Ultrahardcore 91 Trials Unleash Beeby


Here is the basics launch on Stefan Heine’s ultrahardcore 91, which quickly runs my two assisting solvers out of humanly practical removals. A Single Alternate Sue de Coq  trial reduces two clusters two one, and produces  two critical clues. In the next posts, solvers Beeby and Sudokuwiki can then carry the trial results along their distinctive solution paths. The challenging “exercise for the reader”, a logical solution path without trial, remains in effect for 47 and 91.

By the way, Sysudoku will refer to Phillip Beeby’s solver at philsfolly.net.au as “Beeby” because Phil’s work, and the intent guiding it, is far from folly. Phil joins Andrew Stuart in providing an accessible solver based on human technique. It may go beyond practical for most of us, but it’s very usable without going there.

Two clues and a cluttered grid greet us in Stefan Heine’s ultrahardcore 91.

No room for intuition here.

So imagine finding this ALS_59. It’s humanly feasible, but is the search through so many ALS a humanly practical thing to?

 

 

 

While Beeby finds this one,

 

Andrew Stuart’s Sudokuwiki finds the equivalent grouped Almost Nice Loop on the right, and the grouped ALS boomer below, before giving up,

 

Beeby finds the same removal by its version of the digit forcing chain, starting from 5r2c9 !  Assuming this candidate false, it’s the same southern forcing chain to 5r9c9 as above. Beeby’s northern chain, assuming 5r2c9 is true, is different, but the important point is the absence of any rationale for starting there. To find this, you must be searching for two chains from every candidate on the grid. Humanly possible, but totally impractical. But that’s OK, you don’t have to go where the code takes you, like your computer does.

With both solvers, we have definitely reached trial territory.

First though, on my 7-panel,   there’s a finned swordfish in columns 238. Victims in r8 see the fin, 7r7c8. If they were true, the fin would be removed and they would be false.  Potential victim 7r5c9 confirms the fin. That means, if this candidate is true, the swordfish that threatens it does not exist.

The removals in r7 are from boxline SEr7.

Now my problem is, the solvers are directed by clues, not candidates. I don’t edit their candidates. Fortunately there’s two small and disconnected slink networks, i.e. coloring clusters, and a potential Single Alternate Sue de Coq for a trials plan.

The blue/green and red/orange clusters interact in the East box. Only one of the four colors can be true, but red and blue together is impossible.

Logically,

not(red and blue) => orange or green

Just keep that in mind, because right now,

 

there are no uncolored 7’s to removed for seeing both orange and green.  The wall Cr5 sees one alternate 26 and is logically, 4(2+6)(1+9) if 2 or 6 is present and a naked triple 149 if 2 and 6 are absent.  We get to start the SASdC trial with a triple implying red (r4) and green (c8).   Are you coloring yet?

Together, SW5 and the coming NE5 will remove the last 5 from the SE box. The SASdC failure does not cancel any colors. It only confirms that

Cr5 = 4(1+9)(2+6).

But now the matching (2+6) must come from r4c1), removing 6r4c3 and leaving a naked triple 179 iin r4.

The SASdC trial was indecisive, but made progress. Red implies 9r4c4 and 1r4c3. And since red also implies green, a red trial tests a single 7 pattern.

 

 

 

 

 

 

Here is the grid as the red trial begins, with soon to be removed slink partners are left on.

 

 

 

 

 

When the follow up trace stalls, an XY ANL gets it going again,

 

 

By enabling a second XY ANL

 

 

 

 

The trial drags on, until red forces two 9’s in c7

It was a lot of work for two clues. In case you’ve forgotten, it was two orange 7s

After getting a restart from the trial clues, the two solvers go in distinctly different directions, both with instructive examples, but both winding up in the jaws of the expanded blue/green cluster. Next time, we’ll lean first on Beeby, without the ALS-wings, for a clinic on ALS_XZ and AIC.

 

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Ultrahardcore 47 Pattern Trial


This post bypasses Sudokuwiki’s  forcing chain parade with a value pattern trial, strengthened by the Single Alternate Sue de Coq analysis of the last post.  One pattern is tried and fails, leaving a coloring cluster. When the trial of the chosen color proves indecisive, the opposing color is taken to a contradiction, promoting the chosen color to the true one before continuing to the solution.

