## Thinna Morgana (side doors for a nasty puzzle)

Everything about Sudoku that doesn't fit in one of the other sections

### Thinna Morgana (side doors for a nasty puzzle)

Following the theme of taking difficult puzzles and messing with them. Here's a different take on one of the current crop.

Arthur called for Merlin and asked "What can be done about Fata Morgana. I can't wait for someone to invent the computer so that I can run Sudoku Explainer"
Merlin said "Sire, When I was looking into the future I saw the solution posted on a Forum, but I'm afraid old age means can only remember some of the details."

Here's the starting grid.

Code: Select all
`. . .|. . .|. . 3. . 1|. . 5|6 . .. 9 .|. 4 .|. 7 .-----+-----+-----. . .|. . 9|. 5 .7 . .|. . .|. . 8. 5 .|4 . 2|. . .-----+-----+-----. 8 .|. 2 .|. 9 .. . 3|5 . .|1 . .6 . .|. . .|. . .`

Here's what he could remember.
Rotating the puzzle he could restore the givens using 4 nontrivial and 3 trivial digit pair swaps.
Outside the centre cell just one of them was valid for the completed grid as well.
For row 2 just one more of the ones he could see was true leaving him 4 digits to account for. For this row these 4 digits also paired up on nontrivially on rotation and their pairings weren't observable in the givens.

This was enough to help Arthur solve it using just a naked pair and one case of locked candidates. Can you?
Glyn

Posts: 357
Joined: 26 April 2007

### Re: Thinna Morgana (side doors for a nasty puzzle)

You forgot to mention this important bit of the story:

Arthur, initially dumbfounded by the senile babbling of Merlin, sought help from Merlin's colleague from a distant land named Oz. Below is an extract of the reply from the other wizard:

Merlin wrote:Rotating the puzzle I could restore the givens using 4 nontrivial and 3 trivial digit pair swaps.

This is the original puzzle and its 180-degree rotation:
Code: Select all
`........3..1..56...9..4..7......9.5.7.......8.5.4.2....8..2..9...35..1..6................6..1..53...9..2..8....2.4.5.8.......7.5.9......7..4..9...65..1..3........`

To restore the rotated grid to the original puzzle, we need 2 different stages of operations:

Stage 1 - all blocks except b5:
3 trivial digit swaps: 1-1, 5-5, 9-9.
3 non-trivial digit swaps: 2-4, 3-6, 7-8.

Stage 2 - b5:
1 non-trivial digit swap: 4-9.
1 clue movement: the 2-clue @ r4c4 is moved to r6c6. (Merlin must have forgotten about this important bit.)

So the 4 non-trivial digit swaps Merlin talked about were 2-4, 3-6, 7-8, 4-9, and his 3 trivial digit swaps were 1-1, 5-5, 9-9.

Merlin wrote:Outside the centre cell just one of them was valid for the completed grid as well.

At this stage we can place a hidden single of 5 @ r5c5.

According to Merlin one of the 7 digit swaps above (except the one involving the centre cell, i.e. 5-5) must also be applied globally in the solution grid (i.e. all cells involving those digit(s) must be 180-degree symmetrical to each other).

Since r5c5=5, 1-1 & 9-9 can't be possibly applied globally (considering b5).

Also, r37c5=[42] & r4c6+r6c4=[94], so 2-4 & 4-9 can't be global swapping pairs.

This leaves us only 2 possible swaps: 3-6 & 7-8. One of these two must be a global 180-degree symmetrical swapping pair in the final solution grid.

Merlin wrote:For row 2 just one more of the ones I could see was true leaving me 4 digits to account for.
For this row these 4 digits also paired up on nontrivially on rotation and their pairings weren't observable in the givens.

Now we look at r28: from the given clues, 1-1 & 5-5 are already established as local 180-degree symmetrical swapping pairs in these 2 rows.

Also, the global swapping pair (3-6 or 7-8, we don't know which one yet) must also be applied in these 2 rows.

This leaves us 5 unaccounted digits. But Merlin told us that there was one more local swapping pair leaving him 4 unaccounted digits, so this swapping pair must only involve 1 digit, i.e. the remaining trivial swap, 9-9.

The 4 unaccounted digits must pair up in manners different to the 4 non-trivial swaps specified above (2-4, 3-6, 7-8, 4-9).

However, since r2c7+r8c3=[63], we know 3 & 6 are not one of these 4 digits, so 3-6 is the global swapping pair specified above.

And the 4 unaccounted digits must be {2,4,7,8}, and their pairings must be [2-7|4-8] or [2-8|4-7].

