1 Balancing Vectors in Any Norm Aleksandar (Sasho) Nikolov University of Toronto Based on joint work with Daniel Dadush, Kunal Talwar, and Nicole Tomczak-Jaegermann Sasho Nikolov (U of T) Balancing Vectors 1 / 25

2 Introduction Outline 1 Introduction 2 Volume Lower Bound 3 Factorization Upper Bounds 4 Conclusion Sasho Nikolov (U of T) Balancing Vectors 2 / 25

3 . ·‖ U ‖ Natural to consider arbitrary norms: any norm can be written as ∞ Introduction Discrepancy − 1 1 1 1 1 1 1 0 0 0 0 0 0 − 1 0 1 1 0 1 1 0 0 0 0 1 = 0 0 0 0 1 0 0 1 0 0 − 1 − 1 0 0 0 1 1 1 1 1 1 1 1 − 1 ε ‖·‖ ‖ U ) = ‖ , disc( U min ∞ ∞ N ε ∈{± 1 } Sasho Nikolov (U of T) Balancing Vectors 3 / 25

4 Introduction Discrepancy 1 − 1 1 1 1 1 1 0 0 0 0 0 0 1 − 0 1 1 0 1 1 0 0 0 0 1 = 0 0 0 0 1 0 0 1 0 0 1 − 1 − 0 0 0 1 1 1 1 1 1 1 1 1 − U , ‖·‖ ‖ disc( min U ‖ ε ) = ∞ ∞ N } ∈{± ε 1 Natural to consider arbitrary norms: any norm can be written as ‖ U ·‖ . ∞ Sasho Nikolov (U of T) Balancing Vectors 3 / 25

5 n n B : For any u , there exist ,..., u 1] ∈ , Implied by 1 = [ − 1 N ∞ √ + . ‖ u ε ... + u ε ‖ s.t. } +1 , 1 n . ε ∈{− ,...,ε ∞ 1 1 1 N N N n N × U ∈{ 0 , 1 } with at most t [Beck and Fiala, 1981]: For any matrix t ones per column, disc( ) ≤ 2 U − 1 n Implied by u , there exist ,..., u : For any ∈ B 1 N 1 < 2. ,...,ε ‖ ε } u +1 , ... ‖ u ε ε + 1 ∈{− + s.t. 1 1 1 ∞ N N N Most combinatorial discrepancy bounds are implied by geometric vector balancing arguments. Introduction Basic Bounds n × N 1 , } , 0 ∈{ U [Spencer, 1985; Gluskin, 1989]: For any matrix √ . disc( n ) U Sasho Nikolov (U of T) Balancing Vectors 4 / 25

6 × n N ∈{ 0 , 1 } [Beck and Fiala, 1981]: For any matrix U with at most t 2 − t ≤ ) U ones per column, disc( 1 n : For any Implied by , there exist ,..., u B ∈ u 1 N 1 ,...,ε ε ∈{− 1 2. < ‖ , u +1 ε + ... + } u ε ‖ s.t. 1 1 ∞ 1 N N N Most combinatorial discrepancy bounds are implied by geometric vector balancing arguments. Introduction Basic Bounds n × N , 1 } U [Spencer, 1985; Gluskin, 1989]: For any matrix ∈{ , 0 √ U n . ) disc( n n ∈ u B , there exist u : For any Implied by 1] = [ − 1 , ,..., 1 N ∞ √ . n ‖ . ‖ ε u ,...,ε ε + ... + ∈{− u 1 ε , s.t. } +1 ∞ 1 1 1 N N N Sasho Nikolov (U of T) Balancing Vectors 4 / 25

7 n Implied by u , there exist ,..., u : For any ∈ B 1 N 1 < ε ,...,ε ∈{− 1 , +1 } s.t. ‖ ε u + ... + ε u ‖ 2. 1 1 1 ∞ N N N Most combinatorial discrepancy bounds are implied by geometric vector balancing arguments. Introduction Basic Bounds n × N 1 } ∈{ [Spencer, 1985; Gluskin, 1989]: For any matrix , U , 0 √ U disc( ) . n n n 1 u − ,..., , there exist , : For any Implied by 1] B ∈ = [ u 1 N ∞ √ . n ε u ,...,ε ∈{− 1 , +1 } s.t. ‖ ε u + . ... ‖ + ε ∞ 1 1 1 N N N × n N t with at most ∈{ 0 1 U } [Beck and Fiala, 1981]: For any matrix , ones per column, disc( ≤ t 2 − 1 ) U Sasho Nikolov (U of T) Balancing Vectors 4 / 25

8 Most combinatorial discrepancy bounds are implied by geometric vector balancing arguments. Introduction Basic Bounds N n × } ∈{ 1 U 0 [Spencer, 1985; Gluskin, 1989]: For any matrix , , √ n U ) . disc( n n u : For any ∈ B u Implied by = [ − 1 , 1] ,..., , there exist 1 N ∞ √ n . ‖ 1 s.t. ‖ ε ε u +1 + ... + ε ,...,ε u ∈{− } . , ∞ 1 1 1 N N N × n N t with at most U ∈{ 0 , 1 } [Beck and Fiala, 1981]: For any matrix ≤ 2 t − 1 ) U ones per column, disc( n Implied by ,..., u : For any ∈ B u , there exist 1 N 1 ,...,ε + ∈{− 1 , +1 } s.t. ‖ ε 2. u < ε ... + ε ‖ u ∞ 1 1 1 N N N Sasho Nikolov (U of T) Balancing Vectors 4 / 25

9 Introduction Basic Bounds n × N 1 } , ∈{ 0 U , [Spencer, 1985; Gluskin, 1989]: For any matrix √ ) . n U disc( n n Implied by ,..., u , there exist ∈ B : For any u = [ − 1 , 1] 1 N ∞ √ n . ‖ ∈{− , +1 } s.t. ‖ ε ε u ,...,ε + ... + ε . u 1 ∞ 1 1 1 N N N n × N with at most t } ∈{ U [Beck and Fiala, 1981]: For any matrix , 1 0 ones per column, disc( U ) ≤ 2 t − 1 n ∈ , there exist ,..., u Implied by u B : For any 1 N 1 ε 2. ,...,ε < ∈{− 1 , +1 } s.t. ‖ ε ‖ u u + ... + ε 1 1 ∞ 1 N N N Most combinatorial discrepancy bounds are implied by geometric vector balancing arguments. Sasho Nikolov (U of T) Balancing Vectors 4 / 25

10 N t ‖ x ‖ ). = inf { t : x ∈ tK } ; : = disc(( u ‖·‖ ) Minkowski Norm , i K K =1 i Vector Balancing Constant : worst case over sequences in C } { N , ) u = ( disc( U , ‖·‖ U ) : N ∈ N , u C ,..., u ∈ C ) = sup K , vb( 1 i N K i =1 Introduction The Vector Balancing Problem n n ,..., u ), ∈ R ⊂ , and symmetric convex body K u R Given ( K = − K 1 N find the smallest t such that ... + ∈ u } ε : u +1 , 1 ∈{− tK ,...,ε ε ε ∃ + 1 1 1 N N N − u + u u + u 2 1 2 1 u − u 2 1 − u u − 2 1 Sasho Nikolov (U of T) Balancing Vectors 5 / 25

