2D Computer Graphics Resultants and implicitization Two types of - PowerPoint PPT Presentation

Summer 2020 Diego Nehab IMPA 1 2D Computer Graphics

Resultants and implicitization

Two types of renderers and their applications • Traditional vs. vector textures • Amortized vs. random access • In both cases, parallelism is key Amortized renderers need actual point of intersection Random access only need to count them Can count using implicit tests For that, we will use resultants 2 Why we need resultants

Two types of renderers and their applications • Traditional vs. vector textures • Amortized vs. random access • In both cases, parallelism is key Amortized renderers need actual point of intersection Random access only need to count them Can count using implicit tests For that, we will use resultants 2 Why we need resultants

Two types of renderers and their applications • Traditional vs. vector textures • Amortized vs. random access • In both cases, parallelism is key Amortized renderers need actual point of intersection Random access only need to count them Can count using implicit tests For that, we will use resultants 2 Why we need resultants

Two types of renderers and their applications • Traditional vs. vector textures • Amortized vs. random access • In both cases, parallelism is key Amortized renderers need actual point of intersection Random access only need to count them Can count using implicit tests For that, we will use resultants 2 Why we need resultants

Two types of renderers and their applications • Traditional vs. vector textures • Amortized vs. random access • In both cases, parallelism is key Amortized renderers need actual point of intersection Random access only need to count them Can count using implicit tests For that, we will use resultants 2 Why we need resultants

Two types of renderers and their applications • Traditional vs. vector textures • Amortized vs. random access • In both cases, parallelism is key Amortized renderers need actual point of intersection Random access only need to count them Can count using implicit tests For that, we will use resultants 2 Why we need resultants

Two types of renderers and their applications • Traditional vs. vector textures • Amortized vs. random access • In both cases, parallelism is key Amortized renderers need actual point of intersection Random access only need to count them Can count using implicit tests For that, we will use resultants 2 Why we need resultants

x p y p f p t g p t Polynomials f p and g p have a common root at t . one-variable polynomials f p and y p have a common root. 3 t can be rewritten as p that vanishes if and only if two We need a bivariate polynomial 0 y p y t 0 x p x t x t y t The condition p ? Given expressions for x and y , how do we obtain an expression for if Implicit vs. explicit We say that a bivariate polynomial Γ( u , v ) is the implicit form of a � � parametric polynomial curve γ ( t ) = x ( t ) , y ( t ) Γ( p ) = 0 ⇐ ⇒ ∃ t | p = γ ( t )

x p y p f p t g p t Polynomials f p and g p have a common root at t . one-variable polynomials f p and y p have a common root. 3 t can be rewritten as p that vanishes if and only if two We need a bivariate polynomial 0 y p y t 0 x p x t y t x t The condition p if Implicit vs. explicit We say that a bivariate polynomial Γ( u , v ) is the implicit form of a � � parametric polynomial curve γ ( t ) = x ( t ) , y ( t ) Γ( p ) = 0 ⇐ ⇒ ∃ t | p = γ ( t ) Given expressions for x and y , how do we obtain an expression for Γ ?

Polynomials f p and g p have a common root at t . one-variable polynomials f p and y p have a common root. 3 if p that vanishes if and only if two We need a bivariate polynomial Implicit vs. explicit We say that a bivariate polynomial Γ( u , v ) is the implicit form of a � � parametric polynomial curve γ ( t ) = x ( t ) , y ( t ) Γ( p ) = 0 ⇐ ⇒ ∃ t | p = γ ( t ) Given expressions for x and y , how do we obtain an expression for Γ ? � � The condition p = ( x p , y p ) = x ( t ) , y ( t ) = γ ( t ) can be rewritten as  f p ( t ) = x ( t ) − x p = 0  g p ( t ) = y ( t ) − y p = 0 

one-variable polynomials f p and y p have a common root. 3 if p that vanishes if and only if two We need a bivariate polynomial Implicit vs. explicit We say that a bivariate polynomial Γ( u , v ) is the implicit form of a � � parametric polynomial curve γ ( t ) = x ( t ) , y ( t ) Γ( p ) = 0 ⇐ ⇒ ∃ t | p = γ ( t ) Given expressions for x and y , how do we obtain an expression for Γ ? � � The condition p = ( x p , y p ) = x ( t ) , y ( t ) = γ ( t ) can be rewritten as  f p ( t ) = x ( t ) − x p = 0  g p ( t ) = y ( t ) − y p = 0  Polynomials f p and g p have a common root at t .

