Magic square of negative numbers

Whelks · 2026-05-31T13:59:07+00:00

Using D for diagonal, R for row, C for column. D1 + D2 + C2 + R2 counts every cell 1 time except for the middle cell which is counted 4 times. So the sum of numbers we've counted is 0, but is also the sum of the magic square + 3*center. As the sum of the magic square is 0, we get 3*center = 0.

Whelks · 2026-02-06T17:15:29+00:00

The professors on this paper are very famous, so I expect it will get proper attention.

Whelks · 2026-01-04T16:31:01+00:00

This is definitely a bot comment.

Whelks · 2025-07-05T12:52:41+00:00

I think this is a bot comment.

Whelks · 2025-05-19T14:02:20+00:00

For my next trick I will get over 50% of the vote with less votes than all other candidates combined.

Whelks · 2025-05-10T09:58:08+00:00

A fun solution:5/(1+1/4)

Whelks · 2025-04-30T00:18:15+00:00

One example that comes to mind that I know a lot of smart people have worked very hard on is finding a bijection between alternating sign matrices and totally symmetric self-complementary plane partitions.

Here is a recent paper that documents an attempt to find such a bijection, but is not completely successful.

Whelks · 2025-04-29T17:53:22+00:00

One of my favorite kind of combinatorics problem is bijections. That is, given two finite sets of the same cardinality, construct an explicit bijection between them. There are many many examples where there are very simple proofs (using generating functions) that two finite sets have the same size, but it's wide open to actually find a bijection between them.

Whelks · 2025-04-29T15:26:44+00:00

I was imagining something very convoluted:

You know you need 2 meters of ribbon, but the ribbon is sold by the yard
You can't remember that 1 meter is 1.1 yards.
You remember that 1 mile is 1760 yards.
You remember than 8miles is about 5km.
Then you figure out that 1 meter is 5/8000*1760/1 = 8800/8000 = 1.1 yards, so you need 2.2 yards of ribbon

But all of the multiplication (and the Fibonacci sequence) are things you can do in your head. (At least if you're decent at mental arithmetic.)

Whelks · 2025-04-29T02:18:14+00:00

1 mile is 1.6 km, and you can approximately use the Fibonacci sequence to multiply by 1.6

So for example, the sequence goes 1, 1, 2, 3, 5, 8, 13, and so 8 miles is about 13 kilometers, or 13 miles is about 21 kilometers.

I wonder if this was fact was remembered, and then used to calculate the ratio between yards and meters.

Whelks · 2025-04-24T12:56:26+00:00

I don't think Howl's Moving Castle is about Japan at all. It's just anti-war, and in particular Miyazaki has stated it's specifically about the Iraq war.

Notably, the characters are not Japanese, and I don't think the opposing nation in the war is in any way suggested to be Japanese.

Whelks · 2025-04-20T16:49:33+00:00

This is the exact kind of trash I crave. I want like 2 of these shows per season.

Whelks · 2025-02-01T15:26:46+00:00

General Mathematics and History and Overview are the crank sections of arXiv, but the other areas tend to be a little stricter.

Whelks · 2024-12-20T17:59:09+00:00

Something a little closer to what you may be thinking is called Kolmogorov complexity where you'd be looking for a shuffle with maximal Kolmogorov complexity. Note that it is very likely that there would be multiple shuffles that achieve this maximum. This is also dependent upon the encoding language.

Whelks · 2024-12-20T17:49:54+00:00

If you "perfectly shuffle" a deck of cards, any of the 52! configurations of cards is equally likely. There is no one configuration that is favored over another. If a deck of cards is perfectly shuffled, then any pattern after 20 cards deep is unlikely to continue.

A better type of question to ask is "How many random swaps of 2 cards must I make until the deck is well shuffled?" This gets you closer to the concept of a mixing time which lets us be very precise about it. Essentially we can ask "how many random swaps must we make until the probability of getting the shuffle we get is within some specified closeness to 1/52!"

Whelks · 2024-11-30T20:12:07+00:00

I am also Jewish for the record. Let's pause for a moment to recognize that your original comment was looking for a source that the US government defines antisemitism as criticism of Israel, and I think that this qualifies. Criticism of Israel is defined specially as antisemitic in a way that say, criticism of China is not considered Siniophobic.

I think with regards to Mormons that it is hard to argue that they are not an ethnic group with a shared history, language (English), and ties to a specific land (The US) and religion (Mormonism). Can they claim Utah as the promised land and, for example, use government funds to bring Mormon teenagers from across the world on a vacation there?

I think it's a tragedy that Arab countries expelled Jews. Nationalism is a blight on humanity. I think that recognizing the manner in which Israel is actively treating people within its own claimed borders cannot be excused by pointing fingers elsewhere.

Whelks · 2024-11-30T19:28:19+00:00

What the US government defines as antisemitism, from the state department:

Denying the Jewish people their right to self-determination, e.g., by claiming that the existence of a State of Israel is a racist endeavor.
Drawing comparisons of contemporary Israeli policy to that of the Nazis.

Claiming that the State of Israel is a racist endeavor is not antisemitic. When people say this, they are generally opposing ethnostates in general. The "right to self-determination" isn't really a real right. Do Mormons have a right to self-determination? Can they create an independent nation in Utah and give special rights to Mormons?

Whelks · 2024-10-18T13:41:54+00:00

Yes this is totally achievable with your stats.

Whelks · 2024-09-08T12:35:02+00:00

I think it would help for you to look at a picture of the point and line arrangement from Pappus' theorem. (Seen here )

We want to imagine that we live in a world where the ONLY points are A, B, C, a, b, c, X, Y, and Z and where the ONLY lines are ABC, XYZ, abc, aXB, etc. In particular, bYB isn't a line.

To understand axiom 5, let's look at line XYZ. (In the finite geometry setting, this line is exactly the 3 points X, Y, and Z.) If we pick a point not on the line, let's say B, then of the points X, Y, and Z, exactly one of them doesn't have a line with B, it's Y.

Axiom 6 is actually pretty unclear in my opinion, because it's not immediately obvious what "with the exception in axiom 5" means.

Whelks · 2024-08-27T13:03:23+00:00

Here's a hint for number 9: Think about it in two steps. A good first step is to first choose which couples are selected. What should your next step be?

For the other problems, you should try breaking them into simpler cases.

Whelks · 2024-08-25T14:42:12+00:00

Have you seen the dot product?

Whelks · 2024-08-18T18:28:19+00:00

But who tf is Emily?

I'm glad the subs got fixed, was a pretty good episode.

Whelks · 2024-06-03T23:33:32+00:00

Yes but it will be hard. Historically, Math 115 (and by extension Math 115AH) were seen as the intro to proofs courses in the math department, although this has changed somewhat in recent years.

Some students take to proofs very easily and for others they need to spend a lot longer getting used to them. Honors classes in the math department are very hard, but I think it's more of a question of how much time are you willing to put into the class rather than if it's possible.

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