11 C : worst case over sequences in Vector Balancing Constant } { N , u ) disc( U , ‖·‖ = ( ) : N ∈ N U u , ,..., u C ∈ K , C ) = sup vb( 1 i K N i =1 Introduction The Vector Balancing Problem n n K Given ⊂ ∈ R ,..., , and symmetric convex body K u R u ( K = − ), 1 N such that find the smallest t +1 ∈ tK ... + ε : } ε , 1 ∈{− u ,...,ε u ε ∃ + 1 1 1 N N N u + u − u + u 1 2 2 1 u − u 1 2 u u − − 1 2 N x ). ‖·‖ , ) Minkowski Norm u = disc(( t ; } tK ∈ : : t { = inf ‖ ‖ x i K K i =1 Sasho Nikolov (U of T) Balancing Vectors 5 / 25

12 Introduction The Vector Balancing Problem n n u Given ∈ R u , and symmetric convex body K ⊂ R ,..., ( K = − K ), 1 N find the smallest t such that ε tK ,...,ε ∈ ∈{− 1 , +1 ∃ : ε u u ε + ... + } 1 1 1 N N N − u + u u u + 1 2 2 1 u u − 2 1 − u u − 1 2 N ‖ x ‖ ). = inf { t : x ∈ tK } ; t = disc(( u Minkowski Norm ) : ‖·‖ , i K K =1 i C Vector Balancing Constant : worst case over sequences in } { N , ‖·‖ u ) : disc( ∈ N , u U ,..., u ) ∈ C , U = ( N , K ) = sup vb( C 1 i N K i =1 Sasho Nikolov (U of T) Balancing Vectors 5 / 25

13 √ n n n B . ) , [Spencer, 1985; Gluskin, 1989] vb( B ∞ ∞ n n < B , ) B [Beck and Fiala, 1981] vb( 2 ∞ 1 n ( B has , [Banaszczyk, 1998] ) vb ≤ 5 if K K 2 1 ≥ Gaussian measure ( γ K ) n 2 n n , ) B : Prove or disprove vb( . Koml ́os Problem 1. B ∞ 2 √ n n , . B log 2 Banaszczyk’s theorem implies vb( ) n B . ∞ 2 Introduction Questions and Prior Results )? K , K [Dvoretzky, 1963] “What can be said” about vb( ) K , . [B ́ar ́any and Grinberg, 1981] vb( K K for all n ≤ Sasho Nikolov (U of T) Balancing Vectors 6 / 25

14 n ( , ≤ ) B K vb [Banaszczyk, 1998] 5 if K has 2 1 ) ( γ ≥ Gaussian measure K n 2 n n B : Prove or disprove vb( Koml ́os Problem 1. B , . ) ∞ 2 √ n n n , B . log 2 ) . Banaszczyk’s theorem implies vb( B ∞ 2 Introduction Questions and Prior Results K , K )? [Dvoretzky, 1963] “What can be said” about vb( [B ́ar ́any and Grinberg, 1981] vb( K . K ) ≤ K n for all , √ n n ) , [Spencer, 1985; Gluskin, 1989] vb( B n B . ∞ ∞ n n 2 B < , B [Beck and Fiala, 1981] vb( ) ∞ 1 Sasho Nikolov (U of T) Balancing Vectors 6 / 25

15 Introduction Questions and Prior Results , K )? [Dvoretzky, 1963] “What can be said” about vb( K , K ) ≤ n for all K . [B ́ar ́any and Grinberg, 1981] vb( K √ n n B [Spencer, 1985; Gluskin, 1989] vb( ) . B n , ∞ ∞ n n , B [Beck and Fiala, 1981] vb( 2 ) < B ∞ 1 n , B [Banaszczyk, 1998] ( vb K ) ≤ 5 if K has 2 1 ≥ ( K ) Gaussian measure γ n 2 n n B . 1. , B Koml ́os Problem : Prove or disprove vb( ) ∞ 2 √ n n , B Banaszczyk’s theorem implies vb( . ) . B log 2 n ∞ 2 Sasho Nikolov (U of T) Balancing Vectors 6 / 25

16 Introduction Vector Balancing and Rounding N N , 1] C , any U = ( w For any ) ∈ [0 , and any symmetric convex , u ∈ u i i =1 i N , there exists a x ∈{ 0 , 1 } K such that C ‖ − Uw ‖ . ≤ vb( Ux , K ) K Sasho Nikolov (U of T) Balancing Vectors 7 / 25

17 C , K ) is tight up to a A natural volumetric lower bound on vb( ) factor. O (log n N The proof implies an efficient algorithm to compute ε 1 , 1 } ∈{− given K , ). C ) vb( n (1 + log . ∈ u ,..., u ‖ C , so that ‖ ε u + ... + ε u N 1 N 1 1 N K Also rounding version. ) is tight up to K An efficiently computable upper bound on vb( C , . n factors polynomial in log Based on an optimal application of Banaszczyks’ theorem. ). Implies an efficient approximation algorithm for vb( C , K The results extend to hereditary discrepancy with respect to arbitrary norms. Prior work [Bansal, 2010; Nikolov and Talwar, 2015] implies bounds which . deteriorate with the number of facets of K Introduction Our Results lower bounds ) and K , C on vb( We initiate a systematic study of and upper its computational complexity: Sasho Nikolov (U of T) Balancing Vectors 8 / 25

18 An efficiently computable upper bound on vb( C , K ) is tight up to factors polynomial in log n . Based on an optimal application of Banaszczyks’ theorem. C , K ). Implies an efficient approximation algorithm for vb( The results extend to hereditary discrepancy with respect to arbitrary norms. Prior work [Bansal, 2010; Nikolov and Talwar, 2015] implies bounds which K deteriorate with the number of facets of . Introduction Our Results K lower bounds on vb( C , We initiate a systematic study of upper and ) and its computational complexity: ) is tight up to a K , C A natural volumetric lower bound on vb( n (log O ) factor. N 1 The proof implies an efficient algorithm to compute ∈{− given } 1 , ε , K ). u ∈ C , so that ‖ ε u u + ... + ε u ‖ . (1 + log ,..., ) vb( C n N 1 N 1 N 1 K Also rounding version. Sasho Nikolov (U of T) Balancing Vectors 8 / 25

19 The results extend to hereditary discrepancy with respect to arbitrary norms. Prior work [Bansal, 2010; Nikolov and Talwar, 2015] implies bounds which deteriorate with the number of facets of K . Introduction Our Results , and lower bounds on vb( C upper K We initiate a systematic study of ) and its computational complexity: A natural volumetric lower bound on vb( C , K ) is tight up to a n (log ) factor. O N given } 1 , 1 ε The proof implies an efficient algorithm to compute ∈{− ε u ,..., + ... + ε u ‖ . (1 + log n ) vb( C , K ). u ∈ u , so that ‖ C N 1 N 1 N 1 K Also rounding version. ) is tight up to An efficiently computable upper bound on vb( C , K . factors polynomial in log n Based on an optimal application of Banaszczyks’ theorem. ). K , C Implies an efficient approximation algorithm for vb( Sasho Nikolov (U of T) Balancing Vectors 8 / 25

