An Improved Algorithm for Recovering Exact Value from its Approximation

0.13 MB·English

by  Jingzhong Zhang

An Improved Algorithm for Recovering Exact Value from
its Approximation ∗
Jingzhong Zhang Yong Feng
LaboratoryforAutomatedReasoningand LaboratoryforAutomatedReasoningand
Programming Programming
7 ChengduInstituteofComputerApplications ChengduInstituteofComputerApplications
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ChineseAcademyofSciences ChineseAcademyofSciences
0
610041Chengdu,P.R.China 610041Chengdu,P.R.China
2
[e-mail] [e-mail]

n
a
J ABSTRACT factor obtained is an approximate one. After then, numer-
4 Numerical approximate computation can solve large and ical methods have been studied to get approximate factors
2 complex problems fast. It has the advantage of high effi- of a polynomial[4][11][15][16] [17][18]. In themeantime, nu-
ciency. However it only gives approximate results, whereas merical methods are applied to get approximate greatest

] we need exact results in many fields. There is a gap be- commondivisorsofapproximatepolynomials[1][2][12][5],to
G tweenapproximatecomputationandexactresults. Abridge compute functional decompositions[6], to test primality[10]
A overriding the gap was built by Zhang, in which an ex- and to find zeroes of a polynomial[13]. In 2000, Corless
act rational number is recovered from its approximation et al. applied numerical method in implicitization of para-
.
h by continued fraction method when the error is less than metric curves, surfaces and hypersurfaces[3]. The resulting
t 1/((2N +2)(N −1)N), where N is a bound on absolute implicit equation is still an approximate one.
a
value of denominator of the rational number. In this pa-
m
per, an improved algorithm is presented by which a exact There is a gap between approximate computations and ex-
[ rational number is recovered when the error is less than act results[19]. People usually use rational number compu-

1/(4(N −1)N). tationstooverridethegap[7]. Infact,thesearenotapproxi-
1
matecomputationsbutbignumbercomputations,whichare
v
also exact computations. In 2005, Zhang et al proposed an
2 Keywords
7 algorithm to get exact factors of a multivariate polynomial
Numericalapproximatecomputation,Symbolic-numericalcom-
6 by approximate computation[20] but they did not discuss
putation, Continued fraction.
1 howtooverridethegap. In[9],ChezeandGalligodiscussed
0 how to obtain an exact absolute polynomial factorization
7 1. INTRODUCTION fromitsapproximateone,whichonlyinvolvesrecoveringan
0 Numerical approximate computations have the advantage integer from its approximation. They did not discuss how
/ of being fast, flexible in accuracy and being applicable to toobtain anexactrational numberfrom itsapproximation.
h
large scale problems. They only give approximate results Command convert in maple can obtain an approximate ra-
t
a and are applied in many fields. However, some fields such tional number from a float if we set variable Digits to a
m as theorem proving, need exact results and symbolic com- positive integer. When variable Digits is taken to different
: putations are used to obtain the exact results. Symbolic positive integers, a different rational numbersare obtained.
v computationsareprincipallyexactandstable. Theyhavea Whichoneistherationalnumberwewant? Wedonotknow.
Xi high complexity. They are slow and in practice, are appli- Insomecases,what’smore,noneoftherationalnumbersis
cable only to small systems. In recent two decades, numer- the one we want. For example, if we want to get 1/7, we
ar ical methods are applied in the field of symbolic computa- take 0.1196013289 as its approximation. However, no mat-
tions. In1985,Kaltofenpresentedanalgorithmforperform- terwhattakingvariableDigitsto,wecannotobtain1/7by
ing the absolute irreducible factorization, and suggested to command convert. In [22], Zhang and Feng systematically
perform his algorithm by floating-point numbers, then the discussed how to obtain the exact result from its approxi-
mation. Theyprovedthattheexactrationalnumbercanbe
∗The work is partially supported by China 973 Project obtainedfromitsapproximationwhentheerrorislessthan
NKBRPC-2004CB318003. 1/(2N(N−1)),butinpractice, thealgorithm requiresthat
the approximate error is less than 1/((2K +2)N(N −1)),
where N is an bound on absolute value of denominator of
the exact rational number and K ≥ N. In this paper, we
proposeanimprovedalgorithm whichcanrecovertheexact
rational number from its approximation when the error is
less than 1/(4N(N −1)).
