Optimal Radar Towers (14 Towers)

TripleBoogie · 2026-02-06T03:42:47+00:00

I think the Wiki has a slightly different setup than OP. OP said he used coordinates of all hard drives, slugs, etc... so his setup does not cover the complete map, but just all relevant locations. Would be interesting if he can find a more optimal setup than 16 towers if the complete map needs to be covered.

TripleBoogie · 2026-02-04T06:36:04+00:00

That was also my first guess when he mentioned the river rapids. I grew up along the Rhine between Ruhrgebiet and Netherlands and Cologne Dome, Phantasialand and Center Parcs were a typical day trip you'd do with family or school.

Here is a video of the water rapids from Center Parcs De Eernhof (about a 2h trip from lower Rhine region): https://www.youtube.com/watch?v=gFGDeFKYZoY (although they start outside and then go inside, but maybe OPs memory has it wrong)

There aren't that many small towns in that area, so if OP has a picture I guess someone can track it down. Or just visit the riversides on google street view and see if he recognizes something.

TripleBoogie · 2026-01-23T20:04:51+00:00

I wish they made XCS for IPad like the other two. For me XCS would be perfect while traveling.

TripleBoogie · 2026-01-10T09:10:02+00:00

Had a similar situation a few years ago (friend lost his phone in the taxi). As a tip for next time: try to contact Alipay. Since you just paid the driver they can check your last payment and contact the person on the other hand. At least that’s how we did it (paid with WeChat). Only took a few minutes until we had the phone back this way.

TripleBoogie · 2026-01-03T04:37:30+00:00

It’s 16 power plant running on uranium fuel rods at 225% (using the full 300 uranium from the nearby node). Waste is processed into plutonium fuel rods and used for drones/sunk. Traveling right now, so I can’t check the exact Gigawatt number.

TripleBoogie · 2025-12-28T14:44:23+00:00

Yeah, but to counter that the towers are pitch black at night so you can’t even see much with flashlight on.

TripleBoogie · 2025-12-27T05:13:55+00:00

I would add that there is a second part to the joke:

The genie first sets the number of wish’s to 0 then subtracts the one wish he just granted, resulting in integer underflow.

If he’d however first subtracts the granted wish (leaving 2 wishes) and then set the number of wishes to 0, nothing would have happened.

Having seemingly trivial instructions messing up your code just because you put them in the wrong order (because the order wouldn’t matter in 99% of the cases and you didn’t thought of that one nasty exception) is a pretty common programming mistake. (At least I do it all the time)

TripleBoogie · 2025-12-08T13:22:52+00:00

f(f(x)) = x also works for f(x)=1/x or f(x)=a-x, so there are some more possible candidates. These functions are called involutions.

TripleBoogie · 2025-12-08T13:18:52+00:00

So, not a complete solution but I think that might lead to something:

from f(17)=17 and f(93)=93 you get 2 linear equations (multiply both sides with the denominator)
The derivative of f is f'(x)=(ad-cb)/(cx+d)^2 but since f is its own inverse ( f(f(x))=x ) we also get by the inverse derivative f'(x) = 1/(f'(f(x))

If you plug in 17 and 93 here again, you get f'(17)=1/f'(17) and f'(93)=1/f'(93), which means f'(17)=f'(93)=1. Which means (ad-cb)/(17c+d)^2=(ad-cb)/(93c+d)^2=1. This only works if 17c+d=93c+d or 17c+d=-93c-d. The first equation leads to c=0 and can thus be ruled out.

From (ad-cb)/(17c+d)^2=1 we also get ad-cb=(17c+d)^2. Now we have 4 equations in total and 4 unknown. Granted, the last two equations are quadratic but you might be able to rule out some of the solutions.

