A Long Path Begins in ultrahardcore 311

This post begins a series of three, reporting moves by the human oriented review solvers Sudokuwiki and Beeby on Stefan Heine’s UHC 311.  Maybe you can shortcut this long path and a trial, but it allows me to illustrate human engineered ALS XZ finding and ultimately,  a new pattern analysis trial technique.

Ultrahardcore 311 yields no clue in to the bypass, and doubles down in box marking, leaving basic with two naked singles, a naked pair, and a more typical number of slinks and bv.

The line marked grid is remarkably playable, considering the very long fill strings.

Sudokuwiki starts advanced with a grouped “digit forcing chain”, more accurately classified by Beeby as a 1-way from 4r3c6. .  The starter sees the victim, and if true, it removes 4r4c6 directly. If 4r3c6 is false, the inference chain confirms 6r4c6, with the same effect.

Another dfc by Sudokuwiki notes. It is an AIC ANL, in Sysudoku Advanced sequence, and would come in AIC building. It could start at either terminal.  For simpler graphics we  omit the closing winks on the almost nice loops.

This grouped boomer from 4r3c6 implies N4.   Beeby has already found N4 in basic, where explanations are not given.  Want to go back to the line marked grid and explain how Beeby did that with basic operations? I missed it.

Another dfc AIC ANL, or what  Beeby calls a “simple chain”.

Sudokuwiki finds this ALS_65!   And then gives up. 

The ALS spotting problem for the human solver is the large number of ALS, and the difficulty of seeing ALS_XZ pairs at once

A systematic approach is to maintain an ALS_XZ map showing only the ALS having two value groups that are single or aligned for possible Z and X with other ALS.

Here’s an ALS_XZ map at this point. Every ALS is contained in a line or box so three or four colors are enough to distinguish them within their unit.

The Z value groups in each ALS must see each other completely. The X value groups must be seen completely by some outside candidate.

In r1, no curves are drawn for r1c47 ALS 236 or r1c457 ALS 2356. Each as a Z=6, but no X restricted common with r8c1579 ALS14568.

Instead of redrawing the ALS_XZ map on each move, a more practical approach is to bring the updated ALS_XZ slide along as you solve, updating it on each move, and keeping an ALS on the map as long as a potential Z and X value groups are available. 

Sudokuwiki found single 5 of r1c247 ALS 2356. Its ALS XZ algorithm must do some kind of limited search. Beeby’s ALS search is more exhaustive.

Next week, we continue from here with Beeby on this long November march.

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Ultrahardcore 267 Downed By An SASdC Surprise

After two interestingly matching sequences in this post, Sudokuwiki and Beeby toss in their towels. This allows an exploration of an offbeat version of the exclusively Systematic Sudoku Single Alternative Sue de Coq. The trial formulation is different, and the trial is the first in the review to capture the solution instead of a contradicion based removal.

The later right pages of ultrahardcore review have become tougher on the bypass. And line marking doesn’t usually start with 5f: lines. Fortunately, a hidden pair clears a little of the congestion

Advanced methods start with a 5-chain ANL, by both solvers, and a Sudokuwiki 597-wing.

Beeby does XYZ-wings as ALS_XZ. Here’s the ALS_57.

Beeby’s next step is an ALS full Nelson, its ALS Wing It’s not the imbedded AIC that’s hard to spot, its the 3 or ore ALS with single or aligned value groups.

Sudokuwiki does it as a Death Blossom. The victims grab 9’s from both ALS which in turn grab both values from the stem r2c6.

Beeby flashes its ALS advantage by going one ALS value group better in the same ALS for clue C3

The assault on the 9’s continues with an AIC  boomer from 2r7c5 and a unque rectangle, type 4. In Beeby notes the boomer is a simple discontinuous loop, implying a unforced 1-way from 9r6c5

The next  Beeby ALS wing is a ANL with ALS value group of four candidates. How do you define the problem to have the code look for such a thing?

