## 22 (Clues) / 7 (Columns) / Boolean Algebra

Post puzzles for others to solve here.

### Re: 22 (Clues) / 7 (Columns) / Boolean Algebra

Hi Mith

A further question since the link between (16) r4c789 and 16r4c56 is (16) r4c789-16r4c56, is it correct to say that does the chain 2r4c6 = r6c6-r6c12 = r5c1- (2 = 4,5,1,6) r5c3479- (16 ) r4c789- (16) r4c56 imply the deletion of 7r4c6? Since even if it is not an AIC, when 2r4c6 is false whether 16r4c56 is true or false it always sees 7r4c6, in fact the negation of (16) r4c56 which means (not 1r4c56 or not 6r4c56), that is (r4c6 = 1 and not r4c5 = 6) or (not r4c5 = 6 and not r4c6 = 1) or (r4c5 = 1 and not r4c5 = 6) sees 7r4c6.

Paolo
Ajò Dimonios

Posts: 201
Joined: 07 November 2019

### Re: 22 (Clues) / 7 (Columns) / Boolean Algebra

eleven wrote:Is it really that hard to understand ??

Yes.

(I assume you're talking about the May 17 puzzle. Is there a higher purpose to why you're repeatedly keeping your comments as vague as possible?)

The point is not, if the link is true, but if it is proved.
Any elimination chain, AIC, implication, what else, must prove, that the elimination is valid.
But SpAce's link is not proved.

Is that a fact?

Depending on, what (additional) links are used, negating one side, you can follow, that the other side is false or true.

Did Paolo hack your account? I can't honestly tell the difference if I read comments like this.

Otherwise in the current puzzlei can write
(3r1c3=3r5c1) => -3r6c3, stte,
because (looking at the solution) the link is true.

The two situations have nothing in common. Nothing at all. My link does not depend on any external logic whatsoever. If it did, I would have written it as a derived link with '==' like you should have done above. Yes, there's some implicit logic in my link, but that's commonly accepted if it happens within the nodes and between the candidates that are explicitly written in the link. That's certainly true here. It is YOU, and YOU ONLY, who thinks some extra logic is needed for that link.

I never denied that my link is not as readable as it could be, but that doesn't make it ambiguous (i.e. invalid). Besides, it's irrelevant in this discussion because you denied the validity of even the explicit form of it. It means you don't accept my logic at all, so it's not just a notation issue.

If you're right, then at least Steve, mith, and I have a serious problem in our understanding of AICs. Are you really sure you alone are right about this? That too sounds a lot like our other friend here. Granted, you could afford such a level of confidence much better (and for sure, the majority is not always right), but in this case you should probably reconsider your position. At the very least, try to find arguments that make any sense.

PS. Just to make sure...

because (looking at the solution) the link is true.

Do you think I find my strong links by looking at the solution? I have to ask because you already hinted that you misinterpreted my comments about the strong-link definition like that. If you really did, I'm speechless. Such a misinterpretation would take a lot of bad will and total disrespect for my skills and solving methods. Unfortunately I can't rule it out, because you've proved to be quite capable of both.

SpAce

Posts: 2573
Joined: 22 May 2017

### Re: 22 (Clues) / 7 (Columns) / Boolean Algebra

mith wrote:Ok, interpreted that way, yes, the logic [of the NOR-chain] is fine.

Thanks for confirming it!

I'm not a fan of the non-binary operators, though I shudder to think what monstrosity that would be without.
(Just because you *can* write any boolean expression using only NOR gates doesn't mean you *should*.)

I agree on all three accounts. That said, the Apollo guidance computer (which I mentioned in the intro) used three-input NOR-gates.

As I said, the logic of the May 17 link is clearly sound.

Thank you for the reconfirmation. Unfortunately eleven still doesn't agree.

I got the impression eleven was quibbling about the notation / lack of detail on why that link is strong, since it is not immediately obvious just from the cells involved?

Something like that, but I don't understand his argument at all. All the logic needed for that link is contained within the four cells that are listed with the link. Besides, I think his problem goes deeper than that, because he denied the validity of the explicit form as well (link to it in my previous reply to him).

There's a difference between "ambiguous" and "not obvious". The former is invalid (i.e. it wouldn't compile if it were a piece of code), while the latter is just harder to understand. eleven specifically called my link "ambiguous" and has never pulled it back. So, it's not a question of obviousness but correctness.

But I'm not entirely clear even on which puzzle he was talking about you being right, so.

Yeah, he tends to keep things vague for some reason. Nevertheless, I think it's pretty clear at this point that it was not the May 17 puzzle, which leaves just one possibility.

