Discovering the roots: Uniform closure results for algebraic classes - PowerPoint PPT Presentation

Theorem 1 i f i • Assumption: is algebraically closed and characteristic=0 Any factor g of a polynomial f computed by a circuit of size s has size poly s deg(rad f . • The degree of square-free part is polynomially bounded size “any” factor is!(and factor conjecture is true in this case!) • This subsumes both the results of Kaltofen 9 Relating Squarefree part to Complexity • For f = ∏ i f e i i , define radical to be rad ( f ) = ∏

Theorem 1 i f i Any factor g of a polynomial f computed by a circuit of size s has size poly s deg(rad f . • The degree of square-free part is polynomially bounded size “any” factor is!(and factor conjecture is true in this case!) • This subsumes both the results of Kaltofen 9 Relating Squarefree part to Complexity • For f = ∏ i f e i i , define radical to be rad ( f ) = ∏ • Assumption: F is algebraically closed and characteristic=0

i f i Any factor g of a polynomial f computed by a circuit of size s has • The degree of square-free part is polynomially bounded size “any” factor is!(and factor conjecture is true in this case!) • This subsumes both the results of Kaltofen 9 Relating Squarefree part to Complexity • For f = ∏ i f e i i , define radical to be rad ( f ) = ∏ • Assumption: F is algebraically closed and characteristic=0 Theorem 1 size poly ( s , deg(rad ( f )) .

i f i Any factor g of a polynomial f computed by a circuit of size s has size “any” factor is!(and factor conjecture is true in this case!) • This subsumes both the results of Kaltofen 9 Relating Squarefree part to Complexity • For f = ∏ i f e i i , define radical to be rad ( f ) = ∏ • Assumption: F is algebraically closed and characteristic=0 Theorem 1 size poly ( s , deg(rad ( f )) . • The degree of square-free part is polynomially bounded = ⇒

i f i Any factor g of a polynomial f computed by a circuit of size s has size “any” factor is!(and factor conjecture is true in this case!) • This subsumes both the results of Kaltofen 9 Relating Squarefree part to Complexity • For f = ∏ i f e i i , define radical to be rad ( f ) = ∏ • Assumption: F is algebraically closed and characteristic=0 Theorem 1 size poly ( s , deg(rad ( f )) . • The degree of square-free part is polynomially bounded = ⇒

Factoring Reduces to Root Approximation

This is root finding as f x g 1. guess a good starting point x 0 10 approximation ? What is the starting point ? • Can we do similar thing to find g ? If yes, what is the notion of f x n f x n x n 1 2. calculate x n the following: 0. Idea is 0. Assume f x approximation of x such that f x • (Newton Iteration) Suppose we want to find “good enough“ 0. g ? Finding linear factor • Suppose f ( x , y ) = ( y − g ( x )) · u ( x , y ) where y − g ∤ u . Can we find

1. guess a good starting point x 0 2. calculate x n approximation ? What is the starting point ? • Can we do similar thing to find g ? If yes, what is the notion of f x n f x n x n 1 10 the following: 0. Idea is 0. Assume f x approximation of x such that f x • (Newton Iteration) Suppose we want to find “good enough“ Finding linear factor • Suppose f ( x , y ) = ( y − g ( x )) · u ( x , y ) where y − g ∤ u . Can we find g ? This is root finding as f ( x , g ) = 0.

1. guess a good starting point x 0 • (Newton Iteration) Suppose we want to find “good enough“ the following: 2. calculate x n 1 x n f x n f x n • Can we do similar thing to find g ? If yes, what is the notion of approximation ? What is the starting point ? 10 Finding linear factor • Suppose f ( x , y ) = ( y − g ( x )) · u ( x , y ) where y − g ∤ u . Can we find g ? This is root finding as f ( x , g ) = 0. approximation of x such that f ( x ) = 0. Assume f ′ ( x ) ̸ = 0. Idea is

• (Newton Iteration) Suppose we want to find “good enough“ the following: 2. calculate x n 1 x n f x n f x n • Can we do similar thing to find g ? If yes, what is the notion of approximation ? What is the starting point ? 10 Finding linear factor • Suppose f ( x , y ) = ( y − g ( x )) · u ( x , y ) where y − g ∤ u . Can we find g ? This is root finding as f ( x , g ) = 0. approximation of x such that f ( x ) = 0. Assume f ′ ( x ) ̸ = 0. Idea is 1. guess a good starting point x 0