The two solvers push their bailout buttons after one more jab at ultrahardcore 267. Beeby ends with an ordinary 1-way in which finds 4r6 is false when it sees a true 4r6c4, when a false 4r6c4 starts an AIC making it false.

The Sudokuwiki contribution is the 5-pattern orphan 5r1c1.

Here is the 5 panel showing many West to East freeforms from 5r3c1 and the failure of any from 5r1c1. Reason through the why not and you’re onto patterns by freeform.

The solvers having had their say and not much to color, I’m on the hunt for a trial. And theres a possible Single Alternate Sue de Coq at Wr5 as marked above. The contents formula is

Wr5 = 8(2+4)(7+9) + [7(89+98) +897 ]

That second term is a bit awkward to test. The progress already made on r5, and tracing technology, gives us a way out. Either Wr5 = 8(2+4)(7+9) with 2 or 4 present, or if 2 and 4 are both absent,  then in r5, Wr5 is a naked triple, r5c6 is 5, r5c9 is 4 and r5c4 is 2. A trial failure doesn’t gain much, only a 4r5c9 removal, but given the investment so far, we give it a shot. And wouldn’t you know it, this time, we reach no contradiction, only the solution.

Next week, we start ultrahardcore 311.

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A Coloring Merge Overwhelms Heine’s ultrahardcore 223

Continuing from last week’s post, Beeby’s  parade of ALS chains merges the clusters and a near BUG (Bi-Value Universal Grave) wrap finishes UHC 223.

Beeby does an ALS_12 with an aligned Z value group in r1 and a single Z value in the green ALS. In a DIY hidden unique rectangle, were 3r3c5 to be true, so would the adjacent corners be 8. Then 8r3c6 forces 7r8c6,   a value interchangeable rectangle, and an obvious double solution.

Continuing on a second 2-chain ANL we believe this may end, after all.

Beeby finds two ALS-wings, black and red, packaged in three ALS. This remarkable structure raises the prospect that other ALS aided ANL are invoked in the Beeby solver, beyond ALS wings.

Beeby follows this with two more ALS-wings. Here, one ANL terminal ALS value group is a single candidate, and the other terminal ALS value group is a box group. In the previous one, all terminal value groups were line groups.

In a surprising remap of the grid, a four member ALS-chain removes 3r3c3, with ANL terminals being a bv candidate and a box group. The removal triggers a Wc3 boxline.

Cluster expansions now show a merge:

red =>blue and orange => green.

The two clusters merge

Beeby doesn’t do color logic. On the merged grid Beeby finds this removal as a discontinuous loop, from 5r5c5. In coloring it’s a regular trap. 5r5c7 is false regardless of which color is true.  If blue, it sees a blue 5. If green it loses its cell to green 8.

The spotting rule is seeing one color and living with the other, out you go.  A   bit like marriage. As 6r5c7 turns blue, 6r5c9 is trapped.

With this extensive cluster, a color trial would resolve 223 immediately. Beeby brings back a variation of that four node ALS-wing, and a NWr1 boxline. Actually, you can leave out the orange ALS and make it an “Other” type of ALS-wing. The NWr1 boxline does the damage.

The follow up brings the cluster to the state below.

where NW3 erased 2r1c2, turning 2r9c2 green and removing 1, and turning 2r9c7 blue, springing the “home with blue and seeing green” trap on 7r9c8.

From there, 6r9c4, 6r7c9 and 6r3c7 are trapped, 7r9c3 turns blue, removing 1r9c4, for a blxline 1r7c9, establishing SE7 and wrapping green in r7c4.

The fully expanded cluster places everything. This time the human oriented  solvers of the review were stymied, until a simply conceived but arduous trial enabled one  solver to reach a coloring solution to ultrahardcore 223.

