Tough Sudoku for 5/August/2010

Today's puzzle is an ER 9.0. Should be a tough one. I may have a quick solution that involves one complex step. I will probably blog the step.

Provided I have not erred in my haste, the step is a continuous network using the following "at least one is true" sets:
(4)rows 3 and 7. (349) columns e and g. (3) column c. (34)b4. (46)a8. (69) a9.
=> 16 eliminations => Simple Sudoku Technique Set => UP 81.

Hopefully, I can demonstrate how to "easily" (a relative term) find this nice 12 SIS continuous network.

Lovely place.

Another nice photo Wendy!

quite tough for me: UP=26
1.Kraken Cell(169)e9=> g8<>4, UP=28
(9)e9-d7=a7-HP(34)ab7=(4)i7
||
(6)e9-e12=f2d1-HP(27)f2d1=(7)df3-(347=5)b347-(568=4)fhi7
||
(1)e9-gh9=g8
2.loop (9=6)a9-(6=4)a8-c8=c2-e2=e6-HP(39)e46=(9)e9
=> c9<>9,a7e6<>6,a27f2i8<>4,e6<>8; UP=37
3.(1=269)aeh9-(2=6)i8-i1=e1 => e1<>1; UP=81.

1) Start 23 UP 26
2) IMO, the continuous network is an efficient approach to this one:
[(4=6)a8 - (6=9)a9 - (9)e9 = HP(93)e46 - HP(34)e46 = (4)e2 - (4*)f3 = (4*)a3 loop] = (4*)b3 -
[(4=3)b4 - (3)c45 = (3)c1 - (3)g1 = HP(39)g56 - HP(94)g56 = (4)g8 - (4*)i7 = (4*)b7 loop] = (4*)a7 LOOP

=> 16 eliminations:
(4)a24, (4)b8, (4)f2, (4)i8, (3)a4, (3)b1, (9)c9, (6)a7, (68)e6, (18)e4, (58)g5, (8)g6
From here, Simple Sudoku can finish the puzzle. Thus, SSTS => UP 81

probable loop of step 2 is a poor cousin of Steve's massive continuous network - is there an "easy" algorithm for finding these and if so is the method cloaked in a masonic veil of secrecy?

for me - chaining over 5 sets is too deep; plus not one but two AAIC's in one chain - geeze (sans the SECRET METHOD).

farpointer: I am not sure about the algorithm perspective. I have some clue about pattern matching perspectives, but that often is a self-fullfilling prophecy - one finds what one seeks, etc.

I would hardly call your step 2 a "poor" cousin. In fact, it is, imo, a clue that the "massive loop" must exist. One could build a recursive algorithm of Almost Continuous Animals to find the actual Continuous Animals.

The Masons have hidden the rest. We cannot allow such information to be exposed to the general public.

Farpointer wrote: "for me - chaining over 5 sets is too deep; plus not one but two AAIC's in one chain"

"sets" seems to be a rather subjective term. One can see the entire mega-chain of 12 native SIS as simply two sets. That, imo, is no different than using ALS, etc. It just happens that these sets are in differently orthodox containers.

Farpointer,
Complexity is always a bit subjective. So, although you obviously think the continuous network is more complex, the following objective data exists:
Your kraken step uses more "native sets". It also uses more "derived sets" if one counts the deriviations within the groups of box, column, row as you presentation implies.
It would be difficult to argue that it is less complex than the large continuous network from either of those perspectives. I suspect that it also uses more weak inference groups, both native and derived.

I can see one saying that my step uses 2 krakens, and your step just one. However, your step must find the elimination from (3) endpoints. My step has no endpoints.

IMO, complexity is completely in the eye of the beholder - especially in this comparison.

Steve: there are usually 200..300 native sets to deal with. The combinations 12 in 300 is astronomic - ie. you have some method you are unwilling to impart (saying that it is a 'pattern' is a gloss which evades the question).