• (Newton Iteration) Suppose we want to find “good enough“ the following: • Can we do similar thing to find g ? If yes, what is the notion of approximation ? What is the starting point ? 10 Finding linear factor • Suppose f ( x , y ) = ( y − g ( x )) · u ( x , y ) where y − g ∤ u . Can we find g ? This is root finding as f ( x , g ) = 0. approximation of x such that f ( x ) = 0. Assume f ′ ( x ) ̸ = 0. Idea is 1. guess a good starting point x 0 2. calculate x n + 1 = x n − f ( x n ) f ′ ( x n )

• (Newton Iteration) Suppose we want to find “good enough“ the following: • Can we do similar thing to find g ? If yes, what is the notion of approximation ? What is the starting point ? 10 Finding linear factor • Suppose f ( x , y ) = ( y − g ( x )) · u ( x , y ) where y − g ∤ u . Can we find g ? This is root finding as f ( x , g ) = 0. approximation of x such that f ( x ) = 0. Assume f ′ ( x ) ̸ = 0. Idea is 1. guess a good starting point x 0 2. calculate x n + 1 = x n − f ( x n ) f ′ ( x n )

f x y t f x y t . Can we say that y t is an approximation • If f x y t is invertible, then one can show that g mod x 2 t g mod x 2 t • f x y t is invertible f 0 • If f 0 1 where deg g calculating y log d 0 11 0 and f 0 0 0. Then, one can find g by d . 0 1 x f x y t 1 y t y t of g ? y t 1 • Define y t Finding linear factor Continued • Initial starting point y 0 = µ where µ : = g ( 0 )

• If f x y t is invertible, then one can show that g mod x 2 t g mod x 2 t • f x y t is invertible f 0 • If f 0 1 where deg g calculating y log d 11 0 d . 0. Then, one can find g by 0 and f 0 0 f x y t 0 x 1 1 y t y t of g ? Finding linear factor Continued • Initial starting point y 0 = µ where µ : = g ( 0 ) • Define y t + 1 = y t − f ( x , y t ) f ′ ( x , y t ) . Can we say that y t is an approximation

• f x y t is invertible f 0 • If f 0 1 where deg g calculating y log d 11 0 d . 0. Then, one can find g by 0 and f 0 0 0 x f x y t of g ? Finding linear factor Continued • Initial starting point y 0 = µ where µ : = g ( 0 ) • Define y t + 1 = y t − f ( x , y t ) f ′ ( x , y t ) . Can we say that y t is an approximation • If f ′ ( x , y t ) is invertible, then one can show that y t ≡ g mod ⟨ x ⟩ 2 t = ⇒ y t + 1 ≡ g mod ⟨ x ⟩ 2 t + 1

• If f 0 1 where deg g calculating y log d 11 d . of g ? 0. Then, one can find g by 0 and f 0 Finding linear factor Continued • Initial starting point y 0 = µ where µ : = g ( 0 ) • Define y t + 1 = y t − f ( x , y t ) f ′ ( x , y t ) . Can we say that y t is an approximation • If f ′ ( x , y t ) is invertible, then one can show that y t ≡ g mod ⟨ x ⟩ 2 t = ⇒ y t + 1 ≡ g mod ⟨ x ⟩ 2 t + 1 � • f ′ ( x , y t ) is invertible ⇐ ⇒ f ′ ( x , y t ) ⇒ f ′ ( 0 , µ ) ̸ = 0 � ̸ = 0 ⇐ � � x = 0

11 of g ? Finding linear factor Continued • Initial starting point y 0 = µ where µ : = g ( 0 ) • Define y t + 1 = y t − f ( x , y t ) f ′ ( x , y t ) . Can we say that y t is an approximation • If f ′ ( x , y t ) is invertible, then one can show that y t ≡ g mod ⟨ x ⟩ 2 t = ⇒ y t + 1 ≡ g mod ⟨ x ⟩ 2 t + 1 � • f ′ ( x , y t ) is invertible ⇐ ⇒ f ′ ( x , y t ) ⇒ f ′ ( 0 , µ ) ̸ = 0 � ̸ = 0 ⇐ � � x = 0 • If f ( 0 , µ ) = 0 and f ′ ( 0 , µ ) ̸ = 0. Then, one can find g by calculating y log d + 1 where deg ( g ) = d .

