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HATMAN
PostPosted: Mon Jan 08, 2024 9:03 am 
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Rainbow Killer 8 and KiMo 8
For those of you with long memories:
When I first came up with the KiMo idea Udosuk did all of them, but eventually complained that the were very dependent on pairs to solve, so I largely shelved them.
On making these rainbows I noticed that there was also a tendency to be pair driven (particularly for SS) so I thought of combining the two, as maybe the faults would cancel out.

Rainbow Killer 8
This is a simple one, set-up to be the basis for the KiMo - about the same difficulty as Rain 7.

Image

Rainbow KiMo 8
This is the KiMo which is unique.
The green cells form a _5(9) cage.
I have not solved it but am close. From where I am JSudoku gives 10 solutions and with one placement solves on naked singles.
I will put my current position up in a few days to see if anyone here wants to tag-solve it.
Now solved - I'll post the W-T shortly.

Image

I have solved Rain KiMo 9 and will post it next week.


Last edited by HATMAN on Sun Jan 21, 2024 5:52 pm, edited 1 time in total.

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Andrew
PostPosted: Sun Jan 14, 2024 5:46 am 
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Thanks HATMAN for Rainbow Killer 8. It felt to me a bit harder than RK 7 as for RK 8 I used two short forcing chains, steps which some will see as contradictions, and it took me a while to spot good ones, which were both immediately effective.

Here's how I solved Rainbow Killer 8:
Prelims

a) R1C12 = {19/28/37/46}, no 0,5
b) R12C3 = {07/16/25/34}, no 8,9
c) R12C4 = {06/15/24}, no 3,7,8,9
d) R1C56 = {07/16/25/34}, no 8,9
e) R1C89 = {06/15/24}, no 3,7,8,9
f) R23C1 = {19/28/37/46} no 0,5
g) R23C6 = {49/58/67}, no 0,1,2,3
h) R23C9 = {19/28/37/46} no 0,5
i) R3C23 = {08/17/26/35}, no 4,9
j) R3C78 = {19/28/37/46} no 0,5
k) R4C12 = {08/17/26/35}, no 4,9
l) R45C3 = {29/38/47/56}, no 0,1
m) R4C45 = {39/48/57}, no 0,1,2,6
n) R45C6 = {03/12}
o) R4C78 = {08/17/26/35}, no 4,9
p) R45C9 = {19/28/37/46} no 0,5
q) R56C1 = {07/16/25/34}, no 8,9
r) R56C4 = {69/78}
s) R7C12 = {19/28/37/46} no 0,5
t) R78C3 = {19/28/37/46} no 0,5
u) R7C45 = {04/13}
v) R78C6 = {06/15/24}, no 3,7,8,9
w) R78C7 = {59/68}
x) R7C89 = {19/28/37/46} no 0,5
y) R89C1 = {05/14/23}, no 6,7,8,9
z) R89C4 = {19/28/37/46} no 0,5
aa) R89C9 = {05/14/23}, no 6,7,8,9
ab) R9C23 = {08/17/26/35}, no 4,9
ac) R9C56 = {69/78}
ad) R9C78 = {03/12}

45 rule on the whole grid shows that R258C258 total 45 but that’s no help as some numbers in cells which don’t “see” each other may be repeated.

Each row column and nonet must contain 0 and be missing one of 1-9 -> totals in each must be between 36 and 44 with each row, each column and each nonet having a different total.
1a. 45 rule on N1 cages total 35 -> no 0 in R2C2
1b. 45 rule on N2 cages total 35 -> no 0 in R2C5
1c. 45 rule on N3 cages total 35 -> no 0 in R2C8
1d. 45 rule on N4 cages total 35 -> no 0 in R5C2
1e. 45 rule on N5 cages total 39 -> max R5C5 = 5
1f. 45 rule on N7 cages total 33 -> min R8C2 = 3
1g. 45 rule on N8 cages total 35 -> no 0 in R8C5
1h. 45 rule on N9 cages total 32 -> min R8C8 = 4

