Fast Laplacian Solvers by Sparsification

Transcription

1 Spectral Graph Theory Lecture 19 Fast Laplacan Solvers by Sparsfcaton Danel A. Spelman November 9, 2015 Dsclamer These notes are not necessarly an accurate representaton of what happened n class. The notes wrtten before class say what I thnk I should say. I sometmes edt the notes after class to make them way what I wsh I had sad. There may be small mstakes, so I recommend that you check any mathematcally precse statement before usng t n your own work. These notes were last revsed on November 9, Overvew We wll see how sparsfcaton allows us to solve systems of lnear equatons n Laplacan matrces and ther sub-matrces n nearly lnear tme. By nearly-lnear, I mean tme O(m log c (nκ 1 ) log ɛ 1 ) for systems wth m nonzero entres, n dmensons, condton number κ. and accuracy ɛ. Ths algorthm comes from [PS14] Today s noton of approxmaton In today s lecture, I wll fnd t convenent to defne matrx approxmatons slghtly dfferently from prevous lectures. Today, I defne A ɛ B to mean e ɛ A B e ɛ A. Note that ths relaton s symmetrc n A and B, and that for ɛ small e ɛ 1 + ɛ. The advantage of ths defnton s that A α B and B β C mples A α+β C The Idea I begn by descrbng the dea behnd the algorthm. Ths dea won t qute work. But, we wll see how to turn t nto one that does. 19-1

2 Lecture 19: November 9, We wll work wth matrces that look lke M = L + X where L s a Laplacan and X s a non-zero, non-negatve dagonal matrx. Such matrces are called M-matrces. A symmetrc M-matrx s a matrx M wth nonpostve off-dagonal entres such that M 1 s nonnegatve and nonzero. We have encountered M-matrces before wthout namng them. If G = (V, E) s a graph, S V, and G(S) s connected, then the submatrx of L G ndexed by rows and columns n S s an M- matrx. Algorthmcally, the problems of solvng systems of equatons n Laplacans and symmetrc M-matrces are equvalent. The sparsfcaton results that we learned for Laplacans translate over to M-matrces. Every M- matrx M can be wrtten n the form X + L where L s a Laplacan and X s a nonnegatve dagonal matrx. If L ɛ L, then t s easy to show (too easy for homework) that X + L ɛ X + L. In Lecture 7, Lemma 7.3.1, we proved that f X has at least one nonzero entry and f L s connected, then X + L s nonsngular. We wrte such a matrx n the form M = D A where D s postve dagonal and A s nonnegatve, and note that ts beng nonsngular and postve semdefnte mples Usng the Perron-Frobenus theorem, one can also show that D A 0 D A. (19.1) D A. (19.2) Multplyng M by D 1/2 on ether sde, we obtan Defne I D 1/2 AD 1/2. B = D 1/2 AD 1/2, and note that nequaltes (19.1) and (19.2) mply that all egenvalues of B have absolute value strctly less than 1. It suffces to fgure out how to solve systems of equatons n I B. One way to do ths s to explot the power seres expanson: (I B) 1 = I + B + B 2 + B 3 + However, ths seres mght need many terms to converge. We can fgure out how many. If the largest egenvalue of B s (1 κ) < 1, then we need at least 1/κ terms. We can wrte a seres wth fewer terms f we express t as a product nstead of as a sum: (I + B 2j ). 0 B = j 1 To see why ths works, look at the frst few terms (I +B)(I +B 2 )(I +B 4 ) = (I +B +B 2 +B 3 )(I +B 4 ) = (I +B +B 2 +B 3 )+B 4 (I +B +B 2 +B 3 ).

3 Lecture 19: November 9, We only need O(log κ 1 ) terms of ths product to obtan a good approxmaton of (I B) 1. The obstacle to quckly applyng a seres lke ths s that the matrces I + B 2j are probably dense. We know how to solve ths problem: we can sparsfy them! I m not sayng that flppantly. We actually do know how to sparfy matrces of ths form. But, smply sparsfyng the matrces I + B 2j does not solve our problem because approxmaton s not preserved by products. That s, even f A ɛ  and B ɛ B,  B could be a very poor approxmaton of AB. In fact, snce the product  B s not necessarly symmetrc, we haven t even defned what t would mean for t to approxmate AB A symmetrc expanson We wll now derve a way of expandng (I B) 1 that s amenable to approxmaton. We begn wth an alternate dervaton of the seres we saw before. Note that and so Takng the nverse of both sdes gves (I B)(I + B) = (I B 2 ), (I B) = (I B 2 )(I + B) 1. (I B) 1 = (I + B)(I B 2 ) 1. We can then apply the same expanson to (I B 2 ) 1 to obtan (I B) 1 = (I + B)(I + B 2 )(I B 4 ) 1. What we need s a symmetrc expanson. We use (I B) 1 = 1 2 I (I + B)(I B 2 ) 1 (I + B). (19.3) We wll verfy ths by multplyng the rght hand sde by (I B): (I + B)(I B 2 ) 1 (I + B)(I B) = (I + B)(I B 2 ) 1 (I B 2 ) = I + B; so 1 [ I + (I + B)(I B 2 ) 1 (I + B) ] (I B) = 1 [(I B) + (I + B)] = I. 2 2 Ths expresson for (I B) 1 plays ncely wth matrx approxmatons. If M 1 ɛ (I B 2 ), then you can show (I B) 1 1 [ ɛ I + (I + B)M (I + B)]. If we can apply M 1 1 quckly and f B s sparse, then we can quckly approxmate (I B) 1. You may now be wonderng how we wll construct such an M 1. The answer, n short, s recursvely.

4 Lecture 19: November 9, D and A Unfortunately, we are gong to need to stop wrttng matrces n terms of I and B, and return to wrtng them n terms of D and A. The reason ths s unfortunate s that t makes for longer expressons. The analog of (19.3) s (D A) 1 = 1 2 [ D 1 + (I + D 1 A)(D AD 1 A) 1 (I + AD 1 ) ]. (19.4) In order to be able to work wth ths expresson nductvely, we need to check that the mddle matrx s an M-matrx. Lemma If D s a dagonal matrx and A s a nonnegatve matrx so that M = D A s an M-matrx, then M 1 = D AD 1 A s also an M-matrx. Proof. As the off-dagonal entres of ths matrx are symmetrc and nonpostve, t suffces to prove that M 1 0 and M 1 0. To compute the row sums set d = D1 and a = A1, and note that d a 0 and d a 0. For M 1, we have whch s nonnegatve and not exactly zero. (D AD 1 A)1 = d AD 1 a d A1 = d a, We wll apply transformaton lke ths many tmes durng our algorthm. To keep track of progress, I say that (D, A) s an (α, β)-par f a. D s postve dagonal, b. A s nonnegatve (and can have dagonal entres), and c. αd A and βd A. For our ntal matrx M = D A, we know that there s some number κ > 0 for whch (D, A) s a (1 κ, 1 κ)-par. At the end of our recurson we wll seek a (1/4, 1/4)-par. When we have such a par, we can just approxmate D A by D. Lemma If M = D A and (D, A) s a (1/4, 1/4)-par, then M 1/3 D.

5 Lecture 19: November 9, Proof. We have and M = D A (1 + 1/4)D e 1/4 D, M = D A D (1/4)D = (3/4)D e 1/3 D. Lemma If (D, A) s an (α, α)-par, then (D, AD 1 A) s an (α 2, 0)-par. Proof. From Lecture 14, Lemma 3.1, we know that the condton of the lemma s equvalent to the asserton that all egenvalues of D 1 A have absolute value at most α, and that the concluson s equvalent to the asserton that all egenvalues of D 1 AD 1 A le between 0 and α 2, whch s mmedate as they are the squares of the egenvalues of D 1 A. So, f we start wth matrces D and A that are a (1 κ, 1 κ)-par, then after applyng ths transformaton approxmately log κ tmes we obtan a (1/4, 0)-par. But, the matrces n ths par could be dense. To keep them sparse, we need to fgure out how approxmatng D A degrades ts qualty. Lemma If ɛ 1/3, a. (D, A) s a (1 κ, 0) par, b. D A ɛ D Â, and c. D ɛ D, then D Â s an (1 κe 2ɛ, 3ɛ)-par. Proof. Frst observe that Then, compute (1 κ)d A D A κd. D Â e ɛ (D A) e ɛ κd e 2ɛ κ D. For the other sde, compute e 2ɛ D e ɛ D e ɛ (D A) ( D Â). For ɛ 1/3, 3ɛ e 2ɛ 1, so 3ɛ D (e 2ɛ 1) D Â.