n such that g 1 g 1 x g 2 x g e u ? We can differentiate e apply NI on f e 1 . • What about f x y 1 x y k c 0 x 12 y k 1 c k u where k 1? 1 times and • What if f y 1. Pick 4. apply Newton Iteration (NI) g 2 y g 1 0 we will get y 3. when we put x y y y 2. f x g 2 • What if f = ( y − g 1 )( y − g 2 ) but g 1 ( 0 ) = g 2 ( 0 ) ?

2. f x g 1 x g 2 x g e u ? We can differentiate e apply NI on f e 1 . • What about f x y 1 x y k c 0 x 12 1? k u where 1 c k y k y 1 times and • What if f 4. apply Newton Iteration (NI) g 2 y g 1 0 we will get y 3. when we put x y y y • What if f = ( y − g 1 )( y − g 2 ) but g 1 ( 0 ) = g 2 ( 0 ) ? 1. Pick α ∈ F n such that g 1 ( α ) ̸ = g 2 ( α )

g e u ? We can differentiate e apply NI on f e 1 . • What about f x y 1 x y k c 0 x 12 1? k u where 1 c k y k 1 times and y • What if f 4. apply Newton Iteration (NI) g 2 y g 1 0 we will get y 3. when we put x • What if f = ( y − g 1 )( y − g 2 ) but g 1 ( 0 ) = g 2 ( 0 ) ? 1. Pick α ∈ F n such that g 1 ( α ) ̸ = g 2 ( α ) 2. f ( x + α , y ) = ( y − g 1 ( x + α )) ( y − g 2 ( x + α ))

g e u ? We can differentiate e apply NI on f e 1 . • What about f x y 1 x y k c 0 x 1? k u where 1 c k 12 y k 1 times and y • What if f 4. apply Newton Iteration (NI) • What if f = ( y − g 1 )( y − g 2 ) but g 1 ( 0 ) = g 2 ( 0 ) ? 1. Pick α ∈ F n such that g 1 ( α ) ̸ = g 2 ( α ) 2. f ( x + α , y ) = ( y − g 1 ( x + α )) ( y − g 2 ( x + α )) 3. when we put x = 0 we will get ( y − g 1 ( α )) ( y − g 2 ( α ))

g e u ? We can differentiate e apply NI on f e 1 . • What about f x y 1 x y k c 0 x 1? k u where 1 c k 12 y k 1 times and y • What if f 4. apply Newton Iteration (NI) • What if f = ( y − g 1 )( y − g 2 ) but g 1 ( 0 ) = g 2 ( 0 ) ? 1. Pick α ∈ F n such that g 1 ( α ) ̸ = g 2 ( α ) 2. f ( x + α , y ) = ( y − g 1 ( x + α )) ( y − g 2 ( x + α )) 3. when we put x = 0 we will get ( y − g 1 ( α )) ( y − g 2 ( α ))

apply NI on f e 1 . • What about f x y 1 x y k c 0 x c k 1? k u where 1 12 y k 1 times and We can differentiate e 4. apply Newton Iteration (NI) • What if f = ( y − g 1 )( y − g 2 ) but g 1 ( 0 ) = g 2 ( 0 ) ? 1. Pick α ∈ F n such that g 1 ( α ) ̸ = g 2 ( α ) 2. f ( x + α , y ) = ( y − g 1 ( x + α )) ( y − g 2 ( x + α )) 3. when we put x = 0 we will get ( y − g 1 ( α )) ( y − g 2 ( α )) • What if f = ( y − g ) e · u ?

