The strength of a geometric simplex is a key ingredient in a polynomial-time classification of unordered point clouds by Lipschitz continuous invariants

The strength of a geometric simplex is a k ey ingredient in a polynomial-time classification of unordered point clouds by Lipschitz continuous in v ariants Olga Anosov a, V italiy Kurlin School of Computer Science and Informatics, University of Liverpool, Ashton str eet, Liverpool, L69 3BX, , United Kingdom Abstract The basic input for many real shapes is a finite cloud of unordered points. The strongest equiv alence between shapes in practice is Euclidean motion. The recent polynomial-time classification of point clouds required a Lipschitz contin- uous function that vanishes on degenerate simplices, while the usual volume is not Lipschitz. W e define the strength of any geometric simple x and prove its continuity under perturbations with e xplicit bounds for Lipschitz constants. K e ywor ds: unordered point cloud, rigid motion, complete classification, Lipschitz continuity, geometric simplex 2000 MSC: 51F20, 51N20, 51M25, 15A15 1. The importance of Lipschitz continuity for distinguishing mirr or images under noise Many applications deal with point configurations or clouds of points obtained as edge pixels or feature points of objects across all scales from galaxies to molecules [1]. Positions of points in a Euclidean space, such as atomic cen- tres, are al ways uncertain due to measurement noise or thermal vibrations. Molecular dynamics simulates trajectories of atomic clouds, while machine learning tries to predict molecular properties that depend on atomic geometry . The robustness of algorithmic predictions means that an output only continuously changes under perturbations of giv en points. The classical ε − δ concept of continuity is very weak in the sense that all standard functions are continuous on domains, where they are defined. For example, f ( x ) = 1 x is continuous for any x , 0 because, for any ε > 0, there is δ = | x | 2 min { 1 , ε | x |} such that if | x − y | < δ , then | f ( x ) − f ( y ) | = | x − y | | xy | ≤ δ | xy | ≤ 2 δ x 2 ≤ ε . Howe ver , for small x > 0, the chosen delta δ is much smaller than ε , so f ( x ) = 1 x grows too fast close to 0. The Lipschitz continuity below is much more practical by restricting the gro wth of a function via a constant and an amount of perturbation. Definition 1.1 (Lipschitz continuity) . A function f : R m → R is Lipschitz continuous if there is a Lipschitz constant λ > 0 such that | f ( x ) − f ( y ) | ≤ λ | x − y | for any x , y ∈ R n , where | x − y | denotes the Euclidean distance. Then f ( x ) = 1 x is not Lipschitz continuous because for an y λ > 0, we can set c = max { 1 , λ } , x = 1 2 c and y = 1 c such that | f ( x ) − f ( y ) | = c ≥ 1 > λ 2 c = λ | x − y | . Though the Lipschitz continuity makes sense for maps between metric spaces, we consider only scalar functions on simplices in R n . A simplex can be defined as a finite set of elements whose ev ery subset is also a simplex. W e consider only geometric realizations of a simplex, still called a simple x. Definition 1.2 (a geometric simplex A on n + 1 points in R n ) . The (geometric) simplex A on any n + 1 ordered points p 0 , . . . , p n ∈ R n is the subset A = ( n P i = 0 t i p i t i ∈ [0 , 1] , n P i = 0 t i = 1 ) ⊂ R n with ordered v ertices p 0 , . . . , p n . An orientation of