Brain Teaser

[Anonymous No.9094464]

Let's see how many people on /biz/ can solve this problem

[Anonymous No.9094471]

Fuck off, this is not school

[reluctant kneepad girl No.9094478]

>>9094464
10

[Anonymous No.9094480]

The answer is sage

[Anonymous No.9094483]

>>9094464
1

[Anonymous No.9094490]

>>9094464
= 1

[Anonymous No.9094499]

>inb4 200+ replies

[Anonymous No.9094498]

>>9094478
Where is the link to your selfies

[Anonymous No.9094514]

7

[Anonymous No.9094515]

>>9094464
42 you fucking idiot

[Anonymous No.9094521]

>>9094464
1

[Anonymous No.9094533]

>this shit again
at least try something not for brainlets

[Anonymous No.9094534]

how is everyone getting 1

[Anonymous No.9094538]

>>9094464
9

[Anonymous No.9094544]

>>9094464
1 and you call Vergens retard. Blame on you

[Anonymous No.9094547]

1.000000090000001

[Anonymous No.9094564]

>>9094464
its 9 idiots

[Anonymous No.9094568]

>>9094533
6

[Anonymous No.9094571]

>>9094533
6?

[Anonymous No.9094573]

1

[Anonymous No.9094585]

>>9094464

8 dumbasses

[Anonymous No.9094587]

>>9094464
The answer is 9.

[Anonymous No.9094588]

9

[Anonymous No.9094591]

Im 50% in TNB should i kill myself biz?

[Anonymous No.9094590]

7

[Anonymous No.9094600]

>>9094464
Doesn't matter because it's not an oracle problem

[Anonymous No.9094601]

>>9094568
>>9094571
no

[Anonymous No.9094615]

>>9094533
Make 5 races, pick the fastest horse in each group, and make a sixth race to establish the top three.
Am I hired?

[Anonymous No.9094618]

9

[Anonymous No.9094631]

7. Because if you devide 3 by a third you end up with 3÷1 which is 3. Pemdas. 9-3+1

[Anonymous No.9094642]

>>9094464
9-3/1/3+1
Negative 3 divided by 1 is negative three.
9-3/3+1
Negative three divided by three is negative 1.
9-1+1
It's 9.

[Anonymous No.9094645]

>>9094615
Wrong. What if two or three of the top three are in a single group?

[Anonymous No.9094655]

>>9094533
10

[Anonymous No.9094656]

>>9094533
4. Group 3 races of 8+8+9 horses, top 2 of each go in final race.

[Anonymous No.9094659]

>>9094533
11

[Anonymous No.9094666]

>>9094591
When the other 60% is not XVG
>YES

[Anonymous No.9094671]

>>9094656
Oh, up to 5. So yeah 6 right? Horse racing is subjective though as differnet horses have better races and worse races, so who knows you'd have to do some complicated probabilistic formula right?

[Anonymous No.9094673]

PEMDAS you faggots - my fuck, it’s negative 1

[Anonymous No.9094675]

>>9094615
>>9094655
>>9094659
wrong
>>9094656
can only race 5 at a time

[Anonymous No.9094676]

>>9094568
>>9094571
>>9094615
>>9094656
It is 11

[Anonymous No.9094683]

>>9094533
12?

[Anonymous No.9094686]

>>9094464
9

[Anonymous No.9094699]

>>9094676
How is this achieved, by cross section?

[Anonymous No.9094701]

>>9094675
How would you introduce the subjectiveness and other factors however, a probabilistic formula of some sorts? I've been to horse races and the fastest horses can have off runs, so you'd need to do more to account for that right?

[Anonymous No.9094702]

This shouldn't surprise me but most of you have not been taught simple math properly. It's 1.

[Anonymous No.9094706]

>>9094673
PEMDAS would mean the answer is 9, not -1, Kenneth.

[Anonymous No.9094708]

>>9094699
Run off races for the top three of each race, til you have a final three.

[Anonymous No.9094717]

>>9094701
assume the horses are mechanical and in working condition

[Anonymous No.9094721]

Brainlets
9 – 3 ÷ (1/3) + 1
3 divided by a third is 9
9-9+1=1

[Anonymous No.9094731]

>>9094721
this

[Anonymous No.9094735]

>>9094590
Oops I fucked up. The answer is 1.

