The New Sudoku Players' Forum

[eleven]

As there are 4 target cells, the base digits 6789 are all true in the two single JExocets so can be eliminated in the non-'S' cells in their respective cover rows; (6) in r3 & r9, (7) in r8 & r9, (8) in r1 & r7, and (9) in r1 & r3.

Ah, that's it, thanks.
I had tried to search for double (J)Exocets to learn more about the forced eliminations, but i ended nowhere. Maybe you can add a link in the first post of your JExocet definitions thread.

[David P Bird]

Eleven, I'm conscious that this is off-topic here, so will keep it brief. Because there are so many possible configurations those eliminations are far from 'forced'. In the definition document, section 8 covers double JExocets and subsection 8.3 covers the rule concerned, but I'll consider your suggestion.

[champagne]

As there are 4 target cells, the base digits 6789 are all true in the two single JExocets so can be eliminated in the non-'S' cells in their respective cover rows; (6) in r3 & r9, (7) in r8 & r9, (8) in r1 & r7, and (9) in r1 & r3.

Ah, that's it, thanks.
I had tried to search for double (J)Exocets to learn more about the forced eliminations, but i ended nowhere. Maybe you can add a link in the first post of your JExocet definitions thread.

As far as i know, all double exocets are "JExocets".

The generic basic rule is very simple
Any cell seeing both targets can not contain one digit of the base
Any cell seeing the four targets can not contain any digit of the base.

I think that most of the old discussions on that pattern can be found



here

[eleven]

Sorry, i don't have the time to read all the threads (partially i had followed them).
The point seems to be, that - apart from the "generic basic rule" - for each digit common constraints of both exocets can be used for eliminations.

So the 5th puzzle (ER 10.5), which is the same as the last one in Mladens first list with 3 backdoors, has a double exocet with following eliminations, then pairs and w-wing solve it.

Hidden Text: Show

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.....1..2...2...3...4.5.6....9....766...9...554....9....5.6.7...1...8...8..3.....
+-------------------------+-------------------------+-------------------------+
| 79-3    56789-3 3678    | 46789   3478    1       | 458     4589    2       |
| 79-1    56789   1678    | 2       478     4679    | 1458    3       4789-1  |
| 12379   23789   4       | 789     5       379     | 6       189     1789    |
+-------------------------+-------------------------+-------------------------+
|#123    #238     9       | 1458   @12348   2345    | 12348   7       6       |
| 6       2378   @12378   | 47-18   9       47-23   |#12348   1248    5       |
| 5       4       7-1238  | 1678   #12378   2367    | 9      @128    @138     |
+-------------------------+-------------------------+-------------------------+
| 2349    239     5       | 149     6       249     | 7       12489   13489   |
| 2479-3  1       2367    | 4579    247     8       | 2345    24569   49-3    |
| 8       2679    267     | 3       1247    24579   | 1245    24569-1 49-1    |
+-------------------------+-------------------------+-------------------------+

The second puzzle seems to be harder after the exocet (i found no additional eliminations here), i needed a couple of (symmetry) moves.