The remainder of the paper is organized as follows. Sec-
tion 2 gives a review of some properties of continued frac-
tion, which are used to prove our theorems later. Section Theorem 3. Each convergent is nearer to the nth con-
3 proves that the rational number we want is the only one vergent than any of the preceding convergents. In symbols,
satisfying the error control ǫ ≤ 1/(2N(N −1)), and then if the nth convergent is taken to be [a0;a1,a2,··· ,an] = x,
shows that a kind of rational numbers can not be obtained then
from its approximation bycontinuedfraction method when
the error 1/(4N(N −1))<ǫ≤1/(2N(N −1)); and finally, |[a0;,a1,a2,··· ,ar]−x|>|[a0;,a1,a2,···,as]−x|
an improved algorithm is proposed. Section 4 gives some
for all r<s<n.
experimental results. The final section makes conclusions.
2. PROPERTIESOFCONTINUEDFRACTION Proof: Denote x=[a0;a1,a2,··· ,ar,xr+1], where xr+1 =
A continuedfraction representation of a real numberis one [ar+1,··· ,an]. From theorem 1, it holds that
of theforms:
1 x= xr+1hr+hr−1
a0+ a + 1 , (1) xr+1kr+kr−1
1 a2+a3+1··· Accordingly, we can deduceas follows:
where a is an integer and a ,a ,a ,··· are positive inte-
gers. On0e can abbreviate the1 ab2ove3 continued fraction as x(xr+1kr+kr−1)=xr+1hr+hr−1
[a0;a1,a2,···]. For finitecontinued fractions, note that ⇒xr+1(xkr−hr)=−(xkr−1−hr−1)
[a0;a1,a2,··· ,an,1]=[a0;a1,a2,··· ,an+1].
Dividing aboveequation byxr+1kr yields
So,foreveryfinitecontinuedfraction,thereisanotherfinite
continued fraction that represents thesame number. Every x− hr =(− kr−1 )(x− hr−1)
finite continued fraction is rational number and every ra- kr xr+1kr kr−1
tional number can be represented in precisely two different
waysasafinitecontinuedfraction. Theotherrepresentation Since xr+1≥1 and kr >kr−1>0, we have
is one element shorter, and the final term must be greater 0<( kr−1 )<1
than 1 unless there is only one element. However, every xr+1kr
infinitecontinued fraction is irrational, and every irrational
numbercanberepresentedinpreciselyonewayasaninfinite Therefore, it is provedthat
continued fraction. An infinite continued fraction represen-
tation for an irrational numberis mainly useful because its |x− hr|<|x− hr−1|.
initialsegmentsprovideexcellentrationalapproximationsto kr kr−1
the number. These rational numbers are called the conver- The proof is finished
gentsofthecontinuedfraction. Even-numberedconvergents
are smaller than the original number, while odd-numbered
onesarebigger. Ifsuccessiveconvergentsarefound,withnu- Theorem 4. Let x=[a0,a1,a2,···] and hn =anhn−1+
merators h1,h2,···, and denominators k1,k2,···, then the hn−2, kn =ankn−1+kn−2. Then it holds that
relevant recursiverelation is: ˛ ˛
˛˛hn − hn−1˛˛= 1
hn =anhn−1+hn−2, kn =ankn−1+kn−2. ˛kn kn−1˛ knkn−1
The successive convergentsare given by theformula and
˛ ˛
hknn = aannhknn−−11++hknn−−22, kn(kn+11+kn) <˛˛˛x− hknn˛˛˛< knk1n+1
where h−1 = 1, h−2 = 0, k−1 = 0 and k−2 = 1. Here are 3. AN IMPROVED ALGORITHM FOR RE-
some useful theorems[8]:
COVERINGTHEEXACTNUMBERFROM
ITS APPROXIMATION
Theorem 1. For any positive x∈R, it holds that
Inthissection,wewillsolvesuchaproblem: Someonehasa
[a0,a1,··· ,an−1,x]= xxhknn−−11++hknn−−12 (2) panosuitpivpeerrabtoiounnadlnNumofbderenmnomininhaitsomr oinfdt,haenrdatyioounaolnnlyumknboewr,
andhecannottellyoutherationalnumberbutanapproxi-
mationoftherationalnumberatanyaccuracy. Howdoyou
Theorem 2. Theconvergentsof[a ,a ,a ,···]aregiven computetherational numberfrom oneof these approxima-
0 1 2
tions? Let’sattackthisproblem. Withoutlossofgenerality,
by
we always assume that m,n are positive numbers. At first,
[a0,a1,··· ,an]= hknn we havea lemma as follows:
and
Lemma 1. m,n,p,qareinteger,andpn>0. If|m−q|<
n p
knhn−1−kn−1hn =(−1)n. 1 , then m = q.
