## January 3, 2015

Post puzzles for others to solve here.

### Re: January 3, 2015

SteveG48 wrote:
daj95376 wrote:However, initially assuming two values not true does not allow him to conclude r8c4<>46.

Ordinarily, no. However, if neither of r8c56 is a 4 or a 6, then the conclusion that r23c4 is a 46 pair holds. ....

It seems to me, the first conclusion from the removal of all 4s and 6s from r8c56, is that there is then an 8 at r8c6 and a 3 at r8c5, so how can you argue that there can also be a 3 at r8c4? So this one of your two alternatives (the two sides of your initial '=' sign) leads to a contradiction and proves that your other alternative must be true. With a 4/6 in one cell, and an 8 in the other, putting a 3 at r8c4, NOT because the 4 and 6 have been eliminated (they never were) but simply because there is no other place for the 3 in r8. Only AFTER this 3 is placed, do the 4 and 6 get eliminated. Tricky, and confusing, yes.
gurth

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Location: Cape Town, South Africa

### Re: January 3, 2015

I just hope, that there is common agreement now, that the link (46=8)r8c56 is definitely wrong, as is (46=38)r8c56.
It is another matter, that the deduction here can be argumented with it as a part of a network (covering the open cases), which depends on the 2 strong links for 3 and 8 in row 8.
eleven

Posts: 1915
Joined: 10 February 2008

### Re: January 3, 2015

Code: Select all
` 5       3       4      |a17     9     b18     | 6      78     2       67      679     8      |a467    2      5      | 3      49     1       1       679     2      |a467    468    3      | 789    5      47     ------------------------+----------------------+--------------------- 36      145     7      | 8      1346   9      | 2      13     56      2       14      36     | 1346   5      146    | 78     78     9       8       15      9      | 2      136    7      | 4      13     56     ------------------------+----------------------+--------------------- 367     2       36     | 5      146    146    | 79     49     8       9      c67      5      | 346    3468  c468    | 1      2     c47      4       8       1      | 9      7      2      | 5      6      3      `

This seems to have become unnecessarily complicated. Can't it just be viewed as two ALS wings:
(4=1)r123c4 - (1=8)r1c6 - (8=4)r8c269 => -4 r8c4 }
(6=1)r123c4 - (1=8)r1c6 - (8=6)r8c269 => -6 r8c4 } => -46 r8c4

Phil
pjb
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### Re: January 3, 2015

Well, I'm still sticking with (46=38)r8c56. Another perspecive on using it:

Code: Select all
`             (3)r8c5                       - (3 )r8c4            /                                        \ (46)r8c56 =                                          r8c4 = [empty] ; contradiction            \                                        /             (8)r8c6 - (8=1467)r1c46,r23c4 - (46)r8c4`

Thus, one of 46 must be true in r8c56. Since one of r8c56 must be 8, we can conclude -3r8c5.

The eliminations in r8c4 follow from the SL on 3 forcing r8c4=3.

Bottom Line: If 8 must be true in r8c56, and combining it with 3 in r8c5 leads to a contradiction, then 3 can be eliminated from r8c5. It doesn't matter what other values are present in r8c56.

_
daj95376
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### Re: January 3, 2014

SteveG48 wrote:
Code: Select all
` *------------------------------------------------------------* | 5     3     4     | b17    9    b18    | 6     78    2     | | 67    679   8     | b467   2     5     | 3     49    1     | | 1     679   2     | b467   468   3     | 789   5     47    | *-------------------+--------------------+-------------------| | 36    145   7     |  8     1346  9     | 2     13    56    | | 2     14    36    |  1346  5     146   | 78    78    9     | | 8     15    9     |  2     136   7     | 4     13    56    | *-------------------+--------------------+-------------------| | 367   2     36    |  5     146   146   | 79    49    8     | | 9     67    5     |  3-46 a3468 a468   | 1     2     47    | | 4     8     1     |  9     7     2     | 5     6     3     | *------------------------------------------------------------*`

(46=8)r8c56 - (8=46)r123c4,r1c6 => -46 r8c4

So maybe SteveG48's logic really was..
if r8c56 doesn't contain (4 or 6), then r8c6=8 - (8=46)r123c4, r1c6 => r8c4 <> 46
if r8c56 does contain (4 or 6 with digit 8), then - 3r8c5 = 3r8c4 => r8c4 <> 46

but this (46=8)r8c56 is not only confusing,it also doesn't make sense.
I have to agree with eleven. is simply wrong.
7b53
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### Re: January 3, 2015

daj95376 wrote:Well, I'm still sticking with (46=38)r8c56. Another perspecive on using it:

Code: Select all
`             (3)r8c5                       - (3 )r8c4            /                                        \ (46)r8c56 =                                          r8c4 = [empty] ; contradiction            \                                        /             (8)r8c6 - (8=1467)r1c46,r23c4 - (46)r8c4`

Thus, one of 46 must be true in r8c56. Since one of r8c56 must be 8, we can conclude -3r8c5.

