The New Sudoku Players' Forum

[denis_berthier]

.
7x7 Pandiagonal LS DB#1

This is my 1st try at generating manually a 7x7 Pandiagonal. My method was the same as for my previous 13x13: start from one of the (4) full grids and delete all the givens for each number except one.
Here I could delete all the givens except 6:

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3 1 . . 6 . .
4 . . . 2 . .
. 7 . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .

31..6..4...2...7.................................

There's no doubt it is minimal.
It is not totally trivial, but not very hard.

[999_Springs]

knowing that all 7x7 pandiagonal solution grids are cyclic makes the puzzle (and perhaps all 7x7 pandiagonals) very easy

in row 2, 4 is 4 cells left of 2 which means by the cyclic property that this holds for row 1 as well, so either r1c3=4 and r1c7=2, or r1c6=4 and r1c3=2. the latter case leaves no space for 7 in row 1 so cannot happen

row 1 is immediately filled, which means i can copy down rows 2 and 3 as cyclic shifts of row 1, and after finding one naked single in rows 5,4,6 the puzzle solves instantly

total time 3 minutes

phone screenshot of notes app with clock: Show

[denis_berthier]

knowing that all 7x7 pandiagonal solution grids are cyclic makes the puzzle (and perhaps all 7x7 pandiagonals) very easy

Yes, that's an easy way to solve it. You could say the same of all the 11x11 grids. Which leads to the conclusion that only 13x13s and above are "interesting". Unfortunately, they are much too large for being really interesting for many manual solvers.


All the 7x7s I've seen are quite easy. It'd be interesting to look for the hardest possible ones (not using cyclicity in the solution).*
This puzzle remains easy even if one doesn't know or want to use cyclicity (basically, one Naked-Pairs):

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Resolution state after Singles (there are no Singles):
3       1       2457    45      6       457     25
4       56      356     1357    2       1357    1567
1256    7       1245    3456    1345    1456    235
125     2345    1356    124567  457     23567   134567
157     345     1234567 1256    57      234567  123456
1567    3456    12457   23567   135     12456   23457
2567    2456    3567    123457  1345    1235    14567

As in many Pandiags, there are lots of whips[1] at the start (I usually skip this part in my Sudoku answers, but people are not yet used to whips[1] in Pandiags):

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whip[1]: c5n4{r7 .} ==> r7c2 ≠ 4
whip[1]: c2n4{r6 .} ==> r5c3 ≠ 4
whip[1]: c2n2{r7 .} ==> r7c6 ≠ 2, r4c6 ≠ 2
whip[1]: c6n2{r6 .} ==> r5c7 ≠ 2, r6c7 ≠ 2
whip[1]: a2n2{r7c1 .} ==> r3c1 ≠ 2, r5c3 ≠ 2, r7c4 ≠ 2
whip[1]: a6n2{r6c4 .} ==> r4c4 ≠ 2
whip[1]: c4n2{r6 .} ==> r6c3 ≠ 2
whip[1]: d1n7{r6c3 .} ==> r6c7 ≠ 7
whip[1]: d4n6{r5c7 .} ==> r5c6 ≠ 6
whip[1]: r5n6{c7 .} ==> r2c7 ≠ 6
whip[1]: r2n6{c3 .} ==> r6c6 ≠ 6
whip[1]: a1n6{r7c7 .} ==> r4c7 ≠ 6
whip[1]: c7n6{r7 .} ==> r6c1 ≠ 6, r7c2 ≠ 6
whip[1]: c2n6{r6 .} ==> r4c4 ≠ 6
whip[1]: r4n6{c6 .} ==> r7c3 ≠ 6
whip[1]: r7n6{c7 .} ==> r3c4 ≠ 6
whip[1]: c4n6{r6 .} ==> r5c3 ≠ 6
whip[1]: c5n7{r5 .} ==> r5c6 ≠ 7, r4c4 ≠ 7, r4c6 ≠ 7
whip[1]: r4n7{c7 .} ==> r2c7 ≠ 7
whip[1]: c7n7{r7 .} ==> r7c3 ≠ 7, r7c4 ≠ 7
whip[1]: r7n7{c7 .} ==> r6c1 ≠ 7
whip[1]: r6n7{c4 .} ==> r5c3 ≠ 7
whip[1]: d4n3{r5c7 .} ==> r5c3 ≠ 3
whip[1]: a4n1{r3c6 .} ==> r3c1 ≠ 1
whip[1]: c1n1{r6 .} ==> r5c7 ≠ 1

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Resolution state after Singles and whips[1]:
3     1     2457  45    6     457   25
4     56    356   1357  2     1357  15
56    7     1245  345   1345  1456  235
125   2345  1356  145   457   356   13457
157   345   15    1256  57    2345  3456
15    3456  1457  23567 135   1245  345
2567  25    35    1345  1345  135   14567

naked-pairs-in-an-anti-diagonal: a6{d1 d7}{n1 n5} ==> d4a6 ≠ 5, d4a6 ≠ 1, d2a6 ≠ 5, d5a6 ≠ 5, d3a6 ≠ 5, d6a6 ≠ 5
stte



(*) Up to now, the hardest 7x7 seems to be 1to9only's puzzle:

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. 1 . . . . .
4 . . . . . .
. . . 3 . . 5
. . . . . . .
. . . 6 . . .
. . . 2 . . .
. . . . . . .

It requires a Naked Triplet (or z-chains[3])