LARA

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revision Previous revision
Next revision
Previous revision
Next revision Both sides next revision
sav08:interval_analysis_and_widening [2008/05/20 19:53]
vkuncak
sav08:interval_analysis_and_widening [2008/05/20 20:34]
vkuncak
Line 40: Line 40:
 \] \]
  
-Another approach: iterate a few times with $H_i$ only (without using $w$), if not a fixpoint at this program point, then widen. +Approaches: 
-  This is not monotonic: if you start at fixpoint, it converges, if start below, can jump over fixpoint+  * always apply widening (we will assume this) 
 +  * iterate a few times with $H_i$ only (without using $w$), if not a fixpoint at this program point, then widen. 
 +       ​this is not monotonic: if you start at fixpoint, it converges, if start below, can jump over fixpoint
  
 Standard iteration: $\bot, \ldots, (F^{\#​})^n(\bot),​ \ldots$ Standard iteration: $\bot, \ldots, (F^{\#​})^n(\bot),​ \ldots$
 +
 Widening: $\bot, \ldots, ((W \circ F)^{\#​})^n(\bot),​ \ldots$ Widening: $\bot, \ldots, ((W \circ F)^{\#​})^n(\bot),​ \ldots$
  
-Narrowing: apply fixpoint iteration after widening to improve precision. 
  
-Why will narrowing improve the result?+Here, $x \sqsubseteq W(x)$ for all $x$ by definition of $W$ 
 + 
 +Narrowing: after finding fixpoint of $(W \circ F)^{\#}$, apply $F^{\#}$ to improve precision. 
 + 
 +Observation:​ if $F^{\#}$ and $W$ are $\omega$-continuous functions and $x \sqsubseteq W(x)$ for all $x$, then narrowing ​will improve the result, that is, if $x_* = lfp (F^{\#})$ and $y_* = lfp (W \circ F^{\#})$, then $x_* \sqsubseteq y_*$ and 
 +\[ 
 +    x_* = F(x_*) \sqsubseteq F^{\#}(y_*) \sqsubseteq (W \circ F^{\#​})(y_*) \sqsubseteq y_* 
 +\]