The New Sudoku Players' Forum

[SpAce]

Speaking of memory chains... I hope you don't mind a bit of friendly feedback on yours.

7r8c6=r4c6-r4c4=r2c4-(7=8)r1c5-(8=6)r1c8-6r1c2=r2c1-r5c1=(6-4*5) r5c9=(5-48)r6c9=48r4c789-(48=7)r4c4=>-7r4c6=>stte

I like the logic and have no problem understanding your chain, but it could be improved. (In general, I would suggest adding white space to make your chains more readable. Marking the chains in the grid would be nice too, to make them easier to follow.)

1. You could cut a few nodes from the start and the end. While this is not my main point, I start with it so we can deal with a bit shorter chain:

Code: Select all

7r4c4 = r2c4 - (7=8)r1c5 - (8=6)r1c8 - 6r1c2 = r2c1 - r5c1 = (6-4*5)r5c9 = (5-48)r6c9 = 48r4c789 => -48r4c4; stte

2. Related to the discussion about eleven's chain and the correct use of '|', your chain would also need it in a couple of places to make it Boolean-correct:

Code: Select all

7r4c4 = r2c4 - (7=8)r1c5 - (8=6)r1c8 - 6r1c2 = r2c1 - r5c1 = (6-4*|5)r5c9 = (5-4|8)r6c9 = 48r4c789 => -48r4c4

(The original needs it in one more place at the end: 48r4c789 - (4|8=7)r4c4 => -7r4c6)

3. It's good that you marked the stored memory, but it should be marked also where it's used. In this case it enables the last strong link, so it's connected to the last node:

Code: Select all

7r4c4 = r2c4 - (7=8)r1c5 - (8=6)r1c8 - 6r1c2 = r2c1 - r5c1 = (6-4*|5)r5c9 = (5-4|8)r6c9 = *48r4c789 => -48r4c4

Otherwise the last strong link looks invalid. That practice also makes it possible to read the chain backwards, as long as it's Boolean-correct (see 2). The most readable style is to mark stored memories to the right of the stored item (like you did) and recalled memories to the left of the affected item (like I did), so the same symbols are facing each other. If more than one memory is used, they should have different symbols.

4. In general, "negative memories" should be avoided, because they make strong links harder to verify and backwards reading more awkward. Your chain stores the "elimination" of 4r5c9 (when read from the left) which later enables the strong link to 48r4c789. It works, but it's generally preferable to store "placements" and then use their weak links remotely. In this particular case the latter style is actually a bit uglier, so I don't mind the negative memory so much. Here's the other option, storing the "placement" of 6r5c9:

Code: Select all

7r4c4 = r2c4 - (7=8)r1c5 - (8=6)r1c8 - 6r1c2 = r2c1 - r5c1 = (6*-5)r5c9 = 5r6c9 - (4|8)r6*5c9 = 48r4c789 => -48 r4c4

(This latter style corresponds with the t-candidates in Denis' chains. He claims to have invented memory chains. Perhaps it's actually better to use negative memories to avoid the risk of having to pay royalties. Just kidding.)

5. There's no need to use a memory chain at all. The simpler and the more elegant option is to use the hidden pair (56)r56c9 to do the job:

Code: Select all

.------------------.-------------------.----------------------.
|  5     e26  3    |  1    c78     9   |  4    d68     2678   |
| f26     1   478  | b478   3478   5   |  278   9      23678  |
|  478    79  4789 |  2     3478   6   |  5     18     1378   |
:------------------+-------------------+----------------------:
|  9      3   5    | a7-48  14678  127 | i128  i1468  i12468  |
| g1246   8   14   |  45    9      3   |  12    7     h12456  |
|  12467  26  147  |  458   1468   12  |  9     3     h124568 |
:------------------+-------------------+----------------------:
|  17     59  2    |  3     17     8   |  6     45     49     |
|  3      4   1789 |  6     5      17  |  178   2      1789   |
|  178    57  6    |  9     2      4   |  3     158    178    |
'------------------'-------------------'----------------------'

Code: Select all

7r4c4 = r2c4 - (7=8)r1c5 - (8=6)r1c8 - r1c2 = r2c1 - r5c1 = (65-4|8)r56c9 = 48r4c789 => -48r4c4

It could also be written (6,5-4|8)r56c9 if you want to make the preceding strong link easier to understand. The link is valid either way.

I really like the ending of that chain. My improvements only make it shine more brightly, if I may say so, but the logic is yours. Nice job.

[eleven]

To formulate it without AIC:

An example: if you have sky and dazn (no comment), you are not locked by sky and you are not locked by dazn, i.e. you are not locked of something shown by sky or dazn.

The same way, if (1r5c7 and 6r6c9), then (not 1r4c789) and (not 6r4c789), is equivalent to not (1r4c789 or 6r4c789) (and if neither - not [the one or the other] - can be there, both must be in r4c56).