20 Prior work [Bansal, 2010; Nikolov and Talwar, 2015] implies bounds which deteriorate with the number of facets of K . Introduction Our Results on vb( K ) and lower bounds C and upper We initiate a systematic study of , its computational complexity: K ) is tight up to a A natural volumetric lower bound on vb( C , (log ) factor. n O N } given The proof implies an efficient algorithm to compute ε ∈{− 1 , 1 . , so that ε u C + ... + ε u ‖ ‖ (1 + log n ) vb( C , K ). ∈ ,..., u u 1 K N 1 N 1 N Also rounding version. K ) is tight up to An efficiently computable upper bound on vb( , C factors polynomial in log n . Based on an optimal application of Banaszczyks’ theorem. ). Implies an efficient approximation algorithm for vb( C , K The results extend to hereditary discrepancy with respect to arbitrary norms. Sasho Nikolov (U of T) Balancing Vectors 8 / 25

21 Introduction Our Results and on vb( C , K ) and upper lower bounds We initiate a systematic study of its computational complexity: A natural volumetric lower bound on vb( K ) is tight up to a C , (log n ) factor. O N ∈{− 1 , 1 } The proof implies an efficient algorithm to compute given ε + ,..., ∈ C , so that ‖ ε u u + ... u ε u ‖ . (1 + log n ) vb( C , K ). 1 N 1 N 1 K N Also rounding version. C , K ) is tight up to An efficiently computable upper bound on vb( factors polynomial in log n . Based on an optimal application of Banaszczyks’ theorem. Implies an efficient approximation algorithm for vb( C , K ). The results extend to hereditary discrepancy with respect to arbitrary norms. Prior work [Bansal, 2010; Nikolov and Talwar, 2015] implies bounds which deteriorate with the number of facets of K . Sasho Nikolov (U of T) Balancing Vectors 8 / 25

22 Volume Lower Bound Outline 1 Introduction 2 Volume Lower Bound 3 Factorization Upper Bounds 4 Conclusion Sasho Nikolov (U of T) Balancing Vectors 9 / 25

23 K . U vb( C , ) is more robust, but not about a specific matrix is a robust analog of discrepancy: Hereditary discrepancy disc( ) = max , ) K K , U hd( , U S S ⊆ [ N ] is the submatrix of . indexed by where U = ( u ) S U i S i ∈ S Observation : } { N u ,..., , u ∈ C ) , N ∈ . U u vb( C , K ) = sup = ( hd( U , K ) : N 1 i N =1 i Volume Lower Bound Hereditary Discrepancy U Issue : disc( U ) is , ‖·‖ , K ) = disc( K not robust to slight changes in U (e.g. repeat each column) hard to approximate [Charikar, Newman, and Nikolov, 2011] Sasho Nikolov (U of T) Balancing Vectors 10 / 25

24 is a robust analog of discrepancy: Hereditary discrepancy , , U hd( ) = max K disc( U K , ) S S ] ⊆ [ N = ( U . u ) is the submatrix of U indexed by S where i i S ∈ S Observation : { } N U ∈ C , U N = ( u ) : hd( ∈ N , ) . u ) = sup , ,..., u vb( C , K K 1 i N =1 i Volume Lower Bound Hereditary Discrepancy ) is : disc( U , Issue ) = disc( U , ‖·‖ K K not robust to slight changes in U (e.g. repeat each column) hard to approximate [Charikar, Newman, and Nikolov, 2011] , vb( C . K ) is more robust, but not about a specific matrix U Sasho Nikolov (U of T) Balancing Vectors 10 / 25

25 : Observation { } N U , C ∈ u ,..., u , N ∈ N ) : K , U hd( . ) = sup K , C vb( ) u = ( 1 i N =1 i Volume Lower Bound Hereditary Discrepancy U ) is U , K ) = disc( ‖·‖ : disc( Issue , K not robust to slight changes in (e.g. repeat each column) U hard to approximate [Charikar, Newman, and Nikolov, 2011] . U K , C vb( ) is more robust, but not about a specific matrix Hereditary discrepancy is a robust analog of discrepancy: disc( U U , K ) , K , ) = max hd( S ⊆ [ N ] S ) u . U where is the submatrix of U indexed by S = ( i S ∈ i S Sasho Nikolov (U of T) Balancing Vectors 10 / 25

26 Volume Lower Bound Hereditary Discrepancy ‖·‖ U ) = disc( U , K , ) is : disc( Issue K (e.g. repeat each column) not robust to slight changes in U hard to approximate [Charikar, Newman, and Nikolov, 2011] C , K ) is more robust, but not about a specific matrix U . vb( Hereditary discrepancy is a robust analog of discrepancy: U , K ) = max hd( , ) K , disc( U S N [ ⊆ ] S U . is the submatrix of u where ) S indexed by U = ( i S i S ∈ Observation : { } N K ) = sup u hd( U , K ) : N ∈ N , vb( C ,..., u , ∈ C , U = ( u ) . 1 i N i =1 Sasho Nikolov (U of T) Balancing Vectors 10 / 25

27 [Lov ́asz, Spencer, and Vesztergombi, 1986]: ⋃ N ( = hd( , K ), then [0 , 1] t ⊆ If x + tL ). U N x ∈{ } 1 , 0 [Banaszczyk, 1993]: N N tL ) ) ≥ vol( , ) = t 1 = vol([0 vol( L 1] Volume Lower Bound The Volume Lower Bound N ”. x : the set of “good } K ∈ Ux L Define R ∈ x { = : N U . ∅} = 6 disc( } 1 , 1 ∩{− tL : t { ) = min K , Sasho Nikolov (U of T) Balancing Vectors 11 / 25

28 [Banaszczyk, 1993]: N N vol( 1] , 1 = vol([0 ≥ tL ) = t ) vol( L ) Volume Lower Bound The Volume Lower Bound N } L ”. = { x : the set of “good Define K ∈ Ux : ∈ R x N ∅} disc( , K ) = min { t : tL ∩{− 1 , 1 } U 6 = . [Lov ́asz, Spencer, and Vesztergombi, 1986]: ⋃ N , 1] ( x + tL ). If ⊆ t = hd( U , K ), then [0 N } 1 , 0 ∈{ x Sasho Nikolov (U of T) Balancing Vectors 11 / 25

29 Volume Lower Bound The Volume Lower Bound N ∈ x ∈ R = : L { K } : the set of “good x ”. Define Ux N , K ) = min { t : tL ∩{− 1 , 1 } disc( 6 = ∅} . U [Lov ́asz, Spencer, and Vesztergombi, 1986]: ⋃ N = hd( U , K ), then [0 , 1] If ⊆ t ). tL + x ( N } 1 , 0 ∈{ x [Banaszczyk, 1993]: N N 1] ) 1 = vol([0 ≥ vol( tL ) = t , vol( L ) Sasho Nikolov (U of T) Balancing Vectors 11 / 25

