SB CSE 675
Fall 2000 |
Program Transformation and Program Analysis
Annie Liu Solution 10 |
Handout S10
Dec. 13, 2000 |

**Problem 1.**

rules: dx, strict op. \dE-s := {} E':= {} s with:= y E'(f(y)) +:= 1 s less:= y E'(f(y)) -:= 1 f :={} E':= {} f(y):=om if y in s then E'(f(y)) -:=1 f(y):=z if y in s then E'(z) +:= 1 Note that, same as in rule M2 in paper, we assume that any use of E'(z) for which z is outside the domain of E' has the value 0 and that any assignment of 0 to E(z) removes z from the domain of E. Otherwise, we just need to add the corresponding code, in particular, for s with:=y, we have if E'(f(y))=om then E'(f(y))=1 else E'(f(y)) +:= 1. for s less:=y, we have if E'(f(y))=1 then E'(f(y))=om else E'(f(y)) -:= 1.

**Problem 2.**

No. Either way we get \dE2,E1= \dE2<\dE1 > \dE2<\dE1 >

**Problem 3.**

Change the expression [[U,X],W] to [[U,W],X].

**Problem 4.**

It is easy to see how this cycle test problem is similar to the tree center problem and the topological sort problem. given: program cycle1; ... s := domain e; while exists x in s| e{x} * s = {} loop s less:= u; end loop; ... end; canonical forms: s*t --> {x in s| x in t} s={} --> #s = 0 exists x in s --> exists x in {x in s} B: while exists x in {x in s| #{y in e{x}| y in s} = 0} s less:= x; expensive computations: E1 = {y in e{x}| y in s} nodes in s that are successors of x based on e E2 = #E1 number of above E3 = {x in s| E2=0} nodes in s that have no successors auxiliary expressions: E1 is needed for any x in s E2 too E4 = {[y,x]:[x,y] in e} maintain: for E1: succ = {[x,y] in e| y in s} so succ{x} = E1 for E2: nsucc= {[x,#succ{x}]: x in domain succ} so nsucc(x) = E2 for E3: tail = {x in s|nsucc(x)=0} for E4: pred = {[y,x]:[x,y] in e} FD: \d succ, nsucc, tail, pred<B> =\d nsucc, tail, pred<\d succ<B>> -- chain rule 1. insert \d-succ<s less:=x> before s less:=x = for u in domain e| [u,x] in e -- rule D2 succ{u} less:= x; = for u in pred{x} -- can be viewed \d pred succ{u} less:= x; 2. replace {y in e{x}| y in s} by succ{x} B1: while exists x in {x in s | #succ{x} = 0} for u in pred{x} succ{u} less:= x; s less:= x; =\d nsucc, tail<B1> =\d tail<\d nsucc<B1>> -- chain rule 1. insert \d-nsucc<succ{u} less:=x> before succ{u} less:=x = nsucc(u) -:= 1 -- rule M2 2. replace #succ{x} by nsucc(x) B2: while exists x in {x in s| nsucc(x) = 0} for u in pred{x} nsucc(u) -:= 1; succ{u} less:= x; s less:= x; =\d tail<B2> 1. insert \d-tail<nsucc(u) -:= 1> before nsucc(u) -:= 1 = if nsucc(u)=1 then tail with:= u; insert \d-tail<s less:= x> before s less:= x = if nsucc(x)=0 then tail less:= x; 2. replace {x in s| nsucc(x) = 0} by tail B3: while exists x in tail for u in pred{x} if nsucc(u)=1 then tail with:= u; nsucc(u) -:= 1; succ{u} less:= x; -- stmt and succ not needed (see later) if nsucc(x)=0 then -- cond always true, since x in tail tail less:= x; s less:= x;