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HATMAN
 Post subject: Square Wisdom 23-01
PostPosted: Thu Nov 30, 2023 9:01 am 
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 694
Location: Saudi Arabia
Square Wisdom 1 for 2023

I've not made one of these for a few years. The name is a pun by Jean-Christophe (http://jcbonsai.free.fr/sudoku/JSudokuUserGuide/) as I cannot remember what we call them on this forum.
No sign means addition.

Note that on the repeats I have followed standard KenKen rules - so repeats are allowed in all cages. No repeats in rows, columns or nonets (which standard KenKen does not use).

If you remember there was a whole legal argument about copyrighting the term "KenKen" so everyone used different names for it.

Minus and divide work either way. Some versions of this puzzle type such as Dr Sudoku's (FKA motris) allow longer minus and divide cages with the largest number always first.

Medium hard, about 1.0 on the assassin scale. Easy start and finish with some tricky bits in the middle.

Image

Solution:
492163785
578294631
613785942
285417396
937526418
146839527
854972163
329651874
761348259


Last edited by HATMAN on Fri Jan 05, 2024 6:30 am, edited 1 time in total.

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Andrew
 Post subject: Re: Square Wisdom 23-01
PostPosted: Sun Dec 10, 2023 11:27 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
At first I got stuck and asked HATMAN where I'd gone wrong. That was when he told me that repeats are allowed for some cages according to KenKen rules, which he's now explained in the puzzle post. It was the first time I'd come across them and I expect that some other users of this forum may not have either, so thanks HATMAN for explaining them.

Here's how I solved Square Wisdom 1 for 2023, the middle bit isn't too hard:
Some cages are regular killer cages, some multiples of the numbers in the cages while a couple are differences between values in the two cells; cages which overlap two nonets can have repeated numbers.

Prelims

a) R4C34 = {18/27/36/45}, no 9
b) R45C8 = {19/28/37/46}, no 5
c) R6C56 = {39/48/57}, no 1,2,6
d) R7C56 = {18/27/36/45}, no 9
e) R89C7 = {19/28/37/46}, no 5
f) R9C23 = {16/25/34}, no 7,8,9
g) R7C89 = 18x(2) cage = 1x2x3x3 = {29/36}
h) R89C6 = 8x(2) cage = 1x2x2x2 = {18/24}
i) 60x(3) cage at R1C6 = 1x2x2x3x5 = {256/345}
j) 15x(3) cage at R1C9 = {135}
k) 108x(3) cage at R3C7 = 1x2x2x3x3x3 = {269/349}
l) 96x(3) cage at R3C9 = 1x2x2x2x2x2x3 = {268/348}
m) 10x(3) cage at R6C7 = {125}
n) 60x(3) cage at R8C5 = 1x2x2x3x5 = {256/345}
o) 720x(4) cage at R6C2 = 1x2x2x2x2x3x3x5 = {2589/3568/4459}, no 1,7
Note that the 14(4) cage at R2C3, which would be in prelims for regular killers, may include repeated numbers.

1a. 45 rule on N9 3 innies R7C789 = 10 can only be {136} -> R7C7 = 1, R7C89 = {36}, locked for R7 and N9, clean-up: no 8 in R7C56, no 4,7,9 in R89C7
1b. Naked pair {28} in R89C7, locked for C7 and N9 -> R6C78 = [52], clean-up: no 8 in R45C8, no 7 in R6C56, no 6 in 108x(3) cage at R3C7
1c. Killer pair 2,4 in R7C56 and 60x(3) at R8C5, locked for N8
1d. Naked pair {18} in R89C6, locked for C6 and N8, clean-up: no 4 in R6C5
1e. Naked triple {135} in 15x(3) at R1C9, locked for N3
1f. Naked pair {49} in R3C78, locked for R3, N3 and 108x(3) at R3C7 -> R4C7 = 3, clean-up: no 6 in R4C34, no 7 in R45C8, no 4 in 96x(3) cage at R3C9
1g. Naked pair {67} in R12C7, locked for C7 and N3 -> R1C8 = 8, R3C9 = 2
1h. Naked pair {68} in R45C9, locked for N6, 6 locked for C9 -> R7C89 = [63], clean-up: no 4 in R45C8
1i. Naked pair {15} in R12C9, locked for C9 and N3 -> R2C8 = 3
1j. Naked pair {19} in R45C8, 9 locked for C8 and N6 -> R3C78 = [94], R5C7 = 4, R6C9 = 7
1k. 60x(3) cage at R1C6 = {256/345}, 5 locked for C6 and N2, clean-up: no 4 in R7C5
1l. 2-(2) at R6C3 = {13/46/68}, no 9
1m. Combined cages 2-(2) at R6C3 and R6C56 = {13}{48}/{46/68}{39}, 3 locked for R6
1n. 60x(3) cage at R8C5 = {256/345}, 5 locked for N8, clean-up: no 4 in R7C6
1o. Naked pair {27} in R7C56, locked for R7 and N8, clean-up: no 6 in 60x(3) cage at R8C5
1p. Naked triple {345} in 60x(3) cage, 3 locked for N8 -> R78C4 = [96], clean-up: no 4,8 in R6C3
1q. Naked triple {458} in R7C123, locked for N7, clean-up: no 2,3 in R9C12
1r. Naked pair {16} in R9C23, locked for N7, 1 locked for R9 -> 8x(3) cage at R8C6 = [18], R89C7 = [82]
1s. 2-(2) cage at R5C3 = {13/35/57}/[97] (cannot be [86] which clashes with R5C9), no 2,6,8
1t. 9 in C6 only in R456C6, locked for N5, clean-up: no 3 in R6C6
1u. 6 in R6 only in R6C123, locked for N4
1v. 720x(4) cage at R6C2 = {2589/3568/4459}, 5 locked for N7
1w. 8 of {2589/3568} must be in R7C23 -> no 8 in R6C2