• What about f x y 1 x y k c 0 x 12 1? k u where 1 y k c k 4. apply Newton Iteration (NI) • What if f = ( y − g 1 )( y − g 2 ) but g 1 ( 0 ) = g 2 ( 0 ) ? 1. Pick α ∈ F n such that g 1 ( α ) ̸ = g 2 ( α ) 2. f ( x + α , y ) = ( y − g 1 ( x + α )) ( y − g 2 ( x + α )) 3. when we put x = 0 we will get ( y − g 1 ( α )) ( y − g 2 ( α )) • What if f = ( y − g ) e · u ? We can differentiate e − 1 times and apply NI on f ( e − 1 ) .

4. apply Newton Iteration (NI) 12 • What if f = ( y − g 1 )( y − g 2 ) but g 1 ( 0 ) = g 2 ( 0 ) ? 1. Pick α ∈ F n such that g 1 ( α ) ̸ = g 2 ( α ) 2. f ( x + α , y ) = ( y − g 1 ( x + α )) ( y − g 2 ( x + α )) 3. when we put x = 0 we will get ( y − g 1 ( α )) ( y − g 2 ( α )) • What if f = ( y − g ) e · u ? We can differentiate e − 1 times and apply NI on f ( e − 1 ) . ( y k + c k − 1 ( x ) y k − 1 + . . . + c 0 ( x ) ) • What about f ( x , y ) = · u where k > 1?

1 3 1 3 2 1 3 2 1 3 2 • g 2 x 2 3 x 3 1 3 3 mod x 4 3 13 2 3 2 x 3 1 2 1 z • So g is a root of f x x z as f x 1 g 0 • Note that z 2 x z g 3 z g x u x 1 z x 3 can apply NI. • Consider f x z z 2 x 3 u x z z x 3 2 z x 3 2 u • One can assume that z 2 u as otherwise we can We would like to relate non-linear factors to linear factors so that we differentiate appropriately many times and work with the new polynomial • f x 1 z z 2 x u x 1 z z x z x Non linear factor

1 3 1 3 2 1 3 2 1 3 2 • g 2 x 2 3 x 3 1 3 3 mod x 4 z • So g is a root of f x 2 3 2 g x 1 z 3 z as f x 2 x 1 g 0 • Note that z 2 x z g 3 13 3 1 can apply NI. • One can assume that z 2 x 3 u as otherwise we can differentiate appropriately many times and work with the new polynomial • f x 1 z z 2 x u x 1 z z x z x u x 1 z x We would like to relate non-linear factors to linear factors so that we Non linear factor • Consider f ( x , z ) = ( z 2 − x 3 ) · u ( x , z ) = ( z − x 3 / 2 )( z + x 3 / 2 ) · u

1 3 1 3 2 1 3 2 1 3 2 • g 2 x 2 3 x 3 1 3 3 mod x 4 13 3 2 • So g is a root of f x 1 z 1 g x z as f x 3 0 • Note that z 2 x z g 3 z g 2 1 2 x 3 can apply NI. differentiate appropriately many times and work with the new polynomial • f x 1 z z 2 x u x 1 z z x z x u x 1 z x We would like to relate non-linear factors to linear factors so that we Non linear factor • Consider f ( x , z ) = ( z 2 − x 3 ) · u ( x , z ) = ( z − x 3 / 2 )( z + x 3 / 2 ) · u • One can assume that z 2 − x 3 ∤ u as otherwise we can

1 3 2 • g 2 x 2 3 x 3 1 3 3 mod x 4 z 3 g z x 2 0 g 1 g z as f x x 1 z • So g is a root of f x • Note that z 2 13 3 polynomial 2 3 2 x 3 1 x can apply NI. We would like to relate non-linear factors to linear factors so that we differentiate appropriately many times and work with the new Non linear factor • Consider f ( x , z ) = ( z 2 − x 3 ) · u ( x , z ) = ( z − x 3 / 2 )( z + x 3 / 2 ) · u • One can assume that z 2 − x 3 ∤ u as otherwise we can • f ( x + 1 , z ) = ( z 2 − ( x + 1 ) 3 ) · u ( x + 1 , z ) = ( z − ( x + 1 ) 3 / 2 ) ( z + ( x + 1 ) 3 / 2 ) · u ( x + 1 , z )