Value patterns are another avenue for decisive trials.  A value pattern is a possible placement of the remaining candidates of the value among the units. It is revealed as a freeform segmented line crossing the grid and hitting every candidate just once in each box and crossing line. Only one pattern is true. In a coloring cluster, the true pattern candidates are always the same color.

Here is the 3-panel as the forcing parade steps off. Going West to East, the freeforms show four possible patterns. We’ve colored the candidates to reveal the single blue/green cluster that is compatible with the possible patterns.  The red dashed freeform is one of two starting on blue, and prevents any other pattern starting green. The solid line blue pattern permits one of the two patterns starting green to complete a cluster. The cluster pair leave two orphans, candidates outside any pattern.

In the SASdC analysis of the last post, we learned that failure of the NWr2 = 374 would leave 37 in r3c1. That means tests of blue patterns will start with an extra clue 7r3c1.

Let’s test the NW SASdC, then the red dashed pattern, then with the blue/green cluster confirmed, the blue pattern.

Going back to the SASdC analysis grid of last post, here is the breath first trace of the NW single alternate Sue de Coq, testing NWr2 = 374. It fails with two 6s forced to the same row.

The trial failure leaves   NWr2 = (5+6)(3+7) and the SdC removal of 56r3c1.

Next is the set up for the trial of the cluster blocking pattern:

 

 

 

 

 

 

 

 

That pattern leaves both 4’s in W and r4.

 

 

Now in the confirmed blue/green cluster, we choose blue  for its extra clue. The follow up ends with a naked triple.

After Sudokuwiki pulls in a grouped AIC boomer, we go back to try green, just in case that test is more decisive. It is, reaching a contradiction.

 

 

Now back on the naked triple and grouped AIC boomer, it’s no longer a trial.

 

 

 

 

 

 

 

Next, Sudokuwiki finds this ALS boomer, starting with 7r6c7, around to the 5 group, then to ALS 5, out to 9r6c7 for an AIC wink.

 

 

 

Then a finned swordfish, plus boxline.  Victim 9r7c5 is in the fin box, and its removal triggers the SWr8 boxline. After SW57 => SW49, we get

 

a BARN(Bent Almost Restricted n-Set).

 

 

The remaining bv field invites coloring and the cluster expands by two traps, reaching another naked triple and E6.

 

 

 

The follow up confirms orange, and it’s over.

Unless a reader comes up with a trial free solution of 47, the report on Heine’s ultrahardcore 91 begins next week.

 

 

 

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Ultrahardcore 47 Single Alternate SdC Trials


This post does the basics on the second Stefan Heine ultrahardcore right page under review.  Then it shows eight ways to do the five advanced moves up to forcing chains, two of them duplicated by boomerangs. It closes by  identifying removals for three Single Alternate Sue de Coq trials, holding the trials in reserve.

Sysudoku Basic begins with 4-fill which resolves to a 3-fill.

 

Advanced begins with a busy grid.

The 4-chain ANL, has interesting consequences. The C4 and S7 clues trigger a E7t aligned triple boxline,  and a SEc7 boxline removing three more candidates.

 

 

The 4 removal also allows a naked quad in c6 to accompany the naked triple in the S box allowed by S7 removals.

Next, a grouped 6 -ANL removing 6r4c1 is extended into a second ANL removing 6r7c1.

 

Then a section starting at 5r1c9 is re-used for one of Phillip Beeby’s one-way chains. If the starter candidate 5r1c1 is true, 5r4c9 is false, of course. But if the starter is false, the chain forces 8r4c9, making 5r4c9 false, regardless.

The one-way is also a grouped boomerang, starting at 8r4c1. One-ways belong in the Sysudoku AIC building phase.

There are at least two more ways to remove 5r4c9.  Both Beeby and Sudokuwiki found the grouped boomer at left, starting on 8r4c9.

By applying ALS first, both see the ALS_65. The restricted common between ALS 56 and 235679 is value 6, and 5r7c9 sees all 5’s in both ALS. Since one ALS is missing 6 in the solution, a 5 in one of the value groups is true.

 

 

 

 


Another interesting cover is that of the ALS Death Blossom, at left. The victim  5r8c9  would lock 67 in ALS 679 and 9 in ALS 569, stripping the stem r7c5.

 

 

Sudowiki’s digit forcing chain is an ALS ANL. One terminal is the ALS 5 value group.

