This post returns to the Stefan Heine’s utrahardcore  3 solution path just before cell and unit forcing chains. It turns out that an immediate coloring trial reaches a solution without them. Then we report how two Sysudoku readers solve it without trial.

Boomer 6, the last one before cell and unit forcing,  expands the cluster slightly, turning 6r5c7 blue. Looking at the blue/green cluster, the green trial goes nowhere.  Only candidates in the same unit of the same value are removed.

But, as in the revised blue/green trial of the previous post, the removal and added blue 6 makes blue a wrecking ball by comparison.

Combing through the wreckage, there’s that regular 154 wing and a 5-chain ANL

You also might like to have this 1-wing.

And  do take along the BARN, as well.

And this follow up trace to an early trial solution, on a modest coloring cluster, without forcing methods.

In my post of June 23 readers were invited to do the color trial after the Sudokuwiki  cell and unit forcing chains, ahead of last week’s post. Instead, two of my Sudoku friends came up with ultrahardcore 3 solution paths without forcing methods or trials!  First, an unexpected collapse of ultrahardcore 3, courtesy of Dov Mittelman,  as the trial is to begin.

Two slinks connect in 4r4c7. We can get a boomer or a confirming ANL if we can get from 4r4c3 to 4r7c7. In SW, two 4 groups are strongly linked, and the SWc12 group teams with a 5 box slink to form a hidden pair SW45. That doesn’t get us to 4r7c7, but if 4r4c7 is false, the pair excludes 8r7c2 and confirms 8r1c2.

But wait. The chain branches at 4r4c3 to a NWc3 4 group that 8r1c2 makes false, along with 4r1c2. Merge the two into a NW group of three, with a group slink to the 4 group  NWc1. If 4r4c7 is false, NWc1 is true and 4r9c1 false. The hidden pair confirms 4r7c3 and 4r7c7 is false.

So if 4r4c7 is false, so is 4r7c7. No 4’s for c7 unless 4r7c7 is true.

Following up, we now have the orange trial candidates of the last post, and an extra clue E4.

After the cluster expands with a host of traps,

there is a multi-leg collapse which curiously, puts off the wrap of green as long as possible.

Dov’s branching AIC is a bit extraordinary, but entirely logical. If the starting candidate is false, the team-up hidden pair, and the merge of a group for a slink on the fly to rob another group, all combine to show that the starting candidate cannot be false.

The other solution, from Gordon Fick, offers a very long path generated by Phillip Beeby’s solver at philsfolly.net.au. Rather than the point of trial, Gordon had to start at the beginning, since no clues were generated by Sudokuwiki to that point.

Sudokuwiki has a fixed sequence of methods, which it repeats from the beginning on every move. On Beeby’s solver, the user chooses a category and level of method for each move, and gets a report if nothing is available at that point. The Sysudoku advanced repertoire is covered by categories Deadly, AICs, Fish,  ALSs and maybe lower level POMs. Medusa coloring and Sysudoku pattern analysis are not included, but ALS methods are well supported.

Gordon stayed with AICs and ALSs and largely covered the chains and boomerang parade just seen in ultrahardcore 3, but then, the list began to go beyond these with a unidirectional chaining technique that Beeby calls a complex  discontinuous loop.  With Phil’s solver, you can reproduce a similar Eureka list, but the 44 grid diagrams of Gordon’s list are way too much for posting here. Like Dov’s solution, the Beeby path is fodder for analysis, for innovative humanly practical methods, and for spottable entry points to them.

Here is an example from the list that Beeby describes as a simple discontinuous loop, listed as

(4=7)r4c3 – (7=5)r4c8 – r3c8 = (5-9)r1c9 = r5c9 – r4c7 = r4c4 => -4 r4c4.

From the Eureka descriptions, the AIC generally start on bv, and end on a candidate seen by the starting candidate. The AIC is unidirectional,  giving more options, since links do not have to be reversable. So what is the governing theme for spotting these? If the starter is false, the victim is false by AIC, and we know ahead of time that if the starter is true, the victim is false. So the victim is logically false.

Does this method require a bv to start? No. It requires only a slink partner. Is there always a potential victim? Yes, the starter always sees at least one potential victim. Don’t ask how many times the starter fails before the next one succeeds, on average. This is a solver method.  Of course this is also a boomerang, starting on 3r4c2.

Phil’s complex discontinuous loop uses the same final logic, but presses the unidirectional advantages further. Not as far as Dov does in the above, but significantly far.

Here’s an example that bumfuzzled yours truly until I caught on to the complex discontinuous loop, which Gordon and Beeby descrbe as

(7=4)r4c3 – r4c7 = r7c7 – r7c2 = (4-8)r1c2 = r1c3 => -7 r1c3.

When the AIC gets around to 4r7c2, we can  selectively ignore 4r5c2 and 4r6c2 and use the one way slink to 4r1c2 to show that if 7r4c3 is false, so is 7r1c3. But 7r1c3 is just as false if starter 7r4c3 is true, so out it goes.

In Phil’s  complex continuous loop, we can use any of the candidates along the AIC that are true when the starter is false. Or not use them, whatever. The logic is still valid.

Well, Stefan’s challenge has been met with ultrahardcore 3, whether you count trials or not. I’m happy we went there, and there’s more to follow up on, such as: Can the 44 diagram list be shortened?

Before moving on with the review, I have to spend one last post on 3.  In an attempt to shorten the Beeby solution list, I prescribed a sequence of Deadly, AICs, Fish, and ALSs on each move, consistent with the Sysudoku Order of Battle. It provides interesting alternatives to the AICs and ALSs Gordon reported, and shortens the parade of solution grids to 13, before ending on a coloring bridge!