[Anonymous No.9094736]

>>9094683
wrong

[Anonymous No.9094737]

>>9094533
5

use your smartphone to time them

[Anonymous No.9094744]

>>9094675
shit it gets complicated into 1v1 round robin after 10, 19 would be enough

[Anonymous No.9094750]

>>9094533
>>9094568
>>9094571
>>9094615
>>9094671
The problem is that the 2nd fastest and 3rd fastest horses could've been "eliminated" if they were in the same group as the fastest horse.

So after your 6 races (5 elimination rounds + 1 final round) to get the fastest horse:
you need to do another race of these 5 horses:
- 2nd fastest horse in the final race + 4 horses from the elimination round the top 1 horse came from
This will give you the 2nd fastest horse.

THEN, you have 2 possible outcomes:
a. the winner is from the horses that were in the "elimination round" with horse #1, or
b. the winner is from the final round

In case of (a), the final race you need to do is:
3 remaining horses from the "elimination round" horse #1 and #2 came from + 2nd fastest horse in the "final round" earlier that determined horse #1

In case of (b), it's possible that the 3rd fastest horse was eliminated in the elimination rounds by horse #1 or horse #2.
So you'll need 2 races more:
- 4 horses from the elimination round of horse #1 + the 3rd horse in the "final round" earlier
- 4 horses from the elimination round of horse #2 + the 3rd horse in the "final round" earlier

So you can do it in 8 or 9 races.

[Anonymous No.9094755]

>>9094744
no

[Anonymous No.9094768]

>>9094750
wrong

[Anonymous No.9094770]

>>9094533
11.. You have to race 5 at a time, each time taking the top 3 and racing them against another pair

[Anonymous No.9094776]

>>9094533
25 horses divides into 4 groups of 4 and 1 group of 5
race the group of 5 first, then take the winner and race it against a group of 4, take the winner and race it against the next group of 4, do this until all horses have raced
5 races total

[Anonymous No.9094782]

>>9094755
the final 6 horses have to battle to the death. 22 races MUST be more than enough

[Anonymous No.9094788]

>>9094776
US maths? Damn boy, its hard to sit next to some Niggers.

[Anonymous No.9094789]

>>9094656

not hired due to poor reading comprehension

[Anonymous No.9094791]

>>9094533
Why the hell would I trouble myself with horse races?

[Anonymous No.9094793]

>>9094782
it's asking for the minimum

[Anonymous No.9094802]

>>9094721
Who needs math when you have digits.

[Anonymous No.9094805]

>>9094706
PEMDAS
parentheses - none
exponents - none
multiplication and division - expression becomes 9 - 9 + 1
addition and subtraction - expression becomes 1

[Anonymous No.9094809]

>>9094736
1. 25/5=5
5x3=15
2.15/5=3
3x3=9
3.Group into 5 and 4
6 remaining
4.Race 4 and have the two remaining in a final race with the top three.
Rounds:5+3+ 2+2=12

[Anonymous No.9094814]

Multiplication or division goes first, then addition and subtraction, moving from left to right. 3 divided by 1/3 = 3*3 =9, 9-9=0, +1 =1.

Some of us made it past basic 7th grade arithmetic.

[Anonymous No.9094828]

>>9094533
6

[Anonymous No.9094831]

>>9094805
My God.

[Anonymous No.9094834]

>>9094809
>>9094828
nope

[Anonymous No.9094841]

>>9094834
>t. does not work at Google
Lol. Yeah it is. You can't figure out the top three in less than 12 and mine would

[Anonymous No.9094845]

Add before subtract faggots, that means -1

[Anonymous No.9094861]

>>9094841
nope
solution:
https://www.youtube.com/watch?v=i-xqRDwpilM

[Anonymous No.9094864]

6 races faggot

[Anonymous No.9094866]

>>9094861
13?

[Anonymous No.9094889]

>>9094464
9-3=6
6 /1/3 --flip the denominator--> 6*3/1=18/1=18
18+1=19

[Anonymous No.9094896]

>>9094864
>>9094866
wrong

[Anonymous No.9094900]

>>9094768
Oh, I get it... I was including ALL the horses in each group.

So I guess after the 6 races to get the winner, you can do one more race:

- 2nd and 3rd horse from the elimination round horse #1 came from
- 2nd and 3rd horse from the final round
- 2nd horse from the elimination round that final round's horse #2 came from (the only horse that was possibly eliminated but could've been 3rd fastest)

[Anonymous No.9094906]

>>9094789
>implying I come here to read and didnt just look at the picture
>not purposely getting the answer wrong so you dont have to work for jewgle in a $9k/month rent cuckshack

Lmao. Brainlets these days

[Anonymous No.9094933]

>>9094861
Damn

[Anonymous No.9094981]

>>9094845
It's left to right. Type it in a calculator. 9-9+1=1.