pn n p
Proof. |m − q| = |pm−qn|. Noticing |pm−qn| is a non- So,wehaver=[0,1,n−2,2,n,n−1]. Itisobviousthatwe
n p pn
negative integer, and |m − q| < 1 , yields |pm−qn| < 1. cannotobtainn−1/nfrom[0,1,n−2,2,n,n−1]. Therefore,
n p pn
Hence|pm−qn|=0. Thatis m = q. Theproofisfinished. we need smaller neighborhood so as to recover the exact
n p rational numberbycontinued fraction method.
And now, we discuss how to obtain rational number from
Corollary 1. m,n,p,q are integers and p > 0,n > 0. itsapproximationbycontinuedfractionmethod. Letn2/n1
Let N ≥max{p,n,2}. If |m − q|< 1 , then m = q. be a rational number and r0 its approximation. Their con-
n p N(N−1) n p
tinued fraction representations are n2/n1 =[a0,a1,··· ,aL]
and r0 =[b0,b1,···,bM] respectively. Wewish that ai =bi
Proof: When p 6= n, it holds that pn≤ N(N −1). Hence for i = 0,1,2,··· ,L−1 and for the last term of the con-
|mn − pq| < N(N1−1) ≤ p1n. According to lemma 1, it is tinued fraction representations of n2/n1, either aL = bL or
obtained that m = q. When p=n, we have aL−1=bL,sothatwecangetn2/n1from[b0,b1,··· ,bL+1].
n p
This is thefollowing theorem:
m q |m−q| 1
| − |= <
n p n N(N −1) Theorem 6. Let n /n be a rational number and r its
2 1 0
n 1 approximation. Assumethatn ,n arecoprimepositivenum-
⇒|m−q|< ≤ ≤1 2 1
N(N −1) N−1 bers, where n < n ,and n > 1. The representations of
2 1 1
So, it holds that m = q. The proof of the corollary is fin- n2/n1 andr0 are[a0,a1,··· ,aL]and[b0,b1,··· ,bM]respec-
tively. If |r −n /n | < 1/(4n (n −1)), then one of the
ished. 0 2 1 1 1
following statements must hold.
Theorem 5. Letx= mn beareducedproperfraction,and • ai=bi (i=0,1,··· ,L);
N ≥max{n,2}. Assume that |x−w|<1/(2N(N −1)). If
we get positive rational number p/q such that |p/q−w| < • ai=bi (i=0,1,··· ,L−1),aL−1=bL,and bL+1 =
1.
1/(2N(N −1)) , where q≤N , then it holds that x=q/p.
According to assumption of n < n , we have that a =
2 1 0
Proof: From the assumption of the theorem, we have |x− 0, and b = 0. Hence a = b . In order to finish the
0 0 0
q/p|<1/(N(N−1)). Accordingtocorollary1,itholdsthat proof of theorem 6, we need two lemmas. Due to n /n =
2 1
q/p=m/n=x. The proof of the theorem is finished. [a0,a1,···,aL] and r0 = [b0,b1,··· ,bM], we have the fol-
lowing expansions:
Theorem 5 shows us as follows. Onehas a rational number
n n n n
mn in his mind, if he tells you an approximation w such n1 =a1+ n3, n2 =a2+ n4, ··· ,
that |m −w| < 1/(2N(N −1)), then there is an unique 2 2 3 3
nraetigiohnbanolrhnouomdb(ewr w−h1o/s(e2Nde(nNom−in1a)t)o,wr i+s l1e/s(s2Nth(aNn −N1i)n))t.he nnLL−1 =aL−1+ n1L, nL =aL (3)
and
The remaining question is as follows. How do we fetch out 1 1
the unique rational number in the neighborhood? We wish r =b1+r1, r =b2+r2, ··· ,
0 1
get it by continued fraction. Unfortunately, we can not al-
1 1
ways fetch out the unique rational number in the neigh- =bL+rL, ··· , =bM (4)
borhood (w−1/(2N(N −1)),w+1/(2N(N −1))) by con- rL−1 rM−1
tinued fraction method. One counterexample is the ratio- Denotingdi=ri−ni+2/ni+1,wehavealemmaasfollows:
nal number such as (n − 1)/n for n > 1. Let us show
this: set N = n, and its approximation r := (2n+2n3 − Lemma 2. Let n /n be a rational number and r its ap-
4n2 −1)/(2n2 −2n+1)/n. One can check its error d = 2 1 0
proximation. Assume that n ,n are coprime positive in-
1/(2n(n−1)+1)<1/(2N(N −1)). However, one can not 2 1
tegers, where n < n ,and n > 1. The representations
recover rational number (n−1)/n from its approximation 2 1 1
by continued fraction method. In fact. First we can easily of n2/n1 and r0 are [a0,a1,··· ,aL] and [b0,b1,··· ,bM] re-
compute (n−1)/n=[0,1,n−1], continued fraction repre- spectively. And assume that ai = bi for i ≤ k < L(k is a
sentationofn−1)/n. Andthencomputecontinuedfraction positive integer). Then when |dk| < nk+1(n1k+1−1), it holds
representation of r as follows. that ak+1 =bk+1 for k<L−1; when |dL−1|< nL(n1L+1), it
holds that aL=bL or aL−1=bL.
