Maverick 1

[mhparker]

Hi Para,

I see all this talk about x-wing and x-cycles going on. When i solved this i used our basic killer techniques. There's no need really to go looking for these techniques in this puzzle really, but they are a nice shortcut.

Good to see that my posts provoked you into writing a WT again! BTW, what did you think of my estimated rating of 1.5?

In case you misunderstood me, I wasn't suggesting that this puzzle can't be done with regular killer techniques. I just wanted to use the opportunity to showcase other techniques that we don't see very often, and which might not crop up again for some time. The idea is that, if I ever end up writing an article on chain-based techniques ("I have a dream..."), and need some examples for X-Cycles, I can just provide a link to these posts and won't need to re-invent the wheel.

Don't get me wrong, regular killer techniques are great. It's just that we see them so often. So, if I wanted to showcase them, I wouldn't need to use this puzzle as the basis for my examples.

As far as what to look for is concerned, I guess we're back to personal preferences again, so the most I/we can really do is just present the various techniques as neutrally as possible, and let people decide for themselves. Also, what to look for will depend on people's strengths. Some people (greetings, Richard!) love heavy permutation work ("mere" combination crunching being just child's play ), whereas others (hi Cathy!) can see such stuff as unique rectangle and coloring moves from miles away.

I know from your previous posts that your personal preference is to stick to the regular killer techniques for killers. Maybe other people on this forum do, too, and possibly regard other techniques as shortcuts and/or "cheating" to some extent. However, I've made the experience that if someone finds a shortcut that makes a puzzle significantly easier, it's usually quite a challenge (read: just about impossible!) to convince them that they really shouldn't have used it and to (re-)solve the puzzle pretending it's not there. A prime example of this was goooders' breakthrough combination for the A64V2, which (despite being strictly-speaking T&E) got a lot of support from other forum members at the time. So although one can try insisting: "No, don't go through that door, go through this one!", many people will still end up taking the path of least resistance.

That said, one does get used to the different styles that people use around here, and would probably miss them if they were to be abandoned. Good on you all!

[CathyW]

Hi Mike!

You're right there - I'd much rather use URs, colouring and other techniques like x-wings, grouped x-wings, in addition to standard 45s etc, than lots of heavy combo crunching when I'm solving.

I haven't had a chance to have a go at Maverick 1 yet (hopefully it will be quiet at work tomorrow!!)

Edit: Wow - I'm now a master

[Para]

Hi Mike

I just wanted to show another way of going through it. Especially because you posted a second x-cycle after Afmob's WT to break open Andrews position. I rather use other moves but if needed i'll use anything i have to work with . There have been (grouped) x-wings, (killer) xy-wings and xyz-wings in my walk-throughs as well. But it was just to show the other side.

greetings

Para

[Andrew]

In case you misunderstood me, I wasn't suggesting that this puzzle can't be done with regular killer techniques. I just wanted to use the opportunity to showcase other techniques that we don't see very often, and which might not crop up again for some time. The idea is that, if I ever end up writing an article on chain-based techniques ("I have a dream..."), and need some examples for X-Cycles, I can just provide a link to these posts and won't need to re-invent the wheel.

Don't get me wrong, regular killer techniques are great. It's just that we see them so often. So, if I wanted to showcase them, I wouldn't need to use this puzzle as the basis for my examples.

As far as what to look for is concerned, I guess we're back to personal preferences again, so the most I/we can really do is just present the various techniques as neutrally as possible, and let people decide for themselves. Also, what to look for will depend on people's strengths. Some people (greetings, Richard!) love heavy permutation work ("mere" combination crunching being just child's play ), whereas others (hi Cathy!) can see such stuff as unique rectangle and coloring moves from miles away.

I know from your previous posts that your personal preference is to stick to the regular killer techniques for killers. Maybe other people on this forum do, too, and possibly regard other techniques as shortcuts and/or "cheating" to some extent. However, I've made the experience that if someone finds a shortcut that makes a puzzle significantly easier, it's usually quite a challenge (read: just about impossible!) to convince them that they really shouldn't have used it and to (re-)solve the puzzle pretending it's not there. A prime example of this was goooders' breakthrough combination for the A64V2, which (despite being strictly-speaking T&E) got a lot of support from other forum members at the time. So although one can try insisting: "No, don't go through that door, go through this one!", many people will still end up taking the path of least resistance.

That said, one does get used to the different styles that people use around here, and would probably miss them if they were to be abandoned. Good on you all!

Interesting thoughts Mike!

If you do get round to writing that article, then linking to those messages is a great idea.

I'm certainly one of the group that prefer to stick to normal killer techniques although, of course, the definition of normal has expanded quite a lot in the last year. Others, particularly Cathy and yourself, like to use "advanced" Sudoku techniques when you see them. Provided that their outcomes are given clearly, so that I and others can find other ways to check those steps, that's fine by me. Don't get me wrong. I'd like to learn "advanced" techniques and I did buy Andrew Stuart's book but there isn't time to study them; solving puzzles comes first. Even if/when I learn then, my current feeling is that I'll still look for normal killer techniques first.

I hope we will never reach the state where original Assassins and the majority of forum puzzles need the use of such techniques to solve them. If they are needed in a few of the harder puzzles, as an alternative to massive hypotheticals or T&E, then that's no problem so long as it doesn't become widespread.

I'm sure you are right that anyone who finds a shortcut isn't going to go back and re-solve the puzzle pretending it's not there, apart from your JSudoku example for A72V2 which I haven't yet looked at but will probably do so soon as I don't think I'm going to be able to finish that puzzle.

I don't think I'd ever think that a shortcut was "cheating" unless it was used so early that it didn't appear that the person had made a serious attempt to solve the puzzle by normal methods.