Hidden Text: Show

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.....7.6...4.3.5...7.6....3..1....32....1....45....1..5....8.9...3.5.2...8.9.....
+----------------------+----------------------+----------------------+
| 12389  1239   589    | 12458  2489   7      | 89     6      14     |
| 12689  1269   4      | 128    3      129    | 5      127    17     |
| 1289   7      589    | 6      2489   12459  | 89     124    3      |
+----------------------+----------------------+----------------------+
| 6789   69     1      | 578    6789   569    | 4      3      2      |
| 36789  369    89     | 24     1      24     | 67     578    56789  |
| 4      5      2      | 378    6789   369    | 1      78     6789   |
+----------------------+----------------------+----------------------+
| 5      124    67     | 12347  2467   8      | 367    9      1467   |
| 19     149    3      | 147    5      146    | 2      1478   14678  |
| 12     8      67     | 9      2467   12346  | 367    1457   14567  |
+----------------------+----------------------+----------------------+
6r4c2 or
9r4c2->8r5c3->6r5c7 => r5c12<>6
9r4c2->9r6c5/7r4c5 => r4c5<>6
9r4c2 or
6r4c2->6r2c1/8r9c8->6r8c6->6r6c5->9r5c3 => r45c1,r5c2<>9
6r4c2->6r2c1/8r9c8->6r8c6->6r6c5/8r3c5 => r3c5<>9
+----------------------+----------------------+----------------------+
| 12389  1239   589    | 12458  2489   7      | 89     6      14     |
| 12689  1269   4      | 128    3      129    | 5      127    17     |
| 1289   7      589    | 6      2489   12459  | 89     124    3      |
+----------------------+----------------------+----------------------+
| 678    69     1      | 578    78     569    | 4      3      2      |
| 3678   36     89     | 24     1      24     | 67     58     5689   |
| 4      5      2      | 378    69     369    | 1      78     689    |
+----------------------+----------------------+----------------------+
| 5      124    67     | 12347  2467   8      | 367    9      1467   |
| 19     149    3      | 147    5      146    | 2      1478   14678  |
| 12     8      67     | 9      2467   12346  | 367    1457   14567  |
+----------------------+----------------------+----------------------+
6r5c7=r6c5 => r6c9<>6
+----------------------+----------------------+----------------------+
| 3      129    589    | 12458  2489   7      | 89     6      14     |
| 12689  1269   4      | 128    3      129    | 5      127    17     |
| 1289   7      589    | 6      2489   12459  | 89     124    3      |
+----------------------+----------------------+----------------------+
| 67     69     1      | 578    78     59     | 4      3      2      |
| 78     3      89     | 24     1      24     | 67     5      69     |
| 4      5      2      | 37     69     369    | 1      78     89     |
+----------------------+----------------------+----------------------+
| 5      124    67     | 12347  2467   8      | 367    9      1467   |
| 19     149    3      | 147    5      146    | 2      1478   14678  |
| 12     8      67     | 9      2467   12346  | 367    147    5      |
+----------------------+----------------------+----------------------+
6r8c9 or 6r8c6->6r6c5->9r5c3->6r6c9 => r7c9<>6
then 7r2c9 or 7r2c8->14r12c9->7r7c9 => r8c9<>7
triples/x-wings/pairs
+----------------------+----------------------+----------------------+
| 3      12     589    | 12458  2489   7      | 89     6      14     |
| 68     69     4      | 128    3      129    | 5      12     7      |
| 12     7      589    | 6      2489   1245   | 89     124    3      |
+----------------------+----------------------+----------------------+
| 67     69     1      | 58     78     59     | 4      3      2      |
| 78     3      89     | 24     1      24     | 67     5      69     |
| 4      5      2      | 37     69     36     | 1      78     89     |
+----------------------+----------------------+----------------------+
| 5      124    67     | 1234   2467   8      | 367    9      14     |
| 9      14     3      | 147    5      146    | 2      78     68     |
| 12     8      67     | 9      2467   12346  | 367    14     5      |
+----------------------+----------------------+----------------------+
1r1c2=1r3c1 => r9c1/r1c9<>1

Again better solutions are welcome.

[David P Bird]

Eleven, for your 2nd grid you have missed the common non-'S' cell eliminations here:

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 *-----------------------*-----------------------*-----------------------*
 | 123-89 123-9  589     | 1245-8 2489   <7>     | 89     <6>    14      |
 | 12689  1269   <4>     | 128    <3>    129     | <5>    127    17      |
 | 12-89  <7>    589     | <6>    2489   1245-9  | 89     124    <3>     |
 *-----------------------*-----------------------*-----------------------*
 | 6789   69     <1>     | 578    6789   569     | 4      <3>    <2>     |
 | 36789  369    89      | 24     <1>    24      | 67     578    56789   |
 | <4>    <5>    2       | 378    6789   369     | <1>    78     6789    |
 *-----------------------*-----------------------*-----------------------*
 | <5>    124    67      | 1234-7 2467   <8>     | 367    <9>    14-67   |
 | 19     149    <3>     | 147    <5>    146     | <2>    1478   14678   |
 | 12     <8>    67      | <9>    2467   1234-6  | 367    145-7  145-67  |
 *-----------------------*-----------------------*-----------------------*
Now naked pairs (12)r39c1 and (14)r17c9 produce a series of tuple eliminations to leave
 *-----------------*-----------------*-----------------*
 | 3    12   589   | 1245 2489 <7>   | 89   <6>  14    |
 | 68   69   <4>   | 128  <3>  129   | <5>  12   7     |
 | 12   <7>  589   | <6>  2489 1245  | 89   124  <3>   |
 *-----------------*-----------------*-----------------*
 | 678  69   <1>   | 578  6789 569   | 4    <3>  <2>   |
 | 678  3    89    | 24   <1>  24    | 67   5    689   |
 | <4>  <5>  2     | 378  6789 369   | <1>  78   689   |
 *-----------------*-----------------*-----------------*
 | <5>  124  67    | 1234 2467 <8>   | 367  <9>  14    |
 | 9    14   <3>   | 147  <5>  146   | <2>  78   68    |
 | 12   <8>  67    | <9>  2467 1234  | 367  14   5     |
 *-----------------*-----------------*-----------------*