A is the sign of the determinant of the n × n matrix with the columns p 1 − p 0 , . . . , p n − p 0 , and is denoted by sign( A ). In dimension n = 1, the simplex on any points p 0 , p 1 ∈ R is the line segment connecting p 0 and p 1 . If points p 0 , . . . , p n ∈ R n are a ffi nely independent , i.e. there is no ( n − 1)-dimensional a ffi ne subspace of R n containing all p 0 , . . . , p n and hence the simplex A , then A is n -dimensional. Ho wev er , Definition 1.2 makes sense for an y points. The volume v ol( A ) of a simplex A or an arbitrary polyhedron is often used as a shape descriptor and detects a ffi ne independence in the sense that A is deg enerate if and only if vol( A ) = 0. Howe ver , the v olume and all other distance-based descriptors do not distinguish mirror images, which hav e di ff erent signs of orientation. When A goes through a degenerate configuration, an orientation of A can discontinuously change the sign. This discontinuity is an obstacle to recognizing simplices (or more general clouds) that are nearly mirror-symmetric. One attempt to resolve this discontinuity is to consider the signed v olume sign( A )vol( A ), because v ol( A ) vanishes only on degenerate simplices. Howe ver , vol( A ) is not Lipschitz continuous for any n ≥ 2, as illustrated below . Example 1.3 (the area of a triangle is not Lipschitz continuous) . For any large l > 0 and real ε close to 0, let T ( l , ε ) ⊂ R 2 be the 2D simplex (triangle) on the vertices (0 , ε ) and ( ± l , 0). The signed area l ε of T ( l , ε ) distinguishes mirror images T ( l , ± ε ) but is not Lipschitz continuous under perturbations. Indeed, as ε → 0, the triangle T ( l , ε ) degenerates to a straight line, while the area drops to 0 too quickly so that vol( T ( l , ε )) − vol( T ( l , 0)) ε − 0 = l ε ε = l is not bounded. Hence, if giv en points are not restricted to a fixed bounded region, a small change in their positions may lead to a large change for the area of a triangle, and similarly for the v olume of a high-dimensional simplex in R n . Problem 1.4 (Lipschitz continuous detection of de generate simplices) . Find a Lipschitz continuous real-v alued func- tion f ( A ) for all simplices A on n + 1 points p 0 , . . . , p n ∈ R n such that f ( A ) = 0 if and only if A is degenerate. 2. The strength is a Lipschitz continuous in variant W e solv e Problem 1.4 by introducing the strength function whose Lipschitz continuity is proved in Theorem 2.4. Definition 2.1 (the str ength σ ( A ) and signed str ength s ( A ) of a simplex A ⊂ R n ) . Let A ⊂ R n be the simplex on any n + 1 points p 0 , p 1 , . . . , p n in R n . The half-perimeter p ( A ) = 1 2 P i , j | p i − p j | is one half of the sum of all distances between the vertices of A . The str ength of A is defined as σ ( A ) = vol 2 ( A ) p 2 n − 1 ( A ) . The signed strength is s ( A ) = sign( A ) σ ( A ). Example 2.2 (the strength of a line segment) . For n = 1, the simplex A on any points p 0 , p 1 ∈ R is the line segment with vol( A ) = | p 1 − p 0 | = 2 p ( A ), strength σ ( A ) = vol 2 ( A ) p ( A ) = 2 | p 1 − p 0 | , and signed strength s ( A ) = 2( p 1 − p 0 ). Example 2.3 (the strength of a triangle) . (a) Let A ⊂ R 2 be a triangle with sides a , b , c . Heron’ s formula for the area vol( A ) = p p ( p − a )( p − b )( p − c ), where p = a + b + c 2 = p ( A ) is the half-perimeter, giv es the strength σ ( A ) = vol 2 ( A ) p 3 ( A ) = ( p − a )( p − b )( p − c ) p 2 . The triangle T ( l , ε ) in Example 1.3 has p = l + ε + √ l 2 + ε 2 2 . Using l ≤ √ l 2 + ε 2 ≤ l + ε , the strength σ = ( l − ε + √ l 2 + ε 2 )( ε + √ l 2 + ε 2 − l )( l + ε − √ l 2 + ε 2 ) 2( l + ε + √ l 2 + ε 2 ) 2 ≤ 2 l · 2 ε · ε 2(2 l ) 2 ≤ ε 2 2 l ≤ ε 2 for any 0 ≤ ε ≤ l . Hence, the strength of T ( l , ε ) is Lipschitz continuous with respect to perturbations of ε . (b) T o visualize the strength σ ( A ), we assume that 0 < a ≤ b ≤ c and normalize A by the larger side c to get the sides ˜ a = a c ≤ ˜ b = b c ≤ ˜ c = 1. The resulting space of normalized triangles is parametrized by the coordinates x = a c and y = 1 − b c in the triangular region ∆ = { ( x , y ) ∈ R 2 | x ∈ [0 , 1] , x ≥ y , x + y ≤ 1 } , where x + y ≤ 1 means that a ≤ b , while x ≥ y is equiv alent to the triangle inequality a + b ≥ c , see Fig. 2 (left). The half-perimeter is ˜ p = 1 2 ( ˜ a + ˜ b + ˜ c ) = 1 2 ( x + (1 − y ) + 1) = 1 2 (2 + x − y ). Then ˜ p − ˜ a = 1 2 (2 − x − y ), ˜ p − ˜ b = 1 2 ( x + y ), ˜ p − ˜ c = 1 2 ( x − y ), and the strength is σ ( x , y ) = (2 − x − y )( x 2 − y 2 ) 2(2 + x − y ) 2 , see Fig. 2 (right). In the region ∆ , the horizontal side { x ∈ (0 , 1] , y = 0 } represents all (normalized) isosceles triangles with ˜ a ≤ ˜ b = ˜ c = 1 and strength σ = (2 − x ) x 2 2(2 + x ) 2 for x ∈ [0 , 1]. The right 2 hand side { x ∈ (0 . 5 , 1] , x + y = 1 } of the region ∆ represents all (normalized) isosceles triangles with ˜ a = ˜ b ≤ ˜ c = 1 and strength σ = 2 x − 1 2(2 x + 1) 2 for x ∈ [ 1 2 , 1]. The verte x ( x , y ) = (1 , 0) represents all equilateral triangles with side 1 of strength σ = 1 18 . An y equilateral triangle with sides a = b = c has the strength σ = ( a 2 √ 3 / 4) 2 (3 a / 2) 3 = a 18 . Figure 1: Left : Example 2.3(b) parametrizes the space of (normalized) triangles with sides 0 < a ≤ b ≤ c by x = a c and y = 1 − b c over the re gion ∆ = { ( x , y ) ∈ R 2 | x ∈ [0 , 1] , x ≥ y , x + y ≤ 1 } . Right : the strength σ ( x , y ) = (2 − x − y )( x 2 − y 2 ) 2(2 + x − y ) 2 of a normalized triangle over the re gion ∆ . Figure 2: V ertical sections of the strength σ of normalized triangles parametrized by x , y in Fig. 2. Left : isosceles triangles with ˜ a = ˜ b ≤ ˜ c = 1 hav e y = 0 and σ = (2 − x ) x 2 2(2 + x ) 2 for x ∈ [0 , 1]. Right : isosceles triangles with sides ˜ a ≤ ˜ b = ˜ c have y = 1 − x and σ = 2 x − 1 2(2 x + 1) 2 for x ∈ [ 1 2 , 1]. Recall that an isometry is any distance-preserving transformation of R n , which decomposes into translations and orthogonal maps from the group O( R n ). A rigid motion is any composition of translations and rotations from the special orthogonal group SO( R n ) The strength of a simplex was essentially used to define a Lipschitz continuous metric on in variants of n -dimensional clouds of m unordered points, which are complete under rigid motion in R n and can be computed in a polynomial time of m , for a fixed dimension n [2, Theorem 4.7], see detailed proofs in [3]. Theorem 2.4 (strength properties: in v ariance, computational time, and Lipschitz continuity) . (a) The strength σ ( A ) and signed strength s ( A ) of any simplex A ⊂ R n are in variant under isometry and rigid motion in R n , respecti vely , and can be computed in time O ( n 3 ). The uniform scaling of R n by any factor c > 0 multiplies both σ ( A ) and s ( A ) by c . (b) Fix any dimension n ≥ 1. Then there is a constant λ n > 0 such that, for any ε > 0, if a simplex B is obtained from another simplex A ⊂ R n by perturbing e very verte x of A within its ε -neighborhood, then | σ ( A ) − σ ( B ) | ≤ 2 λ n ε , where λ 1 = 2, λ 2 = √ 3, and λ n ≤ 2 n + 0 . 