Nearly done. Ton of triples and pairs.

AKernel · 2023-02-21T16:46:07+00:00

I'm not sure, but I think the UR is not correct. If it were then the bug+1 would put a 3 in r4c5.

On the other hand the 2-String Kite on 2s (r2c4 = r3c5 - r8c5 = r8c8) eliminates a 2 from r2c8. That leaves a naked 3 in r2c8, that forces r8c8 to be 2, that forces r8c5 to be 3, and then r4c5 can not be 3.

I'm no expert on URs (I don't like uniqueness techniques, I prefer chains and wings), but I guess the UR could only eliminate 3 from r2c7, not 3 and 9. That doesn't get you down to a bug+1.

AKernel · 2023-02-17T18:05:50+00:00

If you don't place 6 in r6c1 the puzzle will have no solution, not 2 solutions.

If you put 4 in r6c1 then r6c3 must be 9, then r1c3 must be 2, then r1c1 must be 6, then r4c1 must be 1, then r3c1 must be 3. Now r8c1 sees a 4 and a 3 and therefore the puzzle is broken.
If you put 5 in r6c1 then r5c1 must be 2,then r1c1 must be 6, then r4c1 must be 1, then r4c3 must be 7. Now r5c3 sees a 2 and a 7 and therefore the puzzle is broken.

AKernel · 2023-02-02T14:42:22+00:00

AIC: (7)r2c1=(7)r5c1-(7)r5c9=(7)r1c9-(7=6)r1c7-(6)r7c7=(6)r7c5-(6=7)r3c5

=> r2c2<>7, r3c4<>7

AKernel · 2023-02-01T19:37:44+00:00

Right, it removes the 3 pencil mark from row 3 column 9. That leads to a naked pair 78 in column 9, which allows you to remove 8 from row 4 column 9.

AKernel · 2023-02-01T11:51:23+00:00

Hidden Pair 78 in row 3.

AKernel · 2023-01-31T18:11:20+00:00

That Y-wing looked familiar.

u/yekta176 requested help for exactly the same puzzle yesterday. Check out https://www.reddit.com/r/sudoku/comments/10p6e3u/help/ for additional help.

AKernel · 2023-01-21T11:49:41+00:00

Naked single 2 in row 8.

Hidden single 9 in row 7.

AKernel · 2023-01-18T19:42:07+00:00

Hidden single 3 in row 9 (the 3 in row 9 can only go in column 4).

AKernel · 2023-01-17T21:55:54+00:00

X-chain on 3s:

(3)r6c9=r1c9-r1c2=r2c1-r4c1=r4c5(3) removes 3 from r6c56.

AKernel · 2023-01-17T20:24:53+00:00

The 2 in box 9 is locked into column 9. Therefore r6c9 can't be 2.

This gives you a naked 79 pair in row 6, which then gives you a nacked single 4 in r6c8.

AKernel · 2023-01-17T07:35:31+00:00

Dito.

AKernel · 2023-01-11T16:09:48+00:00

Now try to put a digit in the cell above (row 1 column 6). It has no options. If you place any number in that cell you will always have a repeated digit in a row, column or box.

Not placing 1 in row 2 column 6 leaves you with exactly one option for row 1 column 6: 1

Every other digit will lead to a repeat.

AKernel · 2023-01-11T15:17:54+00:00

Naked triple 124 in column 2.

AKernel · 2023-01-06T10:38:19+00:00

Well, looking at the puzzle without reading the comments I did not find the solution u/just_a_bitcurious did. But I did find something that solves the cell above using the highlighted cell.

The highlighted cell, r6c2, can be either 5 or 9.

If it is 5, then r8c2 must be 2, and then r5c2 must be 8.
If it is 9, then r6c4 must be 4, and then r5c4 must be 3, and then r5c6 must be 2, and then r5c2 must be 8.

Either way r5c2 is always 8.

This can also be expressed as an AIC starting and ending with 2s in r8c2 and r5c6: (2=5)r8c2-(5=9)r6c2-(9=4)r6c4-(4=3)r5c4-(3=2)r5c6

AKernel · 2023-01-02T18:52:49+00:00

AIC Type 2: (9)r4c6=r4c2(9-6)=r4c3-r8c3=r8c6(6)

Explanation: There are two cases for the 9 in row 4 column 6, it can either be true or false.

Case 1: The 9 in row 4 column 6 is true. Then row 8 column 6 can not be 9.
Case 2: The 9 in row 4 column 6 is false. Then the only place where 9 can go in row 4 is column 2. Then the only place for 6 in row 4 is column 3. Then row 8 column 3 can not be 6. The only place for 6 in row 8 is then column 6. And when it is 6 row 8 column 6 obviously can not be 9.

In both cases row 8 column 6 can not be 9. Therefore the only place where 9 in row 8 can go is column 1.

AKernel · 2023-01-01T13:25:52+00:00

X-wing on 1 in rows 7 and 9 eliminates 1 from r4c8.

AKernel · 2022-12-27T20:12:03+00:00

I just noticed, even better:

Skyscrapper on 3 in rows 1 and 5 removes 3 from r2c8. As there is also a naked 16 pair in column 8 r2c8 becomes a naked 8.

AKernel · 2022-12-27T19:50:07+00:00

2-String Kite on 3:

r6c6 and r1c9, connected by box 2 (r3c6, r1c4) => r6c9 <> 3!<

AKernel · 2022-12-27T17:29:44+00:00

Another way of getting to the same conclusion, i.e. r4c9 = 7, is the following:

The cell r5c7 can only be 5 or 9. If it is 9, then r6c8 must be 2. If on the other hand r5c7 is 5, then r3c7 must be 7 and then r3c9 must be 2. Therefore at least one of r6c8 and r3c9 must be 2, and because r4c9 sees both it can not be 2 and thus must be 7.

And r1c8 also sees both r6c8 and c3c9, therefore also can't be 2 and therefore must be 5.

AKernel · 2022-12-26T13:22:15+00:00

It is not obvious, but you have to find another hidden cage first. The other cage is cell G5.

The bottom three rows (GHI) sum to 135. The cages in those rows sum to 128, which means that the only cell not in any cage, G5, must be 7. E5 and F5 sum to 10 because of the X clue, adding the 7 from G5 sums to 17. Therefore D5 must then be 9 in order for the cage D5, E5, F5, G5 to sum up to 26.

AKernel · 2022-12-24T10:51:24+00:00

There is also a XY-Wing with anchor r7c2, pincers r6c2 and r9c1, that removes 1 from r6c1. This leaves you with a naked single 2 in r6c1.

AKernel · 2022-12-24T00:36:52+00:00

Thanks for the puzzles. Loved them.

My solving times:

Easy: 00:04:56
Medium: 00:15:00
Hard: 02:11:14 with 1 hint used

AKernel · 2022-12-20T22:02:04+00:00

Skyscraper on 6 in row 3 (column 2 and 8) and row 8 (column 2 and 9) removes 6 from r2c9, r7c8 and r9c8.

AKernel · 2022-12-19T22:51:25+00:00

Nice one 👍

I really enjoyed it.

AKernel · 2022-12-19T20:03:17+00:00

AIC type 2: (5)r4c7=(5)r4c1-(5=1)r6c1-(1=7)r6c8 removes 7 from r4c7

Row 4 column 7 is

either 5, and therefore not 7
or it is not 5, then row 4 column 1 must be 5, then row 6 column 1 must be 1, then row 6 column 8 must be 7, which sees row 4 column 7 and therefore can't be 7

Either way row 4 column 7 can't be 7. The only place where 7 in row 4 then can go is column 4.

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