Sharing Hanabi’s Secrets

This post details a fireworks analysis on “Hanabi”, shye’s fifth example. The first four examples used matching elbows alone, along with bi-value cells. This one uses forcing chains to identify the value winning placement in the neck cell shared by four elbows and a firework.

Hanabi allows the bypass a few clues, a naked pair, and two 3-fills.

Box marking adds box slinks (strong links), and line marking, line slinks and bv borders.

For elbows analysis, the line marked grid

is dissected by value in X-panels. X-Panels 1 – 4 display only show empty X-wings, but panels 5 – 8 reveal many elbows.

Fireworks proposer shye selects firework r9, SW, c1 on values 56789. There are no 9 – elbows. Here are the selected elbows and the 9-panel firework:

Returning to our “values’ version of the fireworks principle,

if the four 9 candidates outside the box are false, 9 is placed in r9c1. The four elbows compete with the 9 firework for placement in this common neck cell.  By including the 9 firework, we get five fireworks placing five values on r9c1 and four elbow cuffs.

In his forum post, shye designates r1c9 and r6c4 as base cells and r9c1 as a target cell and states that values missing from the base cells can be removed from the target cell, removing 56r9c1. That’s leaving out a lot.

The definition of shye’s firework base cell is unclear, but for an elbow, let’s say the base cell is the cell outside of the box that sees both cuffs. All the elbows above have the same neck cell. In every elbow above, if the base candidate is true, it is false in both cuffs and is therefore true in the neck of every elbow above.

The set of five cells, four cuff cells and the neck cell, must solve to the four elbow and 9 firework values.

Using a term introduced in Sysudoku to explain BARNs, this is a 5-set, an n-set with n = 5. An n-set is a naked set, but not a subset because more than one house is involved. The most common BARN is a 4-set in a bent area, the intersection of a box and a line.

Highlighting the cuff and neck cells, if the base cell candidate 8r1c9 is true, the two forcing chains shown here force the neck candidate 8r9c1 to be true. That means the base candidates of the other elbows must be false, because if true, they would force their value in the common neck cell r9c1.

Goodbye, 5r6c9, 6r1c4, and 7r6c4.

Starting with this, and working in trial tracing order, here is the trace to the point where we are resisted by

a locked rectangle of 59 bv cells.  However, with 7 already placed, and 9 in its cell, you can place 5r6c1 and break the rectangle to finish the solution.

Next week we continue with the full Sysudoku solver path, because it’s worth seeing, and to highlight the DIY benefit of examining your elbows.

Here is the solution, with some coloring from next week’s final wrap. In shye’s well chosen example, the elbows route is much easier.

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Fireworks 4 Stretches the Matched Elbow Rule.

After a challenging basic, we get two matched pairs on two values again, but this time, linked by a single bv. The second challenge is the application of the matched elbow rule of the last post in a more hypothetical situation. The fireworks result is not an immediate collapse, and we get  interestingly  parallel solution paths of ALS_XZ and BARN.

The basic is easier tracked than performed.


The 1-wing removals are made in line marking, before X-panels are constructed.  

Doing the X-panels for the fireworks elbows scan, we note the dead swordfish joining the now dead 1-wing, but what is the significance of having four matched elbows nested around the center cells of four boxes? I’ll say at the end of the post, but more on topic is the two matching elbow pairs on 2 and 8.ic is the two matching elbow pairs on 2 and 8.

Plotting the matching pairs on the grid with curves, there’s a 28 bv on the row and column between the cuff cells of the two matched pairs of elbows.

By the matched elbow rule, at least one cuff cell on each matching pair solves to  2 or 8.  The bv switches the true 2 or 8 to the opposite true value, so the cell on the other end must have neither 2 nor 8. By the matched elbow rule, that means the opposite cuff cell has 2 or 8, and the bv forces it to be the same as the true 2 or 8 we started with.  Regardless of which cuff cell we start with, corner cells r1c9 and r9c1 are limited to 2 or 8.

Limiting two neck cells to two values is not necessarily decisive. But solver Beeby takes us through some interesting steps to a solution. 8r1c5 is removed by ALS_98 (in black), or a BARN on 2789.

The 8’s attack continues with an ALS _28, or if you prefer, a BARN on 2368.

The cherry on top is another ALS_28, 

and a rare 5 – pole BARN on 23578, for C8 =>(NW8, SW8).

