The New Sudoku Players' Forum

[Leren]

Here is a list of easy Trigadon Puzzles from the Phil's Folly site.

Hidden Text: Show

Code: Select all

........1......234.....5.6.....7......72.8.96.2956...8.96.52...2.86.7...75.98....
........1.12.34....561.7..........85..8...6.9.65.8.12..89....165.1...9.262.9..85.
..............1.23....45.67.27....8986....37.9.38..2.6.39......28.......7.6.9..32
........1.....2.34....56....17......2.8......65..27..8.2678.1..17..65...8.52.17..
........1.....2.3.....45..6.1457.8..5.28.17..78..241...51..8...42..57..88.7......
........1....12.3....456782....2596....6......561.92...19.6...45.2.94..664...1...
.......12.....13.4..1.2356.....14.53.1.36.42..4.2.51.61.......57...3....8.9.56...
........1.....2.3..45...267....48.....83.5.9..5492..83.895.43..4.328....52..39...

My question is - When can you use the Remote Triples idea? It only applies in about half of these puzzles.

Leren

[eleven]

Not sure, what you are out for. Only in puzzles 4,5,8 the single extra candidate is in the rectangle of the tridagon, which gives a remote triple of the other 3.

[Leren]

Hi eleven. It would be helpful if you provide an example of what you are saying. The remote pairs apply in puzzles 4,5,6,8.

Leren

[totuan]

Hi Leren,

Code: Select all

........1.....2.3..45...267....48.....83.5.9..5492..83.895.43..4.328....52..39...  No.8
 *-----------------------------------------------------------------------------*
 | 236789  3679    267     | 4678    5679    367     | 4589    45      1       |
 | 16789   1679    167     | 14678   15679   2       | 4589    3       4589    |
 | 1389    4       5       | 18      19      13      | 2       6       7       |
 |-------------------------+-------------------------+-------------------------|
 | 39      39     *1267    |*167A    4       8       | 1567    1257    256     |
 | 1267   *167     8       | 3      *167     5       | 1467    9       246     |
 |*167     5       4       | 9       2      *167     | 167     8       3       |
 |-------------------------+-------------------------+-------------------------|
 |*167     8       9       | 5      *167     4       | 3       127     26      |
 | 4      *167     3       | 2       8      *167     | 15679   157     569     |
 | 5       2      *167A    |*167A    3       9       | 14678   147     468     |
 *-----------------------------------------------------------------------------*

Tridagon (167) * marked cells with one guardian 2r4c3 that is form rectangle
=> r4c3=2 and we have RT (remote triple) “A” marked cells r49c4/r9c3

Code: Select all

 *-----------------------------------------------------------*
 | 2     3679  67    | 4678  5679  367   | 589   45    1     |
 | 89    1679  167   | 4678  5679  2     | 589   3     48    |
 | 389   4     5     | 18    19    13    | 2     6     7     |
 |-------------------+-------------------+-------------------|
 | 39    39    2     |A167   4     8     |#167  #17    5     |
 | 167   17    8     | 3     167   5     | 4     9     2     |
 | 167   5     4     | 9     2     167   |#167   8     3     |
 |-------------------+-------------------+-------------------|
 | 17    8     9     | 5     17    4     | 3     2     6     |
 | 4     167   3     | 2     8     167   | 157   157   9     |
 | 5     2    A167   |A167   3     9     | 8-17  147   48    |
 *-----------------------------------------------------------*

Whichever (1|6|7)r4c3 lead to (1|6|7)r6c7 by B6 => r6c7/r9c34 form RT (167) => r9c7<>17

Hope this help by my too bad English
totuan

[Leren]

Code: Select all

*--------------------------------------------------------------------------------*
| 236789  3679    267      | 4678    5679    367      | 4589    45      1        |
| 16789   1679    167      | 14678   15679   2        | 4589    3       4589     |
| 1389    4       5        | 18      19      13       | 2       6       7        |
|--------------------------+--------------------------+--------------------------|
| 39      39      2        |c167     4       8        | 1567    1257    256      |
|a167b   a167b    8        | 3      b167     5        | 1467    9       246      |
| 167b    5       4        | 9       2      a167      | 167     8       3        |
|--------------------------+--------------------------+--------------------------|
|b167     8       9        | 5      b167     4        | 3       127     26       |
| 4      a167     3        | 2       8      a167      | 15679   157     569      |
| 5       2      c167      |c167     3       9        | 14678   147     468      |
*--------------------------------------------------------------------------------*

Thanks totuan and eleven. I think I've finally worked out what eleven meant. By rectangle, he meant box, and it became clear.

