 3
-> HS 3 in c5 -> r5c5 = 3
-> HS 5 in c5 -> r1c5 = 5
Also HS 2 in c6 -> r9c6 = 2
-> HS 2 in c5 -> r8c5 = 2
-> HS 2 in c4 -> r7c4 = 2
-> HS 2 in C3 -> r6c3 = 2
Also HS 2 in r4 -> r4c1 = 2
-> HS 2 in r5 -> r5c2 = 2

5. 3 in c6 in r234c6.
-> HS 2 in c7 -> r1c7 = 2
-> HS 2 in r3 -> r3c9 = 2
-> r2c8 = 2

6. 4 in r12. At least one 4 in r1c89,r2c9
-> 29(5) = [47747]

7. HS 3 in c7 -> r4c7 = 3
-> HS 3 in c6 -> r3c6 = 3

etc.


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PostPosted: Fri Mar 16, 2018 11:39 pm 
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Grand Master
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Thanks HATMAN. The second of your puzzles from early last year that I've just done this week, the other being Human Solvable 26 in the Killer forum.

There was one tricky step which made this puzzle feel a bit harder than HS26, where the way in was more obvious.

In this puzzle my key step was in the same place as in wellbeback's walkthrough, but we saw this in very different ways. I liked the was NOT 2 was used.

Here is my walkthrough:
Cages are maximum repeat, no unrepeated numbers, NC and FNC. There are no nonets.

Prelims

a) 15(6) cages must be 2 pairs of 3 digits (only way to get odd total with each group of at least 2 digits) = {111444} (cannot be {222333} NC)
b) 29(5) cage must be 3 of an odd digit and 2 of another digit = {55577/77744/99911}, no 2,3,6,8
c) R3C8 must be the digit repeated 3 times = {579}

Steps resulting from Prelims
1a. Naked pair {14} in R3C34, locked for R3
1b. Naked pair {14} in R4C24, locked for R4
1c. Naked pair {14} in R5C38, locked for R5
1d. Naked pair {14} in R6C67, locked for R6
1e. Naked pair {14} in R7C57, locked for R7
1f. Naked pair {14} in R35C3, locked for C3
1g. Naked pair {14} in R34C4, locked for C4
1h. Naked pair {14} in R27C5, locked for C5
1i. Naked pair {14} in R68C6, locked for C6
1j. Naked pair {14} in R67C7, locked for C7
1k. Naked pair {14} in R3C34, no 2,3,5 in R2C34 + R4C3 (FNC,NC)
1l. Naked pair {14} in R6C67, no 2,3,5 in R5C67 + R7C6 (FNC,NC)
1m. Naked pair {14} in R34C4, no 2,3,5 in R34C5 (FNC,NC)
1n. Naked pair {14} in R67C7, no 2,3,5 in R67C8 (FNC,NC)

2a. Naked pair {14} in R2C5 + R5C3 (because 15(6) cages contain 3 each of 1,4), also for the same reason naked pair {14} in R2C5 + R4C2, CPE no 1,4 in R2C2
2b. 15(6) cage at R2C5 must be either [114414] or [441141] because of naked pair {14} in R2C5 + R5C3) -> naked pairs {14} in R2C5 + R3C4, R3C3 + R4C2 and R4C4 + R5C3, CPE no 2,3,5 in R3C2 + R5C4 (FNC,NC)
2c. Naked pair {14} in R5C8 + R8C6 (because 15(6) cages contain 3 each of 1,4), CPE no 1,4 in R8C8
2d. 15(6) cage at R5C8 must be either [114414] or [441141] because of naked pair {14} in R5C8 + R8C6) -> naked pairs {14} in R5C8 + R6C7, R6C6 + R7C5 and R7C7 + R8C6, CPE no 2,3,5 in R6C5 + R8C7 (FNC,NC)