Then Beeby supplies a finned swordfish.

Asked for another fish, Beeby gets a “sashimi X-wing” with fin at r1c5. I shoulda asked for a simple AIC.

 Finally, the collapse. In the solution, you can verify what you concluded from the four matched elbows on the 1-panel. One set of opposing necks are 1.

Next time, we examine Hanabi, proponent shye’s fifth fireworks example in

It suggests there are additional ways in which combinations of elbows can be used to find candidate removals. There are no matching elbows, but elbows of four values share the same box and intersection cell. And there is a firework sharing this cell which is not an elbow.

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Fireworks 3 Matches Elbows and Bv

This post develops a property of a 2 value elbow match, and uses it with matching bv. Before we start on that, here are the elbow panels for the last week’s second example, just in case you want to compare it to your homework. 

The number of elbows is easy to count, and matches are easy to make, just looking at the horizontal or the vertical lines.

Here is the basic trace for the third fireworks example, leading to the line marked grid below.

We may want to do immediate X-panel only for very difficult puzzles, but in addition to a quick fireworks analysis, the panels will access X-chains, regular fish, XYZ, XY railway, freeform pattern analysis, and coloring assessment from the start.

On Fireworks 3 we have one elbow match, values 1 and 9. We could anticipate more as removals come in

Looking back at the line marked grid, there is something to note about the matching values 1 and 9, and their matching elbows. The the elbow cells outside the intersection box, the cuff cells,both see bv of values 1 and 9! These are on r1 and c9. Each elbow cell outside the box sees one of these bv. At least one of these, r1c1 or r9c9 or both, must have a 1 or 9 in the solution. The bv on the same line must have the opposite 1 or 9 in the solution.

That means 9r1c9 sees a true 9, either in an  elbow cell or a bv, hence its removal.

For future reference, let’s encapsulate the critical fireworks property at work here in the matched elbow rule:

The rule doesn’t mention bv. That’s because the rule is a general property of matched elbows that we might apply without matching bv.  Why is it true? It’s because two of the three cells of the elbow, the two cuff cells and the intersection cell of the fireworks rule, solve to the two matched elbow values. If one of the cuffs doesn’t, then the other cuff does.

By the way the removal of 9r1c9 collapses fireworks 3.

The fourth example in shye’s introductory post uses bv in a similar way, but instead of a single pair of matching elbows and two bv, there’s two pairs of elbows and one bv. The firework is easily spotted, but the logic applies the matched elbow rule in a hypothetical way. Next time.

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X-panel Elbows for Fireworks

This post illustrates the use of the Sysudoku X-panel for fireworks analysis, as introduced  in forum corresponent shye’s first two examples.

Here is shye’s statement of the fireworks principle:

What does this actually mean? Given a row, column and intersecting cell, if a value in the cell cannot appear in the row or column outside the box containing the intersection, that value is placed in the intersection cell. This holds because no other candidate of that value in the box can provide that value to both row and column. Shye’s formulation of his fireworks principle uses the word “candidate”, when he means “value”.

A practical way to apply this fact is to find row/box/column combinations with only single candidates of an intersecting cell value in the row and column outside the box. Then by the fireworks principle above, the two single candidates and the intersection cell mark three cells, one of which contains the true candidate. For a reason soon to be apparent, we’ll call this combination of row, column and intersection cells, an elbow. We’ll call the single row and column cells containing the value of the elbow,  the cuff cells. To go along with that, the intersection cell will be the neck of the elbow.

Commentator shye refers to elbows as fireworks, also uses that term for other forms restricting intersection box values.

Shye’s first example in illustrates the idea that, if you find three elbows matching three values at the neck and cuffs, you have a type of hidden triple.  The three cells must be reserved for these three values, so candidates of other values are removed.

Here is the Sysudoku line marked grid for shye’s first example, with matching  neck and cuffs for values 1, 2. There is no 3-candidate in c1 outside of the NE box, so value 3 is also limited to the three cells of the hidden triple, and candidates other than 1, 2, and 3  are removed. We must consider one armed elbows.

We depict elbows on grids by freeforms because they are not made of strong links.

How hard is it to spot matching elbows? It’s much easier once you have all elbows of a value on an X-panel, with all 9 X-panels side by side. For that, you mark outside singles having outside singles from the same box in the crossing direction. After finding two matching elbows, you could look for a third one – armed elbow to match them. Here’s an elbow map for shye’s first example.

