this post was submitted on 30 Nov 2023
2489 points (99.6% liked)

Programmer Humor

19423 readers
233 users here now

Welcome to Programmer Humor!

This is a place where you can post jokes, memes, humor, etc. related to programming!

For sharing awful code theres also Programming Horror.

Rules

founded 1 year ago
MODERATORS
 
you are viewing a single comment's thread
view the rest of the comments
[–] lingh0e 33 points 11 months ago (4 children)

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.

[–] [email protected] 7 points 11 months ago* (last edited 11 months ago) (1 children)

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

[–] [email protected] 4 points 11 months ago* (last edited 11 months ago) (2 children)

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...


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```
[–] [email protected] 3 points 11 months ago

Cool, thanks. I’m not the best at formatting when using my phone.

[–] [email protected] 2 points 11 months ago* (last edited 11 months ago)

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.

[–] [email protected] 7 points 11 months ago (1 children)

Where is the piped bot when you need it

[–] [email protected] 3 points 11 months ago* (last edited 11 months ago) (1 children)

You can just replace the domain of the url with piped.video:

Https://Piped.video/Mgqqzt6Iah4

[–] [email protected] 2 points 11 months ago (2 children)

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.

[–] [email protected] 0 points 11 months ago (1 children)

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.

[–] [email protected] 3 points 11 months ago (1 children)

You don't know if the person is lighter or heavier yet.

[–] [email protected] 1 points 11 months ago* (last edited 11 months ago) (1 children)

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

[–] [email protected] 2 points 11 months ago (1 children)

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.

[–] [email protected] 0 points 11 months ago (2 children)

The question is to figure out who is different, not how they are different. That takes one more step, half the time.

[–] [email protected] 2 points 11 months ago

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.

[–] [email protected] 0 points 11 months ago (1 children)

The question was to find who doesnt weigh the same and if its heavier or lighter. Watch the clip again.

[–] [email protected] 1 points 11 months ago (1 children)

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.

[–] [email protected] 1 points 11 months ago (1 children)

Thats 4 uses of the seesaw. It has to be 3.

[–] [email protected] 1 points 11 months ago (1 children)
[–] [email protected] 2 points 11 months ago* (last edited 11 months ago) (1 children)

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

[–] [email protected] 1 points 11 months ago* (last edited 11 months ago) (1 children)

Well I meant to write 3 groups of four. Same general thought just adjust the logic somewhat

[–] [email protected] 1 points 11 months ago* (last edited 11 months ago)

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.

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
[–] [email protected] -1 points 11 months ago

If 1234 and 5678 don't weigh the same youd need 4 seesaws in some cases

[–] [email protected] 1 points 11 months ago (1 children)

Here is an alternative Piped link(s):

the see-saw riddle on Brooklyn 99.

Piped is a privacy-respecting open-source alternative frontend to YouTube.

I'm open-source; check me out at GitHub.

[–] lingh0e 1 points 11 months ago