Lower Bound on the Optimal Access Bandwidth of (K+2,K,2)-MDS Array Code with Degraded Read Friendly
Ting-Yi Wu, Yunghsiang S. Han, Zhengrui Li, Bo Bai, Gong Zhang, Liang Chen, Xiang Wu
aa r X i v : . [ c s . I T ] F e b Lower Bound on the Optimal AccessBandwidth of ( K + 2 , K, )-MDS Array Codewith Degraded Read Friendly Ting-Yi Wu ∗ , Yunghsiang S. Han †∗ , Zhengrui Li ‡∗ , Bo Bai ∗ , Gong Zhang ∗ , Liang Chen ∗ , Xiang Wu ∗∗ Huawei Technologies Co., Ltd. † Dongguan University of Technology ‡ University of Science and Technology of China
I. P
RELIMINARY
Notation: • F q denotes the field of size q . • ∅ denotes empty set. • [ N ] , { , , , . . . , N } and [0] , ∅ . • I , . • O , . • R , β β : β i ∈ F q for all i ∈ [2] and β β = 0 . • L , β β : β i ∈ F q for all i ∈ [2] and β β = 0 . • M , B = β β β β
6∈ {L ∪ R ∪ O } : β i ∈ F q for all i ∈ [4] • M inv , B = β β β β ∈ M , and B is invertible : β i ∈ F q for all i ∈ [4] We let the parity-check matrix of the ( K + 2 , K )-MDS array be H = I I · · · IA A · · · A N , (1) February 5, 2021 DRAFT where N = K + 2 and these upper identities make the degraded read friendly. To satisfy MDS property, I IA i A j must be invertible for any i = j , which is equivalent to that ( A i + A j ) must be invertible [1] for any i = j , Forany node i ∈ [ N ] , it stores two symbols as a vector α ( i ) , α ( i )1 α ( i )2 .To repair the i th node, we have to find two check equations from the row space of H in order to retrieve α ( i ) .Let M be a repair matrix of size × , MH denotes two check equations from the row space of H such that MH , h H H · · · H N i , (2)and H α (1) + H α (2) + · · · + H N α ( N ) = , (3)where H i = M IA i for all i ∈ [ N ] . If M can be used to recover α ( i ) , we have to make sure that H i is invertibleand recover α ( i ) as α ( i ) = H − i X j ∈ [ N ] \{ i } H j α ( j ) . (4)To be able to repair each node, we have to design the corresponding repair matrix for each α ( i ) . Since somerepair matrices can be used to recover multiple nodes, i.e., M IA i and M IA j are both invertible for some i = j ,we are not restricted to find N different repair matrices for recovering all N nodes. Let R be the number of repairmatrices, ≤ R ≤ N , and {I (1) , I (2) , . . . , I ( R ) } be a partition of [ N ] , a repair process of size R is formed by R repair matrices, M (1) , M (2) , . . . , M ( R ) , where M ( r ) can be used to recover α ( i ) is i ∈ I ( r ) . The formal definitionof a repiar process of size R is given below. Definition 1.
A repair process P R of R repair matrices is defined as P R , n(cid:16) M ( r ) , I ( r ) (cid:17) : r ∈ [ R ] o , (5) such that I (1) , I (2) , . . . , I ( R ) form a partition of [ N ] and M ( r ) IA i is invertible for i ∈ I ( r ) . We then explain the process of recovering α ( i ) by a P R , where i ∈ I ( r ) . Let M ( r ) H , h H ( r )1 H ( r )2 · · · H ( r ) N i , (6)where H ( r ) i = M ( r ) IA i . By Definition 1, H ( r ) i is invertible, therefore α ( i ) = h H ( r ) i i − X j ∈ [ N ] \{ i } H ( r ) j α ( j ) . (7) February 5, 2021 DRAFT
Definition 2.
The repair bandwidth for repairing node i , where i ∈ I ( r ) , by using P R is denoted by B i ( P R ) = X j ∈ [ N ] \{ i } B i,j ( P R ) , (8) and B i,j ( P R ) = , if i = j or H ( r ) j = O ;1 , if H ( r ) j ∈ L or H ( r ) j ∈ R ;2 , otherwise. (9)Note that B i,j ( P R ) denotes the number of symbols needed to be download from j to repair node i by usingrepair process P R . Lemma 1.
Given a P R , B i,j ( P R ) = B j,i ( P R ) = 2 if i = j and i, j ∈ I ( r ) for r ∈ [ R ] .Proof: By Definition 1, H ( r ) i and H ( r ) j are both invertible since i, j ∈ I ( r ) . Therefore, B i,j ( P R ) = B j,i ( P R ) = 2 by (9). Lemma 2.
Given a P R , B i ( P R ) = B j ( P R ) when both i, j ∈ I ( r ) for some r .Proof: By Definition 2, B i ( P R ) = P k ∈ [ N ] \{ i } B i,k ( P R ) and B i,k ( P R ) = B j,k ( P R ) for all k ∈ [ N ] \ { i, j } .Since B i,j ( P R ) = B j,i ( P R ) , we have B i ( P R ) = X k ∈ [ N ] \{ i } B i,k ( P R ) (10) = X k ∈ [ N ] \{ i,j } B i,k ( P R ) + B i,j ( P R ) (11) = X k ∈ [ N ] \{ i,j } B j,k ( P R ) + B j,i ( P R ) (12) = X k ∈ [ N ] \{ j } B j,k ( P R ) = B j ( P R ) . (13)From Lemma 2, we have B i ( P R ) = B j ( P R ) for all i, j ∈ I ( r ) and i = j . Hence, we define B ( r ) ( P R ) as below. Definition 3.
Let B ( r ) ( P R ) be the repair bandwidth for repairing the node in I ( r ) , i.e., B ( r ) ( P R ) = B i ( P R ) forall i ∈ I ( r ) . Lemma 3.
The total repair bandwidth of P R is B ( P R ) , P i ∈ [ N ] B i ( P R ) = P r ∈ [ R ] |I ( r ) | B ( r ) ( P R ) , where |I ( r ) | is the number of distinct elements in set I ( r ) . February 5, 2021 DRAFT
Proof:
By Definition 3, X i ∈ [ N ] B i ( P R ) = X r ∈ [ R ] X i ∈I ( r ) B i ( P R ) (14) = X r ∈ [ R ] X i ∈I ( r ) B ( r ) ( P R ) (15) = X r ∈ [ R ] |I ( r ) | B ( r ) ( P R ) . (16)(17) Example. • H = , where N = 3 . • P = (cid:0) M ( r ) , I ( r ) (cid:1) r ∈ [2] , where M (1) = and I (1) = { } , M (2) = and I (2) = { , } . • M (1) H = h H (1)1 H (1)2 H (1)3 i = , h B R R i , where B ∈ M inv and R i ∈ R forall i ∈ [2] . M (2) H = h H (2)1 H (2)2 H (2)3 i = , h L B B i , where B i ∈ M inv for all i ∈ [2] and L ∈ R . • Repairing node 1: H (1)1 α (1) + H (1)2 α (2) + H (1)3 α (3) = ⇒ α (1) = h H (1)1 i − (cid:16) H (1)2 α (2) + H (1)3 α (3) (cid:17) (18) ⇒ α (1) = − α (2) + α (3) . (19) • Repairing node 2: H (2)1 α (1) + H (2)2 α (2) + H (2)3 α (3) = ⇒ α (2) = h H (2)2 i − (cid:16) H (2)1 α (1) + H (2)3 α (3) (cid:17) (20) ⇒ α (2) = − α (2) + α (3) . (21) • Repairing node 3:
February 5, 2021 DRAFT H (2)1 α (1) + H (2)2 α (2) + H (2)3 α (3) = ⇒ α (3) = h H (2)3 i − (cid:16) H (2)1 α (1) + H (2)2 α (2) (cid:17) (22) ⇒ α (2) = − α (1) + α (2) . (23) • B i,j ( P R ) = , B ( P R ) = 2 , B ( P R ) = 3 , and B ( P R ) = 3 . II. S
OME PROPERTIES
Lemma 4.