From here I'm not sure of the quickest way to complete the solution.
(Now that Champagne has shown he doesn't mind his thread going off-topic, I don't feel so guilty!)

[eleven]

I see, in the first exocet (r4c12) the 8 leaves an x-wing in c37, in the second 8 can't be both in r4c5 and r5c3, so there is an x-wing either in c37 or in c57. Similar for 9.
In the grid, which is left, my last move is enough: 1r1c2=1r3c1 => r9c1/r1c9<>1

[Added:]Had a look now at the remaining two of the 5 puzzles (same as on page 1), both solve easily after the double exocet too, e.g. nr 4 this way

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+----------------------+----------------------+----------------------+
| 689   #34     679    | 5689   23     1      | 5789   2-4    89     |
| 689    17     2      | 4      68     5689   | 5789   15     3      |
| 13     5      1349   | 389    7      289    | 2489   6      12489  |
+----------------------+----------------------+----------------------+
| 123    9      134    | 135    1234   245    | 6      8      7      |
| 7      1234   134    | 68     9      68     | 234    1234   5      |
| 5      6      8      | 137    1234   247    | 234    9      124    |
+----------------------+----------------------+----------------------+
| 123    8      1369   | 169    5      469    | 2349   7      2469   |
| 4      27     5679   | 6789   68     3      | 1      25     2689   |
| 69     1-3    5679   | 2      14     6789   | 589   #34     689    |
+----------------------+----------------------+----------------------+
(3=4)r1c2/r9c8 =>r1c8<>4,r9c2<>3

[blue]

Eleven, for your 2nd grid you have missed the common non-'S' cell eliminations here:

Code: Select all

 *-----------------------*-----------------------*-----------------------*
 | 123-89 123-9  589     | 1245-8 2489   <7>     | 89     <6>    14      |
 | 12689  1269   <4>     | 128    <3>    129     | <5>    127    17      |
 | 12-89  <7>    589     | <6>    2489   1245-9  | 89     124    <3>     |
 *-----------------------*-----------------------*-----------------------*
 | 6789   69     <1>     | 578    6789   569     | 4      <3>    <2>     |
 | 36789  369    89      | 24     <1>    24      | 67     578    56789   |
 | <4>    <5>    2       | 378    6789   369     | <1>    78     6789    |
 *-----------------------*-----------------------*-----------------------*
 | <5>    124    67      | 1234-7 2467   <8>     | 367    <9>    14-67   |
 | 19     149    <3>     | 147    <5>    146     | <2>    1478   14678   |
 | 12     <8>    67      | <9>    2467   1234-6  | 367    145-7  145-67  |
 *-----------------------*-----------------------*-----------------------*

There are 4 more eliminations from the double exocet: r4c4<>7, r4c6<>6. r6c4<>8 and r6r6<>9.
Each of those can see all occurences of the digit in the base cells of one exocet and the target cells of the other.