After this, Sudokuwiki manages three cell forcing chains and four unit forcing chains before it gives up.

Instead of going there, let’s look at a Sue de Coq trial variation proposed in  Sysudoku, the Single Alternative Sue de Coq, or SASdC. Here are three examples.

In Sue de Coq, the contents of a box/line intersection is dictated by two ALS in the two bent remainders. For a Sue de Coq example, suppose r7c5 became S2.

Then the bv 57 in the r9 remainder would prevent both 5 and 7 in SEr9.  ALS 589 in the SE remainder would prevent both 8 and 9 in SEr9. SEr9 would have to contain 6 and two alternatives, 5 or 7, and 8 or 9. Neither could be missing, because each ALS can supply only one value.

The contents of Sue de Coq SEr9 can be described as:  SEr9 = 6(5+7)(8+9).

It removes 57 from r9c2 without trial, because the 5 or 7 matching that alternative must come from r9c3. Similarly the 8 or 9 must come from the ALS 589, so the other SE 9’s are removed.

But S2 didn’t happen. Instead, in the grid above, SEr9 sees only one alternative ALS, the r9c3 bv 57. The logical description of SEr9 reads:  SEr9 = 6(5+7)(8+9) + 6(89+98).

But that is enough for an often effective trial. The bv does narrow the possibilities. SEr9 cannot contain 5 and 7, so either it contains 5 or 7  (the (5+7) above), or both 5 and 7 are missing, hence the second term. In a trial of SEr9 = 6(89+98)  a contradiction removes that second term, and a Sue de Coq is in effect, for the single alternative and its remainder. If no contradiction, the puzzle may be solved. The SASdC assumption is very specific.

Looking back to the full grid above, there are three SASdC’s available, and we can calculate the removals by contradiction for all three. But just keep them in mind. We don’t need to perform a trial unless it turns out that we can use one of those removals.  A red trial  failure removes 56r3c1, a blue one removes 5r7c1 and a green one,  57r9c2.  Work these out and you’re in command of SASdC.

Next time, ultrahardcore 47 is solved with a pattern trial. Want to do it again? A reader solution to ultrahardcore 47 without trials? Have your trial free solution path in for the post after next. Identify any solver used. I hope it’s short enough to publish here for you.

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Beeby on Heine’s ultrahardcore 3


This post highlights several moves from Phillip Beeby’s solver, which are alternatives to those appearing in the five earlier posts on Stefan Heine’s ultrahardcore 3. One result is a solution without the second blue green trial, left as a project for you. Just so you know, the post of June 30 has been revised to correct a significant error. Diligent reader Dov Mittelman detected it, and the modest blue/green cluster saved the day.

To Sudokuwiki’s APE of June 9, Beeby finds two alternatives. On the left, the APE  uses a bv and an ALS to eliminate combinations of 7r5c9 with candidates of r6c9. Under AICs, Beeby finds the AIC slink. If 7 is false, 9 in r5c9 is true, eliminating 7. The spotting significance is clear. Bv partner 7 does simple win-win logic for any 7 you can slink to, in E, r4 and c7.

On the right, 7r5c9 also falls victim to an ALS_XS. Note that the enlarged ALS has a single 5 and the restricted common with bv57. Its 7s align with 7r5c9,

Single candidate and aligned value sets are spotting keys in dealing with ALS.

Getting into the AIC building, the June 9 post reported the Beeby duplication Sudokuwiki’s Boomer 2. In the ALS category, Beeby finds an alternative that Phil labels an ALS–wing. It’s an ANL with links embedded in an ALS chain. Here is the first example in Gordon’s series, with Beeby description

ALS-wing of

(7=6)green – (6=2)red – (2=7)orange => -7 r1c9

I finally got the message. From now on, ALS value groups are represented as thick curves. The AIC slinks are internal slinks in the three ALS. All winks are group(restricted common) weak links. Here, one ANL terminal is the 7 value group in the green ALS. The other is the 2 to 7 internal slink in the orange ALS.

But why call it a wing? Structurally, it’s a chain, an ALS ANL of length not necessarily three.

Shortly after, on the next selection of the ALS category, Beeby recombines two of these ALS with a third one. The 7 trap is from cluster expansion, and 9r5c7 is an ALS-wing victim. Note the overlapped ALS and value groups.

