Puzzle Description:

This is the symbol for the Guild. This means the place is as safe as the Flagon's cistern. If you see this shadowmark, someone from the Guild is nearby for certain.

#1: CB Paul (cbpaul) on Jul 31, 2021 [SPOILER]

Which Guild?

what game?

The Thieve's Guild in Elder Scrolls V: Skyrim is headquartered in an underground bar called the Ragged Flagon.
There are marks like these scratched onto various shops, buildings, and houses all over the country by thieves for other thieves so you know which are worth sneaking into, or dangerous to sneak into.
This is the mark for the guild, but others tip off thieves guild members to other things that I expect we'll see if this turns out to be a series

I love me some Skyrim! Right now I'm playing its precursor, Elder Scrolls IV: Oblivion, but Skyrim is great. I cannot wait for Elder Scrolls VI: Hammerfell but that is years away, since Bethesda is releasing Starfield this year and I hear that Fallout:London is on the horizon as well.

I haven't played Skyrim in a coon's age. I didn't remember this symbol at all. :)

You'll tend not to notice it if you're not in the thieves guild, and actively looking for places to rob.
Playing as a mage or fighter, you'd never know that these are all over the place, as they are very subtle if you're not actually looking for them.

I can see that. (or, can I?)

I relied on the symmetry for this one. Not sure if that counts as lookahead or not. I deduced that the 1s in R1 and R25 must be in the middle, and that helped get the rest moving along.

Perfect symmetry on both axes makes it clear where things need to be for a single number.

Found to be solvable with deep lookahead by infrapinklizzard.

We do not count symmetry as logical on this site.

So far as we know, symmetry logic probably holds if ALL the clues are symmetrical and there is only one solution. In that case, using symmetry is considered "cheating" only because there is no actual proof that it is true, just empirical evidence.

However, whether the puzzle has only one solution is not specified as part of the clues, and so is not really part of the puzzle.

Note that you cannot use symmetry to outright place pixels, but you can use it to find places to use logic.

Symmetrical puzzles are often much harder to solve using pure logic.



This is my solve, which shows that it can be solved with deep lookahead. It is not necessarily the best way; it is merely the way I did it. I welcome people to post a specific solve if they find a better way.


After LL gets two black pixels,
Internal EL 11 c4:
If it were to go in r1-11, then:
> the left 4 in r12-16 would be forced to occupy c7, making a block of five.
= That conflict in c7 means that c4r1 must be white.

Now you can use symmetry to find that the same *logic* applies to the 11 in c4 when it is all the way down, and also to the 11 in c12 when it is at the extremes of that column as well. So c4r25, c12r1, and c12r20 must also be white.

Again, internal EL 11 c4:
If it were to go in r2-12, then:
> the left 4 in r13-16 would be forced to occupy c8, making a block of five.
= That conflict in c8 means that c4r2 must be white.

Again, symmetry lets you look for places to apply the same logic, leading to c4r24, c12r2, & c12r24 being white.

LL

Once more, IEL on the 11 in c4 (but in a different direction):
If it were to start in r3, then:
> it would force c1 and c7-15 to be white in that row
> which would force all the rows above those to be white because of the crossing clues
= which would not leave room for the 3 in r2, So c4r3 must be white

Again, symmetry lets you look for places to apply the same logic, leading to c4r23, c12r3, & c12r23 being white.

LL

IEL 9 c3:
If it were in r1-9, then :
> It would trigger the 3 in r2
> which would trigger the 3 and 5 in c1&2
= creating a block of 3 in r4 (a conflict) so c3r1 must be white

Again, symmetry lets you look for places to apply the same logic, leading to c3r25, c13r1, & c13r25 being white.

Some unusual two-way logic on the 11 in c4:
whether it goes up into r11, or down into r15 (and obviously if it does both) it will cross more than 3 horizontal 4 clues. The maximum clue size of 3 in c5 will force at least one of those 4s into c3, thus making both r2 and r24 out of reach, and thus white.

The same on c13.

LL

EL 3 c1 = r3, r23 w
(same on c15)
LL

Look at those seven rows filled with 4s.
if the top of the 3 in r1 is anywhere in r4-9, it will do two things:
> it will force the 9 in c3 to cross all seven rows and
> it will force c1 in at least four of the rows to be white.
= therefore c5 will have a block of four or more creating a conflict. So c1r4-9 must be white

Again, symmetry lets you look for places to apply the same logic, leading to c1r17-22, c15r4-9, & c15r17-22 being white
LL

Same with the 5 in c2; If the top of the 5 is anywhere from r4 to r8
> it will force the 11 in c4 to cross all seven rows and
> it will force c2 in at least four of the rows to be white.
= therefore c5 will have a block of four or more creating a conflict. So c1r4-8 must be white

Again, symmetry lets you look for places to apply the same logic, leading to c2r18-22, c14r4-8, & c14r18-22 being white
LL

Same again with the 3 in c1 again; If it were in r10-12,
> it will force the 9 to cross four rows of 4s
> it will make c1 of those 4 rows white
= therefore c5 will have a block of four creating a conflict. So c1r10 must be white

Again, symmetry lets you look for places to apply the same logic, leading to c1r16, c15r10, & c15r16 being white
LL

If the 9 in c3 were in r6 or above, then
> both c3&4 in r17 will be white causing a conflict (there will be no room for the left 2 in that row)
= So c3r5 must be white

Again, symmetry lets you look for places to apply the same logic, leading to c3r20-21, c13r5-6, & c13r20-21 being white
LL

If the 11 in c4 were all the way up, it would make the rest of it's column white which would:
> make r15c2-3 white
> make r16c2-3 black
= making a conflict in both c2&c3 (more than one block). So c4r5 must be white

Again, symmetry lets you look for places to apply the same logic, leading to c4r21, c12r5, & c12r21 being white
LL

Deep lookahed (not that some of the previous might not be, but this definitely is):
If the 5 in c2 were in r9-13, then
> it would trigger the 9 in c3, which would make c3r18-19 white
> since the rest of c2 would be white, the 4s in r14-15 would extend into c5
.> which would make c5r17-25 white (since the bottom 3 of c5 would be fixed)
= Which would leave no room for the left 2 in r18. Therefore c2r9 must be white

Again, symmetry lets you look for places to apply the same logic, leading to c2r17 being white. However, you don't need to look at the other side, because LL will finish it from here.

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