2a. Combined cages R89C9 + R9C78 = {05}{12}/{14}{03}, 1 locked for N9, no 2,3 in R89C9, clean-up: no 9 in R7C89
2b. Combined cages R7C45 + R78C6 = {04}{15}/{13}{06} (cannot be {13}{24} which clashes with R89C4), 1 locked for N8, no 2,4 in R78C6, clean-up: no 9 in R89C4
[Alternatively combined cages R7C45 + R78C6 = {04}{15}/{13}{06} must contain 0 for N8. I originally had these two cages as steps 1a and 1b, then realised how important the 45s on nonets are so moved those to become step 1.]
2c. R89C4 = {28/37} (cannot be {46} which clashes with combined cages), no 4,6
2d. R9C56 = {69} (cannot be {78} which clashes with R89C4), locked for N8, 6 locked for R9, clean-up: no 0 in R78C6, no 2 in R9C23
2e. Naked pair {15} in R78C6, locked for C6 and N8, clean-up: no 2,6 in R1C5, no 8 in R23C6, no 2 in R45C6, no 4,8 in R6C5, no 3 in R7C45
2f. Naked pair {03} in R45C6, locked for C6 and N5, clean-up: no 4,7 in R1C5, no 9 in R4C45, no 6,9 in R6C56
2g. R56C4 = {69} (cannot be {78} which clashes with R4C45), locked for C4, clean-up: no 0 in R12C4, no 0,3 in R3C5
2h. Naked pair {04} in R7C45, 4 locked for R7 and N8, clean-up: no 6 in R7C12, no 6 in R7C89, no 6 in R8C3
2i. Killer pair 6,9 in R23C6 and R9C6, locked for C6, clean-up: no 1 in R1C5
2j. R1C56 = [07/34] (cannot be [52] which clashes with R12C4), no 2,5
2k. R1C12 = {19/28/46} (cannot be {37} which clashes with R1C56), no 3,7
2l. Killer pair 4,7 in R1C6 and R23C6, locked for C6 and N2, clean-up: no 2 in R12C4, no 2,5 in R3C45, no 2,5 in R6C5
2m. Naked pair {15} in R12C4, locked for N2, 5 locked for C4, clean-up: no 8 in R3C45, no 7 in R4C5
2n. R1C89 = {06/24} (cannot be {15} which clashes with R1C4), no 1,5
2o. R1C12 = {19/28} (cannot be {46} which clashes with R1C89), no 4,6
2p. R23C9 = {19/28/37} (cannot be {46} which clashes with R1C89), no 4,6
2q. R3C78 = {19/28/37} (cannot be {46} which clashes with R1C89), no 4,6
2r. Naked pair {03} in R1C5 + R3C4, 3 locked for N2
2s. Killer pair 6,9 in R23C6 and R3C45, locked for N2

3a. 0 in N1 only in R12C3 = {07} or R3C23 = {08} -> R3C23 = {08/26/35} (cannot be {17}, locking-out cages), no 1,7
3b. 0 in N3 only in R12C7 = {09) or R1C89 = {06) -> R12C7 = {09/18/27/45} (cannot be {36}, locking-out cages), no 3,6
3c. 0 in N7 only in R89C1 = {05} or R9C23 = {08} -> R9C23 = {08/17} (cannot be {35}, locking-out cages), no 3,5
3d. Killer pair 0,1 in R9C23 and R9C78, locked for R9, clean-up: no 4,5 in R8C1, no 4,5 in R8C9
3e. Killer pair 0,4 in R1C56 and R1C89, locked for R1, clean-up: no 3,7 in R2C3, no 5,9 in R2C7

[Time to start looking at forcing chains.]
4a. R1C89 = {06/24}
4b. Consider combinations for R1C12 = {19/28}
R1C12 = {19}, 9 locked for R1, no 9 in R1C7 => no 0 in R2C7, 0 in N3 only in R1C89 = {06}
or R1C12 = {28}, 2 locked for R1 => R1C89 = {06}
-> R1C89 = {06}, locked for R1 and N3, R1C5 = 3 -> R1C6 = 4, R3C4 = 0 -> R3C5 = 9, R7C45 = [40], R9C56 = [69], clean-up: no 9 in R1C7, no 1 in R2C1, no 4 in R2C3, no 1 in R2C9, no 8 in R3C23, no 1 in R3C78, no 8 in R4C5
4c. R2C3 = 0 (hidden single in N1) -> R1C3 = 7, clean-up: no 3 in R23C1, no 2 in R2C7, no 4 in R45C3, no 2,9 in R6C2, no 3 in R78C3, no 1,8 in R9C2
4d. Killer pair 2,3 in R3C23 and R3C78, locked for R3, clean-up: no 8 in R2C1, no 7,8 in R2C9