30 Volume Lower Bound The Volume Lower Bound N ∈ x ∈ R L : Ux { K } : the set of “good x ”. Define = N , disc( ) = min { t : tL ∩{− 1 , 1 } U 6 = ∅} . K [Lov ́asz, Spencer, and Vesztergombi, 1986]: ⋃ N , = hd( ), then [0 , 1] t ⊆ If K ). ( x + tL U N ∈{ 0 , 1 } x [Banaszczyk, 1993]: N N − 1 / N vol( tL ) = t , vol( L ) ⇐⇒ hd( U , K ) ≥ vol( L ) 1] ) 1 = vol([0 ≥ . Sasho Nikolov (U of T) Balancing Vectors 11 / 25

31 Lower Bound on vb( C , K ): { } N u K , . ) : ,..., u ∈ C K , ) volLB( ≥ ) K C vb( u volLB(( , ) = sup C 1 i N i =1 Theorem n , C R , U K matrix N × n For any ⊂ , and any symmetric convex ) , U hd( ≤ ) . (1 + log n ) · K volLB( U U , K K ) volLB( , volLB( , K · ) n (1 + log . ) K C C vb( ≤ ) K , C volLB( ) , Volume Lower Bound A Hereditary Volume Lower Bound A simple strengthening: | S | / 1 − S } K ∈ x vol( U : ) R ∈ x { . hd( U , K ) ) = max K , U volLB( ≥ S S ⊆ [ N ] Sasho Nikolov (U of T) Balancing Vectors 12 / 25

32 Theorem n For any R ⊂ K , , , and any symmetric convex U matrix N × n C ≤ ) K ) U volLB( K K , U volLB( · ) n (1 + log . ) , U hd( , n (1 + log . ) K , C vb( ≤ · ) K , C volLB( volLB( ) C , K ) Volume Lower Bound A Hereditary Volume Lower Bound A simple strengthening: | S | S / 1 − , K ) ≥ volLB( U , K ) = max ∈ R vol( : U x ∈ . hd( K U { x ) } S ⊆ N [ ] S K , C Lower Bound on vb( ): } { N . u , ,..., u K ∈ C ) : ) u volLB(( C , ) ≥ vb( K , K ) = sup C volLB( 1 i N =1 i Sasho Nikolov (U of T) Balancing Vectors 12 / 25

33 Volume Lower Bound A Hereditary Volume Lower Bound A simple strengthening: | S | / 1 − S , hd( K . U , K vol( { x ∈ R U : U ) = max x ∈ K } ) volLB( ≥ ) S ] N [ ⊆ S C , K ): Lower Bound on vb( } { N . u ) : K , C ∈ u ,..., u ) volLB(( C ) = sup K , , volLB( ≥ vb( ) K C 1 i N =1 i Theorem n C N matrix For any , and any symmetric convex n , K ⊂ R × , U volLB( U , K ) ≤ hd( U , K ) . (1 + log n ) · volLB( U , K ) (1 + log volLB( , K ) ≤ vb( C , K ) . C n ) · volLB( C , K ) Sasho Nikolov (U of T) Balancing Vectors 12 / 25

34 Volume Lower Bound Rothvoß’s Algorithm n K ⊂ R , Algorithm [Rothvoß, 2014]: given 1 G ∼ N (0 , I Sample a standard Gaussian ); n 2 Output X G 2 n [ ‖ {‖ X : x ∈ K ∩ − − 1 , 1] G } . = arg min x 2 Goal : |{ i : X . ∈{− 1 , +1 }}|≥ α n for a constant α i X is a partial coloring .) ( n : If K is “big enough,” then in an average direction ∂ [ − 1 , 1] Intuition is closer to the origin than ∂ K and is more likely to be hit by X . Sasho Nikolov (U of T) Balancing Vectors 13 / 25

35 Volume Lower Bound Rothvoß’s Algorithm n K , R ⊂ Algorithm [Rothvoß, 2014]: given 1 ∼ N (0 , I Sample a standard Gaussian ); G n 2 Output X G 2 n G ‖ = arg min {‖ : x ∈ K ∩ [ − 1 , 1] x } . X − 2 Goal : |{ i : X . ∈{− 1 , +1 }}|≥ α n for a constant α i is a partial coloring .) X ( n K is “big enough,” then in an average direction ∂ [ − 1 , 1] is Intuition : If ∂ and is more likely to be hit by X . K closer to the origin than α there is a δ so that [Rothvoß, 2014] For any small enough − δ n γ , then ( K ) ≥ e if K has Gaussian measure n , |{ i : X . ∈{− 1 with high probability +1 }|≥ α n i Sasho Nikolov (U of T) Balancing Vectors 13 / 25

36 Volume Lower Bound Rothvoß’s Algorithm n ⊂ R , K Algorithm [Rothvoß, 2014]: given 1 ∼ N (0 , I Sample a standard Gaussian ); G n 2 Output X G 2 n G ‖ [ X : x ∈ K ∩ = arg min − 1 , 1] {‖ } . x − 2 α |{ i : X Goal ∈{− 1 , +1 }}|≥ : n for a constant α . i ( X is a partial coloring .) n : If K is “big enough,” then in an average direction ∂ [ − 1 , 1] Intuition is closer to the origin than ∂ X . K and is more likely to be hit by there is a δ so that [Rothvoß, 2014] For any small enough α (1 − δ ) n subspace W for which if there exists a dimension − δ n has Gaussian measure γ , then ( K ∩ W ) ≥ e K ∩ W W , |{ i : X . ∈{− 1 with high probability +1 }}|≥ α n i Sasho Nikolov (U of T) Balancing Vectors 13 / 25

37 Proof by an algorithm: ) and recurse. Find a partial coloring with discrepancy . volLB( U , K 1 I = ; Preprocess so that N = n , U n 2 , t , tK Apply Rothvoß’s algorithm to volLB( ); K I n If conditions hold , gives a partial coloring X ∈ tK ; S 3 on K ; Project } 1 < X < 1 − : { = S and recurse. i R i Need a “recentered” variant of Rothvoß’s algorithm. k 1 n k so that 1 + log . iterations, we have X After ,... X n k 1 ∈{− } ; X ... + 1 1 X , + k 1 ‖ X ) ≤ kt . (1 + log n + ) volLB( I ... , K + . X ‖ n K : Show that the conditions of Rothvoß’s algorithm are Main Challenge satisfied. Volume Lower Bound Tightness of the Volume Lower Bound N n × n and symmetric convex K ⊂ R R ∈ U Need to show: for any K , U . hd( ) K , U volLB( · ) n (1 + log . ) Sasho Nikolov (U of T) Balancing Vectors 14 / 25