If a shortcut is used when a puzzle has reached a very difficult stage, for example if it's either a lot of combination/permutation crunching or a short cut if one spots it, then in my book the shortcut is justified. Mike's shortcut in A72V2 is a clear example of this.

Returning to Maverick 1, I managed to finish it yesterday evening and will post my full walkthrough once I've checked it and had at least a glance at Para's walkthrough and Mike's analysis.

Mike asked Para what he thought of the estimated rating of 1.5. No Way! I'm pleased to see that Ed has rated M1 as 1.75. That's what I was going to put it at.

I was going to post my walkthrough in normal text but since Cathy is about to start the puzzle I'll use TT unless Cathy has already finished it. It took me a week, on and off and with breaks to do A74 and A74V2. If I hadn't received encouragement off-forum from Mike and then seen that Para had posted his partial walkthrough from my position, now expanded to a full walkthrough, I might well have given up on this one.

Good luck Cathy! Unless you also find a shortcut using an "advanced" plain vanilla technique, it will take you a long time.

[CathyW]

Don't worry about the TT Andrew - I'm unlikely to be doing a WT for it myself and I won't peek at yours until I've finished or given up! Got first placement during break today but basics haven't yielded much else so far!

[Para]

Good luck Cathy! Unless you also find a shortcut using an "advanced" plain vanilla technique, it will take you a long time.

I think the way i went through it with the regular sudoku techniques may be one of the only paths through this puzzle with those techniques. But feel free to prove me wrong. I think that is probably what makes the puzzle harder. This puzzle felt more like a 1.5 rating to me. But that is mostly because it fell fairly quickly compared to say A60RP-Lite which i rated a 1.75. For me it wasn't in the same league as that puzzle.

Para

[mhparker]

Hi folks,

Wow - I'm now a master

Congratulations, Cathy! You're keeping me on my toes, there. I clearly need to keep ardently posting just to stay in front!

I just wanted to show another way of going through it. Especially because you posted a second x-cycle after Afmob's WT to break open Andrews position.

I included it because of it being a second type of AIC, where the chain begins and ends in the same cell, forcing a placement there. As you know, there's also (for Killers) a third type, where the chain begins and ends in different cells, but within the same cage. Maybe I/we can find another example for that sometime.

BTW, I hope that no-one thinks I'm trying to show off by writing those sort of posts. It's just that I like writing, and see this as a way in which I can give something back to the forum in return for all the great stuff I've learned from you all in the past. Also, by trying to analyze and consolidate information for technical articles and posts, I end up learning things myself that I would probably never have appreciated if I had continued to approach the subject in an "ad-hoc" manner.

I'm certainly one of the group that prefer to stick to normal killer techniques although, of course, the definition of normal has expanded quite a lot in the last year. Others, particularly Cathy and yourself, like to use "advanced" Sudoku techniques when you see them. Provided that their outcomes are given clearly, so that I and others can find other ways to check those steps, that's fine by me.

I hope you like my approach for AIC-based moves of including both the formal Eureka notation and an informal verbose step-by-step explanation of it, which I hope and expect everyone here will readily understand.

I hope we will never reach the state where original Assassins and the majority of forum puzzles need the use of such techniques to solve them.

I expect that there will never be such puzzles on this forum, because (in practice) we will probably always be able to resort to "informal" hypotheticals on puzzles that would otherwise require advanced "formal" techniques. For those writing automated solvers, the situation is radically different, because hypothetical or T&E approaches are generally avoided (or not even implemented) here, in favor of nibbling away at the candidates using formalized, smaller steps. So, from a SudokuSolver and JSudoku perspective (for example), there definitely will be some puzzles that can (currently) only be solved using chains, UR moves, etc., and which cannot be solved by intensive combinational/permutational analysis alone. Therefore, for people like Richard and Jean-Christophe (and anyone else wanting to write a state-of-the-art solver), a full understanding of both regular killer and chain-based techniques is mandatory, not optional (as it is for the rest of us). Note that, although SudokuSolver is more powerful than JSudoku in general (in terms of the percentage of extreme Killers it can solve), one can still occasionally find examples of puzzles that JSudoku can complete, but on which SudokuSolver gives up. Although I haven't done an intensive analysis on this, I strongly suspect that the bulk of these puzzles are the ones that are very "chain-intensive".

I'm sure you are right that anyone who finds a shortcut isn't going to go back and re-solve the puzzle pretending it's not there, apart from your JSudoku example for A72V2...

That was different, because in the one case it was 100% me, and in the other case it was a 95% JSudoku-based WT. Although I was perfectly happy with the shortcut I used to solve the puzzle, I wasn't sure at the time whether any other WTs would be published (which may indeed turn out to be the case). Therefore, I felt that, to do justice to Para's puzzle, there needed to be at least one "traditional" WT, even if it was only an "android" one (i.e., automated solver WT dressed up in human WT form).

Returning to Maverick 1, ...Mike asked Para what he thought of the estimated rating of 1.5. No Way! I'm pleased to see that Ed has rated M1 as 1.75. That's what I was going to put it at...If I hadn't received encouragement off-forum from Mike and then seen that Para had posted his partial walkthrough from my position, now expanded to a full walkthrough, I might well have given up on this one.

I'm glad you persevered and finally succeeded. I asked Para for the rating because of his comment about the puzzle being "solvable by just sticking to the basic killer techniques as 45-test cage combos and Killer Subsets", which made it sound about as difficult as taking a Sunday stroll through the park

P.S. Just seen Para's latest post and agree with him completely that the A60RP-Lite was harder than this one. We shouldn't forget that the rating should reflect the difficulty of the optimum path through the puzzle.

