The New Sudoku Players' Forum

[daj95376]

I noticed that neither you nor JC started the chain at r3c8. Please show it without any special symbols or networking.

Modern style: (3=9)r3c8 - (9=8)r7c8 - (8=67)r7c1 - (67=3)r1689c2
Personally i would prefer (3=9)r3c8 - (9=8)r7c8 - (8=67)r7c1 - (6&7=4|5)r89c2 - (4&5=3)r16c2, because it is more like you could find it.
(i spare you to write it in my notation)

[Edit: Removed original response. I need time to clear my thoughts and see if I "stepped in it". ]

I was mistaken !!! I misread eleven's second chain. He did an excellent job of converting my forcing chain into a bidirectional chain. However, I wish he had added one more connector:

(3=9)r3c8 - (9=8)r7c8 - (8=6|7)r7c1 - (6&7=4|5)r89c2 - (4&5=3)r16c2

_

[eleven]

I'm trying to derive a new connector that would be bidirectional for a pair of values in two cells. A symbol that would work on said pair as if they were a single item.

For this case you need 2 symbols, one for (a|b=c&d), and one for (a&b=c|d). The point here is, that you have to distinguish a&b and a|b in 2 cells. If you define ab as a&b (for 2 or more cells), you need a single | to make it clear (this is, what i suggested months ago). The connection between left and right is the same as usual, an OR. So why introduce new connectors, which then tell you, if the left side has to be read as 'or' or as 'and' ?

[David P Bird]

Now, I could go on and discuss how your chain is appropriate in terms of what I call a DAJC, but I'm afraid that DPB would throw another tantrum.
....
[Edit: Removed original response. I need time to clear my thoughts and see if I "stepped in it". ]

DAJ, by winging it you've been 'stepping in it' all the way through this thread. That's the reason there isn't a single toy left in my pram.
As a final attempt to get you to consider things properly please tell us
a) are your DAJC chains going to be uni- or bi- directional?
b) what improvements will they provide over the classic methods (that I reminded you about on page 1)?

.

[daj95376]

I'm trying to derive a new connector that would be bidirectional for a pair of values in two cells. A symbol that would work on said pair as if they were a single item.

For this case you need 2 symbols, one for (a|b=c&d), and one for (a&b=c|d). The point here is, that you have to distinguish a&b and a|b in 2 cells. If you define ab as a&b (for 2 or more cells), you need a single | to make it clear (this is, what i suggested months ago). The connection between left and right is the same as usual, an OR. So why introduce new connectors, which then tell you, if the left side has to be read as 'or' or as 'and' ?

There is a single symbol that meets my needs.

"The statement A ∧ B is true if A and B are both true; else it is false".

Now, consider the Easter Monster grid:

Code: Select all

 +--------------------------------------------------------------------------------+
 | *1       48+7    34578   |  3567    3689    5678    |  3489    39+6   *2       |
 |  38+2    9       38+7    |  4       12368   12678   |  38+1    5       38+6    |
 |  23458   48+2   *6       |  1235    12389   1258    | *7       39+1    3489    |
 |--------------------------+--------------------------+--------------------------|
 |  2468    5       1478    |  9       1246    3       |  128     1267    678     |
 |  234689  12468   13489   |  126     7       1246    |  123589  12369   35689   |
 |  2369    1267    1379    |  8       5       126     |  1239    4       3679    |
 |--------------------------+--------------------------+--------------------------|
 | *7       48+1    14589   |  1235    12348   12458   | *6       39+2    3459    |
 |  45+6    3       45+1    |  12567   1246    9       |  45+2    8       45+7    |
 |  45689   48+6   *2       |  3567    3468    45678   |  3459    39+7   *1       |
 +--------------------------------------------------------------------------------+
 # 179 eliminations remain


If both <4> and <8> are not true in r13c2, then both <2> and <7> must be true in r13c2.
If both <2> and <7> are not true in r13c2, then both <4> and <8> must be true in r13c2.

This translates into an SL with one special connector being used: ( 4 ∧ 8 = 2 ∧ 7 ) r13c2

Since my text editor can't replicate the connector, my last proposal was to use: ( 4 ' 8 = 2 ' 7 ) r13c2


Earlier, I posted:

The format of the SK-Loop and V-Loop for Easter Monster are now represented by:

Code: Select all

    SK-Loop:

    (2'7)r13c2 = (2'7 - 1'6)r56c2 = (1'6)r79c2 - (1'6)r8c13 = (1'6 - 2'7)r8c45 = (2'7)r8c79 -
    (2'7)r79c8 = (2'7 - 1'6)r45c8 = (1'6)r13c8 - (1'6)r2c79 = (1'6 - 2'7)r2c56 = (2'7)r2c13 - loop


    V-Loop:

    (4'8 = 2'7)r13c2 - (2'7 = 3'8)r2c13 - (3'8 = 1'6)r2c79 - (1'6 = 3'9)r13c8 -
    (3'9 = 2'7)r79c8 - (2'7 = 4'5)r8c79 - (4'5 = 1'6)r8c13 - (1'6 = 4'8)r79c2 - loop


These loops are bi-directional ...