1 3 3 mod x 4 • Note that z 2 1 z x z as f x 1 g 0 13 x We would like to relate non-linear factors to linear factors so that we z g 3 z g • So g is a root of f x 2 3 polynomial can apply NI. 2 3 differentiate appropriately many times and work with the new Non linear factor • Consider f ( x , z ) = ( z 2 − x 3 ) · u ( x , z ) = ( z − x 3 / 2 )( z + x 3 / 2 ) · u • One can assume that z 2 − x 3 ∤ u as otherwise we can • f ( x + 1 , z ) = ( z 2 − ( x + 1 ) 3 ) · u ( x + 1 , z ) = ( z − ( x + 1 ) 3 / 2 ) ( z + ( x + 1 ) 3 / 2 ) · u ( x + 1 , z ) • g : = ( x + 1 ) 3 / 2 = 1 + 3 2 ) x 2 + ( 3 ) x 3 + . . . 2 x + (

1 3 3 mod x 4 13 We would like to relate non-linear factors to linear factors so that we g z 3 g z x • Note that z 2 2 3 2 3 polynomial can apply NI. differentiate appropriately many times and work with the new Non linear factor • Consider f ( x , z ) = ( z 2 − x 3 ) · u ( x , z ) = ( z − x 3 / 2 )( z + x 3 / 2 ) · u • One can assume that z 2 − x 3 ∤ u as otherwise we can • f ( x + 1 , z ) = ( z 2 − ( x + 1 ) 3 ) · u ( x + 1 , z ) = ( z − ( x + 1 ) 3 / 2 ) ( z + ( x + 1 ) 3 / 2 ) · u ( x + 1 , z ) • g : = ( x + 1 ) 3 / 2 = 1 + 3 2 ) x 2 + ( 3 ) x 3 + . . . 2 x + ( • So g is a root of f ( x + 1 , z ) ∈ F [[ x ]][ z ] as f ( x + 1 , g ) = 0

13 We would like to relate non-linear factors to linear factors so that we can apply NI. 2 3 differentiate appropriately many times and work with the new polynomial 2 3 Non linear factor • Consider f ( x , z ) = ( z 2 − x 3 ) · u ( x , z ) = ( z − x 3 / 2 )( z + x 3 / 2 ) · u • One can assume that z 2 − x 3 ∤ u as otherwise we can • f ( x + 1 , z ) = ( z 2 − ( x + 1 ) 3 ) · u ( x + 1 , z ) = ( z − ( x + 1 ) 3 / 2 ) ( z + ( x + 1 ) 3 / 2 ) · u ( x + 1 , z ) • g : = ( x + 1 ) 3 / 2 = 1 + 3 2 ) x 2 + ( 3 ) x 3 + . . . 2 x + ( • So g is a root of f ( x + 1 , z ) ∈ F [[ x ]][ z ] as f ( x + 1 , g ) = 0 • Note that z 2 − ( x + 1 ) 3 = ( z − g ≤ 3 )( z + g ≤ 3 ) mod x 4

Power Series Split Theorem (DSS’18) 14 g i x • f x 1 1 y x n n y makes f monic in y • For irreducible h , one can show that h x c deg h i 1 y g i where k i g i r x i x i i y i , where i i , deg(rad( f y d 0 , f x k i d 0 Power Series Split Theorem

14 • For irreducible h , one can show that g i y 1 i deg h c x h n y makes f monic in y x n 1 y • f x 1 Power Series Split Theorem Power Series Split Theorem (DSS’18) τ : x i �→ x i + α i y + β i , where α i , β i ∈ r F , deg(rad( f )) = d 0 , f ( τ x ) = k · ∏ ( y − g i ) γ i i ∈ [ d 0 ] where k ∈ F × , g i ∈ F [[ x ]]

14 x g i y 1 i deg h c h • For irreducible h , one can show that Power Series Split Theorem Power Series Split Theorem (DSS’18) τ : x i �→ x i + α i y + β i , where α i , β i ∈ r F , deg(rad( f )) = d 0 , f ( τ x ) = k · ∏ ( y − g i ) γ i i ∈ [ d 0 ] where k ∈ F × , g i ∈ F [[ x ]] • f ( x 1 + α 1 y , . . . , x n + α n y ) makes f monic in y