The Beeby description is

ALS-wing of

(9=6)green – (6=2)red – (2=9)orange => -7 r5c7, -9 r5c7

This time, the ALS-wing is a confirming ANL. The green ALS 9 value group is confirmed. It’s a true group. One of its members is in the unique solution. Beeby’s solver misses that. The confirmation removes 9r5c7, yes, but it also removes both 9r4c7 and 9r1c9.

And Beeby’s 7r5c7 removal?  Beeby doesn’t color.

It’s justified, but the  attribution to the ALS-wing is misleading. Instead, it’s an ALS_37 defined by the overlapping green and orange ALS. 7r5c7 sees all 7s in both, so where is the restricted common? Its between the orange and green 3 value groups, also overlapped. One ALS must lose its 3s because 3r7c7 is true or false. Beeby should issue a separate Eureka for that removal.

 

 

When using Beeby, the elimination follow up, including boxlines, is done automatically and without documentation.  But some unusual follow up boxlines are recovered on your Deadly scan immediately after the move. After E9 => C9 below, invoking all Deadly obtained the boxline removing 3r7c12. The Deadly category includes, and is named after unique rectangles.

Here’s a Sysudoku project for you. Using a recommended fast browser, start Beeby with the uhc 3 grid after the red trial in the June 30 post. Load the clues, including orange. On every move, go through the Deadly, Simple ALS, Swordfish Fish, Simple AICs, Continuous AICs, Discontinuous AICs, Complex AICs sequence. Drawing curves is optional.

What I got was 3 chains catching up, to the diagram,  a hidden unique rectangle, an X-ANL, 3 AIC ANL, 2 simple discontinuous boomers, 2 complex discontinuous boomers, and 4 ALS_XZ.

That’s 13 diagrams, arriving here, where the blue green expansion traps 4r6c1, and calls for a second, red/tan cluster.

 

 

 

 

Adding the cluster, there’s bridging logic in cells:

r7c2: not (red and green) => tan or blue,    and r5c1:    not (red and blue) => tan or green

Read it either way, not red and (blue or green) =>  not red   or

tan or (blue and green) => tan,   red is wrapped and tan is confirmed.

The steep collapse confirms blue and it’s done, without a second trial.  If you’re going ultrahardcore, take up coloring. The strong link network is the human defense against seemingly endless searching.

Next time we finally get to Stefan’s ultrahardcore 47, shown at the end of the June 23 post.

 

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Readers Solve Ultrahardcore 3


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!

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Ultrahardcore 3 Falls to Coloring Trials


This post reports coloring trials that gain the solution after forcing chains when Sudokuwiki gives up.   This includes an example of a grid diagram illustrating the route to contradiction in a breath first trial trace. Then we close with a comment  distinguishing Sysudoku trials from the less than logical practice of trial-and-error.

Red is our choice for the trial, based on the color bridge of the previous post. Since blue and red is impossible, by assuming red, we can include the green candidates with them.

Red with green fails, and our diagram shows the shortest inference paths to the contradiction, two SW8 clues. Red forces 8r7c3, but creates a naked triple that wants 8r7c2. The arrows show how one clue leads to another.

Reading the trial diagram, red E7=>W4, and because red E9=>SE4=>SW4, and green SW3 removes 3r9c3, W4=>SW2->8r7c7. But W4 also removes the 4 group to create a naked triple removing 2, 4 and 5 fromr7c2, leaving the second 8 in W.

The diagram  is easy to derive from the breath first trial trace. Just follow the paths directly to the two results in conflict, adding supporting paths as needed.

The red/green contradiction confirms orange, but careful now,  not blue. Surprisingly, with the additional orange clues, Sudokuwiki remains stalled.

The original post found a grouped ANL and two irregular XYZ wings expanding the blue green cluster for the remaining trial. But alert reader Dov Mittelman pointed out that the wings depend on the elimination of a candidate by graphic typo.

To correct that, we go with the base grid for a blue/green trial, as left by the red wrap Blue or green?  Green removes few candidates and creates no clues. So the trial choice is blue.

After the breadth first follow up,

 

a real 154-wing, a 5 – ANL, and a 1-wing.

This path is duplicated by in the bypassed trial of the next post, but not with trial marking and breadth first grid.

