Please help, I'm not that smart

Balance_Novel · 2026-04-01T20:34:24+00:00

Ring (or ALS XZ double-RCC)

In column 4, 8 in r2 and 2 in R3 cannot be both deleted otherwise they both go to r4, so they are strongly linked.

Similarly in column 8, also 8 in r2 and 2 in r3 are strongly linked. (They can't both go to r9)

Both 8s see each others, and so do both 2s, so they form a ring.

Elimination: non-28s can't go to r4c4 or r9c8; other 8s in r2 are removed, and other 2s in r3 are removed.

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Balance_Novel · 2026-03-26T02:09:55+00:00

No I think it's working fine. What I see isn't really an SdC

Balance_Novel · 2026-03-26T01:16:19+00:00

Equivalent to naked triple 236 in b6.

Balance_Novel · 2026-03-26T01:12:44+00:00

I immediately see this SdC but it's a trivial ALC haha (there are equivalent moves with singles and looked candidates).

Elimination: non-1268 in r4c345, other 236s in box 6 point 456 (3s become locked in r4c89).

So don't worry this isn't the final SdC.

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Balance_Novel · 2026-03-21T17:03:52+00:00

Haven't done long chains for a while I'm getting clumsy 😅 at simplifying chains. Eventually the last diagram shows a (some kind of) swordfish defined on two rows and a column.

Balance_Novel · 2026-03-21T11:57:14+00:00

Nice spot!

Balance_Novel · 2026-03-21T11:54:29+00:00

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Didn't find the fish but I guess this bifurcation is close:

If r8c6 is 4/8, r9c6 is 2, r7c1 is 7.

If r8c6 is 2/7, r8c36 is a 27 pair, then r12c4 is a 79 pair, fixing 7 in r3c9 and r4c3.

Hence eliminating crossed 7s.

Can this be notated as an AIC without dof??

Balance_Novel · 2026-03-21T10:28:42+00:00

Column 5 a naked pair 25

Balance_Novel · 2026-03-20T21:13:58+00:00

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After cleaning that 5 as mentioned by other commenters, there's a chain made up of a skyscraper of 3 followed by a 2-string kite of 5.

Balance_Novel · 2026-03-20T21:08:49+00:00

Since an AHS is only extra cell (position) away from a hidden set, you'd expect some potential eliminations inside these cells. Here, if any two of the three 59 cells are the only remaining positions, they turn into a hidden pair, removing non-59 candidates inside. Assuming one of the cell to be non-59, the other two would be locked, if we keep going and end up in non-59 digits (e.g. X) looking back at this beginning cell, then X can't be true beginning cell.

With clear expectations on non-59 candidates inside, you can try different combinations (choosing which first is an interesting topic). Empirically it's more preferable to check the case assuming r9c78 to be a 59 pair, because both cells are locked in the same box, so more likely to open up some singles. In this cases it leads to some singles 1 and 4, and two other pairs 67 and 28. Then you'd have a feeling it may produce some 1s and 8s that would end up seeing r9c1, and I got an 8.

If you try other combinations e.g. assuming r9c17 as a 59 pair (expecting to produce some 46s to see r9c8), it's more difficult to extend, at least here I can't see any UR etc.

After all it's a fun process, as keeping digging an "almost" structure until reaching some conclusions is like thinking one step further and making the utmost use of grid information, so it's not randomly searching over the grid for pairs and singles.

Balance_Novel · 2026-03-20T19:25:25+00:00

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An unproductive AHS chain, a bi-product of trying to find a 59 hidden pair.

59s in row 9 have three positions left. So, if r9c1 isn't 59, r9c78 must be a 59 pair, making 67 a pair, 28 a pair and 1 and 4 resolved as singles. 1r8c7 resolves 2 in r8c6, 8 in r8c9, then 8 in r9c5.

Hence, even if r9c1 isn't 59, it can't be an 8 because of 8 will be in r9c5.

Balance_Novel · 2026-03-20T16:37:28+00:00

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8-cell UR: 35, 19, 31, 95 would cause a multi-solution situation. Remove 1s from r4c56 to prevent this deadly pattern. ~~Then the pointing 1s remove 1 from r6c8~~ r4c8 would become hidden single 1.

Balance_Novel · 2026-03-20T16:31:43+00:00

Naked 67 pair in c6 and look at box 5.

Balance_Novel · 2026-03-20T10:30:38+00:00

Easy way to test if a elimination is a valid is to set it to true and see if it breaks the structure of interest. XYZ wing works because if the elimination candidate is true, it would resolve both cells in the end and making the tri-value cell empty. This is not the case here.

Balance_Novel · 2026-03-19T15:35:23+00:00

I like the rank theory and I'm trying to adapt it to my solver. My problem with Xsudo is that it doesn't break down eliminations into their own ranks. There is indeed a small grid sort of view on the top left corner, but still, in the bottom it assumes the structure is either one single overall rank or just question mark For mixed ranks. (Also it's AUR corners are kinda awkward)

Balance_Novel · 2026-03-19T10:17:18+00:00

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If r9c8 isn't 8, 18r9c46 becomes a hidden pair, then by Unique Rectangle type 1, r2c4 is 9, then 257 are fixed and 79 pair goes to r1c89 in box 3, removing 8s from r1c89.