Given a P R and r ∈ [ R ] , if there is a B i,j ( P R ) = 0 for some i ∈ I ( r ) and j ∈ [ N ] \ I ( r ) , then B ( r ) ( P R ) = B i ( P R ) = 2 K .Proof: Since i = j , B i,j ( P R ) = 0 implies H ( r ) j = O due to (9). Let M ( r ) = h M ( r )1 M ( r )2 i be a repair matrixof P R , we have an invertible matrix H ( r ) i such that H ( r ) i = H ( r ) i + H ( r ) j = M ( r ) IA i + M ( r ) IA j = M ( r ) I + IA i + A j = h M ( r )1 M ( r )2 i OA i + A j = M ( r )1 O + M ( r )2 ( A i + A j ) = M ( r )2 ( A i + A j ) . (24)Since H ( r ) i and ( A i + A j ) are invertible, we conclude that M ( r )2 is also invertible. From (24), we have H ( r ) k = M ( r )2 ( A k + A j ) for all k ∈ [ N ] . Since both M ( r )2 and ( A k + A j ) are invertible if k = j , H ( r ) k is invertible if k = j .Therefore, from (9), we have B ( r ) ( P R ) = B i ( P R ) = P k ∈ [ N ] \{ i } B i,k ( P R ) = P k ∈ [ N ] \{ i,j } N −
2) = 2 K . Example. M ( r ) H = , h B B B O i is okay, but M ( r ) H = , h B B R O i is impossible, where R ∈ R and B i ∈ M inv for all i ∈ [5] . Lemma 5.
Given a P R , then M ( r ) is of rank for all r ∈ [ R ] . Note.
This lemma can be applied to N − K ≥ . February 5, 2021 DRAFT
Proof:
Since rank( AB ) ≤ min { rank( A ) , rank( B ) } and M ( r ) IA i is invertible if i ∈ I ( r ) , rank M ( r ) IA i = 2 ≤ min rank( M ( r ) ) , rank IA i (25) ≤ rank( M ( r ) ) (26) ≤ . (27)Therefore, rank( M ( r ) ) = 2 for all r ∈ [ R ] . Lemma 6.
Given a P R , if H ( r ) i and H ( r ) i are both in L (or R ), H ( r ) j and H ( r ) j are both in L (or R ) for some r , r ∈ [ R ] , i, j ∈ [ N ] \ {I ( r ) ∪ I ( r ) } , and i = j , then M ( r ) M ( r ) is of rank . Note.
This lemma can be applied to N − K ≥ .Proof: Since I IA i A j is invertible and rank( A ) = rank( AB ) if B is of full rank, rank M ( r ) M ( r ) = rank M ( r ) M ( r ) I IA i A j = rank H ( r ) i H ( r ) j H ( r ) i H ( r ) j ≤ . (28)Combining (28) with Lemma 5, we conclude that M ( r ) M ( r ) is of rank for all r , r ∈ [ R ] . Lemma 7. If M ( r ) = TM ( r ) for some r , r ∈ [ R ] , r = r , and invertible matrix T , then both M ( r ) and M ( r ) can be used to repair nodes in I ( r ) ∪ I ( r ) and B ( r ) ( P R ) = B ( r ) ( P R ) .Proof: Since T is invertible, H ( r ) i is invertible if and only if H ( r ) i = TH ( r ) i is invertible. With the factsthat H ( r ) i is invertible when i ∈ I ( r ) and H ( r ) j is invertible when j ∈ I ( r ) , we can conclude that both H ( r ) k and H ( r ) k are invertible when k ∈ I ( r ) ∪ I ( r ) . Therefore, both H ( r ) and H ( r ) can be used to repair nodes in I ( r ) ∪ I ( r ) .From (9), B i,j ( P R ) = 1 , for some i ∈ I ( r ) , denotes that H ( r ) j ∈ {L ∪ R} , so H ( r ) j = TH ( r ) j ∈ {L ∪ R} .Also, B i,j ( P R ) = 1 , for some i ∈ I ( r ) , denotes that H ( r ) j ∈ {L ∪ R} , so H ( r ) j = T − M ( r ) j ∈ {L ∪ R} .Furthermore, B i,j ( P R ) = 0 , for some i ∈ I ( r ) and i = j , denotes that H ( r ) j = O , so H ( r ) j = TH ( r ) j = O .Also, B i,j ( P R ) = 0 , for some i ∈ I ( r ) and i = j , denotes that H ( r ) j = O , so H ( r ) j = T − M ( r ) j = O .We then prove that if B i,j ( P R ) = 2 for some i ∈ I ( r ) and i = j , then H ( r ) j
6∈ { O ∪ R ∪ L} . Assuming H ( r ) j ∈ { O ∪ R ∪ L} , so H ( r ) j = T − H ( r ) j ∈ { O ∪ R ∪ L} . Hence we can conclude that B i,j ( P R ) = 2 from (9). February 5, 2021 DRAFT
By contradiction, B i,j ( P R ) = 2 , for some i ∈ I ( r ) and i = j , implies H ( r ) j
6∈ { O ∪ R ∪ L} . Following the similarway, we can also prove that if B i,j ( P R ) = 2 for some i ∈ I ( r ) and i = j , then H ( r ) j
6∈ { O ∪ R ∪ L} .Combining above results and (9), we can conclude that B i,k ( P R ) = B j,k ( P R ) for some i ∈ I ( r ) , j ∈ I ( r ) , andfor all k ∈ [ N ] \ { i, j } . Therefore, B ( r ) ( P R ) = B ( r ) ( P R ) . Theorem 1.
For any P R such that H ( r ) i and H ( r ) i are both in L (or R ), H ( r ) j and H ( r ) j are both in L (or R )for some r , r ∈ [ R ] , and i, j ∈ [ N ] \ {I ( r ) ∪ I ( r ) } , where r = r and i = j , then the repair process P R − = n(cid:16) M ( r ) , I ( r ) (cid:17) : r ∈ [ R ] \ { r , r } o ∪ n(cid:16) M ( r ) , I ( r ) ∪ I ( r ) (cid:17)o (29) has the same total repair bandwidth as P R has.Proof: By Lemmas 5 and 6, we can have an invertible transform matrix T such that M ( r ) = TM ( r ) . Thenaccording to Lemma 7, M ( r ) can repair nodes in I ( r ) and B ( r ) ( P R ) = B ( r ) ( P R ) . Therefore, B ( P R ) = X r ∈ [ R ] |I ( r ) | B ( r ) ( P R ) (30) = X r ∈ [ R ] \{ r ,r } |I ( r ) | B ( r ) ( P R ) + |I ( r ) | B ( r ) ( P R ) + |I ( r ) | B ( r ) ( P R ) (31) = X r ∈ [ R ] \{ r ,r } |I ( r ) | B ( r ) ( P R ) + |I ( r ) | B ( r ) ( P R ) + |I ( r ) | B ( r ) ( P R ) (32) = X r ∈ [ R ] \{ r ,r } |I ( r ) | B ( r ) ( P R ) + |I ( r ) + I ( r ) | B ( r ) ( P R ) (33) = B ( P R − ) . (34) Lemma 8.
Given a P R , if B i ( P R ) < K for some i ∈ I ( r ) , then B i ( P R ) = 2( N − − (cid:12)(cid:12)(cid:12)n j : j ∈ [ N ] \ { i } and H ( r ) j ∈ {L ∪ R} o(cid:12)(cid:12)(cid:12) (35) = 2( N − − |{ j : j ∈ [ N ] \ { i } and B i,j ( P R ) = 1 }| , (36) where H ( r ) j = M ( r ) IA j .Proof: Due to Lemma 4 and B i ( P R ) < K , we can not have B i,j ( P R ) = 0 for all j ∈ [ N ] \ { i } . By Definition February 5, 2021 DRAFT B i ( P R ) = X j ∈ [ N ] \{ i } B i,j ( P R ) (37) = 2 (cid:12)(cid:12)(cid:12) { j : j ∈ [ N ] \ { i } and H ( r ) j ∈ M} (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12) { j : j ∈ [ N ] \ { i } and H ( r ) j ∈ {L ∪ R}} (cid:12)(cid:12)(cid:12) (38) = 2( N − − (cid:12)(cid:12)(cid:12) { j : j ∈ [ N ] \ { i } and H ( r ) j ∈ {L ∪ R}} (cid:12)(cid:12)(cid:12) (39) = 2( N − − |{ j : j ∈ [ N ] \ { i } and B i,j ( P R ) = 1 }| (40)We then present the properties of repair matrices of any P R . Lemma 9.