Next week it’s Stefan Heine’s UHC 267. It may have only one post on the trial schedule, because a trial solution comes next week. You can win it more space by sending in your description of a trial free solution. But in case you don’t have one, here is the grid for the following review puzzle, Stefan Heine’s ultrahardcore 267.

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Beeby Carries UHC 223 to a Second Cluster

This post reports how the Beeby solver comes to life after the Single Alternate Sue de Coq trial , and nibbles Heine’s ultrahardcore 223 to a second coloring cluster.

Beeby notes the ANL isas a simple chain, but it’s AIC building class, with two cell wink nodes.

Then a relatively simple ALS_63,

is followed by the ALS_42, whose construction challenges drawing, and much moreso, spotting.

Now imagine spotting this double ALS_68, so called because there’s two restricted commons. Each ALS gives up a value, one 6, the other, 8. Other value sets in each ALS contain a true candidate.     



The naked pair C49 implies W4, and a “discontinuous loop” takes another nibble. As in ANL, we can omit  the weak link  closure that breaks the AIC sequence,. The wink  from 1r2c6 is direct, “seeing” in c6 when it is true..  The wink from 9r6c7 comes from the 1-way AIC when 1r2c6 is false. This chain is used again later, when intermediate removals make it more damaging.

Meanwhile Beeby’s “complex” chains get a workout. These have to be  1-way’s because the branches that make way for the chain exist in one direction. It is Phil Beeby’s baby. Humanly possible, under sufficient pressure.     

Leaving from 1r4c6 one branch removes 1r9c4, and the other, 2r9c2, allowing the chain to confirm 1r9c2, if 1r4c6 is false, that is. Either 1r4c6 or 1r9c2 is true.

A second discontinuous 1-way nibble,


and another discontinuous 1-way,

and Beeby throws another complex curve ball. No need to trace this one out for anyone.

The complex removal enables the 2 chain ANL,

and the discontinuous 1-way from 5r8c5, whose 5r8c2 removal leaves a NW c1 boxline removing three 5s in NW, which picks up a 5-wing removing 5r9c4.

Now the hognose chain returns as an ANL, but with a second cluster. The clash of red and green in r4c2 means that orange or blue or both are true. The clusters share value 3. Any 3 seeing orange and blue is false.

Let’s save the finish of this overlong path for next week. If you go back to the last clue, you can run Beeby(philsfolly.com) and see how the action looks in text form.

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Single Alternate Sue Unlocks ultrahardcore 223

This post reports how both review solvers run out of humanly practical options, and the arduous SASdC trial that unties the knot.

At the halfway point of the review UHC 223 starts tough, conceding four clues in basic.

Those 6f: to 8f: lines and their correspondingly long fill strings map onto a severely crowded grid.



Sudokuwiki leads off with two of its digit forcing chains, first a group mittened AIC ANL,





then a digit forcing chain boomer from 4r4c1. Then, being restricted from using cell or unit forcing chains, Sudokuwiki runs out of options.




Similarly, Beeby dries Similarly, Beeby dries up after this unlikely ALS_68.






There is a ray of hope in a nearby Single Allternate Sue de Coq, namely

NEc2 = 4(2+3)(6+9) +649.

The bv23 match in both remainders brings a clue and 8 candidate remainders. Having slogged through the review this far,  you know what comes next.

It’s a trial of the second term 649, which adopts blue and gathers enough clues to get the solvers working.

The trial continues with this grouped ANL, and a band of helpers.







First an XY ANL (black) removes 1r9c2, creating the naked pair SE25.

Then the ANL is extended to make the terminals 8 (red), removing 8r7c3.

Then the ANL is extended to make the terminals 8 (red), removing 8r7c3.

And of course,  a BARN.


The beat goes on with another XY ANL,


and after a brief follow up,










yet another one.



Trials don’t normally require a series of  AIC eliminations, but the ultrahardcore are not normal.


It takes a bit  more follow up to reach a contradiction, a second 2 in the NW box.