[Para]

I asked Para for the rating because of his comment about the puzzle being "solvable by just sticking to the basic killer techniques as 45-test cage combos and Killer Subsets", which made it sound about as difficult as taking a Sunday stroll through the park

No, didn't mean it like that. I solved the A60RP-Lite with "only" those techniques as well. That doesn't mean it isn't hard. Using those techniques can make a puzzle awfully difficult as well. You just need the right insight to use those techniques appropriately.
I think i just said it like that because some people might not be too skilled in AIC's and wanted to make sure they didn't give up on it yet.

As you know, there's also (for Killers) a third type, where the chain begins and ends in different cells, but within the same cage. Maybe I/we can find another example for that sometime.

I am not sure what you mean by this type of chain, but i assume you mean a chain that eliminates a certain combination from a cage without making specific eliminations. Because if a chain makes eliminations it always has to end in a particular cell otherwise it can't make any eliminations in that cell.
How would you qualify the chain i used in my Brick Wall WT at the end(step 46)(well i really think it is the middle because there doesn't seem to be an end in sight, but near the end of my WT so far). Although it isn't really written down as a chain but it could be.

greetings

Para

[Para]

Hi Mike

You've been talking about AIC's etc. a lot. Maybe you know this. I was just wondering something about my Brick Wall chain. I used a link i can't really place. I think there should be a third type of link for Killer Sudoku.
Normally you have strong and weak links.
Strong link: If A is wrong, B is right.
Weak Link: If A is right, B is wrong.
In Killer Sudoku we can also have: If A is right, B is right or if A is wrong, B is wrong. The second, which i used in my chain, is probably more likely to happen. This can happen because certain values only occur with another value in a cage.
Don't know if you mentioned this before. But this doesn't really fit into the strong/weak link view of AIC's.

Para

[mhparker]

Hi Para,

Firstly, thanks for the clarification regarding your comment about "only" using basic killer techniques. I understand you fully now (he said, hopefully).

As you know, there's also (for Killers) a third type, where the chain begins and ends in different cells, but within the same cage.

I am not sure what you mean by this type of chain, but i assume you mean a chain that eliminates a certain combination from a cage without making specific eliminations.

Absolutely right. I was thinking of eliminating cage combos there. Thus, if the starting premise that a cell (or group of cells) within a cage does not contain a particular digit ends up proving (by following the chain) that another cell (or group of cells) in the same cage must contain that digit, then clearly that cage must contain that digit (somewhere). Thus, we can eliminate all combinations that don't contain that digit. Of course, this could result in candidate eliminations within the cage if one or more other digits are no longer present in any remaining combination.

You've been talking about AIC's etc. a lot. Maybe you know this. I was just wondering something about my Brick Wall chain. I used a link i can't really place. I think there should be a third type of link for Killer Sudoku.
Normally you have strong and weak links.
Strong link: If A is wrong, B is right.
Weak Link: If A is right, B is wrong.
In Killer Sudoku we can also have: If A is right, B is right or if A is wrong, B is wrong. The second, which i used in my chain, is probably more likely to happen. This can happen because certain values only occur with another value in a cage.
Don't know if you mentioned this before. But this doesn't really fit into the strong/weak link view of AIC's.

Yes, I've discussed this before here, where I reached the same conclusion that you did, in that a new link type is needed. I called it a direct link at the time, and introduced the comma notation. However, I'm now thinking about calling them neutral links instead, because (unlike strong and weak links) they don't change the truth state of the premise at the current end of the chain.

How would you qualify the chain i used in my Brick Wall WT at the end(step 46)

Yes, I'd already noticed that. Excellent move! Here's my analysis:

46. Killer XY-Chain with 9 links (4 strong, 3 weak, 2 direct) on 7 as follows:

(7=1)r6c7-(1)r456c9=(1)r1c9,(68)r1c78-(68=3)r2c89|r3c9-(3)r123c7=(3)r7c7,(7)r6c7 => r6c7=7

This can be explained in verbose form as follows:
(i) Either r6c7 = 7 or...
(ii) ...r6c7 <> 7
(ii) => r6c7 = 1 (strong link, bivalue cell r6c7)
(iii) -> r456c9 <> 1 (weak link, n6)
(iv) => r1c9 = 1 (strong link, c9), r1c78 = {68} (direct link, combinations 15(3))
(v) -> 17(3) at r2c89+r3c9 <> {(6/8)..} (weak link, n3)
(vi) => 17(3) at r2c89+r3c9 = {3..} (strong link, combinations 17(3))
(vii) -> r123c7 <> {3..} (weak link, n3)
(viii) => r7c7 = 3 (strong link, c7), r6c7 = 7 (direct link, combinations 14(3))
In other words, "if r6c7 is not a 7, then ... it must be a 7" (contradiction)
Conclusion: r6c7 = 7

Notes:

1. Nomenclature: "XY-Chain" because not all links are on the same digit, "Killer XY-Chain" because some links are based on cage combinations (Note: JSudoku uses the term "Complex" instead of "Killer" here).
2. Chain does not make use of side-effects of previous links, and thus is fully bidirectional.
3. Chain contains 1 (more complex) link that is based on 2 digits ({68}) rather than the usual 1.

(well i really think it is the middle because there doesn't seem to be an end in sight...).

I'm really impressed with how far you've managed to get already. Afmob's doing really well, too. BTW, I gave up after taking a peek your WT, where I realized that you'd already found everything that I had, and more besides. Maybe I'm still suffering from a kind of paranoia when it comes to so many "L" shaped cages after our experience with TJK18?

BTW, if you and Afmob get completely stuck on the Brick Wall, maybe we should merge yours and Afmob's progress and convert it to a team (tag) effort? I guess that's what you would do anyway. I personally would have preferred it to be a team solution right from the start, but you and Afmob have come so far now, I'm rather hoping that one of you can carry on and claim victory. Good luck!

[Andrew]

Don't worry about the TT Andrew - I'm unlikely to be doing a WT for it myself and I won't peek at yours until I've finished or given up! Got first placement during break today but basics haven't yielded much else so far!

