Gap Property Zahra Shojaee Yazd University December 7, 2015 1 / - PowerPoint PPT Presentation

wt ( E ) < (1 + 2 w ) . wt ( MST ( S )) . log n ◮ size of E / E ′ is at most m 2 and it satisfies the w-gap property. ◮ By the induction hypothesis: 18 / 67

wt ( E ) < (1 + 2 w ) . wt ( MST ( S )) . log n ◮ size of E / E ′ is at most m 2 and it satisfies the w-gap property. ◮ By the induction hypothesis: ◮ wt ( E / E ′ ) < (1 + 2 w ) . wt ( MST ( S )) . log m 2 18 / 67

wt ( E ) < (1 + 2 w ) . wt ( MST ( S )) . log n wt ( E ) = wt ( E ′ )+ wt ( E / E ′ ) < (1+ 2 w ) . wt ( MST ( S ))(1+log m 2 ) 19 / 67

wt ( E ) < (1 + 2 w ) . wt ( MST ( S )) . log n wt ( E ) = wt ( E ′ )+ wt ( E / E ′ ) < (1+ 2 w ) . wt ( MST ( S ))(1+log m 2 ) ◮ It can be used in any other metric space. 19 / 67

A Lower Bound Theorem 6.2.1: ◮ Let w be a real number with 0 < w < 1 ,and k ≥ 2 be an integer. ◮ let n = 3 k − 1 20 / 67

A Lower Bound Theorem 6.2.1: ◮ Let w be a real number with 0 < w < 1 ,and k ≥ 2 be an integer. ◮ let n = 3 k − 1 ◮ there exists a set S of n points on the real line and a set E of directed edges, such that E satisfies the strong w-gap property and : 20 / 67

A Lower Bound Theorem 6.2.1: ◮ Let w be a real number with 0 < w < 1 ,and k ≥ 2 be an integer. ◮ let n = 3 k − 1 ◮ there exists a set S of n points on the real line and a set E of directed edges, such that E satisfies the strong w-gap property and : wt ( E ) = Ω( wt ( MST ( S ))) log n 20 / 67

wt ( E ) = Ω( wt ( MST ( S )) log n )( proof ) ◮ We partition the interval [0 , 1] into 3 i interval (0 ≤ i < k ), each having length 1 3 i . 21 / 67

wt ( E ) = Ω( wt ( MST ( S )) log n )( proof ) ◮ We partition the interval [0 , 1] into 3 i interval (0 ≤ i < k ), each having length 1 3 i . ◮ The j -th interval is 3 i , j + 1 [ j ] 3 i 21 / 67

wt ( E ) = Ω( wt ( MST ( S )) log n )( proof ) ◮ We partition the interval [0 , 1] into 3 i interval (0 ≤ i < k ), each having length 1 3 i . ◮ The j -th interval is 3 i , j + 1 [ j ] 3 i ◮ We divide the j -th interval into three subintervals of equal length. 21 / 67

wt ( E ) = Ω( wt ( MST ( S )) log n )( proof ) ◮ We partition the interval [0 , 1] into 3 i interval (0 ≤ i < k ), each having length 1 3 i . ◮ The j -th interval is 3 i , j + 1 [ j ] 3 i ◮ We divide the j -th interval into three subintervals of equal length. ◮ e ij ;middle of these three subintervals 21 / 67

wt ( E ) = Ω( wt ( MST ( S )) log n )( proof ) e ij = [ j 3 i +1 , j 1 2 3 i + 3 i + 3 i +1 ] 22 / 67

wt ( E ) = Ω( wt ( MST ( S )) log n )( proof ) e ij = [ j 3 i +1 , j 1 2 3 i + 3 i + 3 i +1 ] E i = { e ij : j = 0 , 1 , ..., 3 i − 1 } 22 / 67

wt ( E ) = Ω( wt ( MST ( S )) log n )( proof ) e ij = [ j 3 i +1 , j 1 2 3 i + 3 i + 3 i +1 ] E i = { e ij : j = 0 , 1 , ..., 3 i − 1 } E = E 0 ∪ E 1 ∪ .... ∪ E k − 1 22 / 67

wt ( E ) = Ω( wt ( MST ( S )) log n )( proof ) | E i | = 3 i so | E | = � 3 i = 3 k − 1 2 23 / 67

wt ( E ) = Ω( wt ( MST ( S )) log n )( proof ) | E i | = 3 i so | E | = � 3 i = 3 k − 1 2 | S | = n and since endpoints of all edges are pairwise distinct: 23 / 67

wt ( E ) = Ω( wt ( MST ( S )) log n )( proof ) | E i | = 3 i so | E | = � 3 i = 3 k − 1 2 | S | = n and since endpoints of all edges are pairwise distinct: n = 2 × | E | = 3 k − 1 23 / 67