From XSudo:

Code: Select all

+----------------------+------------------------+----------------------+
| 123-89  123-9  5(89) | 1245-8  24(89)  7      | (89)   6      14     |
| 12689   1269   4     | 128     3       129    | 5      127    17     |
| 12-89   7      5(89) | 6       24(89)  1245-9 | (89)   124    3      |
+----------------------+------------------------+----------------------+
| (6789)  (69)   1     | 58-7    (6789)  59-6   | 4      3      2      |
| 36789   369    (89)  | 24      1       24     | (67)   578    56789  |
| 4       5      2     | 37-8    (6789)  36-9   | 1      (78)   (6789) |
+----------------------+------------------------+----------------------+
| 5       124    (67)  | 1234-7  24(67)  8      | 3(67)  9      14-67  |
| 19      149    3     | 147     5       146    | 2      1478   14678  |
| 12      8      (67)  | 9       24(67)  1234-6 | 3(67)  145-7  145-67 |
+----------------------+------------------------+----------------------+

16 Truths = {6789C3 6789C5 6789C7 4N12 6N89}
24 Links = {6r4679 7r4679 8r1346 9r1346 5n3 46n5 5n7 89b4 67b6}
18 Eliminations --> r1c14<>8, r1c12<>9, r3c16<>9, r7c49<>7, r9c69<>6, r9c89<>7, r3c1<>8,
r4c6<>6, r4c4<>7, r6c4<>8, r6c6<>9, r7c9<>6,

[David P Bird]

Blue, I'm on a Sudoku tick-over mode through the summer, so when I found all the first level JE eliminations I stopped. However you are now highlighting further eliminations that result from what I'm now calling 'mirror nodes'. However the way you're describing them doesn't work from a purely JExocet point of view - each of the corner cells in box 5 sees one pair of base cells plus one target from one single JE and a target call from the other.

Mirror Nodes are described in sections 3.1 & 3.2 in the < JE Definition document>. They are two cells one of which will mirror the base digit held in a target cell and the other will contain a non-base digit. For example the digit in target r5c7 will also be true in the base cells in r4c12, so must be true in r6c46 (r6c5 is excluded as it's the other target). Eliminations involving mirror cells are listed in section 5 points 7,8,9.

Comparing (67)r5c7 target and (36789)r6c46 mirror node base digits (89)r6c46 can be eliminated as they don't occur in both nodes
Comparing (6789)r6c5 with (5689)r6c89 base digit (7)r6c5 similarly can be eliminated
Likewise (67)r4c46 and (9)r4c5 can be also be eliminated in the second single JE.

I don’t run Xsudo so can't say if your truth and links set will always work, but I suspect that symmetry will have a role to play in this puzzle.

I've been trying to follow the symmetry discussions, but have found it virtually impossible because the material is so fragmented and the terminology is vague. However briefly looking at tier 2, if r4c2 and r6c8 are a symmetric pair then each set of base cells will contain one digit from (69) and one digit from (78) allowing (6)r4c1 and (8)r6c9 to be eliminated. Of course this deduction depends on the type of symmetry that exists.

DPB

[daj95376]

So the 5th puzzle (ER 10.5), which is the same as the last one in Mladens first list with 3 backdoors, has a double exocet with following eliminations, then pairs and w-wing solve it.

Code: Select all

.....1..2...2...3...4.5.6....9....766...9...554....9....5.6.7...1...8...8..3.....
+-------------------------+-------------------------+-------------------------+
| 79-3    56789-3 3678    | 46789   3478    1       | 458     4589    2       |
| 79-1    56789   1678    | 2       478     4679    | 1458    3       4789-1  |
| 12379   23789   4       | 789     5       379     | 6       189     1789    |
+-------------------------+-------------------------+-------------------------+
|#123    #238     9       | 1458   @12348   2345    | 12348   7       6       |
| 6       2378   @12378   | 47-18   9       47-23   |#12348   1248    5       |
| 5       4       7-1238  | 1678   #12378   2367    | 9      @128    @138     |
+-------------------------+-------------------------+-------------------------+
| 2349    239     5       | 149     6       249     | 7       12489   13489   |
| 2479-3  1       2367    | 4579    247     8       | 2345    24569   49-3    |
| 8       2679    267     | 3       1247    24579   | 1245    24569-1 49-1    |
+-------------------------+-------------------------+-------------------------+

I know that everyone is more interested in the other puzzle mentioned, but I haven't resolved a double exocet in quite awhile, and I wanted to expand on the eliminations associated with this double exocet ... but not its symmetry. Sorry for interrupting your stream of thought!