With these bits of information in mind, we can set out to solve the puzzle:

Code: Select all
`+-------------------------+-------------------------+-------------------------+| 2458    2467    245678  | 126789  16789   1678    | 24589   1248    3       || 2348    2347    1       | 23789   3789    5       | 6       248     249     || 2358    9       2568    | 12368   4       1368    | 258     7       125     |+-------------------------+-------------------------+-------------------------+| 12348   12346   2468    | 13678   13678   9       | 2347    5       12467   || 7       12346   2469    | 136     5       136     | 2349    12346   8       || 1389    5       689     | 4       13678   2       | 379     136     1679    |+-------------------------+-------------------------+-------------------------+| 145     8       457     | 1367    2       13467   | 3457    9       4567    || 249     247     3       | 5       6789    4678    | 1       2468    2467    || 6       1247    24579   | 13789   13789   13478   | 234578  2348    2457    |+-------------------------+-------------------------+-------------------------+`

We know 9-9 must be a local swap in r28, and the only possible way is r2c9=r8c1=9.
Also, since 7-8 can't be a swap in r28, r28c5 can't be {78}, must be [36].
Since 3-6 is a global swapping pair, it must also be applied in b5.
The only possible way is r5c46={36} (naked pair @ r5,b5).

After 10 hidden singles and a box-line intersection of 8:

Code: Select all
`+----------------------+----------------------+----------------------+| 245    6      24578  | 12789  1789   178    | 2458   1248   3      || 24     247    1      | 278    3      5      | 6      248    9      || 3      9      258    | 1268   4      168    | 258    7      125    |+----------------------+----------------------+----------------------+| 1248   3      6      | 178    178    9      | 247    5      1247   || 7      124    24     | 36     5      36     | 9      124    8      || 18     5      9      | 4      178    2      | 3      6      17     |+----------------------+----------------------+----------------------+| 145    8      457    | 137    2      1347   | 457    9      6      || 9      247    3      | 5      6      478    | 1      248    247    || 6      1247   2457   | 1789   1789   1478   | 24578  3      2457   |+----------------------+----------------------+----------------------+`

We know the remaining 8 cells of r28 (r2c1248+r8c2689) must have 4 swapping pairs with the pairings [2-7|4-8] or [2-8|4-7].

Now r2c1 is from {24}, so r8c9 can't be from {24}, must be 7.
Now r8c2 is from {24}, so r2c8 can't be from {24}, must be 8.
Now r8c8 is from {24}, so r2c2 can't be from {24}, must be 7.
Thus r2c14=[42], and we can confirm the pairings are [2-8|4-7].
Hence r8c268=[284].

All naked singles from here.

udosuk

Posts: 2698
Joined: 17 July 2005

udosuk Spot on, identical walkthough to mine. Glad my clues made sense. I struggled to come up with a smaller set in vain. Hope you enjoyed it.
Incidentally the solvers without Merlin's vision are playing with 1,3 and 6 here at the moment.
Glyn

Posts: 357
Joined: 26 April 2007

Unreasonable optimism. I can be 95% sure (with all due respect to other fellow members here) that if I didn't reply the solution above this thread would be unvisited for weeks and then sink to invisibility. Your clues only "made sense" to a very exclusive group of people with the following attributes:

1. being enthusiastic about cracking riddles (e.g. participating in those riddle threads in the Off-topic sub-forum)

2. being competent in solving hard Sudoku puzzles

3. having a good understanding of the concept of automorphism (or symmetry) in Sudoku puzzles

4. having a reasonable high level of the English language

At the moment the only people I know of having all these 4 attributes are RW, you and me. eleven should have attributes 2-4 but I'm not too sure about attribute 1. RW, though, is probably not that actively participating these days.

Don't hold too much hope for your other riddle (Tungston Rod) though, IMO it's way too vague to raise much interest for anyone.

Glyn wrote:Hope you enjoyed it.

Considering it dragged me out from semi-retirement mode to come back posting something, very impressive enjoyment level.
udosuk

Posts: 2698
Joined: 17 July 2005

udosuk I realized the target audience was small on this one, the main worry was you not spotting it.
BTW Tungston Rod reminds of the that other rod problem that kept us amused elsewhere,
Glyn

Posts: 357
Joined: 26 April 2007

Glyn wrote:BTW Tungston Rod reminds of the that other rod problem that kept us amused elsewhere,

Is that supposed to be a hint? I guess I'll have to probe you a bit more later...
udosuk

Posts: 2698
Joined: 17 July 2005

No Matt the hint is in the appropriate thread.
Glyn

Posts: 357
Joined: 26 April 2007

Hi Glyn
Glyn wrote:Incidentally the solvers without Merlin's vision are playing with 1,3 and 6 at the moment.

You should say "players" except for Allan solver/method.

BTW, I tried to catch your point, This is another world. I was highly interested after Mauricio recent publication in "hardest puzzles" and I thought I would find something at the same level of difficulty.

I did not catch the first word in udosuk detailed explanation.

udosuk wrote:Your clues only "made sense" to a very exclusive group of people with the following attributes:

1. being enthusiastic about cracking riddles (e.g. participating in those riddle threads in the Off-topic sub-forum)

2. being competent in solving hard Sudoku puzzles

3. having a good understanding of the concept of automorphism (or symmetry) in Sudoku puzzles

4. having a reasonable high level of the English language

Sorry udosuk, I am missing most of the above qualities, but you did not include me in potential members of that esoteric club.
champagne
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