The New Sudoku Players' Forum

[daj95376]

I'm reintroducing this grid because discussions elsewhere were tangent to a thread's topic, and information was withdrawn.

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 +-----------------------------------------------------------------------+
 |  9      8      124    |  7      125    1235   |  6      345    12345  |
 |  7      3      124    |  8      1256   1256   |  9      45     1245   |
 |  12     5      6      |  9      4      123    |  1278   378    12378  |
 |-----------------------+-----------------------+-----------------------|
 |  8      124    127    |  5      126    12467  |  3      9      1467   |
 |  1234   6      1237   |  124    8      9      |  147    457    1457   |
 |  14     9      5      |  14     3      67     |  78     2      678    |
 |-----------------------+-----------------------+-----------------------|
 | X2345   7      238    |  6      25     245    |  248    1      9      |
 |  1246   124    9      |  1234   7      8      |  5     B346   B234    |
 | Q12456  124    128    | R1234   9      1245   |  2478   34678  23478  |
 +-----------------------------------------------------------------------+
 # 118 eliminations remain

 ### -2346- qExocet   Base = r8c89   Target = r9c1,r9c4   aligned

 NOTE: SL[ 5r7c1 = 5r9c1 ]   (explains lowercase "q" with Exocet)


Values <2> & <4>: If either is true in the base cells, then a SL can be created between target cells r9c4 and hp(n5)r79c1.

Code: Select all

 +-----------------------------------+
 |  .  .  2  |  .  2  2  |  .  . ~2  |
 |  .  .  2  |  .  2  2  |  .  . ~2  |
 |  2  .  .  |  .  .  2  |  2  . ~2  |
 |-----------+-----------+-----------|
 |  .  2  2  |  .  2  2  |  .  .  .  |
 |  2  .  2  |  2  .  .  |  .  .  .  |
 |  .  .  .  |  .  .  .  |  .  2  .  |
 |-----------+-----------+-----------|
 | X2  .  2  |  .  2  2  | ~2  .  .  |
 | ~2 ~2  .  | ~2  .  .  |  .  . =2  |   If a base cell is true for <2>
 | Q2  2  2  | R2  .  2  | ~2  . ~2  |
 +-----------------------------------+

 2r8c9 - r79c7 = r3c7 - 2r3c1*        setup

 2r9c4 = r5c4 - r5c1* = hp(25)r79c1   SL


Code: Select all

 +-----------------------------------+
 |  .  .  4  |  .  .  .  |  .  4  4  |
 |  .  .  4  |  .  .  .  |  .  4  4  |
 |  .  .  .  |  .  4  .  |  .  .  .  |
 |-----------+-----------+-----------|
 |  .  4  .  |  .  .  4  |  .  .  4  |
 |  4  .  .  |  4  .  .  |  4  4  4  |
 |  4  .  .  |  4  .  .  |  .  .  .  |
 |-----------+-----------+-----------|
 | X4  .  .  |  .  .  4  | ~4  .  .  |
 | ~4 ~4  .  | ~4  .  .  |  . =4 =4  |   If a base cell is true for <4>
 | Q4  4  .  | R4  .  4  | ~4 ~4 ~4  |
 +-----------------------------------+

 4r8c89 - r79c7 = r5c7 - 4r5c14*      setup

 4r9c4 = r6c4 - r6c1* = hp(45)r79c1   SL


Within the chute containing the base cells, if <3> or <6> is true in the base cells, then different target cells must be true for the values.

Code: Select all

 +-----------------------------------+
 |  .  .  .  |  .  .  3  |  .  3  3  |
 |  .  3  .  |  .  .  .  |  .  .  .  |
 |  .  .  .  |  .  .  3  |  .  3  3  |
 |-----------+-----------+-----------|
 |  .  .  .  |  .  .  .  |  3  .  .  |
 |  3  .  3  |  .  .  .  |  .  .  .  |
 |  .  .  .  |  .  3  .  |  .  .  .  |
 |-----------+-----------+-----------|
 | X3  .  3  |  .  .  .  |  .  .  .  |
 |  .  .  .  | ~3  .  .  |  . =3 =3  |   If a base cell is true for <3>
 | Q.  .  .  | R3  .  .  |  . ~3 ~3  |
 +-----------------------------------+

