## a^*b^*c^* – {a^n b^n c^n | n ≥ 0} is not regular using pumping lemma

$$L=a^*b^*c^* \setminus \{a^n b^n c^n \mid n \geq 0\}$$ can be proved as context-free by partitioning it as $$L = \{a^nb^mc^* \mid n \neq m\} \cup \{a^*b^nc^m \mid n \neq m\}$$ and further dividing each $$\neq$$ into smaller and larger. You will have four sets. You can give CFG’s for each. Then since CFGs are closed under union, you have your proof.

Now how can I go around proving that $$L$$ is not regular? If you prove that the each of these four sets are not regular, you still can not prove that the union is not regular, can you?

## Decide whether two strings $x, y$ can be split into substrings $a,b,c$ such that $x=abc$ and $y=cba$

What is the fastest algorithm for the following problem?

Given two strings $$x, y \in \Sigma^*$$ as input, decide whether there exists strings $$a, b, c \in \Sigma^*$$, such that $$x=abc$$ and $$y=cba$$.

By calculating all the length of the longest common prefix of $$s = x y$$ and all suffixes of $$s$$, we can compute all candidates for $$a$$ in $$O(n)$$ time.

For each candidate of $$a$$ with length $$k$$, we now just need to check whether $$x_k, x_{k + 1}, \ldots, x_{n}$$ is a rotation of $$y_0, y_1, \ldots, y_{n – k}$$. This can of course be done in $$O(n)$$.

However checking the rotations naively results in a worst-case $$O(n^2)$$ algorithm.

It seems that using the result from either https://arxiv.org/pdf/1601.08051.pdf or https://arxiv.org/pdf/1311.6235.pdf would make the algorithm run in expected $$O(n)$$ time.

Is there a simpler way of speeding up the rotation checking, where it is still faster than $$O(n^2)$$? Is there a way of making it deterministic, so that it still runs in $$O(n)$$ time?

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## Calculate coordinates of a centroid of $ABC$

We have a triangle $$ABC$$, where on the cartesian coordinate system:

$$A$$ lies on $$[-3, -2]$$,

$$B$$ lies on $$[1, 1]$$,

$$C$$ lies on $$[0, -6]$$.

How do we calculate coordinates for the centroid of this triangle?

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## Check which numbers satisfy the condition [A*B*C = A! + B! + C!]

#include <iostream> #include <cmath> using namespace std; int s = 1; //Serial No. 

Should I use recursion for the factorials function?

int Factorial(int n) {     int k=1;     for(int i=1;i<=n;++i)     {         k=k*i;     }     return k; } 

How I can go about doing this in a single for loop instead of the 3 while loops?

int main() {     int a = 1;     int b = 1;     int c = 1;     int Fact1;     int Fact2;     int Fact3;     while (a < 11)     {         Fact1 = Factorial(a);         while (b < 11)         {             Fact2 = Factorial(b);             while (c < 11)             {                 Fact3 = Factorial(c);                 cout << s << " : ";                 int LHS = Fact1 + Fact2 + Fact3;                 if (LHS == a * b * c)                 {                     cout << "Pass:" <<"    "<< a << " & " << b << " & " << c << endl;                 }                 else                 {                     cout << "Fail" /*<<"   "<< Fact1 <<"   "<< Fact2 <<"   "<<Fact3*/<<endl;                 }                     c++;                     s++;             }             c = 1;             b++;         }         b = 1;         a++;     }     return 0; } 

Also I would love some variable naming tips.

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## If $a$ is an $n$-th root of $z$ , $b$ is an $n$-th root of $w$, and $c$ is an $n$-th root of $zw$, then $ab=c$.

For all natural numbers $$n$$ and complex numbers $$a, b, c, z, w$$ if $$a$$ is an $$n$$-th root of $$z$$ , $$b$$ is an $$n$$-th root of $$w$$, and $$c$$ is an $$n$$-th root of $$zw$$, then $$ab=c$$.

Thanks.

## Suppose $a,b,c$ are roots of $x^3-3x^2-2x+5=0$. How to find $a*(b*c)$?

Let $$x,y\in\mathbb{R}$$ and we define $$*$$ with $$x*y = \frac{{x + y}}{{1 – xy}}$$

Suppose $$a,b,c$$ are roots of $$x^3-3x^2-2x+5=0$$.

How do we find $$a*(b*c)$$?

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## Algebraic inequality for positive reals $a,b,c$

The problem is from a previous maths olympiad and the last step is to prove the inequality

$$4a^4bc + a^4c + 9a^3bc^2 + a^2b^4 + 9ab^2c^3 + 4ab^4c + 4b^3c^3 + b^2c^4 +4abc^4 \geq 24a^2b^2c^2 + 10a^3b^2c + 10a^2bc^3 + 10ab^3c^2$$ for all $$a,b,c > 0$$.

I would very much appreciate if you could help me with how I can proceed from this step and finally solving the problem. Thank you

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## Is this $abc’$ conjecture true of the natural numbers?

Here, the notation $$abc’$$ is identical to $$(abc)’$$ and is a name of a conjecture which is (so to speak) a reduction of the familiar $$abc$$ conjecture. This $$abc’$$ conjecture would be described as below.

Let $$P’$$ be the set of primes defined per the following:

1. If there exists an even number $$e$$ greater than two and is not a sum of two primes, then the prime $$p$$ which is the greatest prime less than $$e$$ is called a prime$$‘$$ number. For instance, if $$10$$ were not a sum of two primes then the prime $$7$$ would be a prime$$‘$$ number.

2. $$P’ = \{p | \text{ p is a prime but not a prime ‘ number}\}$$

Naturally $$P’$$ isn’t empty since, for example, $$2, 3, 5, 7 \in$$ $$P’$$.

The $$abc’$$ conjecture is then defined to be the familiar $$abc$$ conjecture except that all the relevant primes must necessarily be in $$P’$$.

Is this $$abc’$$ conjecture true of the natural numbers?