5a. R7C12 = {19/28/37}, R7C78 = {28/37}, R9C23 = [08/71], R9C78 = {03/12}
5b. Consider combinations for R9C23 = [08/71]
R9C23 = [08] => R9C78 = {12}, 2 locked for N9
or R9C23 = [71], R7C12 = {28}, 2 locked for R7
-> R7C78 = {37}, locked for R7 and N9, clean-up: no 0 in R9C78
5c. Naked pair {12} locked for R9, 1 locked for N9, R8C9 = 0 -> R9C9 = 5, R1C89 = [06], R9C3 = 8 -> R9C2 = 0, clean-up: no 8 in R4C1, no 3 in R45C3, no 1 in R6C2, no 9 in R6C3, no no 3,4,9 in R6C8, no 9 in R6C9, no 2 in R7C12, no 2 in R78C3, no 9 in R78C7, no 3 in R8C1, no 2,8 in R8C4
5d. Naked pair {19} in R7C12, locked for N7, 1 locked for R7 -> R78C3 = [64], R78C6 = [51], R7C87 = [86], R89C9 = [23], R89C4 = [37], R4C4 = 8 -> R4C5 = 4, R6C6 = 2 -> R6C5 = 7, R28C5 = [28], clean-up: no 8 in R1C2, no 1 in R12C7, no 8 in R3C1, no 2 in R3C2, no 2 in R3C8, no 8 in R3C9, no 0 in R4C1, no 5,6 in R4C2, no 5 in R45C3, no 0 in R4C7, no 2 in R4C8, no 0,4,5 in R5C1, no 1,2,3,7 in R5C7, no 2,4 in R5C9, no 4,5 in R6C1, no 3,5 in R6C2, no 1,3 in R6C7
5e. R23C9 = [91] (cannot be [37] which clashes with R7C9), clean-up: no 0,8 in R6C8
5f. R45C9 = [28] (cannot be {37} which clashes with R7C9), R45C3 = [92], clean-up: no 6 in R3C2, no 6 in R4C1, no 6 in R4C8, no 1 in R6C8
5g. Naked pair {46}, locked for N1, 6 locked for C1, clean-up: no 1 in R56C1
5h. R56C1 = [70], clean-up: no 1 in R4C12, no 9 in R5C7
5i. R4C12 = [53], R45C6 = [03], R3C23 = [53], R6C3 = 1 -> R6C2 = 8, R2C2 = 1, R7C12 = [19], R1C2 = 2 -> R1C1 = 8, R1C7 = 5 -> R2C7 = 4, R56C7 = [09], R56C4 = [96], R6C8 = 5 -> R6C9 = 4, clean-up: no 7 in R3C78
5j. R3C78 = [28], R23C1 = [64], R23C6 = [76], R2C8 = 3, R9C78 = [12], R4C78 = [71], R5C28 = [46]
5k. 45 rule on the whole grid 1 remaining innie R5C5 = 5

and the rest is naked singles.

Solution:
8 2 7 1 3 4 5 0 6
6 1 0 5 2 7 4 3 9
4 5 3 0 9 6 2 8 1
5 3 9 8 4 0 7 1 2
7 4 2 9 5 3 0 6 8
0 8 1 6 7 2 9 5 4
1 9 6 4 0 5 8 7 3
2 7 4 3 8 1 6 9 0
3 0 8 7 6 9 1 2 5
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HATMAN
PostPosted: Fri Jan 19, 2024 12:10 am 
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My solution was much like Andrew's but I avoided the forcing chains and the end count.

