Can someone solve it?

[Anonymous No.8884398]

Show me your maths

[Anonymous No.8884414]

>>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?

[Anonymous No.8884429]

Yes, you can repeat numbers

[Anonymous No.8884430]

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

[Anonymous No.8884438]

>>8884430
Non linear optimization here will help.

[Anonymous No.8884454]

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

[Anonymous No.8884457]

>>8884454
>what is order of operations

[Anonymous No.8884459]

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

[Anonymous No.8884462]

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

[Anonymous No.8884463]

>>8884462
linear algebra I believe

[Anonymous No.8884465]

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

[Anonymous No.8884468]

>>8884462
You got it, amazing!!!

[Anonymous No.8884470]

>>8884462
11 * 4 = 8

Interdasting

[Anonymous No.8884485]

>>8884470
PEMDAS motherfucker

[Anonymous No.8884487]

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

[Anonymous No.8884488]

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

[Anonymous No.8884490]

>>8884487
Oh, and no zeroes.

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

[Anonymous No.8884492]

>>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.

[Anonymous No.8884498]

Solved it.

[Anonymous No.8884507]

>>8884490

[Anonymous No.8884508]

>>8884498
Fractions and zeroes aren't allowed.

[Anonymous No.8884515]

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

[Anonymous No.8884520]

>>8884507
Close. Try doing it without duplicates.

[Anonymous No.8884533]

>>8884520
That's impossible.

[Anonymous No.8884549]

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

[Anonymous No.8884562]

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.

[Anonymous No.8884583]

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

[Anonymous No.8884585]

>>8884520
Stop moving the goalposts >>8884429

[Anonymous No.8884691]

>>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).

[Anonymous No.8884785]

Fuck it.
Here are a homework solution of some.

[Anonymous No.8885054]

>>8884785
>8-56 = 8
neck yourself

[Anonymous No.8885069]

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

[Anonymous No.8885091]

>>8884470

[Anonymous No.8885096]

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

[Anonymous No.8885112]

>>8884462
its multivariable algebra at best squeebs

[Anonymous No.8885290]

>>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.

[Anonymous No.8886698]

>>8884398
It's impossible

[Anonymous No.8886800]

>>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.

[Anonymous No.8886830]

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

[Anonymous No.8887401]

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

[Anonymous No.8887414]

>>8884470
Please kill yourself. Thank you!

[Anonymous No.8887441]

>>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.

[Anonymous No.8887453]

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

[Anonymous No.8887461]

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

[Anonymous No.8887634]

>>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]

[Anonymous No.8887670]

>>8884430
Just means there are multiple solutions.

[Anonymous No.8888858]

>>8887634
this looks legit enough

bravo

[Anonymous No.8889759]

>>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

[Anonymous No.8889841]

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?