 

 

 

 

 

Simple follow up now brings a decisive BARN, a naked pair, and a final breadth first collapse below

 

There are several types of trial endorsed by Sysudoku that we hope to illustrate in the ultrahardcore review. But one type we don’t endorse. Let’s take a bivalue cell, say r7c5. Now we do a “trial” based on the knowledge that 5 or 6 must be true.  If that is indecisive, forget it and try another bi-value cell. Otherwise try another with the clue and removals we just gained.

 

Repeat until the puzzle collapses then, go on the next puzzle. To this process, Sudoku lovers apply the term “trial-and-error”, to exclude it from logical solving.

We could have ignored all the information gained about ultrahardcore 3, and started trial-and-error as soon as all possible candidates were available. We could be checking Stefan’s answer, and claiming we have solved this one. Doing this, all we would know about ultrahardcore puzzles is that they take longer. It certainly isn’t logical to spend that much time to learn nothing. Computer’s do it, by following logically constructed code, but they are expected to learn nothing.

Next time, we do the coloring trial on ultrahardcore 3, before forcing methods, that we bypassed earlier.  Also we begin what logically follows a solution by trial, the search for a humanly practical solution path without trial. That leads us to several new things.

 

 

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Ultrahardcore 3 Schlagt Forcing Chains


In this post, we continue with Sudokuwiki into forcing chain methods, and their relevance to the ultrahardcore theme. Sudokuwiki stalls, as Stefan predicts.  Then we fashion a trial for a logical, but practical outcome.

On the grid after the red/orange traps of last week,  you’re looking at the three 6’s in c3 for a unit forcing chain. Where do you start?

How about with every forcing chain from 6r1c3? Compile a list of every force out, including the cell and the value forced out. The list starts with r1c3 and every other value in it.

Then from there include all the extended AIC. Start with every wink from r1c3 that is followed by a slink of any kind. You’re lucky there are none. But how about the wink to the other candidates in r1c3, like 8. That gets to three more cells, and to 5, and a bunch more. Does Sudokuwiki include 8 and those chains? I don’t know. Anyway, once you have your list for a second starting value, then you can limit the chain terminations of the third search to the terminations common to the first two lists. That describes the coding of a solver’s exhaustive search for unit forcing chains. With minor modification, it applies to cell forcing chains.

Humans can take advantage of all kinds of special breaks, like watching for 6r3c3 and 6r5c3 to force any of the other values out of r1c3.  Or from  6r5c3, landing on blue, and forcing blue out everywhere, including 3r1c3.   That gives us incentive to find r3c3 guilty of the same.  It is, by virtue of the same blue force out. We have our unit forcing, but there’s more. Two of the 6’s confirm green 3. Does 6r1c3 do that as well? Well, not quite. By the red forcing chain, 6r1c3 forces out 3r7c3, but there’s a third 3 in c3.  But the other two, by confirming 3r8c3, also force out 3r7c3, and we have a second unit forcing chain.

But a finding process is not logical if it is not exhaustive. Humans are not exhaustive in forcing chains, and we can’t be sure  about solvers until we analyze their codes. Just strike the “in forcing chains” for one answer to the Stefan’s “intuition vs. logic” dilemma, and to possible objections to the use of trials here.

We are here to demonstrate what humans can reasonably find and enjoy in Sudoku puzzles. Sudokuwiki was written by Andrew Stuart with much the same intent. It is used here with exceptions and modifications where easier and more likely means are found.

The construction of trials is as logical as other solving techniques in common use. You can logically object to a trial by demonstrating the logic that the trial conceals. In that sense, a use of a trial is inferior, though the construction of the trial is not.

Next on view is a unit forcing with four candidates, a quad forcing chain. The four cooperate, with three forcing chains merging into one. But the point really is, how many possible quad forcing chains are you going to look at before you come upon the next one?

 

 

Sudokuwiki’s next is an r5 forcing chain on the three 6 candidates. The inside killer is 6r5c1. The removal above allows 6r5c3 to reach 6r5c1. The 6r5c7 forcing chain is a bit devious, but you know what you need to do.

 

 

 

 

In a parting cell forcing,  Sudokuwiki uses its new 7 slink in forcing chains from 26r2c1 to 7r2c6. Candidate 7 has the straight shot that might induce you to search out the rest.

Now as Sudokuwiki throws up its electronic hands in disgust, you can lay your plans for a coloring trial. Next week, we’re going to pick a color and do a coloring trial. But remember we have a bridge between the two clusters. Red and blue together is impossible, so orange or green or both is true.

