Question

Using the grammar below: <program> → begin <stmt_list> end <stmt_list> <stmt> | <stmt>; <stmt_list> <stmt> <var> = <expression> <var> → ABC <expression> <var> + <var> | <var> - <var> | <var> 1) show a leftmost derivation and draw a parse tree for each of the statements below: (1) begin A=A-B; B=C; C=A end (2) begin A=B+C; C=C+B end 2) try a rightmost derivation and draw a parse tree for each of the statements in Q1).

Answer 1

Given Grummar is <programs → begin<stmt_list > end <stmt_lists + <stmts] <stmt >; <strot_list> cexpression> < var > — AlBIC <expression > < <Vam >+<vor > I <voor>- svarsl<var> estme > < <var >. = 1 (1) Derivation of begin A = A-B; B=C;C=A end 'Left Most Derivation <programs → begin <stmt_lists end begin <stmt>; <stmt_list > end begin < yours = < expression >; <stmt_list>end begin Aa <expressions; <stmt_listsend > begin A = <vars-var>; <stmt_lists end begin A - A.-<var>; <stmt_list>end → begin A = A - B ; <stmt_list>end → begin A= A-B;; <stmt>; <stmt_list> end → begin A = A-B; evan>:<expression>; <stmt_lists end > begin A= A-B; B = cexpression>; <stmt_list > end begin A=A-B; B=crar>; <stmt_list> end *) begin A = A-B; B=c; <stmta end begin A=A-B; B=C; <vars = cexpression> end 5 begin A=A-B; B-C; C=<erpression> end begin A=A-B; B=C; C=cvac > end > begin A =A-B; B=c; A A; ; C = c = A end *

Parse Tree Camp) < programs end begin <shot_list > <stmt> <shnt_list > estmts. <stmt_list> 1 <strit> <var> <expression> I Kexpression> A <vars 1 А A <von> <vars 1 B B Kvan> <var> ? Kexpression> 1 .C so c А (2) Derivation of begin A =B+C;C=C+B end LMD → begin sprograms → begin<stmt_list> end <stmt>; <stmt_list > end → begin <vars= <expression>; <stmt_l'sts end → begin A = {expression>; <stmt_list send → begin A <var>+<v«i>; <stmt_list send - B+ svar>; <stmt_ust send begin A = B+C;<stont_list send → begin A = B+C; <stmt > end . begin A

3 begin A = B+C; svars= <expression > end begin A = B+C; C: cexpression> end → begin A=B+C; C = <van>+<vars end → begin A =B+C; C = C + cvas> end → begin A=B+C; C = C+B end Parse Tree CLMD) <programs end begin <strot_list> <simt_list> <stmt> 1 <stmt> <var> cexpression> 1 <var> Cexpression> А A <var> 1 <var> + 1 1 B 1 с C svar> <var > с B 1 i 1

2) RMD of begin A = A-B ; B=C;C=A end 1 | . 1 : <programs + begin<stmt_list>end begin <stmt >; <stmt_lists end > begin <stmt>; <stmt>; <stmt_lists end begin estrat>; <stmts; astmt > end > begin <stmt>; <stront>; <varsa cvars= xexpressions and >> begin <stmt>; <stmtz; evar x = <van > end begin <stmt>; <stmt>; evan s'= A end » begin esimt>; <stmt>; C = A end * begin <stmt>; <var >> Kerpression >, C = A end => begin <stmt>; <var> = <vams; C = A end > begin estmt>; <van> = CC = A end => begin <simt>; B=C;C=A end » begin evan>= pexpression >; B=c;c za end es begin <van> = cvari> -< Var>; B-C; CEA end → begin svars - BiB-C; CEA end → begin cvars = A-B; B=C; CEA end → begin Az A-B; B=C;C=A end i z<V01 >

Parse Tree (RMD) 3 <program end begin <shtalists <strmt> <stmt_list> kvar> > <stront> <strst_lists 1 rsht> Kerpression> N. j А <vary 1 svan> <van> 1 <expressions svan> 1 с C B А с c 1 <expression> 1 cvaris I <var> 1 A (3) Rmp of begin A=B+C; C=C+B end eprogram> → begin < stmt_list> end → begin <stmt>; <stmt_list>end * begin <stmt>; <stmts end → begin <stmt>; <van> = <expression > end → begin <strit>; cua> = var>+<vars end in < <simt>; <10 = <var> + Bend Cuarz = C+B end > C =C+ Bend → begin → begin <stmt>; begin <stmt>;

G 1 1 1 begin <vars = rexpression>;C=C+B end s begin cvass = <V001>+cvars; C+C+B end » begin cvars = <V04> + C ;C=C+B end >> begin cvars = B+C;C=C+B end A =B+C; C =C+B end 1 ! i → begin Porse Tree (RMD) <programs <stmt lists end begin <strmt_list> <stmt> 1 . <stint> U <var> Kerpression> + Kexpression> А A <var> 1 B <van> van> 1 с C + с C evar> <var> 1 B
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