38 1 ; I = U , Preprocess so that = N n n 2 tK I volLB( ); t , Apply Rothvoß’s algorithm to K , n If conditions hold , gives a partial coloring X ∈ tK ; S 3 − : { = S and recurse. i R on K ; Project } 1 < X < 1 i Need a “recentered” variant of Rothvoß’s algorithm. k 1 X X ,... so that After iterations, we have n 1 + log . k k n 1 1 } X ; X + ... ∈{− 1 + , 1 k ≤ ‖ , K ) . X n . kt ) volLB( + ‖ I X + ... (1 + log n K : Show that the conditions of Rothvoß’s algorithm are Main Challenge satisfied. Volume Lower Bound Tightness of the Volume Lower Bound × n N n R Need to show: for any ∈ and symmetric convex U R ⊂ K U volLB( · ) n (1 + log . ) K , U hd( . ) K , Proof by an algorithm: U Find a partial coloring with discrepancy . volLB( ) and recurse. , K Sasho Nikolov (U of T) Balancing Vectors 14 / 25

39 1 k . 1 + log n iterations, we have X k ,... X After so that 1 k n + ... + X X ∈{− 1 ; } , 1 1 k ≤ ) + ... + X X ‖ K ‖ kt . (1 + log n ) volLB( I , . n K Main Challenge : Show that the conditions of Rothvoß’s algorithm are satisfied. Volume Lower Bound Tightness of the Volume Lower Bound n × N n and symmetric convex ∈ R K R ⊂ Need to show: for any U U , volLB( ) . (1 + log n ) · hd( U . ) K , K Proof by an algorithm: , U volLB( . Find a partial coloring with discrepancy ) and recurse. K 1 = N n , Preprocess so that = I ; U n 2 , K ); volLB( I , t tK Apply Rothvoß’s algorithm to n ; tK ∈ X , gives a partial coloring If conditions hold S 3 and recurse. { i : − 1 < X S < 1 } ; Project K on R = i Need a “recentered” variant of Rothvoß’s algorithm. Sasho Nikolov (U of T) Balancing Vectors 14 / 25

40 Main Challenge : Show that the conditions of Rothvoß’s algorithm are satisfied. Volume Lower Bound Tightness of the Volume Lower Bound N n n × ⊂ R R ∈ and symmetric convex U Need to show: for any K K . ) ) · volLB( U , (1 + log ) . hd( K , U n Proof by an algorithm: , K ) and recurse. Find a partial coloring with discrepancy . volLB( U 1 n U = I Preprocess so that ; N = , n 2 K I tK , ); , t volLB( Apply Rothvoß’s algorithm to n tK ; ∈ X If conditions hold , gives a partial coloring S 3 : − 1 < X S < 1 } ; Project K on R = and recurse. { i i Need a “recentered” variant of Rothvoß’s algorithm. 1 k k After ,... X n so that iterations, we have X . 1 + log n k 1 X 1 , 1 } ... ; + + ∈{− X k 1 + X X ‖ ‖ ≤ kt . (1 + log n ) volLB( I + , K ) . ... n K Sasho Nikolov (U of T) Balancing Vectors 14 / 25

41 Volume Lower Bound Tightness of the Volume Lower Bound × N n n ⊂ K ∈ R and symmetric convex U R Need to show: for any , K hd( . (1 + log n ) · volLB( U , K ) . U ) Proof by an algorithm: . volLB( U , K ) and recurse. Find a partial coloring with discrepancy 1 Preprocess so that n , U = = N ; I n 2 tK , t volLB( I Apply Rothvoß’s algorithm to , K ); n If conditions hold , gives a partial coloring X ∈ tK ; S 3 = { i : − 1 < X and recurse. < 1 } ; Project K on R S i Need a “recentered” variant of Rothvoß’s algorithm. k 1 n iterations, we have X . ,... k 1 + log so that After X 1 k n X X + ∈{− 1 , 1 } ... ; + 1 k ‖ + ... + X X ‖ . ≤ kt . (1 + log n ) volLB( I ) , K n K Main Challenge : Show that the conditions of Rothvoß’s algorithm are satisfied. Sasho Nikolov (U of T) Balancing Vectors 14 / 25

42 From the definition of volLB( , K ): I n S S ⊆ [ n ] : vol((volLB( I · , K ) K ) ∩ R ∀ ) ≥ 1 . n Theorem (Structural result) = ) For any δ there exists a m m ( δ so that the following holds. S ⊆ S for all 1 ≥ ) R be a symmetric convex body s.t. . ] n [ Let L ∩ vol( L − for which n ) δ (1 of dimension W There exists a subspace − n δ . e ≥ ) W ∩ ) mL (( γ W K K I Apply to , L = volLB( ) · to get that the conditions of Rothvoß’s n algorithm are satisfied. Volume Lower Bound From Volume To Gaussian Measure For Rothvoß’s algorithm, we need that on some subspace of large K , t I volLB( ), has large Gaussian measure. , tK dimension, the body n Sasho Nikolov (U of T) Balancing Vectors 15 / 25

43 Theorem (Structural result) m so that the following holds. ) δ ( For any = m there exists a δ S be a symmetric convex body s.t. Let for all 1 ≥ ) S R ∩ L vol( ⊆ L [ n ] . ) (1 − δ of dimension W There exists a subspace n for which n − δ . ) ∩ W ) γ e (( mL ≥ W Apply to L = volLB( I to get that the conditions of Rothvoß’s , K ) · K n algorithm are satisfied. Volume Lower Bound From Volume To Gaussian Measure For Rothvoß’s algorithm, we need that on some subspace of large dimension, the body tK , t volLB( I ), has large Gaussian measure. , K n From the definition of volLB( I , K ): n S S . ≥ ) ∀ R ∩ ) K · ) K , 1 I ] : vol((volLB( n [ ⊆ n Sasho Nikolov (U of T) Balancing Vectors 15 / 25

44 Volume Lower Bound From Volume To Gaussian Measure For Rothvoß’s algorithm, we need that on some subspace of large K volLB( I dimension, the body , ), has large Gaussian measure. t tK , n , K ): I From the definition of volLB( n S ∀ n ] : vol((volLB( I S , K ) · K ) ∩ R ⊆ ) ≥ 1 . [ n Theorem (Structural result) ( there exists a m = m δ δ ) so that the following holds. For any S L be a symmetric convex body s.t. vol( L ∩ R Let ) ≥ 1 for all S ⊆ [ n ] . There exists a subspace of dimension (1 − δ ) n for which W n δ − e mL W ) ≥ ∩ (( . γ ) W Apply to L = volLB( I to get that the conditions of Rothvoß’s , K ) · K n algorithm are satisfied. Sasho Nikolov (U of T) Balancing Vectors 15 / 25

45 2 Approximate a general convex body L by an appropriate ellipsoid. -ellipsoid, [Milman, 1986; Pisier, 1989]) Theorem (Regular M n L ⊆ R For any symmetric convex there exists an ellipsoid E such that for any ≥ t 1 t cn / ) ) e }≤ tL , E ( N , max { N ( L , tE , where c is a constant. K , . K N ( needed to cover L L ) = number of translates of L . E preserves “large scale” information about S S L has large volume. E ⇒ has large volume = R ∩ R ∩ has large Gaussian W ∩ L ⇒ has large Gaussian measure = W ∩ E measure. Volume Lower Bound Proof Ideas Generally applicable strategy: n 1 ). B ( T = E Prove the theorem for an ellipsoid 2 Reduces to linear algebra! Sasho Nikolov (U of T) Balancing Vectors 16 / 25