2a. 9 in N2 only in 16(3) cage at R1C4 = {169/349}, no 2,7,8
2b. 1 of {169} must be in R1C4 -> no 1 in R12C5
2c. Killer pair 3,6 in 16(3) cage and 60x(3) cage, locked for N2
2d. R2C3 + R3C23 cannot total 7 = {124} because 2,4 only in R2C3 -> no 7 in R2C4
2e. 7 in N2 only in R3C45 = {17/78}, 7 locked for R3
2f. 18(4) cage at R3C4 = 8+10/15+3 = {17}{28}/{17}[46]/{17}[73]/{78}{12}, no 5,6 in R4C5, no 5,7 in R5C5
2g. 5 in N5 only in R45C4, locked for C4
2h. 8 in N2 only in R2C4 + R3C34, CPE no 8 in R3C23 (since 14(4) cage at R2C3 cannot repeat 8)
2i. Combined cages 16(3) + 60x(3) = 1{69}{345}/{349}{256}, 4 locked for N2
2j. 2-(2) at R6C3 (step 1l) = {13}/[68] (cannot be [64] which clashes with R6C2 + R6C6, killer ALS block), no 4
2k. Killer pair 3,8 in 2-(2) at R6C3 and R6C5, locked for R6

3a. 16(3) cage at R1C4 (steps 2a, 2b) = 1{69}/{349}, 60x(3) cage at R8C5 = {256/345}, R3C45 (step 2e) = {17/78}, 18(4) cage at R3C5 (step 2f) = {17}{28}/{17}[46]/[71][73]/{78}{12}
3b. Consider permutations for R6C56 = [39/84]
R6C56 = [39] => blocks 16(3) cage = {349} + 60x(3) cage which have {3459} in R1289C5 => 16(3) cage = 1{69} => R3C45 = {78} = 15, R45C5 = 3 = {12}
or R6C45 = [84]
-> 18(4) cage = [71][73]/{78}{12}, no 4,6,8 in R45C5
3c. R12C5 = {69} (hidden pair in C5), 6 locked for N2, R1C4 = 1, R12C9 = [51], 60x(3) cage at R1C6 = {345}, 3,4 locked for C6, R3C45 = {78} -> R45C5 = {12}, 2 locked for C5 and N5 -> R7C56 = [72], R3C45 = [78], R6C56 = [39], R6C4 = 8 -> R6C3 = 6, R45C4 = [45] (hidden pair in N5), R4C3 = 5, R5C3 = {37}, R9C23 = [61], R9C4 = 3

4a. R6C2 = 4 -> 720x(4) cage at R6C2 = {4459} (only remaining combination) -> R7C23 = [54], R8C3 = 9, R679C1 = [187]
4b. R2C4 = 2 (cannot repeat in 14(4) KenKen cage because R2C3 in same row) -> R2C3 + R3C12 = 12 = {138} (only remaining combination) -> R2C3 = 8, R3C23 = [13], R3C6 = 5, R3C1 = 6
4c. R12C1 = [45] (hidden pair in C1), R3C1 = 6 -> R2C2 + R4C1 = 9 = [72]

and the rest is naked singles.
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HATMAN
 Post subject: Re: Square Wisdom 23-01
PostPosted: Mon Dec 11, 2023 2:49 pm 
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Quite different in the middle bit. I liked your r2c4<>7.
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HATMAN
 Post subject: Re: Square Wisdom 23-01
PostPosted: Thu Dec 14, 2023 7:40 am 
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Joined: Wed Apr 30, 2008 9:45 pm
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Andrew asked me to post my middle bit to show different approaches:
After the easy start:
+-------------------------------+-------------------------------+-------------------------------+
| 12345679  12345679  12345679  | 134       3469      23456     | 67        8         15        |
| 12456789  12456789  12456789  | 12478     469       2456      | 67        3         15        |
| 135678    135678    135678    | 1378      13678     356       | 9         4         2         |
+-------------------------------+-------------------------------+-------------------------------+
| 1245789   1245789   124578    | 124578    1245678   24679     | 3         19        68        |
| 1235789   1235789   13579     | 1357      1235678   23679     | 4         19        68        |
| 134689    469       136       | 1348      38        49        | 5         2         7         |
+-------------------------------+-------------------------------+-------------------------------+
| 48        58        458       | 9         27        27        | 1         6         3         |
| 2379      2379      239       | 6         345       1         | 8         57        49        |
| 379       16        16        | 345       345       8         | 2         57        49        |
+-------------------------------+-------------------------------+-------------------------------+


from this point:
r3c23<>8 8 pointing from N2 and 14/4 not 88..
1 in N5 point to r3c4 <>1 as 18/4 no 9 so <>11..
I note that this is too clever as 1 in C5 locked in 18(4) - annoying!
C6 3 in 60x -> r6c56 = [39] or 3 in R5c6 no other 3 in N5
innies r6 c1234=19 8 in r6c45 locked N5
r6c2=6 ->r8c3=3 blocked r6c3 720x = 9{58}2 or 4549 9 <> r5c3, r8c2
49 locked r6c26
r2c4<>7 as per Andrew, 7 locked r3c45
346 locked 16/3 60/3
r3c45 = [71]/{78} -> r45c5 = {12}/[46]
[46] -> 16/3 = 4[39] but N8 60x now has 4 in r9c4 fail 18/4 = {78}{12}
easy-peasy from here {I've never understood the Japanesy bit}. It comes from WWII British troops first then probably Americans in the Pacific, later becoming a children's rhyme.

Probably not more difficult than Andrew's approach, but I found it tricky.
corrected as per Andrew


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