5 ( n !) n 2 n − 4 for any n ≥ 3, e.g. λ 3 < 0 . 37. 3 3. Proofs of str ength properties in all dimensions Proof of Theorem 2.4(b) for dimension n = 1 and λ 1 = 2 . For n = 1, a simplex A ⊂ R has two vertices p 0 , p 1 at a distance d = | p 0 − p 1 | and the strength σ ( A ) = 2 d , which has Lipschitz constant 2 λ 1 = 4. Indeed, perturbing each of p 0 , p 1 up to ε changes the distance d up to 2 ε . W e use v ertical lines to denote the determinant | M | of a matrix M and the Euclidean length | v | of a vector v ∈ R n . Proof of Theorem 2.4(b) for dimension n = 2 and λ 2 = √ 3 . Let a triangle A ⊂ R 2 hav e pairwise distances a , b , c . Using the half-perimeter p = a + b + c 2 , the variables ˜ a = p − a , ˜ b = p − b , ˜ c = p − c are expressed via a , b , c , so a = ˜ b + ˜ c , b = ˜ a + ˜ c , c = ˜ a + ˜ b , p = ˜ a + ˜ b + ˜ c . The Jacobian of this change of v ariables is       ∂ ( a , b , c ) ∂ ( ˜ a , ˜ b , ˜ c )       =         0 1 1 1 0 1 1 1 0         = 2. If ev ery point of A is perturbed up to ε , then any pairwise distance between the vertices of A changes by at most 2 ε . By the mean value theorem [4], this bound 2 ε gives | σ ( A ′ ) − σ ( A ) | ≤ 2 ε sup |∇ σ | , where ∇ σ = ∂σ ∂ a , ∂σ ∂ b , ∂σ ∂ c ! is the gradient of the first order partial deriv atives of σ ( A ) with respect to the three distances between points of A . Since |∇ σ | =       ∂ ( a , b , c ) ∂ ( ˜ a , ˜ b , ˜ c ) · ∂σ ∂ ˜ a , ∂σ ∂ ˜ b , ∂σ ∂ ˜ c !       ≤ 2       ∂σ ∂ ˜ a , ∂σ ∂ ˜ b , ∂σ ∂ ˜ c !       , it remains to estimate the first order partial deriv a- tiv es of the strength σ = ˜ a ˜ b ˜ c ( ˜ a + ˜ b + ˜ c ) 2 with respect to the v ariables ˜ a , ˜ b , ˜ c . Since σ is symmetric in ˜ a , ˜ b , ˜ c , it su ffi ces to consider ∂σ ∂ ˜ a = ˜ b ˜ c ( ˜ a + ˜ b + ˜ c ) 2 − 2 ˜ a ˜ b ˜ c ( ˜ a + ˜ b + ˜ c ) 3 = ˜ b ˜ c ( ˜ b + ˜ c − ˜ a ) ( ˜ a + ˜ b + ˜ c ) 3 = ( p − b )( p − c )(2 a − p ) p 3 = 1 − b p ! 1 − c p ! 2 a p − 1 ! . By the triangle inequalities for the sides a , b , c , we hav e max { a , b , c } ≤ p = a + b + c 2 . Then a p , b p , c p ∈ (0 , 1], and 1 − b p , 1 − c p ∈ [0 , 1), and 2 a p − 1 ∈ ( − 1 , 1], so      ∂σ ∂ ˜ a      ≤ 1. The similar bounds      ∂σ ∂ ˜ b      ,      ∂σ ∂ ˜ c      ≤ 1 imply that |∇ σ | ≤ 2 √ 1 2 + 1 2 + 1 2 = 2 √ 3, so | σ ( A ′ ) − σ ( A ) | ≤ 2 sup |∇ σ | ε ≤ 2 λ 2 ε for λ 2 = √ 3. Theorem 2.4(b) for dimesions n ≥ 3 will be prov ed by using Lemmas Lemma 3.1 (edge ratios) . For an y simplex A on points p 0 , . . . , p n ∈ R n , we hav e | p i − p j | p ( A ) ≤ 2 n for all i , j = 0 , . . . , n . Pr oof. Set d i j = | p i − p j | and use triangle inequalities: 2 p ( A ) = n P k , l = 0 d kl ≥ d i j + P k , i , j ( d ki + d k j ) ≥ d i j + ( n − 1) d i j = nd i j . Definition 3.2 (Cayley-Menger determinant, [5, Chapter II, p.98]) . Let the simplex A on any points p 0 , . . . , p n ∈ R n hav e the ( n + 1) × ( n + 1) matrix D i j of squared Euclidean distances d 2 i j = | p i − p j | 2 for i , j = 0 , . . . , n . The ( n + 2) × ( n + 2) matrix ˆ D is obtained from D by adding the top row (0 , 1 , . . . , 1) and the left column (0 , 1 , . . . , 1) T . The Cayle y-Menger determinant [6, 7] expresses the squared volume of the simple x A as vol 2 ( A ) = ( − 1) n − 1 2 n ( n !) 