Can you solve this?

Can you solve this?

Other urls found in this thread:

wolframalpha.com/input/?i=|(x^2)+((sqrt.3)(y^2))-((sqrt.5)(z^2))|<(1/(10^12))
play.golang.org/p/i-QBcwovds
twitter.com/NSFWRedditVideo

shouldn't this have infinitely many solutions?

wolframalpha.com/input/?i=|(x^2)+((sqrt.3)(y^2))-((sqrt.5)(z^2))|<(1/(10^12))

Hopefully I got all of that right

>one inequality of three variables

Fuck outta here.

You typed it in right but WolframAlpha can't find a nontrivial integer solution for you

Zero for all of them

Nice bait.

>each fruit represents a nonzero integer

>set 2 variables to 1
>equation’s solution must lie between (-1/10^12 ;1/10^12)
>essentially solve 2 inequations with the results of the two ANDed

It’d still have infinite solutions

>set 2 variables to 1
>when all 3 are different
You're not a smart person.

it is not said that they are all different

Are you a literal retard? They have different pictograms in the picture.

Nowhere does it say the variables are unique.

It's implied, mongoloid. If given variables named X, Y, Z, do you also need to be told that they're supposed to be unique?

Eh no? Are you saying x,y,z must always be different in an equation or something?
Cause that’s wrong.

...

that's an olympiad problem isn't it?

no, I can't solve it. maybe some MOSP whiz can, but I honestly doubt they waste their time lurking Sup Forums

PS I realized this is Sup Forums and OP probably meant solve it programmatically, with an efficient algorithm

[spoiler]False.[/spoiler]

>x2=y2 only has one solution
Retard.

They are constrained to be integers though so there can only be finite solutions

unless explicitly excluded x = y = z is valid and assuming them to be different thought they might be the same value might be the very reason why you can't solve the equation.

I remember this, you need elliptic curves to solve it. So it's more like 99.99% of people can't solve it.

Each fruit is 1x10^-500

It’s still nonzero

>If given variables named X, Y, Z, do you also need to be told that they're supposed to be unique?
Yes.
Just because X is 5 doesn't mean Y can't also be 5.
Are you fucking retarded?

Apple=infinitesimal
Strawberry=a slightly different infinitesimal
Grape=yet another different infinitesimal

That is not an integer

Quite an intriguing problem. Superficially it would seem that might be right, but exhaustive search does not converge really fast. The best approximation I have stumbled upon so far is 22125*22125+sqrt(3)*208466*208466-sqrt(5)*184069*184069, but this (-.0003675553389) is still some 9 decades off from required.

It could be that at least trivial C data types are running out of steam before this is solved by brute force.

answer is 0

B-but my nonzero integer!

-10^-12 < x^2 + 3^1/2 y^2 - 5^1/2 z^2 < 10^-12

-10^-12 - x^2 < 3^1/2 y^2 - 5^1/2 z^2 < 10^-12 - x^2

x^2 + 10^-12 > 3^1/2 y^2 - 5^1/2 z^2 > + x^2 - 10^-12

x^2 + 10^-12 > (3^1/4 y + 5^1/4 z)(3^1/4 y - 5^1/4 z) > (x - 10^-6)(x + 10^-6)

Not sure where I'd go from here. Or even if I made it better

Triplets such as
66526, 202195, 183431 -> -.0001462303517
or
3447, 52511, 46273 -> -.0001480605165
appear within number space below 1M, but are still magnitudes off.

The program I am running has got some (serious) problems with accuracy. It detects at least some (small) values, but might ignore others, so this search can not be considered fully exhaustive. What is needed is operating with at least the 10^12 level, which corresponds to around 40 bits and I cannot reliably figure out the precision of my long double operations and if the typecasts get executed correctly or not - for now.

Does not feel like a math olympiad problem. More likely something like said, although I am a bit skeptical about this being directly related with any elliptic curve problem.