Thanks Cathy. I'll do that. I'm just taking a break from Maverick 1 and looking at walkthroughs for A74 and A74V2, then I'll get back to it, edit my earlier posts and post my walkthrough.

Congratulations Cathy on reaching Master! Mike thinks you'll catch him up soon. You'll pass me before that.

My first placement in Maverick 1 came early but it was almost at the end before I got any more.

I think the way i went through it with the regular sudoku techniques may be one of the only paths through this puzzle with those techniques. But feel free to prove me wrong. I think that is probably what makes the puzzle harder. This puzzle felt more like a 1.5 rating to me. But that is mostly because it fell fairly quickly compared to say A60RP-Lite which i rated a 1.75. For me it wasn't in the same league as that puzzle.

I haven't tried A60RP-Lite or looked at any of the walkthroughs; must find time to do that. However it's clearly hard since that's the one that is always used as the comparison when deciding whether other puzzles should be 1.75.

It's easy to forget that each of these ratings represent a range. Obviously, from what people say about A60RP-Lite, it must be at the very top of the 1.75 rating range.

Maverick 1 felt to me at least as hard as any of the other 1.75s that I've done. I spent longer on it than on any of those others. It's definitely my longest walkthrough yet. Therefore I've no hesitation in rating Maverick 1 as 1.75.

Ratings by humans do, of course, depend on the difficulty of finding moves as well the moves themselves and the number of steps to finish a puzzle.

I haven't yet gone through Para's walkthrough. I'll do that before I post my own one. Para's is clearly shorter. I'll be interested to see if I missed anything important.

BTW, I hope that no-one thinks I'm trying to show off by writing those sort of posts. It's just that I like writing, and see this as a way in which I can give something back to the forum in return for all the great stuff I've learned from you all in the past. Also, by trying to analyze and consolidate information for technical articles and posts, I end up learning things myself that I would probably never have appreciated if I had continued to approach the subject in an "ad-hoc" manner.

I hope nobody would think that. When you post "tutorial style" analyses (I think you've used that phrase in our off-forum discussions) they are very interesting. Maybe I don't study them as much as I ought to; it's just like I've not yet found time to study Andrew Stuart's book.

You've been talking about AIC's etc. a lot. Maybe you know this. I was just wondering something about my Brick Wall chain. I used a link i can't really place. I think there should be a third type of link for Killer Sudoku.
Normally you have strong and weak links.
Strong link: If A is wrong, B is right.
Weak Link: If A is right, B is wrong.
In Killer Sudoku we can also have: If A is right, B is right or if A is wrong, B is wrong. The second, which i used in my chain, is probably more likely to happen. This can happen because certain values only occur with another value in a cage.
Don't know if you mentioned this before. But this doesn't really fit into the strong/weak link view of AIC's.

I've got an example of Para's third type of link in my walkthrough for Maverick 1, in one of the later steps not yet posted. It's not in a chain, just in clean-up.

I think that's all I've got to comment about all the interesting discussion that has appeared here in the last 24 hours.

[CathyW]

My first placement in Maverick 1 came early but it was almost at the end before I got any more.

I presume that was at r3c7. Haven't had much time to look at it today so no progress.

I managed to mess up today's Times puzzle. Somehow, I'd made 3+7 = 11 Note to self: check the basics before trying the advanced stuff. (Not that the Times puzzles often need anything other than basic killer moves!)

[mhparker]

Hi Andrew,

Maverick 1 felt to me at least as hard as any of the other 1.75s that I've done. I spent longer on it than on any of those others. It's definitely my longest walkthrough yet. Therefore I've no hesitation in rating Maverick 1 as 1.75.

Ratings by humans do, of course, depend on the difficulty of finding moves as well the moves themselves and the number of steps to finish a puzzle.

I'm sure you're correct in rating the puzzle as 1.75 is correct as far as the difficulty of your solving path is concerned. However, there's a strong argument that states that the published rating for a puzzle should relate to the difficulty of the optimum path, not the difficulty of the actual path taken by any particular person. Of course, this would mean that, given two puzzles with the same published rating, one of which has a narrow solving path and the other a much broader one, the puzzle with the narrower path will often be perceived as being significantly harder than the one with the broader path. This is simply because there are fewer alternative routes available through the puzzle with close-to-optimum difficulty that one can pick up on if one loses track of the ideal route. In short, a puzzle with a narrow solving path penalizes one's oversights more than a puzzle with a broader solution path.

In practice, this will mean that, say, a puzzle with a published rating of 1.5 that has a narrow solving path can easily be perceived as being more difficult than another puzzle with a published rating of 1.75, but which has a much broader solving path. But what does one do in the above case? Increase the rating of the first puzzle to 2.0 to reflect the perceived difficulty? If so, then one runs the risk of someone really finding the optimum path and then complaining that the puzzle is not anywhere near as difficult as it was cracked up to be. The other (possibly better) approach is to avoid this whole mess completely, and just publish the difficulty of the optimum path through the puzzle. This latter approach is the one I took when estimating the rating of the M1 at 1.5. Neither of our approaches is necessarily "wrong", it's just a matter of definition.

Another argument for not including the narrowness of the solving path in the ratings is that it cannot easily be measured. So, if the resulting fudge factor cannot be estimated with any reasonable degree of accuracy, it's arguably better to ignore it completely, and just inform everybody that the published ratings do not consider this parameter.

A good example of the effect of solving path on the rating was the A64V2, which received an SScore of a mere 1.28 fractionally lower than the SScore for the A64 (1.29), even though I originally rated it as 1.75 and you had said:

I'll be interested to see what rating Ed eventually comes up with for this puzzle; it's clearly much, much harder than A64.