The point to note is that the puzzle was set up with 5 pairs in each nonet.
So after Andrews 2q
2q. R3C78 = {19/28/37} (cannot be {46} which clashes with R1C89), no 4,6

N3: the pair R2C8 + missing = 10(2) if [46] -> r1c89 = {06} so six not missing or if [64] -> r1c89={24} so four not missing
so R2C8 + missing cannot be {46} so = {19/28/37} same as r3c78 and r23c9 so r1c789r2c7={0,4,5,6} 6(2) = {06} and 9(2) = {45}.
This places {06} for Andrew's step 4b

and at Andrew's step 5b - which is neat and clever:
consider from above r12c7={45} -> r78c7={68}->r7c89={37}

At the end the counting in Andrew's step 5k can be avoided as:
5k. 45 rule on the whole grid 1 remaining innie R5C5 = 5
At this point R5C5=1/5 but 1 is place in all other nonets so must be missing in N5 ->R5C5 =5
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HATMAN
PostPosted: Sun Jan 21, 2024 5:51 pm 
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Rain KiMo 8 B

I realised I could make this a bit easier quite neatly.
When solving it I found I had missed something which I then used to solve Rain KiMo 8 above.

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Andrew
PostPosted: Wed Jan 24, 2024 12:04 am 
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Thanks HATMAN for your comments about my walkthrough and what I'd missed.

I agree that's the point of the puzzle. I have used 45 rule giving the total of uncaged cell(s) + missing number but forgot about it this time, so only used 45 rule to remove lowest/highest numbers in the uncaged cells. As it happened, after step 1, I didn't really need to use those results apart from nonets which required 0 in cages. There have been previous puzzles, probably Triankles, where I have kept track of missing numbers and used the fact that a number missing from one nonet could then be used for its only position in a different nonet.

Using HATMAN's intended method my step 1 should have been
1a. 45 rule on N1 cages total 35 -> R2C2 + missing number = 10, no 0,5 in R2C2
1b. 45 rule on N2 cages total 35 -> R2C5 + missing number = 10, no 0,5 in R2C5
1c. 45 rule on N3 cages total 35 -> R2C8 + missing number = 10, no 0,5 in R2C8
1d. 45 rule on N4 cages total 35 -> R5C2 + missing number = 10, no 0,5 in R5C2
1e. 45 rule on N5 cages total 39 -> R5C5 + missing number = 6 -> R5C5 = {1245}
1f. 45 rule on N7 cages total 33 -> R8C2 + missing number = 6 -> R8C2 = {345789}
1g. 45 rule on N8 cages total 35 -> R8C5 + missing number = 10, no 0,5 in R8C5
1h. 45 rule on N9 cages total 32 -> R8C8 + missing number = 13 -> R8C8 = {456789}

HATMAN's clever step after my step 2q has the first part reading like a forcing chain; I think it could be rephrased as R2C8 + missing number in N3 = {19/28/37} (cannot be {46} which clash with R1C89), no 4,6, then R1C789 + R2C7 = {0456} (hidden quad in N3) -> R1C89 = {06}, R12C7 = {45}

Then, as HATMAN pointed out, at the start of my step 5, using R12C7 = {45}, actually [54] at that point, leads directly to R7C89 = {37}

Finally, I agree that the point of the puzzle was which numbers were missing from rows, columns and nonets but my step 5 had flowed so well that I wasn't taking note of the missing numbers; using 45 rule on the complete grid was just as easy as checking for them.