How am I not rich yet trading against you brainlets?

[Anonymous No.9094994]

>>9094533
6 races. 5 races for each group of 5, and then race 6 would for the 5 winners of each group. You would then be able to discover 1st, 2nd, and 3rd place horses.

[Anonymous No.9095016]

>>9094981
Because biz is for brainlets and semibrainlets.
Youre just one semibrainlet

[Anonymous No.9095030]

>>9094994
what if the 3 fastest are all in one race?

[Anonymous No.9095102]

>>9095030
The fastest of a group weeds out the rest among the 5 groups. The remaining 5 fastest horses of each 5 groups compete for 3 titles while the other 2 go back to their stables. If I'm wrong then I'm sorry because I don't even deserve to be considered for the position.

[Anonymous No.9095165]

>>9095030
this is fucking gay just race all 25 at once you cunts the first three to the finish are fastest

[Anonymous No.9095191]

>>9095102
That's the problem, in the initial 5 "elimination rounds", once a horse loses, it's eliminated.

So if the 2nd fastest or the 3rd fastest (or both) are in an elimination round with the fastest horse, they don't get to compete in the "finals", they're eliminated early.

[Anonymous No.9095212]

>>9094994
That will only get you the top three of winners from each race. Not the top 3 of the whole 25, you fucking faggot.

You are thinking like most championship brackets but that isn't always fair. Imagine if one of those first five races were stacked with elites and the other four races were chumps. The 2nd place of the elite race would smoke any of the "winners" of the other four but it doesn't get to race in your 6th race. Try again, chump

[Anonymous No.9095247]

Race 5 sets of 5 and eliminate the bottom 2 of each race. This leaves 15. Do the same thing with 3 sets of 5 to leave 9. Race one set (a) of 5 and then a second (b) with the 3rd place getter and the remaining 4. Run the last race with the top 2 of a and the top 3 of b. 11 races total.

[Anonymous No.9095266]

>>9094464

Numbers aren't real, only approximations.

[Anonymous No.9095276]

>>9094464

Simple - the answer is: 19.000000000000000000000000000018

> 9 minus 3 = 6
> 1 divided by 3 = 0.33333333333333333333333333333333
> 6 divided by 0.33333333333333333333333333333333
> = 18.000000000000000000000000000018
> + 1
> = 19.000000000000000000000000000018

[Anonymous No.9095323]

>>9095191
>>9095212
My method is correct except after the last part that comprises race 6. The answer is 7.

[Anonymous No.9095347]

Here's a view after labeling of the last race (7). Makes sense when you remove the slowest of the winners.

[Anonymous No.9095353]

>>9094564
underrated

Is the american educational system so poor?

[Kekovicius !!seZ2sXABmE3 No.9095354]

>>9094533
Split horses in 5 groups and have them race (5 races). Take top three horses from each group and make 5 new groups, take 1st horse from each and race them, repeat 3 times (3 races). The winner of each group is the fastest horse. Therefore, 8 races.

[Anonymous No.9095386]

>>9095165
replies like this make my day

[Anonymous No.9095419]

>>9095323
> My method is correct
Everyone and their mom figured out that method.
That's the entire point of the question, to get beyond that step, and realize that the 2nd and 3rd fastest could've been prematurely eliminated. Hence the 7th race.

[Anonymous No.9095438]

>>9094464
3

[Anonymous No.9095446]

>>9094533
race 5 groups of 5 - record rankings of 1st to 3rd of each group. (5 races)
Race the winners. Get a top 3 of the winners. Remember the groups of each winner - denoted 1f, 2f, 3f (1 race)
take 2nd (1f2) and 3rd (1f3) place from original group of winner and 2nd (2f2) place of the bracket from second finalist
race 1f2, 1f3, 2f1, 2f2, 3f1 (1 race)
The two winners there are second and third place overall.

Total 7 races.

This is faster than some strats posted above by avoiding repeat races with the 2nd and 3rd runners from dumpster brackets. but I think it should still guarantee an accurate top 3.

[Anonymous No.9095524]

>>9095419
The way you put it is as if I'm patting myself on the back. Yes, I understand I was stating what was obvious to being with. My intention wasn't to sound pompous, but to highlight that I wasn't that far off.

You got a problem with that?