46 S S ∩ L has large volume = ⇒ E ∩ R has large volume. R E ∩ W has large Gaussian measure = ⇒ L ∩ W has large Gaussian measure. Volume Lower Bound Proof Ideas Generally applicable strategy: n 1 ( B = T Prove the theorem for an ellipsoid E ). 2 Reduces to linear algebra! 2 L Approximate a general convex body by an appropriate ellipsoid. Theorem (Regular M -ellipsoid, [Milman, 1986; Pisier, 1989]) n L For any symmetric convex ⊆ R there exists an ellipsoid E such that for 1 ≥ t any t / cn tE ) , N ( E , tL ) }≤ e max { N , ( L , where is a constant. c needed to cover L ) = number of translates of . K ( N , K L E L . preserves “large scale” information about Sasho Nikolov (U of T) Balancing Vectors 16 / 25

47 Volume Lower Bound Proof Ideas Generally applicable strategy: n 1 ). ( B Prove the theorem for an ellipsoid = E T 2 Reduces to linear algebra! 2 Approximate a general convex body L by an appropriate ellipsoid. M -ellipsoid, [Milman, 1986; Pisier, 1989]) Theorem (Regular n L R there exists an ellipsoid E such that for ⊆ For any symmetric convex t ≥ 1 any cn / t ( L , tE ) , N ( E , tL ) }≤ e max { N , where c is a constant. K ( , L ) = number of translates of N needed to cover K . L E preserves “large scale” information about L . S S R has large volume = ⇒ E ∩ R L has large volume. ∩ E ∩ W has large Gaussian measure = ⇒ L ∩ W has large Gaussian measure. Sasho Nikolov (U of T) Balancing Vectors 16 / 25

48 volLB( U , K )? Is the hereditary discrepancy of partial colorings The hereditary discrepancy of partial colorings is U . volLB( , K ). A lower bound would follow from Conjecture n K 1 Suppose ⊂ R ≤ is a symmetric convex body of volume . Then there √ S | exists a . ) diam . R ∩ K ( | S s.t. S ⊆ [ n ] ` 2 True for ellipsoids and reduces to the Restricted Invertibility Principle. S K R True for general bodies with an arbitrary subspace if we replace . | S | with dim W W and Volume Lower Bound Partial Colorings . K , U ) volLB( n (1 + log ) is in general tight. ) K , U The bound hd( Sasho Nikolov (U of T) Balancing Vectors 17 / 25

49 ). K , The hereditary discrepancy of partial colorings is . volLB( U A lower bound would follow from Conjecture n K 1 . Then there Suppose is a symmetric convex body of volume ⊂ R ≤ √ S | R ∩ K ( diam s.t. ] n [ ⊆ S ) exists a . | S . ` 2 True for ellipsoids and reduces to the Restricted Invertibility Principle. S True for general bodies K if we replace R with an arbitrary subspace W . | S | with dim W and Volume Lower Bound Partial Colorings ) volLB( The bound hd( U , K ) . ) is in general tight. K , U (1 + log n )? Is the hereditary discrepancy of partial colorings volLB( U , K Sasho Nikolov (U of T) Balancing Vectors 17 / 25

50 A lower bound would follow from Conjecture n R is a symmetric convex body of volume Suppose . Then there ⊂ 1 K ≤ √ S ] n [ ⊆ . | S | S . ) ( R ∩ K exists a diam s.t. ` 2 True for ellipsoids and reduces to the Restricted Invertibility Principle. S True for general bodies K if we replace R with an arbitrary subspace . | S | with dim W W and Volume Lower Bound Partial Colorings ) is in general tight. The bound hd( U , K ) . (1 + log n ) volLB( U , K Is the hereditary discrepancy of partial colorings volLB( U , K )? The hereditary discrepancy of partial colorings is ). . volLB( U , K Sasho Nikolov (U of T) Balancing Vectors 17 / 25

51 True for ellipsoids and reduces to the Restricted Invertibility Principle. S R if we replace K True for general bodies with an arbitrary subspace and W | S | with dim W . Volume Lower Bound Partial Colorings K , K ) is in general tight. U ) volLB( n (1 + log . ) The bound hd( U , K U )? Is the hereditary discrepancy of partial colorings , volLB( U , K . The hereditary discrepancy of partial colorings is ). volLB( A lower bound would follow from Conjecture n 1 Suppose ≤ is a symmetric convex body of volume . Then there R ⊂ K √ S | S | . ⊆ [ n ] s.t. diam exists a S ( . R ∩ K ) ` 2 Sasho Nikolov (U of T) Balancing Vectors 17 / 25

52 Volume Lower Bound Partial Colorings ) is in general tight. , ) . (1 + log n ) volLB( U , U K The bound hd( K volLB( U , K )? Is the hereditary discrepancy of partial colorings . volLB( U , K ). The hereditary discrepancy of partial colorings is A lower bound would follow from Conjecture n ⊂ R is a symmetric convex body of volume Suppose ≤ 1 . Then there K √ S ⊆ [ n ] s.t. diam . | S | exists a S R . ∩ K ( ) ` 2 True for ellipsoids and reduces to the Restricted Invertibility Principle. S K R True for general bodies if we replace with an arbitrary subspace W and | S | with dim W . Sasho Nikolov (U of T) Balancing Vectors 17 / 25

53 Factorization Upper Bounds Outline 1 Introduction 2 Volume Lower Bound 3 Factorization Upper Bounds 4 Conclusion Sasho Nikolov (U of T) Balancing Vectors 18 / 25

54 K ). , C We do not know how to efficiently compute volLB( We need a natural K , C ). on vb( upper bound Recall [Banaszczyk, 1998]: 1 n n ≥ γ 5. R ⊂ K For any convex ≤ ( ) such that ) , vb( B K , K n 2 2 : Observations 1 I . 1 for G ≥ ) K (2 ≤ γ ), then ‖ ‖ , (0 N ∼ If E G n n K 2 n vb( ‖ ) . G K , ‖ B E . K 2 ). vb( diam · ) C ‖ G ‖ C ( ( . ) K , E K ` 2 Last bound can be very loose! Can we do better? Factorization Upper Bounds Upper Bounds from Banaszczyk’s Theorem We showed how to efficiently compute near optimal signs u ε ,...,ε ∈{− 1 , 1 } . for any u ,..., 1 1 N N But what if we want to compute vb( C K ) or hd( )? U , , K Sasho Nikolov (U of T) Balancing Vectors 19 / 25

55 Recall [Banaszczyk, 1998]: 1 n n K , For any convex R B , vb( K ≤ 5. ) K ( ⊂ γ such that ) ≥ n 2 2 : Observations 1 (0 If . E ) (2 ‖ γ ), then G I , K N ∼ G 1 for ≤ ‖ ≥ n n K 2 n . . B vb( E ) K , G ‖ ‖ K 2 E C ( diam · vb( ‖ G ‖ ). ( . ) K , C ) K ` 2 Last bound can be very loose! Can we do better? Factorization Upper Bounds Upper Bounds from Banaszczyk’s Theorem We showed how to efficiently compute near optimal signs ε ,..., ,...,ε u ∈{− 1 , 1 } for any u . 1 1 N N K , C But what if we want to compute vb( )? K , U ) or hd( C , K ). We do not know how to efficiently compute volLB( upper bound We need a natural ). on vb( C , K Sasho Nikolov (U of T) Balancing Vectors 19 / 25

