COP 5021 meeting -*- Outline -*-

* constraint based analysis (1.3.2 and 1.4)

      Based on
       section 1.4 of Nielson, Nielson and Hankin's book
       Program Analysis (Springer, 1999)
      and the article:
       Aiken, Alexander. "Introduction to set constraint-based program
       analysis." Science of Computer Programming 35, no. 2-3 (1999):
       79-111.
       https://www.sciencedirect.com/science/article/pii/S0167642399000076/pdf
** goals

  The main goal is a modular analysis...

  Q:  What's the difference between a data flow analysis
      and a constraint-based analysis?
------------------------------------------
  DATA FLOW VS. CONSTRAINT-BASED ANALYSIS

Main differences?







------------------------------------------
    ... for the data flow analysis the constraints
        are the equivalent of two subset constraints
           (i.e., == is <= and >=)
        so are: - more restrictive vs. more expressive)
                - have to be complete all in one place (vs. compositional)

** example
------------------------------------------
   REACHING DEFINITIONS USING CONSTRAINTS

For the program:

   [y := x]^1;
   [z := 1]^2;
   while [y>1]^3
   do ([z:=z*y]^4;
       [y:=y-1]^5);
   [y:= 0]^6


























------------------------------------------
...
RDexit(1) \supseteq RDentry(1)
                     \ {(y,l)|l in Lab*}
RDexit(1) \supseteq {(y,1)}
RDexit(2) \supseteq RDentry(2)
                     \ {(z,l)|l in Lab*}
RDexit(1) \supseteq {(z,2)}
RDexit(3) \supseteq RDentry(3)
RDexit(4) \supseteq RDentry(4)
                     \ {(z,l)|l in Lab*}
RDexit(4) \supseteq {(z,4)}
RDexit(5) \supseteq RDentry(5)
                     \ {(y,l)|l in Lab*}
RDexit(5) \supseteq {(y,5)}
RDexit(6) \supseteq RDentry(6)
                     \ {(y,l)|l in Lab*}
RDexit(5) \supseteq {(y,5)}

RDentry(1) \supseteq {(x,?),(y,?),(z,?)}
RDentry(2) \supseteq RDexit(1)
RDentry(3) \supseteq RDexit(2)
RDentry(3) \supseteq RDexit(5)
RDentry(4) \supseteq RDexit(3)
RDentry(5) \supseteq RDexit(4)
RDentry(6) \supseteq RDexit(3)


    Note that there is for an assignment there is:
       - a constraint that takes out the variable being assigned
       - a (separate) constraint that adds in variable being assigned

    Note that the entry constraints directly encode the control flows
       and there are two for the entry into 3.

** Solutions
   Also want the least solution, which works here too
   as it's more precise.

   The least solution exists and is the same