wt ( E ) = Ω( wt ( MST ( S )) log n )( proof ) 24 / 67

wt ( E ) = Ω( wt ( MST ( S )) log n )( proof ) e ij = [ j 3 i +1 , j 1 2 3 i + 3 i + 3 i +1 ] e kl = [ l 3 k +1 , l 1 2 3 l + 3 k + 3 k +1 ] ◮ if i = k then put l = j + 1 to have the nearest sources so we 1 have: | S e ij − S e ij +1 | = 3 i +1 Done! 1 ◮ else if i < k ( or k < i ) then | S e ij − S e ij +1 | = 3 i +1 Done! 25 / 67

wt ( E ) = Ω( wt ( MST ( S )) log n )( proof ) wt ( E i ) = 1 3 wt ( E ) = k × 1 3 26 / 67

wt ( E ) = Ω( wt ( MST ( S )) log n )( proof ) wt ( E i ) = 1 3 wt ( E ) = k × 1 3 n = 3 k − 1 ⇒ k = log( n + 1) 26 / 67

wt ( E ) = Ω( wt ( MST ( S )) log n )( proof ) wt ( E i ) = 1 3 wt ( E ) = k × 1 3 n = 3 k − 1 ⇒ k = log( n + 1) wt ( E ) = 1 3 (log( n + 1)) = Ω( logn ) 26 / 67

wt ( E ) = Ω( wt ( MST ( S )) log n )( proof ) ◮ min element of S is S = 1 3 k ◮ max element of S is S = 1 − 1 3 k 27 / 67

wt ( E ) = Ω( wt ( MST ( S )) log n )( proof ) ◮ min element of S is S = 1 3 k ◮ max element of S is S = 1 − 1 3 k ◮ Since MST is consist of the sorted elements sequence: wt ( MST ( S )) = 1 − 2 3 k < 1 27 / 67

wt ( E ) = Ω( wt ( MST ( S )) log n )( proof ) ◮ min element of S is S = 1 3 k ◮ max element of S is S = 1 − 1 3 k ◮ Since MST is consist of the sorted elements sequence: wt ( MST ( S )) = 1 − 2 3 k < 1 ◮ By applying these bounds : wt ( E ) × wt ( MST ( S )) < wt ( E ) Ω(log n ) × wt ( MST ( S )) < wt ( E ) × wt ( MST ( S )) < wt ( E ) 27 / 67

wt ( E ) = Ω( wt ( MST ( S )) log n )( proof ) ◮ min element of S is S = 1 3 k ◮ max element of S is S = 1 − 1 3 k ◮ Since MST is consist of the sorted elements sequence: wt ( MST ( S )) = 1 − 2 3 k < 1 ◮ By applying these bounds : wt ( E ) × wt ( MST ( S )) < wt ( E ) Ω(log n ) × wt ( MST ( S )) < wt ( E ) × wt ( MST ( S )) < wt ( E ) ◮ wt ( E ) = Ω( wt ( MST ( S ))) log n 27 / 67

Definitions d Π ◮ c d = 2 Γ( d / 2+1) 28 / 67

Definitions d Π ◮ c d = 2 Γ( d / 2+1) ◮ If d is large then C dw = 1 + 2 2 d +1 C d W d 28 / 67

Definitions d Π ◮ c d = 2 Γ( d / 2+1) ◮ If d is large then C dw = 1 + 2 2 d +1 C d W d ◮ Γ( x + 1) = x ! ◮ Γ( x ) = ( x − 1)Γ( x − 1) ◮ Γ(1 / 2) = √ π/ 2 28 / 67

An Upper Bound(cont.) Theorem 6.3.1: ◮ Let S be the set of n points in the d-dimensional unit cube [0 , 1] d ◮ And E be the set of directed edges that satisfies the w-gap property 29 / 67

An Upper Bound(cont.) Theorem 6.3.1: ◮ Let S be the set of n points in the d-dimensional unit cube [0 , 1] d ◮ And E be the set of directed edges that satisfies the w-gap property ◮ for d ≥ 2 and we have: 29 / 67