However, the discrepency was arguably mainly (or even completely) due to the fact that we were all attacking the wrong part of the puzzle (namely the middle tower) when it would have been better to go up the left hand side, as goooders and SudokuSolver did.

So, again, one can claim that the SScore for the A64V2 of 1.28 is actually about right (for the optimum path), even though most people will find it much harder than a typical 1.25-rated puzzle in practice.

[mhparker]

Hi Cathy,

Somehow, I'd made 3+7 = 11

The trick I use in such cases is to put the bigger number first (i.e., 7+3 instead of 3+7). That way, it's easier (seven...eight...nine...ten! => you see, only three steps to count), and one doesn't run out of fingers or have to switch hands so frequently.

[Andrew]

Sorry for the delay in posting my walkthrough. I've now worked through Afmob's and Para's walkthroughs and Mike's analysis messages.

Thanks to Para and Mike for their encouragement which helped me finish this puzzle. If Para hadn't posted his partial walkthrough from my diagram, since expanded to a full walkthrough, and Mike hadn't given me off-forum encouragement, then I might have given up at the diagram stage. The fact that Mike pointed out that I'd missed a clean-up move in my diagram was also useful because, when I'd worked that out, it led directly to the next few steps.

Maverick 1 felt to me at least as hard as any of the other 1.75s that I've done. I spent longer on it than on any of those others. It's definitely my longest walkthrough yet. Therefore I've no hesitation in rating Maverick 1 as 1.75.

It's easy to forget that each of these ratings represent a range. Obviously, from what people say about A60RP-Lite, it must be at the very top of the 1.75 rating range.

I think the only important move that I missed was part of Para's step 33a. If I'd spotted that then I'd have been down to two combinations for the 20(4) cage in N6, instead of still having three, and would have been able to lock a second number in that cage. It took me a further 20 steps, from my step 46, to eliminate that combination for a completely different reason. Even if I'd spotted that extra move, I think I'd still have rated this puzzle 1.75 because of the time and thought I'd put in up to that stage.

Para's step 46 was impressive to spot that those innies/outie can be so useful. I also missed his step 50 which would have speeded up my later steps.

Afmob's breakthrough shortcut, his step 11a, was a nice use of simple chains. It depended, of course, on spotting that as well as two options for 4 in C1, he also spotted that there were two options for 4 in C9.

Mike's analysis showing a similar breakthrough, from the diagram that I posted earlier, was interesting. However I must admit it does look to me like a formalised way to express what is essentially a T&E step, starting from either R2C7 or R3C8 must be 4, using a long chain.

BTW That diagram was after step 42, but excluding sub-steps 24c and 36e to 36h which were added later.

I’ve tried as much as possible to avoid using hypotheticals and T&E to solve this problem. Some moves, for example step 42, are really combination analysis but are easier to present as if they were hypotheticals.

I must admit that at one stage, when I was stuck, I did look at hypotheticals based on the two permutations for the innies of N69 but the resulting long chains looked too much like T&E to me.

Here is my walkthrough for Maverick 1. I hope it's of interest. I did find a few things not in the other walkthroughs, for example the relationships between pairs of cages in C7, although only the one for R45C7 (step 40) proved useful. Many thanks to Ed for the comments and corrections.

Prelims

a) R1C34 = {59/68}
b) R1C67 = {49/67} (cannot be {58} which clashes with R1C34)
c) R2C23 = {49/58/67}, no 1,2,3
d) R2C78 = {39/48/57}, no 1,2,6
e) R34C1 = {17/26/35}, no 4,8,9
f) R34C4 = {39/48/57}, no 1,2,6
g) R34C6 = {29/38/47/56}, no 1
h) R34C9 = {69/78}
i) R5C34 = {18/27/36/45}, no 9
j) R5C67 = {59/68}
k) R67C1 = {69/78}
l) R67C4 = {18/27/36/45}, no 9
m) R67C6 = {18/27/36/45}, no 9
n) R67C9 = {17/26/35}, no 4,8,9
o) R8C23 = {49/58/67}, no 1,2,3
p) R8C78 = {39/48/57}, no 1,2,6
q) R9C34 = {19/28/37/46}, no 5
r) R9C67 = {39/48/57}, no 1,2,6
s) 6(3) cage in N3 = {123}, locked for N3, clean-up: no 9 in R2C78
t) 8(3) cage at R6C7 = 1{25/34}
u) 6(3) cage in N7 = {123}, locked for N7, clean-up: no 7,8,9 in R9C4

Must be getting close to a record number of prelims!!

1. Killer pair 6,9 in R1C34 and R1C67, locked for R1

2. R2C23 = {49/67} (cannot be {58} which clashes with R2C78), no 5,8

3. Killer pair 4,7 in R2C23 and R2C78, locked for R2

4. Killer triple 1,2,3 in R34C1 and R89C1, locked for C1

5. Killer triple 1,2,3 in R12C9 and R67C9, locked for C9

6. 45 rule on C1234 2 innies R28C4 = 7 = {16/25}/[34], no 7,8,9, no 3 in R8C4

7. 45 rule on C6789 2 innies R28C6 = 10 = {19/28}/[37/64], no 5, no 3,6 in R8C6

8. 45 rule on C12 4 innies R2378C2 = 26 = {2789/3689/4589/4679/5678}, no 1

9. 45 rule on C789 3 innies R159C7 = 24 = {789}, locked for C7, clean-up: no 7,9 in R1C6, no 4,5 in R2C8, no 8,9 in R5C6, no 3,4,5 in R8C8, no 7,8,9 in R9C6
9a. R34C6 = {29/38/47} (cannot be {56} which clashes with R5C6), no 5,6

10. 45 rule on C89 4 innies R2378C8 = 24 = {1689/2589/2679/3489/3579/3678/4578} (cannot be {4569} because R2C8 only contains 7,8)
10a. All combinations require two of 7,8,9 which must be in R28C8 -> no 7,8,9 in R3C8

11. Hidden triple 7,8,9 in N3 -> R3C9 = {789}, clean-up: no 9 in R4C9
[Alternatively, as Para pointed out, there is naked triple {456} in R2C7 + R3C78.]

