CC BY
31
UDC 519.612
ESTIMATES OF THE RUNNING TIME AND MEMORY REQUIREMENTS OF THE NEW ALGORITHM OF SOLVING LARGE SPARSE LINEAR SYSTEMS OVER THE FIELD WITH TWO ELEMENTS
© Vasiliy Vadimovich Astakhov
Moscow State University named after M.V. Lomonosov, Leninskie Gory, 1, Moscow, 119991, Russia, Numbers Theory Department of Information Protection Branch, Student of Mechanics and Mathematics Faculty, e-mail: astvvas89@mail.ru
Key words: linear sparse systems; pade approximations; corank distribution.
A new algorithm of solving large sparse linear systems over field with two elements is considered in this work. Algorithm was proposed by M.A. Cherepniov. Algorithm uses the construction of matrix Pade approximations over finite fields. It is supposed that elements of approximation polynomials are independent and are identically distributed.
Method for finding distributions of coranks of random symmetric, antisymmetric and common matrices is constructed. Lower and upper bounds for number of previous approximations sufficient to construct the next one are obtained. The logarithmic dependence for sufficient number of keeping approximations on every step for successful completion of algorithm with probability of 0.99 is found. Using the computer program exact values of estimates of running time and memory requirements are found, results are given in this work.
1 Introduction
A new algorithm of solving sparse linear systems over Z2 is considered in [1]. Algorithm uses the construction of matrix Pade approximations over Z2 . Let
a = £°=0a'A-', a G F(n x n), n G N, n ^ 64,
where A is transcendental variable.
Matrix polynomials Q(s)(A) G F(n x n)[A] that satisfy
a(A)Q“>(A) - P(s)(A) = E£,+1r,A-i
degQ(s) ^ s, degP(s) ^ s,
for some P(s)(A) G F(n x n)[A], are said to be matrix Pade approximations with number s,
a(A)
Denote the coefficients of approximation according to the formula:
q(s)(a) = E^qS” ■ A'.
Let QSs) vanishes. Consider following transformation. With the help of elementary transformations of the columns of matrix polynomial Q(s) (A) we reduce its leading term to the form where left columns are linearly independent and the rest are zeros. Then we multiply columns of reduced
matrix polynomial Q(s)(A) corresponding to the zero columns of its leading term on A , Repeat such procedures until the leading term become nonsingular. Denote the necessary number of such iterations by (s, There was shown in [1], that for s > k — 1 fulfillment of conditions:
Zs ^ k — 1,Cs-i ^ k — i,i = 1,..., k — 1,
is sufficient for constructing approximation with the number s + 1, with the help of coefficients of approximations with numbers s — k + 1,..., s .
In this paper we obtain estimates on the expectation of the running time and memory requirements for the successful completion of the algorithm with high probability.
In the second part of the work we obtain the distribution of coranks of random square matrices over finite fields. In the third part estimates on the distribution of stohastic variable Z are obtained. In the fourth part we make the estimates on the expected value and the distribution of auxiliary stochastic variable t . In the fifth and sixth parts we obtain final results and give some experimentally established facts,
2 Distribution of coranks of random matrices over Zp
Consider a matrix, whose elements are independent identically distributed random variables with values in Zp, where p is a prime number, Denote q = p ,
anm[r] - probability that the matrix of size n x m with values in Zp has rank r, tn[r] -probability that the matrix of size n x n with values in Zp has cora nk r .Let bn[r] = ann[r], tn[r] = bn[n — r]
Bn(x) - generating function for the distribution of ranks of matrices of size n x n . Tn(x) -generating function for the distribution of coranks of matrices of size n x n . Bn(x) = xn ■ Tn( X) sn[r] - probability that the symmetric matrix of size n x n with values in Zp have rank r ■ fn[r] - probability that the symmetric matrix of size n x n with values in Zp have corank r. sn[r] = fn[n — r]
Sn(x) - generating function for the distribution of ranks of svmmetrie matrices of size n x n , Fn(x) - generating function for the distribution of coranks of svmme trie matrices of size n x n , Sn(x) = xn ■ Fn(X)
P(A) - probability of the event A.
Statement 1
ttnm[r] = a(n-i)m[r] ■ pr-m + a(n-i)m[r — 1] ■ (1 — pr-m-1) n, m, r > 0 (1)
Proof 1 If the matrix E 'size n x m has rank r, then its submatrix without the last row (of size n — 1 x m) may have a rank r or r — 1. In the first case n -th row is in the linear span of the previous rows with probability ^ . In the second it is not in the linear span with probability
r — 1
1 — ppm • obtain the required equality. □
Now consider the method that will help us not only to solve this problem for square matrices, but also for symmetric matrices.
Consider the matrix E of size n x n, rankE = r and discard its last column and row. The resulting matrix may have a rank r, r — 1 or r — 2, Denote the last row without the last
element for a*, the last column (without the last element) for /, an angular element for e, and the matrix (n — 1) x (n — 1) for E',
Lemma 1 Let the matrix E' has rank j .
1. /3 is not in the linear span of its columns, and a* is not in the linear span of its rows. Then, regardless of the value of e matrix E has rank j + 2 .
2. ft is in the linear span of its colu, mns, but a * is not in the linear span of its rows. Then,
regardless of the value of e matrix E has rank j + 1.