Since NEr2 =  649  fails,

NEr2 = 4(2+3)(6+9) . The (6+9)  alternate comes from r2c2; the (2+3)alternate,  from r2c4 and r1c8.

Next week, we go on without these 7 candidates to a solution to 223. If you already have your trial free solution, then after sending in your guidepost description to have it published here, you may want to get started with ultrahardcore  267.


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Two Clues and Pattern Coloring Beat 179

This post carries a pair of clues, gained a trial of an alternative pair of clues,  to a solution to Stefan Heine’s ultrahardcore 179. The resolution requires a pattern analysis on the 5-panel, aided by two coloring clusters.

Going back two posts, here is the full grid before the trial. So happens, there’s an empty BARN that Sudokuwiki did not report, because there are no victims.






The follow up to the new clues 8r2c9 and 8r5c8 includes a boxline elimination.


In the unique rectangle type 1, Beeby removes 7r1c6, but we can also remove 5r1c6. Either 5 or 7 being true generates a deadly rectangle.







Beeby continues with the ALS_43 below





and the overlapped ALS-wing to the right.

The complex 1-way takes less time than a regular AIC on Beeby. I think it’s because the full use of one way AIC makes much more of each opportunity, and more are available.  When you ask for one, you get the first one Beeby comes to. This one starts on 5r1c9, suppresses two 5’s to make 5-slink in the SW box. It looks like an ANL, but is not. It depends on the 1-way AIC. If the starting value is true, both victims are false, and if it is false the AIC turns on 5r3c4, and both are false.

There’s much the same structure in the second complex 1-way. There’s some shifting because we’ve added a cluster and added slinks.

Coloring brings a trap in r1c3.




Next is the ALS_36, with an aligned 6 group,





and an overlapped ALS_57.





One more complex 1-way,  requiring only one stiff arm block to get the 1-way AIC to 9r5c4, which if false regardless of 1r6c6 being true or false. The Cr6 boxline removes another 1.






Finally a chain ANL prompts us to look for a second cluster, since the blue/green one isn’t growing.


  1. it has blue/green and red/orange 5’s and
  2. there’s a bridge. Blue and red together in r1c7 mean that blue and red are not both true. So either green or orange or both is true.


All of the candidates of the true pattern are the same color. For any combination of two colors to be true the two colors must eventually merge.

In the 5-panel, the combination of green and orange is possible for only one pattern. That’s a very specific case to try.

Removing in order, blue. red, seeing 5, seeing green, seeing orange,  the  normal breadth first trace leaves 6,9 in 3 cells of c6.



The combination of green and red leaves four patterns available. The trial is less specific.




The contradiction, traced below, is unusual.




9r5c5 forces a hidden pair in the West box, to join with NW79 for a deadly 79 rectangle.








The trials leave blue and orange freeforms to define possible 5 patterns.






On the grid, the 5 pattern restrictions produce a naked quad and naked quint placing N3.

That’s followed by







and a hidden unique rectangle  as 9r4c6 forces 6 in two corners, and the slink forces 9 in  the opposite corner.

The contest is ends with a Beeby ALS_46, and the SWr9 boxline triggering  the collapse with S9.










Next week, its on to  ultrahardcore 223.






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A Partial Pattern Trial on ultrahardcore 179

This post starts with the 5-panel freeform  analysis to compare with yours and then moves on with the partial pattern trial suggested in the last post, of 8 patterns containing 8r2c9.

My 5-panel patterns are much less promising  than the 8-panel ones.

Too many patterns in both sets.

On the 8-panel there were two ways to start in the first 2 columns, while the 5-panel has three. Or to put it a better way, on the 8-panel it was 2 clues vs 2 other clues, giving us  2 clues either way. Remember how two clues were enough to get us past solver stalls on UHC 135?

Well we could have decided  that with a lot less work. One thing you get with a full analysis is the orphans, the candidates not on any pattern. Do we have any on the 5-panels?  I think I have one. All 5 pattern freeforms start in r1c9 or r9c9, and none include 5r9c2.  Not much help.