2 det ˆ D . Proof of Theorem 2.4(a) . Any isometry preserves all distances and hence the strength σ ( A ) expressed via distances in Definition 2.1. Any special orthogonal matrix M ∈ SO( R n ) keeps the sign of a simple x A , i.e. multiplies sign( A ) by det M = 1. Then any rigid motion preserves the signed strength s ( A ) = sign( A ) σ ( A ). T o compute σ ( A ) and s ( A ), we need the half-perimeter p ( A ), which requires a quadratic time O ( n 2 ), and then sign( A ) and vol 2 ( A ) by using determinants of sizes up to n + 2, which can be calculated in time O ( n 3 ) by matrix diagonalization [8, section 11.5]. The r encontre number r n = n ! n P k = 0 ( − 1) k k ! counts all permutations of 1 , . . . , n without a fixed point [9] and equals the integer nearest to n ! e , e.g. r 2 = 1, r 3 = 2, r 4 = 9, r 5 = 44. 4 Lemma 3.3. In the notations of Definition 3.2, we hav e det ˆ D p 2 n ( A ) ≤ r n + 2 2 n ! 2 n for any n ≥ 1. Pr oof. The determinant formula det ˆ D = P ξ ∈ S n + 2 ( − 1) sign( ξ ) ˆ D 1 ,ξ (1) . . . ˆ D n ,ξ ( n ) ov er all permutations ξ ∈ S n + 2 excludes all zeros on the diagonal. Then k , ξ ( k ) for k = 0 , . . . , n + 1, i.e. we can consider only permutations ξ of 0 , . . . , n + 1 that hav e no fixed elements. The number of such permutations is the rencontre number r n + 2 . Then det ˆ D is a sum of r n + 2 non-zero terms, each being a product of n squared distances d i j . The upper bound d i j ≤ 2 n p ( A ) in Lemma 3.1 implies that each term in the sum of det ˆ D is at most 2 n p ( A ) ! 2 n . Then det ˆ D p 2 n ( A ) ≤ r n + 2 2 n ! 2 n as required. For n = 1 in Lemma 3.3, we have det ˆ D = 2 d 2 01 and p ( A ) = d 01 2 , so det ˆ D p 2 ( A ) = 8 ≤ r 3 (2 2 ) = 8 as expected. Lemma 3.4. In the notations of Definition 3.2, we hav e       ∂ det ˆ D ∂ d i j       1 p 2 n − 1 ( A ) ≤ 4( r n + r n + 1 ) 2 n ! 2 n − 1 for any fix ed i , j . Pr oof. Since det ˆ D has d 2 i j in exactly two cells in di ff erent rows and columns, det ˆ D is a quadratic polynomial α d 4 i j + β d 2 i j + γ for some α, β, γ that depend on other fixed distances d kl , d i j . Then ∂ det ˆ D ∂ d i j = 4 α d 3 i j + 2 β d i j . The coe ffi cient α is the determinant of the n × n submatrix ( b i j ) obtained from ˆ D by removing two ro ws and columns index ed by i + 2 , j + 2. For example, fix i = 0 and j = 1. If n = 2, then α =       0 1 1 0       = − 1. If n = 3, then α =         0 1 1 1 0 d 2 23 1 d 2 32 0         = 2 d 2 23 . Since the matrix ( b i j ) has zeros on the main diagonal, its determinant α = P ξ ∈ S n ( − 1) sign( ξ ) b 1 ,ξ (1) . . . b n ,ξ ( n ) is a sum ov er all permutations ξ ∈ S n with no fix ed points. Then the sum α has r n non-zero products b 1 ,ξ (1) . . . b n ,ξ ( n ) and the total degree 2( n − 2) in all distances d kl , d i j . After dividing the polynomial α d 3 i j of the degree 2 n − 1 by p 2 n − 1 ( A ), we use the upper bound d kl p ( A ) ≤ 2 n in Lemma 3.1 to get | α | d 3 i j p 2 n − 1 ( A ) ≤ r n 2 n ! 