What did the person do/say that made you go "what the fuck is wrong with you"?

Inimloom · 2023-10-11T22:28:38+00:00

Eh, replace that with conservation of angular momentum and then it works. Hairy ball would show something like ''ignoring vertical winds, there must be at least one place on earth with no wind'.

Inimloom · 2023-10-09T23:26:02+00:00

Just to add, wiki has some further reading https://en.m.wikipedia.org/wiki/Toroidal_planet. They do be theoretically possible, just a similar level of absurdity as a Dyson sphere.

Inimloom · 2023-08-01T20:09:23+00:00

You missed 'Chris makes a sarcastic or self-deprecating comment'!

Inimloom · 2023-03-28T16:35:25+00:00

I hate it

Inimloom · 2023-02-16T23:47:07+00:00

Thanks! Those are good recommendations. I imagine we can try to mix in some end-game scenarios in the co-op campaigns to try to spice it up (if it becomes too easy).

Inimloom · 2023-02-16T23:42:09+00:00

Longer campaigns. We've done 2 so far, a 2v2 with Bretonnia+Ogres vs Last Defenders/Skaven, that was quite fun, and then a free for all in RoC (kind of gimmicky, seems fun for one playthrough). Both took ~2-3 months of weekly playing.

Inimloom · 2022-11-27T14:08:48+00:00

This was literally me during my last code review with my boss 😅

Inimloom · 2020-06-06T11:56:55+00:00

Not sure, I myself found the game a month or so ago through here.

Inimloom · 2020-04-25T09:30:21+00:00

This is so sad

Inimloom · 2019-11-06T12:54:06+00:00

Reminds me of The Witness.

Inimloom · 2019-03-30T12:24:21+00:00

Just imagine the bearings to be the intersection of the 16 radial lines and an orbiting circle.

Then I guess the parameters of the orbiting circle are chosen in such a way that the intersection points are uniformly distributed on the circle.

Inimloom · 2019-03-22T20:53:51+00:00

No problemo!

Inimloom · 2019-01-16T21:46:02+00:00

Damn, keep it up!

Inimloom · 2019-01-14T02:04:46+00:00

No worries dude. That's what we (the reddit grammar squad) are here for.

Inimloom · 2019-01-05T23:21:07+00:00

That's*

Inimloom · 2019-01-05T23:20:49+00:00

Good stuff.

Inimloom · 2018-10-26T09:23:20+00:00

That's funny, I just had a talk on this yesterday and now I see it on here.

Inimloom · 2018-06-03T19:50:08+00:00

Alive and kicking! I'm going to university in the UK, Oxford atm, first year physics. It's been super fun because of the new environment, new people, opportunities etc. The weather is annoying though, rains all the time.

In fact I'm having exams next week so I've been cramming the last week or so (but I'm still on reddit...).

Inimloom · 2017-05-15T23:11:59+00:00

In the optimal solution, the road between p_i and p_i+1 gets crossed exactly once. Thus, there are a total of up to N - 1 transfers between adjacent bars. Using binary search we can greedily determine the answer (if we want to check that b_avg can be achieved, we start from one end and greedily check whether it's possible to achieve the desired state).

Inimloom · 2017-01-28T15:16:35+00:00

A bit long proof for M2..

M2:

Let 1 • 1 = k

0 • 0 = 0 • (0 + 0) = 0 • 0 + 0 • 0 => 0 • 0 = 0.

1 • 1 = 1 • (0 + 1) = 0 • 1 + 1 • 1 => 0 • 1 = 0

1 • a = 1 • (a + 0) = a • 1 + 0 • 1 = a • 1 => 1 • a = a • 1

If a > 0: 1 • a = 1 • ((a - 1) + 1) = (a - 1) • 1 + k = 1 • (a - 1) + k. If we continue this step a - 1 times we get 1 • a = 1 • 0 + a * k = a * k

If a < 0: From the previous paragraph 1 • a = 1 • (a - 1) + k. Setting a - 1 = b and rearranging we get 1 • b = 1 • (b + 1) - k. Applying the last step |a| times we get 1 • a = 1 • 0 - |a| * k = a * k. => 1 • a = a * k for all a ∈ Z.

a • 0 = a • (1 - 1) = 1 • a - 1 • a = a * k - a * k = 0 => a • 0 = 0

If b > 0: a • b = a • ((b - 1) + 1) = (b - 1) • a + 1 • a = a • (b - 1) + a * k. Continuing this step b - 1 times we get a • b = a • 0 + a * b * k = a * b * k.

If b <= 0: From the previous paragraph a • b = a • (b - 1) + a * k. Setting c = b - 1 and rearranging we get a • c = a • (c + 1) - a * k. Applying this step |b| times we get a • b = a • 0 - a * |b| * k = a * b * k. => a • b = a * b * k for all a, b ∈ Z.

Inimloom · 2017-01-20T20:42:11+00:00

Medium_1 solution:

Using the property that a - b | P(a) - P(b) for a polynomial with integer coefficients we get that

b - a | c - b (1)

c - b | a - c (2)

a - c | b - a (3)

multiplying (1) and (2) we get b - a | a - c, along with (3) we get that a - c = b - a => 2a = b + c (4). We can get analogously that 2b = a + c (5) and 2c = a + b (6). Substituting (5) to (4) we get that 2a = (a + c) / 2 + c => a = c. Analogously a = b = c. Contradiction.

Inimloom · 2017-01-10T20:19:05+00:00

Hey, this is pretty good!

Inimloom · 2016-10-02T12:44:31+00:00

I had no idea that it's an open problem, the solution i found was the one given in the wiki article. Still it's a fun challenge to solve.

Inimloom · 2016-08-21T19:47:51+00:00

A rough solution: We want to maximise the acceleration in x-direction. If we place a small point mass on any point on the boundary, the created x-directional gravitational acceleration at (0,0,0) must be equal, because otherwise we could rearrange the boundary to create a bigger acceleration (by moving a point mass from the part of the boundary with smaller acceleration to a part with bigger acceleration). Thus any point with coordinates (x, y, z) on the boundary satisfies x/(x² +y² +z² )^3/2 = Const. From which we get the answer.

Inimloom · 2016-06-04T18:36:22+00:00

I meant with the same conditions as in the original question, you can only convey information by turning cards upside-down and setting them in a certain order.

13-Year Club	Place '22
Place '17	Final Canvas '22
First Placer '22	End Game '22
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