2n+2n3−4n2−1 1 2n2−n+1
r= ⇒ =1+
(2n2−2n+1)n r 2n+2n3−4n2−1
Proof: At first, we show that under the assumption of the
2n2−n+1 1 1+n2−n
r = ⇒ =n−2+ lemma if we have
1 21n++n22n−3−n4n2−11 r1 n−1 2n2−n+1 ˛˛˛ n2k+1dk ˛˛˛< 1 (5)
r2= 2n2−n+1 ⇒ r =2+ 1+n2−n ˛nk+2(nk+2+nk+1dk)˛ nk+2
2
n−1 1 1 then, it holds that ak+1 = bk+1 for k < L−1, and ak+1 =
r = ⇒ =n+
3 1+n2−n r3 n−1 bk+1 orak+1−1=bk+1 fork=L−1. Wediscussit intwo
cases: Lemma 3. Let n /n be a rational number and r its ap-
2 1 0
Case1(k<L−1): Fromdk =rk−nk+2/nk+1,itholdsthat proximation, where n2,n1 are coprime positive integers, and
rk =nk+2/nk+1+dk. Hencewe havethat n2 <n1,andn1 >1. Thecontinuedfractionrepresentations
1 − nk+1 =− n2k+1dk sopfenct2i/vnel1y.anDdern0otaered[ia0=,ar1i,·−··n,ia+L2]/nain+d1[bfo0,rb1i,=···0,,b·M··],rLe-.
rk nk+2 nk+2(nk+2+nk+1dk) Assumethat ai=bi fori≤k<L−1(k isapositiveinteger
⇒ 1 = nk+1 − n2k+1dk ). Then when |dk|< nk+1(n1k+1−1), it holds that
⇒ rr1kk =ankk++12+ nnnkkk+++322(−nkn+k2++2(nnknk++2k2+1d1+kd)knk+1dk) |dk+1|< nnkk++12((nnkk++12−−11))|dk| (11)
=bk+1+rk+1
Proof: Undertheassumptionthatai =bifori=0,1,··· ,k,
Hence, obviously,ak+1 =bk+1 if and only if fromequation(5),wegetdk+1 =−nk+2(nnk2k++21+dnkk+1dk). Hence
0≤ nk+3 − n2k+1dk <1. (6) we deducea relation as follows:
Therefore, if innekq+u2alitynk(5+)2(hnokl+d2s,+thnekn+1adbko)ve inequality is |dk+1| = n2k+2+n2kn+k1+|d1kn|k+2dk = nk+2n(knn+kk1++|12dk+| dk)
guaranteed.
Case 2:(when k=L−1) Wehave = nk+1(nk+1−1)|dk|
rL1−1 =aL− nL+1(nLn+2L1dL+−n1LdL−1) =bL+rL = nk+2(nk+2n(nk+kn+1k1+−11()n+k+(1n−k+11)−|dk1|)dk)
From theabove equation, if nk+2(nk+2−1+ nk+n1k−+n1k+2 +(nk+1−1)dk)
| n2LdL−1 |< 1 =1 When|dk|< nk+1(n1k+1−1),itholdsthat nk+n1k−+n1k+2+(nk+1−
nL+1(nL+1+nLdL−1) nL+2 1)dk >0. Hencewe havea relation between dk+1 and dk:
TthheenreafLore=, wbLefhoarvdeLs−h1ow<n0t,haantdifaiLne−qu1a=litybL(5f)orhdolLd−s1, t≥he0n. |dk+1|< nnkk++12((nnkk++12−−11))|dk|
ak+1 = bk+1 for k < L−1, and either ak+1 = bk+1 or
The proof of thelemma is finished.
ak+1−1=bk+1 for k=L−1.