Given a P R , if B i ( P R ) < K for some i ∈ I ( r ) , then M ( r )2 is not invertible, where M ( r ) = h M ( r )1 M ( r )2 i .Proof: Since B i ( P R ) < K , we have B i ( P R ) = 2( N − − (cid:12)(cid:12)(cid:12)n j : j ∈ [ N ] \ { i } and H ( r ) j ∈ {L ∪ R} o(cid:12)(cid:12)(cid:12) < K, (41)which induces (cid:12)(cid:12)(cid:12)n j : j ∈ [ N ] \ { i } and H ( r ) j ∈ {L ∪ R} o(cid:12)(cid:12)(cid:12) > N − − K = 2 . (42)By pigeonhole principle, there must be j , j ∈ [ N ] \ { i } , such that j = j and both H ( r ) j , H ( r ) j ∈ L (or R ).WLOG, we assume H ( r ) j , H ( r ) j ∈ L and then we have H ( r ) j + H ( r ) j = M ( r ) IA j + M ( r ) IA j (43) = h M ( r )1 M ( r )2 i OA j + A j (44) = M ( r )2 ( A j + A j ) ∈ L . (45)Since M ( r )2 ( A j + A j ) is not invertible and A j + A j is invertible, we can conclude that M ( r )2 is not invertible. Lemma 10.
Given a P R = (cid:8)(cid:0) M ( r ) , I ( r ) (cid:1) : r ∈ [ R ] (cid:9) and ˆ r ∈ [ R ] such that B i ( P R ) < K for some i ∈ I (ˆ r ) . Byletting ˆ M (ˆ r ) = h H (ˆ r ) i i − M (ˆ r ) , (1) we can construct an repair process ˆ P R as ˆ P R = n(cid:16) M ( r ) , I ( r ) (cid:17) : r ∈ [ R ] \ { ˆ r } o ∪ n(cid:16) ˆ M (ˆ r ) , I (ˆ r ) (cid:17)o , (46) where B i ( P R ) = B i ( ˆ P R ) . February 5, 2021 DRAFT (2)
For ˆ P R , ˆ M (ˆ r ) = β (ˆ r )1 β (ˆ r )2 β (ˆ r )3 β (ˆ r )4 , (47) if ∃ j , j ∈ [ N ] and j = j such that ˆ H (ˆ r ) j , ˆ H (ˆ r ) j ∈ L , where ˆ H (ˆ r ) j = ˆ M (ˆ r ) IA j for all j ∈ [ N ] and β (ˆ r ) k ∈ F q for all k ∈ [4] . (3) For ˆ P R , ˆ M (ˆ r ) = β (ˆ r )1 β (ˆ r )2 β (ˆ r )3 β (ˆ r )4 , (48) if ∃ j , j ∈ [ N ] and j = j such that ˆ H (ˆ r ) j , ˆ H (ˆ r ) j ∈ R .Proof: (1) is followed directly from Lemma 7.Next we prove (2). Since ˆ M (ˆ r ) = h H (ˆ r ) i i − M (ˆ r ) , ˆ H (ˆ r ) i , ˆ M (ˆ r ) IA i = h H (ˆ r ) i i − M (ˆ r ) IA i = h H (ˆ r ) i i − H (ˆ r ) i = I . (49)Since B i ( ˆ P R ) < K , ˆ M (ˆ r )2 is not invertible due to Lemma 9.We first prove that ˆ H (ˆ r ) j = β for some β ∈ F q if ˆ H (ˆ r ) j ∈ L . Let ˆ H (ˆ r ) j = β ′ β ∈ L for some β, β ′ ∈ F q ,then we have ˆ H (ˆ r ) i + ˆ H (ˆ r ) j = I + β ′ β = β ′ β = ˆ M (ˆ r )2 ( A i + A j ) . (50)Since ˆ M (ˆ r )2 is not invertible, then β ′ β is not invertible as well, which implies that the determinant of β ′ β is . Therefore, × (1 + β ′ ) = 0 implies β ′ = 1 .We next prove that ˆ M (ˆ r )2 = β (ˆ r )3 β (ˆ r )4 for some β (ˆ r )3 , β (ˆ r )4 ∈ F q , if ∃ j , j ∈ [ N ] and j = j such that ˆ H (ˆ r ) j , ˆ H (ˆ r ) j ∈ L . Let ˆ H (ˆ r ) j = β and ˆ H (ˆ r ) j = β , we have ˆ H (ˆ r ) j + ˆ H (ˆ r ) j = β + β (51) = ˆ M (ˆ r )2 ( A j + A j ) . (52)Since ( A j + A j ) is invertible, then ˆ M (ˆ r )2 = β + β ( A j + A j ) − = β (ˆ r )3 β (ˆ r )4 , (53) February 5, 2021 DRAFT0 for some β (ˆ r )3 , β (ˆ r )4 ∈ F q . Lastly, we prove that ˆ M (ˆ r )1 = β (ˆ r )1 β (ˆ r )2 for some β (ˆ r )1 , β (ˆ r )2 ∈ F q if ˆ M (ˆ r )2 = β (ˆ r )3 β (ˆ r )4 for some β (ˆ r )3 , β (ˆ r )4 ∈ F q . Since ˆ H (ˆ r ) i = h ˆ M (ˆ r )1 ˆ M (ˆ r )2 i IA i = ˆ M (ˆ r )1 + ˆ M (ˆ r )2 A i = I (54)and ˆ M (ˆ r )2 = β (ˆ r )3 β (ˆ r )4 , ˆ M (ˆ r )1 = ˆ M (ˆ r )2 A i + I = β (ˆ r )1 β (ˆ r )2 for some β (ˆ r )1 , β (ˆ r )2 ∈ F q .Therefore, we can conclude that ˆ M (ˆ r ) = β (ˆ r )1 β (ˆ r )2 β (ˆ r )3 β (ˆ r )4 , if ∃ j , j ∈ [ N ] and j = j such that ˆ H (ˆ r ) j , ˆ H (ˆ r ) j ∈ L .A similar proof can be made for (3). Theorem 2.