Hidden Text: Show

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 +--------------------------------------------------------------------------------+
 |  379     356789  3678    |  46789   3478    1       |  458     4589    2       |
 |  179     56789   1678    |  2       478     4679    |  1458    3       14789   |
 |  12379   23789   4       |  789     5       379     |  6       189     1789    |
 |--------------------------+--------------------------+--------------------------|
 | B123    B238     9       |  1458   q238-14  2345    |  12348   7       6       |
 |  6      q238-7  r1238-7  |  1478    9       2347    | R1238-4 Q128-4   5       |
 |  5       4       12378   |  1678   Q128-37  2367    |  9      b128    b138     |
 |--------------------------+--------------------------+--------------------------|
 |  2349    239     5       |  149     6       249     |  7       12489   13489   |
 |  23479   1       2367    |  4579    247     8       |  2345    24569   349     |
 |  8       2679    267     |  3       1247    24579   |  1245    124569  149     |
 +--------------------------------------------------------------------------------+
 # 176 eliminations remain

 ### -1238- QExocet   Base = r4c12 <12>   Target = (Q)r6c5==r5c8,(R)r5c7  =>  -4 r5c78; -37 r6c5

 ### -1238- QExocet   base = r6c89 <38>   Target = (r)r5c3,(q)r4c5==r5c2  =>  -14 r4c5; -7 r5c23


Code: Select all

 *** double exocet forces additional eliminations  =>  -18 r5c4; -23 r5c6; -1238 r4c7,r6c3
 +--------------------------------------------------------------------------------+
 |  379     356789  3678    |  46789   3478    1       |  458     4589    2       |
 |  179     56789   1678    |  2       478     4679    |  1458    3       14789   |
 |  12379   23789   4       |  789     5       379     |  6       189     1789    |
 |--------------------------+--------------------------+--------------------------|
 | B123    B238     9       |  1458   q238-14  2345    |  4-1238  7       6       |
 |  6      q238-7  r1238-7  |  47-18   9       47-23   | R1238-4 Q128-4   5       |
 |  5       4       7-1238  |  1678   Q128-37  2367    |  9      b128    b138     |
 |--------------------------+--------------------------+--------------------------|
 |  2349    239     5       |  149     6       249     |  7       12489   13489   |
 |  23479   1       2367    |  4579    247     8       |  2345    24569   349     |
 |  8       2679    267     |  3       1247    24579   |  1245    124569  149     |
 +--------------------------------------------------------------------------------+
 # 176 eliminations remain


_

[Leren]

blue wrote :
There are 4 more eliminations from the double exocet: r4c4<>7, r4c6<>6. r6c4<>8 and r6r6<>9.
Each of those can see all occurences of the digit in the base cells of one exocet and the target cells of the other.

The eliminations are valid but not for the reason stated. Just to clarify what David is saying about mirror nodes here are the the eliminations in r4c46:

Code: Select all

*--------------------------------------------------------------------------------*
| 12389   1239    589      | 12458   2489    7        | 89      6       14       |
| 12689   1269    4        | 128     3       129      | 5       127     17       |
| 1289    7       589      | 6       2489    12459    | 89      124     3        |
|--------------------------+--------------------------+--------------------------|
| 6789    69      1        |*58-7   t6789   *59-6     | 4       3       2        |
| 36789   369    T89       | 24      1       24       | 67      578     56789    |
| 4       5       2        | 37      6789    369      | 1      B78     B6789     |
|--------------------------+--------------------------+--------------------------|
| 5       124     67       | 12347   2467    8        | 367     9       1467     |
| 19      149     3        | 147     5       146      | 2       1478    14678    |
| 12      8       67       | 9       2467    12346    | 367     1457    14567    |
*--------------------------------------------------------------------------------*

I see these as a dual secondary equivalence of the single Exocet r6c8 r6c9 r5c3 r4c5 6789.

Since one of r4c46 must contain 5 (a non-base digit) one of two possible secondary equivalences r4c4==r5c3 or r4c6==r5c3 must exist and the one that does not exist must solve to 5.