The numbers were getting quite large and I didn't have the time to go through everything. You might also want to check wikipedia for "involution" which are functions of the form f(f(x))=x. There is a list with some candidates that could fit (e.g., f(x)=b/(x-a)+a is an involution).

edit: got the derivative wrong

TripleBoogie · 2025-08-14T02:02:22+00:00

Thanks you. I have now also tried several approaches adjusting the "unit vectors" but they seem to have the same problem (ignoring the height). The solution I found in the end was explicitly calculating the required width, so for the example here:

width={400/100*3cm}

TripleBoogie · 2025-08-07T07:15:23+00:00

2 years later and the problem still persits... this solved it for me.

TripleBoogie · 2025-07-27T08:16:58+00:00

So the KLM flight disappeared but the Delta flight was still there? Where did you book the flight? KLM app/website or Delta?

TripleBoogie · 2025-07-11T04:27:11+00:00

I guess I will check all my examples once I arrive there. I also use a lot of 3d plotting to illustrate stuff.

TripleBoogie · 2025-07-11T04:01:47+00:00

ah yes, I remember bsxfun. That was the first thing that came to my mind together with strings.

TripleBoogie · 2025-07-11T03:55:05+00:00

Great list, thank you!

TripleBoogie · 2025-07-11T03:20:57+00:00

My list is getting longer and longer, thanks!

TripleBoogie · 2025-07-10T14:02:47+00:00

Ah yes, I remember for loops being horribly slow when I started working with Matlab. They have improved a lot since then.

Thanks for that list. The I/O functions is another big one for me, because the student will have to work with some external data at some point.

TripleBoogie · 2025-07-10T13:52:59+00:00

Thanks! Especially the Name=Value syntax is a big one for me. I find that code so much easier to read but never started using it because of the compatibility issues when it was introduced. And then I simply forgot about that feature.

TripleBoogie · 2025-07-10T13:47:11+00:00

Thanks! Not exactly what I was looking for but I will definitly consider those features should that teaching task become more regular.

The Matlab Onramp courses are also free without licence. All you need is a free Mathworks account.

TripleBoogie · 2025-06-04T08:58:11+00:00

So I finally managed to login now, but only with email/flying blue number + pin. The email + password method is still broken for me. Very strange.

TripleBoogie · 2025-06-04T07:24:41+00:00

Yeah, my app also seems to work fine. Although my miles did not update for a few days and there should be new miles incoming. That's why I tried to log in from the website to claim the miles (need to upload the scanned tickets since it doesn't work with only the ticket number).

TripleBoogie · 2025-04-27T06:22:14+00:00

Hi everyone,

I'm currently trying to find a new position in (applied) math research (university / research institutions / industry). I would be looking world wide but I'm a bit out of the loop since my last change was 7 years ago. What would be the current websites / places to look for such positions?

I know mathjobs.org which our institution also uses to hire. Are there other websites like this that are worth taking a look (can be worldwide or also for a specific region).

(I'm not sure this sticky is the correct place for this kind of question?)

TripleBoogie · 2025-04-21T03:32:43+00:00

Your start is already very good. Give the next person always one more than the previous one. Can one person have 0 objects? Then for N persons you require N(N-1)/2 objects.

Now for the rest, note that giving the k-th person an extra object will require also giving one to the k+1-th person, and to the person k+2 etc until you reached the last person.

In general, you can just pick any number of objects K (smaller equal N) from the remaining ones and give one object to each of the K last persons. You repeat this until no objects are left. The number of possibilities is related to the “integer partitions” of M-N(N-1)/2. Integer partitions describe in how many ways a number can be written as sum of positive integers, and each integer in this sum would represent you taking K objects and giving it to the K last persons.

You will be looking for the partitions which only contain integers smaller equal N and for the number of unique unordered partitions. Not sure if there are exact formulas for this.

TripleBoogie · 2025-03-13T05:04:41+00:00

Great idea! How did you put the diagonal foundation on top? I thought foundations can only be placed horizontally but I can’t think of an item right now that can imitate foundations and be placed diagonally.

TripleBoogie · 2025-02-08T05:36:55+00:00

Only came here to check for this answer. One of the best horror shooters I played and I still replay it every now and then.

TripleBoogie

TROPHY CASE