Given a P R and let J i , { j : j ∈ [ N ] and B i,j ( P R ) = 1 } , if B i ( P R ) < K for some i ∈ I ( r ) , theneither H ( r ) j ∈ L for all j ∈ J i or H ( r ) j ∈ R for all j ∈ J i .Proof: From Lemma 8 and B i ( P R ) < K , we can find at least different indexes from [ N ] \ I ( r ) , j , j ,and j , such that H ( r ) j , H ( r ) j , H ( r ) j ∈ {L ∪ R} . Hence, |J i | ≥ . By pigeonhole principle, there must be two of n H ( r ) j , H ( r ) j , H ( r ) j o are either in L or R .WLOG, we assume H ( r ) j , H ( r ) j ∈ L . By Lemma 10, we can construct a repair matrix ˆ M ( r ) = β ( r )1 β ( r )2 β ( r )3 β ( r )4 , (55)where ˆ M ( r ) = h H ( r ) i i − M ( r ) and β ( r ) k ∈ F q for all k ∈ [4] . Suppose there is a ˆ j ∈ [ N ] \ { j , j } such that H ( r )ˆ j ∈ R , then H ( r )ˆ j = M ( r ) IA ˆ j = H ( r ) i ˆ M ( r ) IA ˆ j = H ( r ) i β β = H ( r ) i ˆ H ( r )ˆ j (56)for some β , β ∈ F q . Since ˆ H ( r )ˆ j ∈ {L ∪ R} , we have β = 0 and ˆ H ( r )ˆ j ∈ L . Therefore, H ( r )ˆ j = H ( r ) i ˆ H ( r )ˆ j ∈ L ,which contradicts to the assumption of H ( r )ˆ j ∈ R . Therefore, we can prove that, for all j ∈ J i , H ( r ) j ∈ L .A similar proof can be made for H ( r ) j ∈ R for all j ∈ J i when we assume H ( r ) j , H ( r ) j ∈ R . February 5, 2021 DRAFT1
III. L
OWER BOUND ON B ( P R ) Let P R be the set of the all effective repair processes of size R , our goal is to minimizes the total repairbandwidth over P R for a given R , i.e., B ∗ ( R ) , min P R ∈P R B ( P R ) . (57)Let J r ( P R ) , n j : j ∈ [ N ] and H ( r ) j ∈ {L ∪ R} o , (58)as stated in Lemmas 3 and 8, the total repair bandwidth of P R is B ( P R ) = X r ∈ [ R ] |I ( r ) | × h N − − |J r ( P R ) | i (59) = X r ∈ [ R ] |I ( r ) | × h N − i − X r ∈ [ R ] |I ( r ) | × |J r ( P R ) | (60) = N × h N − i − X r ∈ [ R ] |I ( r ) | × |J r ( P R ) | . (61)Therefore, B ∗ ( R ) = min P R ∈P R B ( P R ) = N × h N − i − max P R ∈P R X r ∈ [ R ] |I ( r ) | × |J r ( P R ) | , (62)which means that minimizing B ( P R ) over P R ∈ P R is equivalent to maximizing P r ∈ [ R ] |I ( r ) | × |J r ( P R ) | over P R ∈ P R .For any repair process P R , we define H P R as H P R , M (1) M (2) ... M ( R ) H = H (1)1 H (1)2 · · · H (1) N H (2)1 H (2)2 · · · H (2) N ... ... . . . ... H ( R )1 H ( R )2 · · · H ( R ) N . (63)From Theorem 2, we know that either H ( r ) i ∈ L for all i ∈ J r ( P R ) or H ( r ) i ∈ R for all i ∈ J r ( P R ) . Hence wecan categorize all repair matrices into types. Since we can swap any M ( i ) and M ( j ) , i, j ∈ [ R ] , without effectingthe total repair bandwidth, WLOG, we assume that there is a ≤ R L ≤ R such that H ( r ) k ∈ L if k ∈ J r ( P R ) and r ∈ [ R L ] (64)and H ( r ) k ∈ R if k ∈ J r ( P R ) and r ∈ [ R ] \ [ R L ] . (65) From Theorem 1, we know that P R can be reduced to P R − if there exists different r , r ∈ [ R ] and different i, j ∈ [ N ] such that H ( r ) i , H ( r ) i ∈ L (or R ) and H ( r ) j , H ( r ) j ∈ L (or R ). For those repair processes of size R satisfy the above constraint are said to be noteffective and excluded from P R to avoid duplicate computation. February 5, 2021 DRAFT2
Since we can also swap any A i and A j in H without effecting the total repair bandwidth of P R , WLOG, wecan assume that any i ∈ I ( r ) is less than any j ∈ I ( r +1) for all r ∈ [ R − . Let ℓ r , |I ( r ) | for all r ∈ [ R ] , then {I ( r ) } r ∈ [ R ] can be uniquely decided by ~ℓ , ( ℓ , ℓ , . . . , ℓ R ) as I (1) = { , , . . . , ℓ } = [ ℓ ∗ ]; (66) I (2) = { ℓ ∗ + 1 , ℓ ∗ + 2 , . . . , ℓ ∗ } = [ ℓ ∗ ] \ [ ℓ ∗ ]; (67)... (68) I ( r ) = (cid:8) ℓ ∗ r − + 1 , ℓ ∗ r − + 2 , . . . , ℓ ∗ r (cid:9) = [ ℓ ∗ r ] \ [ ℓ ∗ r − ]; (69)... (70) I ( R ) = (cid:8) ℓ ∗ R − + 1 , ℓ ∗ R − + 2 , . . . , N (cid:9) = [ N ] \ [ ℓ ∗ R − ] , (71)where ℓ ∗ r , P k ∈ [ r ] ℓ k .By summarizing Theorems 1 and 2, we can have the following constraints on H P R . • Constraint 1-L : There exists no distinct r , r ∈ [ R L ] and distinct i, j ∈ [ N ] such that H ( r ) i , H ( r ) i , H ( r ) j , H ( r ) j ∈L ; • Constraint 1-R : There exists no distinct r , r ∈ [ R ] \ [ R L ] and distinct i, j ∈ [ N ] such that H ( r ) i , H ( r ) i , H ( r ) j , H ( r ) j ∈R ; • Constraint 2-L : If r ∈ [ R L ] , H ( r ) j ∈ L for all j ∈ J r ( P R ) ; • Constraint 2-R : If r ∈ [ R ] \ [ R L ] , H ( r ) j ∈ R for all j ∈ J r ( P R ) .We then turn our attention to the block matrices which satisfy above constraints without considering the repairprocess P R .Given R , R L , and ~ℓ = ( ℓ , ℓ , . . . , ℓ R ) such that ≤ R ≤ N , ≤ R L ≤ R , and P r ∈ [ R ] ℓ r = N , we define aset of block matrices as H ( R, R L , ~ℓ ) = H (1)1 H (1)2 · · · H (1) N H (2)1 H (2)2 · · · H (2) N ... ... . . . ... H ( R )1 H ( R )2 · · · H ( R ) N : H ( r ) i is invertible if i ∈ [ ℓ ∗ r ] \ [ ℓ ∗ r − ]; The first R L rows satisfy Constraints 1-L and 2-L ; The last R − R L rows satisfy Constraints 1-R and 2-R , (72)where ℓ ∗ r = P k ∈ [ r ] ℓ k . For every ¯ H = h ¯ H ( r ) i i r ∈ [ R ] ,i ∈ [ N ] ∈ H ( R, R L , ~ℓ ) , similar with the definitions of J r ( P R ) and B ( P R ) , we define the J r ( ¯ H ) and B ( ¯ H ) as J r ( ¯ H ) , n j : j ∈ [ N ] and ¯ H ( r ) j ∈ {L ∪ R} o , (73) B ( ¯ H ) , N × h N − i − X r ∈ [ R ] ℓ r × |J r ( ¯ H ) | . (74)It should be noted that, as discussed at the beginning of this section, we can construct a P ′ R for any repairprocess P R such that B ( P R ) = B ( P ′ R ) and H P ′ R ∈ H ( R, R L , ~ℓ ) ; on the contrary, there is no guarantee that wecan construct a P R such that H P R = ¯ H for any ¯ H ∈ H ( R, R L , ~ℓ ) . Therefore, we can conclude that B ∗ ( R ) = min P R ∈P R B ( P R ) ≥ min ¯ H ∈H ( R ) B ( ¯ H ) , ¯ B ∗ ( R ) , (75) February 5, 2021 DRAFT3 where H ( R ) , [ ~ℓ ∈{ ( ℓ ,...,ℓ R ): P r ∈ [ R ] ℓ r = N } [ ≤ R L ≤ R H ( R, R L , ~ℓ ) . (76)We then turn our attention to ¯ B ∗ ( R ) rather than B ∗ ( R ) . Theorem 3. ¯ B ∗ ( R = 2) = 2 N ( N − − j N k .Proof: Given an ~ℓ = ( ℓ , ℓ ) , we consider an ¯ H ∈ H (2 , , ~ℓ ) such that ¯ H = I (1)1 · · · I (1) ℓ L (1) ℓ +1 · · · L (1) N R (2)1 · · · R (2) ℓ I (2) ℓ +1 · · · I (2) N , (77)where I ( r ) i ∈ M inv , L ( r ) i ∈ L , and R ( r ) i ∈ R . Since there must be at least ℓ ′ r invertible matrices in the r thblockwise row of any ¯ H ∈ H (cid:16) R, R L , ~ℓ ′ = ( ℓ ′ , . . . , ℓ ′ R ) (cid:17) , we have |J r ( ¯ H ) | ≤ N − ℓ ′ r for any ¯ H ∈ H ( R, R L , ~ℓ ′ ) .Since |J r ( ¯ H ) | = N − ℓ r for all r ∈ [2] , the total repair bandwidth of any ¯ H ∈ H (2 , R L , ~ℓ ) is lower bounded as B ( ¯ H ) = 2 N ( N − − ℓ |J ( ¯ H ) | − ℓ |J ( ¯ H ) | (78) ≥ N ( N − − ℓ ( N − ℓ ) − ℓ ( N − ℓ ) (79) = min ¯ H ′ ∈H (2 ,R L ,~ℓ ) B ( ¯ H ′ ) = B ( ¯ H ) (80)for any ≤ R L ≤ . Therefore, ¯ B ∗ ( R ) = min ~ℓ ∈{ ( ℓ ,ℓ ): ℓ + ℓ = N } min ≤ R L ≤ min ¯ H ∈H (2 ,R L ~ℓ ) B ( ¯ H ) (81) = min ~ℓ ∈{ ( ℓ ,ℓ ): ℓ + ℓ = N } B ( ¯ H ) (82) = min ~ℓ ∈{ ( ℓ ,ℓ ): ℓ + ℓ = N } N ( N − − ℓ ( N − ℓ ) − ℓ ( N − ℓ ) (83) = min ~ℓ ∈{ ( ℓ ,ℓ ): ℓ + ℓ = N } N ( N − − ℓ ( N − ℓ ) − ( N − ℓ ) ℓ (84) = min ~ℓ ∈{ ( ℓ ,ℓ ): ℓ + ℓ = N } N ( N −
1) + 2 ℓ − N ℓ (85) = min ~ℓ ∈{ ( ℓ ,ℓ ): ℓ + ℓ = N } N ( N −
1) + 2 (cid:18) ℓ − N (cid:19) − N (86) = 2 N ( N − − (cid:22) N (cid:23) . (87) Lemma 11. ¯ B ∗ (2) ≥ ¯ B ∗ (3) February 5, 2021 DRAFT4
Proof:
Given an ¯ H = I (1)1 L (1)2 · · · L (1) ℓ B (1) ℓ +1 · · · B (1) N − L (1) N L (2)1 I (2)2 · · · I (2) ℓ L (2) ℓ +1 · · · L (2) N − L (2) N R (3)1 R (3)2 · · · R (3) ℓ I (3) ℓ +1 · · · I (3) N − I (3) N ∈ H (3 , , ~ℓ ) , (88)where ~ℓ = (1 , ℓ − , ℓ ) , ℓ = ⌈ N ⌉ , and ℓ = N − ℓ . We then have ¯ B ∗ (3) ≤ B ( ¯ H ) (89) = 2 N ( N − − ℓ − ( ℓ − ℓ + 1) − ℓ ( ℓ ) (90) = 2 N ( N − − ℓ − ( ℓ − N − ℓ + 1) − ( N − ℓ ) ℓ (91) = 2 N ( N − − ( N − ℓ + 1)(2 ℓ − (92) = N ( N − − N − N + 1 if N is even ;2 N ( N − − N − N if N is odd (93) ≤ N ( N − − N (94) ≤ ¯ B ∗ (2) . (95) Lemma 12.
