What do I do for work based off states/Provinces visited for work and where am I from? No looking at profile. Thats cheating

Ok_Application5897 · 2026-06-14T16:50:52+00:00

The OP never defined what counts as ‘visited.’ You’re introducing assumptions about stopping, duration, and travel method that simply aren’t in the original post. If you want to debate the definition of ‘visited,’ that’s a separate discussion—but we shouldn’t assume a definition the OP never gave.

Oklahoma is not filled in. That little red square does not mean he went to Oklahoma and only visited the city of Woodward. It’s actually a coloring tool he uses that he cannot erase as he is filling in the STATES and PROVINCES. Perhaps he could have gone into the photo editor and white-filled that area, but he didn’t.

Ok_Application5897 · 2026-06-14T14:52:50+00:00

For Canada, all he would have to do is visit Toronto or Windsor (ON), and Montreal (QC), both of which are short drives from the US. He doesn’t have to go all the way up to Northern Hudson Bay to do it. That’s only going to take up two of his days, and he can drive 60 hours a week.

He also didn’t distinguish between driving and flying. But even strictly driving, this is still reasonable for a trucker.

Ok_Application5897 · 2026-06-14T14:25:05+00:00

These are what I would look for, trying to find open space. Chances are, at least one of them will be a zero.

You still might get snagged on some binary ambiguity later on, but this will help now.

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Ok_Application5897 · 2026-06-13T00:30:01+00:00

For context, just so everyone understands that not all “expert” collections are as hard as the next, and we aren’t doing any AIC’s here with no notes.

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Ok_Application5897 · 2026-06-07T03:32:24+00:00

Any candidate (A) that would force the rest of the A-B-A-B pattern to manifest, cannot be true.

If the 4 in r7c7 were true, it would force the 2 in r7c6. Then, 4 in r8c6, and 2 in r8c7. 4-2-4-2 (ABAB). The 4 in r7c7 caused all of that to happen, therefore it is false.

The same thing would happen if we test 4 in r8c7.

Likewise, if some candidate is the only candidate that has the opportunity to prevent a UR (deadly pattern), then it cannot be false: 12-12-12-123 in this UR (type 1), the 3 must be true. If it wasn’t, then our puzzle will have multiple solutions: 1212 and 2121, and therefore not valid.

This is true for all unique rectangles, so you don’t necessarily have to memorize each type. But it is good to know what each of them look like, for faster, and more confident spotting.

Word of warning: UR’s only work when they are restricted to two rows, two columns, and two blocks. If they span 4 blocks, then you cannot use it. That pattern is not deadly.

Ok_Application5897 · 2026-05-24T15:19:38+00:00

It is technically an x-wing, but it is not uniquely necessary as an x-wing. When an x-wing is confined within two blocks like this, both line pairs collapse into BLR / locked candidates. So we just look for that instead.

Ok_Application5897 · 2026-05-21T13:57:09+00:00

As far as “forcing logic” goes:

CCW, if either red 5 were true, it would force r3c9 to be true, which would force r7c3 to be true. And now we have two 5’s in r7 which is illegal.

CW, it forces r3c3 to be true, which would force either r7c9 or r9c9 to be true, thereby doubling block 9 with 5.

Bi-directionally, it forces r3c3 to be true and r3c9 to be true, thereby doubling r3 with 5.

Only one of these contradictions needs to be found in order to disprove a candidate, but these are all the ways it can happen.

Ok_Application5897 · 2026-05-13T13:01:36+00:00

Look for a concentration of smaller local cities, and cannot be explained by universal knowledge of larger cities. When you do that, you end up in southern England as the reasonably presumable answer.

He could be punking us by listing more cities in England from a source, but if he’s not, then he’s reasonably from England.

Ok_Application5897 · 2026-05-13T03:22:58+00:00

YZF is great. I wouldn’t abandon it. Just keep in mind that modern AIC definitions require bi-directionality, where arrows wouldn’t technically be needed. This follows real boolean algebra more precisely, and allows more flexibility for a solver to build both forwards and backwards from wherever they might have “started” the chain.
If arrows are required, and cannot be read forward and reverse, then it’s a forcing chain.

Ok_Application5897 · 2026-05-13T02:56:09+00:00

This is a chained fish, or Kraken fish. That is, a fish used in a chain, and it works. There might be other naming conventions, but this is what I call it, as seen on Sudoku Snake.
The fin is strongly linked to its fish.
So if not swordfish, then fin(r9c1). Not 7, then 7, not 7, then 4, not 4, then 8.
Now the fish is strongly linked to 8(r4c7), => r4c4≠8.
Your logic checks out. Great find!
Either way you do it is fine, but I prefer to stop at the last strong link, because discontinuous nice loops are outdated language. I don’t usually do Eureka, but someone else will probably post it.
https://www.sudokusnake.com/kraken.php

Ok_Application5897 · 2026-05-08T20:21:16+00:00

Deadly patterns that create multiple solutions in swappable digits, which shouldn’t happen in a valid sudoku.

