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[JoshuaFH]

I've gotten into Helldivers 2. It has these things called stratagems which are air strikes and stuff you can call down. To do that you need to input a combination of up, down, left, and right. There are something like 20 or 30 stratagems in the game, but it occurred to me that the devs have a serious developmental problem: there's only so many unique combinations you can have without any of them overlapping. The stratagems cannot overlap, or players won't be able to bring both at the same time. It seems like a math problem, but I have no clue how I'd even get started on it, and considering my post on the Helldivers 2 Reddit got instantly deleted for being a "trivial question" I'm not sure who else I'd bring the question to but here.

Not important, just really curious.

[bloop_bleep]

If you have no limit on how many inputs are required for a strategem you can make it so you never run out of combinations. If you have a limited number of inputs you can figure out the maximum number of unique combinations: it's 4 to the power of the number of inputs.

By the way, a correspondence between combinations of characters and items you want to convey such that no combination is a prefix of another is called a Huffman encoding. There is a way to make such an assignment based on how often you use each item to minimize the average number of characters used to describe items.

[Ulfarr]

The closest analog I can think are the codons (genetic code) where you have four "bases" A, G, C, U (for dna) and they are combined in groups of 3, resulting in 64 different combinations (4^3 = 64).

In a similar vein if the helldivers startegems use groups of four then there are already 4^4=256 different combinations. I guess that should be more than enough for the entire game.

[JoshuaFH]

If you have no limit on how many inputs are required for a strategem you can make it so you never run out of combinations. If you have a limited number of inputs you can figure out the maximum number of unique combinations: it's 4 to the power of the number of inputs.

By the way, a correspondence between combinations of characters and items you want to convey such that no combination is a prefix of another is called a Huffman encoding. There is a way to make such an assignment based on how often you use each item to minimize the average number of characters used to describe items.

There is a limit. (Sorry to get back to this so much later). For most stratagems, that's 6. And the minimum seems to be 3. I'm not sure how to go about this huffman encoding, but googling it, it seems to be a method of compression?

[da_nang]

Not really help, just a curiosity. At first I thought it was just a coincidence, but now I've nerd-sniped myself.

How many square-free integers a>1 exist such that the cube of a, and the squares nearest to that cube, are separated by the square roots of the other squares?

By that I mean, is there a natural number b such that a³ - b² = b+1 and (b+1)² - a³ = b, or alternatively, a³ - (b+1)² = b and b² - a³ = (b+1)? Let's call the first set of conditions H, and the alternative set of conditions G.

All I've been able to deduce is that H implies 4a³-3 must be a square number, and G is impossible since it implies b = (√(4a³+5)+1)/2 and b = (√(4a³+5)-3)/2, which results in 1/2 = -3/2.

Other than that, 7 is the only one of the square-free integers from the list on OIES that satisfies H (7³ - 18² = 19 and 19²-7³ = 18). In fact, I've checked all square-free integers below ten million with some Python code, and 7 is the only one of those that satisfies H.

Considering the problem and the necessary conditions are so simple, I'm surprised there are no other examples. Am I too blind to see something obvious?

[Magmacube_tr]

What is 6 x 3?

[zhijinghaofromchina]

根据乘法表口诀，三六一十八， so， I guess 3*6=18

[NJW2000]

By that I mean, is there a natural number b such that a³ - b² = b+1 and (b+1)² - a³ = b, or alternatively, a³ - (b+1)² = b and b² - a³ = (b+1)? Let's call the first set of conditions H, and the alternative set of conditions G.

Very interesting question. I'd like to see the python code too, mine was rubbish and chugged the laptop hard when checking a up to 10,000.

Had a look at this but haven't made much progress on the why of it. I'm not a number theorist though. Here's a few things:

Facts about squares tell us that the first part of H is enough, i.e. a^3 = b^2 + b + 1 means that a^3 + b is (b+1)^2.

Wolfram gave me 4 integer solutions for a^3 - b^2 - b - 1 = 0 , but the other 3 rely on negative or 0 a or b.

I think you can just reformulate the problem as: 4 a^3 = k^2 +3 for some integers a,k, but I don't know why that only has the 4 integer solutions.


You're right that G is impossible for positive integer a and b. There's a nice way of seeing why:

suppose G holds for some integer a, b

Then a^3 = b (b+1)

b and b+1 share no common factors... so must be cubes of integers themselves, as otherwise a would not be a cube.

The only integer cubes with a difference of 1 are -1, 0 and 1. So we obtain b=0 or b+1 = 0, so a=0 and b=-1 or b=0

Would love to see other people chime in on this, but I'd advise you to check if it's a case of an unsolved problem in number theory before getting too invested. There's a lot of those, and some can be stated very simply.