2B) Program to recognize whether a given sentence is simple or compound. [VTU SS&OS]

Code:

Algorithm:

Step1: Start

Step2: Read Input file Name

Step3: Split the input into tokens

Step4: Check if conjunctive is present, then proceed to Step 5 else goto Step6

Step5: Print the message as Compound Statement

Step6: Print the message as Simple Statement

Step7: Stop

Explanation:
This program tells whether the input given is compound sentence or simple one. Its easy, just check the scanned symbol with some conjunctives and if they are found, then its compound sentence. Else a simple one. We use a flag variable in this code which is initially 0 indicating a simple sentence. If any conjunctives are matched, set the flag to 1. at the end, if flag=1 then compound sentence, else simple sentence.

%{
int flag=0;

%}

Declaration section with flag variables set to 0.

%%
.+[ \t]+[Aa][Nn][Dd][ \t]+.+ {flag=1;}
.+[ \t]+[Oo][Rr][ \t]+.+ {flag=1;}
.+[ \t]+[Bb][Uu][Tt][ \t]+.+ {flag=1;}
.+[ \t]+[Bb][eE][Cc][Aa][Uu][Ss][Ee][ \t]+.+ {flag=1;}
.+[ \t]+[Nn][Ee][Vv][Ee][Rr][Tt][Hh][Ee][Ll][Ee][Ss][Ss][ \t]+.+ {flag=1;}
.  ;

%%

Scan the input symbol with any of the Conjunctives like and, or, but, because, nevertheless. if matched, set flag to 1.

regex for these conjunctives are self explanatory. see basics of Lex article to learn what is regex.

main()
{
printf(“Enter text:\n”);
yylex();
if (flag==1)
printf(“Compound Sentence”);
else
printf(“Simple Sentence”);
}

This is user subroutine. ask user for input sentence. and call yylex() for tokenising.

At the end, if flag=1, print compound. else, print simple sentence.

Author : Unknown

He is a simple passionate tech freak who himself is an engineering student at Canara Engineering college. He likes App Development, Web designing, Blogging, Youtubing, Debugging and also is a CodeGeek!

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