[Ajò Dimonios]

Hi Space

Space wrote:
I told you the reason right off the bat. The last strong link is not valid that way, which is easily verified because both sides are false in the solution. There are rarely situations where two ANDed terms are strongly linked anyway, certainly not here. Usually one side is ANDed and the other ORed. In this case both (1|6)r4c789 = 16r4c56 and 16r4c789 = (1|6)r4c56 are valid strong links, but 16r4c789 = 16r4c56 is not.

The weak link actually works with 16r4c789 too, but then the last node should be written (1|6)r4c56 to make the strong link valid. That obviously doesn't give us the conclusion we need, because both 1 AND 6 are needed in r4c56 to eliminate 7r5c6. Thus the only way to get the elimination and write the strong link correctly is to have the ANDed node last and the ORed node before it, just like eleven wrote it. Fortunately the weak link works with that arrangement as well.

When we talk about Boolean logic there are only two hypotheses about A (A true or A false) and two hypotheses about B (B true or B false). this means that if A true and B true respectively mean that 1 and 6 are present in r4c789 and 1 and 6 are present in r4c45, A false means all the other 3 hypotheses (1 true; 6 false), (1 false; 6 true ) and (1 false; 6 false). The same thing can be said of B, if B true means that 1 and 6 are present in r4c56, B false means all the other hypotheses (1 true; 6 false), (1 false; 6 true) and (1 false; 6 false). Consequence of this A = B means that at least one of the two hypotheses is true. This is always verified because when A is false all the possible hypotheses of 1 and 6 on r4c789 are true except clearly the one that makes A false, a consequence of this will be present also the one relating to the solution of the puzzle r4c9 = 1, therefore (1 true; 6 false), therefore in the implication A false => B true, even if the hypothesis B true leads to a false conclusion (in r4c56 there are 1 and 6) A = B is logically true because with A false the hypothesis r4c9 = 1 and 6 not present in r4c789 is true.

Paolo

[eleven]

A=1r5c7, B=6r6c9, C=1r4c789, D=6r4c789
(A and B) => (not C) and (not D) <=> not (C or D)

[SpAce]

When we talk about Boolean logic there are only two hypotheses about A (A true or A false) and two hypotheses about B (B true or B false). this means that if A true and B true respectively mean that 1 and 6 are present in r4c789 and 1 and 6 are present in r4c45, A false means all the other 3 hypotheses (1 true; 6 false), (1 false; 6 true ) and (1 false; 6 false). The same thing can be said of B, if B true means that 1 and 6 are present in r4c56, B false means all the other hypotheses (1 true; 6 false), (1 false; 6 true) and (1 false; 6 false). Consequence of this A = B means that at least one of the two hypotheses is true. This is always verified because when A is false all the possible hypotheses of 1 and 6 on r4c789 are true except clearly the one that makes A false, a consequence of this will be present also the one relating to the solution of the puzzle r4c9 = 1, therefore (1 true; 6 false), therefore in the implication A false => B true, even if the hypothesis B true leads to a false conclusion (in r4c56 there are 1 and 6) A = B is logically true because with A false the hypothesis r4c9 = 1 and 6 not present in r4c789 is true.

Give up. Once again you're trying to prove total nonsense. If we assume that the strong link A=B is valid, then either A or B (or both) must be TRUE in the solution, by definition. In this case both A and B are FALSE in the solution, which is a contradiction. It proves directly that our assumption is false, i.e. the strong link A=B can't be valid. End of story.

Any logic that appears to deny that FACT is obviously unsound and can be ignored (and I do). It should be obvious for other reasons too, but this is the most trivial proof. I thought even you could not deny it, which is why I replied at all. Yet, even after all I've seen, I still underestimated your ability to generate imaginary logic just to avoid admitting even the most obvious mistake.

A slightly more elaborate way to prove the same:

Code: Select all

(A=B) <-> (A OR B)

A : 16r4c789, true iff both (1 AND 6) in r4c789
B : 16r4c56,  true iff both (1 AND 6) in r4c56

The solution has: (841r4c789 AND 62r4c56)

841r4c789 -> NOT 16r4c789 <-> NOT A
62r4c56   -> NOT 16r4c56  <-> NOT B

-> ((NOT A) AND (NOT B)) <-> (A NOR B) <-> NOT(A OR B) <-> NOT(A=B)


NOT(A=B) => the strong link is invalid; QED

--
This is as far as I go. If you still want to argue about it, you'll have to do it with someone else.

[Ajò Dimonios]

Hi Space

Spacewrote:
Give up. Once again you're trying to prove total nonsense. If we assume that the strong link A=B is valid, then either A or B (or both) must be TRUE in the solution, by definition. In this case both A and B are FALSE in the solution, which is a contradiction. It proves directly that our assumption is false, i.e. the strong link A=B can't be valid. End of story.