What’s your pick?

The next Stefan Heine ultrahardcore is coming in July. Got the book? Not yet, but need a head start?OK, one more time, here’s ultrahardcore 47.

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Cell and Unit Forcing Begins in Ultrahardcore 3


This post reveals yet another way to start a boomerang, and reconsiders the practicality of Andrew Stuart’s Cell  and Unit Forcing Chains, this time facing an ultrahardcore right page. We get to a choice between Sudokuwiki and a coloring trial enhanced by cluster bridging.

Returning to the last uhc 3 AIC building result, 4r4c2 is removed because the 4 value group in the ALS is confirmed. Sudokuwiki  has constructed a confirming ANL . The group, and at least one of its members, is true. The result isn’t a boomer, but the current issue is, can a human solver discover this ANL by scanning boomer starting cells?

The answer is  yes, but we have to be looking for a different type of boomer. We could start an AIC chain with the internal ALS slink in r4c7, anticipating a boomerang back to 4r4c2. That boomerang would have a starting line, not a starting cell. Do we still call it a boomerang?  It would be up to us to know, as we build the chain, that a wink into any other value set would close an ANL confirming the starting value set.

The new bv from the 4r4c2 removal enables a shortcut  XY-chain. Yeah, it’s a show off shortcut.

Speaking of shortcuts, think of the host of blue/green slinks a coloring cluster adds to the boomer starting pool. Sudokuwiki happily supplies one below, with a group filling out r4 for the second slink in boomer 6.

Sudokuwiki didn’t use coloring to find this, but the example suggests we put clusters in and carry them along as strong link network markers in very hard puzzles.

 

This cluster may be decisive in a trial, but let’s pass it up now, and. reconsider it later as an alternative to the forcing chain methods Sudokuwiki goes into next

In the cell fc method, find a cell such that any of its candidates being true forces a candidate elsewhere to be false. In this case, forcing chains from 2, 4 and 6 of r7c7 force 2r9c1 out. Cell r7c7 must have a placement, so 2r9c1 is indeed out.

Is this humanly practical? Even if you limit candidates to 3, how many cells do you have to examine to find one whose 3 candidates see the same victim? Solvers like Sudokuwiki don’t need to care how many.

Let’s color code the forcing chains to suggest a feasible order. Here I’d start with cells having a colored candidate, and a matching candidate in another unit seeing possible victims. That includes the black arrow. Then I would look for a forcing chain from another candidate in the starting cell to one those victims, the red chain.  Then with a cell in hand with two fc’s I’d go all out for the third candidate, the green chain requiring the ALS node.

Sudokuwiki’s next elimination is equally or more impractical, Andrew Stuart’s  Unit Forcing Chain.

In the Unit Forcing Chain, all candidates of a value in a row or column force that value in a target cell to be false. You  scan unit by unit,  looking for more than one of, let’s say, 3 forcing chains  of the same value to target the same victim. That event then triggers an intense search for another from the unit. Sudokuwiki does try for as many as four candidates. Such a case is called a quad forcing chain.

In this case, r9 has three 4-candidates targeting a single 5-candidate. The set of 5’s also qualifies for a unit forcing chain. How many such sets of three on this grid?

As a new cluster is added, netting two traps, a bridge between the two slink networks is possible. A scan for  common values finds 7 !  Red and blue 7 in the c8 make the  Sudoku logic assertion:

Not(red and blue), so logically,

(not red) or (not blue).

Whoa. Orange or green is true!! Maybe both.

If a candidate sees orange and green, it gets removed. As to trials,  if we try orange and it fails, the red and green armies are both true. I’m so tempted to look there.

So what do you want to do? The orange trial, or do we turn back to Sudokuwiki? Experience is that cell fc and unit fc are generally followed by more of the same. It may be different this time. The general rule is to hold the trial in reserve, while we go as far as we can without it. It depends on your tolerance for search frustration, but I’m bypassing that with Sudokuwiki, so here goes. It will tell us something.

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Utrahardcore 3 Lifts Off


The review of Sudoku ultrahardcore 1 right pages, a separate collection from (and tougher than) its left pages, begins with number 3, shown here post before last.  In this post, we follow Sudoku Basic and the Andrew Stuart’s Sudokuwiki solver along 3’s boomerang parade, noting its challenge to the human solver.  The solver, and possibly, comments from readers, back me up on humanly practical methods. Thanks again to reader Alex (review page) for pointing out this collection. And of course, Stefan Heine for creating it.

