Dorkson, Chatdanai

Ng, Siaw

English

Royal Holloway - Pure

Multi-Message Private Information Retrieval usingProduct-Matrix Minimum Storage Regenerating CodesChatdanai Dorkson and Siaw-Lynn NgDepartment of Mathematics, Royal Holloway, University of LondonAbstractMulti-message PIR scheme allows a userto download multiple messages from thedatabase without revealing the identity ofdesired messages. Obviously, the user canrepeatly use a single-message PIR scheme,but we wish for a more efficient way. Inthis work, we design a multi-message PIRusing a product-matrix MSR codes from[1] achieving cPoP equals p+2p when p isthe number of desired records. The useof regenerating codes beneficially reducesrepair cost when a node failure occurs inthe system. This work is the generalisationof our result on the single-message PIR [2].Regenerating CodesAn (n, k, r, α, β,B) regenerating code storesB symbols among n nodes. Each node storesα symbols satisfying:•The B symbols can be recovered from anyk nodes• If one node fails, we can connect to some setof r remaining nodes where k ≤ r < n, andgets β symbols from each of these r nodesto regenerate α symbols in the failed node.Regenerating codes optimally trade repairbandwidth (rβ) with the amount of datastored per node (α). One interesting extremalpoint on the optimal trade-off curve is calledthe minimum storage regeneration (MSR)which minimises α first and then minimise rβ.Contact InformationChatdanai.Dorkson.2016@rhul.ac.ukReferences[1] K. Rashmi, N. Shah, P. KumarOptimal Exact-Regenerating Codes forDistributed Storage at the MSR andMBR Points via a Product-MatrixConstruction.IEEE Trans. Inf. Theory 2011,pp.5227–5239.[2] C. Dorkson and S. NgPrivate Information Retrieval usingProduct-Matrix Minimum StorageRegenerating Codes.http://arxiv.org/abs/1805.07190.(Manuscript submitted for publication.)System Model•A database X consists of m records, each of length ` over Fq, denoted by X1, X2, . . . , Xm.•Each record is encoded and distributed across n non-communicating nodes using the same product-matrix MSR code.•A user who wants to download the record Xf1, Xf2, . . . , Xfp submits a d×mα query matrix Qi over GF (q) to node i•Node i responds with an answer Ai = QiSi where Si the column vector consisting of all symbols stored in node i.•A scheme is perfect information-theoretic if for i ∈ [n], Qi gives no information about which records are beingretrieved, and Ai ensures that the user can recover the desired records Xf1, . . . , Xfp with no errors.An Example of Our SchemeIn our construction, we use the product-matrix MSR code from [1] with n = (p + 2)(k − 2), over the finite field Fq.The parameters of the MSR code are(n, k, r, α, β,B) = ((p + 2)(k − 2), k, 2k − 2, k − 1, 1, k(k − 1)).Assume ` = k(k − 1) in this construction. We give an example to motivate our scheme.Example: Suppose that we have 3 records over the finite field F13, each with size 6, so we can use an (8, 3, 4, 2, 1, 6)product-matrix MSR code over F13 to encode each record. We writeX i = {xi1, xi2, xi3, xi4, xi5, xi6}. Choose the encodingmatrix Ψ8 to be the Vandermonde matrix, and the message matrixMi for the record i, i ∈ {1, 2, 3} as described in [1]:Ψ8 =1 1 1 1 1 1 1 11 2 3 4 5 6 7 81 4 9 3 12 10 10 121 8 1 12 8 8 5 5T, Mi =xi1 xi2xi2 xi3xi4 xi5xi5 xi6.Hence, each node storesnode 1 node 2 node 3 node 4 node 5 node 6 node 7 node 8x11 + x12 + x14 + x15 x11 + 2x12 + 4x14 + 8x15 x11 + 3x12 + 9x14 + x15 x11 + 4x12 + 3x14 + 12x15 x11 + 5x12 + 12x14 + 8x15 x11 + 6x12 + 10x14 + 8x15 x11 + 7x12 + 10x14 + 5x15 x11 + 8x12 + 12x14 + 5x15x12 + x13 + x15 + x16 x12 + 2x13 + 4x15 + 8x16 x12 + 3x13 + 9x15 + x16 x12 + 4x13 + 3x15 + 12x16 x12 + 5x13 + 12x15 + 8x16 x12 + 6x13 + 10x15 + 8x16 x12 + 7x13 + 10x15 + 5x16 x12 + 8x13 + 12x15 + 5x16x21 + x22 + x24 + x25 x21 + 2x22 + 4x24 + 8x25 x21 + 3x22 + 9x24 + x25 x21 + 4x22 + 3x24 + 12x25 x21 + 5x22 + 12x24 + 8x25 x21 + 6x22 + 10x24 + 8x25 x21 + 7x22 + 10x24 + 5x25 x21 + 8x22 + 12x24 + 5x25x22 + x23 + x25 + x26 x22 + 2x23 + 4x25 + 8x26 x22 + 3x23 + 9x25 + x26 x22 + 4x23 + 3x25 + 12x26 x22 + 5x23 + 12x25 + 8x26 x22 + 6x23 + 10x25 + 8x26 x22 + 7x23 + 10x25 + 5x26 x22 + 8x23 + 12x25 + 5x26x31 + x32 + x34 + x35 x31 + 2x32 + 4x34 + 8x35 x31 + 3x32 + 9x34 + x35 x31 + 4x32 + 3x34 + 12x35 x31 + 5x32 + 12x34 + 8x35 x31 + 6x32 + 10x34 + 8x35 x31 + 7x32 + 10x34 + 5x35 x31 + 8x32 + 12x34 + 5x35x32 + x33 + x35 + x36 x32 + 2x33 + 4x35 + 8x36 x32 + 3x33 + 9x35 + x36 x32 + 4x33 + 3x35 + 12x36 x32 + 5x33 + 12x35 + 8x36 x32 + 6x33 + 10x35 + 8x36 x32 + 7x33 + 10x35 + 5x36 x32 + 8x33 + 12x35 + 5x36Let Y aij denote the jth symbol stored in node i of the record a.In the retrieval step, suppose the user wants to download X1 and X2. The query Qi is a (3 × 6) matrix which we caninterpret as 3 subqueries submitted to node i for each i ∈ [8]. To form the query matrices, the user generates a (3× 6)random matrix U = [uij] whose elements are chosen uniformly at a random from F13. LetV1 =1 0 0 0 0 00 1 0 0 0 00 0 0 0 0 0, V2 =0 0 0 0 0 01 0 0 0 0 00 1 0 0 0 0, V3 =0 1 0 0 0 00 0 0 0 0 01 0 0 0 0 0, V4 =0 0 1 0 0 00 0 0 1 0 00 0 0 0 0 0, V5 =0 0 0 0 0 00 0 1 0 0 00 0 0 1 0 0, V6 =0 0 0 1 0 00 0 0 0 0 00 0 1 0 0 0.For node i ∈ [6], the query matrix is Qi = U +Vi and Q7 = Q8 = U . Then each node computes and returns the length-3vector Ai = QiSi where Si is a length-6 vector of symbols stored in node i. Write Ai = (ri1, ri2, ri3)T . Letw1 = (x11, x12, x21, x22, x31, x32)T , w2 = (x12, x13, x22, x23, x32, x33)T ,w3 = (x14, x15, x24, x25, x34, x35)T , w4 = (x15, x16, x25, x26, x35, x36)T .Consider first subquery 1, we obtainY 111 + I1 + I2 + I3 + I4 = r11 (1)I1 + 2I2 + 4I3 + 8I4 = r21 (2)Y 132 + I1 + 3I2 + 9I3 + I4 = r31 (3)Y 243 + I1 + 4I2 + 3I3 + 12I4 = r41 (4)I1 + 5I2 + 12I3 + 8I4 = r51 (5)Y 264 + I1 + 6I2 + 10I3 + 8I4 = r61 (6)I1 + 7I2 + 10I3 + 5I4 = r71 (7)I1 + 8I2 + 12I3 + 5I4 = r81 (8)where Il = u1wl, l = 1, 2, 3, 4, and u1 is the first row of U . The user can solve for I1, I2, I3, I4 from (2), (5), (7), (8) asthey form the equation with (4×4) submatrix of Ψ8 which is invertible. Therefore, the user gets Y 111, Y 132 for record 1 andY 243, Y264 for record 2. Similarly, from subquery 2, the user obtains Y 112, Y 121 for record 1 and Y 244, Y 253 for record 2. Lastly,from subquery 3, the user obtains Y 122, Y 131 for record 1 and Y 254, Y 263 for record 2. Hence, the user has all the symbols ofX1 which are stored in the node 1, 2, 3 and all the symbols of X2 which are stored in the node 4, 5, 6. From the propertyof regenerating codes, the user can reconstruct X1 and X2 as desired.node 1 node 2 node 3 node 4 node 5 node 6 node 7 node 81 2 32 3 11 2 32 3 1Table 1: Retrieval pattern for a (8,3,4,2,1,6) MSR code

Multi-Message Private Information Retrieval using Product-Matrix Minimum Storage Regenerating Codes

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Multi-Message Private Information Retrieval using Product-Matrix Minimum Storage Regenerating Codes

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