Back to the plan, let’s see if we can get to coloring with two clues from the 8-panel value pattern analysis. It’s one trial to eliminate three or four more. We start by restarting at the stall point with the givens, plus one clue from the bypass, and now with two clues  from one of the 8-pattern sets, 8r2c9 and 8r5c8.

Following up (NE8, E8) with NW5, Beeby pulls in an ALS_75, and we’re off again, but  in trial mode.








Next is an  ANL with an ALS value set on one terminal. The victim must see the whole value set.


Beeby follows with another hat, even more extreme. The 2 value set is crosswise, but there’s still a victim, and it comes with a clue.












We get NE1, and NW1m and W14, which show up here with a hidden UR, type 2.


Next, a couple of Beeby specials, down the middle.

First, a discontinuous 1-way: 5r5c4 is false if 5r3c4 is true, or if it is false, and the slink chain forces it out with a 4 in r5c4. Beeby notes call it a discontinuous loop.Put in the winks, and there is a loop. It’s the alternation of inferences that is discontinuous.









The other special is a complex 1-way.  Complex 1-ways allow slink partners, true when 3r1c6 is false, erase candidates interfere with  the AIC.  You don’t search for these. You just build them with the hope that something will happen. And often enough, it does.


Now a very similar pair of 1-ways, another force out on the left, and a complex ANL on the right.  It’s not a regular ANL, because the chain is 1-way. Two bonus removals are a NWc4 boxline, for a naked pair r2s79, which removes 9r2c4.











Remember, all of this in a trial to see if it’s this or another pair of clues, fits the solution.


Anyway, spot two long ALS in parallel lines with two matching singles and you might have this ALS_64.

In the same neighborhood, a hidden unique rectangle. 3r8c5 would force 2 in two corners, and the slink would force a 3 in the opposite  corner.






The examples keep teaching. Next is an ALS 43 in which both Z sets are not singles, and only one is aligned with the victim. The other one is in the box with it.







Next is a simple AIC ANL whose removal triggers a productive naked pair.




The hidden UR is a direct result, but the removal from a crowded cell is slow progress.



Finally, it happens.  An ANL brings a naked pair in c9, in turn bringing E2. Coloring is on,  and traps leave a single 9 in c4, wrapping green.

The blue army then reaches a contradiction, when it removes the last 9 from W and from c3.





In the breadth first trace below, the blue army then reaches a contradiction,

when it removes the last 9 from W and from c3.


The big picture summary:  We found that the UHC 179 8-patterns allowed only two placements of two 8’s in columns c89. We now know that one of those placements, and all the patterns including it, are false. We can put in the two clues of the other placement, go for a solution next week. And you’re invited.

Could it be time for your next attempt to get there first with the second documented trial free solution of the Stefan Heine ultrahardcore right page Sysudoku review? If you get one, I’ll draw the grids and interpret your moves and guess at how you spotted them for Sysudoku readers, and you can order a box of  copies to hand out to your friends, relatives and other admirers. Here’s the next one, ultrahardcore 223.





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Preparing a Pattern Trial on UHC 179

This post does Basic and reports humanly practical results from the two review solvers on  ultrahardcore 179. Then for a stall breaking trial, we turn to  the limited value patterns, and their discovery through the X-panel view and freeform drawing.  These become easy  value pattern trials aided by coloring in the following post.

UHC 179 gets the stingiest bypass award in the ultrahardcore review, and is in contention for the busiest line marked grid.

The possible hidden UR’s in c78 lack the necessary slinks. But Beeby starts it off with a finned 2-wing in c58, with a 2 cell fin. The victim sees both cells of the fin, so if it were true, it would remove the fin and therefore validate the X-wing that makes it false. The finned fish victims literally  can’t win for losing.

