Differences
This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision Next revision | Previous revision | ||
|
sav08:qe_for_presburger_arithmetic [2008/04/21 21:46] david |
sav08:qe_for_presburger_arithmetic [2012/03/09 10:44] vkuncak |
||
|---|---|---|---|
| Line 4: | Line 4: | ||
| We consider elimination of a quantifier from a conjunction of literals (because [[QE from Conjunction of Literals Suffices]]). | We consider elimination of a quantifier from a conjunction of literals (because [[QE from Conjunction of Literals Suffices]]). | ||
| + | |||
| + | Running example: | ||
| + | \[ | ||
| + | \exists y. 3 y - 2 x + 1 > - x \land 2y - 6 < z \land 4 \mid 5y + 1 | ||
| + | \] | ||
| ===== Normalizing Conjunctions of Literals ===== | ===== Normalizing Conjunctions of Literals ===== | ||
| Line 29: | Line 34: | ||
| $t_1 < t_2$ becomes ++| $0 < t_2 - t_1$++ | $t_1 < t_2$ becomes ++| $0 < t_2 - t_1$++ | ||
| - | We obtain a disjunction of conjunctions of literals of the form $0 < t$ and $K \mid t$ where $t$ are | + | We obtain a disjunction of conjunctions of literals of the form $0 < t$ and $K \mid t$ where $t$ are of the form $K_0 + \sum_{i=1}^n K_i \cdot x_i$ |
| ===== Exposing the Variable to Eliminate ===== | ===== Exposing the Variable to Eliminate ===== | ||
| - | By previous transformations, we are eliminating $y$ from conjunction $F(y)$ of $0 < t$ and $K \mid t$ where $t$ is a linear term. To eliminate $\exists y$ from such conjunction, we wish to ensure that the coefficient next to $y$ is one or minus one. Note that $0 < t$ is equivalent to $0 < c\, t$ and $K \mid t$ is equivalent to $c\, K \mid c\, t$ for $c$ a positive integer. If $K_1,\ldots,K_n$ are all coefficients next to $y$ in the formula, let $M$ be a positive integer such that $K_i \mid M$ for all $i$, $1 \le i \le n$ (for example, let $M$ be the least common multiple of $K_1,\ldots,K_n$). Multiply each literal where $y$ occurs in subterm $K_i y$ by constant $M/|K_i|$. | + | By previous transformations, we are eliminating $y$ from conjunction $F(y)$ of $0 < t$ and $K \mid t$ where $t$ is a linear term. |
| + | |||
| + | **Coefficient next to** $y$: | ||
| + | To eliminate $\exists y$ from such conjunction, we wish to ensure that the coefficient next to $y$ is one or minus one. | ||
| + | |||
| + | Observation: $0 < t$ is equivalent to $0 < c\, t$ and $K \mid t$ is equivalent to $c\, K \mid c\, t$ for $c$ a positive integer. | ||
| + | |||
| + | If $K_1,\ldots,K_n$ are all coefficients next to $y$ in the formula, let $M$ be a positive integer such that $K_i \mid M$ for all $i$, $1 \le i \le n$ (for example, let $M$ be the least common multiple of $K_1,\ldots,K_n$). Multiply each literal where $y$ occurs in subterm $K_i y$ by constant $M/|K_i|$. | ||
| What is the coefficient next to $y$ in the resulting formula? ++| either $M$ or $-M$ ++ | What is the coefficient next to $y$ in the resulting formula? ++| either $M$ or $-M$ ++ | ||
| Line 42: | Line 53: | ||
| What is the coefficient next to $y$ in the resulting formula? ++| either $1$ or $-1$ ++ | What is the coefficient next to $y$ in the resulting formula? ++| either $1$ or $-1$ ++ | ||
| + | **Lower and upper bounds:** | ||
| Consider the coefficient next to $x$ in $0 < t$. If it is $-1$, move the term to left side. If it is $1$, move the remaining terms to the left side. We obtain formula $F_1(x)$ of the form | Consider the coefficient next to $x$ in $0 < t$. If it is $-1$, move the term to left side. If it is $1$, move the remaining terms to the left side. We obtain formula $F_1(x)$ of the form | ||