BTW - my solution uses no 'derived' weak inferences, the chains from each possible of the kraken cell use only native strong sets, the two chains which complete the solution use only native strong and weak links. Max chain depth is 5, in the loop - something achievable by a mere mortal before he despairs.

Farpointer,
I hardly wish this to become acrimonious. However, please be consistent. You can choose some arbitrary measure and say something about complexity, if you wish. But, if we are making comparisons, some consistency in your implied measures is required. By your accounting then, for example:
A=B-C=D-E=F
is length 3, but

C - B=A
||
D- E=F
is length 1?

If I use a similar structure as your kraken structure, and count as you count, then max depth 3 for my continuous network.

Finally, HP(27) uses a derived set in your chain, for example. If not so, please show me how this is not a derivation. Also, show how there is no implicit derived weak inference in that statement.

If you hate my continuous network, so be it. I can live with that subjective statement. I remember now why I should like to no longer do sudoku.

I have written a blog page regarding the step. I am fain to say one should do this, or do that. I would hate being preached at like that. So, I provide the information and leave it up to the solvers to develop their own search methods. The method I used to find the chain is simple - I brainstorm short chain fragments, without regard to targets. I combine them later. Usually, something falls out. The hints as to where they may fall out are also available in the blog, if one reads carefully.

It is all about "patterns" for me. To say that evades the question means I cannot give you an answer.

Steve wrote my "step 2 ... is, imo, a clue that a massive loop must exist." Step 2 is not possible until step 1 is complete and so can not be a clue, because it doesn't yet exist.

Farpointer wrote:
Step 2 is not possible until step 1 is complete and so can not be a clue, because it doesn't yet exist.

I have never yet, nor shall I ever, create a step. They all exist, already. We just find them. To argue that it "does not yet exist" is, well, interesting.

Imo, of course the step exists. It expresses a truth, or a set of truths, that are always in existence for this puzzle. How one uncovers that set of truths has no effect upon their existence.

Steve - I don't recall claiming the kraken's depth was 1, the static chains stemming from each possible have a look-ahead depth max of 4 sets. I never made a judgement good/bad on your solution; in fact it is a scintillating curiosity! (fain?)
As to 'derived' strong sets vs. 'native'; I have no idea of your meaning. HP(27)f2d1=(7)df3 is static - "native" in the position at that solution point (no different from say a pair in a cell being a "native", non-derived, strong set (a=b)cell. (ditto for weak links).

Steve: saying that all is current in the present implies knowledge of the end - mortals are not granted this omniscience and must proceed stochastically step by step OR reverse engineer the solution.

Not solve this tough, just read Steve’s and farpointer’s comment
As I understand about “length” or “dept” on Steve’s count is meant that: ALL STRONG SET on that move, right? => All Kraken must be at least “length” 4.
Example: kraken cell (ABC) => ... <>A
A
||
B-(D=A)
||
C-(E=A)

Steve: I would like to invite & require you come to new PF then have some idea on hardest list – especially for colBF2 326 puzzle

Hi Gath, Are you able to provide asbestos web pages to prevent my PC burning up? Best regards, Neil

Neil: surely the acrimony is one-sided. I made two light-hearted comments and incurred a firestorm from Steve (why? I have my stochastic guess) - naturally I defended.

farpointer wrote: “...and incurred a firestorm from Steve (why?...)...”
Hmm..., I think that you should be read careful Steve’s comments first before saying that
Everyone always think that – even me in sometimes : “I’m the best”. But IMO, the best is only ONE, so if you like to become “the best”: at first you must be follow “the best”

, Wendy!

Play nice children

First, I apologize for the apparent tone in my replies to Farpointer. I truly do not mean any offense.

Secondly, I would like to be granted the right to explain myself. I have, over the years (yes, it has been years - yikes!) had discussions about some of the same stuff more than once. I may have mistakingly read into some of Farpointer's comments things which are not subjedively there.

