I mean, in the era of VHS this won't work because ultimately you're fast forwarding and rewinding. So you're gonna watch it anyway. but in the digital era I thought this would be what any Police officer did?
Like... they're not even gonna spend 10 minutes on a theft?
Afterwards I found a chatroom thread among Cambridge computer scientists, one of whom had also been told that unless he could pin down the moment of theft no one would look at the footage. He said he had tried to explain sorting algorithms to police - he was a computer scientist, after all. You don’t watch the whole thing, he said. You use a binary search. You fast forward to halfway, see if the bike is there and, if it is, zoom to three quarters of the way through. But if it wasn’t there at the halfway mark, you rewind to a quarter of the way through. It’s very quick. In fact, he had pointed out, if the CCTV footage stretched back to the dawn of humanity it would probably have only taken an hour to find the moment of theft. This argument didn’t go down well.
It’s got huge amounts of applicability in many lifestyles and situations that most people never realize until the moment arrives. I once played a fun game that had you guess a number between 1 and 1 Billion with them telling you higher or lower to earn your freedom. Takes a couple of minutes at most.
The final project in my instrumentation class was to tune a PID controller for a hot/cold mixing valve. I (CS/ENG) was paired up with an engineering student and a lot of it was throwing parameters in, seeing if weird shit happened, and then turning down or up based on the result. I had a programming final and something else I was supposed to be studying for, so I just started doing a binary search with the knobs. We got the thing tuned relatively fast and my partner acted like I was a wizard.
How do you do a binary search for an open-end scale (are PID params open-end?) and three knobs at the same time when they interdepend in their influence? I need to know since i have a PID tuning on my personal projects plate
It’s been ages, but we’d done rough calculations for the three controls so we roughly knew what we needed. Our teacher was big on manually tuning instead of just using formulas since he thought just running numbers “lacked artfulness.”
So we grabbed a point and started searching around manually. I think we were just tuning the derivative portion at that point, trying to get a fast response without the system without it going chaotic and noisy.
I once had a friend who was robbed of all kinds of stuff including a PS3, and that the guy was signed into his Netflix changing account profiles the very same day. I told him he can just get a tracking number by calling Playstation and that the active police officer can use it to track them. Thing is, the officer ghosted him for like 8 months despite having everything they needed to immediately find the exact location of the perpetrator actively using the stolen property.
They don’t care really. As has been my experience anyway.
I once had my car window smashed, a mix of gear taken…some was expensive, some was personal to me. I felt violated. Called the police, explained, gave S/Ns to what I could, told them exactly who did it. He didn’t give a shit. Actually made me feel like I was wasting his time. I think Seinfeld covered this…
“We’ll let you know if we find anything” “Do you ever find anything?” “No”
But oh, my reg is out of date and the plate scanner picked it up? Boom, they really kick it into gear. So that’s $130… i could just go take care of the tags immediately with a friendly warning but now don’t even want to. And in the end I end up pretty fucked.
If only they put that effort into other things I just might have gotten my linear power amps back. Props to anyone who knows that product.
I’m sure it didn’t go well. If it was somehow framed in a sycophantic way where the police were led to believe it was their idea, I’m sure it would have gone better. Wait that might not be too difficult to do.
Na. If it’s British police it’s just an excuse. All they’re there for after all these years of Tory cuts is to give you a reference number so you can make an insurance claim.
This would be easier to parse with a monospaced font. I’m not sure how that works in lemmy so this might take an edit or two…
<span style="color:#323232;">
</span><span style="color:#323232;">round 1: L L L L R R R R — — — -
</span><span style="color:#323232;">
</span><span style="color:#323232;">round 2: L L R R R — — — L R L -
</span><span style="color:#323232;">
</span><span style="color:#323232;">round 3: L R R — — L R — L L — R```
</span>
Oh i get it. So if in round 1 it tilted down on the right. Round 2 it was even then round 3 it tilted down on the right then it was person G and they are heavier. However if it was reversed and tilted on the left then even then left then it was still person G but they are lighter. Because that pattern only occurs once. This is brilliant. Thankyou to you and the person you corrected the formatting of.
How do you solve that? I saw a solution in the comments where it says to start with numbering all the people and butting 1234 and 5678 on the see saw, then it says if they weight the same then continue and that seems to work. But if they dont weigh the same it doesnt work and it doesnt say what to do in that case.
That’s not the question. Either the scales balance, and the third is heavier or lighter, or the scales don’t balance and you get both answers, but the question is purposely framed this way
That’s easy enough to answer, but he really should work on his grammar. In that case you just do 3 groups of three, weigh two of them. If they’re even, the third group is different. Weigh 2 membres of the third group, they’ll either be even or one heavier. Weight the last member against the heavier one from step 2 to see if they’re even or not for your answer.
I’ve had a look into it, and it doesn’t work if you try to do it mathmatically. You always need more than 3 gos on the seesaw.
There is a solution in the replies to my original comment that is the actual solution, and it works every time and is much simpler than any grouping method.
It involves assigning a letter to each person and then aligning that with a grid of positions “left” or “right” or “none” on the seesaw. Over the three rounds. So, person A is on the right all three rounds person b is on the right for 2 rounds then on the left for the 3rd round.
You end up with a list of 12 patterns that do not repeat or mirror any other pattern like “LLL” “LLR” “LRR” “LR-” etc. Then you do all three rounds and compare the position the seesaw was in with those patterns.
If the seesaw was down on the left 2 times the down on the right the third time then you look for which person had that pattern in this case it was person B. So they are the one with a different weight and they were heavier.
Equally, if the opposite pattern occurred. It was down on the right 2 times, then down on the left for round, then that is the opposite pattern of person B and does not occur anywhere else, so it was person B, and they were lighter.
<span style="color:#323232;">person: A B C D E F G H I J K L
</span><span style="color:#323232;">
</span><span style="color:#323232;">round 1: L L L L R R R R — — — -
</span><span style="color:#323232;">
</span><span style="color:#323232;">round 2: L L R R R — — — L R L -
</span><span style="color:#323232;">
</span><span style="color:#323232;">round 3: L R R — — L R — L L — R
</span>
Yes, I’m aware. But with 12 people you can’t simply divvy the groups in threes constantly, because if you weigh and the groups are unequal, then you don’t know in which group the different person is (yet). E.g., weighing ABCD - EFGH can tell you the different person is in IJKL if the groups are even, but if they’re uneven you don’t know in which of the other two groups the different person is.
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