14 • For irreducible h , one can show that Power Series Split Theorem Power Series Split Theorem (DSS’18) τ : x i �→ x i + α i y + β i , where α i , β i ∈ r F , deg(rad( f )) = d 0 , f ( τ x ) = k · ∏ ( y − g i ) γ i i ∈ [ d 0 ] where k ∈ F × , g i ∈ F [[ x ]] • f ( x 1 + α 1 y , . . . , x n + α n y ) makes f monic in y deg ( h ) ∏ h ( τ x ) = c · ( y − g i ) i = 1

g i e i g i b i for b i x mod x d h b i mod x d h 1 on h 15 x x h 1 • h g c y d h d h i 1 • Apply x to get back h x . • Hence h deg h x • • Suppose h f . Apply on f • f x k y x y is UFD h x c y e i • If deg h d h Factoring reduces to Root Approximation Notation : g ≤ k ≡ g mod ⟨ x ⟩ k + 1 .

g i e i g i b i for b i x mod x d h b i mod x d h 1 on h 15 c h 1 • h x g y • Hence h d h i 1 • Apply x to get back h x . x x d h y is UFD • f x k y • x h x c y e i • If deg h d h deg h Factoring reduces to Root Approximation Notation : g ≤ k ≡ g mod ⟨ x ⟩ k + 1 . • Suppose h | f . Apply τ on f

g i b i for b i x mod x d h b i mod x d h 1 on h 15 1 • h x c g y x d h i 1 • Apply x to get back h x . h • Hence h c • x deg h d h • If deg h e i y d h x h y is UFD x Factoring reduces to Root Approximation Notation : g ≤ k ≡ g mod ⟨ x ⟩ k + 1 . • Suppose h | f . Apply τ on f • f ( τ x ) = k · ∏ ( y − g i ) e i

x mod x d h b i mod x d h 1 on h 1 x to get back h x . • Apply 1 i d h g y c x • h 15 h x • Hence h d h x deg h d h • If deg h Factoring reduces to Root Approximation Notation : g ≤ k ≡ g mod ⟨ x ⟩ k + 1 . • Suppose h | f . Apply τ on f • f ( τ x ) = k · ∏ ( y − g i ) e i • F [[ x ]][ y ] is UFD = ⇒ h ( τ x ) = c · ∏ ( y − g i ) b i for b i ≤ e i

x mod x d h b i mod x d h 1 on h 15 x x to get back h x . • Apply 1 i d h g y c • h 1 h x • Hence h Factoring reduces to Root Approximation Notation : g ≤ k ≡ g mod ⟨ x ⟩ k + 1 . • Suppose h | f . Apply τ on f • f ( τ x ) = k · ∏ ( y − g i ) e i • F [[ x ]][ y ] is UFD = ⇒ h ( τ x ) = c · ∏ ( y − g i ) b i for b i ≤ e i • If deg ( h ) = d h = ⇒ deg ( h ( τ x )) = d h

b i mod x d h 1 on h 15 y x to get back h x . • Apply 1 i d h g c x • h Factoring reduces to Root Approximation Notation : g ≤ k ≡ g mod ⟨ x ⟩ k + 1 . • Suppose h | f . Apply τ on f • f ( τ x ) = k · ∏ ( y − g i ) e i • F [[ x ]][ y ] is UFD = ⇒ h ( τ x ) = c · ∏ ( y − g i ) b i for b i ≤ e i • If deg ( h ) = d h = ⇒ deg ( h ( τ x )) = d h • Hence h ( τ x ) = h ( τ x ) mod ⟨ x ⟩ d h + 1

1 on h • Apply i x to get back h x . 15 Factoring reduces to Root Approximation Notation : g ≤ k ≡ g mod ⟨ x ⟩ k + 1 . • Suppose h | f . Apply τ on f • f ( τ x ) = k · ∏ ( y − g i ) e i • F [[ x ]][ y ] is UFD = ⇒ h ( τ x ) = c · ∏ ( y − g i ) b i for b i ≤ e i • If deg ( h ) = d h = ⇒ deg ( h ( τ x )) = d h • Hence h ( τ x ) = h ( τ x ) mod ⟨ x ⟩ d h + 1 • h ( τ x ) = c · ∏ ( y − g ≤ d h ) b i mod ⟨ x ⟩ d h + 1