The eliminations in r8c4 follow from the SL on 3 forcing r8c4=3.

Bottom Line: If 8 must be true in r8c56, and combining it with 3 in r8c5 leads to a contradiction, then 3 can be eliminated from r8c5. It doesn't matter what other values are present in r8c56.

_

I agree with daj.
In plain English, (46=38)r8c56 means "If there is no 4 or 6 in the two cells r8c56, then there must be a 3 and an 8 in these two cells" which is perfectly correct and undeniable. This is not where the trouble started. In fact there is no problem with this solution at all, if we follow the lines as set out by me and daj in our last two posts.
gurth

Posts: 358
Joined: 11 February 2006
Location: Cape Town, South Africa

### Re: January 3, 2015

gurth wrote:In plain English, (46=38)r8c56 means "If there is no 4 or 6 in the two cells r8c56, then there must be a 3 and an 8 in these two cells" which is perfectly correct and undeniable.

No, thats wrong.

In the notation you are using for years, what means 12r12c1 ?
r12c1 has the PAIR 12, i.e. one of the cells is 1 AND the other is 2.

What means (12=X)r12c1 ?
If r12c1 does not have the pair 12, then it contains X. If it does NOT have the pair, then NOT BOTH 1 AND 2 are in the cells.
This means that AT LEAST ONE of the cells has another digit.

Then (46=38)r8c56 is wrong.
If r8c56 does not have the PAIR 46, it does NOT follow, that it contains 38 (it only follows, that AT LEAST ONE OF 4 and 6 is NOT in the cells, it therefore can also be 34, 36, 48 or 68) and vice versa.
eleven

Posts: 1915
Joined: 10 February 2008

### Re: January 3, 2015

"When I use a word," Humpty Dumpty said, in rather a scornful tone, "it means just what I choose it to mean—neither more nor less."
Lewis Carroll's Through the Looking-Glass (1872)

The same seems to go with contributors' notations on this forum.

I like to consider that our posts are being read by newcomers to Sudoku and as they get drawn into its finer arts they will join us. Being cliquey and clever with our notations is the way to ensure that won't happen and the forum will suffer a lingering death.

This notation:
(3)r8c3 = (3-8)r8c5 = (8)r8c6 - (8=1)r1c6 - (1=7)r1c4 - (7=46)r23c4 = r8c3 <> 46
Is easy to follow but look at the debate caused by writing this in a way that tries to be clever and amusing but then proves to be wrong!

In the Eureka notation:
(a)cells12 is true when (a) is true in either cell.
(ab)cells12 is true when both digits are true in the two cells.
(a=bc)cells12 splits into (a)cells12 = (bc)cells12 and therefore makes sense when there are exactly three candidates in total.

Now (ab=cd)cells12 splits into (ab)cells12 = (cd)cells12 but that isn't a strong link, its a weak one; (ab) and (cd) obviously can't both be true, but they could both be false.
(ab-cd)cells12 is therefore OK but (ab=cd)cells12 isn't.
To notate a strong link the notation should be (ab#2 = cd#1) or (ab#1=cd#2) now one side or the other must be true.

In other circumstances writers must clarify what is needed to consider the Boolean terms they use to be true.
Personally I favour (abc#N) to indicate when N of them must be true. I think this is more versatile than the alternative of using OR symbols such as (a|b|c).

 Second thoughts! The blue text replaces what I originally wrote which was:
(ab=cd)cells12 splits into (ab)cells12 = (cd)cells12 and usually also makes sense when there are exactly four candidates in total. However this depends on the context as, taken alone, the cells could contain (ac) when neither side of the strong link would be true (which is at the heart of Eleven's point).
Last edited by David P Bird on Mon Jan 05, 2015 4:14 pm, edited 1 time in total.
David P Bird
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### Re: January 3, 2015

eleven wrote:
gurth wrote:In plain English, (46=38)r8c56 means "If there is no 4 or 6 in the two cells r8c56, then there must be a 3 and an 8 in these two cells" which is perfectly correct and undeniable.

No, thats wrong.
...

It seems to me that the majority of posters have been using this notation in agreement with my meaning. There is simply a difference of opinion between me and eleven on the correct interpretation.
The answer to this dilemma is not to avoid the use of groups where possible, but to reach a consensus on the interpretation. So: does (46= mean "if neither 4 nor 6" or does it mean "if not the intact pair 46, but possibly part of it" ? Your vote is required please.
May I say that I think adopting eleven's interpretation would have disastrous results.
gurth

Posts: 358
Joined: 11 February 2006
Location: Cape Town, South Africa

### Re: January 3, 2015

Gurth,

either you are the meaning, that 46r8c56 means 4 OR 6 (or both) are in the 2 cells (which i find disastrous) or you are lacking the very basics of logic.
eleven

Posts: 1915
Joined: 10 February 2008

### Re: January 3, 2015

Gurth, I don't quite understand the "disastrous consequences" bit. What we are after is clarity not any shift in logic!