Using puzzle 8 consider cells marked a, b and c in Boxes 5, 8 and 7. Both a's and b's in Boxes 5 and 7 can see at least one 167 cell in the Trigadon Box 4. The c's can't see a 167 cell.

So that's 2 Passes and one Fail. However, from Mith's post on the Loki puzzle, 2 Passes is all you need. The third RT is forced by the existence of the other two. I've checked all 8 puzzles and this works correctly in every case.

Thanks to both of you for your forebearence and your explanations. Leren

[denis_berthier]

.
All this is quite unsatisfactory to me.

The 1st RT is defined unambiguously by the unique rectangle in the tridagon pattern in case there's a single guardian (no, Leren, rectangle doesn't mean block; it means the only set of 4 cells in the tridagon pattern that forms a rectangle).
Unfortunately, this RT is sterile.

The 2nd RT is defined by some ad hoc piece of reasoning that amounts to T&E, e.g.

Whichever (1|6|7)r4c3 lead to (1|6|7)r6c7 by B6

You can write this in a compact form if you like(*), but the intended meaning is;
- suppose n1r4c3 then after some reasoning n1r6c7
- suppose n6r4c3 then after some similar reasoning n6r6c7
- suppose n7r4c3 then after some similar reasoning n7r6c7
(*)Fortunately, you didn't write (1|6|7)r4c3 => (1|6|7)r6c7, which would be logically false (with "|" understood as "OR")

Then you add still more ad hoc reasoning. And then you conclude that there's a 2nd RT

=> r6c7/r9c34 form RT (167) => r9c7<>17

I think if these 2 RTs are in any way related to the tridagon pattern instead of ad hoc positions of candidates on the grid, their eliminations should somehow be the consequences of another pattern (probably one in the 630 list).

[jco]

"Remote triple is a set of 3 cells C1,C2,C3 each containing the same triplet {a,b,c}, and having each a different solution within the triplet {a,b,c}" (see this post by Cenoman).
"If in a TH pattern one of the rectangle cells is solved, the other 3 always are a remote triple."
nicely proven by eleven and Marek.
Based on this fact, the correct digits in the three remaining cells of the rectangle r4c34,r9c34 must be different.
I see totuan's move as follows

Code: Select all

.-----------------------------------------------------------.
| 2     3679  67    | 4678  5679  367   | 589   45    1     |
| 89    1679  167   | 4678  5679  2     | 589   3     48    |
| 389   4     5     | 18    19    13    | 2     6     7     |
|-------------------+-------------------+-------------------|
| 39    39    2     |*167a -4---- 8---. | 167   17    5     |
| 167   17    8     | 3    *167   5   | | 4     9     2     |
| 167   5     4     | 9     2    *167 '-| 167a  8     3     |
|-------------------+-------------------+-------------------|
|*17    8     9     | 5    *17    4     | 3     2     6     |
| 4    *167   3     | 2     8    *167   | 157   157   9     |
| 5     2    *167c  |*167b  3     9     | 8-17  147   48    |
'-----------------------------------------------------------'

Remote Triple (167)r4c4,r9c4,r9c3 noticing that the correct
digit "a" at r4c4 is the same at r6c7 (due to the ER, box 6).

=> - 17 r9c7 [18 eliminations and basics]

[denis_berthier]

.
I have no problem with the 1st RT.
The 2nd remains ad hoc.

[jco]

The second RT follows immediately from the first since the correct digit at r4c4 must be the same
as the correct digit at r6c7 due to the "L" at box 6: (a)r4c4 - (a)r4c78 = (a)r6c7.
(a)r4c4 is "transported" to (a)r6c7, where "a" is the correct digit at r4c4.