On 3, bypass is short and box marking is shorter, leaving a hard line marking and a monster level line marked grid.

 

 

A small network of strong links stand out in a relatively clear area, and coloring produces a trap.Nothing else stands out until we get to the last X-panel.

First, spot the slinks on the 9-panel. There are three. Now find winks connecting slinks.

 

 

There are two ANL on the 9 chain. All of this goes beyond the bv scan, because the few bv are connected only within r8.

Turning to ALS and bent regions, an APE turns up.

 

 

 

 

 

It’s an aligned pair of cells whose candidates are seen by pairs of ALS value sets. No combination of values across the cells are possible when seen by a pair of value sets, because such a combination would remove both values from the ALS.  Combinations 74 and 76 are contained in ALS 3467 and 75 in the bv r4c8, so 7r5c9 must go.

Finding nothing else in ALS territory, it’s time for AIC building, a stage dominated by the boomerang. It’s search intensive, because boomer starting candidates are numerous, and a human solver must try every type of AIC node to extend branching chains to unpredictable destinations. There are the 18 strong links on the grid, representing 36 possible starting points for boomerangs.

Scanning box by box in our normal order, we start with 1r2c2 looking for a wink from 1r3c3 to a slink. Possibles are one of the two slinks as a wink or an ALS containing 1 as a value set. Then we try the reverse with starting cell r3c3, looking for a wink from 1r2c2. Then it’s starting with 1r3c3 and a wink form 1r4c4. The search ends with a wink back to the starting cell.

The starting cell is critical in forming a boomerang. A candidates can be on a other boomerangs and fail to be a starting candidate for its cell. Here is Sudokuwiki’s first boomerang. We know it doesn’t scan the boxes in our preferred order, because the first one starts in East, on 9r5c9, and

the second one starts in NE, from 5r1c9. If we start a boomer from r3c8 and move along the path of boomers 1 or 2 , it fails on r5c9 (1) or r1c9 (2).

The boomer 2  7r1c9 removal turns 7r7c9 blue. The cluster expansion traps 7r5c7, as it sees green by cell wink.

Sudowiki’s boomer 3 is from 2r5c7. The trap shows up on it’s selfie.

In boomer 4, grouping creates a starting slink, showing that our list of 36 possible boomer starting candidates is incomplete.

 

 

The AIC is shortened to illustrate the shortcut that is possible because of the slink between every pair of opposite colors.

 

 

The removal brings an event of interest only, a dead BARN. The bent value is 4, and there is no victim. To get reported by Sudokuwiki, you need a victim.

Continuing in AIC building, the solver finds this AIC nice loop, built around an ALS. Would it be found on the AIC building boomer scan? The answer is yes, given some flexibility in the solver’s goal. Reaching 5r3c8, you would realize that the boomer wink into r1c9 not there.  It’s OK, though if you have in mind that a slink into the starting cell closes a nice loop.

Of course, you automatically examine every link in a nice loop for a victim seeing both ends. That nets 5r6c8. And you have to check every elimination for a boxline. This one is SEc9, from the 3 removal.

The nice loop structure is now  re-used in  boomer 5 from 9r4c4.

This week, we’ve passed up a trial, and our reward has been an exploration of the searching task awaiting the unaided human solver in a right page ultrahardcore  Sudoku.

Next week  with this picture, ultrahardcore 3 brings news that an AIC building search plan based on starting cells is not quite complete. Why is 4r4c3 removed, and is this a boomerang?

Next week also,Ultra 3 brings  Andrew Stuart’s Cell and Unit Forcing chain methods, and the searching task they present for the human solver.  According to Stefan Heine predictions, there will be trials.

 

 

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A Bracket Busting Dave Green 4-Star


Here’s the  basic trace on the Saturday Dave Green 4-star shown at the end of last week’s post. It survived the three Sysudoku Basic  phases: bypass, box marking and line marking. It’s the first I remember.

Let’s talk through the Sysudoku marking of the bypass. It’s short enough to bring up every attention point, but it’s best to read it as you mark your own grid.

