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( 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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Sudokuwiki Finishes ultrahardcore 91

This post shows the Sudokuwiki solver path to the coloring finish of trial wounded ultrahardcore 91. Last week it was with Phillip Beeby’s solver. You could use either solver as a model, depending on how you would rather spend your searching effort.

Sudokuwiki advanced begins with two trial won clues, two X-chain ANL, and the same small blue/green cluster.







Next is this AIC confirming ANL. Beeby claims this clue prior to AIC building by eliminating its slink partner 2r7c3 by ALS_XZ.







This APE is also duplicated by a Beeby ALS_XZ.  The two cells of the the aligned pair cannot contain a combination of values they see in an ALS.


Sudokuwiki notes present this next grouped AIC boomerang as a digit forcing chain from any candidate along the chain to 6r5c9. More accurately, it is a boomer from 7r5c9. Here is a little pop quiz on recent posts. Can you present it as a Beeby 1-way? Yeah, it’s a stretch.



Now it’s an easy ANL,







and after a little follow up,


a rarity, a regular 148-wing.

Next, the first of two BARNs, or in Sudokwiki notes, a WXYZ-wing. It’s a bent region naked set, with one value bent, the others in one unit.

Another rare entity in ultrahardcoreville, the simplest AIC, an XY ANL.










Then a second BARN, this one missed by Sudokuwiki.


Almost there, we get a hidden UR, type 1. The slinks opposite the extra free corner carry 8 to two corners around it.






Sudokuwiki’s  triple  grouped ANL is impressive,









but a regular Sysudoku Sue de Coq

SEc7 = 1(2+9)(7+8)    gets the removal and more.

SEc7 must contain one from each alternative, and a 1. From the SEbox remainder, only the the bv 18 can supply a missing 8.

The coloring cluster expansion overtakes one XY chain, but a second one keeps it going.  We’ve reached the same state of expansion seen in the previous post on Beeby’s solution.

At the finish, the blue wrapping cluster demonstrates how close uhc 91 comes to two solutions. This form of multiple solution is tough on logical solving methods. It doesn’t help with intuition either.

Next week continues the Heine ultrahardcore review with the fourth puzzle, uhc 135.



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Beeby Finishes ultrahardcore 91

The Beeby series continues on trial wounded ultrahardcore 91 until the expanding coloring cluster can  take it home. Is your trial-less solution ready?

Next up t is  a 1-way from 1r6c3 that is exended back to 3r2c3 for a second removal from r3c3.  Or you can call it a boomer from 4r3c with wink into 1 to a wink into 3.







Another 1-way from 3r2c3 slinking into r2c5. The shortcut version slinks in (red) to green 8r2c5 from blue 7r5c9.

The boomer interpretation starts from green 8r2c5 and winks in to r2c3. On the removal, the boxline Nc6 picks off two more, including 3r5c6 that enables the green XY chain and boxline.


Then pick the interpretation for Beeby’s thing on the right: an ALS-18, a 148-wing, or a Bent Almost Restricted 3-set ( 3-BARN).

Back in ALS land, the long red one stretches out for an ALS_81,




then shortens up below for an  ALS_18, bringing with it  with a naked triple.


I’m watching for a cluster expansion. A new 8 slink was just added.  Although the cluster is not expanding, the increasing number of bv is a sign of progress.

Here is a bv style ALS_81, creating a clue, for




and the hidden unique rectangle at the left.

It then takes the AIC ANL





and the ALS_29 to the right,


and another hidden unique rectangle removing 8r3c7, to release  a coiled spring of cluster expansion.







The UR removal turns 8r1c7 green, trapping 1 and 3. The 3 trap brings clues in NE and r1, but let’s hold the clues and let the candidates spread the cluster. The 1r1c7 trap greens 1r1c2.

As the cluster spreads south, a SEc7 boxline removes 1r9c9, and 6r5c9 is removed by green or blue.








A timely XY ANL keeps it going,




as follow up clues and cluster expansion  bring trap after trap.


Coloring finishes the cluster, demonstrating one of the reasons why ultrahardcores are so hard. This one gets ever so close to having two opposing solutions.

The finish is the blue wrap traced here:





One more post on ultrahardcore 91, for you to compare Beeby’s ALS_XZ intensive path with the Sudokuwiki alternative.