[Leren]

OK Denis and eleven, I accept my badge of shame for today, the rectangle cell is the 4th one in the trigadon box that completes the rectangle with the proposed RT cells. Not my first and won't be my last. My revised diagram for puzzle 8 :

Code: Select all

*--------------------------------------------------------------------------------*
| 236789  3679    267      | 4678    5679    367      | 4589    45      1        |
| 16789   1679    167      | 14678   15679   2        | 4589    3       4589     |
| 1389    4       5        | 18      19      13       | 2       6       7        |
|--------------------------+--------------------------+--------------------------|
| 39      39     a2        |a167     4       8        | 1567    1257    256      |
|b167     167     8        | 3      b167     5        | 1467    9       246      |
| 167    c5       4        | 9       2      c167      | 167     8       3        |
|--------------------------+--------------------------+--------------------------|
|b167     8       9        | 5      b167     4        | 3       127     26       |
| 4      c167     3        | 2       8      c167      | 15679   157     569      |
| 5       2      a167      |a167     3       9        | 14678   147     468      |
*--------------------------------------------------------------------------------*

jco quoted : "If in a TH pattern one of the rectangle cells is solved, the other 3 always are a remote triple."

I assume this means that the rectangle cell is solved to a non-trigadon digit, or may not be solved but contains only non-trigadon candidates.

If this is correct the 4 cells marked a and the 4 cells marked c form RT's of the form a-b-c-d. From what mith said here if you have 2 RT's you must have 3.

So the 4 cells I've marked b don't seem to prove anything by themselves, but taken with a, c, and mith, they must form a third RT which may be of the form a-b-c-d or a-b-c-a.

Leren

[Leren]

OK, so here is puzzle 6:

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*--------------------------------------------------------------------------------*
| 2478    2678    4578     | 3789    378     378      | 456     459     1        |
| 478     678     4578     | 789     1       2        | 456     3       59       |
| 139     39      13       | 4       5       6        | 7       8       2        |
|--------------------------+--------------------------+--------------------------|
| 14     b378     14       |b378     2       5        | 9       6       378      |
| 29      29     c378      | 6      c378     378      | 13458   1457    3578     |
|a378     5       6        | 1       4      a9        | 2       47      378      |
|--------------------------+--------------------------+--------------------------|
|a378     1       9        | 25      6      a378      | 358     257     4        |
| 5      b378     2        |b378     9       4        | 138     17      6        |
| 6       4      c378      | 25     c378     1        | 358     2579    35789    |
*--------------------------------------------------------------------------------*

The cells marked a prove an RT of the form a-b-c-d. The cells marked b and c don't seem to prove anything but in the event they form RT's of the form a-b-c-a.

So my question is, is 1 RT of the form a-b-c-d enough to prove that there must be 3, or do you have to prove 2 RT's ?

Leren

[marek stefanik]

I don't know what you mean by a-b-c-d and a-b-c-a and I don't see the RT in your last example (you have all diagonals going the same way, for anything TH-related you need a 3+1 split), so this may not apply there.
But with TH 1 RT is enough.

Code: Select all

*--------------------------------------------------------------------------------*
| 2478    2678    4578     | 3789    378     378      | 456     459     1        |
| 478     678     4578     | 789     1       2        | 456     3       59       |
| 139     39      13       | 4       5       6        | 7       8       2        |
|--------------------------+--------------------------+--------------------------|
| 14     b378     14       |#378     2       5        | 9       6       378      |
| 29      29     c378      | 6       378    #378      | 13458   1457    3578     |
|a378     5       6        | 1      #4       9        | 2       47      378      |
|--------------------------+--------------------------+--------------------------|
|a378     1       9        | 25      6      a378      | 358     257     4        |
| 5      b378     2        |b378     9       4        | 138     17      6        |
| 6       4      c378      | 25     c378     1        | 358     2579    35789    |
*--------------------------------------------------------------------------------*

Here, the b-marked cells form the TH rectangle with r4c4 while the a- and c-marked cells form the 8-loop with b5p68 (obstructed by 4r6c5).
Even though we cannot get any RTs, we can still say something about b478.
The permutations in each of b478 (abc in this order) all have the same parity.
Boxes 47 and b78 cannot have the same permutation, therefore either b48 have the same permutation (and there are no RTs), or b478 each have a different permutation of the same parity (so all permutations are used), leading to all three RTs.