In the 1 scan, the NW crosshatch has 3 cells for 1, the E dublex 2, and the S dublex 2. We’ll mark the box slinks in box marking. The 2 scan will create slinks in 4 boxes. Then for 3, a crosshatch into SW creates a slink and dublex to complete a wall with NW3 and a 4-fill in c3. We follow the slinks in NE and E, and return to scout the new NW wall.  What new dublex will it create? The value 7 is not in the wall, and the limited crosshatch produces NW7.

The 1 crosshatch into NW now leaves one cell for NW1. Now there’s a NW 3-fill, missing 5, 6 and 9. No sweepers, so we continue the 3 scan with cross hatches into E and SW. Slinks are bypassed for 4 and 5. The single 6 can take no action.

In the 7 scan, first a slink in W, and then clue C7 generating a double crosshatch for W7. Now looking in the other direction, a new 4-fill in c5. As we go, more and more 3 and 4-fills occur. Look to see if values in the side lines are missing in other boxes. Values 1 and 7 are not missing. The N6m slink is noted.

On every new clue, an automatic check for a 3-fill. W7 yields c3[689]. The rule is two hits on a cell  or two values on two cells. When a line 3-fill is not immediate, we post a fill string reminder.  Continuing the 7 scan, the last box is double crosshatched for SE7, Two 8 crosshatches for nought. On the 9 value, the NE crosshatch dublexes SE9, which supplies that second 9 seeing the 3-fill cells. We put the effects list in brackets to say it resolves a 3-fill.

Box marking is a routine posting of the slinks we saw in the bypass, until we get to the SE box on the 6: list.  As the 6 slink is added, my count verifies the match of cells and values. With no more additions possible, 5r8c9 is obviously true.  But the more direct analysis seizes on the new hidden pair 16, and two accompanying clues.

The effects of that list is extensive, but somehow the 4-star survives. Maybe there’s a side effect overlooked along the way. Read it out on your grid and if so, let me know.

Another  possibility is a slip up in line marking. Here’s my line marked grid after the close. Did I miss a subset?

Now from here, you can enter advanced in several ways.

 

 

 

My choice is unique rectangles, because a readily spotted pattern can be looked up in a table for corresponding eliminations. Here we have a type 3 where the extra contents of the ceiling corners forms a hidden pair eliminating 8r6c9, and a type 1 where either 6r1c9 or 6r7c8 eliminates 6r1c8 to create a double solution rectangle that Dave Green would certainly not publish.  The collapse from (E6, E8, NE1, SE1) is immediate.

Now advanced readers have waited long enough. The review of Stefan Heine’s Sudoku ultrahardcore 1 begins next post, with the ultra 3 shown in the previous post.

In his book of 500 puzzles, Stefan informs us, puzzles on the left page can only be solved with the most complicated methods. But no one will find a logical way through even one of the 250 on the right. To me, this sounds like a perfect setting in which to explore the quite logical Sysudoku concept of trials, so I’ve  committed  to 10 from the collection of upper right page puzzles.

In his subtitle “Intuition schlagt Logik”,  what does Stefan mean by “intuition”?  The term lacks the objectivity of “logic”.  In Sysudoku, “intuition” is not taken to mean “trial and error”, the repeated trials of largely arbitrary choices, confirmed only by the contradictions reached by alternatives. This technique does, in time, solve the most difficult of puzzles, but without insight into any puzzle’s logic.

Instead, we’ll take “intuition” of ordinary meaning to have a role in what has been identified as trials, and seldom invoked, in the blog.  A trial is two mutually exclusive partial solutions, one of which must be true and the other false. In a trial, intuition guides the decisions that our methods are not enough, and how we will construct a trial. Experience refines our intuition as to when a trial is going to be decisive. We want enough evidence on both sides, for the contradiction of the wrong choice to come soon, and the right choice to go all the way.

Sysudoku employs a different  trace method for trials. The normal trace is depth first, the squirrel going as far out as possible before turning back to a branching limb. A trial trace is breath first, an army of squirrels advancing out in inference lock step, until a contradiction is reached. Breadth first tracing provides a measure of just how obvious the contradiction is.  Also in trial tracing, when we get a solution but suspect there may be more, we try the alternative. There are several instances o occurring here.

After the coming long review, you’ll have 240 more ultrahardcore right page Ratsel to refine your intuition. We’re not saying that a solution by trial is as good as a logical solution without trial. On the other hand, when right page ultrahardcores fall by Sysudoku trial, we don’t expect to hear that logic has been somehow defeated.

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