3. ft is in the linear span of its colu,mns, and <3* is in the linear span of its rows. Then for
e E j
j+1 Proof 2
1. Consider the first n — I rows of the matrix E, due to the fact, that /3 is not in the linear span of the columns of Erank of such submatrix is equal to rankE' + 1 = j + 1. <3* is not in the linear span of the rows of E', hence the last row of E is not in the linear span of the first n — 1 row. Therefore rank of is equal to j + 2 .
n — 1 E /3
span of the columns of E', rank of such submatrix is equal to rankE' = j . <3* is not in
E' E first n — 1 rows. Therefore rank of is equal to j + 1.
n — 1 E /3
span of the columns of E', rank of such submatrix is equal to rankE' = j . <3* is in the
E'
E
with the first n — 1 elements identical to a*, and the last one is equal to some value ei. If e = ei then last row is in the linear span of the first n — 1 and we obtain rankE = j . Now we will show, that if e is not equal to ei rank of E is j + 1. Assume the oppose, rankE = j n — 1
row H = (0, 0...0,e — ei), it is also a linear combination of the first n — 1 rows of E,
E
n — 1 E H n — 1
rows of E have rank rankE' + 1, but it has to be equal to j = rankE'. Contradiction ends the proof. □
Theorem 1
Mr] = bn-i[r] ■ p2^-n)+i + bn-i[r — 1] ■ (p2-(r-n)-i ■ (p — 1) + 2 ■ pr-n ■ (1 — pr-n))+
+bn-i[r — 2] ■ (1 — pr-n-i) ■ (1 — pr-n-i), n,r> 0. (2)
Proof 3 With the help of Lemma 1 we obtain:
E r E' r — 2
r_2 ^
pi = (1 — pn—r)2, because it is possible, if and only if the statements of the first item are fulfilled.
E r E' r — 1
p2 = 2 ■ (1 — pn—t ) ■ pn—r + (pn—r )2 ' P~pT > because it is possible, if and only if statements of the
e = ei
Probability, that E has rank r, under condition, that E' has rank r, is equal to p3 = ()2 ■ p, because it is possible, if and only if statements of the third item are fulfilled and
e = ei
We have:
bn[r] = bn-i[r — 2] ■ pi + bn-i[r — 1] ■ P2 + bn-i[r] ■ P3.
Substituting pi,p2,p3 with found expressions we obtain the required equality. □
Corollary 1
B„[x] = Bn-i(x ■ p2) ■ pi-2^n + x ■ Bn-i(x ■ p2) ■ (pi-2^n ■ (p — 1) — 2 ■ p2-2^) + x ■ Bn-i(x ■ p) ■ 2 ■ pi-n+
+x2 ■ (Bra-i(x) — 2 ■ Bra-i(x ■ p) ■ p1 n + Bra-i(x ■ p2) ■ p2 2^n) =
= pi-2^n ■ (x ■ p — 1) ■ (x — 1)Bn-i(x ■ p2) + 2 ■ pi-n ■ x ■ (1 — x) ■ Bn-i(x ■ p) + x2 ■ Bn-i(x) (3)
Proof 4 Consider the coefficient of xr in the right hand side. bn-i[r] ■ p2^r ■ pi-2^n in the first summand, bn-i[r — 1] ■ p2^r-2 ■ (pi-2^n■ (p — 1) — 2 ■ p2-2^n) - in the second, bn-i[r — 1] ■ pr-i ■ 2■ pi-n
- in the third, bn-i[r — 2] ■ (1 — 2 ■ pr-2 ■ pi-n + p2^-4 ■ p2-2^n - in the fourth. It is identical to the expression in (2) for bn[r] . □
Corollary 2
Sn[r] = sra-i[r] ■ pr-n + sra-i[r — 1] ■ pr-n-i ■ (p — 1) + Sn-i[r — 2] ■ (1 — pr-i-n) (4)
Proof 5 Proof is similar to the Theorem 1, but we have to consider the fact, that some of
Er
that E' has rank r — 2 is equal to pi = (1 — pr-i-n), because it is possible, if and only if the
n — 1
columns (without last elements), then due to the simmetry the last column is not in the linear span of first n — 1 rows (without last elements))
E r E' r — 1 p2 =
pr-n-i ■ (p — 1), because it is possible, if and only if the statement of the third item is fulfilled and e = ei.
Probability, that E has rank r, under condition, that E' has ra nk r is equal to p3 = pr-n, because it is possible, if and only if the statement of the third item is fulfilled and e = ei. We have:
sn[r] = sn-i[r — 2] ■ pi + sn-i[r — 1] ■ p2 + sn-i[r] ■ p3.
pi, p2, p3 □
Corollary 3
Sn(x) = (qn + x ■ (1 — q) ■ qn-i — qn-i ■ x2)S„-i(x ■ p) + x2 ■ Sn-i(x) =
= qn-i ■ (1 — x) ■ (q + x) ■ Sn-i(x ■ p) + x2 ■ Sn-i(x). (5)
Proof 6 Proof is similar to the proof of the Corollary 1. Consider the coefficient of xr in the
left and right hand sides. In the left it is equal to sn[r], and in the rig ht to qn-r ■ sn-i[r] + (1 — q) ■ qn-r ■ sn-i[r — 1] + (1 — qn-r+i) ■ sn-ir — 2 . Required equality is the result of Corollary 2. □
Due to the equalities Bn(x) = xn ■ Tn(1/x) h Sn(x) = xn ■ Fn(1/x), we obtain
x ■ Tn(x) = (q ■ x — 1) ■ (x — 1) ■ Tn-i(x ■ q2) + 2 ■ (x — 1) ■ Tn-i(x ■ q) + Tn_i(x), (6)
x ■ Fn(x) = (x — 1) ■ (q ■ x + 1) ■ Fn-i(x ■ q) + Fn-i(x). (7)
Define formal power series T, F from the following equations:
T(x) = (q ■ x — 1) ■ T(x ■ q2) + 2 ■ T(x ■ q), (8)
F(x) = (q ■ x + 1) ■ F(x ■ q). (9)
They are obtained from(8), (9) by replacement of Tn,Tn-i and Fn,Fn-i by T and F respectively.