Any logic that appears to deny that FACT is obviously unsound and can be ignored (and I do). It should be obvious for other reasons too, but this is the most trivial proof. I thought even you could not deny it, which is why I replied at all. Yet, even after all I've seen, I still underestimated your ability to generate imaginary logic just to avoid admitting even the most obvious mistake.

A slightly more elaborate way to prove the same:

CODE: SELECT ALL
(A=B) <-> (A OR B)

A : 16r4c789, true iff both (1 AND 6) in r4c789
B : 16r4c56, true iff both (1 AND 6) in r4c56

The solution has: (841r4c789 AND 62r4c56)

841r4c789 -> NOT 16r4c789 <-> NOT A
62r4c56 -> NOT 16r4c56 <-> NOT B

-> ((NOT A) AND (NOT B)) <-> (A NOR B) <-> NOT(A OR B) <-> NOT(A=B)


NOT(A=B) => the strong link is invalid; QED

--
This is as far as I go. If you still want to argue about it, you'll have to do it with someone else.

No, as I demonstrated in the previous post we are talking about Boolean logic (true; false)
The strong inference is always true whatever the solution of the puzzle is because if A is automatically false, norA is true, the same applies to B. In practice, the untruth of hypothesis A always contains the solution of the puzzle which is the truth. In A = B, all possible hypotheses are contained, consequently the strong inference is always true. In this case the untruth of A allows me to decide unequivocally on the untruth of B, this is crucial in establishing that it is a strong inference. Different is the case in which in a cell I have 3 possible candidates (A; B; C). The inference A = B is invalid because the non-truth of A does not allow me to decide on the truth or non-truth of B.

P.S. To make you understand even better I reverse the hypotheses (this doesn't change anything because they are booleans). Assumption A = (1 true; 6 false), (1 false; 6 true) and (1 false; 6 false) in r4c789, Assumption B = (1 true; 6 false), (1 false; 6 true) and (1 false; 6 false) in r4c56
A = B is a strong inference because when A is false => B true and when B is false => A true.


Paolo

[SpAce]

No

I stopped reading at that. Bye.

[Ajò Dimonios]

Idem. Bye

[SpAce]

Find the error.

Code: Select all

.------------------.----------------------.----------------------.
|  5     f26  3    | 1     78        9    |  4    g68     2678   |
| e26     1   478  | 478   3478      5    |  278   9      23678  |
|  478    79  4789 | 2     3478      6    |  5    g18     1378   |
:------------------+----------------------+----------------------:
|  9      3   5    | 478  a1486-7  am17-2 | b128  b1468  b12468  |
| d1246   8   14   | 45    9         3    |  12    7     c12456  |
|  12467  26  147  | 458   1468      12   |  9     3      124568 |
:------------------+----------------------+----------------------:
| i17     59  2    | 3    j17        8    |  6     45     49     |
|  3      4   1789 | 6     5        k17   |  178   2      1789   |
|  178   h57  6    | 9     2         4    |  3    g158    178    |
'------------------'----------------------'----------------------'

(16)r4c56 = (16)r4c789 - r5c9 = r5c1 - r2c1 = r1c2 - (6=815)r139c8 - (5=7)r9c2 - r7c1 = r7c5 - r8c6 = (7)r4c6 => -2 r4c6, -7 r4c5; stte

[Ajò Dimonios]

I'll give you another even better clarifying example. A and B are two backdoors of a puzzle and there is also a strong inference between them, as in the August 1 puzzle. So the inference A = B is certainly a strong inference, but it is also a strong inference Nor A = Nor B, since Nor A = A = B = Nor B. If you analyze the inference Nor A = Nor B you will realize that even being bound by strong inference Nor A and Nor B are both false. This case is practically identical to our case (16) r4c789 = 16r4c56.

[SpAce]

Did you even look at the AIC above? Does it look correct to you? If not, where is the error?

Here's the same chain written backwards (the middle ALS broken into two to keep it as simple as possible), in case it's easier to read that way:

Code: Select all

(7)r4c6 = r8c6 - r7c5 = r7c1 - (7=5)r9c2 - (5=18)r93c8 - (8=6)r1c8 - r1c2 = r2c1 - r5c1 = r5c9 - (61)r4c789 = (61)r4c56 => -2 r4c6, -7 r4c5

What is wrong with that, if anything?

(I'm not reading anything you write until you answer that question.)

[Cenoman]

2r4c6 = r6c6 - r6c12 = 2r5c1 - (2=4,5,1,6)r5c3479 - (1|6)r4c789 = 16r4c56 => -7r4c6, stte

Beautiful solution !