Andnow,letusprovethetheorem. If|d |=|r −n /n |<
On the otherhand, we have 0 0 2 1
1/(4n (n −1)), From lemma 3, we can get
˛˛˛˛nk+2(nkn+2k2+1+dknk+1dk)˛˛˛˛= |nk+2(nnk2k++21+|dkn|k+1dk)| 1 1 |di|<1/(4ni+1(ni+1−1))
= nk+1|dk| ≤ nk+1|dk| for i=0,··· ,L−1. Note that nL >nL+1 =1 and
nk+2(|nnkk++12 +dk|) nk+2|(nnkk++21 −|dk|)| 1 < 1 < 1
4ni+1(ni+1−1) ni+1(ni+1+1) ni+1(ni+1−1)
So, in order to ensure inequality (5), we only need it holds
that when ni+1 >1. So, it holds that
nk+2|n(nnkkk+++112|d−k||dk|)| < nk1+2 (7) |di|< 4ni+1(n1i+1−1) < ni+1(ni1+1+1)
fori=0,··· ,L−1. Accordingtolemma2,theproofofthe
Solving inequality (7) yields
theorem is finished.
|dk|< nk+2 (8)
nk+1(nk+1+1) For an unknown rational number n2/n1 and its approxi-
When k<L−1, we have that nk+2 >1. So, it holds that mn2a/tnio1n=r[0b,0,tbh1e,o·r·e·m,b6L,s1h]owwhsenth|art0−n2n/2n/1n=1|<[b01,/·(·4·n,1b(Ln]1o−r
1 ≤ nk+2 1)). However,forpractical purpose,wehopetherestriction
nk+1(nk+1−1) nk+1(nk+1+1) on n1 >1 and n1 >n2 can be lifted. So we have following
theorem:
Accordingly, it is obtained that
1
|dk|< nk+1(nk+1−1) (9) Theorem 7. Letn0/n1 beareducedrationalnumberand
ritsapproximation. Assumethatn ,n arepositiveintegers
0 1
When k = L−1, we have that nL+1 = 1, so it is obtained and N ≥ max{n1,2}. The continued fraction representa-
that tionsofn0/n1 andr are[a0,a1,··· ,aL]and[b0,b1,··· ,bM]
1 respectively. If|d|=|r−n0/n1|<1/(4N(N−1)), then one
|dL−1|< nL(nL+1) (10) of the following two statements must hold
The proof of lemma 2 is finished. • ai=bi for i=0,··· ,L;
• ai=bi fori=0,··· ,L−1, and bL =aL−1, bL+1= Algorithm 1. Input: a nonnegative floating-point num-
1. ber r and a positive number N;
Output: a rational number b.
Proof: Weprovethetheorem in threecases:
Case1(n1 >1, n0 <n1): From1/(4N(N−1))≤1/(4n1(n1− Step 1: Set i = 0, tem = r, h−1 = 1, h−2 = 0,
1)) and theorem 6, thetheorem holds. k−1 =0, and k−2 =1;
Case 2(n = 1): We have that a = n /n . If d = r −
1 0 0 1 Step 2: Get integral part of tem and assigning it
n /n >0,thenb =a andr =r−a <1/(4×2(2−1)).
0 1 0 0 0 0 to a, assigning its remains to b.
If d=r−n /n <0, then
0 1
Step 3: Compute hi = a∗hi−1 +hi−2 and ki =
r=a0−|d|=a0−1+1−|d|=b0+1−|d| a∗ki−1+ki−1. If ki>N,then goto Step 6;
⇒r =r−b =1−|d|
0 0
Step 4: Set i:=i+1;
|d|
⇒1/r =1+
0 1−|d| Step 5: Set tem= 1 and goto Step 2;
b
On the otherhand, we have Step 6: Computing hi−1/ki−1 and assigning it to
b.
|d|<1/8⇒2|d|<1⇒|d|<1−|d|
Step 7: return b.
|d|
⇒0< <1⇒b =1
1−|d| 1
The correctness of algorithm 1 is obvious from theorem 7
and theorem 8.