Denote f [r] and t[r] as coefficients of xr in F(x) and T(x), respectively. Obtain them, considering coefficients of xr in (8), (9):
q2r-1
i|r| = i|r — 11 ' (1- )2 • (10)
f [r] = f [r — 1] = f [r — 1. (11)
J 1 — qr pr — 1 v '
Whence we obtain the expressions:
4 2 k2 k
tv ^_+ n i q■x i q ■x i i q ■x ^ /n^
T(x) = + ((1 — q)(1 — q2))2 + ... + ((1 — q)(1 — q2 )... (1 — qk))2 + .^ (12)
where we set = (1 — q)(1 — q2)... (1 — qk)....
2k x x x
F(x) = Cî' (1 + p—T + (p — 1)(p2 — 1) + +(p — 1)(p2 _ 1) ... (pk _ 1) + ■■■)' f13’
where we set cq , such that F(1) = 1. To find cq we prove auxiliary Lemma.
Lemma 2
1
s“»n;=i —I = nr=01—q2^i+r (w)
Proof 7 Let prove, that
V 2^fc+i_______1______ = V k+ipi2 . _________________________________________________________1_ . _1_ fl c,')
¿=o nj=i(pj — 1) ¿=-p nk=o(p2^j+i — 1) nkzi+i(p2^j — 1) 1 ;
by the mathematical induction. Base case for k = 0, 1 + p—- = p—-. To prove the inductive
step note the following equality:
,2 1 1 ,2 1 1
p ■ rrk / 2. ----7T ■ ^k-,+1 , ----tt + P
nk=o(p2^j+i — 1) nti+V-j — 1) nk+0i(p2j+i — 1) nk=i+i (p2j — 1)
pi2+2-k+3 1
nk+0(p2j+i — 1) nk=i+i (p2^j — 1)
= p(i+i)2 —1------------------^-------+ p(i+i)2 —1--------------r-+r^----------------. (16)
nk+0i(p2j+i—1) n—(p2^^ — 1) nk+0i(p2j+i — 1) n^v*—1)
Now with the help of induction hypothesis:
V 2-fc+3________________1_ = v k+ipi2 ____________________________________1_ _1_I
i=0 nj=i(pj — 1) ¿=-p " nk=o(p2j+i — 1) ■ nkzi+i(p2^j — 1) +
1 1
+ , • TT +
n2=fc+2 (pj — 1) n2=+3 (p — 1)
From (16) with i = 0 we obtain:
= V fc+V2 •____________1__________•___________1___________I___________p_________•___________1_______h
i=i p n=o(p2-j+i — 1) nk=i+i(p2j — 1) n^V-j^ — 1) nk=i (p2j — 1)
h________p_______________________= yfc+ipi2 ._1_________.__________1__________h
n2=fc+3(pj' — 1) ¿=2 p nk=o(p2^j+i — 1) ng+V-j — 1)
+ p4__________________________^_ + V2 * 1_________________________________1
nk+oi(p2j+i — 1) n-v- — 1) ¿=ip nk+oi(p2j+i — 1) nk=i+2 (p2^j — 1)
With the help of (16) with i = 2,..., l — 1 we get:
= = yfc+i p2 ._1___.___________1____________h
... ¿=p nk=o(p2^j+i — 1) nk=i+i(p2^j — 1)
p12 1 , w „¿2 1 1
+ TTk+1 /..o.o_i_i 1 \ tt k+1 — //..o.-i 77 + V i=ip
nk+d(p2^1 — 1) n+i-V-j — 1) ¿=^ nk+oi(p2j+i — 1) nk=i+2(p2^j — 1)
And in the result:
= = v k+2pi2_________________________________1__1_______
... ¿=-p nk+oi(p2j+i — 1) n^+v-j — 1).
In the left side of (15) are partial sums of the series from (14). And in the right: vk+1pi2 .__________________________________________________________________________1_._1_ > p(k+i)2 ._1_
¿=-p nk=o(p2j+- — 1) nkZi+1(p2^j — 1) > p nk=o(p2^j+i — 1).
j=o^ / ^^=
y k+lpi2 . _______________________ . __________
TTfc _(-2j+1_ 1) rrk-i+1/'
nk=o (—2j+1 - 1) nk=i+1(p2^j - 1)
j=U^ )
p(k+1)2 . _______1__________+ p(k+1)2 . __________________1_ . yk p(k+1)2-i2 . ______1________ <
- nk=0(p2^j+1 - 1) +- nk=0(p2^j+1 - 1) *=1- nk=i+1(p2^j - 1) <
^ p(k+1)2 ■ nk ( 21j+i IN ■ (! + k ■ p-(2^fc+1)).
nk=o(p2 j+1 - 1N
WTience we obtain, that expression in the right side of (15) tends to the right side of (14)-Lemma is proved. □
Therefore cq = (1 — q)(1 — q3)... (1 — q2 • k+1)___________
Using (2) and (4) and equalities tn[r] = bn[n — r], sn[r] = /n[n — r] (or equating coefficients of xr+1 in (6), (7)) we obtain:
i«[r] = q2 r 1i«-1[r — 1] — ((q + 1) ■ q2 r — 2 ■ qr) ■ in-1[r] + (q2 r+2 — 2 ■ qr+1 + 1) ■ in-1[r + 1], (17)
fn[r] = qr ■ fn-1[r — 1] +(1 — q) ■ qr ■ fn-1[r] +(1 — qr+1) ■ fn-1[r +1]. (18)
Consider the ratios /¿[r] = ft] and ¿¿[r] = ^ ■
Lemma 3 For each, value of i /¿[r],ti[r] do not increase.