Given an ~ℓ = ( ℓ , ℓ , ℓ ) such that P r ∈ [3] ℓ r = N and ℓ ≤ ℓ . Let ¯ H = I (1)1 · · · I (1) ℓ ∗ L (1) ℓ ∗ +1 · · · L (1) ℓ ∗ B (1) ℓ ∗ +1 · · · B (1) N − L (1) N L (2)1 · · · L (2) ℓ ∗ I (2) ℓ ∗ +1 · · · I (2) ℓ ∗ L (2) ℓ ∗ +1 · · · L (2) N − L (2) N R (3)1 · · · R (3) ℓ ∗ R (3) ℓ ∗ +1 · · · R (3) ℓ ∗ I (3) ℓ ∗ +1 · · · I (3) N − I (3) N ∈ H (3 , , ~ℓ ) , (96) where ℓ ∗ r = P k ∈ [ r ] ℓ k , I ( r ) i ∈ M inv , B ( r ) i ∈ M , L ( r ) i ∈ L , and R ( r ) i ∈ R . Then, B ( ¯ H ) = min ≤ R L ≤ R min ¯ H ∈H (3 ,R L ,~ℓ ) B ( ¯ H ) . (97) Proof:
According to (72), the constraints (1-L and 2-L) on the first R L rows of ¯ H are different from theconstraints (1-R and 2-R) on the last R − R L rows of ¯ H . Therefore, the design of the first R L row is irrelevantto the design of the remaining R − R L rows and vice versa. Also, for any ¯ H ∈ H ( R, R L , ~ℓ ) , we can replace thematrices in L with the matrices in R and the matrices in R with the matrices in L without effecting the total repairbandwidth of ¯ H , hence we have min ¯ H ∈H ( R,R L ,~ℓ ) B ( ¯ H ) = min ¯ H ∈H ( R,R − R L ,~ℓ ′ ) B ( ¯ H ) , (98)where ~ℓ = ( ℓ , ℓ , . . . , ℓ R ) and ~ℓ ′ = ( ℓ R L +1 , ℓ R L +2 , . . . , ℓ R , ℓ , ℓ , . . . , ℓ R L ) . WLOG, we assume R L ≥ R/ . February 5, 2021 DRAFT5
We first prove that, for any ¯ H ∈ H (3 , , ~ℓ ) , we can find an ¯ H ′ ∈ H (3 , , ~ℓ ) such that B ( ¯ H ) ≥ B ( ¯ H ′ ) . If ¯ H iswith R L = 3 , we can construct ¯ H ′ by replacing ¯ H (3) k , for all k ∈ [ ℓ ∗ ] , with matrices in R without violating anyconstraint. Then B ( ¯ H ) = 2 N ( N − − X r ∈ [3] ℓ r |J r ( ¯ H ) | (99) ≥ N ( N − − X r ∈ [2] ℓ r |J r ( ¯ H ) | − ℓ ( N − ℓ ∗ ) = B ( ¯ H ′ ) . (100)Therefore, we only need to consider those ¯ H ∈ H (3 , , ~ℓ ) while minimizing the total repair bandwidth.Let ¯ H be with R L = 2 , as discussed before, we can have ¯ H (3) k ∈ R for all k ∈ [ ℓ ∗ ] to minimize |J ( ¯ H ) | to be N − ℓ . We then consider the first two rows of ¯ H . All those ¯ H (1) k for all k ∈ [ ℓ ∗ ] \ [ ℓ ∗ ] and ¯ H (2) k for all k ∈ [ ℓ ∗ ] can be in L without violating Constraint 1-L. Hence, we have ¯ H = I (1)1 · · · I (1) ℓ ∗ L (1) ℓ ∗ +1 · · · L (1) ℓ ∗ ¯ H (1) ℓ ∗ +1 · · · ¯ H (1) N L (2)1 · · · L (2) ℓ ∗ I (2) ℓ ∗ +1 · · · I (2) ℓ ∗ ¯ H (2) ℓ ∗ +1 · · · ¯ H (2) N R (3)1 · · · R (3) ℓ ∗ R (3) ℓ ∗ +1 · · · R (3) ℓ ∗ I (3) ℓ ∗ +1 · · · I (3) N , (101)where its submatrix ¯ H (1) ℓ ∗ +1 · · · ¯ H (1) N ¯ H (2) ℓ ∗ +1 · · · ¯ H (2) N (102)must satisfy Constraint 1-L. Let J r , |{ k : k ∈ [ N ] \ [ ℓ ∗ ] and ¯ H ( r ) k ∈ L}| , if J + J ≥ ( N − ℓ ∗ ) + 2 = ℓ + 2 ,then the submatrix in (102) must violate Constraint 1-L according to the pigeonhole principle, which induces J + J ≤ ℓ + 1 . Therefore, B ( ¯ H ) = 2 N ( N − − ℓ ( ℓ + J ) − ℓ ( ℓ + J ) − ℓ ( N − ℓ ) (103) ≥ N ( N − − ℓ [ ℓ + ( ℓ + 1 − J )] − ℓ ( ℓ + J ) − ℓ ( N − ℓ ) (104) = 2 N ( N − − ℓ ℓ − ℓ ℓ − ℓ − ℓ ( N − ℓ ) + ( ℓ − ℓ ) J , (105)where (105) can be minimized by maximizing J when ℓ ≤ ℓ . Since J ≤ ℓ by (102) and J + J ≤ ℓ + 1 ,we can minimize (105) by assigning J = ℓ and J = 1 as ¯ H did in (88). Theorem 4. ¯ B ∗ (3) ≥ min ~ℓ ∈{ ( ℓ ,ℓ ,ℓ ): ℓ + ℓ + ℓ = N and ℓ ≤ ℓ } N ( N − − N + ℓ + ℓ + ℓ + ℓ ( ℓ − . (106) Proof:
We have learned from Lemma 12 that min ≤ R L ≤ min ¯ H ∈H ( ,R L ,~ℓ =( ℓ ,ℓ ,ℓ ) ) B ( ¯ H ) (107)can be achieved by the ¯ H in (88) while ℓ ≤ ℓ . For those ¯ H ′ ∈ H (3 , R L , ~ℓ ′ ) , where ~ℓ ′ = ( ℓ ′ , ℓ ′ , ℓ ′ ) = ( ℓ , ℓ , ℓ ) ,we can achieve min ≤ R L ≤ min ¯ H ∈H ( ,R L ,~ℓ ′ =( ℓ ,ℓ ,ℓ ) ) B ( ¯ H ) (108) February 5, 2021 DRAFT6 by ¯ H ′ = I (1)1 · · · I (1) ℓ ′∗ L (1) ℓ ′∗ +1 · · · L (1) ℓ ′∗ L (1) ℓ ′∗ +1 · · · L (1) N − L (1) N L (2)1 · · · L (2) ℓ ′∗ I (2) ℓ ′∗ +1 · · · I (2) ℓ ′∗ B (2) ℓ ′∗ +1 · · · B (2) N − L (2) N R (3)1 · · · R (3) ℓ ′∗ R (3) ℓ ′∗ +1 · · · R (3) ℓ ′∗ I (3) ℓ ′∗ +1 · · · I (3) N − I (3) N (109)and B ( ¯ H ′ ) = B ( ¯ H ) .Therefore, we conclude that ¯ B ∗ (3) = min ~ℓ ∈{ ( ℓ ,ℓ ,ℓ ): ℓ + ℓ + ℓ = N } min ≤ R L ≤ R min ¯ H ∈H (3 ,R L ,~ℓ ) B ( ¯ H ) (110) = min ~ℓ ∈{ ( ℓ ,ℓ ,ℓ ): ℓ + ℓ + ℓ = N and ℓ ≤ ℓ } B ( ¯ H ) (111) = min ~ℓ ∈{ ( ℓ ,ℓ ,ℓ ): ℓ + ℓ + ℓ = N and ℓ ≤ ℓ } N ( N − − ℓ ( ℓ + 1) − ℓ ( N − ℓ ) − ℓ ( N − ℓ ) (112) = min ~ℓ ∈{ ( ℓ ,ℓ ,ℓ ): ℓ + ℓ + ℓ = N and ℓ ≤ ℓ } N ( N − − ℓ ( N − ℓ − ℓ + 1) − ℓ ( N − ℓ ) − ℓ ( N − ℓ ) (113) = min ~ℓ ∈{ ( ℓ ,ℓ ,ℓ ): ℓ + ℓ + ℓ = N and ℓ ≤ ℓ } N ( N − − ( ℓ + ℓ + ℓ ) N + ℓ + ℓ + ℓ + ℓ ( ℓ − (114) = min ~ℓ ∈{ ( ℓ ,ℓ ,ℓ ): ℓ + ℓ + ℓ = N and ℓ ≤ ℓ } N ( N − − N + ℓ + ℓ + ℓ + ℓ ( ℓ − . (115)We then look into ¯ B ∗ ( R ) when R > . From Constraints 1-L, 1-R, 2-L, and 2-R we notice that, among all ¯ H ∈ H ( R, R L , ~ℓ ) , we can minimize the total bandwidth by separately minimizing bandwidth of the first R L rowsin ¯ H and the bandwidth of the remaining R − R L rows in ¯ H , i.e., min ¯ H ∈H ( R,R L ,~ℓ =( ℓ ,...,ℓ R ) ) B ( ¯ H ) (116) = min ¯ H ∈H ( R,R L ,~ℓ =( ℓ ,...,ℓ R ) ) 2 N ( N − − X r ∈ [ R ] ℓ r |J r ( ¯ H ) | (117) = min ¯ H ∈H ( R,R L ,~ℓ =( ℓ ,...,ℓ R ) ) 2 R L ( N − − X r ∈ [ R L ] ℓ r |J r ( ¯ H ) | (118) + min ¯ H ∈H ( R,R L ,~ℓ =( ℓ ,...,ℓ R ) ) 2( R − R L )( N − − X r ∈ [ R ] \ [ R L ] ℓ r |J r ( ¯ H ) | . (119)Also as discussed in (62), we can turn the minimization problem in to a maximization program. Therefore, we firstfocus on the following problem, max ¯ H ∈H ( R,R L ,~ℓ =( ℓ ,...,ℓ R ) ) X r ∈ [ R L ] ℓ r |J r ( ¯ H ) | . (120)The following lemma further simplifies (120). Lemma 13. max ¯ H ∈H ( R,R L ,~ℓ =( ℓ ,...,ℓ R ) ) X r ∈ [ R L ] ℓ r |J r ( ¯ H ) | = max ¯ H ∈H (cid:16) R L +1 ,R L ,~ℓ L =( ℓ ,...,ℓ R L ,N − ℓ ∗ R L ) (cid:17) X r ∈ [ R L ] ℓ r |J r ( ¯ H ) | , (121) February 5, 2021 DRAFT7 where ℓ ∗ r = P i ∈ [ r ] ℓ i .Proof: Since the summation of P r ∈ [ R L ] ℓ r |J r ( ¯ H ) | only considers the first R L rows, the maximization isrelevant to P r ∈ [ R ] \ [ R L ] ℓ r but irrelevant of those individual ℓ r for r ∈ [ R ] \ [ R L ] . Lemma 14.
Given R L ≥ and ~ℓ = { ℓ , ℓ , . . . , ℓ R L , N − ℓ ∗ R L } , for any ¯ H ∈ H (cid:16) R L + 1 , R L , ~ℓ (cid:17) , we can alwaysconstruct a ¯ H ′ ∈ H (cid:16) , , ~ℓ ′ = ( ℓ ′ , ℓ ′ , N − ℓ ∗ R L ) (cid:17) such that X r ∈ [ R L ] ℓ r |J r ( ¯ H ) | ≤ X r ∈ [2] ℓ ′ r |J r ( ¯ H ′ ) | . (122) Proof:
To simplify the problem formulation, for any ¯ H ∈ H (cid:16) R L + 1 , R L , ~ℓ = ( ℓ , . . . , ℓ R L +1 ) (cid:17) , we define J ( r ) i ( ¯ H ) as J ( r ) i ( ¯ H ) , (cid:12)(cid:12)(cid:12)n j : j ∈ [ ℓ ∗ i ] \ [ ℓ ∗ i − ] and ¯ H ( r ) j ∈ L o(cid:12)(cid:12)(cid:12) , (123)which denotes the number of block matrices in h ¯ H ( r ) ℓ ∗ i − +1 ¯ H ( r ) ℓ ∗ i − +2 · · · ¯ H ( r ) ℓ ∗ i i belongs to L . By definition, wehave J ( r ) r ( ¯ H ) = 0 and |J r ( ¯ H ) | = P i ∈ [ R +1] J ( r ) i ( ¯ H ) . Table I J ( r ) i (¯ H ) FOR ¯ H ∈ H (cid:16) , , ~ℓ = ( ℓ , ℓ , ℓ , N − ℓ ∗ ) (cid:17) . ℓ ℓ ℓ N − ℓ ∗ J (1)1 (¯ H ) = 0 J (1)2 (¯ H ) J (1)3 (¯ H ) J (1)4 (¯ H ) J (2)1 (¯ H ) J (2)2 (¯ H ) = 0 J (2)3 (¯ H ) J (2)4 (¯ H ) J (3)1 (¯ H ) J (3)2 (¯ H ) J (3)3 (¯ H ) = 0 J (3)4 (¯ H ) Let us consider the case of R L = 3 , i.e., ¯ H ∈ H (cid:16) , , ~ℓ = ( ℓ , ℓ , ℓ , N − ℓ ∗ ) (cid:17) . WLOG, we asumme ℓ ≤ ℓ ≤ ℓ . The corresponding J ( r ) i ( ¯ H ) for all r ∈ [3] and i ∈ [4] are listed in Table I, and we have X r ∈ [3] ℓ r |J r ( ¯ H ) | = ℓ h J (1)2 ( ¯ H ) + J (1)3 ( ¯ H ) + J (1)4 ( ¯ H ) i + ℓ h J (2)1 ( ¯ H ) + J (2)3 ( ¯ H ) + J (2)4 ( ¯ H ) i + ℓ h J (3)1 ( ¯ H ) + J (3)2 ( ¯ H ) + J (3)4 ( ¯ H ) i . (124)Combining Constraint 1-L and pigeonhole principle, we can have the following inequalities from Table I, J (1)2 ( ¯ H ) + J (1)4 ( ¯ H ) + J (3)2 ( ¯ H ) + J (3)4 ( ¯ H ) ≤ ℓ + ( N − ℓ ∗ ) + 1 (125) J (2)1 ( ¯ H ) + J (2)4 ( ¯ H ) + J (3)1 ( ¯ H ) + J (3)4 ( ¯ H ) ≤ ℓ + ( N − ℓ ∗ ) + 1 (126) J (1)3 ( ¯ H ) + J (1)4 ( ¯ H ) + J (2)3 ( ¯ H ) + J (2)4 ( ¯ H ) ≤ ℓ + ( N − ℓ ∗ ) + 1 (127) J (1)3 ( ¯ H ) + J (2)3 ( ¯ H ) ≤ ℓ + 1 (128) J (1)2 ( ¯ H ) + J (3)2 ( ¯ H ) ≤ ℓ + 1 (129) J (2)1 ( ¯ H ) + J (3)1 ( ¯ H ) ≤ ℓ + 1 . (130) February 5, 2021 DRAFT8
