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 Post subject: Rainbow Killer 4
PostPosted: Tue Sep 26, 2023 11:34 am 
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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
Rainbow Killer 4
Rainbow as in "All the numbers in the rainbow"
The numbers are from 0-9
There is a zero in every nonet, row and column. Zero is chosen as it increases the killer combinations.
Each other number is missing from one and only one nonet, row and column.
This means that
the nonet where a missing number row and the column with the same number missing cross, has that same number missing. I know this by checking examples but cannot see how to prove it. I used this in my solution but doubt that it was necessary. 999_springs gives a simple proof on the players forum.
Relatively easy - about 0.7 on the assassin scale.

Note the hidden cage - which I did not use.

JSudoku puzzle attached.


Image


Attachments:
Rainbow K4 Puzzle.zip [2.86 KiB]
Downloaded 674 times


Last edited by HATMAN on Sun Oct 01, 2023 4:21 am, edited 2 times in total.
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 Post subject: Re: Rainbow Killer 4
PostPosted: Thu Sep 28, 2023 6:02 pm 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks HATMAN for a fun puzzle! I used one nice step, which included a short forcing chain, at the start of my step 2; that would count higher than 0.7 but the rest of my solving path was fairly straightforward. I did use the hidden cage, although it may not have been necessary, but didn't use what HATMAN put in small print.

Here's how I solved Rainbow Killer 4:
Zero is in every row, column and nonet. One of 1-9 is missing in each row, column and nonet.

Prelims

a) R12C5 = {39/48/57}, no 0,1,2,6
b) R12C9 = {02}
c) R3C12 = {59/68}
d) R34C9 = {05/14/23}, no 6,7,8,9
e) R45C7 = {29/38/47/56}, no 0,1
f) R56C9 = {29/38/47/56}, no 0,1
g) R7C56 = {05/14/23}, no 6,7,8,9
h) R78C7 = {07/16/25/34}, no 8,9
i) R78C9 = {69/78}
j) R8C56 = {79}
k) R9C34 = {08/17/26/35}, no 4,9
l) R9C56 = {29/38/47/56}, no 0,1
m) 9(3) cage at R3C8 = {018/027/036/045/126/135/234}, no 9
n) 7(3) cage at R6C5 = {016/025/034/124}, no 7,8,9
o) 4(3) cage at R7C1 = {013}
p) 8(3) cage at R9C7 = {017/026/035/125/134}, no 8,9

1a. Naked pair {02} in R12C9, locked for C9 and N3, clean-up: no 3,5 in R34C9, no 9 in R56C9
1b. Naked pair {14} in R34C9, locked for C9, clean-up: no 7 in R56C9
1c. Naked pair {79} in R8C56, locked for R8 and N8, clean-up: no 0 in R7C7, no 6,8 in R7C9, no 0,2 in R8C34, no 2,4 in R9C56
1d. Naked triple {013} in 4(3) cage at R7C1, locked for N7, clean-up: no 6,8 in R8C4, no 6,8,9 in R9C12, no 5,8 in R9C4
1e. Killer pair 6,8 in R56C9 and R8C9, locked for C9
1f. R9C34 = {26}/[71/80] (cannot be [53] which clashes with R9C56), no 3,5
1g. 8(3) cage at R9C7 = {017/134} (cannot be {026} because R9C9 only contains 3,5,7, cannot be {035} which clashes with R9C56, cannot be {125} which clashes with R9C12), no 2,5,6, 1 locked for R9 and N9, clean-up: no 6 in R78C7, no 7 in R9C3
1h. R9C9 = {37} -> no 3,7 in R9C78

2a. 45 rule on N78, sum of two missing numbers = 14 = {59/68}
2b. Consider combinations for R9C56 = {38/56}
R9C56 = {38}, locked for R7 and N8 => R9C34 = {26}, locked for R9 => R9C12 = {45}, locked for N7 => R8C34 = [81] => R7C56 = {05}
or R9C56 = {56}
-> 5 in R79C56, locked for N8, clean-up: no 4 in R8C3
2c. R789C56 contain both of 5,9 -> missing numbers in N78 must be 6,8
2d. R9C56 contains one of 6,8 -> no other 6,8 in N8, clean-up: no 2 in R9C3
2e. R9C3 = {68} -> no other 6,8 in N7 -> R9C34 = [54], R9C12 = {27}, locked for R9 and N7, R9C34 = [80], R7C56 = {23}, locked for R7, R7C4 = 1, R7C1 = 0, R9C9 = 3, R9C78 = {14}, 4 locked for N9, R56C9 = {56}, locked for N6, 6 locked for C9, R8C9 = 8 -> R7C9 = 7, R7C7 = 5 -> R8C7 = 2, clean-up: no 9 in R45C7
[Cracked, fairly straightforward from here.]
2f. N9 must contain 0 -> R8C8 = 0, R67C8 = 15 = [96]
2g. 9(3) cage at R3C8 = {135/234}, no 7,8
2h. 9(3) cage = {234} (cannot be {135} because 5{13} combined with R4C9 = {14} clashes with R45C7), 3,4 locked for C8 -> R9C78 = [41], clean-up: no 7 in R45C7
2i. Naked pair {38} in R45C7, locked for C7, 3 locked for N6, naked pair {24} in R45C8, 4 locked for C8 and N6 -> R3C8 = 3, R34C9 = [41], R6C7 = 0
2j. Min R1C8 = 5 -> max R1C67 = 7, no 8,9 in R1C67, no 7 in R1C6 (because no 0 in R1C7)
2k. Min R2C8 = 5 -> max R2C67 = 8, no 8,9 in R2C67 (because no 0 in R2C7)
2l. 9 is already missing from N9 -> all other nonets must contain 9 -> R3C7 = 9, clean-up: no 5 in R3C12
2m. Naked pair {68} in R3C12, locked for R3 and N1
2n. R3C7 = 9 -> R3C56 = 8 = {17}, locked for R3 and N2, clean-up: no 5 in R12C5
2o. 11(3) cage at R4C1 = {128/146/236/245} (cannot be {137} which clashes with R8C1), no 7,9
2p. 9 already missing from C9 -> all other columns must contain 9 -> 9 in C1 only in R12C1, locked for N1

