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 Post subject: Killer Low Clue 1 and 2
PostPosted: Wed Jul 24, 2019 8:44 pm 
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Grand Master
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Killer Low Clue 1

I argue that this is not a variant; it is just a killer with a few cages and numbers missing.

It is based on the Torygraph uncaged killer for 070719.

The usual uncaged killer rules. the number is in the top left corner with top dominating left as usual. There are no single cages. Note each question mark represents a digit so for example a double cage with a single ? does not contain a 9.

I was just in the UK for a couple of weeks and everyday I bought the Times and the Telegraph cut out the harder or more interesting puzzles and threw the rest of the paper away. The people in my local cafe may have thought me a bit strange.

I've tried making uncaged puzzles in the past and have found them difficult to structure successfully. The ones in the paper are of course too easy so I came up with the idea of adding question marks for the digits.

I am back in Nigeria now so if any of you takes the Sunday Telegraph please consider scanning the uncaged puzzle and sending it to me. I promise to use it to make one of these low clue killers.


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PostPosted: Sun Jul 28, 2019 9:56 pm 
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Grand Master
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Killer Low Clue 2

Uniqueness: not only must the solution number set be unique; the cages must also be unique.

In creating this one, one potential solution had two cage combination possibilities.

This one is based on the 300619 Telegraph.

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PostPosted: Mon Jul 29, 2019 5:57 pm 
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Grand Master
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HATMAN wrote:
Uniqueness: not only must the solution number set be unique; the cages must also be unique.

Even though I've only just started on the first of these puzzles
I can see that:
because of the surrounding cages, 13 at R2C1 can be either a 2 or a 3 cell cage, with 21 at R2C2 a 4 or a 3 cell cage. Hopefully when I get into solving the puzzle I'll be able to decide which they are.


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PostPosted: Sun Aug 04, 2019 3:49 am 
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Grand Master
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Thanks HATMAN for these fun puzzles. You keep coming up with types of Sudokus which I've never seen before.

Killer Low Clue 1 is an easy one; still worth posting, I assume that the original newspaper version gave all the totals so would have been too easy for this site.

Most, but not all, of the cages can be placed before one starts working on the numbers.

Here is my walkthrough for Killer Low Clue 1:
“The usual uncaged killer rules, the number is in the top left corner with top dominating left as usual. There are no single cages. Note each question mark represents a digit so, for example, a double cage with a single ? does not contain a 9.”

I assume that there are no diagonally-connected cages.

Prelims, based just on positions of totals.
a) R1C12 must be 5(2) cage = {14/23}
b) R1C78 must be 8(2) cage = {17/26/35}, no 4,8,9
c) R2C45 must be 3(2) cage = {12}
d) R1C345 must be 21(3) cage = {489/579/678}, no 1,2,3
e) R12C6 must be 12(2) cage = {39/48/57}, no 1,2,6
f) R23C3 must be 5(2) cage = {14/23}
g) There must be a cage starting at R7C3 including R7C4
h) R6C34 must be 11(2) cage = {29/38/47/56}, no 1
i) R5C45 must be ?(2) cage, no 9
j) R3C45 must be 8(2) cage = {17/26/35}, no 4,8,9
k) R34C6 must be 8(2) cage = {17/26/35}, no 4,8,9
l) R4C45 must be 14(2) cage = {59/68}
m) R45C3 must be ??(2) cage
n) R56C6 must be ??(2) cage
o) There must be 18(?) cage in R9C123, possibly also including R9C4
p) R8C12 must be 14(2) cage = {59/68}
q) R7C12 must be 8(2) cage = {17/26/35}, no 4,8,9
r) R56C1 must be ??(2) cage
s) R56C2 must be ?(2) cage, no 9
t) R23C7 must be 12(2) cage = {39/48/57}, no 1,2,6
u) R4C78 must be ?(2) cage, no 9
v) R5C78 must be 14(2) cage = {59/68}
w) R45C9 must be 9(2) cage = {18/27/36/45}, no 9
x) R7C67 must be 12(2) cage = {39/48/57}, no 1,2,6
y) R6C78 must be ?(2) cage, no 9
z) R67C9 must be ??(2) cage
aa) There must be a 17(?) cage in R8C34, possibly also including R9C4
ab) R678C5 must be 21(3) cage {489/579/678}, no 1,2,3
ac) R7C34 must be ?(2) cage, no 9
ad) R9C56 must be ?(2) cage, no 9
ae) R8C67 must be ?(2) cage, no 9
af) R78C8 must be ?(2) cage, no 9
ag) R89C9 must be 10(2) cage = {19/28/37/46}, no 5
ah) R9C78 must be 14(2) cage = {59/68}
and two more based on a clash
ai) 17(3) cage in R8C34 + R9C4 (cannot be 17(2) cage in R8C34 which would clash with R8C12)
aj) 18(3) cage at R9C1
This just leaves a pair of cages in N14 which can be 13(2) and 21(4) or 13(3) and 21(3) and a pair of cages in N3 which can be 9(2) and 16(3) or 9(3) and 16(2)

