SOLVING EQUATIONS IN FINITE ALGEBRAS Erhard Aichinger Institute - PowerPoint PPT Presentation

SOLVING EQUATIONS IN FINITE ALGEBRAS Erhard Aichinger Institute for Algebra Austrian Science Fund FWF P29931 0/37

Using equation solving for: Graph coloring Task: Color the 10 vertices of the graph with 3 colors. d 1 No vertices connected by an edge may have the same color. Algebraic task: Find c, s 1 , . . . , s 6 , d 1 , . . . , d 3 ∈ R such that s 1 c, s 1 , . . . , d 3 ∈ { 1 , 2 , 3 } , and s 6 s 2 c � = s 1 , . . . , d 2 � = d 3 . c s 5 s 3 s 4 d 3 d 2 1/37

Using equation solving for: Graph coloring Fact Let ( x − 1)( x − 2)( x − 3) = x 3 − 6 x 2 + 11 x − 6 , p ( x ) := = x 2 + xy − 6 x + y 2 − 6 y + 11 . p ( x ) − p ( y ) q ( x, y ) := x − y Then: � For all z ∈ R : z ∈ { 1 , 2 , 3 } iff p ( z ) = 0 . � For ( u, v ) ∈ { 1 , 2 , 3 } × { 1 , 2 , 3 } , we have u � = v iff q ( x, y ) = 0 . 1/37

Using equation solving for: Graph coloring d 1 Algebraic task: Solve p ( c ) = p ( s 1 ) = · · · = p ( d 3 ) = 0 , q ( c, s 1 ) = q ( c, s 2 ) = · · · q ( d 2 , d 3 ) = 0 . s 1 s 6 s 2 c s 5 s 3 s 4 d 3 d 2 1/37

Using equation solving for: Graph coloring d 1 Algebraic task: Solve p ( c ) = p ( s 1 ) = · · · = p ( d 3 ) = 0 , q ( c, s 1 ) = q ( c, s 2 ) = · · · q ( d 2 , d 3 ) = 0 . s 1 Solution of the algebraic task: The s 6 s 2 Gröbner basis of the system is { 1 } . c (Mathematica, 40 ms). s 5 s 3 s 4 Conclusion: No coloring with 3 colors is possible. d 3 d 2 1/37

Using equation solving for: Geometrical Theorem Proving 2/37

Using equation solving for: Geometrical Theorem Proving 2/37

Using equation solving for: Geometrical Theorem Proving 2/37

Using equation solving for: Geometrical Theorem Proving 2/37

Using equation solving for: Geometrical Theorem Proving 2/37

Using equation solving for: Geometrical Theorem Proving 2/37

Using equation solving for: Geometrical Theorem Proving Theorem (Pappus of Alexandria, ∼ 320) Let A, B, C, D, E, F, H, I, J points in the F plane such that each of the following E triples is collinear: ( A, B, C ) , ( D, E, F ) , ( A, H, E ) , D ( D, H, B ) , ( D, I, C ) , ( A, I, F ) , ( E, J, C ) , J H ( B, J, F ) . I A B C Then ( H, I, J ) are collinear. 2/37

Using equation solving for: Geometrical Theorem Proving Theorem (Pappus of Alexandria, ∼ 320) Let A, B, C, D, E, F, H, I, J points in the F plane such that each of the following E triples is collinear: ( A, B, C ) , ( D, E, F ) , ( A, H, E ) , D ( D, H, B ) , ( D, I, C ) , ( A, I, F ) , ( E, J, C ) , J H ( B, J, F ) . I Assume that ( A, B, D ) and ( A, B, E ) are A B C not collinear. Then ( H, I, J ) are collinear. 2/37

Using equation solving for: Geometrical Theorem Proving Theorem. Suppose that ( A, B, C ) , ( D, E, F ) , ( A, H, E ) , ( D, H, B ) , ( D, I, C ) , ( A, I, F ) , ( E, J, C ) , ( B, J, F ) are collinear, and that ( A, B, D ) and ( A, B, E ) are not collinear. Then ( H, I, J ) is collinear. Proof: � We try to construct a counterexample. � We coordinatize points with pairs of real numbers: A = ( a 1 , a 2 ) , . . . , J = ( j 1 , j 2 ) . � � u 1 u 2 1 � C ( u 1 , u 2 , v 1 , v 2 , w 1 , w 2 ) := det( ) = v 1 v 2 1 w 1 w 2 1 − u 2 v 1 + u 1 v 2 + u 2 w 1 − v 2 w 1 − u 1 w 2 + v 1 w 2 has the property: C ( u 1 , u 2 , v 1 , v 2 , w 1 , w 2 ) = 0 iff (( u 1 u 2 ) , ( v 1 v 2 ) , ( w 1 w 2 )) is collinear. 3/37

