this post was submitted on 30 Nov 2023
2439 points (99.6% liked)

Programmer Humor

19623 readers
92 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@sh.itjust.works 33 points 1 year ago (3 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.

[–] skydivekingair@lemmy.world 7 points 1 year ago* (last edited 1 year 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

[–] drislands@lemmy.world 4 points 1 year ago* (last edited 1 year 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```
[–] skydivekingair@lemmy.world 3 points 1 year ago

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

[–] Mr_Dr_Oink@lemmy.world 2 points 1 year ago* (last edited 1 year 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.

[–] RoyaltyInTraining@lemmy.world 7 points 1 year ago (1 children)

Where is the piped bot when you need it

[–] Venat0r@lemmy.world 3 points 1 year ago* (last edited 1 year ago) (1 children)

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

Https://Piped.video/Mgqqzt6Iah4

[–] Mr_Dr_Oink@lemmy.world 2 points 1 year 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.

[–] adrian783@lemmy.world 0 points 1 year 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.

[–] ChairmanMeow@programming.dev 3 points 1 year ago (1 children)

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

[–] NotSoCoolWhip@lemmy.world -1 points 1 year ago

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