I have often attempted to explain, in various venues, how I find things. This finding process has changed dramatically over the years. I do not think that how I look for steps like this one now is the way that I would/ or even could/ find such a step four years ago.

This is because I have a solved a puzzle or two since four years ago. Consequently, "patterns" jump out to me in a fashion they did not back then. I cannot tell someone else how to properly train their personal pattern-matching machine (their brain).

I have yet to ever explain to a group and not have the following occur:
Someone will suggest that "they could never do it", or "that is assumptive", or "that does not work for mortals" - or something akin to that.

I have sometimes grown weary at trying to explain myself. I apologize to everyone for letting my weariness show through today. I should do better.

Regarding an algorithmic approach: I have suggested algorithms for finding continuous networks in the past. Here is one that will work for all SPM type deductions. SPM is Symmetric Pigeonhole Matrix. It is a simple counting method. I will demonstrate on today's puzzle as an example in my next post.

Finally, Farpointer I cannot agree that I was implying that one need omniscience. The entire concept of using Almosting relies upon utilizing something which is not yet quite proven to exist. To suggest that Almosting requires reverse-engineering is simply false.


Almosting within a chain - and using any Almost entity - whether it be an Almost Hidden Pair, or an almost continuous loop, is, in fact, the use of derivation. One does not need to know that a chain exists - one only needs to know that it has potential to exist.

The concept of counting is the simplest of concepts to find an SPM. In the case of a group of SIS that are entirely Bivalues or BiLocals, the count is rather trivial.

The rules of the count. Add at any time and in any arbitrary fashion any SIS one wishes to the mix. The total count will:
Increase by one for each SIS member not covered by previous SIS.
Decrease by one for each SIS considered.

Coverage: Coverage occurs when a new SIS member is weakly linked to a previous SIS member AND it is in the same group as all previously discounted (covered) SIS. In other words, weak links emanating from a SIS node can be only in one sudoku dimension. Sudoku dimensions are Row, Column, Box, Cell.

If the count reaches zero, one has a continuous network.

Begin with any SIS within the 12 that form the SPM. I will present them in an order, but order is of no importance whatsoever.

(4)row 3: (4)a3, (4)b3,(4)f3 Count: (3-1) = 2
(4)row 7: (4)a7, (4)b7, (4)i7 Count: (2+1-1) = 2
(46)a8: Count: (2+1-1)=2
(34)b4: Count: (2+1-1)=2
(3)c: (3)c5, (3)c4, (3)c1 Count (2+1-1)=2
(69)a9: Count: (2+1-1)=2
(9)e: (9)e9, (9)e6, (9)e4. Count (2+2-1)=3
(3)e: (3)e6, (3)e4. Count (3-1)=2
(4)e: (4)e6,(4)e4, (4)e2. Count (2-1)=1
(4)g: (4)g8,(4)g6, (4)g5. Count (1+2-1)=2
(9)g: (9)g6, (9)g5. Count (2-1)=1
(3)g: (3)g6, (3)g5, (3)g1. Count (1-1)=0

That is an algorithm. I find it less than helpful in finding these type of things. Perhaps others may find differently. It is excellent as a device to confirm that one has in fact located a continuous network. Naturally, one can use derived SIS within the algorithm. For example, one can use the Almost Hidden Pair (9x) in column g to reduce the counting.

Farpointer, I understand that from the "point" of the kraken, you mean that using almost sets and actual sets in the manner you described that one need only "look ahead" 4 links. I would add to that one must also know that the kraken is the locus point for such an investigation.

In the continuous network, consider these facts:
Each individual ACL within the CN is comprised of (6) native sets. Some of these are derived, and I could reduce the count. However, let us not even do that reduction.

Since an ACL form a circle, the max distance from the Kraken for any point on the circle is, well, about half of 6. (A bit less, since the Kraken is in fact part of the ACL. Not sure how to count a distance of 2.5 though).

