Can someone solve it?

>>8884398
do i have to use the numbers 1-9? if so, the left column is impossible
if not, what numbers can i use?
can i repeat numbers?

Yes, you can repeat numbers

>6 equations
>8 variables
I'm gonna say no.

>>8884430
Non linear optimization here will help.

going from top left.
4,2,4
6,0,4,
-3,1,2

>>8884454
>what is order of operations

>>8884454
Forgot to say they can only be positive integers. Sorry.

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>>8884398
Is there a general theory behind solving these "array" equations?

>>8884462
linear algebra I believe

>>8884459
Why? Says who? Are there any other arbitrary restrictions missing from the question?

>>8884462
You got it, amazing!!!

>>8884462
11 * 4 = 8

Interdasting

>>8884470
PEMDAS motherfucker

>>8884465
Nope. No others. It's piss easy if you can use negatives.

>>8884470
12- (1*4)=8
12-4=8

>>8884487
Oh, and no zeroes.

So
>>8884462
is out of luck. Sorry, mate. It's supposed to be difficult.

>>8884463
Linear algebra wouldn't help you with this problem. The closest thing I can think of is multilinear algebra, but that's not quite what's going on here either.

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Solved it.

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>>8884490

>>8884498
Fractions and zeroes aren't allowed.

>>8884492
Actually nevermind, this is just linear algebra but with multiplication allowed.

>>8884507
Close. Try doing it without duplicates.

>>8884520
That's impossible.

>>8884533
It's not. If you can't figure it out, I'll post the solution in an hour.

Call
Top left = x
Top middle = y
Middle middle = z
then
Top right = (x - 8)/y
Middle right = 4 + z
Bottom left = -1 + x/6
Bottom middle = 3 - y + z
Bottom right = 6 - (x - 8)/y + z = 1 + x/6 - y + z

Clearly x must be of the form 6k with an integer k. But then if you solve
6 - ((6k) - 8)/y + z = 1 + k - y + z
for y you get
y = k/2 - 5/2 + (1/2) * sqrt(k^2 + 14k - 7)
which is only an integer for k=2 and therefore y=1.

So you have
Top left = 12
Top middle = 1
Top right = 4
Middle left = 6
Middle middle = z
Middle right = 4 + z
Bottom left = 1
Bottom middle = 2 + z
Bottom right = 2 + z

There is already a duplicate no matter how you choose z.

>>8884562
Ahem. You're not thinking big enough.

>>8884520
Stop moving the goalposts >>8884429

>>8884515

Multiplication makes this non-linear. This sort of problem falls under the category of non-linearly constrained optimization problems. In particular, this specific problem has no objective and only constraints (it is a feasibility problem as opposed to maximization or minimization).

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Fuck it.
Here are a homework solution of some.

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>>8884785
>8-56 = 8
neck yourself

>>8884785
>8-7*8=8
>8-(56)=8
nigger are you FUCKING HIGH?

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>>8884470

>>8884785
>8-18=6
Alright, then.

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>>8884462
its multivariable algebra at best squeebs

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>>8884430
kind of I guess, I didn't do the hardest last one because I'm lazy.

>>8884398>>>/b/
OP is a supreme cunt for making this puzzle up as she goes.

>>8884398
It's impossible

>>8884398
[48][8][5]
[6][9][13]
[7][4][10]
There are infinite solutions.
[48][8][5]
[6][x+5][x+9]
[7][x][x+6]
There are even more with different first columns & first rows.

>>8886800
>There are even more with different first columns & first rows.
By that I mean >>8884562's solution. (Which is the only other one.)

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Comp-sci-god here

The simplest way to solve it would be to just brute force it. Try every number in a given range with every number in the range with every number in the range... etc checking each time to see if it satisfies the equation.

Of course that algorithm would run in O(n^6) time, where n is the test range. This is unacceptable. And there are some evident ways to improve the algorithm.

We will label the unknowns from a to h, going from left to right top to bottom.

Condition 1: Given a value, a, it is evident that the value f is always fixed. Simple algebra tells us that f is always equal to (a-6)/6. And, assuming that all numbers have to be integers, if (a-6)/6 isn't an integer, we can skip over that value of a.

Condition 2: the values of b and c both have to be factors of (a-8). We only need to iterate over the factors for the loops b and c, not the whole range

Condition 3: e's value will always be fixed at (d+4). So that loop can be left out.

Condition 4: h's value will always be fixed at (1 - f - g)* -1

Taking these into account, we can make an algorithm that has a time complexity of O(n^2)

Testing on the range -200 to 200 yields over 3000 valid answers.
On the right are some randomly selected values from the answer set

>>8884470
Please kill yourself. Thank you!

>>8887401

A "comp sci God" would know that, when we analyze the complexity of algorithms, n is the number of bits required to represent a problem instance, not the decimal value of inputs. Thus, if your algorithm runs in time O(r^2) where r is the maximum absolute value, your algorithm runs in time O(2^{2w}), where w is the word size required to represent numbers in the solution (for n taken as the total input size, your suggested algorithm has similarly abysmal performance).

A "comp sci God" would further seek to give a general-case algorithm to solve problems of this sort. What you've given is an asymptotically horrendous, exponential-time "improvement" to a brute-force attempt to solve a single instance.

You are not a "comp sci God." The fact that you call yourself such while spouting such absolute garbage is the reason why everybody on this board thinks our field is a meme, and you should be ashamed. Have some humility, and, for heaven's sake, stop being so lazy and actually invest energy into understanding the work of great computer scientists.

>>8884414
(9 - 6) × 2 = 6 you brainlet

>>8887453
B8/10 meme. Now fuck off.

>>8884398
Plug in any numbers for "A" and "B" evaluate.

It was easy to reason through until there was one space left. Then I just set the formula of that space considering only the row equal to it's formula when considering only the column, and made that into a quadratic equation.
There are a few other configurations with slightly different quadratic equations. But I'm too tired to work them all out.

[math]C=\frac{-4+A±\sqrt{A^2+16A+8}}{2}[/math]

[math]
\left[
\begin{array}{ccc}
6(A+1), & C, & \frac{6A-2}{C}\\
6, & C+B-A-2, & \frac{6A-2}{C}+B-2\\
A, & B-A+1, & B
\end{array}
\right][/math]

>>8884430
Just means there are multiple solutions.

>>8887634
this looks legit enough

bravo

>>8887461

PEMDAS
(9 - 6) x 2 // evaluate parentheses
3 x 2 // evaluate exponents
3 x 2 // evaluate multiplication/division
6 // evaluate addition / subtraction
6

(9 - 6) x 2 = 6 QED, git gud brainlet

I wrote a program to solve these in high school
Best I could do at the time was O(n^4). Now I'm wondering if there is a better way. Perhaps solving it as a system of diophatine equations might yield a faster solution?