The net result of this is that you can remove non-common digits between r4c46 and r5c3 (but not the linking digit 5 in r4c46).

How common are these dual secondary equivalences ? About as common as ordinary (single) scondary equivalences, and they appear to be very common in double Exocet patterns.

In fact, in this puzzle there were 4 dual secondary equivalences (a record I think), one of which did not result in any eliminations.

Leren

[blue]

There are 4 more eliminations from the double exocet: r4c4<>7, r4c6<>6. r6c4<>8 and r6r6<>9.
Each of those can see all occurences of the digit in the base cells of one exocet and the target cells of the other.

For David and Leren,

This is a general rule for a double exocet, unless I'm wrong about what a double exocet is. It's two exocets with <= 4 digits each, and 4 digits total, where the target cells for one exocet can see the base cells of the other, and vica-versa (?).
As a result, each of the 4 digits must occur in the base and target cells of exactly one exocet.

Here then, for 7r4c4, for example: If 7 doesn't occur in the base cells of the r4c12 exocet, it must occur in the base and target cells of the r6c89 exocet. Since r4c5 is the only target for that exocet, that actually has a candidate for '7', and since r4c4 sees r4c5, it follows that r4c4 can't contain a 7.

--

P.S.: I see that things have gotten a little off track. The original puzzle didn't have r4c7=4 and r6c3=2. All 4 candidates were present in r5c3 and r5c7, and the double exocet eliminations were only these (AFAIK):

Code: Select all

+----------------------+----------------------+----------------------+
| 123-89  123-9 2589   | 1245-8 2489   7      | 489     6     14-89  |
| 12689   1269  4      | 128    3      129    | 5       1278  1789   |
| 12-89   7     2589   | 6      2489   1245-9 | 489     124-8 3      |
+----------------------+----------------------+----------------------+
| 6789    69    1      | 4578   6789-4 4569   | 4-6789  3     2      |
| 236789  2369  6789-2 | 24-78  1      24-69  | 6789-4  4578  456789 |
| 4       5     2-6789 | 2378   6789-2 2369   | 1       78    6789   |
+----------------------+----------------------+----------------------+
| 5       124-6 267    | 1234-7 2467   8      | 3467    9     14-67  |
| 1679    1469  3      | 147    5      146    | 2       1478  14678  |
| 12-67   8     267    | 9      2467   1234-6 | 3467    145-7 145-67 |
+----------------------+----------------------+----------------------+

Added: The elimnations for -78r5c4 and -67r5c6 are for reasons similar to above: each one can see all occurences of its digit in the 4 target cells.
Also: Thank you. I can see now, that if you add in the mirror nodes with thier locked candidates (in B5), you can get -4r4c46 and -2r6c46 as well.

[daj95376]

There are 4 more eliminations from the double exocet: r4c4<>7, r4c6<>6. r6c4<>8 and r6r6<>9.
Each of those can see all occurences of the digit in the base cells of one exocet and the target cells of the other.

In the case of this puzzle, each elimination can be derived using only one exocet.

Code: Select all

after first double exocet and subsequent Basics
+----------------------+----------------------+----------------------+
| 12389  1239   589    | 12458  2489   7      | 89     6      14     |
| 12689  1269   4      | 128    3      129    | 5      127    17     |
| 1289   7      589    | 6      2489   12459  | 89     124    3      |
+----------------------+----------------------+----------------------+
| 6789   69     1      | 578    6789   569    | 4      3      2      |
| 36789  369    89     | 24     1      24     | 67     578    56789  |
| 4      5      2      | 378    6789   369    | 1      78     6789   |
+----------------------+----------------------+----------------------+
| 5      124    67     | 12347  2467   8      | 367    9      1467   |
| 19     149    3      | 147    5      146    | 2      1478   14678  |
| 12     8      67     | 9      2467   12346  | 367    1457   14567  |
+----------------------+----------------------+----------------------+

7r6c45 = 7r6c89; exocet r6c89 -> =7r4c5  =>  -7 r4c4
6r6c56 = 6r6c 9; exocet r6c89 -> =6r4c5  =>  -6 r4c6

8r4c45 = 8r4c1 ; exocet r4c12 -> =8r6c5  =>  -8 r6c4
9r4c56 = 9r4c12; exocet r4c12 -> =9r6c5  =>  -9 r6c6


_

[David P Bird]

This is a general rule for a double exocet, unless I'm wrong about what a double exocet is. It's two exocets with <= 4 digits each, and 4 digits total, where the target cells for one exocet can see the base cells of the other, and vica-versa (?).
As a result, each of the 4 digits must occur in the base and target cells of exactly one exocet.