So, we havethat b =a −1, b =1.
0 0 1
Case 3(n > n > 1): From n /n = a +n /n , it holds
0 1 0 1 0 2 1 4. EXPERIMENTAL RESULTS
that n /n −a =n /n . Ontheotherhand,we havethat
0 1 0 2 1
|n /n −r| < 1/(4N(N −1)) ≤ 1/n . So, we can deduce ThefollowingexamplesrunintheplatformofMaple10and
0 1 1
that a < r < a +1. Accordingly, it holds that b = a . PIV 3.0G, 512M RAM. They take little time for obtaining
0 0 0 0
Hence, we have exactrationalnumbersfromtheirapproximations,sowedo
not show time.
|d| = |r−n /n |=|b +r −a −n /n |=|r −n /n |
0 1 0 0 0 2 1 0 2 1
= d <1/(4N(N −1))<1/(4n (n −1)) Example 1. Let a be unknown rational number. We only
0 1 1
know a bound of its denominator N = 170. According to
And now, we have n1 > 1 and n2 < n1, which is case 1. theorem 7, Computing rational number a as follows: Com-
Therefore, theproof is finished. pute d = 1/(4∗N ∗(N −1)) = 1/114920. Assume that
we use some numerical method to get an approximation
Forsimplicity,sets=Lwhen[a0,a1,··· ,aL]=[b0,b1,··· ,bL] b=.8106421859 such that |a−b|<1/d. Calling algorithm
and s = L+ 1 when [a0,a1,··· ,aL] = [b0,b1,··· ,bL,1]. 1 yields as follows.
For an unknown rational number n /n and its approxi-
0 1 4 13 17 30 107 137 518
mation r, theorem 7 shows that n0/n1 = [b0,··· ,bs] when 0,1, , , , , , ,
5 16 21 37 132 169 639
|r−n /n | < 1/(4N(N −1)). However, we do not know
0 1
what thenumbersis. Thefollowing theoremshowsushow Whenalgorithm1findsthatthedenominatorof 518 isgreater
639
to get [b0,··· ,bs]. than N, it outputs 113679.
Example 2.Let a be unknown rational number. We only
Theorem 8. Letn /n beareducedrationalnumberand know a bound of its denominator N = 1790. According to
0 1
r itsapproximation. Assumethatn ,n arepositiveintegers theorem 7, Computing rational number a as follows: Com-
0 1
andN ≥max{n1,2}. Wehaver=[b0,b1,··· ,bM]and|r− pute d = 1/(4∗N ∗(N −1)) = 1/12809241. Assume that
n0/n1| < 1/(4N(N −1)). Denote n0/n1 = [b0,b1,··· ,bs]. we use some numerical method to get an approximation
Then, for any positive integer s<t≤M, the denominator b = .178870799516605 such that |a − b| < 1/d. Calling
of rational number g=[b0,b1,··· ,bt] is greater than N. algorithm 1 yields as follows:
1 1 2 5 17 22 149 171 320 1131
0, , , , , , , , , ,
5 6 11 28 95 123 833 956 1789 6323
Proof: Proof is given by contradiction. Denote by m the
Whenthealgorithmfindsthatthedenominatorof1131/6323
denominatorofg. Assumem≤N. Fromtheorem3,itholds
is greater than N, it outputs320/1789.
that |r−[b0,b1,···,bt]|<|r−[b0,b1,··· ,bs]|. Noting that
|r−[b0,b1,··· ,bs]| = |r−n0/n1| < 1/(4N(N −1)) yields
Example 3.Let a be unknown rational number. We only
|r−[b0,b1,···,bt]|<1/(4N(N−1)). Accordingtotheorem
knowaboundofitsdenominatorN =18. Accordingtothe-
5, it should hold that [b0,b1,··· ,bs] =[b0,b1,··· ,bt]. This
orem 7, Computing rational numbera as follows: Compute
contradict to that t>s. The proof is finished.
d=1/(4∗N∗(N−1))=1/1225. Assumethatweusesome
numericalmethodtogetanapproximationb=1.881536615
Basedontheorem7andtheorem5,analgorithmforobtain-
suchthat|a−b|<1/d. Callingalgorithm1yieldsasfollows.
ing the exact numberis as follows:
1,2,15/8,32/17,111/59
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is greater than N, 32/17. Zhi,QR Factoring to ComputetheGCD of Univariate
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