Proof 8 We will use mathematical induction on i. Base case: i = 0 we obtain: /¿[r] = /¿[1] =
0, r > 0 . Consider the induction step.
Let /¿-1[r] = A, /¿-1[r + 1] = B, B ^ A. We have:
?[ ] ?[ l1] = fi[r — 1] ■ qr ■f [r — 1] + fi[r] ■(1 — q) ■ qr ■f [r] + /¿[r +1] ■(1 — qr+1) ■f [r +1]
f [r]—f [r+1] = /jr]
fi[r] ■ qr+1 ■ /[r] + /i[r + 1] ■ (1 — q) ■ qr+1 ■ /[r + 1] + /¿[r + 2] ■ (1 — qr+2) ■ /[r + 2]
/ [r + 1]
A ■ qr ■ / [r - 1] + A ■ (1 - q) ■ qr ■ / [r] + B ■ (1 - qr+1) ■ / [r + 1]
f [r]
A ■ qr+1 ■ / [r] + B ■ (1 - q) ■ qr+1 ■ /[r + 1] + B ■ (1 - qr+2) ■ / [r + 2]
I.
/ [r + 1]
From (1) multiplicating by x - 1 and considering coefficients of xr+1 we obtain equ,ality /[r] = qr ■ / [r - 1] + (1 - q) ■ qr ■ / [r] + (1 - qr+1) ■ / [r + 1] and using equality (11) f+p = pr+1 - 1 we have:
qr+p
I = (A - B)(qr ■ (-r - 1) + (1 - q) ■ qr - ^ 0.
— — 1
Step is proved. Proof for U[r] is simila r. □
Statement 2 tn[r],/n[r] converge to the t[r],/[r], when n tends to the infinity.
1
Proof 9 From Lemma 3, (17) and (18) we have:
/¿[0] = (1 — q) ■ Zn-1[0] + (1 — q) ■ (1 — q)/n-1[1] ^ fn—1[0],
ti[0] = —((q + 1) — 2) ■ ^n-1[0] + (q — 1)2 ■ yz—q s2^n-1[1] ^ ^n— 1[0],
(1 — q)2
therefore tn [0] and /n [0] decrease according to i, and hence converge to some limits / and t, hence tn [0] and /n [0] have limits. Now, using mathematical induction on r , we will prove that tn[r] and /n[r] converge to some limits. Base case for r = 0 is proved. Let prove the induction step. From (17), (18) we have:
(1 — qr+1) ■ in— 1[r + 1] = yn[r] — qr ■ A—1[r — 1] — (1 — q) ■ qr ■ A—1[r]
(q2^r+2 — 2 ■ qr+1 + 1) ■ tn— 1 [r + 1] = tn[r] — q2^r—1tn— 1 [r — 1] + ((q + 1) ■ q2'r — 2 ■ qr) ■ tn—1 [r]
WTience , convergence of the right hand side, we obtain that left side also converges. Step is proved. Limits satisfy (8) and (9), hence are equal to c1 ■ t[r],c2 ■ /[r] Because of the choice of tq and cq constants are equ,al to 1. □
Following results were computationly obtained, if p = 2: t[0] « 0.2887881,t[1] « 0.5775762, t[2] « 0.1283503, t[3] « 0.0052388 . Also we have t[i] ^ c ■ q-2 , for some constant c.
Lemma 4
t [] r2 (nn=r+1(1 — q-))2 ,10,
tn[r] = q ■ T-Tn-r/-,-----(19)
IL=1(1 — q-)
Proof 10 We will prove using mathematical induction on n . Base case for n = 0 t0 [0] = 1. Let prove the induction step, from (17):
tn[r] = q2 r 1 ■ tn—1[r — 1] — ((q + 1) ■ q2 r — 2 ■ qr) ■ tn—1[r] + (q2 r+2 — 2 ■ qr+1 + 1) ■ tn—1[r + 1] =
using induction hypothesis:
= </' r—1 ■ q(r—1)2 ■ ^?n^r<,11— q'))2 — ((«+1) ■ qir — 2 ■ qr) ■ qr2 ■ ^i*2+
IL=1 (1 — q-) ni=1 (1 — q-)
+ („2 • r+2 2 qr+1 + q(r+1)2 (nn=r1+2(1 — q-))2 = qr2 (nn=r1+1(1 — q-))2
+(q — 2^q +1)^q ■ nn=—r2(1 — q-) =q ■ nn=-1r(1 — q-) ■
■ ((1 — qr)2 — (1 — qn—r) ■ ((q + 1) ■ q2r — 2 ■ qr) + q2^1 ■ (1 — qn—r) ■ (1 — qn—r—1)) =
r2 (nn=r+1(1 — q-))2
= q •---------!-------------.
q nn=—1r (1 — q-)
□
Statement 3 When p = 2 at a fixed r0 not equal to 0 sequence tn[r0] increases, when n = r0, r0 + 1,... and decreases in case r0 = 0 .
Proof 11 From Lemma 4-'
tn+1 [r 0] = (1 — qn+1)2 = 1 — qn + q2^n+2 tn [r 0] = (1 — qn+1—ro) = 1 — qn+1—ro .