Using equation solving for: Geometrical Theorem Proving Theorem. Suppose that ( A, B, C ) , ( D, E, F ) , ( A, H, E ) , ( D, H, B ) , ( D, I, C ) , ( A, I, F ) , ( E, J, C ) , ( B, J, F ) are collinear, and that ( A, B, D ) and ( A, B, E ) are not collinear. Then ( H, I, J ) is collinear. Proof: � A counterexample has to satisfy C ( a 1 , a 2 , b 1 , b 2 , c 1 , c 2 ) = · · · = C ( b 1 , b 2 , j 1 , j 2 , f 1 , f 2 ) = 0 , C ( a 1 , a 2 , b 1 , b 2 , d 1 , d 2 ) � = 0 , C ( a 1 , a 2 , b 1 , b 2 , e 1 , e 2 ) � = 0 , C ( h 1 , h 2 , i 1 , i 2 , j 1 , j 2 ) � = 0 . 3/37

Using equation solving for: Geometrical Theorem Proving Theorem. Suppose that ( A, B, C ) , ( D, E, F ) , ( A, H, E ) , ( D, H, B ) , ( D, I, C ) , ( A, I, F ) , ( E, J, C ) , ( B, J, F ) are collinear, and that ( A, B, D ) and ( A, B, E ) are not collinear. Then ( H, I, J ) is collinear. Proof: � A counterexample has to satisfy C ( a 1 , a 2 , b 1 , b 2 , c 1 , c 2 ) = · · · = C ( b 1 , b 2 , j 1 , j 2 , f 1 , f 2 ) = 0 , C ( a 1 , a 2 , b 1 , b 2 , d 1 , d 2 ) · z 1 = 1 , C ( a 1 , a 2 , b 1 , b 2 , e 1 , e 2 ) � = 0 , C ( h 1 , h 2 , i 1 , i 2 , j 1 , j 2 ) � = 0 . 3/37

Using equation solving for: Geometrical Theorem Proving Theorem. Suppose that ( A, B, C ) , ( D, E, F ) , ( A, H, E ) , ( D, H, B ) , ( D, I, C ) , ( A, I, F ) , ( E, J, C ) , ( B, J, F ) are collinear, and that ( A, B, D ) and ( A, B, E ) are not collinear. Then ( H, I, J ) is collinear. Proof: � A counterexample has to satisfy C ( a 1 , a 2 , b 1 , b 2 , c 1 , c 2 ) = · · · = C ( b 1 , b 2 , j 1 , j 2 , f 1 , f 2 ) = 0 , C ( a 1 , a 2 , b 1 , b 2 , d 1 , d 2 ) · z 1 = 1 , C ( a 1 , a 2 , b 1 , b 2 , e 1 , e 2 ) · z 2 = 1 , C ( h 1 , h 2 , i 1 , i 2 , j 1 , j 2 ) · z 3 = 1 . � The theorem holds if and only if this system of equations has no solution in the real numbers. 3/37

Using equation solving for: Geometrical Theorem Proving We use the computer algebra system “Mathematica”. 4/37

Using equation solving for: Geometrical Theorem Proving Conclusions � There is no counterexample to Pappus’s Theorem, not even in the complex plane C 2 . � Hence (this version) of Pappus’s Theorem holds. � Similar proofs for: Desargues, Ceva, Menelaus, . . . . � Algebraic way to decide which first order formulae hold in the relational structure L = ( C 2 , IsCollinearTriple ( x, y, z )) by solving systems of polynomial equations . What about other axiomatizations or calculi for this structure L ? 5/37

Using equation solving for: Boolean Satisfiability Given: Equations over the algebraic structure ( B = { 0 , 1 } ; ∧ , ∨ , ¬ , 0 , 1) , e. g. x 1 ∨ x 2 ∨ x 3 = 1 x 2 ∨ ¬ x 3 = 1 x 1 ∨ ¬ x 3 = 1 . Asked: Does this system have a solution? 6/37