Nevertheless: thanks for OP for getting me to refresh a bit of C data type madness. I have a couple of ideas to develop this further, but this thread will be long gone, if I happen to chase them. Any actual real-world significance for this?

...

You can do a faster brute force search by only iterating through all possible values of a and b and solving for c algebraically. That's the best I got.

I got 99462, 83639, 99211. I don't think that's right, but it's as far as I can go with floating point: https://www.google.com/search?q=(99462^2)%2B((sqrt(3)(83639^2))-((sqrt(5)(99211^2))&oq=(99462^2)%2B((sqrt(3)(83639^2))-((sqrt(5)(99211^2))

99462*99462+sqrt(3)*83639*83639-sqrt(5)*99211*99211 = .0003449604411
Still off by factor of 100 000 000 or so, but quite good. (Checked on Linux bc, with scale=13)

That is right. Only possible Z is sqrt((X*X + sqrt(3)*Y*Y)/sqrt(5)) - or that +1, because typecasting to integers simply truncates. Looping the outer variables is still N^2, though, taking quite long for any interesting space. The problem is also embarrassingly parallel, if somebody is interested.

OK, ok... the bc part of Linux actually is GNU.

Here's my code and other found solutions. I don't know how to get around floating point imprecision without making the code unbearably complex and slow. Still nowhere near 10^-12.

I wonder if a SAT solver might work on this.

I bet my pants the solution is way above uint64 range.

I'm sorry I cannot show you mine, since I got a marvelous stroke of ingenuity and tried to convert the whole problem on integer arithmetic. In that process the functionality (what was left of it anyway) somehow escaped the program. No backup. But I am not foolish enough to admit, that 1457 538 648 is less than 100 000 000.

Could well be. As can be seen, opportunities to match the squares dwindle rapidly. Yet I have a gut feeling that there could be infinitely many solutions.

A little correction to my earlier claim: increasing the scale factor gives
99462*99462+sqrt(3)*83639*83639-sqrt(5)*99211*99211 = .0000028001096111926920594333639461804862, which is only about 3 000 000 times too large difference. Quite a bit better than I claimed and possibly the best fit so far, so congrats!

x=0
y=0
z=0

x

wrong

explain

>No backup
datfeel.png

>only about 3 000 000 times too large
Still the search seems to slow down exponentially (or quadratically?) So getting those last few zeros is going to be hell.

Use continued fractions to find rational approximations for the roots. Then multiply through by the denominators to get integers.

Belongs in /sci/

why would we even try to solve problem that's represented in speudonumbers instead of actual expressions like X Y Z I?

maybe not a math olympiad problem, but an informatics olympiad one.

if it was a math olympiad problem, it'd probably involve inequalities manipulations (I'm always amazed at the contestants knowing exactly what to put on the other side of an inequality... you can figure out an expression that will fit your less than or greater than, but it's hard to know find one which will advance your problem)

>what is algebra?, im a retard
The post

since there are infinite numbers
yes

>b-but the pictures is difrent
dunno what to tell you man

play.golang.org/p/i-QBcwovds

lol no generics

Can you post that not as an image so we can test your solution? Also I don't speak go, how does that work?

No, it gets spam filtered; copy it from the playground page after clicking run.
It's using Newton's method starting from 1e80 for all the fruits, 1e80 being the value high enough that the solution didn't go towards zero.

Hey retards,
[spoilers]\sqrt(3) = 3/\sqrt(3)[#endspoilers]
You want to end up with 1/(x*10^12) ∃ x(fruit)
t. Russian

5^1/4 1 1
Just fucking end me

*2^1/4

As far as I can tell your solution is way off, but perhaps I have made a mistake. Here it is calculated with bc to a thousand digits of precision.

My solution is wrong, but also b and c aren't squared in the picture of a screen you posted

Ugh my mistake. I posted the wrong screenshot.

Anyway it's a cool idea but I don't see how it could possibly work. Newton's method works on continuous problems and this is definitely not continuous. Any set of integers could be a solution.

When you divide a fruit, you cut it into smaller pieces.
The solution is fruit salad.