When r0 = 0 we have qn+1 + q2^n+2 < qn ; hence ratio is smaller than 1 and sequence tn [0] decreases. When r0 ^ 1 we have qn+1—ro + q2^n+2 > qn ; hence ration is bigger than 1 and sequence tn[r0] increase s. □
3 Distribution of stohastie variable Z
Consider following stohastie process. On the zero step matrix A0 over Z2 with size n x n is constructed, each element of A0 is 0 or 1 with probability 2 , With the help of elementary
A0
A1
filling zero-columns of A0, and reducing it to the form with corankA1 zero-columns at the
A2, A3 . . . A- . . .
reducing it to the form with corank zero-columns at the end.
Denote for Z = minjijcorankAj = 0} , Constructed stohastie variable is identieal to Z from
A- k A-+1
n — k A- A-+1
random matrix with size k x k, Denote for ,n,k = P(Z = k). Then ,n,0 = tn [0],
^n,k S,= 1tn[i] ■ ^¿;k—1. (20)
Now we obtain lower and upper bounds for ,n,k,
,n,k = tn[n]k ■ tn[0] + E^Sn—itnM* ■ tn[j] ■ ,j,k—i— 1.
Where i is number of zero-matrices, and j is a corank of the first none-zero. Then ^1 k = (2 )k+1. ’
,2, = |^(16)k + l!^k=01(116)i^ (1)k—- =
= 3 (l)k + 9 ■ ((2)k — (116)k) =1 (Uk_ 15 (^)k
8 (16) 32 ■ (2 — 15) 14 (2) 56 (16) .
Similarly we have
= 1 (1)^ 105 (l)k + _5H (^)k
,3,k 10 (2) 248 (16) 9920 (512) .
Now we obtain lower and upper bounds for ,n,k, n > 3, We do this by selecting the first moment when corank is lesser or equal 3, From Statement 3 a = £°=4t[i] - is upper bound for probability of random matrix to have corank greater than 3, Using Statement 3 we obtain:
,n,k ^ (a)k ■ t4 [0] + Sk==01S3=1(a)i ■ t[j] ■ ,j;k—¿—1 ^
Replacing ,1)k, ,2,k, ,3,k by obtained values and using known tresholds for t we have :
^ 0.75001 ■ (1)k — 0.58599 ■ ^-1)k + 0.14156 ■ (-^ + 0.00207 ■ (a)k, (21)
2 16 512
a =1 — S4=0t[i] ^ 0.0000467,
,n,k ^ (t4 [4])k ■ t[0] + £k=01S3U(t4[4])- ■ t4[j] ■ ,j,k—i—1 ^
Replacing ,1)k, ,2,k, ,3,k by obtained values and using known thresholds for t we have :
^ 0.72580 ■ (1)k — 0.50404 ■ ¿)k + 0.09126 ■ (-^ — 0.02425 ■ ^^^)k (22)
2 16 512 65536
hence when k ^ 2 we obtain:
(0.72580 — 0.50404 ■ -1) ■ (V = 0.717924 ■ (^)k S ,nk S 0.75001 ■ (V. (23)
64 2 2 2
Now we will found bounds for Ek = — E°=k/n(P(Z S i)) (13), auxiliary value, which we will
use later. From (23):
1 — 0.75001 ■ (1)k S 1 — P(Z > k) =
= P(Z <= k) S 1 — (0.72580 — 0.50404 ■ ^) ■ (1)k S 1 — 0.717924 ■ (1)k, k ^ 2, (24)
0.62036 S P(Z S 1) S 0.62744. (25)
Then when k ^ 2 we have:
E* = —E|=k.ln(P(Z S i)) = S“kS“=1(1 — PjZ S i))j = S»1S£k(1 — P(Z S i))j.
We obtain the bounds:
E ^ ESk(0.717924)j ■ (2)'j _ (0.717924 ■ £)j ^
Ek * EiT1 j = EiT1 j' (1 — j *
1 4 1
> 2 ■ (0.717924) ■ ^ + - ■ (0.717924)2 ■ ^, (26)
E S E£k(0.75001)' ■ (1 )'j = (0.75001 ■ 2k)j S
Ek S E, = j = E, = a — j S
1 4 1 1
S 2 ■ (0.75001) ■ -T + - ■ (0.75001)2 ■ —----------;--------- —r. (27)
v ; 2k 3 v ; 4k 1 — (0.75001) ■ 2k { ’
4 Expected value and distribution of auxiliary stohastie variable t
There was shown in [1] , that for r > t — 1 > 0 fulfillment of inequalities
Zr s t — 1, Zr—1 s t — 1, Zr—2 s t — 2,..., Zr—*+1 s 1
is sufficiently, for constructing approximation with number r + 1 with the help of coefficients
r, r — 1 , . . . , r — t
£1 = Zr, . . . , Ci = Zr—1+1, . . . .
Let:
9 = min{t ^ 1 : £1 S t — 1, £ S t — i + 1, i = 2,..., t}, t = min{t ^ 1: £ S t — i + 1, i = 1,... , t}.
Let n(i) = n*k=1P(£ S k), n(0) = ^^d /(s) is generating function of distribution of t.
Lemma 5
^ _ " Vjn,aieNU{0}
P(t = k) = Sniai£Nu{0},ai+...+a„=k(—1)n—1 ■ nn=1n(a-), k e N (28)
Proof 12 We will prove using mathematical induction on k. Base case for k = 1: P(t = 1) = P(£ S 1) = n(1). Induction step:
k—1
P(t = k) = P({£- S k—i+1, i =1, 2,..., k}\y {t = j, £'+- S k—j — i+1,i =1, 2,..., k—j}) =
j=1
(it is true, that £ S k — i + 1, i = 1, 2,..., k, but it is not true that t < k J
= n(k) — Ek=1n(k — j) ■ P(t = j) = n(k) +
+ E«n+11 = 1( —n(an+1)) ■ En,a;eNU{0},ai+...+an=k—an+i (— 1)n 1 ■ ^=1^ =
= En,a;eN U{0},ai+...+an=k (— 1)n 1 ■ nn=1n(a').