_

[eleven]

However, I wish he had added one more connector:

(3=9)r3c8 - (9=8)r7c8 - (8=6|7)r7c1 - (6&7=4|5)r89c2 - (4&5=3)r16c2

Yes, no problem, if you write out the relation between all multiple digits.
There are several ways to express the same, all being correct, if it is defined, what means what:
(3=9)r3c8 - (9=8)r7c8 - (8=6|7)r7c1 - (6&7=4|5)r89c2 - (4&5=3)r16c2
(3=9)r3c8 - (9=8)r7c8 - (8=67)r7c1 - (6&7=4|5)r89c2 - (4&5=3)r16c2 (obviously 67r7c1 must mean (6|7)r7c1)
(3=9)r3c8 - (9=8)r7c8 - (8=67)r7c1 - (67=4|5)r89c2 - (45=3)r16c2 (with the convention, that ab in 2 or more cells means a&b)
(3=9)r3c8 - (9=8)r7c8 - (8=67)r7c1 - (67=3)r1689c2 (because of the ALS with 5 digits in 4 cells "NOT both 6&7" or "NOT 3" implies that all other digits must be in the cells - only those of interest are listed)
etc. etc.

There is a single symbol that meets my needs.

"The statement A ∧ B is true if A and B are both true; else it is false".

Why don't you take & then, if you mean it ?

If both <4> and <8> are not true in r13c2, then both <2> and <7> must be true in r13c2.
If both <2> and <7> are not true in r13c2, then both <4> and <8> must be true in r13c2.

This translates into an SL with one special connector being used: ( 4 ∧ 8 = 2 ∧ 7 ) r13c2

You adhere to the same mistake all the time:
"both <4> and <8> are not true in r13c2" means "(NOT 4) AND (NOT 8)", which is equivalent to NOT (4 OR 8) (not the one or the other).
So "If both <4> and <8> are not true in r13c2, then both <2> and <7> must be true in r13c2" has to be written as
(4|8 = 2&7)r13c2

( 4 ∧ 8 = 2 ∧ 7 ) r13c2
would state: "if NOT (4 AND 8) in r13c2 (not both, at least one missing, not the one or not the other), then (2 AND 7)", which is simply wrong.
In other words, it is wrong that the 2 cells must hold either both 4 and 8 or both 2 and 7.

[Added:]What you would need here, is not a new symbol for & between digits, but for the '=' relation, with the following meaning:
E.g. if we take '|=|' for it,
(48 |=| 27)r13c2 means (4|8 = 2&7)r13c2 AND (4&8 = 2|7)r13c2 (note that the second one is different to the first one, read from right to left)
(if you have only 4 candidates in the cells like in the V-loop, both relations are obvious)
If you reach a loop consisting of only such relations, connected with '-' (NAND) relations, the known eliminations of the 2 common digits in all units of the loop cells (and in case of SK loop the digits not present in a pair of the loop cells) are proven (see e.g. blues' way).

[ronk]

I noticed that neither you nor JC started the chain at r3c8. Please show it without any special symbols or networking.

Modern style: (3=9)r3c8 - (9=8)r7c8 - (8=67)r7c1 - (67=3)r1689c2
Personally i would prefer (3=9)r3c8 - (9=8)r7c8 - (8=67)r7c1 - (6&7=4|5)r89c2 - (4&5=3)r16c2, because ...

I agree with your preference except I wouldn't use the AND and OR symbols.

I misread eleven's second chain. He did an excellent job of converting my forcing chain into a bidirectional chain.

It looks much like a reversed version of the inference stream I posted.

(3=45)r16c2 - (45=67)r89c2 - (67=8)r7c1 - (8=9)r7c8 - (9=3)r3c8 ==> r1c789<>3, r3c123<>3

[eleven]

I agree with your preference except I wouldn't use the AND and OR symbols.

This is ok for you, if you like to read the meaning from the context, but was the cause of many troubles and head ache, when at least the OR was left out.
For a stand-alone link (ab=cd) in 2 cells you then cannot say, what it means - and people are mislead to believe, it means both (a|b=cd) and (ab=c|d), or even (a&b=c&d), which is impossible in 2 cells .

[Edit]
To give an example.
AIC's tell you, that (A=B)-(C=D) implies (A=B)
Now look at this one:
(X=12)r12c1-(12=34)r56c1
Now you would follow Xr12c1=34r56c1.
This is fine, if X=3 (a single digit).
But if X=34, it would mean that you have a pair in one of the cell pairs and could eliminate 34 in the rest of c1.
In fact you only can follow 34r12c1=(3|4)r56c1, if the original chain is (34=1|2)r12c1-(12=3|4)r56c1

[daj95376]

[Withdrawn: I seem to be unable to define a new connector without hitting a wall. Done.]

_