Thus, the Max look ahead from the kraken to get to a point on the other kraken is (3). Thus, if I were to make a dual Kraken diagram, the max look-ahead would be 3.

The point being not that I think the continuous network is more or less complex than your kraken step. The point being that sometimes measures of complexity are tricky depending on how well one defines them. That is all I was trying to point out. A (3) SIS bivalue chain would have a "look ahead" of (1) is one starts in the middle, I sh

Continued:
A three SIS bivalue chain would have a "look ahead" of one if one starts in the middle. That was the intent of my abstract demonstration. You did confirm that by your answer. You seemed to deny it, but what you wrote is consistent with agreement, unless I am missing something.

In other words, a bivalue/bilocal AIC of length (3) has a "krakenized look ahead" value of (1).

ttt:

It may take some considerable time for me to properly attach the puzzles you mentioned. I have a completely different mode of deduction in mind for those puzzles. Unfortunately, I have not completely developed the theory yet. I can say that it looks promising to use "minimality" as a constraint. In fact, I think the harder a puzzle is, the better "minimality" should work as a constraint. If I do not finish my research, I might email you my work. I have no doubt you will solve the puzzles before me, but just in case, you might find a new approach interesting!

Steve: one must also know that the 12 of 300 or so sets is a locus of investigation for your matrix - in your example 2..3 intertwined krakens -- at least an order of magnitude larger search space than the 3 points of one kraken (but those magic twelve, all conveniently linkable just popped out as a pattern?, I have known idiot savants capable of such seeming magic with numbers and I grant that it is remotely possible - very remotely.) You skew the intent of my argument by claiming I assert my solution is not complex. A kraken (whether two points or three) is a very natural place to start a search in a puzzle such as this one where chains offer little at the start - in fact non-chainers do this all the time, but only examine UP on each possible. I have neither the time nor the patience to do such things - the talent which I possess is most assuredly not solving sudoku puzzles. I post solutions only as examples of what is possible, making no claim other than that they are valid and the simplest I could find in any given puzzle. If a mere mortal can do better - Bravo!

ttt: calm yourself! I tilt not at Sudoku Gods and make no pretense at being one. My considerable talent as a computer programmer is another matter entirely!

Farpointer:
Please review what you are writing more carefully. I never claimed that you said your kraken was not complex. I have merely asserted, repeatedly, that measures of complexity are subjective. If you truly find another meaning, then either I have misspoken, or you have misread.

You asked for an algorithm. I provided one. I did not claim to use that algorithm.

I cannot possibly explain everything that I do to come to a find a step. I can outline it. I have done so:

I look at fragments. I look at as many as I can as quickly as I can. I connect them.

Practice makes the looking not stochastic at all. Practice makes the looking an investigation of patterns.

Many years ago when chess computers first came out, one could easily beat them using unsound moves by merely being able to overcome the "move horizon" of the machine.

The point of that digression is: Human beings are best left to use their parallel processor as pattern matchers. Digital machines are much better at analysis methods that will not miss anything.

Finally, any reasonable attempt to analyze sudoku by a computer will undoubtedly reveal the following:

If one uses all the weak links available, the number of SIS one must consider is only 81 - # of givens. If one uses uniqueness, that number is lower. If one uses minimality and likely patterns, that number is even lower.

I can go much further with problems with your interpretation of the utility of the algorithm. Please, remember that contempt prior to investigation is rarely a good idea.

Even though I personally hate the algorithm, I can certainly defend it from rather shallow investigations into its possible utility.

Steve: I have been "around the block" many times in the half century (yikes) the earth has circled it's sun. You err if you assume I am naive. BTW - computers are astonishingly good at pattern matching and parallel processing as they also are at processing with incomplete and noisy data - try not limit your argument to matters you well understand. (max 81 weak links ?? 81*9 - givens to start, add weak links to groups and the number explodes)

Hey Men, this is a very interesting discussion and one that my computer programmer mate will take on, if I can locate him somewhere in the ether!! Maybe Gaia also would have some bearing on it!

about 1 hour

Hi Jyrki,
Thanks for your note yesterday - I have replied on that page. Best regards, Neil

Neil, you're welcome!