2 n − 1 . The coe ffi cient β in det ˆ D = α d 4 i j + β d 2 i j + γ is the sum of products from the determinants of the two submatrices obtained from ˆ D by removing row i + 2 and column j + 2 (for one submatrix), then row j + 2 and column i + 2 (for another submatrix). If n = 3, i = 0, j = 2, the determinants are            0 1 1 1 1 d 2 10 0 d 2 13 1 d 2 20 d 2 21 d 2 23 1 d 2 30 d 2 32 0            +            0 1 1 1 1 d 2 01 d 2 02 d 2 03 1 0 d 2 12 d 2 13 1 d 2 31 d 2 32 0            . Since each ( n + 1) × ( n + 1) submatrix includes one entry d 2 i j , we exclude all products with this entry , which are multiplied by the remo ved d 2 i j from ˆ D and hence were counted in α d 4 i j . Hence we can replace d 02 = d 02 with 0 and get β =            0 1 1 1 1 d 2 10 0 d 2 13 1 0 d 2 21 d 2 23 1 d 2 30 d 2 32 0            +            0 1 1 1 1 d 2 01 0 d 2 03 1 0 d 2 12 d 2 13 1 d 2 31 d 2 32 0            . Up to a permutation of indices, each submatrix can be rewritten with the diagonal that has one original d 2 i j (now replaced with 0), while all other diagonal elements are initial zeros. The example above gives β = −            0 1 1 1 1 0 d 2 21 d 2 23 1 d 2 10 0 d 2 13 1 d 2 30 d 2 32 0            −            0 1 1 1 1 0 d 2 12 d 2 13 1 d 2 01 0 d 2 03 1 d 2 31 d 2 32 0            . Similarly to the argument for 5 the determinant α , the sum β d i j contains 2 r n + 1 products of the total degree 2 n − 1, so | β | d i j p 2 n − 1 ( A ) ≤ 2 r n + 1 2 n ! 2 n − 1 by Lemma 3.1. Then the required inequality follo ws:       ∂ det ˆ D ∂ d i j       1 p 2 n − 1 ( A ) = | 4 α d 3 i j + 2 β d i j | p 2 n − 1 ( A ) ≤ (4 r n + 4 r n + 1 ) 2 n ! 2 n − 1 . Pr oof of Theorem 2.4(b) for n ≥ 3 . For n = 3, the squared volume is v ol 2 ( A ) = 1 288               0 1 1 1 1 1 0 d 2 01 d 2 02 d 2 03 1 d 2 10 0 d 2 12 d 2 13 1 d 2 20 d 2 21 0 d 2 23 1 d 2 30 d 2 31 d 2 32 0               . Similarly to the case n = 2, the mean v alue theorem [4] for the strength σ ( A ) = vol 2 ( A ) p 2 n − 1 ( A ) implies that | σ ( A ) − σ ( B ) | ≤ 2 ε sup |∇ σ | ≤ 2 ε s P i , j sup       ∂σ ∂ d i j       2 ≤ 2 ε r n ( n + 1) 2 max i , j sup       ∂σ ∂ d i j       . T o find an upper bound of       ∂σ ∂ d i j       , we initially ignore the numerical factor in the square volume vol 2 ( A ) = ( − 1) n − 1 2 n ( n !) 2 det ˆ D and di ff erentiate det ˆ D · 1 p 2 n − 1 ( A ) by the product rule: ∂ ∂ d i j det ˆ D p 2 n − 1 ( A ) ! = ∂ det ˆ D ∂ d i j · 1 p 2 n − 1 ( A ) − det ˆ D p 2 n ( A ) · 2 n − 1 2 . Lemmas 3.3 and 3.4 belo w imply the upper bound       ∂ ∂ d i j det ˆ D p 2 n − 1 ( A ) !       ≤ (4 r n + 4 r n + 1 ) 2 n ! 2 n − 1 + n − 1 2 ! r n + 2 2 n ! 2 n < ( 4 r n + 4 r n + 1 + 2 r n + 2 ) 2 n ! 2 n − 1 . T aking into account the f actors ( − 1) n − 1 2 n ( n !) 2 in vol 2 ( A ) and r n ( n + 1) 2 for estimating the length of the gradient ∇ σ of n ( n + 1) 2 first order partial derivati ves ∂σ ∂ d i j , we get | σ ( A ) − σ ( B ) | ≤ 2 ε r n ( n + 1) 2 max i , j sup       ∂σ ∂ d i j       ≤ 2 ε c n for the upper bound c n = √ n ( n + 1) / 2 2 n ( n !) 2 ( 4 r n + 4 r n + 1 + 2 r n + 2 ) 2 n ! 2 n − 1 = (2 r n + 2 r n + 1 + r n + 2 ) 2 n − 0 . 5 √ n + 1 ( n !) 2 n 2 n − 1 . 5 . The estimate r n ≤ n ! 2 giv es c n < n ! 2 (2 + 2( n + 1) + ( n + 1)( n + 2)) 2 n − 0 . 5 √ n + 1 ( n !) 2 n 2 n − 1 . 5 = ( n 2 + 5 n + 6) √ n + 1 2 n − 1 . 5 ( n !) n 2 n − 1 . 5 = 1 + 5 n + 6 n 2 ! r 1 + 1 n 2 n − 1 . 5 ( n !) n 2 n − 4 . F or n ≥ 3, we get 1 + 5 n + 6 n 2 ≤ 10 3 and r 1 + 1 n ≤ 2 √ 3 , so c n < 20 3 √ 3 · 2 n − 1 . 5 ( n !) n 2 n − 4 = 5 √ 2 3 √ 3 2 n ( n !) n 2 n − 4 < 2 n + 0 . 5 ( n !) n 2 n − 4 = b n . References [1] O. Anosov a, V . 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