Both solvers find the 3 – ANL, and Sudokuwiki gets the finned 2-wing  removal with a grouped ANL.








Next we’ll take Beeby’s simple 1-way from 7r1c8, and Sudokuwiki’s dfc ALS boomerang from 8r2c9, which Beeby duplicates with an ALS-wing.







With Beeby getting into ALS wings and Sudokuwiki into forcing chains, we get this ANL with one terminal an ALS value group. It’s also a form of 1-way.



One more bv is gained by this ALS -wing, with bv r9c6 as the third ALS.

But it is also time to look for reasonable trials. I was scanning the X-panels for new X-chains and fish, but now it’s time to  scour the  panel edges for limited value patterns.

Below is the so far undisturbed 8-panel. We’ll walk through the process that uncovers a limited set of patterns that can be the basis of a trial.







A pattern is a set of candidates, each one being the single candidate in a box and two lines, that provides a candidate for every box and line. Each pattern can pictured as a segmented line drawn from one side of the panel to the opposite side, which crosses every candidate of the pattern. We call these freeforms, for the graphic element that is a segmented line.

The possible patterns are most clearly seen when freeforms are drawn from the side that most limits how they can start. On the 8-panel,  North to South freeforms start the first two lines in 4 ways.   South to North, i’ts 5 ways.  East to West freeforms start two lines in 2 ways.  Also notice that many candidates are left out of patterns starting either way.  When all patterns are identified, any candidate not on a pattern can be removed. I call them orphans.

It requires patience, but there is a systematic method for mapping out all patterns. Let’s say we start with 8r2c9, going East to West.  We pick a favored cross direction to go for the next segment, say North. Starting in r2c9, for column c8, we must go to r5 because NE has an 8. Then we veer back in the favored direction for c6. For c4 it’s r7 to avoid a second 8 in the C box, and to get an 8 for the S box. For c3, the first two rows have their 8’s, and we go to r7 and back up to r3 for c2

When we see that no  8 is available for c1 and the  SouthWest box, let’s note there is an interchange between columns 2 and 3 on the last two rows, because there is an alternate 8 in both remaining rows. That makes it easy to see a second freeform getting just as far.

To find the first pattern, we work back along the columns and down the rows in each column looking to continue from every unused candidate. In c3, we can take r8 in place of r6. What happens in c2, though is, r3 leaves no landing place in on r6 and, and r6 leaves  no landing place in r3




Next back continuation is r8c4. It produces two freeforms to c1r7 and two more failures.


Finally the patterns through r6c6 and r8c6, for a total of 6 patterns.

So what have we learned by mapping out the patterns from 8r2c9? For one thing, there are no more candidates other than 8r5c8to be added to the trial, because they join these two in every pattern. Also, we have the information to make inferences from any addition or removal of 8 candidates in the trial.

We can make single trial of a pair of patterns in the diagrams by coloring the two candidates that differ, but it still takes up to four trials. Each is likely to be decisive, since in each trial a large number of 8 candidates are removed. Only eight 8 candidates remain in each trial. But if all four trials fail, we likely have up to four more on the patterns which include 8r2c9.

Next post, we’re going to explore a pattern trial alternative that is often better, but in the meantime, you get  homework. The 5-panel at this point has a limited number of freeforms starting from the East side. Your job is to enumerate them, lay them out, so that the number of patterns can be counted, and they can be assembled into fewer trials.

Of course, the longer term challenge to find a solution of another right page ultrahardcore without trials is still on the books.

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A Pattern Trial Defeats ultrahardcore 135

This post reports on the value pattern trial of the last post, its results, and the relatively easy march to a two cluster coloring solution of Stefan Heines ultrahardcore 135.

The trial starts with the three 7 clues and naked pair E18. New clues are left in pencil marks, because they may be taken back. The cluster expansion can be taken back, as well. Still at the wheel, Beeby drives the trial on, with another hidden unique rectangle and a simple AIC ANL.