3 if Implicit vs. explicit We say that a bivariate polynomial Γ( u , v ) is the implicit form of a � � parametric polynomial curve γ ( t ) = x ( t ) , y ( t ) Γ( p ) = 0 ⇐ ⇒ ∃ t | p = γ ( t ) Given expressions for x and y , how do we obtain an expression for Γ ? � � The condition p = ( x p , y p ) = x ( t ) , y ( t ) = γ ( t ) can be rewritten as  f p ( t ) = x ( t ) − x p = 0  g p ( t ) = y ( t ) − y p = 0  Polynomials f p and g p have a common root at t . We need a bivariate polynomial Γ( p ) that vanishes if and only if two one-variable polynomials f p and y p have a common root.

We call R f p g p the resultant of f p g p knowledge of the roots of f p and g p ? 4 depend on p , of course, we could write for sums of products of roots! r It makes sense that there should be! Think about the Vieta formulas s Is there an expression for the resultant that does not require The resultant If we knew the roots of a 1 , a 2 , . . . , a r of f p and b 1 , b 2 , . . . , b s of g p , which � � R ( f p , g p ) = ( a i − b j ) i = 1 j = i

knowledge of the roots of f p and g p ? 4 Is there an expression for the resultant that does not require depend on p , of course, we could write for sums of products of roots! r It makes sense that there should be! Think about the Vieta formulas s The resultant If we knew the roots of a 1 , a 2 , . . . , a r of f p and b 1 , b 2 , . . . , b s of g p , which � � R ( f p , g p ) = ( a i − b j ) i = 1 j = i We call R ( f p , g p ) the resultant of f p , g p

4 s depend on p , of course, we could write for sums of products of roots! r It makes sense that there should be! Think about the Vieta formulas Is there an expression for the resultant that does not require The resultant If we knew the roots of a 1 , a 2 , . . . , a r of f p and b 1 , b 2 , . . . , b s of g p , which � � R ( f p , g p ) = ( a i − b j ) i = 1 j = i We call R ( f p , g p ) the resultant of f p , g p knowledge of the roots of f p and g p ?

4 s depend on p , of course, we could write for sums of products of roots! r It makes sense that there should be! Think about the Vieta formulas Is there an expression for the resultant that does not require The resultant If we knew the roots of a 1 , a 2 , . . . , a r of f p and b 1 , b 2 , . . . , b s of g p , which � � R ( f p , g p ) = ( a i − b j ) i = 1 j = i We call R ( f p , g p ) the resultant of f p , g p knowledge of the roots of f p and g p ?

5 g t g t r t f t s t These polynomials are identical , so all coeffjcients must be the same g t r t h t r t s t f t s t We can eliminate h from the equations by noticing that h t s t and Let f and g have a common root and let h t r t f t 1 such that h There is h with and The Sylvester form for the resultant deg( f ) = m deg( g ) = n

5 h t r t s t g t r t and f t s t These polynomials are identical , so all coeffjcients must be the same g t r t and Let f and g have a common root and let We can eliminate h from the equations by noticing that f t s t The Sylvester form for the resultant deg( f ) = m deg( g ) = n There is h with deg( h ) = 1 such that f ( t ) = h ( t ) r ( t ) g ( t ) = h ( t ) s ( t )

Let f and g have a common root and let and and We can eliminate h from the equations by noticing that These polynomials are identical , so all coeffjcients must be the same f t s t g t r t 5 The Sylvester form for the resultant deg( f ) = m deg( g ) = n There is h with deg( h ) = 1 such that f ( t ) = h ( t ) r ( t ) g ( t ) = h ( t ) s ( t ) f ( t ) s ( t ) = h ( t ) r ( t ) s ( t ) = g ( t ) r ( t )

Let f and g have a common root and let and and We can eliminate h from the equations by noticing that These polynomials are identical , so all coeffjcients must be the same 5 The Sylvester form for the resultant deg( f ) = m deg( g ) = n There is h with deg( h ) = 1 such that f ( t ) = h ( t ) r ( t ) g ( t ) = h ( t ) s ( t ) f ( t ) s ( t ) = h ( t ) r ( t ) s ( t ) = g ( t ) r ( t ) f ( t ) s ( t ) = g ( t ) r ( t )

r 0 r 1 1 of r and s 0 s 1 1 of s , but we know the coeffjcients f i , g j of f and g g 2 r 0 g 1 r 1 g 0 r 2 . . 6 . f 1 s 1 f m s n 1 g n r m 1 f 0 s 2 g 1 r 0 f 2 s 0 g 0 r 1 These polynomials are identical , so all coeffjcients must be the same f 0 s 1 f 1 s 0 g 0 r 0 f 0 s 0 The coeffjcient equations are s n r m We do not know the value of t , so we don’t know the coeffjcients of The Sylvester form for the resultant f ( t ) s ( t ) = g ( t ) r ( t )