i 15 Factoring reduces to Root Approximation Notation : g ≤ k ≡ g mod ⟨ x ⟩ k + 1 . • Suppose h | f . Apply τ on f • f ( τ x ) = k · ∏ ( y − g i ) e i • F [[ x ]][ y ] is UFD = ⇒ h ( τ x ) = c · ∏ ( y − g i ) b i for b i ≤ e i • If deg ( h ) = d h = ⇒ deg ( h ( τ x )) = d h • Hence h ( τ x ) = h ( τ x ) mod ⟨ x ⟩ d h + 1 • h ( τ x ) = c · ∏ ( y − g ≤ d h ) b i mod ⟨ x ⟩ d h + 1 • Apply τ − 1 on h ( τ x ) to get back h ( x ) .

Simultaneous Root Approximation (allRootsNI)

g e u , to find g , we have to differentiate e (wrt y ). What is the size of f e 1 ? • We know factoring reduces to root approximation. • We know standard newton iteration would give us approximation. • Are we done? • If f y 1-times 16 Are we done?

g e u , to find g , we have to differentiate e (wrt y ). What is the size of f e 1 ? • We know factoring reduces to root approximation. • We know standard newton iteration would give us approximation. • Are we done? • If f y 1-times 16 Are we done?

g e u , to find g , we have to differentiate e (wrt y ). What is the size of f e 1 ? • We know factoring reduces to root approximation. • We know standard newton iteration would give us approximation. • Are we done? • If f y 1-times 16 Are we done?

• We know factoring reduces to root approximation. • We know standard newton iteration would give us approximation. • Are we done? 16 Are we done? • If f = ( y − g ) e · u , to find g , we have to differentiate e − 1-times (wrt y ). What is the size of f ( e − 1 ) ?

Proof Idea. keep track of u u 1 u k size circuit Compute inductively from bottom to top calculating upto k -th derivative i.e. at some node calculating u in the actual circuit, we instead! 17 Derivative Computation ∂ k f f computed by size s circuit = ⇒ ∂ y k can be computed by O ( k 2 s )

size circuit Compute inductively from bottom to top calculating upto k -th derivative i.e. at some node calculating u in the actual circuit, we 17 Derivative Computation ∂ k f f computed by size s circuit = ⇒ ∂ y k can be computed by O ( k 2 s ) Proof Idea. keep track of ( u , u ( 1 ) , . . . , u ( k ) ) instead!

w u v 18 + w ( i ) = u ( i ) + v ( i )

w u v i 18 × ( i ) w ( i ) = u ( i − µ ) v ( µ ) ∑ µ µ = 0

y k can be computed by poly log k s • Can one show log dependency on k in the size of the derivative circuit? • If k f permanent can be computed by a polynomial size circuit 19 Can we do better for derivative computing? Observation: size ( f ′ ) = O ( s ) where size ( f ) = s

y k can be computed by poly log k s • Can one show log dependency on k in the size of the derivative circuit? • If k f permanent can be computed by a polynomial size circuit 19 Can we do better for derivative computing? Observation: size ( f ′ ) = O ( s ) where size ( f ) = s

• Can one show log dependency on k in the size of the derivative circuit? computed by a polynomial size circuit 19 Can we do better for derivative computing? Observation: size ( f ′ ) = O ( s ) where size ( f ) = s • If ∂ k f ∂ y k can be computed by poly ( log k , s ) = ⇒ permanent can be

g e u , then if we define e f y t g mod x 2 t g mod x 2 t 20 • Does this help? No! 1 1 y t Then f y t and y t • Can we avoid exponential many derivatives? y t 1 y t y • One can show that f Modified Newton Iteration : Does this help?

e f y t g mod x 2 t g mod x 2 t 20 • Does this help? No! 1 1 y t Then and y t • Can we avoid exponential many derivatives? f y t y t 1 y t Modified Newton Iteration : Does this help? • One can show that f = ( y − g ) e · u , then if we define

g mod x 2 t • Can we avoid exponential many derivatives? Then y t 1 1 • Does this help? No! 20 Modified Newton Iteration : Does this help? • One can show that f = ( y − g ) e · u , then if we define y t + 1 = y t − e f ( y t ) f ′ ( y t ) and y t ≡ g mod ⟨ x ⟩ 2 t