3a. R3C3 = 0 (hidden single 0 in R3) -> R4C23 = 15 = {69}/[87]
3b. R5C2 = 0 (hidden single 0 in N4)
3c. 45 rule on complete grid, taking into account that 1-9 are each missing in one nonet -> R1C1 + R5C2 = 9, R5C2 = 0 -> R1C1 = 9, clean-up: no 3 in R2C5
3d. R4C5 = 0 (hidden single 0 in C5) -> R45C6 = 17 = {89}, locked for N5, 9 locked for C6 -> R8C56 = [97], R3C56 = [71], clean-up: no 3 in R1C5
3e. Naked pair {48} in R12C5, locked for N2, 4 locked for C5
3f. 8 in R6 only in R6C12, locked for N4 -> R4C23 = {69}, locked for R4 and N4 -> R45C6 = [89], R45C7 = [38]
3g. 9 already missing from C9 -> C4 must contain 9 -> R2C4 = 9, R34C4 = 7 = {25}, locked for C4

4a. 18(3) cage at R6C2 = {378/468} (cannot be {567} which clashes with R6C9 -> R6C2 = 8, R6C34 = {37} (cannot be [46] because R6C349 = [465] clashes with 7(3) cage at R6C5), locked for R6
4b. 7(3) cage = [160]/{25}0, no 4, no 6 in R6C5
4c. Killer pair 5,6 in 7(3) cage and R6C9, locked for R6
4d. 11(3) cage at R4C1 (step 2o) = {245} (only remaining combination), locked for C1, 2,4 locked for N4 -> R9C12 = [72]
4e. 8 missing from C4 -> C1 must contain 8 -> R3C12 = [86], R4C23 = [96], R7C23 = [49]
4f. 13(3) cage at R5C3 = {157} (cannot be {256} because 2,5 only in R5C5) = [175], 18(3) cage = [873], R34C4 = [52], 7(3) cage = [160], R56C9 = [65], 9(3) cage at R3C8 = [342], 11(3) cage = [542], R9C56 = [65]
4g. 13(3) cage at R2C6 = {067/238/256} (cannot be {058} because R2C7 only contains 1,6,7, cannot be {157} because R2C6 only contains 0,2,3
4h. 12(3) cage at R2C1 = {147/345} (cannot be {237} which clashes with 13(3) cage) -> R2C3 = 4, R12C5 = [48]
4i. R1C4 = 6 -> R1C23 = 9 = [72], 12(3) cage = {345} = [354], R12C9 = [02], R8C12 = [13]
4j. R1C67 = [31] -> R1C8 = 8 (cage sum), 13(3) cage at R2C6 = [067], R7C56 = [32]

Solution:
9 7 2 6 4 3 1 8 0
3 5 4 9 8 0 6 7 2
8 6 0 5 7 1 9 3 4
5 9 6 2 0 8 3 4 1
4 0 1 7 5 9 8 2 6
2 8 7 3 1 6 0 9 5
0 4 9 1 3 2 5 6 7
1 3 5 4 9 7 2 0 8
7 2 8 0 6 5 4 1 3


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 Post subject: Re: Rainbow Killer 4
PostPosted: Fri Sep 29, 2023 7:35 am 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 694
Location: Saudi Arabia
I had the same breaking point but my approach was less clear.


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 Post subject: Re: Rainbow Killer 4
PostPosted: Fri Sep 29, 2023 8:09 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
HATMAN wrote:
how to prove it.

This is the way it makes sense to me but I'm easily tricked.

Take for example the 1s. There are eight of them on the grid, one only is missing. However, one row and one column are both missing a 1, ie two houses. How can you get two for one? Must be by missing a 1 from the crossover nonet.

Hence, the missing 1 from the row and column must also be missing from the crossover nonet.

Cheers
Ed


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 Post subject: Re: Rainbow Killer 4
PostPosted: Sun Oct 01, 2023 4:10 am 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 694
Location: Saudi Arabia
I think that is correct. 999_Springs has put it more clearly on the players forum.
He has also proved that W X cannot be Anti-Knight.


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