[Anonymous No.9095533]

>>9094464
-1

[Anonymous No.9095535]

>>9094533
naively, you'd race 5, take the top 3 and race them vs. 2 more and then take top 3 out of that and race them vs. 2 more for a total of 11 races

but this is inefficient so let's race five, take the top three and try to eliminate them vs. horses we didn't test yet

if one of the top 3 makes it in the top three of the next race with unknown horses, then we have eliminated two potentials, but importantly we have RANKED the top three out of this group

let's see what happens when we consistently throw away a horse because it ranks poorly vs. the rest

now we have 2 horses from the new set, 2 horses from the old set that are slower so they are also thrown away (since we tested the fastest horse)

so we test the fastest from the two and four more, and if it gets throw away again and again this case is 6 comparisons where at that point you found the 3 fastest so it's kind of the best scenario

let's say you see the horse from the first set is the fastest in the second set too - you will probably need more comparisons since you now have a ranking of 5 horses but you don't know how the two sets compare

so maybe try the second fastest horse from the first set - if it's top 2 of the new set, you can eliminate bottom 3

if it's not, you can eliminate it AND the slower horse and the slowest horse in the set, so also eliminating three

if it's #2, then you don't know whether the previous set's fastest or the new set's fastest is the best horse, but you know it's the slowest

if it's the fastest, then your fastest horse is #1 still, you just don't know which set's #3 horse is the slowest of the top 4

we do the middle one again, now we have a set of three to waddle through with 16 left and we can take two at a time for a total of... 9 comparisons since we always can eliminate 2

that's the best I can do right now

[Anonymous No.9095538]

>>9094533
Best Case: 7
Worst Case: 8

[Anonymous No.9095540]

>>9094533
race each row, 5 races (we eliminate 4ths 5ths)
assuming collumn 1 is the winners, we race them 6th race (eliminate 4th 5th placed rows)


A1 is the fastest horse, A2 got second, A3 got third (eliminate B3, C3, C2)

race 7 is B2, B3, A2, B2, A3

[Anonymous No.9095574]

>>9095354
you could miss the second fastest horse because it's the #2 best in the second round, but you eliminated it

[Anonymous No.9095575]

>>9095540
edit on race 7 (B1, C1, A2, B2, A3)

[Anonymous No.9095584]

>>9094464
1

anyone who says otherwise is trolling

[Anonymous No.9095598]

>>9095524
> highlight that I wasn't that far off.
You were, your step was the step EVERYONE came up with. Try reading ALL the responses to that problem here.

If I was the one holding the hiring test (which I HAVE done several times for our software development firm), everyone who realized that the 2nd and 3rd fastest could've gotten eliminated early, but didn't get the MOST OPTIMIZED solution to narrow it down (like my very first response up top) gets partial credit.

Everyone who just decided to do 6 races (which is what EVERYONE thought was the solution) doesn't get any credit and doesn't get hired.

Because this problems tests your critical thinking, and the 6 races answer is the answer with none of that.

Basically anyone whose answer is more than 6 has a chance to be hired, while everyone who answered 6 gets disqualified, is how these questions work.

If you want to know why, google about programmer interview questions/or google/microsoft interview questions, and you'll see it's full of "trap" like questions like this, and answer the generic (wrong) answer everyone comes up with is what will instantly disqualify you.

[Anonymous No.9095620]

>>9095446
better than my solution, I think this can't be beat

[Anonymous No.9095658]

>>9095540
Wrong. All five horses in row 1 could be faster than the other 20 horses.

[Anonymous No.9095688]

>>9094776
>25 horses divides into 4 groups of 4 and 1 group of 5

That's it, this post here has solidified my decision to cash out of crypto completely.

[Anonymous No.9095689]

>>9094533
I thought about it a bit more and really the "minimum" time need is 0 races. You could get lucky and just happen to identify the top 3 immediately through pure luck without even attempting to race horses.
Kind of like if you had to choose between 25 random cryptos and got ETH, BAT, and LINK.

The odds are pretty bad though, with so many shitcoins out there.

[Kekovicius !!seZ2sXABmE3 No.9095703]

>>9094533
I figured it out! Take any 5 horses and have them race. Remove the slowest two and add two new horses from the 20 remaining. Keep doing this until you've run out of horses. The three top horses in the 11th race are the three fastest ones. So, 11 races.

[Anonymous No.9095715]

>>9095703
it's an okay solution, but you can do better

[Anonymous No.9095743]

probably answer is 12
keep discarding all but the top 3 in each race
you'll race 25 and discard 10
then race 15 and discard 6
etc.