While working on something else I realised I had fired from the hip before and have now edited my previous post.
David P Bird
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Location: Middle England

### Re: January 3, 2015

Hmmm!!! Consider the Easter Monster puzzle:

Code: Select all
`Steve Kurzhal's original presentation w/ RonK's labels:1.......2.9.4...5...6...7...5.9.3.......7.......85..4.7.....6...3...9.8...2.....1     1      A478     34578   | 3567    3689    5678    | 3489   I369     2    L238     9      L378     | 4      K12368  K12678   |J138     5      J368     23458  A248     6       | 1235    12389   1258    | 7      I139     3489    -------------------------+-------------------------+-----------------------     2468    5       1478    | 9       1246    3       | 128    H1267    678     234689 B12468   13489   | 126     7       1246    | 123589 H12369   35689     2369   B1267    1379    | 8       5       126     | 1239    4       3679    -------------------------+-------------------------+-----------------------     7      C148     14589   | 1235    12348   12458   | 6      G239     3459    D456     3      D145     |E12567  E1246    9       |F245     8      F457     45689  C468     2       | 3567    3468    45678   | 3459   G379     1    (27)r13c2=(27-16)r56c2=(16)r79c2-(16)r8c13=(16-27)r8c45=(27)r8c79-    (27)r79c8=(27-16)r45c8=(16)r13c8-(16)r2c79=(16-27)r2c56=(27)r2c13-loop`

Particularly, (27-16)r56c2=(16)r79c2. I'm pretty sure that (-16)r56c2 means the elimination of all 1s and 6s in these cells. Otherwise, =(16)r79c2 would not follow.

The same logic -- removal of all candidates with these values -- is present in the V-Loop eliminations for this puzzle:

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 (48=27)r13c2 - (27=38)r2c13 - (38=16)r2c79 - (16=39)r13c8 - (39=27)r79c8 - (27=45)r8c79 - (45=16)r8c13 - (16=48)r79c2 - loop  =>  r2c5<>38; r5c2<>48; r5c8<>39;  r7c3<>1; r3c1<>2; r8c5<>4; r8c4<>5; r9c1<>6; r1c3<>7; r2c6<>8`

I still support (46=38)r8c56 as a valid SL.

_
daj95376
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### Re: January 3, 2015

Oops, was too angry before, sorry.
16r56c2=16r79c2 indeed is no strong link, i.e. not one of the both sides must be true.
So we are back to my initial comment, that this notation can be confusing. Steve K. should have used another one for his loop. (Normally both sides are definitely wrong - the eliminations are only correct, because it can be shown, that they are valid also for this case).
eleven

Posts: 1915
Joined: 10 February 2008

### Re: January 3, 2015

DAJ: SK Loops don't use pure AIC logic and are a special case that works only when a closed loop is formed.
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       | +--------------------------------------------------------------------------------+`

(48=27)r13c2 - (27=38)r2c13 - (38=16)r2c79 - (16=39)r13c8 -
(39=27)r79c8 - (27=45)r8c79 - (45=16)r8c13 - (16=48)r79c2 - loop =>

r2c5<>38; r5c2<>48; r5c8<>39; r7c3<>1; r3c1<>2; r8c5<>4; r8c4<>5; r9c1<>6; r1c3<>7; r2c6<>8

Specifiying the truths needed to consider a node true can be written two ways as the loop is followed
(27#2=38#1)r2c13 - (38#2=16#1)r2c79 or (27#1=38#2)r2c13 - (38#1=16#2)r2c79
One of these will eliminate (38)r2c5 and (8)r2c6 whichever way the counts are alternated when both the digits in one of the pairs are true.

But there is another possibility namely that every term has one truth. Now it still works because the truth in the repeating pair (38) can't be for the same digit in r2c13 as it is in r2c79.

Steve K's find was therefore brilliant, but you can't hold its notation up as as an example of a valid AIC.

Now does that ring the bells for you? It should do.

PS Just in case: (27#2) will be false when one or both digits are false, (38#1) will be true when one or both digits are true.
David P Bird
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### Re: January 3, 2015

David P Bird wrote:Specifiying the truths needed to consider a node true can be written two ways as the loop is followed
(27#2=38#1)r2c13 - (38#2=16#1)r2c79 or (27#1=38#2)r2c13 - (38#1=16#2)r2c79
One of these will eliminate (38)r2c5 and (8)r2c6 whichever way the counts are alternated when both the digits in one of the pairs are true.

To give an idea, why. We distinguish 3 cases: The pairs on the left side are true, those on the right side, and both are false.
case 1: 38r2c13, 16r2c79
case 2: 27r2c13, 38r2c79
case 3: r2c13<>38, r2c13<>27 and r2c79<>38, r2c79<>16
If r2c13 is neither 38 nor 27, one must be 3 or 8, the other 2 or 7.
If r2c79 is neither 16 nor 38, one must be 3 or 8 too, the other 1 or 6.
So 38 always must be in the cell pairs.
[Edited: a bit simpler]

This can be shown in general, so whenever you find a closed pair loop, it has this property.
eleven

Posts: 1915
Joined: 10 February 2008

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