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Sliva [168]
3 years ago
11

Get the algorithm to remove the indirect left recursion from a grammar from Aho et al. (2006). Use this algorithm to remove all

left recursion from the following grammar: S→ Aa | Bb A→ Aa | Abc | c I Sb B → bb
Mathematics
1 answer:
Advocard [28]3 years ago
6 0

Answer:

Step-by-step explanation:

Here is the gramamr

S -> Aa | Bb

A -> Aa | Abc | c | Sb

B -> bb

-----------------------------------------------------------------------------------------------------------------------

Now for first gramamr step introduce new symbol S'

S -> BbS'

S' -> e | S'aA

----------------------------------------------------------------------------------------------------------------------------------------------------

Now for second gramamr statement...

A -> SbA' | cA'

A' -> e | A'a | A'bc

---------------------------------------------------------------------------------------------------------------------------------------------

For third grammar statement it is not required since it doesn't contain any recursion

So final gramamr will be

S -> BbS'

S' -> e | S'aA

A -> SbA' | cA'

A' -> e | A'a | A'bc

B -> bb

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2 years ago
Determine la ecuación general de la recta que pasa por los puntos (1,4); (- 2, - 5) y grafíquela.
Rudiy27

The equation of the line that passes through the points (1 , 4) and (-2, -5) is y = 3x + 1

<h3>Further explanation</h3>

Solving linear equation mean calculating the unknown variable from the equation.

Let the linear equation : y = mx + c

If we draw the above equation on Cartesian Coordinates , it will be a straight line with :

<em>m → gradient of the line</em>

<em>( 0 , c ) → y - intercept</em>

Gradient of the line could also be calculated from two arbitrary points on line ( x₁ , y₁ ) and ( x₂ , y₂ ) with the formula :

\large {\boxed {m = \frac{y_2 - y_1}{x_2 - x_1}} }

If point ( x₁ , y₁ ) is on the line with gradient m , the equation of the line will be :

\large {\boxed{y - y_1 = m ( x - x_1 )} }

Let us tackle the problem.

Let :

(1 , 4) → (x₁ , y₁)

(-2, -5) → (x₂ , y₂)

To find the straight line equation, the following formula can be used :

\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}

\frac{y - 4}{-5 - 4} = \frac{x - 1}{-2 - 1}

\frac{y - 4}{-9} = \frac{x - 1}{-3}

\frac{y - 4}{3} = \frac{x - 1}{1}

y - 4 = 3 ( x - 1 )

y = 3x - 3 + 4

\large {\boxed {y = 3x + 1} }

<h3>Learn more</h3>
  • Infinite Number of Solutions : brainly.com/question/5450548
  • System of Equations : brainly.com/question/1995493
  • System of Linear equations : brainly.com/question/3291576

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Linear Equations

Keywords: Linear , Equations , 1 , Variable , Line , Gradient , Point

4 0
3 years ago
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statuscvo [17]

It's true. That's how parallelograms are defined. Here's a diagram to show you what the question means. The opposite sides are parallel and there are 2 sets of parallel lines.

3 0
3 years ago
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oee [108]

Answer:

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Step-by-step explanation:

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Marta_Voda [28]

Answer:

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2 years ago
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