□
Statement 4
1 ^ (s) = E~0n(i) ■ s', 0 < s < 1 (29)
Proof 13 Using the fact, that 1 — 1+x = E'=1(—1)'—1 ■ s' we obtain:
1-----------^: = E°=1 (—1)j—1(E°=1n(i) ■ s')j =
1 + E°=1n(i) ■s' j=u ; w )
= E°=1En,aieWU{0},«i+...+an=k ( —1)n 1 ■ nn=1n(ai) ■ s' = /(s)
0 < s < 1 □
Theorem 2
2.382189 S Mt S 2.484705 (30)
12.797192 S Mt2 S 14.300803 (31)
Proof 14 Let n(ro) = n°=1P(£ S i) • Because
E' = —/n(n(^)) + /n(n(i — 1)), (32)
then using (25), (27) (26) we obtain 0.402462 S n(ro) S 0.419781. We have, that if s ^ 1
/(1) = 1
1—s
1s
/(1) _ /(s) = 1 + E“^« — n(i — 1)) ■ s‘. (33)
When s ^ 1 we have:
f or1 = n„. (34)
Hence we obtain lower and upper bounds for Mt = /'(1): 2.382189 S Mt S 2.484705.
From
(1 — s) ■ E“0(nw — ■ s' = 1 — n(^) + E^=1(n(i) — n(i — 1)),
we have:
—1—s-----/’/(1)—1
-----------= E^TO — ■ s'
1—s
reducing left side to the common denominator and proceeding to the limit, when s ^ 1
///(1)
/ () -E~(П(i) — ^«0).
2 ■ //(1)2
Now from (32) and (34) we have:
/"<1) = 2 ■ //<1) ■ <E“1exp<E,) — 1). (35)
From (34) and (35) and using Taylor series for exponent:
f(1) = 2 ■ //<1) ■ ((/(1) — 1) + E~2E~1 j).
From (26) and (27) we know, that c1 ^2—'+1 S E' S c2■ 2—'+1, where c1 = 0.71792, c2 = 0.77884. Now from (30) we obtain inequalities
///(1) ^ 6.58527 + 4.764378 ■ E~ 1 (0.71792 ■ 2—k+1 + 0.5 ■ (0.71792)2 ■ 4—k+1) ^ 10.415003,
///(1) S 7.378108+4.96941 E^=1 (0.77884■ 2—k+1+0.5 ■ (0.77884)2 ■ 4—k+1 + 6 ^ ((10.^7(8087)7388|)3k+0)25) S
S 11.816098.
Mt2 = ///(1) + //(1), using (30) we obtain required inequality. □
Now we will obtain bounds for values of all derivatives. Let 7(s) = fS-—1, g(s) = 1 — E^^i —
1) ■ P(Z > i). From (33) "
1s
-------j- = 1 — E^—1 ■ P(Z > i) ■ s',
1 — / (s)
Y(s) ■ g(s) = 1.
n
E'u(^ ■Y(“-‘)<s) ■ g(l)<s) = 0,
7(„)<s) = ) ■Y^W^).
g(s)
Note, that g(')(s) < 0 < g(s), when 0 S s S 1,i ^ 1 ■ Hence, if for some functions g1 (s),g2(s) the following statements are fullfilled:
g1(s0) = g2(s0) = (37)
gl')(s0) S g(')(s0) S g2')(s0) < 0,i > 1, (38)
then
(-)W(s0) S Y(')(s0) S (-)W(s0),i > 1. (39)
g2 gl
We take g' equal to a' — T3V, a' = g(s0) + -Ao- ,for satisfiyng (37) and ftT = 0.75001, and
1 2 1 2 " ft2 == 0.28893 S 0.71792 ■ ^ro) ,is sufficient due to the defenition of g(s) and (24) to satisfy
(38) , Then
(^)(n)(s0) = (______1____2_)(n)| = 0 = n! . _______2 ■ ^' ■ a__________________ n > 1
(g') (0) (a' — ft' — Of) |2=20 ! a' ■ (2 ■(a' — ft') — a'■ s0)n+T, > .
We obtain, taking a', ft', s0 e {0,1}
0.22416 ■ (0.64446)n ■ n! S Y(n)(0) S 0.42857 ■ (0.87501)n ■ n! (40)
1.37657 ■ (2.37657)n ■ n! S Y(n)(1) S 1.95908 ■ (4.72711)n ■ n! (41)
/(s)
1.37657 ■ (2.37657)n—1 ■ n! S /(n)(1) S 1.95908 ■ (4.72711)n—1 ■ n!
Now we will find bounds for P(t > n),
/(n)(0) = i^ Y(n—T)(0) + Y(n),
/(n) (0) Y(n—t) Y(n)
P (t = n)
n! (n — 1)! n!
hence:
Y(')
P(t > i) = ^. (42)
Now from (40) we obtain:
0.22416 ■ (0.64446)' S P(t > i) S 0.42857 ■ (0.87501)'. (43)
Here we can obtain more precise upper bounds for P(t > i), using polynomials of higher
degree. For example consider function g1 (s) = a — ft■ s — , where a = 1 + 5, 5 = 0.47060, ft =
1 2
0.14434 = 0.37964 — 0.23530 > g/(0) — 0.5 ■ 5, Then g1(0) = g(0^d g1(n) S g(n), and hence
Y(n)(0) < g1(n)(0). From the other side:
1 2 — s cT c2
gi(s) ft ■ s2 — (2 ■ ft + a) ■s + 2 si — s s2 — s
i) ,c2 = ^-(si — S2) ■
wher sT, s2 are roots of the polynomial in denominator, cT = ^.(^SS), c2 = ^.(22)—222) ■ Finding
s1 s2
Y(n)(0) < n! ■ (ci ■ s—n + c2 ■ s—n) S n!(0.52487 ■ (0.78807)n+1 + 6.40328 ■ (0.09157)n+1).