I wish I had been able, first to find it, and then, to write it so clear and elegant !

No need for any accompanying comment.
eleven gave three additional explanations, though:

...there is a hidden pair 16 in r4c56, if both digits cannot be in r4c789. Then the only remaining cells for both are r4c56.

...if (1r5c7 and 6r6c9), then (not 1r4c789) and (not 6r4c789), is equivalent to not (1r4c789 or 6r4c789) (and if neither - not [the one or the other] - can be there, both must be in r4c56).

A=1r5c7, B=6r6c9, C=1r4c789, D=6r4c789
(A and B) => (not C) and (not D) <=> not (C or D)

That's more than needed to close the discussion.

[eleven]

Paolo,
your excursions about boolean logic are interesting.
Good stuff for teachers. The students have to find out, what is wrong.
You seem to have your own definitions, and i don't mind any more. There is no medicine against "alternative facts" in these times.

[SpAce]

Hi Cenoman,

eleven gave three additional explanations, though:
...
That's more than needed to close the discussion.

It should be, but unfortunately it isn't. It seems to me that both of you missed the real question. Paolo claims that the strong link (16)r4c56 = (16)r4c789 is valid. Why can't anyone comment on that directly? (Or similar absurd claims before.)

All of you know just as well as I that it isn't anywhere close to valid. Paolo obviously doesn't, or can't admit it. Why can't any of you back me up and clear up this nonsense once and for all?

I've now proved the invalidity of that link in as many ways as I can, including with the otherwise valid AIC that actually uses that link (and ends up with false conclusions because of it).

That won't make any dent, just as it has never before. It's obvious that Paolo will ignore any and all arguments I present no matter how good they are. I'd be interested in seeing if he would ignore all of you, too. (Assuming, of course, that you guys agree with me, but I can't seriously even doubt that.)

I think nonsense like this should be stopped in its tracks. Prolonged "discussions" like this aren't healthy for anyone, or this forum. I'm already leaving because of that (once I get non-finished business wrapped up). We already live in a post-truth era otherwise, so it would be nice to have at least one environment where facts and logic and sound arguments actually matter.

There is no medicine against "alternative facts" in these times.

Exactly.

[Ajò Dimonios]

Hi Space

Space wrote:
Find the error.

CODE: SELECT ALL
.------------------.----------------------.----------------------.
| 5 f26 3 | 1 78 9 | 4 g68 2678 |
| e26 1 478 | 478 3478 5 | 278 9 23678 |
| 478 79 4789 | 2 3478 6 | 5 g18 1378 |
:------------------+----------------------+----------------------:
| 9 3 5 | 478 a1486-7 am17-2 | b128 b1468 b12468 |
| d1246 8 14 | 45 9 3 | 12 7 c12456 |
| 12467 26 147 | 458 1468 12 | 9 3 124568 |
:------------------+----------------------+----------------------:
| i17 59 2 | 3 j17 8 | 6 45 49 |
| 3 4 1789 | 6 5 k17 | 178 2 1789 |
| 178 h57 6 | 9 2 4 | 3 g158 178 |
'------------------'----------------------'----------------------'

(16)r4c56 = (16)r4c789 - r5c9 = r5c1 - r2c1 = r1c2 - (6=815)r139c8 - (5=7)r9c2 - r7c1 = r7c5 - r8c6 = (7)r4c6 => -2 r4c6, -7 r4c5; stte

Your chain is correct, only the deletions that are not correct r4c6 = 2 and r4c5 = 7, because both r4c6 = 1 and r4c5 = 6 must be connected with weak inference to the candidates to be eliminated. This AIC does not produce eliminations. This also makes me review the Eleven solution with (16) r4c789 = 16r4c56, also in this case the elimination of r4C6 = 7 is not correct because it is not connected to r4c6 = 1 and r4c5 = 6. So at this point, returning to (1 | 6) r4c789 = 16r4c56 in the Eleven resolution, the two options remain: (1) r4c789 = 16r4c56 or (6) r4c789 = 16r4c56. On this I have already answered.

Ajò Dimonios Wrote:
The final part of the chain of your resolution is not clear to me. In my opinion instead of
(1 | 6) r4c789 = 16r4c56 => - 7r4c6, stte you should have written (1 | 6) r4c789 = (16 | 26) r4c56 => - 7r4c6, stte.
If 6r4c789 is false, after several singles we obtain the solution that has r4c5 = 6 and r4c6 = 2 and not r4c5 = 6 and r4c6 = 1.

The first leads with a chain of contradiction to r4c6 = 1 while the second, by inserting singles to the solution with r4c6 = 2, neither of the two options leads to having 16r4c56. Evidently this is enough to say that candidate 7 in r4c6 can be deleted but there is no trace in the AIC of how this is done.

Paolo