56 Observations : 1 ) ‖ G ‖ . ≤ 1 for G ∼ N (0 , E K ), then γ (2 If ≥ I n n K 2 n ) K , ‖ G B vb( ‖ E . . K 2 ‖ C vb( C , K ) . ( E ( G ‖ diam ) · ). K ` 2 Last bound can be very loose! Can we do better? Factorization Upper Bounds Upper Bounds from Banaszczyk’s Theorem We showed how to efficiently compute near optimal signs ε ∈{− . 1 , 1 } ,...,ε u u ,..., for any 1 1 N N But what if we want to compute vb( C )? K , U ) or hd( K , We do not know how to efficiently compute volLB( K , ). C ). upper bound C We need a natural K on vb( , Recall [Banaszczyk, 1998]: 1 n n , vb( ) K ≥ ) K ( , γ such that 5. R ⊂ K For any convex ≤ B n 2 2 Sasho Nikolov (U of T) Balancing Vectors 19 / 25

57 ‖ vb( , K ) . ( E ‖ G C ). ) · diam C ( ` K 2 Last bound can be very loose! Can we do better? Factorization Upper Bounds Upper Bounds from Banaszczyk’s Theorem We showed how to efficiently compute near optimal signs ∈{− 1 , 1 } for any u ε ,..., u ,...,ε . 1 1 N N But what if we want to compute vb( )? C , , ) or hd( K K U We do not know how to efficiently compute volLB( C , ). K , C on vb( upper bound We need a natural ). K Recall [Banaszczyk, 1998]: 1 n n ( K ) ≥ ⊂ R , vb( B For any convex such that , K ) ≤ 5. γ K n 2 2 : Observations 1 ∼ . E ‖ G ‖ ≤ 1 for G If N (0 , I ), then ) K (2 ≥ γ n n K 2 n , ‖ E . ) K . ‖ vb( G B K 2 Sasho Nikolov (U of T) Balancing Vectors 19 / 25

58 Factorization Upper Bounds Upper Bounds from Banaszczyk’s Theorem We showed how to efficiently compute near optimal signs } ∈{− 1 , 1 ε for any u ,..., u ,...,ε . 1 1 N N , K ) or hd( U , K C But what if we want to compute vb( )? C , K ). We do not know how to efficiently compute volLB( We need a natural on vb( C , K ). upper bound Recall [Banaszczyk, 1998]: 1 n n K such that γ 5. ( K ) For any convex ⊂ ≤ , vb( B R ) , K ≥ n 2 2 Observations : 1 If G ‖ . ≤ 1 for G ∼ N (0 , I ‖ ), then γ E (2 K ) ≥ n n K 2 n B ) . , K vb( . E ‖ G ‖ K 2 C , K ) . ( E vb( G ‖ C ) · diam ). ( ‖ ` K 2 Last bound can be very loose! Can we do better? Sasho Nikolov (U of T) Balancing Vectors 19 / 25

59 λ , C ( K achieving T Take a linear map ); ( )) = 1, so G ‖ E ); C ( T ‖ , diam Can assume = λ ( C K ` ( ) T K 2 , ) C ( T ) = vb( K , C vb( )) and apply Banaszczyk’s theorem. K ( T Factorization Upper Bounds A Better Upper Bound n C into B Idea using a linear map. : Map 2 T . } a linear map T )) : ( ( C ) ‖ G ‖ E diam { ) = inf K , C ( λ · ( ` T ( K ) 2 ( C , K ). Claim : vb( C , K ) . λ Sasho Nikolov (U of T) Balancing Vectors 20 / 25

60 C vb( , K ) = vb( T ( C ) , T ( K )) and apply Banaszczyk’s theorem. Factorization Upper Bounds A Better Upper Bound n using a linear map. C into B Idea : Map 2 )) : ( C T a linear map } . ( T · ( ‖ { E ‖ G ) = inf K , C ( λ ) diam ` ) K ( T 2 : vb( C , K ) . λ ( C , K ). Claim Take a linear map T achieving λ ( C ); K , ); )) = 1, so E = λ ( C , K ( ( ‖ G diam Can assume ‖ T C ` ) K ( T 2 Sasho Nikolov (U of T) Balancing Vectors 20 / 25

61 Factorization Upper Bounds A Better Upper Bound n : Map B using a linear map. Idea C into 2 C } a linear map T )) : ( T ( . G ‖ λ ( C , ) · diam K ) = inf { ( E ‖ ` ) K ( T 2 Claim : vb( C , K ) . λ ( C , K ). Take a linear map λ ( C , K ); achieving T diam , ); ( T K C )) = 1, so E ‖ G ‖ Can assume = λ ( C ( ` K ) T ( 2 ) C , K ) = vb( T ( C vb( , T ( K )) and apply Banaszczyk’s theorem. Sasho Nikolov (U of T) Balancing Vectors 20 / 25

62 Proof outline : 1 Formulate λ ( C , K ) as a convex minimization problem; 2 Derive the Lagrange dual: an equivalent maximization problem; 3 Relate dual solutions to the volume lower bound. Factorization Upper Bounds Tightness of the Upper Bound Theorem n C , K ⊂ R For any symmetric convex , λ ) K , C ( C , K . . vb( C , K ) . λ ( ) 2 / 5 (1 + log ) n and a vertex K Moreover, given membership oracle access to ( C , K ) . representation of C , we can efficiently compute λ × n N , and then U ∈ R ± For a matrix } , we can take C = conv {± ,..., u u 1 N C , U ) approximates hd( K , K ( λ ). Sasho Nikolov (U of T) Balancing Vectors 21 / 25

63 Factorization Upper Bounds Tightness of the Upper Bound Theorem n For any symmetric convex R C , ⊂ K , , K ) C λ ( K . , . vb( C , K ) . λ ( C ) 5 / 2 ) n (1 + log Moreover, given membership oracle access to K and a vertex representation of C , we can efficiently compute λ ( C , K ) . n × N {± R For a matrix , we can take C = conv ∈ u , and then ,..., ± u } U 1 N λ ( C , K ) approximates hd( U , K ). Proof outline : 1 Formulate λ ( C , K ) as a convex minimization problem; 2 Derive the Lagrange dual: an equivalent maximization problem; 3 Relate dual solutions to the volume lower bound. Sasho Nikolov (U of T) Balancing Vectors 21 / 25