12. 6 in N3 locked in R3C78, locked for R3 and 13(3) cage at R3C7, clean-up: no 2 in R4C1
12a. 13(3) cage at R3C7 = 6{25/34}, no 1
12b. 2,3 only in R4C7 -> R4C7 = {23}

13. R3C7 = 6 (hidden single in C7)

14. 1 in C7 locked in R67C7 -> no 1 in R7C8

15. 45 rule on C789 3 outies R159C6 = 15 = {456} (only remaining combination), no 3, clean-up: no 9 in R9C7
15a. Naked triple {456} in R159C6, locked for C6, clean-up: no 7 in R34C6, no 3 in R67C6
15b. R28C6 (step 7) = {19}/[37] (cannot be {28} which clashes with R34C6), no 2,8

16. 45 rule on C123 3 outies R159C4 = 17 = {179/269/278/359/368/458/467}
16a. 1 of {179} must be in R9C4 -> no 1 in R5C4, clean-up: no 8 in R5C3
16b. 4 of {458} must be in R9C4
16c. 6 of {467} must be in R1C4 and therefore the 4 must be in R9C4
16d. -> no 4 in R5C4, clean-up: no 5 in R5C3

17. 45 rule on C123 3 innies R159C3 = 16 = {169/178/259/268/349/358/367/457}
17a. 1,2,3 of {169/268/367} must be in R5C3 -> no 6 in R5C3, clean-up: no 3 in R5C4

18. 45 rule on C1 1 innie R5C1 – 1 = 2 outies R19C2, max R19C2 = 8, no 8 in R1C2

19. 45 rule on C9 2 outies R19C8 – 1 = 1 innie R5C9, min R5C9 = 4 -> min R19C8 = 5, no 1 in R9C8

20. 1 in N9 locked in R7C79, locked for R7, clean-up: no 8 in R6C46

21. 45 rule on R9 2 outies R8C19 = 1 innie R9C5, max R8C19 = 9, no 9 in R8C9
[Not sure why I didn’t also do Min R8C19 = 5 -> min R9C5 = 5. It would have avoided the complications of steps 36a to 36d.]

22. 45 rule on N1 2 innies R1C3 + R3C1 – 5 = 1 outie R4C3
22a. IOU no 5 in R3C1, clean-up: no 3 in R4C1
22b. R1C3 + R3C1 cannot total 14 -> no 9 in R4C3

23. 15(3) cage in N1 = {159/249/258/348/357/456} (cannot be {168/267} because R12C1 = [76/86] clashes with R67C1)
23a. 3 of {357} must be in R1C2 -> no 7 in R1C2

24. 45 rule on N7 2 innies R7C1 + R9C3 – 9 = 1 outie R6C3
24a. IOU no 9 in R7C1, clean-up: no 6 in R6C1
24b. Max R7C1 + R9C3 = 17 -> no 9 in R6C3
24c. Deleted.
[Alternatively step 24b comes from min R7C23 = 9 -> max R6C3 = 8.]

25. 45 rule on N9 3 innies R7C789 – 4 = 1 outie R9C6
25a. R9C6 = {45} -> R7C789 = 8,9 and must contain 1 (step 20) = 1{25/26/34/35}, no 7, clean-up: no 1 in R6C9

26. Hidden killer triple 7,8,9 in N9 -> 17(3) cage must contain one of 7,8,9 = {269/359/458/467} (cannot be {278} which contains both of 7,8 (it also clashes with R9C7), cannot be {368} because that leaves no valid cell for 9 in N9)
[Alternatively cannot be {368} because R89C9 = {68} clashes with R34C9. That’s more direct but I saw the other reason first.]
26a. 2 of {269} and 3 of {359} must be in R9C8 -> no 9 in R9C8
26b. If {359} => R8C78 = {48} => R9C67 = [57]
26c. If {458} => R9C67 = [57]
26d. -> 5 of {359/458} must be in R8C9 -> no 8 in R8C9, no 5 in R9C89
26e. R89C9 cannot be {67} which clashes with R34C9

27. 5 in R9 locked in R9C56, locked for N8, clean-up: no 2 in R2C4 (step 6), no 4 in R6C4

28. 1,2,3 in N1 only in R1C2 and R3C123
28a. 45 rule on N1 3 innies R3C123 – 3 = 1 outie R1C4
28b. Min R1C4 = 5 -> min R3C123 = 8 so cannot be {123} -> R1C2 = {123}

29. Naked triple {123} in R1C2 and R1C89, locked for R1

30. 15(3) cage in N1 (step 23) = {159/249/258/348/357} (cannot be {456} which is blocked by R1C2), no 6
30a. Cannot be {357} because R12C1 = [75] => R67C1 = [96] -> no valid combinations for R34C1
30b. 15(3) cage = {159/249/258/348}, no 7
30c. Killer pair 8,9 in R12C1 and R67C1, locked for C1

31. Max R19C2 = 5 -> max R5C1 = 6 (step 18)
[This would also have worked for step 30c but I saw the killer pair first.]
[Ed pointed out R5C34 cannot be [45] because of R5C16. Nice ALS block! I didn’t get this elimination until step 68a but I don’t think that affected the solving path.]