HATMAN's step after my step 2q clearly gave a simpler solving path than mine, although possibly still a bit harder than Rainbow Killer 7.
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HATMAN
PostPosted: Wed Jan 24, 2024 8:32 am 
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Happy with all your points Andrew.
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HATMAN
PostPosted: Mon Feb 12, 2024 7:57 am 
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Having eventually solved them neither is actually that difficult, I would put them on the 1.1 - 1.3 range on the assassin scale - views please.
1 WT Rain KiMo 8B
Zero is present in every row, column and nonet.
Each other number is missing from exactly one row, column and nonet (it has been proven that the nonet is fixed at the crossover of the row and column).
W-T added definitions for Rainbow, as we need a short form to identify the missing numbers, so:
Row 1 missing number is R1m
Column 1 missing number is C1m
Nonet 1 missing number is N1m (I will also use this for Triankles and similar for the repeats there)
1. Nonet Sums
These puzzles are deliberately arranged around nonets.
45 Rule on each nonet:
N1 r2c2+N1m = 10(2)
N2m = 8, N3m = 7, N4m = 6, N5m = 1
N6: R5C8 + N6m = 9(2)
N7m = 5, N8m = 2, N9m = 4
These numbers will be missing from the relevant rows and columns.
Clean up the partners numbers in each two-cage.
2. Fix known cage totals
_8(2) is 8(2), _9(2) is 9(2), _0(2) is 10(2)
_5(9) = 45(9), no 0 in centre spots
Because of the zero there are less than standard KiMo
3. Solve N1 and N6 missing
Remaining missing are:
N16m = {39}
N1m+r2c2 = [91/37]
N6m+r5c8 = [90/36] but R5c8 <> 0
So r2c2 = 1, N1m = 9, r5c8 = 6 N6m = 3
Clean up
N1: r12c3 = 7(2)
N6: 18 locked in 8(2)&10(2)
Note: R123m = {789}, R456m = {136}, R789m = {245}, C123m = {569}, C456m = {128}, C789m = {347}
4. Derive cage totals by simple nonet arithmetic first
N16 done so:
N3: _9(3) = 9(3)
N7: _2(3) = 12(3)
4. Remove KiMo cage candidates
_2(2) <>16, _4(2) <>27, _6(2) <> 38
At this stage all the basics are done saved as Rainbow KiMo 8B s1.jsd.
5 Rows and columns
R3: 5 locked in 8(2), 9(2) r3c45 <>36
R6: set of 9(2) pairs missing from 1/3/6 -something here not sure what?
C9: r167 C9m =20(4) / 10(4) but r45c9 = {19/28} so not 10(4) -> r89c9 = 5(2) = {05/23}
->_3(2) at r9c78 <>{03}
N9: 05,12, 37, 689 / 05, [67], 19, 238 / 05, [67], 28, 139 / 23 19, 58, {06}7 / 23, 19, [67], 058
Consider C9 if r89 = 23 then r45 = {19} but r7c89 = {19} so r89c9 = {05} clean up
N9: 05,12, 37, 689 / 05, [67], 19, 238 / 05, [67], 28, 139 so_3(3) <>7
N3: 2*10(2) is two from 19/28/46 -> 9(2) <>18
C9m = 347 -> 6 locked in r123 so 10(2) = 64 or 9(3) = {03/12}6
R123c7<>6, r123c8r1c9<>4
But r3c78 = 19/28 so 9(3) <>126
N3: 46, 09, 28, 135 / 036, 45, 19/28
C8m = 4
R7: 0 locked in c456
N8: _3(3) <>012 r7c6<>6
Rainbow KiMo 8B s2.jsd.
C9: 8 locked in 10(2)*2 r7 ->r7c9<>2 (probably not much use but I like it)
N7: 08, {2+19/37/46} and 2*{19/37/46} / 039, 2*46/28, 17 / 048, 2*19/37, 26
N1: 05 locked in 7(2), 8(2) so 07, 35, 28/46 / 08, 25, 37/46 46 not in r123c3r3c2
N4: 89, 014, 35, 27 / 89, 023, 17, 45 / 07, 35, 18, 249 / 25, 08, fail 9(2)
/ 25, 09, 17, 348 / 25 078, fail 9(2) / 34, 08, 27, 159 / 34, 09, 17, 258 / 34, 078 fail 9(2)
N4: 89, 014, 35, 27 / 89, 023, 17, 45 / 07, 35, 18, 249 / 25, 09, 17, 348 / 34, 08, 27, 159 / 34, 09, 17, 258 ->_5(3)<>7
Rainbow KiMo 8B s3.jsd.
R4: _2(2) <>57 clash 2*8(2),
R4&N5: _2(2) <>{39} as c36={45} and 9(2), _5(2)<>6 so no 6 in N5
This is a WoW! moment: _2(2) = {02/28} 2*8(2)<>{08} -> r4c78 = {17}, r4c12 = {35}
Easy from here
28 locked in R4c459 R4m = 6, r4c36 ={9,0/4}
N6: r45c9= {28} 2*9(2) = 09/45, r23c9<>28 ->r3c78 = {28}
R3: R3c23 = {35} -> r3c45 = {09} -> R3m = 7
N1: r23c1 = {46}, r12c3 = {07}, R1c12 = {28} -> R1m=9, R2m = 7
N4: 89, 014, 35, 27 / 07, 35, 18, 249
N6&R6: r6c789 from 0459 ->2*9(2) <>0459, R6c23 = [72/81], R6c56 = {27/36}
Cannot be 18&36 as R6m =1/3 27 locked -> r6 c14 = {6/8&0/5/9} as 4 not in c1-6, r56c7<>[45]
Rainbow KiMo 8B s4.jsd.
C2: r9=0 HS, r9c2 = 8, 2*10(2)<>28
R7: c1289 = 19/37 clear r78c3 ={46}, C2m = 6, r5c2 = 4 HS, r45c3 = [92]
Fill some naked singles
Rainbow KiMo 8B s5.jsd.
R56c4 = 15(2) = [96], r4c45 =1292) = {48}, r4c6 =0, r5c6 =3
The rest is NS, HS and fixing cage totals.

