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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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Would each of them be able to be the side lengths of a triangle?
ratelena [41]
Pythagoras Theorem:
hipotenuse²=leg₁²+leg₂²

First posible triangle:
hypotenuse=13    (13²=169)
leg₁=12                ( 12²=144)
leg₂=5                  (5²=25)

13³=144 + 25


Answer:can be side lengths of a triangle

Second triangle:
 hypotenuse=12.6    (12.6²=158.76)
leg₁=6.7                ( 6.7²=44.89)
leg₂=6.5                  (6.5²=42.25)

leg₁²+leg₂²=44.89+42.25=87.14≠158.76

Answer: cannot be side lenghts of a triangle.

third triangle:
hypotenuse=13    (13²=169)
leg₁=12                ( 12²=144)
leg₂=11                  (11²=121)

leg₁²+leg₂²=144+121=265≠169

Answer: cannot be side lenghts of a triangle.

fourth triangle:
hypotenuse=13   (13²=169)
leg₁=6                ( 6²=36)
leg₂=4                  (4²=16)

leg₁²+leg₁²=36+16=52≠169

Answer: cannot be side lenghts of a triangle.
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M = Y + 5.

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