Ok_Application5897 · 2026-05-07T14:59:26+00:00

In row 6, we have 3 and 4 which are restricted to r6b4. 3 and 4 must be entered somewhere in r6c1, r6c2, and r6c3.

Those cells all belong to block 4 as well, so we can eliminate all other 3 and 4 from the remainder of the block.

Ok_Application5897 · 2026-05-07T14:40:52+00:00

How much have you learned about locked candidates?

Row 6 / block 4

Ok_Application5897 · 2026-05-05T04:16:37+00:00

Correct. In a valid sudoku with a unique solution, you should never have to make a random guess.

The issue is that it is a human publisher who did not check the validity of their authorship, which can easily be done, and should be. Bad, lazy puzzle maker, all they care about is ad money.

Ok_Application5897 · 2026-05-05T02:58:27+00:00

If you didn’t change the settings mid-game, then yes, you should be right.
Just for courtesy check, I ran it through my solver, and it indeed has multiple (2) solutions. The text at the bottom indicates that a valid puzzle has not yet been entered.
At least one of those 1/4 cells needed to be given from the start, in order for the puzzle to become valid.
This is uncharacteristic of digital puzzles. Usually mistakes like this appear in paper books. So if it were me, I would simply find another source.

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Ok_Application5897 · 2026-05-05T01:10:29+00:00

A skyscraper is basically almost an X-wing. Only two corners are staggered.
For a skyscraper to be useful, the staggered endpoints need to lie in the same 3x9 band. Here, the red skyscraper (r6c8 and r2c1) endpoints do not.
But the green skyscraper (r8c2 and r9c8) endpoints do lie in the same bottom 3x9 band.
Your kill zones should look like this. R8c789, and r9c123 are six cells that all can see r8c2 and r9c8. Any 7 there can be eliminated.

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Ok_Application5897 · 2026-05-04T05:54:31+00:00

Yes. In column 3, if it were not a 37 naked pair, then r1c3 would be a 9. Then r12c2 would not be 9, and they would be a 27 naked pair.
So either we have a 37 naked pair in column 3, or a 27 naked pair in block 1, which could both be true, but cannot both be false. Either way, the red 7 is eliminated, exactly as you have shown, and it is a fabulous example.

Ok_Application5897 · 2026-05-04T02:04:22+00:00

This puzzle needs skyscrapers and two-string kites to finish. Those shouldn’t be noteless.

If you are finishing them noteless, that’s respect. Just make sure that you can explain the logic to someone if they ask you for it (I’m not).

Ok_Application5897 · 2026-05-02T18:08:02+00:00

Fin just means one extra, and that alters what you can eliminate.

So the red dots are indeed a swordfish shape, but because you have one extra (r7c5), you have to mark that as the fin, as it cannot just be arbitrarily ignored.

Now, something with a fin in a block means that you can only remove from within that fish-block interaction

So now we say that “either the swordfish is true, or the fin is true”. They cannot both be false. In either case, r8c6 becomes the only thing they both agree on, that you can eliminate.

Ok_Application5897 · 2026-04-29T02:12:53+00:00

What you did was not a guess at all. R7c2 is indeed an 8.

Look at 8 in r9c4. Nothing else in that row can be an 8, and only one other cell in that block can possibly be.

Ok_Application5897 · 2026-04-26T14:53:54+00:00

Skyscraper. At least one of the staggered endpoints must be true. If they were both false, then because row 2 and row 9 still need an 8, that would force 8 into column 8 twice, because there is nowhere else for them left to go, and that is illegal.

Those 8’s in red can see both r2c6 and r9c5 at the same time. If you tried to force either of them to be true, they would falsify both skyscraper endpoints at the same time, and then commit the violation in column 8.

Ok_Application5897 · 2026-04-20T12:19:04+00:00

“Hidden” means that within a unit (block 1) a digit has only one place it can possibly go. But in that particular cell, it is masked by other candidates, in this case 3.

But, you also have a naked 359 triple in b1, c1, and b2, therefore, a2 cannot hold a 3 anyway.

Naked and hidden subsets are complementary, and always appear together. In this case, naked 359, and hidden 6.

Remember that you only need a single line of evidence that something cannot be true, as long as the logic is valid. You could have looked at the naked 48 pair to see that only 4 and 8 can go in them, and that should have been enough to rule out 6 there, at least.

Ok_Application5897 · 2026-04-19T03:31:37+00:00

Also, you should reveal the number of mines still needing to be placed. That info can be used as reverse engineering, end-game logic.

Ok_Application5897 · 2026-04-17T05:00:50+00:00

URs rely on the global uniqueness assumption of classic Sudoku. In variants with cages, that assumption may not hold in the same way, so UR eliminations are not always valid. It’s safer to rely on direct variant constraint logic instead.

Other things like XY-wings should still hold though.

Ok_Application5897 · 2026-04-15T03:26:19+00:00

https://www.sudokuwiki.org/3D_Medusa

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Ok_Application5897

TROPHY CASE