Similarly taking gl(s) = a — ft ■ s — e ■ s2 — we have:
" 1 2
1 2- s
g1(s) 2 — 1.7593 ■ s + 0.14435 ■ s2 + 0.21115 ■ s3
and upper bound:
Y(n) (0)
P(t > n) = -—y- < 0.41973 ■ (0.76773)n + 0.19412 ■ (0.07398)n + 0.38590 ■ (0.18590)n. (44) n!
5 Expected value of stochastic variable 6
Now we will found lower and upper bounds for expected value of 6,
Theorem 3
3.63148 S M6 S 3.84696 (45)
22.33992 S M62 S 29.58587 (46)
Proof 15 Let u be an event, in which Zi = i. Then, if Zk S i — k + 1, k = 2,...,i, then
6 = i + r = t(u) + r, where ,,t are variables identically and independently distributed, in other case 6 = t . We have:
M6 = Mt ■ (1 + S~1P(Z = i) ■ n(i — 1)) (47)
M6 > 1 + P<Z = 1) + P<Z = 2) ■ n<1) + P(Z = 3) ■ n<2) + P<Z > 3) ■ n<^}
M6 S 1 + P(Z = 1) + P(Z = 2) ■ n(1) + P<Z = 3) ■ n(2) + P<Z > 3) ■ n<3)
3.63148 < 1.52443Mt < M6 < 1.54826Mt < 3.84696
Now we find bounds for M62 .Let I1 be a stohastic variable equal to 1, if Bk^ = k, Z' S k — i + 1, i = 2,..., k and 0 in other case. Then 6 = t + I1 ■ r . Note, that I1 and r are independent variables. We have:
M62 = Mt2 + 2 ■ Mt ■ r ■ I1 + M(r ■ I1)2 =
= Mt2(1 + M/2) + 2 ■ Mt ■ I1 ■ Mt = Mt2(1 + MI1) + 2 ■ Mt ■ Mt ■ I1 (48)
Note, that 1 + M/t = . Hence, it’s enough to find bounds for Mt ■ /t .
Mt ■ /i = E°=1i ■ P(Z = i) ■ n(i — 1).
Using (21) and (22) we obtain:
Mt ■ /t S E°=1i ■ 0.75001 ■ (1)' = 1.50002 (49)
Mt ■ /i > E°=1i ■ (0.72580 ■ (1)' — 0.50404 ■ (-1)') ■ n(^) = 1.41575 ■ (Mt)—1 (50)
2 16
At the end using (30), (31), (49), (50), inequalities for M/t and equality (48) we have:
22.33992 S M62 S 29.58587
6 Upper bound for algorithm’s memory requirements
For the work of the algorithm from [1], we have to keep 9 previous approximations to construct the new one. Further we talk about probabilities under conditition of successfully completion of first r steps, where r — number of stored approximations.
Theorem 4 For successful execution of the algorithm with probability 0.99 on matrices of size 2s and block-size 2k , where (s — k) > 10, is sufficient to keep 2.622407 ■ (s — k) + 18.805443 previous approximations.
Proof 16 We will find the number of approximations n for successful execution of l steps of
0. 99
1 — l ■ P(6 > n) . Hence it is sufficient, that P(6 > n) < . From (21) and (44)-'
P(6 > n) = P(t > n) + E^P(Z = i) ■ n(i — 1) ■ P(t > n — i), (51)
P(6 > n) S P(t > n) + 0.33866 ■ P(t > n — 1) +
+0.47059En=:212—' ■ (0.41973 ■ (0.76773)n—' + 0.19412 ■ (0.07398)n—' + 0.38590 ■ (0.18590)n—') S
S 0.90382 ■ (0.76773)n—1 + 0.06601 ■ (0.07398)n—1 + 0.33866 ■ (0.5)n—1 + 0.23154 ■ (0.18590)n—t.
Number of steps of the algorithm is equal to l = 2s—k . n will be large, so P(6 > n) <
0.90383 ■ (0.76773)n—1, and it is true, if n = 2.622407 ■ (s — k) + 18.805443 . □
Theorem 5 For successful execution of the algorithm with probability 0.99 on matrices of size 2s and block-size 2k, where (s — k) > 10, is necessary to keep 1.3 ■ (s — k) + 6.22 previous approximations.
Proof 17 Let r be the number of stored approximations. Then {6' > r} and {6'+r > r} are independent events. Than probability of successful execution of the algorithm is not bigger than
1 — P(6 > r) ■ • Using (51), (22) and (43) we obtain:
P(6 > n) > P(t > n) + 0.33139 ■ P(t > n — 1) + 0.05947 ■ P(t > n — 2) >
> 0.22416 ■ (0.64446)n ■ (1 + 0.51421 + 0.14318) = 0.37152 ■ (0.64446)n.