[The tricky step.]
3. R5C4567 and R3456C5 cannot be {6789} (can place 6,7,8,9 NC in either the row or the column, but not both because of FNC, for example R5C4567 = [7968] when there’s no valid placement for R6C5) -> no 6,7,8,9 in R5C5
3a. R5C5 = {35} -> no 4 in R4C4 + R6C6 (FNC)
3b. R4C4 = 1 -> 15(6) cage at R2C5 = [114414], no 2 in R1C456 + R2C2 + R23C6, no 3,5 in R345C1 + R5C2 + R6C234 (FNC,NC)
3c. R6C6 = 1 -> 15(6) cage at R5C8 = [114414], no 2 in R4C789 + R56C9 + R8C8, no 3,5 in R7C4 + R8C45 + R9C567 (FNC,NC)
3d. R15C5 = {35} (hidden pair in C5)
3e. R5C59 = {35} (hidden pair in R5)
3f. 2 in R5 only in R5C12 -> no 3 in R6C1 (FNC,NC)
3g. 2 in C5 only in R89C5 -> no 3 in R9C4 (FNC,NC)
3h. R1C4 = 3 (hidden single in C4) -> R15C5 = [53], R5C9 = 5, no 2 in R1C3 + R4C6 + R6C4, no 6 in R1C6 + R2C46 + R46C89 (FNC,NC)

4. 29(5) cage at R1C8 = {77744} (cannot be {55577/99911} because only one 1 in R1C9 or two 5s in R2C7 + R3C8) = [47747], no 3,5,6,8 in R2C8, no 3,6,8 in R3C9, no 6,8 in R14C7 + R3C67, no 8 in R12C6 + R4C89 (FNC,NC)
4a. R1C67 = [92], no 3 in R2C6 (FNC)
4b. R23C6 = [53], no 6 in R3C5 (FNC)
4c. R9C6 = 2 (hidden single in C6)
4d. R2C8 = 2 (hidden single in C8)
4e. R4C1 = 2 (hidden single in R4)
4f. R3C9 = 2 (hidden single in R3), no 3 in R4C89 (FNC,NC)
4g. R5C2 = 2 (hidden single in R5)
4h. R6C3 = 2 (hidden single in R6), no 3 in R7C23 (FNC,NC)
4i. R7C4 = 2 (hidden single in R7), no 3 R8C3 (FNC)
4j. R8C5 = 2 (hidden single in R8)
4k. R34C7 = [53] (hidden pair in C7), R4C89 = [59], no 6 in R4C8 + R5C7 (FNC)
4l. R6C19 = [53] (hidden pair in R6), no 6 in R5C1 + R6C2 + R7C12 (FNC,NC)
4m. R7C1 = 3 (hidden single in R7) -> R2C2 = 3 (hidden single in R2), R8C8 = 3 (hidden single in R8)
4n. R9C13 = [43] (hidden pair in R9), no 5 in R89C2 (FNC,NC)
4o. R9C4 = 5 (hidden single in R9), no 6 in R8C34 + R9C5 (FNC,NC)
4p. R8C3 = 5 (hidden single in R8) -> R7C2 = 5 (hidden single in R7), no 6 in R7C3 + R8C12 + R9C1 (FNC,NC)
4q. 6 in C4 only in R56C4 -> no 7 in R56C4 + R6C5 (FNC,NC)
4r. R6C2 = 7 (hidden single in R6), no 8 in R5C1 + R7C3 (FNC)
4s. R8C4 = 7 (hidden single in C4), no 8 in R9C5 (FNC)
4t. R5C1 = 7 (hidden single in C1)
4u. R9C5 = 7 (hidden single in R9)
4v. R4C5 = {68} -> no 7 in R4C6 (NC)
4w. R4C3 = 7 (hidden single in R4) -> R7C3 = 9, no 8 in R38C2 + R56C4 (FNC)
4x. R7C6 = 7 (hidden single in R7), no 8 in R8C7 (FNC)
4y. Naked pair {68} in R12C3 -> no 9 in R2C4 (FNC,NC)
4z. R2C134 = [968], no 8 in R13C1 (NC; FNC unnecessary for this step)

and the rest is naked singles, without using FNC,NC.


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PostPosted: Wed Mar 28, 2018 7:10 pm 
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Grand Master
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Location: Saudi Arabia
As you say the not 2 came to me as I was building the puzzle. It minimised any further clues


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