Steps resulting from Prelims
1a. Naked quad {1234} in R1C12 + R23C3, locked for N1
1b. Naked pair {12} in R2C45, locked for R2 and N2, clean-up: no 3,4 in R3C3, no 6,7 in R3C45, no 6,7 in R4C6
1c. Naked pair {35} in R3C45, locked for R3 and N2, clean-up: no 7,9 in R12C6, no 7,9 in R2C7, no 3,5 in R4C6
1d. Naked pair {48} in R12C6, locked for C6 and N2, clean-up: no 4,8 in R7C7

2. 9 in N2 only in R1C45 -> 21(3) cage at R1C3 = {579} (only remaining combination) -> R1C3 = 5, R1C45 = {79}, locked for R1 and N2, clean-up: no 1,3 in R1C78, no 6 in R6C4
2a. R3C6 = 6 -> R4C6 = 2, clean-up: no 7 in R5C9, no 9 in R7C3
2b. Naked pair {26} in R1C78, locked for R1 and N3, clean-up: no 3 in R1C12
2c. Naked pair {14} in R1C12, locked for R1 and N1 -> R12C6 = [84], R23C3 = [32], clean-up: no 8,9 in R3C7, no 8,9 in R7C4

3. R1C9 = 3, no 6 in R2C9 -> 9(3) cage in R123C9 = {135} (only remaining combination) -> R23C9 = [51], clean-up: no 4,6,8 in R45C9, no 7,9 in R89C9
3a. R2C7 = 8 -> R3C7 = 4, clean-up: no 6 in R5C8, no 8 in R9C8
3b. R23C8 must be 16(2) cage = {79}, locked for C8, clean-up: no 5 in R5C7, no 5 in R9C7
3c. R45C9 = [72], clean-up: no 8 in R89C9
3d. Naked pair {46} in R89C9, locked for C9 and N9
3e. R9C7 = 9 -> R67C9 = [98], R9C8 = 5, clean-up: no 3,7 in R7C6
3f. R5C78 = [68] -> R1C7 = 2

4. 13(?) cage at R2C1, min R23C1 = 13 -> 13(2) cage in R23C1 = [67], clean-up: no 1,2 in R7C2, no 8 in R8C2
4a. That leaves R23C2 + R4C12 as the remaining 21(4) cage, R23C2 = [98] = 17 -> R4C12 = 4 = {13}, locked for R4 and N4, clean-up: no 5 in R8C1
4b. R4C78 = [54], clean-up: no 9 in R4C45
4c. Naked pair {68} in R4C45, locked for R4 and N5 -> R4C3 = 9
4d. Naked pair {13} in R6C78, locked for R6, clean-up: no 8 in R7C3
4e. Naked triple {457} in R6C456, locked for R6 and N5
4f. R6C3 = 6 -> R6C4 = 5, R6C6 = 7
4g. R6C5 = 4 -> R78C5 = 17 = [98]
4h. R7C6 = 5 -> R7C7 = 7, clean-up: no 1,3 in R7C1, no 3 in R7C2

5. 45 rule on N7 2 innies R78C7 = 5 = {14}, locked for C7 and N7

and the rest is naked singles.

Solution, showing the cages:
Attachment:
Killer Low Clue 1.jpg
Killer Low Clue 1.jpg [ 56.83 KiB | Viewed 124 times ]


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PostPosted: Sun Aug 04, 2019 3:55 am 
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Grand Master
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Killer Low Clue 2 was only slightly harder. It took longer to place some of the cages, fewer at the start and some only right at the end.

Here is my walkthrough for Killer Low Clue 2:
“The usual uncaged killer rules, the number is in the top left corner with top dominating left as usual. There are no single cages. Note each question mark represents a digit so, for example, a double cage with a single ? does not contain a 9.”

I assume that there are no diagonally-connected cages.

Prelims, based just on positions of totals.
a) R1C12 must be 6(2) cage = {15/24}
b) 7(?) cage at R2C3 must include R3C3
c) R23C2 must be 16(2) cage = {79}
d) R234C1 must be 13(3) cage
e) R4C45 must be 3(2) cage = {12}
f) R5C23 must be 4(2) cage = {13}
g) R4C23 must be 14(2) cage = {59/68}
h) There must be ??(2) cage in R56C1
i) R7C12 must be 4(2) cage = {13}
j) R78C3 must be part of 16(?) cage possibly including R9C3
k) R8C12 must be 6(2) cage = {15/24}
l) 15(?) cage at R3C5 must contain R3C6, 9(?) cage at R2C6 must contain R2C7, one of these cages must contain R3C7
m) R1C67 must be 6(2) cage = {15/24}
n) 17(?) cage at R3C8 must include R3C9, 16(?) cage at R4C7 must include R4C8, one of these cages must contain R4C9
o) R12C9 must be 16(2) cage = {79}
p) R12C8 must be 8(2) cage = {17/26/35}, no 4,8,9
q) R56C7 must be 7(2) cage = {16/25/34}, no 7,8,9
r) 13(?) cage at R7C7 must include R7C8, 9(?) cage at R6C8 must include R6C9, one of these cages must contain R7C9
s) R5C89 must be 14(2) cage = {59/68}
t) R89C9 must be 3(2) cage = {12}
u) 18(?) cage at R8C7 must contain at least R9C67
v) R89C8 must be 7(2) cage = {16/25/34}
w) ??(?) cage at R6C6 must include R7C6
x) R7C45 must be 6(2) cage = {15/24}
Also, to avoid clashes
y) R6C23 must be 6(2) cage (cannot be 6(3) = {123} which clashes with R5C23) = {15/24}
z) 17(?) cage at R3C8 must be 17(3) cage R3C89 + R4C9 to avoid clash with R12C9
aa) R4C78 must be 16(2) cage = {79}