Then we can have an upper bound of (124) by applying (126), (128), and (129) as X r ∈ [3] ℓ r |J r ( ¯ H ) |≤ ℓ h ( ℓ + 1) − J (3)2 ( ¯ H ) + ( ℓ + 1) − J (2)3 ( ¯ H ) + J (1)4 ( ¯ H ) i + ℓ h(cid:16) ℓ + N − ℓ ∗ + 1 (cid:17) − J (3)1 ( ¯ H ) − J (3)4 ( ¯ H ) + J (2)3 ( ¯ H ) i + ℓ h J (3)1 ( ¯ H ) + J (3)2 ( ¯ H ) + J (3)4 ( ¯ H ) i (131) = ℓ h ℓ + ℓ + 2 + J (1)4 ( ¯ H ) i + ℓ h ℓ + N − ℓ ∗ + 1 i +( ℓ − ℓ ) J (3)2 ( ¯ H ) + ( ℓ − ℓ ) J (2)3 ( ¯ H ) + ( ℓ − ℓ ) J (3)1 ( ¯ H ) + ( ℓ − ℓ ) J (3)4 ( ¯ H ) (132) ≤ ℓ h ℓ + ℓ + 2 + J (1)4 ( ¯ H ) i + ℓ h ℓ + N − ℓ ∗ + 1 i +( ℓ − ℓ ) ℓ + ( ℓ − ℓ ) ℓ + ( ℓ − ℓ ) ℓ + ( ℓ − ℓ )( N − ℓ ∗ ) (133) = ℓ h J (1)4 ( ¯ H ) i + ℓ h ℓ + 1 i + ℓ h ℓ + ℓ + N − ℓ ∗ i , (134)where (133) is due to ℓ ≤ ℓ ≤ ℓ , J (3)2 ( ¯ H ) ≤ ℓ , J (2)3 ( ¯ H ) ≤ ℓ , J (3)1 ( ¯ H ) ≤ ℓ , and J (3)4 ( ¯ H ) ≤ N − ℓ ∗ . We theninvestigate the ¯ H achieves (134) as Table II shown. Table II J ( r ) i (¯ H ) FOR THE ¯ H ACHIEVES (134). ℓ ℓ ℓ N − ℓ ∗ J (1)1 (¯ H ) = 0 J (1)2 (¯ H ) = 1 J (1)3 (¯ H ) = 1 J (1)4 (¯ H ) J (2)1 (¯ H ) = 1 J (2)2 (¯ H ) = 0 J (2)3 (¯ H ) = ℓ J (2)4 (¯ H ) = 0 J (3)1 (¯ H ) = ℓ J (3)2 (¯ H ) = ℓ J (3)3 (¯ H ) = 0 J (3)4 (¯ H ) = N − ℓ ∗ Due to inequality (125), we have J (1)4 ( ¯ H ) = 0 . Therefore, we upper bound (124) as X r ∈ [3] ℓ r |J r ( ¯ H ) | ≤ ℓ + ℓ ( ℓ + 1) + ℓ (cid:16) ℓ + ℓ + ( N − ℓ ∗ ) (cid:17) . (135) Table III J ( r ) i (¯ H ′ ) FOR ¯ H ′ ∈ H (cid:16) , , ~ℓ = ( ℓ + ℓ , ℓ , N − ℓ ∗ ) (cid:17) . ℓ + ℓ ℓ N − ℓ ∗ J (1)1 (¯ H ′ ) = 0 J (1)2 (¯ H ′ ) = ℓ J (1)3 (¯ H ′ ) = 1 J (2)1 (¯ H ′ ) = ℓ + ℓ J (2)2 (¯ H ′ ) = 0 J (2)3 (¯ H ′ ) = N − ℓ ∗ Next, we construct a ¯ H ′ ∈ H (cid:0) , , ~ℓ ′ = ( ℓ + ℓ , ℓ , N − ℓ ∗ ) (cid:1) with the corresponding J ( r ) i ( ¯ H ′ ) as shown in TableIII. Therefore X r ∈ [2] ℓ r |J r ( ¯ H ′ ) | = ℓ ( ℓ + 1) + ℓ ( ℓ + 1) + ℓ (cid:16) ℓ + ℓ + ( N − ℓ ∗ ) (cid:17) , (136) February 5, 2021 DRAFT9
By subtracting (135) by (136), we have X r ∈ [3] ℓ r |J r ( ¯ H ) | − X r ∈ [2] ℓ r |J r ( ¯ H ′ ) | ≤ ℓ (1 − ℓ ) ≤ . (137)Therefore, for any ¯ H ∈ H (cid:16) , , ~ℓ = ( ℓ , ℓ , ℓ , N − ℓ ∗ ) (cid:17) , we can always construct an ¯ H ′ ∈ H (cid:16) , , ~ℓ ′ (cid:17) suchthat ~ℓ ′ L = ( ℓ + ℓ , ℓ , N − ℓ ∗ ) and X r ∈ [3] ℓ r |J r ( ¯ H ) | ≤ X r ∈ [2] ℓ r |J r ( ¯ H ′ ) | . (138)Next we consider an ¯ H ∈ H (cid:16) R L + 1 , R L , ~ℓ = ( ℓ , ℓ , . . . , ℓ R L , N − ℓ ∗ R L ) (cid:17) for R L > , which is with thecorresponding J ( r ) i ( ¯ H ) in Table IV. As discussed in (135), we have Table IV J ( r ) i (¯ H ) FOR ¯ H ∈ H (cid:16) R L + 1 , R L , ~ℓ = ( ℓ , ℓ , . . . , ℓ R L , N − ℓ ∗ R L ) (cid:17) . ℓ ℓ · · · ℓ R L − ℓ R L N − ℓ ∗ R L J (1)1 (¯ H ) = 0 J (1)2 (¯ H ) · · · J (1) R L − (¯ H ) J (1) R L (¯ H ) J (1) R L +1 (¯ H ) J (2)1 (¯ H ) J (2)2 (¯ H ) = 0 · · · J (2) R L − (¯ H ) J (2) R L (¯ H ) J (2) R L +1 (¯ H ) ... ... .. . ... ... ... J ( R L − (¯ H ) J ( R L − (¯ H ) · · · J ( R L − R L − (¯ H ) = 0 J ( R L − R L (¯ H ) J ( R L − R L +1 (¯ H ) J ( R L )1 (¯ H ) J ( R L )2 (¯ H ) · · · J ( R L ) R L − (¯ H ) J ( R L ) R L (¯ H ) = 0 J ( R L ) R L +1 (¯ H ) X r ∈{ r ′ ,R L − ,R L } ℓ r |J r ( ¯ H ) | ≤ ℓ r ′ + ℓ R L − ( ℓ R L + 1) + ℓ R L ( N − ℓ R L ) , (139)for all r ′ ∈ [ R L − . Hence, X r ∈ [ R L ] ℓ r |J r ( ¯ H ) | ≤ X r ∈ [ R L − ℓ r + ℓ R L − ( ℓ R L + 1) + ℓ R L ( N − ℓ R L ) = X r ∈ [3] ℓ r |J r ( ¯ H ′ ) | , (140)where ¯ H ′ is with the corresponding J ( r ) i ( ¯ H ′ ) in Table V. Table V J ( r ) i (¯ H ′ ) FOR THE ¯ H ′ IN (140). P r ∈ [ R L − ℓ r ℓ R L − ℓ R L N − ℓ ∗ R L J (1)1 (¯ H ) = 0 J (1)2 (¯ H ) = 1 J (1)3 (¯ H ) = 1 J (1)4 (¯ H ) = 0 J (2)1 (¯ H ) = 1 J (2)2 (¯ H ) = 0 J (2)3 (¯ H ) = ℓ R L J (2)4 (¯ H ) = 0 J (3)1 (¯ H ) = P r ∈ [ R L − ℓ r J (3)2 (¯ H ) = ℓ R L − J (3)3 (¯ H ) = 0 J (3)4 (¯ H ) = N − ℓ ∗ R L Since ¯ H ′ ∈ H , , (cid:18) X r ∈ [ R L − ℓ r , ℓ R L − , ℓ R L , N − ℓ ∗ R L (cid:19) , (141) February 5, 2021 DRAFT0 as proven in (138), we can find an ¯ H ′′ ∈ H (cid:16) , , ( ℓ ′′ , ℓ ′′ , N − ℓ ∗ R L ) (cid:17) such that X r ∈ [ R L +2] ℓ r |J r ( ¯ H ) | ≤ X r ∈ [3] ℓ r |J r ( ¯ H ′ ) | ≤ X r ∈ [2] ℓ r |J r ( ¯ H ′′ ) | . (142)Next, for those remaining R − R L rows of ¯ H , we can have a similar lemma below. Lemma 15.
