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Programmer tries to explain binary search to the police
(startrek.website)
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this post was submitted on 30 Nov 2023
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A police officer being unable to think in such a fashion is exactly why no one could solve the see-saw riddle on Brooklyn 99.
Where is the piped bot when you need it
You can just replace the domain of the url with
piped.video:Https://Piped.video/Mgqqzt6Iah4
For those looking for the handout:
person: A B C D E F G H I J K L
round 1: L L L L R R R R — — — -
round 2: L L R R R — — — L R L -
round 3: L R R — — L R — L L — R
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...
Cool, thanks. I’m not the best at formatting when using my phone.
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.
you can do it like you weight 6v6 then 3v3 then for the last weighing you weight the 2 out of 3.
or you weigh 4v4 to find out which grouping of 4 the light weight person is in, then do 2v2 and 1v1.
You don't know if the person is lighter or heavier yet.
I mean that not knowing it is part of the question, and the proposed solution doesn't work without knowing if the person is heavier or lighter.
If you know if the person is heavier or lighter, the question becomes trivial.
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.
The question was to find who doesnt weigh the same and if its heavier or lighter. Watch the clip again.
Thats 4 uses of the seesaw. It has to be 3.
Im sorry when i read weigh two of them i counted it as two separate weighings of two sets of groups. My bad.
What about the 4th group? There are 12 people
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.
If 1234 and 5678 don't weigh the same youd need 4 seesaws in some cases