An Upper Bound(cont.) Theorem 6.3.1: ◮ Let S be the set of n points in the d-dimensional unit cube [0 , 1] d ◮ And E be the set of directed edges that satisfies the w-gap property ◮ for d ≥ 2 and we have: wt ( E ) ≤ C dw n 1 − 1 d ◮ C dw = 1 + 2 2 d +1 C d W d 29 / 67

wt ( E ) ≤ C dw n 1 − 1 d ◮ | S | = n so | E | is at most n ◮ we partition E according length � E l = ( p , q ) ∈ E : | pq | > n − 1 / d ◮ E = E s = ( p , q ) ∈ E : | pq | ≤ n − 1 / d 30 / 67

wt ( E ) ≤ C dw n 1 − 1 d ◮ In case of E s : − 1 − 1 d = n 1 − 1 d ≤ n × n wt ( E s ) ≤ | E s | × n d 31 / 67

wt ( E ) ≤ C dw n 1 − 1 d ◮ In case of E s : − 1 − 1 d = n 1 − 1 d ≤ n × n wt ( E s ) ≤ | E s | × n d ◮ In case of E l : I j = ( 2 j , 2 j +1 ) 1 1 n n d d l = 2 j 1 n d 31 / 67

wt ( E ) ≤ C dw n 1 − 1 d ◮ In case of E s : − 1 − 1 d = n 1 − 1 d ≤ n × n wt ( E s ) ≤ | E s | × n d ◮ In case of E l : I j = ( 2 j , 2 j +1 ) 1 1 n n d d l = 2 j 1 n d ◮ F j = { ( p , q ) ∈ E l : | pq | ∈ I j } 31 / 67

wt ( E ) ≤ C dw n 1 − 1 d − 1 ◮ if ( p , q ) ∈ E l → n d < | pq | 32 / 67

wt ( E ) ≤ C dw n 1 − 1 d − 1 ◮ if ( p , q ) ∈ E l → n d < | pq | √ dn d − 1 ]] ◮ So j ∈ [0 , [log 32 / 67

wt ( E ) ≤ C dw n 1 − 1 d ◮ | F j | = k ◮ F j = { ( p i , q i ) : 1 ≤ i ≤ k } 33 / 67

wt ( E ) ≤ C dw n 1 − 1 d ◮ | F j | = k ◮ F j = { ( p i , q i ) : 1 ≤ i ≤ k } ◮ L = 2 j → L < | p i q i | ≤ 2 L 1 n d 33 / 67

wt ( E ) ≤ C dw n 1 − 1 d ◮ Edges of E satisfy the w-gap property 34 / 67

wt ( E ) ≤ C dw n 1 − 1 d ◮ Edges of E satisfy the w-gap property ◮ ⇒ | p i p ′ i | > w . min( | p i q i | , | p ′ i q ′ i | ) > wL 34 / 67

wt ( E ) ≤ C dw n 1 − 1 d ◮ Edges of E satisfy the w-gap property ◮ ⇒ | p i p ′ i | > w . min( | p i q i | , | p ′ i q ′ i | ) > wL ◮ If we draw a d-dimensional ball B i of radius wL 2 around each point p i , (1 ≤ i ≤ k ) , then these balls are pairwise disjoint. 34 / 67

wt ( E ) ≤ C dw n 1 − 1 d √ WL 2 ≤ w d ≤ 1 ◮ 2 35 / 67

wt ( E ) ≤ C dw n 1 − 1 d √ WL 2 ≤ w d ≤ 1 ◮ 2 2 ) d of each ball B i is contained in ◮ so at least a fraction of ( 1 the unit cube. 35 / 67

wt ( E ) ≤ C dw n 1 − 1 d √ WL 2 ≤ w d ≤ 1 ◮ 2 2 ) d of each ball B i is contained in ◮ so at least a fraction of ( 1 the unit cube. ◮ volume of d-dimensional ball of radius R = C d R d d Π 2 ◮ c d = Γ( d / 2+1) 35 / 67

wt ( E ) ≤ C dw n 1 − 1 d ◮ So all volume of all portions of the balls B i inside the unit cube, is smaller than or equal to 1 36 / 67

wt ( E ) ≤ C dw n 1 − 1 d ◮ So all volume of all portions of the balls B i inside the unit cube, is smaller than or equal to 1 2 ) d ≤ 1 ◮ k ( 1 2 ) d C d ( WL 2 2 d ◮ ⇒ k ≤ C d W d L d 36 / 67

wt ( E ) ≤ C dw n 1 − 1 d 2 2 d ∀ ( p i , q i ) ∈ F j | p i q i | ≤ 2 L ⇒ wt ( F j ) ≤ 2 LK ≤ 2 L C d W d L d 37 / 67

wt ( E ) ≤ C dw n 1 − 1 d 2 2 d ∀ ( p i , q i ) ∈ F j | p i q i | ≤ 2 L ⇒ wt ( F j ) ≤ 2 LK ≤ 2 L C d W d L d C d W d 2 j ( d − 1) × n 1 − 1 2 2 d +1 wt ( F j ) ≤ d 37 / 67

wt ( E ) ≤ C dw n 1 − 1 d √ 1 ◮ m = ⌊ log d ⌋ dn 38 / 67