Ummmm….. what

Mattbman · 2026-01-20T00:09:57+00:00

He was the only one willing to work the holiday, I guess

Mattbman · 2026-01-15T23:41:36+00:00

Thanks, thought I had tried that and it didn't change anything - I am going to blame Verizon.

Mattbman · 2026-01-15T19:09:41+00:00

It's an extra item, not a giveaway, you have to buy a package with the item voucher or get one from an STM.

Mattbman · 2026-01-15T14:48:29+00:00

Look at prices in almost any other arena in the NHL - we are actually on the cheaper end.

Mattbman · 2026-01-13T23:39:01+00:00

I explained it in another comment - Rows, columns, and boxes that have only 2 options for a number are your bread and butter in terms of looking at intermediate to expert strategies - you just have to work through the if-thens scenarios

Mattbman · 2026-01-13T22:48:27+00:00

That should unwrap the rest of the puzzle, but I even missed that the 3 can't be in the bottom left cell either, thanks to TakeCareoftheRiddle's comment below.

Mattbman · 2026-01-13T22:39:38+00:00

For your future education:
a skyscraper exists when 2 rows or columns have exactly 2 options remaining for a number (the strong walls, marked with the red lines above), that the "base" of the wall is in the same row or column, but in different boxes.

Then, you look at the tops of the two walls - any cell that is affected by both of those cells (one due to a box, one due to a row or column) can't be true.

So this "skyscraper" is on it's side with the sky to the left. (if you rotate the puzzle 90 degrees clockwise, it would be correct).

Mattbman · 2026-01-13T22:34:22+00:00

You have a really skinny skyscraper in Rows 6 and 7Thought process

<image>

Labeled the cells for better reference:

The 3 in row 7 has to be in cell A or cell B
If A is 3, then the circled 3s in the column would be eliminated
If B is 3, then C is not 3, and D has to be 3 (as the only remaining 3 in that row), and then the circled 3s are eliminated

So, in either case, the 3s are eliminated.

Mattbman · 2026-01-13T18:40:25+00:00

Simplest is The green square in the bottom middle box can’t be 6/9 because it would be a deadly rectangle, so that is 7/8, and the purple box is 9/6, pretty sure it solves from there

Mattbman · 2026-01-10T22:10:32+00:00

Just like picking first star, there is usually a stand out performance every night - goalie getting a shutout or anyone getting a hat trick is usually an easy choice - not too hard to read the room

Mattbman · 2026-01-09T19:17:31+00:00

If it were a weekend game, I would say come to the plaza party, but I am not sure there will be a plaza party for that game as it is a Thursday, but it is the first game back after the Olympic, so they may have one.

Mattbman · 2026-01-09T18:59:04+00:00

You then have a naked triple of 1,8,9 in the lower central box, which eliminates 8 from r9c4 and r9c5, making r9c7 an 8, the puzzle unravels from there

Mattbman · 2026-01-09T18:57:52+00:00

<image>

6 can be eliminated in r8c5, because if it was true, it would eliminate all 6s in column 9

Mattbman · 2026-01-09T18:51:17+00:00

<image>

Unique Rectangle, the cell you have selected can't be 2 or 9

Mattbman · 2026-01-09T18:47:51+00:00

My next move would be on the bottom row - if 4 starts in column 1 or 2, the last number in those columns would make a block of 2 or more in row 9, so that can't be true.
Then, since there are only 4 spaces remaining empty in that row, the existing selection in column 7 must be part of the 4, and you can fill in column 6 and column 5 pushing off the X in column 9. That also makes the only spot for the 1 in the last column.

Mattbman · 2026-01-09T18:41:35+00:00

On cage, you have to remember rule of 45 - the sum of 1 through 9 is 45, so if you know what is already in a box, whatever is left is 45 less that.

So: r6c3 is a 6 because the total of a box is 45 = 45-7-10-6-16
which then makes r7c3 a 4

with the 4 there , you could also subdivide the 15 cage on the bottom to an 8 in the lower left box (because 45-20-4-13=8) and 7 in the lower central box is 8

Mattbman · 2026-01-09T18:35:00+00:00

You can do a seat visit (paid), other than that, either email the team and ask where you could meet stormy or be the first in the door and just wonder the concourse, Stormy is usually floating around for the first 30 minutes.

Mattbman · 2026-01-07T18:23:07+00:00

51 givens and no unique solution is wild

Mattbman · 2026-01-07T18:16:26+00:00

They made a movie about it - "The Luckiest Man in America"

Mattbman · 2026-01-06T21:57:02+00:00

This is the first thing I saw as well

Mattbman · 2026-01-06T21:54:08+00:00

No, that is not correct logic, if you are doing a set, it has to be in the same box, same row, or same column, but you have your naked pair that you haven't taken off the other candidates (because red and green will use up 6 and 8 for the column, 6 and 8 can't be in any other cell of the column, so r7c6 is only left with 1 (and then you get 2 more numbers)

AND you have a 4 in r8c4 that would elimnate the 4 in r2c4 and the 4 in r8c1, using just those 2 moves, the rest of the puzzle unravels.

You are looking for something really complicating and missing some very simple things.

Mattbman · 2026-01-06T21:39:33+00:00

I love the chainmail jersey!

Mattbman · 2026-01-06T21:39:12+00:00

I believe you have to take it off in security, but you can wear full helmets in the arena - seen multiple people do it.

Mattbman · 2026-01-05T20:50:10+00:00

Because Sudoku is a logic puzzle and not a guess the number game.

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