SudoCue Users

Well, any method which is logical and makes an elimination is good enough for me.

Steve

Novajlija I don’t think your argument is valid as it stands. What you have shown is that, if r89c1 = (56), r9c1 ≠ 5. That’s fine as far as it goes but you also need to allow for the case r89c1 ≠ (56). In this event r8c1 = 8 so you only need a further line: If r8c1 = 8, r5c1 = 5 s...

I think the nice loop which corresponds to Ron’s pair of almost locked sets is:

r9c3 -2- {r1c3, r2c1} -8- {r7c1, r8c1, r9c2} -2- r9c3

As it happens you could confine yourself to single cells here, if you prefer:

r9c3 -2- r1c3 -5- r2c1 -8- r2c3 =8= r9c3

Steve

The fewest number of fish I can identify to make David’s eliminations from his grid is three: (a) The X-wing for 7 based on rows 5 and 7 (columns 6 and 9; box 9) eliminates from r8c9 and 9c9. (b) The swordfish for 7 based on rows 1, 4 and 6 (columns 5 and 7; box 4) eliminates from r9c5. (c) The swor...

Ruud Yes, this is MO’s Sashimi observation, which is that a finned swordfish, unlike a proper one, does not necessarily degenerate if based on a 2-2-1 pattern. I think this means that, if you rub out the fin, 2-2-1 remains. This is the case here except that, reading downwards, the pattern without th...

Ruud Maybe it’s my fish that are muddled. Do not the cells marked F form a swordfish with fin r3c1, so eliminating 7 from the cell marked X? ------------------------- | 6 . X | . . 3 | 4 . 8 | | . . 1 | . . . | 7 2 . | | F . 9 | 8 F . | . . . | ------------------------- | . . . | 1 . . | . . . | | 5...

Perhaps I have muddled up the puzzles. It wouldn't be the first time.

It seemed to me that the first step was to eliminate 7 from r1c3 using the finned swordfish on rows 3, 5 and 9. The only remaining admissible entry for that cell is 2 and the rest seemed elementary.

Steve

First note the pair (14) in box 2 cells r13c4. Taking this into account there are only two places for 5 in the first row (box 1 and cell r1c6). There are also only two places for 5 in the fifth row, cells r5c16. this finned X-Wing eliminates 5 from r2c1, leaving the possible entries as (79). The XY-...

My way of getting past the hurdle in David's second matrix was to put r5c1 = 7 using the XY-Wing for (46) pivoted on r1c3 with pincers r1c1 and r4c3. The second hurdle (= David's third matrix) took me a long time to resolve. I eventually noticed that there were only two places for 5 in row 3 and in ...

Your suspicions are well founded.
This is just about as tough as I like them: elementary logic plus x- and xy-wing will resolve it.

Steve

The constellations were observed from Munich by someone_somewhere. They include four-star , five-star and other varieties. The constellations are interesting and reasonably common in difficult puzzles but not easily identified – my computer found this one when testing my beginner’s solver program. S...

If you prefer patterns to chains, you might use a four star constellation after Ruud’s first step. The alpha star, cell (6, 9) containing (356), requires 3 to be eliminated from cell (3, 9). It therefore contains 5 and the rest is elementary.

Steve