32. R19C2 = {123} -> 20(4) cage in N4 cannot contain more than one of 1,2,3
32a. 20(4) cage = {1469/1478/1568/2459/2468/2567/3458/3467} (cannot be {1289/1379/2369/2378} which contain two of 1,2,3)
32b. Killer triple 1,2,3 in R19C2 and R456C2, locked for C2

33. Hidden triple {123} in N1 -> R3C13 = {123}, clean-up: no 1 in R4C1

34. 12(3) cage at R3C2, min R3C23 = 5 -> max R4C3 = 7

35. 8 in C7 locked in R59C7
35a. 45 rule on N69 4 innies R4C9 + R459C7 = 25 = [6298/7387] (listed in the order R4C9 + R459C7, to make it easier to work out R459C7, rather than normal cage order, cannot be [6397/7297] which clash with R1C7, cannot be [8287] which would repeat 8 in N6) -> no 8 in R4C9, clean-up: no 7 in R3C9

36. 17(3) cage in N9 (step 26) = {269/359/458/467}
36a. {269/359/458} and {467} with 4 in R9C89 all have one of 2,3,4 in R9C89
36b. If 4 of {467} in R8C9 => R9C89 = {67} => R9C67 = [48]
36c. -> one of 2,3,4 in R9C689
36d. Killer quad 1,2,3,4 in R9C12, R9C34 and R9C689, locked for R9
[Alternatively there’s the simpler Min R8C19 = 5 -> min R9C5 = 5 (step 21) but I only saw that afterward. Strange how it’s sometimes easier to see more difficult steps!]
36e-h. Deleted. These sub-steps have been moved to step 44.

37. 45 rule on R2 5 innies R2C14569 = 20 = {12359/12368}
37a. 8 of {12368} must be in R2C1 -> no 8 in R2C5

38. 17(4) cage in N2 = {1259/1268/1349/1358/1367/2357} (cannot be {1457/2348} because 4,7,8 only in R1C5, cannot be {2456} which clashes with R1C6)
[Ed pointed out that 17(4) cage in N2 must have two of 1,2,3 in R2 which would have simplified the combination eliminations for this step.]

39. R3C7 = 6 -> R3C8 + R4C7 = 7
39a. R2C7 + R3C8 = {45} = 9
39b. -> R2C7 – 2 = R4C7, R24C7 = [42/53]

40. R4C9 + R459C7 (step 35a) = [6298/7387] -> R45C7 = [29/38] = 11
40a. 45 rule on N6 3 remaining innies R4C9 + R6C79 = 14 = {167/257/347/356}

41. If R6C7 = 1 -> R7C9 = 1 (step 20) -> R6C9 = 7
41a. If R6C7 <> 1 -> R7C7 = 1 -> R7C9 <> 1 -> R6C9 <> 7
41b. -> R6C79 = [17] or R6C7 <> 1 and R6C9 <> 7

42. R7C8 cannot be 3, here’s how
42a. R7C8 = 3 -> R67C7 = {14} => R2C7 = 5 clash with R8C7
42b. -> no 3 in R7C8

43. 8(3) cage at R6C7 = 1{25/34}
43a. 4 of {134} must be in R7C8 -> no 4 in R67C7

44. 4 in C7 locked in R28C7
44a. Either R2C78 or R8C78 must be [48] -> 8 locked in R28C8, locked for C8
[Ed commented that each of the 8s locked in R28C8 forces 9 into C9 for the same nonet -> no 9 in R5C9]
44b. 17(3) cage in N9 (step 26) = {269/359/458/467}
44c. 4 of {458} must be in R9C8
44d. 4 of {467} must be in R8C9, here’s how
44da. 4 of {467} in R9C8 => R89C9 = {67} clashes with R4C9
44db. 4 of {467} in R9C9 => R9C67 = [57] clashes with {467}
44e. -> no 4 in R9C9
44f. -> no 7 in R8C9 (step 44d)

45. 8 in N6 locked in R5C79, locked for R5, clean-up: no 1 in R5C3

46. 4 in N6 locked in 20(4) cage = {1469/1478/2459} (cannot be {2468} which clashes with R45C7, cannot be {3458} because R456C8 = {345} clashes with R3C8, cannot be {3467} which clashes with R4C9), no 3
46a. 8 of {1478} must be in R5C9 -> no 7 in R5C9
46b. If {1469} => R5C67 = [68] -> no 6 in R5C89
[Para also eliminated {2459} because cannot have two of {245} in R456C8 which would clash with R37C8.]

47. R4C9 + R6C79 (step 40a) = {167/257/356}
47a. 1 of {167} must be in R6C7 => 7 must be in R6C9 (step 41b)
47b. 6 of {356} must be in R4C9
47c. -> no 6 in R6C9, clean-up: no 2 in R7C9

48. 45 rule on N9 3 outies R6C79 + R9C6 = 12
48a. R4C9 + R6C79 = 14 (step 40a)
48b. -> R4C9 – 2 = R9C6
48c. -> R3C9 – 1 = R9C7

49. From steps 48b and 48c R3C9 = 8 or R9C7 = 8 -> no 8 in R9C9
49a. 17(3) cage in N9 (step 36) = {269/359/467}
49b. 4 of {467} must be in R8C9 (step 44d) -> no 4 in R9C8

50. R159C4 (step 16) = {179/269/278/359/368/458/467}
50a. 8,9 of {359/458} must be in R1C4 -> no 5 in R1C4, clean-up: no 9 in R1C3
50b. If R159C4 = {368} => 9 in C4 locked in R34C4 = {39} clashes with R159C4
50c. -> no {368} in R159C4 = {179/269/278/359/458/467}

51. No 5 in R1C5, here’s how
51a. R1C34 = [59]/{68}
51b. If [59] -> no 5 in R1C5
51c. If {68} => R1C67 = [49] => R1C1 = 5
51d. -> no 5 in R1C5

52. 5 in R1 locked in R1C13, locked for N1

53. 15(3) cage in N1 (step 30b) = {159/249/258/348}
53a. R2C1 = {89} -> no 8 in R1C1