That said, I'm not at all sure what he actually meant by me being right in the other puzzle, because it was so vague too. It could be just one small part, but not the critical point. That's why I would appreciate your opinion on that one too. It's a more complicated issue, but I think it might make it interesting for you. I can explain it in more detail if you're still willing to take a look at it.

I can do logic all day, but if you're looking for an arbiter of AIC notation, you're in trouble.

I don't think either of these is merely a notational issue. There's a deeper logic problem behind. That's why I'm pretty sure you have the skills to be a credible arbiter, even if you don't consider yourself an expert in AIC notations.

SpAce

Posts: 2573
Joined: 22 May 2017

### Re: 22 (Clues) / 7 (Columns) / Boolean Algebra

All the logic needed for that link is contained within the four cells that are listed with the link.

Agreed.

NOT((3,8)r3c6,r7c9) means 8r3c6 OR 3r7c9
8r3c6 => 3r8c6 (3b8p6) => 8r7c5 (8b8p2)
3r7c9 => 8r7c5 (8b8p2) => 3r8c6 (3b8p6)
therefore NOT((3,8)r3c6,r7c9) => (38)b8p26 (specifically, (8,3)b8p26)

(The other direction is also correct; NOT((38)b8p26) => 6b8p2 => 8r7c9 AND 8b8p6 => 3r3c6)

I'll take a look at the other at some point - kids are wearing me out right now.
mith

Posts: 301
Joined: 14 July 2020

### Re: 22 (Clues) / 7 (Columns) / Boolean Algebra

SpAce, a proof is NOT, if somebody sees a link in the way you want, but if no other interpretation is possible.

If r3c6,r7c9 is not 3,8, then it can be 3,3, and no 3 is possible in b8p26. So there is no strong link. dot.
That's enough to show, that it is not proved without writing out the links you use.
Same, if both are 8, then there can:t be an 8 in b8p26.
So the conclusion is, that the writer of the (poor) link only looked at one of 3 possibilities ...
SpAce wrote:My link does not depend on any external logic whatsoever.

This is wrong. It uses the link 8b8p6 = 8b8p2, which is not written out (and if there were another 8 in the box - in a cell not included in the link, it would always be wrong).
eleven

Posts: 2461
Joined: 10 February 2008

### Re: 22 (Clues) / 7 (Columns) / Boolean Algebra

Ok, I can see that side of it. It is true that (3,3)r3c6,r7c9 => (38)b8p26 since we can show (3,3)r3c6,r7c9 is false ((3,3)r3c6,r7c9 => 8r7c5 (8b8p2) => 3r8c6 (3b8p6) => -3r3c6 #); False => A is always true, whatever the A. (Likewise (8,8)r3c6,r7c9 => 3r8c6 (3b8p6) => 8r7c5 (8b8p2) => -8r7c9 #.)

So the logic is ultimately sound, but I can see why it wouldn't be obvious breaking it into three cases rather than using De Morgan; and eleven is correct that it relies on 8b8p6 = 8b8p2 in both cases. (I just viewed this as a hidden single, but it does rely on no 8 candidate elsewhere in the box, which is looking at another cell.)
mith

Posts: 301
Joined: 14 July 2020

### Re: 22 (Clues) / 7 (Columns) / Boolean Algebra

eleven wrote:SpAce, a proof is NOT, if somebody sees a link in the way you want, but if no other interpretation is possible.

If r3c6,r7c9 is not 3,8, then it can be 3,3, and no 3 is possible in b8p26. So there is no strong link. dot.
That's enough to show, that it is not proved without writing out the links you use.
Same, if both are 8, then there can:t be an 8 in b8p26.
So the conclusion is, that the writer of the (poor) link only looked at one of 3 possibilities ...

So, apparently your argument hasn't changed at all since we had the original discussion. In other words, you totally ignored my extensive counterarguments to it. Why should I pay any attention to yours then?

To avoid jumping between threads, here's my original solution for reference:

SpAce wrote:
Code: Select all
`.------------.------------------.----------------.| 7  59   25 | 4    238     1   | 38   6    389  || 3  1    8  | 9    7       6   | 4    5    2    || 4  69   26 | 5    238   ac38* | 1    7    9-38 |:------------+------------------+----------------:| 8  4    1  | 3    9       5   | 7    2    6    || 5  2    3  | 6    1       7   | 9    8    4    || 6  7    9  | 8    4       2   | 5    3    1    |:------------+------------------+----------------:| 1  356  56 | 7  bc38#6*   9   | 2    4  ac38*  || 2  36   4  | 1    5     bc38* | 368  9    7    || 9  8    7  | 2    36      4   | 36   1    5    |'------------'------------------'----------------'`

(3,8)r3c6,r7c9 = (38-6)b8p26 = RP(38)r38c6,r7c59 => -38 r3c9; stte

Let's forget that short version for a second and look at the more explicit one I offered:

eleven wrote:
SpAce wrote:(3,8)r3c6,r7c9 = (3,8)b8p62|(8,3)b8p26 - 6b8p2 = RP(38)r38c6,r7c59

You agree with that, I hope?