 3r8c89 - r9c89 = 3r9c4   target cell must be r9c4


Code: Select all

 +-----------------------------------+
 |  .  .  .  |  .  .  .  |  6  .  .  |
 |  .  .  .  |  .  6  6  |  .  .  .  |
 |  .  .  6  |  .  .  .  |  .  .  .  |
 |-----------+-----------+-----------|
 |  .  .  .  |  .  6  6  |  .  .  6  |
 |  .  6  .  |  .  .  .  |  .  .  .  |
 |  .  .  .  |  .  .  6  |  .  .  6  |
 |-----------+-----------+-----------|
 | X.  .  .  |  6  .  .  |  .  .  .  |
 | ~6  .  .  |  .  .  .  |  . =6  .  |   If a base cell is true for <6>
 | Q6  .  .  | R.  .  .  |  . ~6  .  |
 +-----------------------------------+

 6r8c8 - r8c1 = 6r9c1   target cell must be r9c1


I didn't notice any contradiction to this being an Exocet. In particular, if <6> is true in the base cells and forced true in r9c1, then the only remaining cells in the chute for it to be true is r7c456. The fact that it's already a solved cell in r7c4 is irrelevant.

Others are now welcome to add/revive their comments on this grid.

Note: I'm also planning to reintroduce the <1248> Exocet for discussion in a separate post.

_

[Leren]

As I said in the other thread this Exocet has two secondary equivalences 1. r7c3==r9c4, which reduces those two cells to <23> and 2. r9c6===r7c1 which reduces those two cells to <245>

I've added an extra = sign to the second of these equivalences to indicate it relies on an extra check that if r7c1 = 5 (the non-base S\L digit in r79c1) then r9c6 = 5. This was a new observation (at least for me).

Also blue pointed out that even if the second equivalence did not apply 3 can be removed from r7c1 by noting that if 3 is in a base cell then target cell r9c4 = 3 since there are only two 3's in box 8. If target cell r7c1 was also 3 then 3 would occupy two target cells if it was true in the base, so it could be removed from the base and target cells.

Leren

[champagne]

I'm reintroducing this grid because discussions elsewhere were tangent to a thread's topic, and information was withdrawn.

I didn't notice any contradiction to this being an Exocet. In particular, if <6> is true in the base cells and forced true in r9c1, then the only remaining cells in the chute for it to be true is r7c456. The fact that it's already a solved cell in r7c4 is irrelevant.

Others are now welcome to add/revive their comments on this grid.

Note: I'm also planning to reintroduce the <1248> Exocet for discussion in a separate post.

_

Hi Danny

seen by my solver, this is a standard Jexocet of the "Platinum Blonde" form

base r8c89
target1 r79c1 ('5' is locked in the target)
target 2 r9c4

I did not check eliminations, but when a jexocet is seen, you can apply any rule to find eliminations.
if r8c89 r9c1 r9c4 is an exocet, this means that r7c1 must be 5

Edit relying on the fact that r9c1 r9c4 is an exocet

then r7c1=5; r7c5=2 r7c6=4 r9c4(target=3) .....

[daj95376]

Code: Select all

 Case-by-Case:
                                    common eliminations   forcing
                                    -------------------   -------
 r8c89=32; r9c4=3,   r79c1=25;      r9c14<>1, r7c1<>3     r7c3=3

 r8c89=34; r9c4=3,   r79c1=45;      r9c14<>1, r7c1<>3     r7c3=3

 r8c89=36; r9c4=3,   r79c1=65;      r9c14<>1, r7c1<>3     r7c3=3

 r8c89=24; r9c4=2|4, r79c1=45|25;   r9c14<>1, r7c1<>3     r7c3=3

 r8c89=26; r9c4=2,   r79c1=65;      r9c14<>1, r7c1<>3     r7c3=3

 r8c89=46; r9c4=4,   r79c1=65;      r9c14<>1, r7c1<>3     r7c3=3


In addition, scenarios r8c89={24,26,46} result in an invalid grid because of r7c56 being influenced by r7c1 and r9c4. This forces r9c4=3 and is the only way I could justify the secondary equivalence r9c4==r7c3.

This leaves the following grid for my results:

Code: Select all

 +-----------------------------------------------------------------------+
 |  9      8      124    |  7      125    1235   |  6      345    12345  |
 |  7      3      124    |  8      1256   1256   |  9      45     1245   |
 |  12     5      6      |  9      4      123    |  1278   378    12378  |
 |-----------------------+-----------------------+-----------------------|
 |  8      124    127    |  5      126    12467  |  3      9      1467   |
 |  1234   6      1237   |  124    8      9      |  147    457    1457   |
 |  14     9      5      |  14     3      67     |  78     2      678    |
 |-----------------------+-----------------------+-----------------------|
 | X245-3  7      3-28   |  6      25     245    |  248    1      9      |
 |  1246   124    9      |  1234   7      8      |  5     B346   B234    |
 | Q2456-1 124    128    | R3-124  9      1245   |  2478   34678  23478  |
 +-----------------------------------------------------------------------+
 # 118 eliminations remain


This is sufficient to crack the puzzle.