You can use this information even without the direct RTs, consider puzzle 1:

Code: Select all

.---------------.----------------.--------------.
| 9    6    234 | 7    234   34  | 8    5    1  |
| 5    7    1   | 8    9     6   | 2    3    4  |
| 8    34   234 | 134  1234  5   | 7    6    9  |
:---------------+----------------+--------------:
| 6    8   #5   |a134  7     9   | 134  124  23 |
| 134 #134  7   | 2   b134   8   | 5    9    6  |
|#134  2    9   | 5    6    c134 | 134  7    8  |
:---------------+----------------+--------------:
|a134  9    6   |a134  5     2   | 134  8    7  |
| 2   b134  8   | 6   b134   7   | 9    14   5  |
| 7    5   c34  | 9    8    c134 | 6    124  23 |
'---------------'----------------'--------------'

Here, the guardian is within the TH 8-loop, throwing a wrench into the RTs.
We can however still use the knowledge that b57 have the same parity.
The digit in r4c4 is forced into r6c7 in b6 and into r7c1 in r7.
Since b57 have the same parity, this forces them to have the same exact permutation.
So the digit in r9c3 appears in r6c6 (-1r6c6) and in r1 it is forced into r1c5 (-2r1c5, stte).

Something interesting: In puzzle 8 we can use a broken RT to invalidate a possible TH without a rectangle cell.
After the main TH and a placement in one of its RTs we get here:

Code: Select all

.----------------.----------------.--------------.
| 2    3679  67  | 4678  579  367 | 589  45   1  |
| 89   1679  167 | 4678  579  2   | 589  3    48 |
| 389  4     5   | 18    19   13  | 2    6    7  |
:----------------+----------------+--------------:
| 39   39    2   |#17    4    8   | 6   #17   5  |
| 17   17    8   | 3    #6    5   | 4    9   #2  |
| 6    5     4   | 9     2   #17  |#17   8    3  |
:----------------+----------------+--------------:
| 17   8     9   | 5    #17   4   | 3    2   #6  |
| 4    167   3   | 2     8   #167 | 157 #157  9  |
| 5    2     167 |#167   3    9   |#178  147  48 |
'----------------'----------------'--------------'

TH 167# (with a pre-placed rectangle guardian 2r5c9)
The TH rectangle contains two 6s, so a guardian within the 8-loop is necessary, i.e. 5r8c8 = 8r9c7.
5r8c8 = 8r9c7 – (8=59)r12c7 => –5r1c8, –5r8c7, stte

Marek

[Leren]

Hi Marek, sorry about the abcd confusion. From now on I'll just use abc to indicate which potential RT is being considered, and w-x-y-z or w-x-y-w to represent which "type" of RT it is.

Now according to eleven and others, a w-x-y-z type of RT is easy to prove, because the z is a solved cell that is not a TH digit and is in the Rectangle cell.

So, in the 4 cells marked a (ie proposed RT a, z= 9 is in the rectangle cell), and in the solution r7c1 = w = 8, r6c1 = x = 3, r7c6 = y = 7 and r6c6 = z = 9 and the "type" of the RT is w-x-y-z.

As it turns out, in the 4 cells I've marked b, r8c2 = w = 7, r4c1 = x = 8, r8c3 = y = 3 and r4c3 = w = 7. So two 7's in the 4 cells and the "type" of the RT is w-x-y-w.

In the 4 cells marked c, r9c3 = w = 3, r5c3 = x = 7, r9c5 = z = 8 and r5c5 = w = 3. So two 3's in the 4 cells and the "type" of the RT is w-x-y-w.

You say But with TH 1 RT is enough. If that applies to all TH puzzles then that is all I was ever after.

A simple rule to prove the existence of 3 RT's. Find a w-x-y-z RT using the solved/no TH candidates in the Rectangle cell thing and you are done (I hope).

Leren

[denis_berthier]

.
#8 is a particularly bad example for RTs. After the tridagon, it solves with elementary rules.

Code: Select all

Resolution state after Singles and whips[1]:
   +----------------------+----------------------+----------------------+
   ! 236789 3679   267    ! 4678   5679   367    ! 4589   45     1      !
   ! 16789  1679   167    ! 14678  15679  2      ! 4589   3      4589   !
   ! 1389   4      5      ! 18     19     13     ! 2      6      7      !
   +----------------------+----------------------+----------------------+
   ! 123679 13679  1267   ! 167    4      8      ! 1567   1257   256    !
   ! 1267   167    8      ! 3      167    5      ! 1467   9      246    !
   ! 167    5      4      ! 9      2      167    ! 167    8      3      !
   +----------------------+----------------------+----------------------+
   ! 167    8      9      ! 5      167    4      ! 3      127    26     !
   ! 4      167    3      ! 2      8      167    ! 15679  157    569    !
   ! 5      2      167    ! 167    3      9      ! 14678  147    468    !
   +----------------------+----------------------+----------------------+
170 candidates.