No, if i look at the 4 cells, and r3c6,r7c9 is not 3,8, but 3,3, then b8p6 is 8 and b8p2 is 6.

Does eleven's argument (same as now) make any sense, considering what's written in my chain? Only the left side of the first strong link has the cells r3c6,r7c9. The right side has b8p26 i.e. r7c5,r8c6. Thus it should be pretty obvious that bilocal strong links (+ implicit logic) are used to build the combo strong link. No possible cases are missing from that, as far as I see.

Why would I worry about any alternate digit combos in r3c6,r7c9? My logic is not using them, and it should be obvious from the way it's written. To me eleven's argument is like saying that someone's use of UR externals is invalid if there's some contradicting logic available with the internals. It's ignoring the fact that both sets are valid SIS on their own. What am I missing?

Let's break it down so you can tell me where it's actually ambiguous.

1. Do you agree that:

(3,8)r3c6,r7c9 <-> (3r3c6 & 8r7c9)
(3r3c6 & 8r7c9) = (3r8c6 | 8r7c5)

2. Do you agree that:

3r3c6 = (3-8)r8c6 = 8r7c5
8r7c9 = 8r7c5 - (8=3)r8c6

3. Do you agree that we can (and routinely do) write them (2) as:

3r3c6 = (3,8)b8p62
8r7c9 = (8,3)b8p26

4. Do you agree that (based on 1 and 3):

(3r3c6 & 8r7c9) = (3,8)b8p62|(8,3)b8p26
=>
(3,8)r3c6,r7c9 = (3,8)b8p62|(8,3)b8p26

? If so, you should accept the explicit version. If not, tell me where the error is.

5. Do you agree that:

(3,8)b8p62 <-> (8,3)b8p26
=>
(3,8)r3c6,r7c9 = (3,8)b8p62

? If so, we're pretty close to the short version. I'd be fine leaving it like that if it's easier to read. The original without the latter comma takes a bit more mental acrobatics to see why the strong link is valid.

SpAce wrote:My link does not depend on any external logic whatsoever.

eleven wrote:This is wrong. It uses the link 8b8p6 = 8b8p2, which is not written out (and if there were another 8 in the box - in a cell not included in the link, it would always be wrong).

mith wrote:eleven is correct that it relies on 8b8p6 = 8b8p2 in both cases. (I just viewed this as a hidden single, but it does rely on no 8 candidate elsewhere in the box, which is looking at another cell.)

That is a pretty weird interpretation of external logic. Of course every bilocal link depends on candidates missing in the other cells of the relevant house. If they're counted as part of the logic, then missing digits in bivalue strong links should be counted too. Absurd, no?

SpAce

Posts: 2573
Joined: 22 May 2017

### Re: 22 (Clues) / 7 (Columns) / Boolean Algebra

Do you agree, that my sample link above (3r1c3=3r5c1) is true ?
You see, that one side false implies the other false, not true, as needed ? Bad luck.

Your nice and elegant link is incomplete, ambiguous, and therefore flawed. Swollow it.
eleven

Posts: 2461
Joined: 10 February 2008

### Re: 22 (Clues) / 7 (Columns) / Boolean Algebra

eleven wrote:Do you agree, that my sample link above (3r1c3=3r5c1) is true ?

No. A single '=' implies a direct strong link between the two nodes (possibly with implied internal logic), but that doesn't exist here. Like I said, it should be written 3r1c3==3r5c1. Like I also said originally, I might accept that my compact link needs '==', too, though you haven't argued successfully why. (I can see one such argument myself, but it hasn't been presented.)

That question is not on the table right now, however. We can get back to the problems with the compact link once you answer my questions about the explicit link. I want to hear your reasoning for why you think it's invalid as well. Recycling your old arguments doesn't work, because I've presented counterarguments to them a long time ago. You should refute my counterarguments to make any progress, but as usual, you've simply ignored them and continue to do so.

Your nice and elegant link is incomplete, ambiguous, and therefore flawed. Swollow it.

Whatever its faults, it's infinitely less flawed than your arguments trying to prove them. Then again, your goal was never to convince me, was it? You just wanted to make me look bad, that's all.

--
Three simple questions.