Hence, when r < 1.3 ■ (s — k) + 6.22, 1 — P(6 > r) ■ < 0.99 . □
7 Computed results
Obtained probabilities can be computed for n = 64 using computer. Distribution of coranks
64
Functions ^64,k we compute from (20), up to k S 64, Further members of the product from the right side of formula Mt = (n(<^))—1 can be discarded, because their product differs from 1 lesser than « 2—63, We have Mt ~ 2.39, To compute M6 we use (47): M6 « 3.67,
Then using (29) and (51), we compute P(t > k),P(6 > k) and obtain that, for k > 30
P(t > k) « 0.13816 ■ (0.75471)k+1, P(6 > k) « 0.26795 ■ (0.75471)k+1, hence for k = 60 with probability bigger than 0,99 algorithm will be successfully executed, 60 will require less than
30 gigabyte of memory.
In the proof of theorem 4, expression 1 — l ■ P(6 > k) was used to find lower bound for
P(6 > k)
will be higher. Also on the first step, because of the small size of approximation polynomials we can keep much more approximations. Compute the probability, that using m GB of memory, algorithm for matrix of size N and block-size n > 64 will be successfully executed. We will fix K - maximal number of kept approximations. Let g[i][j] be the probability of successfully execution of i steps with the condition 6' = j, g[0][0] = 1.0. Let find g[i][k]. Memory restriction on the number of kept approximations l' on the i-th step is i — 1 + i — 2 +... + i — l' S m2“3 ■ l'(i — ) S Sm > whence fi nd l' .Let l' = min(l', K). Then run thro ugh k = 1,..., l',
6' = k 6
first inequality, than let it be with number k — j + 1, Then probability of this event is equal to
P <Z = j )-n(j — 1)-(P <Z S k — 2) -nk—.+iP <Z S i — 1) — P <6 S k — j |Z' S k — i,i = 2,...k — j))
, Note, that for i — j + 1 S h S i, will be sufficient memory, and if it will be sufficient for i — j -
i — k + 1 i — j — 1
6'—k S l'—j — j , summarizing obtained probabilities Ez"=0—jg[i — j][z] we obtain the answer in
Z' = k — 1
P(Z = k — 1) ■ (E^n^ — 1) ■ P(Z = z) ■ nk—i+iP(Z S j — 1) ■ Ep=0—zg[i — k][p]+
+n(k — 1)■ P(Z = 0) ■ En=0g[i — k][p]).
Now we compute for Nn ^eps, under condition of successful execution of first g « 100 steps we obtain the probability of successful execution of the algorithm. Using this algorithm was obtained, that for matrix of size N = 226 and block-size n = 26 is sufficient m =26 GB of memory, and block-size n = 211 m = 650 GB of memory.
8 Conclusion
Algorithm from [1] was known, but there wasn’t (or was but not very accurate) theoretically estimates of its efficiency. In this work the lower and upper bound for the expected number of previous approximations is obtained:
3.63148 S M6 S 3.84696.
In the previous works only upper bounds were obtained. In [1] it was some constant C independent from the size of matrix. In [2] the result M6 < 7.233 was obtained, Logarifmieally depended on size of matrix lower and upper bounds are found for memory requirements. There were not any similar results before. Method proposed by A.M. Zubkov was used to obtain the results.
Using this results it can be shown that the algorithm from [1] is better (for exmaple by number of operations and memory usage) than other existing algorithms. Distribution of coranks of random (symmetric and none-symmetrie) matrices were already known(papers: [3],[4]), but the method proposed in this paper is new and may be used for similar calculations in other cases (for example anti-svmmetric matrices).
References
1. Cherepniov М.A. Block Laneosh-based algorithm for solving sparse linear systems // Discrete Mathematics (in Russian). 2008. V. 20. N. 1. P. 145-150.
2. Cherepniov M.A. Algorithms of constructing Pade-approximations // Materials of the Fourth international scientific conference on the problems of security and counter terrorism. Moscow State university named after M.V. Lomonosov, 30-31 october 2008. V. 2. Materials of the seventh all-russian conference "Mathematics and Security of Information Technologies" (MaSIT-2008). Moscow: MCCME, 2009. P. 21-22.
3. Goldman J., Rota G.-C. On the foundations of combinatorial theory IV: Finite vector spaces and Eulerian generating functions // Stud. Appl. Math. 1970. V. 49(3). P. 239-258.
4. Carlitz L. Representations by quadratic forms in a finite field // Duke Math. 1954. J. 21. P. 123-137.
Accepted for publication 7.06.2010.
ОЦЕНКА ОЖИДАЕМОГО ВРЕМЕНИ РАБОТЫ И НЕОБХОДИМОГО ОБЪЁМА ПАМЯТИ ДЛЯ УСПЕШНОГО ЗАВЕРШЕНИЯ АЛГОРИТМА РЕШЕНИЯ БОЛЬШИХ РАЗРЯЖЕННЫХ СИСТЕМ ЛИНЕЙНЫХ
УРАВНЕНИЙ
© Василий Вадимович Астахов
Московский государственный университет им. М.В. Ломоносова, Ленинские горы, 1, Москва, 119991, Россия, кафедра теории чисел, студент механико-математического
факультета, e-mail: astvvas88@mail.ru
Ключевые слова: разряженные линейные системы; аппроксимации Паде; распределение корангов.
В работе рассматривается алгоритм решения больших разреженных линейных систем над Z2, использующий построение матричных аппроксимаций Паде. Предполагается, что элементы аппроксимационных многочленов статистически независимы и равномерно распределены. Строится метод для нахождения распределений корангов для случайных симметричных, кососимметричных и обычных матриц. Даются оценки сверху и снизу на среднее число предыдущих аппроксимаций, необходимых для построения новой аппроксимации. Выявлена логарифмическая зависимость для достаточного числа хранимых аппроксимаций на каждом шагу, для успешного завершения алгоритма с вероятностью 0,99. Средние значения и необходимый объем памяти вычислены при помощи алгоритма, результаты также приведены в работе.