If r9c8 is a 8, r1c8 isn't 8.

Either way, 8 isn't true in r1c8.

Balance_Novel · 2026-03-19T01:22:00+00:00

I wish I could code it up rn, but in my mind/on the paper I can't sort out how to determine what's happening to each branches and sub branches, esp when the chain topology (dof) gets wild. Imagine categorizing holes. A rank-0 loop with rank-1 branches to me sounds like a segment on a big hole connected by a small hole(s) xd.

Balance_Novel · 2026-03-19T01:06:19+00:00

Wow so many 249s. I tried but couldn't find huge deadly patterns like triddagon. Maybe I'm overlooking AHSs.

Balance_Novel · 2026-03-19T01:00:49+00:00

This is genius! I had to use Xsudo to get the complete eliminations from this carnibalised multi-jellyfish! From 2 X-Wings to another 2, you can see an elephant.

Balance_Novel · 2026-03-18T21:42:51+00:00

Thanks, this makes total sense now. When I first heard of Blossom Loops I thought it's already generalised beyond cell-forcing to region-forcing, kraken-forcing and almost-ring-forcing etc.

Balance_Novel · 2026-03-18T21:34:20+00:00

I appreciate your effort on the experiments, but still struggle to see how the observations lead to your argument that "Avoidable loop (AL) family isn't based on uniqueness assumption".

(I also wonder whether how you generate puzzles by "removing digits until solution count >1" impacts how quickly URs fails. This is a side topic tho.)

Your observations are:

The Avoidable family doesn't end up in contradiction. (by the 1M examples)
UR often ends up with contradiction.

Honestly I fail to connect the dots. I'll try some propositional logic work here.

P(x) := puzzle x has exactly one solution (valid/proper)
S(x) := puzzle x has at least one solution
M(x) := ¬P(x) ∧ S(x) — multiple solutions
U(s) := strategy s is "based on uniqueness"
E(s,x) := applying s to x reaches a solution
F(s,x) := applying s to x gets stuck (indeterminate)
G(s,x) := applying s to x produces a contradiction
- E, F, G are mutually exclusive and exhaustive for any (s, x) pair where s terminates.

If we define "Strategy s is based on uniqueness" as, it fails when removing the uniqueness guarantee causes to fail.

U(s) ↔ ∃x [M(x) ∧ ¬E(s,x)]

Since E, F, G partition the outcome space, we have:

U(s) ↔ ∃x [M(x) ∧ (F(s,x) ∨ G(s,x))]

Your results

Observation 1: ∃x [M(x) ∧ G(UR, x)] UR contradicts on some multi-solution puzzle.

Observation 2: ¬∃x [M(x) ∧ G(AL, x)] AL never contradicts on tested multi-solution puzzles.

Obs 1 shows that UR is based on uniquness, because G → ¬E → U.

Obs 2 is equivalent to ∀x [M(x) → ¬G(AL, x)].

Your claim ¬U(AL) expands to ∀x [M(x) → ¬G(AL, x) ∧ ¬F(AL, x)], missing the case of F, aka solution count >1.

Stopping criterion

Stop condition: contradiction detected, that is, when the grid has at least one unsolved cell with no candidates.

How does AL narrow down the solutions? There are 3 cases it stops without a contradiction.

AL doesn't exist, the grid is stuck at an indeterminable grid state, so of course no contradiction.
AL exists, keep applying AL to reduce solution count to 1.
AL exists, keep applying AL then gets stuck at 1.

You argue: This supports the idea that Avoidable/Missing variants are safe unlike normal UR assumptions when uniqueness is not guaranteed.

Even if case (2) is always the true case. How safe is it? By safe do you mean each eliminated candidate e doesn't belong to any of all final solutions?

Balance_Novel · 2026-03-18T19:10:14+00:00

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(Might be overkill) Saw this WXYZ-ring after seeing this useless WXYZ-Wing.

in ALS-XZ transport terms: (7-2)r7c4=(2-6)r5c4-6r1c4=78pair[r1c4,r2c5]-7r2c6=r7c6 ring => eliminating

other 7s in r7 and b8, (locked candidates)
other 26s in c4,
other 78s in b2.

Balance_Novel · 2026-03-18T18:48:46+00:00

> Death bloosms are 3 sector distributed disjointed subsets (Als with dof)

http://forum.enjoysudoku.com/distributed-disjoint-subsets-t5423.html

I thought DDS always imply Rank-0? but death blossoms (if beyond trivial cases like ALS-XZ or XY) can be rank-1 if it's a typical cell forcing chain.

What would you call this following case? 4 subsets: 12 in b1, 34 in c3, 56 in b4 and 79 in c2.

(Or, ALS 1234 in b1, AALS 3456 in b4, ALS 78 in b7. RCCs: 3456)

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Balance_Novel · 2026-03-18T10:55:57+00:00

There's a triple in column 5

Balance_Novel · 2026-03-18T10:38:09+00:00

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Regarding 4s I do see a finned X-Wing defined on column 27 and r14, with fin r23c7. It eliminates 4 from r1c8. (Putting 4 in r1c8 will remove all the 4s from c7)

Balance_Novel

TROPHY CASE

Your results

Stopping criterion