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I have a matrix that is 300x50 big filled with integers. I want to find what number appears the most in that 300x50 matrix and have it returned as an integer.
Just use mode on the linearized matrix:
mode(matrix(:))
Example:
>> matrix = [3 2 3; 2 2 1]
matrix =
3 2 3
2 2 1
>> mode(matrix(:))
ans =
2
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lets suppose
al=['Al','2l','3l','4l','5l','6l','7l','8l','9l','jl','ql','kl']
af=['Af','2f','3f','4f','5f','6f','7f','8f','9f','jf','qf','kf']
ak=['Ak','2k','3k','4k','5k','6k','7k','8k','9k','jk','qk','kk']
an=['An','2n','3n','4n','5n','6n','7n','8n','9n','jn','qn','kn']
are array
let's suppose
there a two-person wait for random numbers to get
but they should not get the same word ( 3 outputs should be to 2 people from 4 arrays without repeating
)
Use Array#sample and then Enumerable#each_slice:
persons, cards = 2, 3
(al + af + ak + an).sample(persons * cards).each_slice(cards).to_a
#⇒ [["qk", "4l", "Ak"], ["6l", "8l", "5l"]]
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I would like to solve a first degree equation with multiple variables (not a system of equations) like :
10x + 5y + 7z = 630
Is there any way to solve it without using bruteforce?
Solutions must be integers.
Regroup the first two terms as 10x+5y = 5(2x+y) = 5t.
Then t/7 + z/5 = 18.
As 5 and 7 are relative primes, t = 7k and z = 5(18-k), where k is abritrary.
Finally, y = t - 2x = 7k - 2x, where x is arbitrary.
As we can check,
10 x + 5 (7k - 2x) + 7 5 (18-k) = 630.
No you can't, you have an infinity of solutions in this case.
To solve such problem you shoud have a system with at least the same number of equations as the number of variables.
Another trick, in some cases you could solve it as an underdetermined system.
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Let X~gamma(2,1).
Find:
P(X>=2)
(X<=10)
I'm struggling to determine how to solve this. I know you must integrate some function from 2 to infinity and 0 to 10 but don't know what to integrate.
There are actually two ways to define Gamma-distribution - one with scaling 'theta' and another with inverse scale 'beta', see https://en.wikipedia.org/wiki/Gamma_distribution
Fortunately, for parameter value of 1 there is no difference.
So you have to integrate PDF
f(x) = (1/G(2)) * x * exp(-x)
Gamma function at 2 is equal to 1. G(2)=1, so
P(x) = S x exp(-x) dx
where S is integration notation.
P(x) = -(x+1)*exp(-x)
So now you have to substitute limits
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I am struggling so answer the following question. I think the reason I am not doing it correctly is because i starts at 0 rather than 1, but I am not sure how to do it.
The sum is equivalent to 3(n+1) + sum_{i=1}^n 2i, which is 3(n+1) + 2*n*(n+1)/2 = 3(n+1) + n(n+1) = (n+3)(n+1).
The result is thus (n+3)(n+1). The whole derivation is here.
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Question moved to: https://math.stackexchange.com/questions/1391780/the-largest-tile-with-2048-with-groups-of-3
A sentence to use up characters.
Dropping 3 and 6 doesn't create the same type of game. It would need to be 3 and 9 (3*3). Or you end up with two completely separate sets to merge, one with base 3 the other with base 6 (2*3).
In case you drop 3 and 9 then any 3^(2^min_moves) would be an acceptable objective just like 2048 = 2^16 = 2^(2^4)
EDIT:
Looks like I missed the part about merging 3 elements. That may need to have 3 and 27 (3^3) drop in. Also the objective would end up being 3^(3^min_moves) as merging 3 tiles is a move