2 WT Rain KiMo 8
I expected the same breakthrough to work so this walkthrough is targeted at getting there early.
1. Nonet Sums
These puzzles are deliberately arranged around nonets.
45 Rule on each nonet:
N1 N1m = 9
N2m = 8, N3m = 7, N4m = 6, N5m = 1
N7m = 5, N8m = 2, N9m = 4
N6: R5C8 + N6m = 9(2) but only 3 un-missing -> N6m=3, R5c8 = 6
These numbers will be missing from the relevant rows and columns.
Clean up the partners numbers in each two-cage.
2. Fix known cage totals
_8(2) is 8(2), _9(2) is 9(2), _0(2) is 10(2)
_5(9) = 45(9), no 0 in centre spots
Because of the zero there are less than standard KiMo
N6: 18 locked in 8(2)&10(2)
Note: R123m = {789}, R456m = {136}, R789m = {245}, C123m = {569}, C456m = {128}, C789m = {347}
3. Derive cage totals by simple nonet arithmetic first
N6 done so:
N1: _8(3) = 8(3)
N3: _9(3) = 9(3)
N7: _2(3) = 12(3)
4. Remove KiMo cage candidates
_2(2) <>16, _4(2) <>27, _6(2) <> 38
At this stage all the basics are done saved as Rainbow KiMo 8 s1.jsd.
R4: _2(2) <>57 conflict 2*8(2), if {39} r56c4&r6c56<>6, r1278 ={08/17} -> R4m = 6 -> r4c6<>6 so no 6 in N5 fail, _2(20 = {02/48} -> r12&r78 <> {08} -> r4c78 = {17} -> r4c12 = {35}
saved as Rainbow KiMo 8 s2.jsd.
Innies C9 r167 C9m =20(4) / 10(4) but 7 locked -> 20(40) -> r89c9 = 5(2) = {05}
N9: 05,12, 37, 689 / 05, [67], 19, 238 so_3(3) <>17
R3: 5 locked in 8(2), 9(2) r3c45 <>36
N7: 1 in C1 locked in N7 8(2)<>17, 26 clashes with _3(2) at r9c78 ->r9c23 = {08}
R9: c56 = {14/69}, c78 = {12}/[67], R9m=2/4 1246 locked, r14 = {3,7/9},
N3: 2*10(2) is two from 19/28/46 -> 9(2) <>18
s3 now
N7: 19, 372, 46, 08 / 192, 37/46, 08, 2 locked in r8c12
N4: 78 locked in _7(2), 9(2)
C9m = 347 -> 6 locked in r123 so 10(2) = 64 or 9(3) = {03/12}6
R123c7<>6, r123c8r1c9<>4
But r3c78 = 19/28 so 9(3) <>126/018 ->r3c78 = {28}, C8m = 4
saved as Rainbow KiMo 8 s3.jsd
N2; 0 of R3 locked in c456 so 09, 15/24, 67, 3-15/24 / 03, 45, 79, 126
R8c2 = 0 HS c2, r12c3 = 0x -> r56c1 = {07} -< r6c23 = [81]
R6m = 3, R5m = 1, r6c56 = {27}, r56c1 = [70], r6c89 = [54], r56c7 = [09], r56c4 = [96]
R4c45 = {48}, r4c3 = 9, r4c6 = 0, r3c45 = [09], r12c4 = 6(2)
R12c7 = {45}, r23c9 = {19}, r1c9 = 6 r12c8 = [03], r23c1 = {46}, r5c56 = [53] from 45(9)
The rest is NS, HS and fixing cage totals.
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