Values 1, 2 and 3 matching elbows, with three cells in common, a hidden triple. Other candidates are removed from these three cells. The 9 x-panels make it easy to verify that these and no more elbows of different values match. In this case, the removals bring an immediate collapse.

For the second example in shye’s fireworks introduction post, the homework of last week, here is a basic trace. This example shows how four elbows on a four cell rectangle,  sharing two pairs of values, makes a hidden quad.

Now in place of the usual line marked grid, here is the corresponding set of X-panels. Find and mark the elbows, including four matching on two values, that allow 1 and 2 in r9c6, SW r1c1 and r4c1; and 3 and 4 in r4c1, C r4c6 and r8c6, removing, 5r4c1, 89r4c6, 951c6 and 56r9c1.

On the line marked grid, the “the matching elbows form a quad.

Values 1 and 2 are limited to r4c1, r9c6, and r9c1, and values 3 and 4 are limited to cells r4c1, r9c6 and r4c6. Thus the four values must be placed in the four corner cells, removing all other candidates, including a rival 8 in the center box.

The “quad” brings an immediate, but long, collapse.

Without the fireworks, this example keeps the ultrahardcore solvers at bay for 30 slides. My general conclusion from these first two examples is that, with X-panels, DIY exhaustive fireworks elbow detection is reasonable to do. A plausible  test of the frequency of results might be to see how many, if any, fireworks matches occur in the 22 ultrahardcore linemarked grids of the right and left page reviews. Volunteers, anybody?

Before you do that, let’s look at shye’s further examples of what to do with them. Next week, we start the new year with shye’s third example.

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UHC 397 Fills the Stocking

After two “simple” ANL, Beeby finds a candidate removal by pattern overlay analysis. On, this is requested under “POMs”.  Then a pattern analysis for The Guide offers many more removals using freeforms.   A finned swordfish gets us to coloring, with two clusters wrapped by a DIY unique rectangle.

Following the complex 1-way of the previous post, Beeby  continues with this very productive ANL.

The follow up includes a NEr1 boxline,

Then a simple ANL removes two 2 candidates.  With fish and other AIC requests denied, Beeby does a pattern analysis removal of 6r1c9, which triggers a NEc3 boxline removing 6r6c7.

Beeby explains the removal this way:

Between them r3c3 and r7c6 include all patterns of 3, so patterns of 6 which include both cannot be true. As a result, no pattern of 6 includes r1c9.

So how does Beeby know that? And WTH is a pattern?

A pattern is a set of remaining candidates of one value, with one member candidate in every row, column and box. You can mark a pattern by drawing a freeform, a connected set of lines from one side of the pattern to the opposite side, going through exactly one candidate per column and box. Similarly, by a freeform from top to bottom or bottom to top, through exactly one row and box.

A first step in duplicating Beeby’s pattern analysis with freeforms is to draw all patterns of 3. I would start them on c6, going left or r8, going up. From c6, we enumerate 6 patterns. Any 3 candidate not on any pattern, an orphan, is removed. On the other hand, any candidate on all patterns is a clue. All patterns go through r6c6 or r6c3, but we want rows and columns to be different, so that patterns of other values crossing all of them can be discarded, removing candidates. The fewer the cells other values must avoid, the more of their patterns we can discard, and the more of their orphans we can remove.  Beeby finds r7c6 as a second cell other value freeforms must avoid, because freeforms through r6c3 start from r7c6.

Now operating DIY, we would have to examine all freeform patterns of 8 other values. We don’t have to find all patterns of these values, just those that don’t include both r3c3 and r7c6.

Starting r1c9, ffs must hit r7c6, r6c7 and r5c4, then avoid r3c3, but then they can’t cover r8. The two patterns hitting both red cells are discarded. That’s Beeby’s removal of 6r1c9

All ff from r5c9 must hit r6c6, so they avoid r7c6, and don’t have to avoid r3c3.

Finally, the 6 freeforms from r6c9 must hit r7c6 and then r5c4. The one that supplies a 6 to r1 can’t get to c1. The one that hits r1c3 and avoids6r3c3 can’t get to c1. And the one that hits r3c3 and gets across hits both the red cells and is eliminated. So our freeform analysis finds another  removal, 6r6c9, plus all the orphans left out of the successful freeforms starting from 6r5c9.