6 Lecture 19: November 9, It remans to confrm that sparsfcaton satsfes the requrements of ths lemma. The reason ths mght not be obvous s that we allow A to have nonnegatve dagonal elements. Whle ths does not nterfere wth condton b, you mght be concerned that t would nterfere wth condton c. It need not. Let C be the dagonal of A, and let L be the Laplacan of the graph wth adjacency matrx A C, and set X so that X + L = D A. Let L be a sparse ɛ-approxmaton of L. By computng the quadratc form n elementary unt vectors, you can check that the dagonals of L and L approxmate each other. If we now wrte L = D Ã, where à has zero dagonal, and set D = D + C and  = à + C You can now check that D and  satsfy the requrements of Lemma You mght wonder why we bother to keep dagonal elements n a matrx lke A. It seems smpler to get rd of them. However, we want (D, A) to be an (α, β) par, and removng subtractng C from both of them would make β worse. Ths mght not matter too much as we have good control over β. But, I don t yet see a nce way to carry out a proof that explots ths Sketch of the constructon We begn wth an M-matrx M 0 = D 0 A 0. Snce ths matrx s nonsngular, there s a κ 0 > 0 so that (D 0, A 0 ) s a (1 κ 0, 1 κ 0 ) par. We now know that the matrx D 0 A 0 D 1 0 A 0 s an M-matrx and that (D 0, A 0 D 1 0 A 0) s a ((1 κ 0 ) 2, 0)-par. Defne κ 1 so that 1 κ 1 = (1 κ) 2 0, and note that κ 1 s approxmately 2κ 0. Lemma and the dscusson followng t tells us that there s a (1 κ 1 e 2ɛ, 3ɛ)-par (D 1, A 1 ) so that and so that A 1 has O(n/ɛ 2 ) nonzero entres. D 1 A 1 ɛ D 0 A 0 D 1 0 A 0 Contnung nductvely for some number k steps, we fnd (1 κ, 3ɛ) pars (D, A ) so that has O(n/ɛ 2 ) nonzero entres, and M = D A M ɛ D A 1 D 1 1 A 1. For the such that κ s small, κ +1 s approxmately twce κ. So, for k = 2 + log 2 1/κ and ɛ close to zero, we can guarantee that (D k, A k ) s a (1/4, 1/4) par. We now see how ths constructon allows us to approxmately solve systems of equatons n D 0 A 0, and how we must set ɛ for t to work. For every 0 < k, we have (D A ) D (I +D 1 A )(D A D 1 A ) 1 (I +A D 1 1 ) ɛ 2 D (I +D 1 A )(D +1 A +1 ) 1 (I +

7 Lecture 19: November 9, and (D k A k ) 1 1/3 D 1 k. By substtutng through each of these approxmatons, we obtan solutons to systems of equatons n D 0 A 0 wth accuracy 1/3 + kɛ. So, we should set kɛ = 1/3, and thus ɛ = 1/(2 + log 2 κ 1 ). The domnant cost of the resultng algorthm wll be the multplcaton of vectors by 2k matrces of O(n/ɛ 2 ) entres, wth a total cost of O(n(log 2 (1/κ)) 3 ) Makng the constructon effcent In the above constructon, I just assumed that approprate sparsfers exst, rather than constructng them effcently. To construct them effcently, we need two deas. The frst s that we need to be able to quckly approxmate effectve resstances so that we can use the samplng algorthm from Lecture 17. The second s to observe that we do not actually want to form the matrx AD 1 A before sparsfyng t, as that could take too long. Instead, we express t as a product of clques that have succnct descrptons, and we form the sum of approxmatons of each of those Improvements The fastest known algorthms for solvng systems of equatons run n tme O(m log n log ɛ 1 ) [CKM + 14]. The algorthm I have presented here can be substantally mproved by combnng t wth Cholesky factorzaton. Ths both gves an effcent parallel algorthm, and proves the exstence of an approxmate nverse for every M-matrx that has a lnear number of nonzeros [LPS15]. References [CKM + 14] Mchael B. Cohen, Rasmus Kyng, Gary L. Mller, Jakub W. Pachock, Rchard Peng, Anup B. Rao, and Shen Chen Xu. Solvng sdd lnear systems n nearly mlog1/2n tme. In Proceedngs of the 46th Annual ACM Symposum on Theory of Computng, STOC 14, pages , New York, NY, USA, ACM. [LPS15] Yn Tat Lee, Rchard Peng, and Danel A. Spelman. Sparsfed cholesky solvers for SDD lnear systems. CoRR, abs/ , [PS14] Rchard Peng and Danel A. Spelman. An effcent parallel solver for SDD lnear systems. In Symposum on Theory of Computng, STOC 2014, New York, NY, USA, May 31 - June 03, 2014, pages , 2014.