[Anonymous No.9095754]

>>9095703
>fastest horse has to race 11 times
>gets exhausted
>loses 11th race

brainlet answer

[Anonymous No.9095792]

>>9094464
PLEASE EXCUSE MY DEAR AUNT SALLY

[Anonymous No.9095800]

>>9094533
Double elimination tourney
Pretty much any other way doesn't guarantee you'll find the fastest 3 horses

[Anonymous No.9095807]

>>9095743
you can do better than this, you're racing too many garb horses because you're not discarding them fast enough

[Anonymous No.9095825]

>>9095800
so, how many races, faggot?

[Anonymous No.9095826]

>>9095689
how can you tell which ones are faster if you dont compare them with the rest?

the answer is infinite, in the scenario where every horse runs at equal speed

[Anonymous No.9095832]

>>9094533

Put horses into groups of 5
Race each group -> (5 races)
Race #1 winner from each group -> (1 race)

At this point, you have determined the absolute fastest horse from race 6. The 2nd and 3rd place from race 6 are still in consideration whether they are fastest overall.

However, we can now eliminate any group that didn't make 2nd and 3rd place in race 6.

So we're left with 3 groups. Lets say they are Group A, Group B, and Group C, with Group A having the fastest horse A1.

Take out A1 because it doesn't need to be raced anymore.

Take out every horse from Group C except C1 (fastest in Group C), because if they are slower than C1 which is in consideration for 3rd place, they aren't going to make it.

Take out every horse from Group B except for B1 and B2. If these horses are slower than A1, B1, and B2, they aren't going to make it.

And take out every horse from Group A except for A1 (removed earlier), A2, and A3. Same reason.

Race the last 5 horses (A2, A3, B1, B2, C1)

[Anonymous No.9095914]

>>9095825
I don't have any brainlet wojacks so...
48. 49 if you want to make the final 2 horses race again

[Anonymous No.9095936]

>>9095826
this is determined in the first 6 races, then the answer you give is top 3 is a tie for all 25 horses since they tied in all of the races

>>9095914
>needing 48 races
image related

[Anonymous No.9095939]

>>9095826
That's the flaw. There would be no way to prove it.

and actually if the horses all tied you'd know it in 6 races.

[Anonymous No.9096086]

>>9094533
If you get lucky with the order of the horses racing, you could do it in as little as 7.

Let's break that picture into rows of horses that all race each other. The first column of each row is the fastest of that row. So in 5 races we establish the winners of each row. You then race the winners of each row in the 6th race to establish your fastest horse. Take the horse that came in second place and race it against the four remaining horses that were in the 1st place horse's row. If that second place horse comes in third place or higher, then the top two horses in that race are the second and third fastest horses of the 25.

[Anonymous No.9096107]

>>9095939
we don't know if horses follow the transitive property. perhaps Horse A > B > C, but horse C really hates horse A and beats him in a race every time despite losing to other, slower horses. they probably want you to make binary tree out of the horses or some shit

[Anonymous No.9096142]

>>9094534
>>9094564
>>9095353
PEMDAS you fucking mongoloids. 3 divided by 1/3 is 9, plus 1 is 10, so 9-10. It's actually -1 but fucking 9? You guys are literally bottomfeeders

[Anonymous No.9096152]

>>9096086
you don't need to compare to all 4 horses in the first place's row, the bottom 2 in that race will never be in the top 3

you only need 2 from there, 2 from the top 2 horse's row (including it as well), and the third best horse

so it's guaranteed to be 7

>>9096107
not binary tree, but I think that question assumes it's transitive, yes

[Anonymous No.9096174]

>>9096152
Good point. 7 is the definite answer.

[Anonymous No.9096233]

>>9096152
yeah my CS is rusty but it seems like some kind of graph data structure would be applicable. with the horses being vertices and their relationships being edges

[Anonymous No.9096256]

>>9096233
you're literally just guessing

[Anonymous No.9096267]

>>9095658
That's accounted for you dolt. Watch the youtube and the read other posts that agree.

[Anonymous No.9096287]

>>9096256
yeah it's late and i don't feel like thinking too hard. but a graph totally seems like the best data structure for that

[Anonymous No.9096289]

>>9094533
you can identify the 3 fastest horses without those 3 actually being in the same race. you just need to time each horse individually, which would take 5 races. one additional race with the winners is unnecessary if you already clocked their times.

[Anonymous No.9096292]

>>9096287
it really isn't, because it's monotonically ordered

an array is the best data structure for this

[Anonymous No.9096314]

>>9096292
welp I guess that's why I don't work at google

[Anonymous No.9096323]

>>9094464
1

[Anonymous No.9096374]

>>9095538
Lol. This is exactly what I got. You must be a fucking genius, props.