We have only verified a particular removal and showed how freeform analysis can generate related removals. Finding all such removals would take magnitudes more calculations.

Here are the freeform analysis removals, along with a Beeby simple 1-way and its  naked pair r8s16 =>W3.

Clue W3 creates naked pair r5s48

A finned swordfish is enabled. Victim 4r6c3 is in the fin box. Potential kraken victims r3c36, r6c7 and r7c3 don’t see the fin, so for them, the swordfish isn’t there.

You know progress is being made when an XY ANL comes in. It’s also time to get get out the crayons.


two small clusters are started.

Starting an AIC from 5r3c5, when red 7 is encountered, any orange 5 on the red/orange cluster is a potential ANL terminal. We know there’s an XY removing chain to 5r7c9 without completing it.

Remarkably, the matter is settled when  a DIY unique rectangle removes 7r3c6, because it confirms 5r3c5 and 2r1c5, stripping  rectangle corners of extra candidates. Removing the offender,

N14 => C7 => orange and blue.

The orange and blue solution is immediate.

This post is the completes 10 left page ultrahardcores, the number selected in all reviews prior to the right page ultrahardcores.

In the multiple cluster solution paths of both reviews, there’s certainly evidence of interlocking slink networks in ultrahardcores that have to be resolved, one by one.

Kudos to Stefan Heine for these superbly constructed puzzles, and to the authors of the two DIY oriented solvers essential to my getting my reviews done

Next week, we look at the first two Sudoku presented on the Enjoy Forum to introduce a solving technique proposed by correspondent shye at on Nov. 3, 2021. Thanks to my friend Gordon Fick for alerted me to this thread.

Forum correspondent shye calls methods based on this technique “fireworks”. Both puzzles are immediately collapsed by a single application of fireworks, but the second one, shown here, turns out to be the equal of a Stefan Heine ultrahardcore right page without it.

Next week’s post illustrates how the fireworks method is supported by Sysudoku X-panels, and why it is best applied immediately after line marking.

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An AIC Parade by ultrahardcore 397

This post follows basic with a variety of Alternate Inference Chains. It is an opportunity to compare their effects and corresponding Sysudoku labeling.

Three bypass clues, naked triple eliminations, and the line slinks marked in cell corners, that’s the  outcome from Sysudoku Basic.

The line marked grid:

Sudokuwiki leads off with a 6 – ANL, a 6-chain almost nice loop. The victims sees the two terminals, one of which has to be true. The diamond marks the victim, and I don’t draw in the two weak links.

Then what Sudokuwiki calls a digit forcing chain, I call a boomerang from 1r7c9, and returning to wink into the starting cell, removing its target. Beeby calls it a discontinuous loop, meaning the strong /weak alternation is broken at one candidate of the loop.

The boomer is a form of almost nice loop, in which one closing wink is an internal wink in the return cell. This ANL is a boomer, but one returning to the starting bent region, not the starting cell.

Next  in UHC 397 is an ANL with groups at both terminals. One group is a value set of the ALS 167 in r8c47. The victim sees the value set edge on, and shares the box with the other group.

Now we get two boomers from 5r6c4, one branching off the other, to hit two candidates in the same cell.

The grouped boomer removing 6r9c9 is an extension of a 1-way removing 3r9c3 and 7r9c9.

With Sudokuwiki running out of moves and Beeby finding nothing simpler, a long complex 1-way keeps it going.

Starting at 3r8c1, a branch wink removes 6r8c7 to enable an internal slink for a long, grouped complex 1-way We’ll go from there next time, unless you can find a simpler way to bypass the 1-way.

We’ll go from there next time, unless you can find a simpler way to bypass the 1-way.

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A BARN Storming UHC 353 Finish

This posts finishes left page ultrahardcore 353, using several notable moves to expand a cluster. 

Sudokuwiki starts it with a grouped terminal ANL with line and box winks,

followed by a grouped confirming ANL barely aided by the cluster.

After the follow up

a regular swordfish

expands the cluster, trapping four candidates.

Expansion creates an XY ANL, that you can also label as a 451-wing, or as a 3-BARN.

Sudokuwiki finds a 4-BARN, a Bent Almost Restricted 4-set. Of the 4 values 1589 confined within the bent region Wc1, 3 are restricted within units, 1 in W and 8,9 in c1. The set is “almost restricted” because Wc1 value 5 is not restricted to a box or a line. An outside candidate  5r9c2 sees both  candidates of the unrestricted  value. The Guide explains the logic.