_

[champagne]

This is sufficient to crack the puzzle.

_

but you can also keep in mind your findings

having established the Jexocet, just show that 5r9c1 is not valid.

As far as I can see, expanding r9c1 true, you came quickly to a contradiction

[Leren]

daj95376 wrote : In addition, scenarios r8c89={24,26,46} result in an invalid grid because of r7c56 being influenced by r7c1 and r9c4. This forces r9c4=3 and is the only way I could justify the secondary equivalence r9c4==r7c3.

I don't quite follow you here - to me it looks like r9c4==r7c3 follows the standard Exocet pattern. Your case by case table shows that, even for the invalid cases, r7c1 <> r9c4 and this is sufficient to establish the secondary equivalence.

Leren

[eleven]

This is sufficient to crack the puzzle.

but you can also keep in mind your findings

having established the Jexocet, just show that 5r9c1 is not valid.

As far as I can see, expanding r9c1 true, you came quickly to a contradiction

As David said, it is an Exocet, but no JExocet (which must not have a base digit in the compagnon cell).

To show r9c1<>5 is not easy, at least for me.

The simplest way for me to crack it, is following blue:
If 3 is not in the base cells, it cannot be in a target cell (especially not in r79c1).
If 3 is in the base cells, it must be in r9c4 (only 3 left in box 8), and cannot be in the other target cells r79c1.

[daj95376]

First off, the Exocet isn't a JExocet because the pattern breaks this constraint.

Code: Select all

 +-----------------------------------+
 |  .  .  .  |  .  .  .  |  .  .  .  |
 |  .  .  .  |  .  .  .  |  .  .  .  |
 |  .  .  .  |  .  .  .  |  .  .  .  |
 |-----------+-----------+-----------|
 |  .  .  .  |  .  .  .  |  .  .  .  |
 |  .  .  .  |  .  .  .  |  .  .  .  |
 |  .  .  .  |  .  .  .  |  .  .  .  |
 |-----------+-----------+-----------|
 |  Q  /  .  |  R  .  .  |  .  .  .  |
 |  .  .  .  |  .  .  .  |  .  B  B  |
 |  Q  .  .  |  R  .  .  |  .  .  .  |
 +-----------------------------------+

 Only one of the QQ (or RR) target pair must/may contain values from the base cells.

 Exception to this rule may exist when QQ (or RR) is a locked pair with a non-base cell value.

 This exception doesn't hold for RR because r9c4=6.


Furthermore, one of my case scenarios for r8c89=24 includes r9c4=4 and r79c1=25. This is a clear contradiction to a secondary equivalence for r9c4==r7c3.

_

[champagne]

As David said, it is an Exocet, but no JExocet (which must not have a base digit in the compagnon cell).

The simplest way for me to crack it, is following blue:
If 3 is not in the base cells, it cannot be in a target cell (especially not in r79c1).
If 3 is in the base cells, it must be in r9c4 (only 3 left in box 8), and cannot be in the other target cells r79c1.

accepted in both cases

a) my code looking for Jexocets is wider open than david's constraints

b) no problem

[blue]

Furthermore, one of my case scenarios for r8c89=24 includes r9c4=4 and r79c1=25. This is a clear contradiction to a secondary equivalence for r9c4==r7c3.

The cases with r9c4=4 can be ruled out though, since they don't leave a candidate for 4 in b7 (or r7) -- in particular, since there's no candidate for 4 in r7c3.

This is just a "normal" secondary equivalence.
Whatever's in r9c4 must be in a base cell too, and with that it's forced into r7c123 (in any solution to the puzzle).
It can't go in r7c1 since it's a target cell, and it cant go in r7c2 since it already contains a 7.
That leaves r7c3 ... r7c3==r9c4 (in any solution to the puzzle).

[ I write "in any solution", but in this case (like usual) there's only one. ]

[daj95376]

This is just a "normal" secondary equivalence.
Whatever's in r9c4 must be in a base cell too, and with that it's forced into r7c123 (in any solution to the puzzle).
It can't go in r7c1 since it's a target cell, and it cant go in r7c2 since it already contains a 7.
That leaves r7c3 ... r7c3==r9c4 (in any solution to the puzzle).

Thanks Leren and blue. I lost sight of the forest for the trees. Yes, the secondary equivalence is certainly true.

_