So perhaps it’s time for a diagram for the next Stefan Heine ultrahardcore in the review, # 135.  If you didn’t order a copy, get it done. After the review, there will be 240 right page ultrahardcores left for you to do all by yourself, perhaps with your favorite solvers and your own uncoded discoveries. Maybe even one of my trials. You’re welcome.



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Beeby Carries the Post Trial Against ultrahardcore 91

This post and the next continue on uhc 91, after last week’s SASdC trial eliminated red candidates of a red/orange cluster and left two clues. Phil Beeby’s solver chisels on the crystalized structure, conducting an ALS_XZ clinic, until the slink network started in the 7’s can get a grip.

Here’s the post trial grid with, in black, a 5-chain ANL with NWr1 boxline removing r1c9.

Then, in red, a 1-way from 3r2c3 or a boomer from 5 r8c3, and a SWr9 boxline.
“1-way” will be my shorthand term for Beeby’s AIC building strategy in which an AIC starting on a slink shows a candidate seen by the starting candidate to be false.

The victim is false if the starting candidate is true, and the AIC proves it false if the starting candidate is false. We’ll call it a complex 1-way when the unidirectional AIC uses forcing branches to clear its way. That’s not the case here.

Beeby’s next is this ALS_72.  In an ALS_XZ, X is the group slinked value in the two ALS, and Z is the common value locked in one of them. One candidate in the two shared Z value sets is true.



Spotting  ALS_XZ depends on recognizing potential X and Z value sets in two repeated values. Here, you might spot the similar values in E and r7, the single 2 and 7 in E when 6s are excluded from red,  and that 7 and 2 can be singles when r7c7 is excluded. Then you’d do the calculations on numbers of cells vs. values for the ALS. You could start with r5c8 in red, and the two cells r7c38 in green, adding cells as necessary to fill out the two ALS. More often, these systematic spotting tasks are much easier.

Next is a hidden UR off my Tools chart. Sure enough, if 5r9c3 is true, 5r8c3 is not, so in r8c9, 5 is and 4 is not, so 4r9c9 is.

A flippable rectangle and two solutions.

Interpreting solver results, the term “hidden” calls for finding the slinks between UR candidates. Candidate cell positions tell where they are not. I have to fix the “Tools”page chart. Is this one “typed”? Somebody fill me in.

Does the spotting method above fit Beeby’s next,  ALS_18?  Kinda? It’s hard to put into words.


Taking a break from ALS, a grouped 1-ANL on the left.


Then a 1-way from 8r7c8, or a boomer from fr9c2.






Okay!  We get a clarifying example of a Beeby complex 1-way!

Starting at 4r7c2, the red branch from 2r4c1 – which is false if the starting 4 is false – destroys an interfering 2r5c9 to insert the green slink, permitting  the 1-way AIC to confirm 5r8c3. So if starting 4 is false, the victim is false. Of course if the starting 4 is true, . .

The beat goes on with a couple of hidden unique rectangles.

The right one is standard slinked opposite the extra free corner, but the left one is another Beeby special: no free corner, but a slink of the opposite value across from the slinked corner. To spot these, you might see the UR slinks first, then see the UR values on that cell rectangle.

It works. Trace it out: if 9r6c4 is true, you get the 5’s on adjacent corners, and the  side 9-slink is necessary.

Beeby accepts this DIY ALS_51 that violates the ALS_XZ rule that the victim value Z be present in both ALS. The reason for the rule is, if X is in the ALS containing Z , that doesn’t necessarily lock the only Z value set. In this case, 5 in the green ALS removes one value and one cell. That leaves  6 cells containing 5 values. The true 1 may be 1r6c4!  When X is not a single candidate, locking depends on where Z candidates are.

Here though, Beeby’s code anticipated something else entirely. Lacking the  5, the 2 and 6  sets in the blue ALS are locked, removing 6r6c3 and therefore, 1r6c4 as well.

On an faulty ALS_XZ with one Z value set missing,  check what happens when the ALS without  Z loses X and is locked. It can be another heads I win; tails, you lose.

It’s not enough for Beeby.  The solver doubles down on the above by excluding  cell r6c5 and one value set 8. Same violation and same saving grace.


Some heavy lifting here? We continue with the Beeby ALS_XZ clinic next week.




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