This was quite tough, so I'm glad to see nobody found simple chains resolving it.

At 26 filled I derived a contradiction, if 4 is in gi8 (after a few steps you will have created a pair 18 in de8 and locked 6 into e12 forcing e9=1, which is a no-no). That took me to 29 filled (notice 3 locked into c45). This step seems to be closely related to farpointer's Kraken step. I simply forced something out of the ALS on column e, and the obvious links coming from a89.

Still not satisfied with the ALSs on column e I used the chain

(4=6)a8-(6=9)a9-e9=(168)e189-(168=4)e2

to eliminate 4 from a2, and then the loop

(4)e2=(168)e128-(16=9)e9-(9=6)a9-(6=4)a8-a3=c2-(4)e2

to do a bunch of eliminations, but disappointingly only to UP 32.

An easy contradiction then shows that g8<>8. After that one quickly sees that both (5)b7 and (5)b8 lead to g9=5 and UP 39.

(7=6)i1-(6=1)e1-e9=(69)ae9-(6)h9=i8-(6=7)i1

finally forces i1=7 and unravels the puzzle.

Jyrki,
Thanks for your post above, but it is time for me to do other things (Sudoku from about 6AM until about 9 or 10AM UK time). I will look at your solution either very late today, or tomorrow morning.
Best regards, Neil

Preface: The interchange between Farponter and I began on August 1, 2010. Farpointer asserted that kraken complexity might be evaluated in a particular fashion. I noted there that such imo such complexity measures are entirely subjective. Most of our interchange above (before we digressed) was an attempt, by me, to make an abstract argument regarding additional faults with the proposed complexity measure. I have yet to completely understand Farpointer's objections to my abstract arguments.

I often think abstractly, and make the mistake of assuming just because I make an abstract argument, that others shall truly understand it. It seems obvious to me that my abstract argument regarding comparing "krakens" to "krakens" was likely not understood. Below, I will illustrate one manner to "krakenize" the continuous loop presentation.

Just like a kraken cell, one can have kraken candidate group. This is no different abstractly then a kraken ALS, which is no different abstractly than a kraken cell.

Loop:

(4)XWab37 - (4)b4,a8
||
(4)i7 - (4)g8 = HP(94-34)g56 = (3)g1 - c1 = c45 - (3=4)b4
||
(4)f3 - (4)e2 = HP(43-39)e46 = (9)e9 - (9=64)a8

For comparison, Farpointer's very clever kraken cell:

(9)e9-d7=a7-HP(34)ab7=(4)i7
||
(6)e9-e12=f2d1-HP(27)f2d1=(7)df3-(347=5)b347-(568=4)fhi7
||
(1)e9-gh9=g8

Presented in this fashion, both krakens are equivalent in maximum "look ahead".

In the loop, one can certainly shorten the apparent "kraken look ahead" depth further by moving some items around. I really am not interested in that academic enterprise. My point being, as it always has been:

Complexity is subjective. Often, measures of complexity are inconsistent with themselves. They often fail to recognize that the measurement may be a function of the presentation more so than a function of the actual deduction.

IMO, to argue that the loop is any more difficult to find than a kraken cell deduction probably assumes that one believes a kraken cell is just a better start point. I freely grant, (yet again), that the conjecture that a kraken cell is a better start point than a kraken group may be objectively true. However, at this point, I am not aware of any proof of this assertion.

Steve: your solution is of course brilliant and better than I could achieve at this point in time (I never denied this) - I haven't yet attempted to implement 'kraken groups' on anything but cells and units, but may do so in a limited way in the future. Also it is quite difficult for me to perceive when an XWing might be usefull as a bool node - two reasons I couldn't reach your mighty loop.

SE=9.0