1. Which one of these is a valid SIS:

a.
Code: Select all
`(3,8)r3c6,r7c9||(3,8)b8p62||(8,3)b8p26`

b.
Code: Select all
`(3,8)r3c6,r7c9||(8,3)r3c6,r7c9||(3,3)r3c6,r7c9||(8,8)r3c6,r7c9`

c.
Code: Select all
`Both.`

2. Which one of them is depicted in the first strong link of my chain?

3. What evidence do you see that it could be (or should be) the other? If you can't demonstrate any, then...

eleven wrote:If r3c6,r7c9 is not 3,8, then it can be 3,3, and no 3 is possible in b8p26. So there is no strong link. dot.

...is the most useless argument ever. You can't claim something is ambiguous if you deliberately refuse to read it as it's written.

SpAce

Posts: 2573
Joined: 22 May 2017

### Re: 22 (Clues) / 7 (Columns) / Boolean Algebra

I tell you, how your chain can be written correctly:
8r3c6 - r8c6 = r7c5
(3-8)r7c9 = r7c5
(3,8)r3c6,r7c9 == (38-6)b8p26

As long as at least the link 8r8c6 = r7c5 is missing in your chain, it is incomplete, ambiguous and flawed, therefore as invalid as my sample link.
Got it now ?
eleven

Posts: 2461
Joined: 10 February 2008

### Re: 22 (Clues) / 7 (Columns) / Boolean Algebra

eleven wrote:I tell you, how your chain can be written correctly

Oh, you tell me? How generous! What makes you think you're such a god that you don't have to argue your point like everyone else? Part of that is responding to other people's counterarguments and questions, which you've totally ignored.

8r3c6 - r8c6 = r7c5
(3-8)r7c9 = r7c5
(3,8)r3c6,r7c9 == (38-6)b8p26

Yeah, that's not confusing at all. Lol. It's not depicting my logic anyway. You know very well that I would never write my chains that way, so you can keep that style all to yourself.

If you want to offer alternatives to my chain, try something reasonable that is at the very least written on one line. But, before you do, answer my questions. I don't give a damn about anything else you write until you do.

As long as at least the link 8r8c6 = r7c5 is missing in your chain, it is incomplete, ambiguous and flawed, therefore as invalid as my sample link. Got it now ?

Just answer my questions and stop playing god. You can't afford it with me. Swallow it.

SpAce

Posts: 2573
Joined: 22 May 2017

### Re: 22 (Clues) / 7 (Columns) / Boolean Algebra

mith, we can actually use the May 17 puzzle as an example of the other dispute too. It was about using chain fragments as nodes. Normally we do everything to avoid such kludges, but that necessarily results in hiding some details of the logic, just like here. If one wants to please eleven (like I desperately do, obviously), the full logic could be written with a chain fragment. The question is, how exactly.

Here's the three main variants that came up in that discussion:

a. (3r3c6 & 8r7c9) = ([(3-8)r8c6 = 8r7c5] | 8r7c5) - 6r7c5 = RP(38)

b. (3r3c6 & 8r7c9) = ([! = (3-8)r8c6 = 8r7c5] | 8r7c5) - 6r7c5 = RP(38)

c. (3r3c6 & 8r7c9) = ((3r8c6 & [(3-8)r8c6 = 8r7c5]) | 8r7c5) - 6r7c5 = RP(38)

Which ones, if any, are correct?

(The chain fragment is colored blue, and the differences purple. '!' is FALSE, i.e. a contradiction.)
Last edited by SpAce on Mon Sep 07, 2020 6:20 pm, edited 1 time in total.

SpAce

Posts: 2573
Joined: 22 May 2017

### Re: 22 (Clues) / 7 (Columns) / Boolean Algebra

SpAce wrote:Just answer my questions and stop playing god. You can't afford it with me. Swallow it.

Can you give me another example of a "valid" AIC, where needed links are missing ?
eleven

Posts: 2461
Joined: 10 February 2008

### Re: 22 (Clues) / 7 (Columns) / Boolean Algebra

I wouldn't be asking if they were. It's funny that you complain about ambiguity in my chains, while you avoid clarity in your own verbal communication like a plague. I have zero interest in trying to decipher your comments. You should understand, having the same problem with my chains. How about you set a good example first?

SpAce

Posts: 2573
Joined: 22 May 2017

### Re: 22 (Clues) / 7 (Columns) / Boolean Algebra

Ok, if it is too hard for you to understand, i will spell it out. a is no valid SIS (without further explanation), b is.
And now you tell me, where i can find the 8 link in box 8 in your chain - or don't you need it ?
I already gave you an example of a link, which is as invalid as yours, what else do you want ?
eleven

Posts: 2461
Joined: 10 February 2008

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