54. Killer triple 4,5,6 in R1C1, R1C34 and R1C6, locked for R1
54a. Killer triple 7,8,9 in R2C1, R2C23 and R2C8, locked for R2
54b. 9 in R2 locked in R2C123, locked for N1

55. 15(3) cage in N1 (step 30b) = {159/249/258/348}
55a. Cannot be {348}, here’s how
55b. R1C2 = 3, R12C1 = [48] => R2C23 = {67} -> R12C1 and R2C23 clash with R3C2
55c. -> no {348} in 15(3) cage = {159/249/258}, no 3
55d. 3 in N1 locked in R3C13, locked for R3, clean-up: no 9 in R4C4, no 8 in R4C6
55e. 3 in R1 locked in R1C89, locked for N3

56. 9 in C4 locked in R13C4, locked for N2, clean-up: no 2 in R4C6, no 1 in R8C6 (step 7)
[The second clean-up step should have been in step 54a but I forgot about it until here.
At this stage I also overlooked that R8C23, R8C6 and R8C8 form killer triple 7,8,9 for R8 which would have been much simpler than step 57 which only achieves this result.]

57. 45 rule on C5 4 innies R1289C5 = 20 = {1289/1379/1478/1568/2378/2567/3458/3467} (cannot be {1469/2369/2459} because R1C5 only contains 7,8, cannot be {2468} because R12C5 = [82] clashes with R3C6)
57a. 1,2,3,4 of {1289/1379/1478/2378/2468/3458/3467} must be in R28C5
57b. 8 of {1568} must be in R1C5
57c. 7 of {2567} must be in R1C5
57d. -> no 7,8,9 in R8C5

58. 3 in N2 locked in 17(4) cage
58a. 45 rule on N2 5 innies R1C46 + R3C456 = 28 with 9 locked in R13C4 = {14689/24589/24679} (cannot be {15679} because R3C6 only contains 2,8)
58b. {14689} must be [64918/96418] because R1C7 = 9 when R1C6 = 4
58c. {24589} must be [84952] because R1C7 = 9
58d. {24679} must be [64972/96472/96742] because R1C7 = 9 when R1C6 = 4
58e. -> no 5,8 in R3C4, no 2,8 in R3C5, clean-up: no 4,7 in R4C4
[Ed commented about step 58b
Actually, the first is blocked: [918] clashes with {89} in R3C9. But this doesn't make any additional elims I don't think. Actually, a nice little chain eliminates both. Innies = {14689} -> R2C6 = 3 => R4C6 = 9 => R3C6 = 2: but no 2 in innies {14689}. This would be nice because it locks 2 in innies and from the end of step 58 this must be R3C6!]

59. 17(4) cage in N2 (step 38) must contain 3 = {1358/1367/2357}
59a. 1 in C6 locked in R26C6
59b. If R2C6 <> 1 => R6C6 = 1 => R7C6 = 8 => R3C6 = 2 => R23C6 = [32]
59c. -> 17(4) cage cannot be {2357}

60. 17(4) cage in N2 (steps 59 and 59c) = {1358/1367}, no 2, 1 locked for N2

61. R3C6 = 2 (hidden single in N2), R4C6 = 9, R8C6 = 7, R67C6 = [18], R2C6 = 3, clean-up: no 6 in R4C1, no 7 in R6C1, no 2 in R6C4, no 7 in R6C9 (step 41b), no 1 in R7C9, no 6 in R8C23, no 5 in R8C7

62. R1C2 = 2 (hidden single in N1)

63. R2C9 = 2 (hidden single in R2), clean-up: no 6 in R7C9
[This should have been naked single in step 60 but I overlooked that 1 was also locked for R2.]

64. R7C7 = 1 (hidden single in C7)

65. Naked pair {35} in R67C9, locked for C9 -> R1C89 = [31]

66. Naked triple {235} in R4C7 + R6C79, locked for N6
[That’s now made the combination elimination from the 20(4) cage that I missed at step 46!]

67. 5 in N9 locked in R7C89, locked for R7

68. 5 in N7 locked in R8C23 = {58}
68a. Naked pair {58} in R8C23, locked for R8 and N7 -> R8C8 = 9, R8C7 = 3, R46C7 = [25], R67C9 = [35], R2C78 = [48], R3C89 = [59], R1C67 = [67], R1C5 = 8, R1C34 = [59], R12C1 = [49], R8C23 = [58], R9C7 = 8, R5C67 = [59], R9C6 = 4, R5C1 = 6, R67C1 = [87], clean-up: no 1 in R3C1, no 7 in R3C4, no 3 in R4C4, no 3,4 in R5C3, no 6 in R7C4, no 6 in R9C3, no 2,3,6 in R9C4

69. R34C1 = [35] (naked pair), R3C3 = 1, R34C4 = [48] (naked pair), R3C5 = 7, R3C2 = 8, R9C34 = [91] (naked pair), R9C12 = [23], R8C1 = 1, R2C45 = [51]
69a. R5C3 = 2 (step 17), R5C4 = 7, clean-up: no 2 in R7C4
69b. R67C4 = [63] (naked pair) -> R8C45 = [26], R8C9 = 4, R9C5 = 5, R5C9 = 8, R7C5 = 9

70. R4C9 = 6, R9C89 = [67]

71. R4C3 = 3 (hidden single in C3), R456C5 = [432]

72. R7C23 = {46} -> R6C3 = 7 (cage sum)

and the rest is naked singles


4 2 5 9 8 6 7 3 1
9 7 6 5 1 3 4 8 2
3 8 1 4 7 2 6 5 9
5 1 3 8 4 9 2 7 6
6 4 2 7 3 5 9 1 8
8 9 7 6 2 1 5 4 3
7 6 4 3 9 8 1 2 5
1 5 8 2 6 7 3 9 4
2 3 9 1 5 4 8 6 7