Given R − ≥ and ~ℓ = { ℓ ∗ R − , ℓ , ℓ , . . . , ℓ R − } , for any ¯ H ∈ H (cid:16) R, R − , ~ℓ (cid:17) , we can alwaysconstruct a ¯ H ′ ∈ H (cid:16) , , ~ℓ ′ = ( ℓ ∗ R − , ℓ ′ , ℓ ′ (cid:17) such that X r ∈ [ R ] \ [1] ℓ r |J r ( ¯ H ) | ≤ X r ∈{ , } ℓ ′ r |J r ( ¯ H ′ ) | . (143) Proof:
The proof is similar to the proof of Lemma 14, hence we omit it here.Combining Theorems 3, 4 and Lemmas 14, 15, we can lower bound ¯ B ∗ ( R ) for all R ∈ [ N ] in the followingtheorem. Theorem 5.
Let ∆ , min ~ℓ ∈{ ( ℓ ,ℓ ,ℓ ): P r ∈ [3] ℓ r = N and ℓ ≤ ℓ } N ( N − − N + ℓ + ℓ + ℓ + ℓ ( ℓ − (144) and ∆ , min ~ℓ ∈{ ( ℓ ,ℓ ,ℓ ,ℓ ): P r ∈ [4] ℓ r = N , ℓ ≤ ℓ , and ℓ ≤ ℓ } N ( N − − ℓ ( ℓ + 1) − ℓ ( N − ℓ ) − ℓ ( ℓ + 1) − ℓ ( N − ℓ ) , (145) then ¯ B ∗ ( R ) ≥ min { ∆ , ∆ } for all R ∈ [ N ] .Proof: As discussed in (98), we only need to consider the case of R L ≥ R − R L ≥ .Lemma 11 and Theorem 4 have proven ¯ B ∗ (1) ≥ ¯ B ∗ (2) ≥ ¯ B ∗ (3) ≥ ∆ for the case of R = 2 . We next considerthe case of R ≥ .For the case of R ≥ and R − R L = 1 , any ¯ H ∈ H { R, R − , ( ℓ , ℓ , . . . , ℓ R ) } can be replaced by some ¯ H ′ ∈ H { , , ( ℓ ′ , ℓ ′ , ℓ R ) } such that B ( H ) ≥ B ( H ′ ) due to Lemma 14. Therefore, ¯ B ∗ ( R ) ≥ ¯ B ∗ (3) ≥ ∆ when R ≥ and R − R L = 1 .The remaining case is of R ≥ and R − R L ≥ . Since R L ≥ R − R L , this case is equivalent to the case of R L ≥ R − R L ≥ . For the case of R L ≥ R − R L ≥ , any ¯ H ∈ H { R, R L , ( ℓ , ℓ , . . . , ℓ R ) } can be replaced bysome ¯ H ′ ∈ H { , , ( ℓ ′ , ℓ ′ , ℓ ′ , ℓ ′ ) } such that ℓ ′ + ℓ ′ = P r ∈ [ R L ] ℓ r , ℓ ′ + ℓ ′ = P r ∈ [ R ] \ [ R L ] ℓ r , B ( H ) ≥ B ( H ′ ) .Therefore, we can conclude that ¯ B ∗ ( R ) ≥ min ~ℓ ∈{ ( ℓ ,ℓ ,ℓ ,ℓ ): P r ∈ [4] ℓ r = N } min ¯ H ∈H (4 , ,~ℓ ) B ( ¯ H ) , (146) February 5, 2021 DRAFT1 when R L ≥ R − R L ≥ . We then prove that (146) can be further lower bounded by ∆ .As discussed in the beginning of this section, we swap rows or columns without effecting the repairing bandwidthof ¯ H . Hence, WLOG, we can assume ℓ ≤ ℓ and ℓ ≤ ℓ for any ¯ H ∈ H { , , ( ℓ , ℓ , ℓ , ℓ ) } . By following thesame trick in proving Lemma 12, we can have X r ∈ [4] ℓ r |J r ( ¯ H ) | = X r ∈ [4] ℓ r X i ∈ [4] \{ r } J ( r ) i ( ¯ H ) (147) ≤ ℓ ( ℓ + 1) + ℓ ( N − ℓ ) + ℓ ( ℓ + 1) + ℓ ( N − ℓ ) , (148)as illustrated in Table VI. Table VI J ( r ) i (¯ H ) FOR ¯ H ∈ H (4 , , ( ℓ , ℓ , ℓ , ℓ )) . ℓ ℓ ℓ ℓ J (1)1 (¯ H ) = 0 J (1)2 (¯ H ) = ℓ J (1)3 (¯ H ) = 1 J (1)4 (¯ H ) = 0 J (2)1 (¯ H ) = ℓ J (2)2 (¯ H ) = 0 J (2)3 (¯ H ) = ℓ J (2)4 (¯ H ) = ℓ J (3)1 (¯ H ) = 0 J (3)2 (¯ H ) = 1 J (3)3 (¯ H ) = 0 J (3)4 (¯ H ) = ℓ J (4)1 (¯ H ) = ℓ J (4)2 (¯ H ) = ℓ J (4)3 (¯ H ) = ℓ J (4)4 (¯ H ) = 0 Therefore, we can rewrite (146) as ¯ B ∗ ( R ) ≥ min ~ℓ ∈{ ( ℓ ,ℓ ,ℓ ,ℓ ): P r ∈ [4] ℓ r = N, ℓ ≤ ℓ , ℓ ≤ ℓ } N ( N − − ℓ ( ℓ + 1) − ℓ ( N − ℓ ) − ℓ ( ℓ + 1) − ℓ ( N − ℓ ) = ∆ , (149)when R L ≥ R − R L ≥ .We finally conclude that ¯ B ∗ ( R ) ≥ min { ∆ , ∆ } for all R ∈ [ N ] .IV. N UMERICAL RESULTS
The derived lower bound, min { ∆ , ∆ } for ≤ N ≤ , is plotted in Figure 1. The values at N = 26 and N = 32 are achieved by the codes provided by the department in Israel, which implies that the derived bound isso tight that it can be achieved. It should be noted that min { ∆ , ∆ } = ∆ for all ≤ N ≤ , except when N = 4 and . R EFERENCES[1] John R. Silvester, “Determinants of Block Matrices,”
The Mathematical Gazette , vol. 84, no. 501, pp. 460–467, Nov. 2000.
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Figure 1. The lower bound min { ∆ , ∆ } from N = 4 to N = 200 ..