64 − 1 ∗ T T ‖ : G ‖ is defined entirely by A = T , because the E Observation K − 1 − 1 G covariance of A T is given by . Formulation : ( λ C ) A ( f ) = inf K , ∈ 〈 s.t. x , Ax 〉≤ 1 C ∀ x A 0 . ∗ 1 − ; f T T T such that T for any A ‖ G = ‖ E ) = A ( K A ; is well defined over positive definite f C diam ( The first constraint encodes 1: ≤ )) T ( ` 2 ∗ 2 = 〈 , x 〈 〉 〉 Ax , x 〈 Tx , Tx Tx 〉 = ‖ Tx ‖ . = T 2 Factorization Upper Bounds Convex Formulation 1 − ‖ ‖ x T = ‖ x ‖ K ) K ( T − 1 ≤ )) C ( T ( } 1 : : inf ‖ G diam First attempt T ‖ E { ` K 2 Not convex : the objective is ∞ for T = 0 and finite for any invertible 1 T − )). + ( ( T , but 0 = T 2 Sasho Nikolov (U of T) Balancing Vectors 22 / 25

65 : Formulation C ) = inf ( λ K , f ( A ) ∀ x x , Ax 〉≤ 1 〈 s.t. ∈ C A 0 . ∗ − 1 for any A ) = E ‖ T = f G ‖ T such that ; A T T ( K is well defined over positive definite A ; f ( diam 1: The first constraint encodes T ≤ C )) ( ` 2 2 ∗ Ax x , . 〉 = 〈 x , T = Tx 〉 Tx , Tx 〉 = ‖ Tx ‖ 〈 〈 2 Factorization Upper Bounds Convex Formulation 1 − ‖ ‖ x T = ‖ x ‖ K ) K ( T − 1 )) } C ( 1 ≤ T ( G E diam First attempt : inf ‖ : T ‖ { ` K 2 T for ∞ : the objective is Not convex = 0 and finite for any invertible 1 T ( T + ( − )). , but 0 = T 2 1 − ∗ , because the Observation : G ‖ E is defined entirely by A = T ‖ T T K − 1 − 1 covariance of G is given by A T . Sasho Nikolov (U of T) Balancing Vectors 22 / 25

66 C The first constraint encodes diam ( T ( 1: )) ≤ ` 2 ∗ 2 , Ax 〉 = 〈 . , T 〈 Tx 〉 = 〈 Tx , Tx 〉 = ‖ Tx ‖ x x 2 Factorization Upper Bounds Convex Formulation 1 − = ‖ ‖ x T x ‖ ‖ K T ( K ) − 1 ( T ( C )) ≤ 1 } G : diam ‖ T First attempt : inf { E ‖ ` K 2 : the objective is Not convex = 0 and finite for any invertible T for ∞ 1 + ( )). T ( T − T , but 0 = 2 ∗ 1 − , because the T : E ‖ T G Observation T ‖ = A is defined entirely by K 1 1 − − covariance of T G is given by A . Formulation : ( A ) λ ( C , K ) = inf f C s.t. 〈 , Ax 〉≤ 1 ∀ x ∈ x A 0 . ∗ − 1 ; ( f ) = T such that T for any A ‖ G = T T ‖ E A K ; A f is well defined over positive definite Sasho Nikolov (U of T) Balancing Vectors 22 / 25

67 Factorization Upper Bounds Convex Formulation − 1 x ‖ x = ‖ T ‖ ‖ K ) ( K T 1 − } ( T ( C )) ≤ 1 ‖ : T E { G ‖ First attempt : inf diam ` K 2 : the objective is ∞ for Not convex = 0 and finite for any invertible T 1 ( T + ( − T )). T , but 0 = 2 − 1 ∗ T Observation E G ‖ , because the ‖ A = T T : is defined entirely by K − 1 − 1 is given by covariance of T G . A Formulation : λ ( C , K ) = inf f ( A ) s.t. 〈 , Ax 〉≤ 1 ∀ x ∈ C x . A 0 − 1 ∗ E ‖ T f ( G ‖ A for any T such that T ) = T = A ; K f is well defined over positive definite A ; The first constraint encodes diam 1: ≤ ( T ( C )) ` 2 ∗ 2 , Ax 〉 = 〈 x , T = Tx 〉 〈 〈 Tx , Tx 〉 = ‖ Tx ‖ x . 2 Sasho Nikolov (U of T) Balancing Vectors 22 / 25

68 ), and, K , C Each dual solution gives a lower bound on volLB( , K therefore, on vb( ); C -convexity, and Sudakov minoration; Tools: K ) gives a lower bound on vb( K , = ( λ ⇒ ). K , C C Computation : The convex optimization problem can be solved using the ellipsoid method, given a membership oracle for K and a vertex . representation of C Factorization Upper Bounds Properties of the Formulation The function f ( A ) is convex in A , and the constraints are also convex; dual maximization equivalent : there exists an Lagrange Duality ); problem, whose value also equals λ ( U , C Sasho Nikolov (U of T) Balancing Vectors 23 / 25

69 Computation : The convex optimization problem can be solved using the ellipsoid method, given a membership oracle for K and a vertex representation of C . Factorization Upper Bounds Properties of the Formulation A The function f ( , and the constraints are also convex; A ) is convex in dual maximization Lagrange Duality equivalent : there exists an ( U , C ); problem, whose value also equals λ ), and, K C Each dual solution gives a lower bound on volLB( , K ); therefore, on vb( C , -convexity, and Sudakov minoration; K Tools: λ , K ). C ( K ⇒ = , ) gives a lower bound on vb( C Sasho Nikolov (U of T) Balancing Vectors 23 / 25

70 Factorization Upper Bounds Properties of the Formulation f ( ) is convex in A , and the constraints are also convex; A The function equivalent dual maximization : there exists an Lagrange Duality problem, whose value also equals ( U , C ); λ Each dual solution gives a lower bound on volLB( C , K ), and, therefore, on vb( C , K ); Tools: -convexity, and Sudakov minoration; K = ⇒ λ ( C , K ) gives a lower bound on vb( C , K ). Computation : The convex optimization problem can be solved using the ellipsoid method, given a membership oracle for K and a vertex representation of C . Sasho Nikolov (U of T) Balancing Vectors 23 / 25

71 Conclusion Outline 1 Introduction 2 Volume Lower Bound 3 Factorization Upper Bounds 4 Conclusion Sasho Nikolov (U of T) Balancing Vectors 24 / 25

72 : Open questions ) give lower bounds on partial colorings? Does volLB( C , K , .) vb( ` )? (True for K K volLB( ) K , K p ) be improved? Can the bounds for λ ( C , K Conclusion Conclusion In this work : Tightness of natural upper and lower bounds for vector balancing. Efficient algorithms to find nearly optimal vector balancing signs, and K , C to compute vb( ), and hereditary discrepancy with respect to any norm. Our results strongly use the geometry of the underlying discrepancy problem. Sasho Nikolov (U of T) Balancing Vectors 25 / 25

73 Conclusion Conclusion : In this work Tightness of natural upper and lower bounds for vector balancing. Efficient algorithms to find nearly optimal vector balancing signs, and to compute vb( C , K ), and hereditary discrepancy with respect to any norm. Our results strongly use the geometry of the underlying discrepancy problem. Open questions : Does volLB( C , K ) give lower bounds on partial colorings? vb( K , K ) volLB( K , K )? (True for ` .) p , λ ( C Can the bounds for K ) be improved? Sasho Nikolov (U of T) Balancing Vectors 25 / 25

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