Steps resulting from Prelims
1a. Naked pair {12} in R4C45, locked for R4 and N5
1b. Naked pair {79} in R4C78, locked for R4 and N6, clean-up: no 5 in R4C23, no 5 in R5C89
1c. Naked pair {68} in R4C23, locked for R4 and N4
1d. Naked pair {68} in R5C89, locked for R5 and N6, clean-up: no 1 in R56C7
1e. Naked pair {13} in R5C23, locked for R5 and N4, clean-up: no 5 in R6C23, no 4 in R6C7
1f. Naked pair {24} in R6C23, locked for R6 and N4 -> R4C1 = 5, clean-up: no 1 in R1C2, no 5 in R5C7, no 1 in R8C2
1g. Naked pair {13} in R7C12, locked for R7 and N7, clean-up: no 5 in R7C45, no 5 in R8C2
1h. Naked pair {24} in R7C45, locked for R7 and N8
1i. Naked pair {24} in R8C12, locked for R8 and N7 -> R89C9 = [12], clean-up: no 5,6 in R89C8
1j. R89C8 = [34], clean-up: no 5 in R12C8
1k. Naked pair {79} in R12C9, locked for C9 and N3, clean-up: no 1 in R12C8
1l. Naked quad {1245} in R1C1267, locked for R1 -> R12C8 = [62], R5C89 = [86], clean-up: no 4 in R1C6
1m. Naked pair {79} in R2C29, locked for R2

[Now some more cages]
2a. Min R789C3 = 18 -> R78C3 must be 16(2) cage = {79}, locked for C3 and N7
2b. Min R7C789 = 18 -> R7C78 must be 13(2) cage = [67/85]
2c. R6C89 + R7C9 must be 9(3) cage = {135} (only possible combination) -> R6C89 = [13], R7C9 = 5, R4C9 = 4, R3C89 = [58], R56C7 = [25], R7C78 = [67], R78C3 = [97], clean-up: no 1 in R1C6
2d. Naked triple {568} in R9C123, locked for R9
2e. R89C7 = [89] -> R8C7 + R9C67 must be 18(3) cage -> R9C6 = 1 (cage sum)
2f. Since 1 now placed for N8 there must be 8(2) cage at R89C4 = [53]
2g. R9C5 = 7 -> there must be 13(2) cage in R89C5 = [67]
2h. R478C6 = [389]

3a. Naked pair {79} in R23C2, locked for N1 (I’d overlooked this earlier, useful now)
3b. R4C1 = 5 -> R23C1 = 8 = [62], R8C12 = [42], R6C23 = [42], R1C12 = [15], R1C67 = [24]
3c. R49C3 = [65] (hidden pair in C3)

4. There must be 9(3) cage in R2C67 + R3C7 (cannot form 9(2) cage with R2C67), R23C7 = {13} -> R2C6 = 5 (cage sum)
4a. R3C56 must be 15(2) cage = [96], R6C56 = [87]
4b. ?(?) cage at R4C6 must be R45C6 = [34]
4c. 7(?) cage at R2C3 must be 7(2) cage R23C3 (since no 1,2 in R23C3) = {34}

5. R2C4 = 8 (hidden single in R2), R3C4 must be part of 12(?) cage at R2C4 -> R23C4 = 12(2) cage = [84]

The remaining ??(?) cages are R1C345 + R2C5, R56C45, R678C6 and R9C123, and the rest is naked singles.

Solution, showing the cages:
Attachment:
Killer Low Clue 2.jpg
Killer Low Clue 2.jpg [ 60.65 KiB | Viewed 123 times ]


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PostPosted: Tue Aug 06, 2019 12:23 am 
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Master
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I will add my thanks to HATMAN. I had not seen this style of puzzle before either. Another type where we get two puzzles for the price of one :)


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PostPosted: Tue Aug 13, 2019 6:56 pm 
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Very cool, looking forward to trying these. (Though, I briefly thought about hunting down the broadsheet versions to get my feet wet. :lol: )

Question: "Top dominating left". What is "usual"?


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PostPosted: Tue Aug 13, 2019 8:58 pm 
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The sum is in the top (first) then left (second) corner. so on the row below the sum (or lower) the cage can go to the left of the sum.


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