A tame looking AIC builder ANL






triggers a near collapse,




and a cluster explosion that wraps green with conflicts in c2, c8, c9, and the West box.

So 7r1c2 fails, leaving two 8’s in r5 and no 8’s in the West box.

We have to give it all back, and restart with the alternative 7r2c2 as a clue, dropping its slink partner, 7r2c8.


The cluster resumes at its pre-trail level. Tracing to the grid below,


NW7 produces three clues and the naked quad, which removes 27r7c9, allowing the hidden triple in r7. Clues are marked as real clues now.







It takes an AIC Almost Nice Loop next. Beeby’s note labels it a simple chain. But when there’s a weak link node, it takes an AIC building level of motivation to search it out. Here it can be a 1-way or boomerang from either bv terminal cell.  The 1-way target is the any 8 seeing the starting candidate. For the boomer, it’s a slink into the starting 8, or a wink into its cell partner.

Without the computer, there’s just too much haystack and not enough needle.

The pay is good, though.  Follow up



produces a host of removals from two naked pairs and a boxline, with accompanying cluster expansion.







This opens a double hidden unique rectangle,  adding two more bv. Look at the effect of each removal, with the slink on the other side, and with the other victim in place. Either 4 causes a reversable rectangle by itself, so both are removed.



Add the second cluster,







then do the follow up without removing the wrapped candidates.

You then see how red is wrapped, and why orange implies blue.


Here’s the solution grid in color


Next time, its on to ultrahardcore 179.


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Deeby Hauls ultrahardcore 135 into Court

In this post, Beeby leads the way, but the instructive methods bring us only to a partial value pattern trial. The trial date is next week.

After the usual slink marking Sysudoku basic,



ultrahardcore 135 has a busy grid:

But like 91, there’s something on it to find.  That tease would prompt me to look for unique rectangles, the most explicit spotting opportunity. There’s the 2 and 4 rectangles r46c29 and r46c49, but the slinks aren’t there.  Then maybe the 8 and 9 rectangle r46c36?  No, too many boxes and also short on slinks.

The review solvers find us the same clue in different ways:  The Sudokuwiki, with a confirming ANL in black, and the Beeby, with a 1-chain eliminating ANL in red.

I’m happy to have a few bv, and mark one connected slink pair,





but Beeby takes the steering wheel with an overlapped ALS_62, and





a grouped ALS_13.




On my part, I filed away this Single Alternate SdC NWr1 =  4(8+9)(2+7) + 274 for a possible trial later.

Beeby expands my pitifully small cluster with a complex 1-way from 1r4c7 with four (count’em) AIC enabling branches.





Fortunately, my attention is drawn to my 7-panel by the hidden unique rectangle below .



On the Limited Pattern Overlay charts below are two very telling sets of West to East value pattern freeforms on the 7-panel, after the 7r9c7 removal. There are only two 7 patterns containing 7r1c2. Can you check me out on what limits them? Both of them also contain 7r2c8 and 7r5c9

After 7 in c2 is placed, there are many patterns containing both c3 7’s.  I’m showing only the 4 from 7r8c3, but you can trace out more. That’s how things are with this form of pattern analysis.

Bottom line, the trial of 7r1c2 runs with three clues, and without 10 of the 7-candidates. If it fails,  it places one clue and removes one candidate.  Is this intuition or logic?  To me it’s logic, based on the earned facts on the 7-panel. Such a trial gains as much information per unit effort as the searching required by the strictly logical methods seen in this review.

Trial logic has another factual component in this case. Sudokuwiki soon turns to cell and unit forcing  chains, then stalls. Beeby offers a MSLS that goes beyond the formula that Phil cites on philsfolly, the Beeby solver site. So we’ll count on more research or reader help, and report on the value pattern trial and its solution for 135 next time.

It’s a possibililty for that ultrahardcore right page trial free solution. Send one in and I’ll draw the grids  and publish it to your credit.

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