That's a particular type of double JE with 4 target cells (a JE4 in my write-up) with only four base digits between them. Other varieties exist with various combinations of base digits, target cells, and cross lines.

Each of those can see all occurences of the digit in the base cells of one exocet and the target cells of the other.

But that could have been better worded because the elimination cells only see one of the target cells of one of the Exocets, and therefore require the digit to be absent from the other one which isn't immediately clear. Once that is understood you've provided an alternative proof of the mirror node inferences for the JE4 pattern, so hats off to you. As you rightly point out, the eliminations aren't immediately available but depend on basic follow-on deductions and that's true for the mirror node logic too.

Note that applying the mirror node inferences also eliminates (9)r4c5 and (7)r6c5 which your Xsudo analysis doesn't show. I guess they would need another formulation of link and cover sets. It would then be interesting to see if there was one formulation which provides all six eliminations in box 5.

[ronk]

... i needed a couple of (symmetry) moves.

Code: Select all

.....7.6...4.3.5...7.6....3..1....32....1....45....1..5....8.9...3.5.2...8.9.....
+----------------------+----------------------+----------------------+
| 12389  1239   589    | 12458  2489   7      | 89     6      14     |
| 12689  1269   4      | 128    3      129    | 5      127    17     |
| 1289   7      589    | 6      2489   12459  | 89     124    3      |
+----------------------+----------------------+----------------------+
| 6789   69     1      | 578    6789   569    | 4      3      2      |
| 36789  369    89     | 24     1      24     | 67     578    56789  |
| 4      5      2      | 378    6789   369    | 1      78     6789   |
+----------------------+----------------------+----------------------+
| 5      124    67     | 12347  2467   8      | 367    9      1467   |
| 19     149    3      | 147    5      146    | 2      1478   14678  |
| 12     8      67     | 9      2467   12346  | 367    1457   14567  |
+----------------------+----------------------+----------------------+


Hi eleven. Please provide the moves, particularly the symmetry moves, you used to get to these pencilmarks.

[David P Bird]

First I'd like to apologise to Mladen regarding the intrusion of JExocet discussions into his symmetry thread (which I misread as being started by Champagne). However it has raised a question in my mind about partial symmetry which may help bring the discussion back on topic.

Having made preliminary eliminations by say using (J)Exocet it could become apparent that the clues in one tier or stack are symmetrical, and this could be recognisable even when the puzzle grid has been scrambled. When is it then safe to use symmetry inferences to reduce the other cells in the same band?

In the grids just discussed the band in question contains clues for 5 digits, [12345], which are symmetrical about the centre cell. Nothing can be assumed about the other 4 digits, but how much evidence is needed to show that that the symmetry for the 5 clued digits must be maintained throughout the rest of the band?

Using the last case discussed it is possible to reduce tier 2 to

Code: Select all

   *-----------------------*-----------------------*-----------------------*
R4 | 678    69     <1>     | 58     678    59      | 4      <3>    <2>     |
R5 | 678    3      89      | 24     <1>    24      | 67     5      689     |
R6 | <4>    <5>    2       | 37     689    36      | <1>    78     689     |
   *-----------------------*-----------------------*-----------------------*

Assuming band symmetry inference (35)r4c4,r6c6 = (35)r4c6,r6c4, produces   
   *-----------------------*-----------------------*-----------------------*
R4 | 67     69     <1>     | 58     78     59      | 4      <3>    <2>     |
R5 | 78     3      89      | 24     <1>    24      | 67     5      69      |
R6 | <4>    <5>    2       | 37     69     36      | <1>    78     89      |
   *-----------------------*-----------------------*-----------------------*

Here it seems the evidence is overwhelming as the clues for (3) & (5) in boxes 2 & 8 in stack 2 are also symmetrical.