Why is vidya such a moron-oriented medium? I sincerely cannot think of a single game that would appeal to someone with an IQ above 90
Why is vidya such a moron-oriented medium? I sincerely cannot think of a single game that would appeal to someone with an IQ above 90
this is literally so simple, what the frick even is this question?
We would need to weigh each coin against each other individually to figure out which ones are fake, it is impossible to do it in 3 weighing session. What a moronic question.
What is your solution, Sherlock?
>What is your solution, Sherlock
Black coins are fake you fricking idiot, god this board is filled with fricking downy homosexuals
Weigh 3v3. If one side goes up, that side has 2 fake coins. Weigh 1v1 for those 3. If one side goes down, that's the real one and the other two are fake. If they're equal, those two are the fake ones.
If the sides in the 3v3 are equal, each side has 1 fake coin. Do one comparison for each 3 group to figure out which one is the fake (if two coins are equal, the third is the fake one).
>average Gankertard intelligence
Ahahaha fricking moron
yikes
You fricking dumb gorilla moron Black person, what if where is a fake on both sides? How the frick do you identity both fakes in just 2 weightings then?
you can't but labcoat girl gives you three weightings so it's fine
Well you've already used one
If you have a group of 3 coins where one is fake and lighter, you can determine it with one comparison. If one is lighter, than that one is the fake. If the two sides are equal, those two are real and the third one is fake.
>They provide only a binary value
>3 possible scale posiyions
lmao pseud
moron, it is either "one side is heavier" or "both sides are equal".
>one side is heavier
Which one?
I dunno, ask the scale
Can't you just keep splitting the groups in half?
Approximately in half*
peepee
Correct
labcoat girls are the best, there can be no other
Tru
Just weigh more than 3 times whats preventing you from doing that
Just uhh use your fricking hands and nerves idiot ahahahhh
>I sincerely cannot think of a single game that would appeal to someone with an IQ above 90
Super Mario 64
Separate the coins in two groups of 3, weight them against each other (1). That leads to two cases.
If they weight the same
>there's a fake on each side
>pick a side, pick 2 random coins and check them (2)
>if the weight's the same, the remaining coin is the fake
>if the weight's different, then the coin that has less weight of the two is the fake
>repeat for the other side (3)
If the weight's different
>that means the two fakes are on the lighter weight side
>pick 2 random coins from that side, weight them against each other (2)
>if they weight the same, you have both fakes
>if they weight different, the lighter coin is a fake, and the one you didn't pick is the other fake
That should be it, right?
Yeah, that seems good to me.
I tried a tournament-like approach, where you weigh them 1 to 1, but I don't think it works 100% of the time.
morons
Weeb site
Easy as shit. Weight 3vs3 to determine where are the fakes in the 2 halves
>If the weights of each are unequal (2 steps)
You now know that both fake coins are on the lighter half. Compare the weight of two of the coins to determine if they are the same weight or not. If so, they are the fakes. Otherwise, the heavier one is real and the remaining coins are fake
>If the weights are equal (3 steps)
You now know that the fake coins are split between halves. Compare the weight of two of the coins on one half to determine if they are the same weight or not. If so, they are real. Otherwise, the lighter one is fake and the remaining coins from that half are real. Repeat for the other half
Since op was a gay and this challenge was so easy, I'll suggest a harder riddle for Ganker to solve with the same premise.
You have 25 coins, of which one coin is fake and is lighter than the others.
How many times do you need to use the scale minimum to identify the fake coin?
24 times because brute force trumps all.
Simple enough.
Exclude one coin and do a 12v12
If the scale is equal, the extra coin is the fake
If one side tips up, that group has the fake
Proceed to a 6v6, then 3v3 to narrow it down
Once you have the group of 3 one more comparison will reveal the fake
4 comparisons at most.
Close, but wrong
The most information you can get out of a single measure in this case is by making 3 groups.
2 groups being measured equal in size and group that's unmeasured.
You want to make the group sizes of the 2 being measured equal and the third group as similar as possible.
The process is to then measure the 2 equal groups against each other, if one weighs less it has the fake coin else the 3rd group has the fake coin.
Then repeat this process for the group with the fake coin.
For 25 this means splitting into 8,8,9.
Then either 3,3,2 or 3,3,3
With groups of size 3 or less being discernable within a single comparison.
So minimum 3 comparisons are needed.
That's correct, 3 times is the minimum
>I sincerely cannot think of a single game that would appeal to someone with an IQ above 90
But people with higher iqs love video games?
He forgot the stereotype of nerds include video gaming as a hobby.
Source of people with higher IQs liking videogames? It's literally a medium enjoyed by literal morons
Put three coins on one side, three coins on the other side.
If one side is lighter than the other, then you know both fake coins are on the lighter side.
>if one side is lighter
Weight one of those three coins to another. If they weigh the same, you've found your two fake coins. If one weighs more than the other, the heavier one is real and the other two are fake.
Done in two measurements.
>if both sides are equal weight
There is one fake coin on each side.
Remove one coin from each side, weight again.
If the sides are now different weights, then the coin you removed from the (now heavier) side is one fake. Weigh the two coins on the lighter side. The lighter of those two coins is the other fake. Done in three measurements.
If both sides are the same weight after removing two coins, then take two coins from one side and weight them together. If they are both the same weight, then all four coins that were on the scale are real and the two first removed are fake. If there is a difference in weight, then the lighter one is the fake. Weigh the lighter one with one of the other two that were still on the scale. If the same weight, you have your two fakes. If not, then the one you didn't weigh is the other fake. Done in three measurements.
I didn't even need to look anything up. Just work the situation out and figure out a minor trick to get a solution in just three measurements for the most extreme example.