Code: Select all

hidden-pairs-in-a-row: r4{n3 n9}{c1 c2} ==> r4c2≠7, r4c2≠6, r4c2≠1, r4c1≠7, r4c1≠6, r4c1≠2, r4c1≠1

   +----------------------+----------------------+----------------------+
   ! 236789 3679   267    ! 4678   5679   367    ! 4589   45     1      !
   ! 16789  1679   167    ! 14678  15679  2      ! 4589   3      4589   !
   ! 1389   4      5      ! 18     19     13     ! 2      6      7      !
   +----------------------+----------------------+----------------------+
   ! 39     39     1267   ! 167    4      8      ! 1567   1257   256    !
   ! 1267   167    8      ! 3      167    5      ! 1467   9      246    !
   ! 167    5      4      ! 9      2      167    ! 167    8      3      !
   +----------------------+----------------------+----------------------+
   ! 167    8      9      ! 5      167    4      ! 3      127    26     !
   ! 4      167    3      ! 2      8      167    ! 15679  157    569    !
   ! 5      2      167    ! 167    3      9      ! 14678  147    468    !
   +----------------------+----------------------+----------------------+

tridagon for digits 1, 6 and 7 in blocks:
        b4, with cells: r4c3 (target cell), r6c1, r5c2
        b5, with cells: r4c4, r6c6, r5c5
        b7, with cells: r9c3, r7c1, r8c2
        b8, with cells: r9c4, r7c5, r8c6
 ==> r4c3≠1,6,7

Code: Select all

singles ==> r4c3=2, r1c1=2, r5c9=2, r7c9=6, r4c9=5, r8c9=9, r5c7=4, r7c8=2
finned-x-wing-in-rows: n1{r7 r3}{c1 c5} ==> r2c5≠1
finned-x-wing-in-rows: n1{r7 r5}{c5 c1} ==> r6c1≠1
whip[1]: b4n1{r5c2 .} ==> r5c5≠1
finned-x-wing-in-rows: n7{r7 r5}{c5 c1} ==> r6c1≠7
singles ==> r6c1=6, r5c5=6, r4c7=6
naked-pairs-in-a-column: c1{r5 r7}{n1 n7} ==> r3c1≠1, r2c1≠7, r2c1≠1
whip[1]: r3n1{c6 .} ==> r2c4≠1
finned-x-wing-in-rows: n6{r9 r2}{c4 c3} ==> r1c3≠6
naked-single ==> r1c3=7
finned-x-wing-in-columns: n7{c6 c7}{r6 r8} ==> r8c8≠7
biv-chain[3]: r8n6{c6 c2} - b7n7{r8c2 r7c1} - r7n1{c1 c5} ==> r8c6≠1
biv-chain[3]: c5n1{r7 r3} - c6n1{r3 r6} - c6n7{r6 r8} ==> r7c5≠7
stte

[Edit]: #4 and #5 are not better. Indeed, this shouldn't be surprising, as they are tagged on Phil's site as easy tridagon puzzles. After applying a tridagon with a single guardian, they have straightforward solutions.
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[marek stefanik]

I still don't completely understand your question.
Firstly, it seems that your 'remote triples' have four cells.
Secondly, I don't see how you got the a-marked RT in puzzle 6, see this placement of 378 consistent with b4578:

Code: Select all

*--------------------------------------------------------------------------------*
| 2478    2678    4578     | 3789    378     378      | 456     459     1        |
| 478     678     4578     | 789     1       2        | 456     3       59       |
| 139     39      13       | 4       5       6        | 7       8       2        |
|--------------------------+--------------------------+--------------------------|
| 14     b#8      14       |b#7      2       5        | 9       6       378      |
| 29      29     c#7       | 6      c#3      #8       | 13458   1457    3578     |
|a#3      5       6        | 1       4      a9        | 2       47      378      |
|--------------------------+--------------------------+--------------------------|
|a#7      1       9        | 25      6      a#3       | 358     257     4        |
| 5      b#3      2        |b#8      9       4        | 138     17      6        |
| 6       4      c#8       | 25     c#7      1        | 358     2579    35789    |
*--------------------------------------------------------------------------------*

Had the b- and c-marked cells formed one 8-loop (for example if they included b5p24 instead of b5p15), you couldn't have 3r6c1&3r7c6, as the 78 pairs in b48 would break either b5 or b7, but in this case you need other parts of the puzzle to prove the RT.

Marek