Now the cluster expands to trap 5 and claim NW7 and SW2,

and a shortcut XY chain wraps green.

An easy wrap, but so hard to get here.

Next is left side ultrahardcore 397.

Want to make it the last one, or do we add two more?

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Softening left page ultrahardcore 353

Here is another puzzle the DIY oriented solvers suggest is too hard for human solving, even as the moves they find are too instructive to ignore.  This post gets quickly to ALS methods that demand exhaustive enumeration from the DIY solver, and gets to coloring. A second post will exploit the expanding cluster in much less demanding ways to finish UHC 353.

Here is the grid after a very productive bypass. The givens are in the more formal font. The less formal is Bradley Hand ITC representing handwriting.

The basic trace shows clues derived from clues, to a point where 3-fill lines appear. One is resolved for the last basic clue.

The first advanced move is a 4 – ANL you could spot in line marking, when you are filling c2 and marking its 4 slink.

Next, spotting this  hidden unique rectangle is very unlikely. The UR slinks permit 9r8c4 to force 6 into adjacent corners, making a solution rectangle in which 6 and 9 can be interchanged for second solution. UR is based on the assumption that a Sudoku composer will not let that happen, and without a solver, that too, is unlikely.

Then the grouped 1-chain Sudokuwiki gets 1r8c7. Beeby gets that with a finned 1-wing.

Then Beeby goes to AIC building for a grouped ALS ANL with a 5 value set as an ANL terminal.

Now both solvers are held off, except for Beeby’s ALS methods.  Either you have the rare talent of spotting them as needed, or you go through the exhaustive ALS incremental mapping procedure outlined previously in this ultrahardcore series.

 The first Beeby find is this ALS_95.

Next, a very related ALS wing, an AIC of ALS nodes strung together just like an XY chain.

Finally, a remarkable pair of  ALS wings. 5r5c1 is a single in blue and in orange ALS. 9r6c1 is a single in orange.

The two chains are detailed below.

The chain (1=5)blue – (5=9)red – (9=1)orange

is a confirming ANL removing 1r8c1. 1r25c1 is a value set in blue and orange ALS.

The chain

(8=5)blue – (5=9)red – (9=8)orange

is an ANL removing 8r3c1. 8r25c1 is a blue value set and 8256 is an orange value set. The Wr2 boxline removes 8r2c9.

Now the solvers’ paths continue more normally.  A second hard-to-spot hidden unique rectangle is followed by an easy 2-chain ANL that wipes the UR away if it comes first. It doesn’t matter, though because the 2-chain clue does the UR removal.

The 2-clue also enables a 9-chain ANL.

Beeby teams the box ALS with a new line ALS for an ALS_98. This might have been a demonstration that the incremental ALS _XZ construction can fail, but this one works even if 9r6c4 has not been removed, because the grouped common is valid.

Sudokuwiki gets to the same results with a different removal, with an APE.

Now we have reached coloring and have a trap, but UHC 353 is not giving up. So let’s close it out from here next week.

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UHC 309 Rounds the Bases

The ultrahardcore review solvers illustrate a diverse solution path, before and after AIC and cluster building.

Here is the grid after the bypass and box marking.

Beeby found a naked single SE4 in r8c7 in its basic.

The Sysudoku Basic doesn’t look for them. See what you have to do to identify it here. Now would you want to do that for every cell?

In Sysudoku they come from construction, not search.

In the basic trace the naked single is the SE4 in the 8: trace.  Beeby’s hidden pair is NW46 in line marking, where it’s a reaction to NE4.

X-wings are caught in line marking. As you reposition candidates to mark line slinks, you glance over at parallel lines for matching pairs. When there’s more than a match, it could be a finned X-wing. Check out Sudokuwiki’s covering X -chain. It’s also made of line slink matches.

After basic, the lowest hanging fruit is usually unique rectangles marked by naked pairs. Of course ultrahardcores are gonna have hidden URs, with extra candidates, and enabling slinks in the UR values.  UHC 309 starts with these two.

For how it works, assume the victim is true and follow the consequences of the slinks.

To find one, watch for the embedded UR values in rectangles over two, not four boxes, and mark the slinks. Then pick the potential victim that produces a rectangle of UR values that can be interchanged without disturbing outside cells.