| \[ | \[ | ||
| Line 59: | Line 71: | ||
| ++++ | ++++ | ||
| + | **Replacing variable by test terms:** | ||
| There is a an alternative way to express the above condition by replacing $F_1(x)$ with $\bigvee_k F_1(t_k)$ where $t_k$ do not contain $x$. This is a common technique in quantifier elimination. Note that if $F_1(t_k)$ holds then certainly $\exists x. F_1(x)$. | There is a an alternative way to express the above condition by replacing $F_1(x)$ with $\bigvee_k F_1(t_k)$ where $t_k$ do not contain $x$. This is a common technique in quantifier elimination. Note that if $F_1(t_k)$ holds then certainly $\exists x. F_1(x)$. | ||
| Line 85: | Line 98: | ||
| We first drop all constraints except divisibility, obtaining $F_2(x)$ | We first drop all constraints except divisibility, obtaining $F_2(x)$ | ||
| \[ | \[ | ||
| - | \bigwedge_{i=1}^D K_i \mid (x_i + t_i) | + | \bigwedge_{i=1}^D K_i \mid (x + t_i) |
| \] | \] | ||
| and then eliminate quantifier as | and then eliminate quantifier as | ||
| Line 94: | Line 107: | ||
| That's it! | That's it! | ||
| - | |||
| - | |||
| - | |||
| ===== Example ===== | ===== Example ===== | ||
| Line 112: | Line 122: | ||
| $\exists res, res',i, i'. \neg F$ | $\exists res, res',i, i'. \neg F$ | ||
| - | We can eliminate quantifiers with equalities: $i'= i-1$ | + | We can eliminate quantifiers with equalities: $i'= i-1$ and $res' = res + 2$ |
| - | $res' + 2i'$ becomes $res' + 2(i-1)$, and $\exists i'$ can be removed | + | $res' + 2i'$ becomes $res + 2 + 2(i-1)$, and $\exists i', res' $ can be removed |
| Finally : | Finally : | ||
| $\exists res, i. \neg (res + 2i = 2x \rightarrow res + 2 + 2(i-1) = 2x)$\\ | $\exists res, i. \neg (res + 2i = 2x \rightarrow res + 2 + 2(i-1) = 2x)$\\ | ||
| - | $\exists res, i. \neg (res + 2i = 2x \rightarrow res + 2i = 2x)$ | + | $\exists res, i. \neg (res + 2i = 2x \rightarrow res + 2i = 2x)$\\ |
| + | $\exists res, i. \neg true$\\ | ||
| + | $false$ | ||
| + | |||
| ===== Some Improvements ===== | ===== Some Improvements ===== | ||
| Avoid transforming to conjunctions of literals: work directly on negation-normal form. | Avoid transforming to conjunctions of literals: work directly on negation-normal form. | ||
| + | * the technique is similar to what we described for conjunctive normal form | ||
| - | See Section 7.2 of the [[Calculus of Computation Textbook]]. | + | This is the Cooper's algorithm |
| + | * {{sav09:reddyloveland78presburgerboundedalternation.pdf|Reddy, Loveland: Presburger Arithmetic with Bounded Quantifier Alternation}} (gives a slight improvement of the original Cooper's algorithm) | ||
| + | * Section 7.2 of the [[Calculus of Computation Textbook]] | ||
| ===== References ===== | ===== References ===== | ||
| Line 133: | Line 150: | ||
| See Section 7.2 of the [[Calculus of Computation Textbook]] for a description of more efficient Cooper's algorithm. | See Section 7.2 of the [[Calculus of Computation Textbook]] for a description of more efficient Cooper's algorithm. | ||
| - | [[http://doi.acm.org/10.1145/135226.135233|A practical algorithm for exact array dependence analysis]] (suitable for a project) | + | [[http://doi.acm.org/10.1145/135226.135233|A practical algorithm for exact array dependence analysis]] |