• Can we avoid exponential many derivatives? Then • Does this help? No! 20 Modified Newton Iteration : Does this help? • One can show that f = ( y − g ) e · u , then if we define y t + 1 = y t − e f ( y t ) f ′ ( y t ) and y t ≡ g mod ⟨ x ⟩ 2 t y t + 1 = g mod ⟨ x ⟩ 2 t + 1

• Can we avoid exponential many derivatives? Then • Does this help? No! 20 Modified Newton Iteration : Does this help? • One can show that f = ( y − g ) e · u , then if we define y t + 1 = y t − e f ( y t ) f ′ ( y t ) and y t ≡ g mod ⟨ x ⟩ 2 t y t + 1 = g mod ⟨ x ⟩ 2 t + 1

• Can we avoid exponential many derivatives? Then • Does this help? No! 20 Modified Newton Iteration : Does this help? • One can show that f = ( y − g ) e · u , then if we define y t + 1 = y t − e f ( y t ) f ′ ( y t ) and y t ≡ g mod ⟨ x ⟩ 2 t y t + 1 = g mod ⟨ x ⟩ 2 t + 1

1 by the modified e f y t iteration the end • Push the division gate and the top and try to remove division at • Compute the whole thing as a circuit with “division” gate allowed f y t y t 1 y t • Recall to recover a factor, it is enough to calculate approximation upto its degree d h . Calculate y log d h • One has to calculate g x x for some g h g f and y • Suppose h 21

1 by the modified e f y t • Recall to recover a factor, it is enough to calculate approximation upto its degree • One has to calculate g d h . Calculate y log d h iteration y t 1 y t f y t • Compute the whole thing as a circuit with “division” gate allowed • Push the division gate and the top and try to remove division at the end 21 • Suppose h | f and y − g | h ( τ x ) for some g ∈ F [[ x ]]

1 by the modified e f y t • Recall to recover a factor, it is enough to calculate approximation upto its degree Calculate y log d h iteration y t 1 y t f y t • Compute the whole thing as a circuit with “division” gate allowed • Push the division gate and the top and try to remove division at the end 21 • Suppose h | f and y − g | h ( τ x ) for some g ∈ F [[ x ]] • One has to calculate g ≤ d h .

• Recall to recover a factor, it is enough to calculate approximation upto its degree iteration • Compute the whole thing as a circuit with “division” gate allowed • Push the division gate and the top and try to remove division at the end 21 • Suppose h | f and y − g | h ( τ x ) for some g ∈ F [[ x ]] • One has to calculate g ≤ d h . Calculate y log d h + 1 by the modified y t + 1 = y t − e f ( y t ) f ′ ( y t )

• Recall to recover a factor, it is enough to calculate approximation upto its degree iteration • Compute the whole thing as a circuit with “division” gate allowed • Push the division gate and the top and try to remove division at the end 21 • Suppose h | f and y − g | h ( τ x ) for some g ∈ F [[ x ]] • One has to calculate g ≤ d h . Calculate y log d h + 1 by the modified y t + 1 = y t − e f ( y t ) f ′ ( y t )

• Recall to recover a factor, it is enough to calculate approximation upto its degree iteration • Compute the whole thing as a circuit with “division” gate allowed • Push the division gate and the top and try to remove division at the end 21 • Suppose h | f and y − g | h ( τ x ) for some g ∈ F [[ x ]] • One has to calculate g ≤ d h . Calculate y log d h + 1 by the modified y t + 1 = y t − e f ( y t ) f ′ ( y t )

• We will be spared with A B and we have to calculate B mod x d h 1 where B is not invertible. • How to eliminate only one division gate at the top? A B A • We don’t know how to calculate this! 22

• We will be spared with A B and we have to calculate B mod x d h 1 where B is not invertible. • How to eliminate only one division gate at the top? A B A • We don’t know how to calculate this! 22 ÷

• How to eliminate only one division gate at the top? A B A • We don’t know how to calculate this! 22 ÷ • We will be spared with A B and we have to calculate B mod ⟨ x ⟩ d h + 1 where B is not invertible.