With that, Beeby begins a series of simple chain ANL with this pair. The black AIC can begin with 6r5c2 or 6r4c8. The black/red AIC may appear to be 1-way as an extension from 6r5c2, but it can be started from 6r8c1, and is two-way.

The AIC leave only 6r5c2 for W6.

This long ANL can start from either end as a 1-way, but finishes as an ANL. The ending is unpredictable. In DIY, all you can do is keep it going, but you might hold special nodes like ALS, or complex 1-way branches for the case where the simple chains produce nothing.

This one produces C2, and in DIY AIC building, it’s definitely better to suspend network branching to exploit the new clue.

The simple chain series ends with two long AIC ANL in the West boxes. First, a reverse bv AIC removes 4r4c1, gaining C4 and S4. C4 leaves a pointing pair r45c6, removing 7r7c6.

Now the  AIC is cut back from r1c1 to r1c4 and extended from 7r4c2 to 6r8c1 for an 6 ANL. If this were found first, the 4 clues of the 1 way would have occurred with the ANL.  We now get NW4 and NW6,

and what Beeby labels as a discontinuous loop. It’s not a 1-way, but another ANL with a group as a terminal. This brings to mind a reason why a Beeby sequence is not going to duplicate your DIY sequence of moves. Beeby’s is governed by its labeling of results, which is controlling what the user asks for. The DIY sequence is governed by the solver’s selection of AIC starting points, made without knowledge of AIC building results

Next is the first true 1-way. The 1-way logic  of 3r4c8 being false regardless of 3r7c8 doesn’t apply to a chain start from 4r4c8.

As  the second 1-way opens two new coloring links, a cluster is finally started, and expands as a finned 3-wing adds another 3-slink.

There’s a a lot to digest on the next grid. The 3-slink connects a second cluster. Beeby’s third 1-way is an ALS AIC. It must start at 6r8c1, with the assumption that 6r8c1 is false. The AIC then confirms 2r9c1 is true to make 6r9c1 false.

We’re not quite done. Beebe finds a long chain from 5r6c8 to 5r8c5 for an ANL, but noticing the cluster entry on blue 8r2c8, we slink directly to green 5r8c5 for a coloring shortcut.

Now the blue/green expansion traps 1r1c5, and Beeby’s long 1-way through the cluster is largely bypassed by coloring shortcut. We could slink directly from blue 9r9c5 to green 7r8c2, but that would leave no reason for starting with such a slink.

Still in search of a wrap, we get a long simple chain ANL with a coloring shortcut. In DIY the advantage of coloring is that once the chain enters a cluster, you get many options for the next slink.

Now, after after W7 implies C7 and SW2, we get three traps and E5 by red/orange expansion.

The 8r6c8 removal will leave blue and orange 8 alone in the West box. They are slinked, merging orange and green.

In the merged cluster, two green 1’s are forced in c3, wrapping green.

Next is 309 + 44 = 353

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The UHC 265 Wrap Up

This post continues on the solution path for left page ultrahardcore 265. We begin where we left off, with the two solvers synchronized and with the first preliminary coloring cluster on the grid.

AIC building begins with this boomer from 5r3c3, followed by

Next, a 3 – ANL creates a productive NE48,

a 7 -ANL brings two clues,

and a 5 – ANL adds two bv. A hidden unique rectangle is carried by 5 – slinks.

With this wide ranging 9 – ANL, I looked for a green cluster candidate seeing a 9 for another removal, or a lite trap.

After Beebe finds this 6 – ANL, it’s clear that green 8r9c2 makes 6r7c3 green lite, and the two removals are lite traps. One creates naked pair r3s49.

Expansion traps 4r4c4 and 3r1c8, then a cluster shortcut ANL removes 1r8c2.

After the expansion, maybe you find the ANL removing 8r8c4.

Or maybe you see it as a green lite trap.

Anyway, the expansion brings a trap and the trap, a SWr7 boxline.

.One more expansion and pick your shortcut from blue 5r6c2. Either to green 1 that sees 1r7c7. Or to 1r8c9 that removes 1r2c1.

With that, the expansion wraps green in r8 and c6.

Next time we start left page ultrahardcore 309. Are we stopping there with 10, or going for 12? How many long, multi-post two solver DIY technique solution paths can you stand?

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