Using equation solving for: Boolean Satisfiability Given: Equations over the algebraic Given: Equations over the algebraic structure ( B = { 0 , 1 } ; ∧ , ∨ , ¬ , 0 , 1) , e. g. structure ( F 2 = { 0 , 1 } ; + , · , 0 , 1) , x 1 ∨ x 2 ∨ x 3 = 1 x 1 + x 2 + x 3 + x 1 x 2 + x 1 x 3 x 2 ∨ ¬ x 3 = 1 + x 2 x 3 + x 1 x 2 x 3 + 1 = 0 x 1 ∨ ¬ x 3 = 1 . x 3 + x 2 x 3 = 0 x 3 + x 1 x 3 = 0 . Asked: Does this system have a solution? Asked: Does this system have a solution? 6/37

Using equation solving for: Boolean Satisfiability Given: Equations over the algebraic Given: Equations over the algebraic structure ( B = { 0 , 1 } ; ∧ , ∨ , ¬ , 0 , 1) , e. g. structure ( F 2 = { 0 , 1 } ; + , · , 0 , 1) , x 1 ∨ x 2 ∨ x 3 = 1 x 1 + x 2 + x 3 + x 1 x 2 + x 1 x 3 x 2 ∨ ¬ x 3 = 1 + x 2 x 3 + x 1 x 2 x 3 + 1 = 0 x 1 ∨ ¬ x 3 = 1 . x 3 + x 2 x 3 = 0 x 3 + x 1 x 3 = 0 . Asked: Does this system have a solution? Asked: Does this system have a solution? 3SAT is known to be computationally hard ( NP -complete). 6/37

Using equation solving for: Boolean Satisfiability Given: Equations over the algebraic Given: Equations over the algebraic structure ( B = { 0 , 1 } ; ∧ , ∨ , ¬ , 0 , 1) , e. g. structure ( F 2 = { 0 , 1 } ; + , · , 0 , 1) , x 1 ∨ x 2 ∨ x 3 = 1 x 1 + x 2 + x 3 + x 1 x 2 + x 1 x 3 x 2 ∨ ¬ x 3 = 1 + x 2 x 3 + x 1 x 2 x 3 + 1 = 0 x 1 ∨ ¬ x 3 = 1 . x 3 + x 2 x 3 = 0 x 3 + x 1 x 3 = 0 . Asked: Does this system have a solution? Asked: Does this system have a solution? 3SAT is known to be computationally Hence solving polynomial systems over hard ( NP -complete). F 2 is also hard ( NP -complete). 6/37

Solving equations over finite fields Given: f 1 , . . . , f s ∈ F 2 [ x 1 , . . . , x N ] . Asked: ∃ a ∈ F N 2 : f 1 ( a ) = · · · = f s ( a ) = 0 . Computational Complexity: Restrictions 1 eqn. s eqns. none none f 1 , . . . , f s in expanded form deg( f i ) ≤ D 7/37

Solving equations over finite fields Given: f 1 , . . . , f s ∈ F 2 [ x 1 , . . . , x N ] . Asked: ∃ a ∈ F N 2 : f 1 ( a ) = · · · = f s ( a ) = 0 . Computational Complexity: Restrictions 1 eqn. s eqns. none none NP -comp. NP -comp. NP -comp. f 1 , . . . , f s in expanded form NP -comp. deg( f i ) ≤ D NP -comp if D ≥ 2 . 7/37

Solving equations over finite fields Given: f 1 , . . . , f s ∈ F 2 [ x 1 , . . . , x N ] . Asked: ∃ a ∈ F N 2 : f 1 ( a ) = · · · = f s ( a ) = 0 . Computational Complexity: Restrictions 1 eqn. s eqns. none none NP -comp. NP -comp. NP -comp. f 1 , . . . , f s in expanded form P NP -comp. deg( f i ) ≤ D NP -comp if D ≥ 2 . 7/37

Solving equations over finite fields Given: f 1 , . . . , f s ∈ F 2 [ x 1 , . . . , x N ] . Asked: ∃ a ∈ F N 2 : f 1 ( a ) = · · · = f s ( a ) = 0 . Computational Complexity: Restrictions 1 eqn. s eqns. none none NP -comp. NP -comp. NP -comp. f 1 , . . . , f s in expanded form P P NP -comp. deg( f i ) ≤ D NP -comp if D ≥ 2 . 7/37

Solving equations over finite fields Given: f 1 , . . . , f s ∈ F 2 [ x 1 , . . . , x N ] . Asked: ∃ a ∈ F N 2 : f 1 ( a ) = · · · = f s ( a ) = 0 . Computational Complexity: Restrictions 1 eqn. s eqns. none none NP -comp. NP -comp. NP -comp. f 1 , . . . , f s in expanded form P P NP -comp. deg( f i ) ≤ D P P NP -comp if D ≥ 2 . 7/37