【机械类毕业论文中英文对照文献翻译】一种约束引导G1连续样条曲线
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机械类毕业论文中英文对照文献翻译
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【机械类毕业论文中英文对照文献翻译】一种约束引导G1连续样条曲线,机械类毕业论文中英文对照文献翻译,机械类,毕业论文,中英文,对照,文献,翻译,一种,约束,引导,G1,连续,曲线
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A constrained guided G1continuous spline curveD.S. Meek*, B.H. Ong1, D.J. WaltonDepartment of Computer Science, University of Manitoba, Winnipeg, Canada R3T 2N2Received 23 December 2001; revised 1 March 2002; accepted 5 March 2002AbstractA method for drawing a guided G1continuous planar spline curve that falls within a closed boundary is presented. The curve is composedof segments of quadratic polynomials (parabolas) and rational quadratics (conics) that join with continuous unit tangent vectors. Theboundary is composed of straight line segments and circular arcs.q 2002 Published by Elsevier Science Ltd.Keywords: Guided G1spline curve; Constrained curve1. IntroductionA method for drawing a guided G1continuous planarspline curve that falls within a closed boundary is presented.The curve is composed of segments of quadratic poly-nomials (parabolas) and rational quadratics (conics) thatjoin with continuous unit tangent vectors. The boundary iscomposed of straight-line segments and circular arcs.There are several problems whose solution requires thistype of method. A user may wish to design a curve that fitsinside a given region as, for example, when one is designinga shape to be cut from a flat sheet of material. A user maywish to design a smooth path that avoids obstacles as, forexample, when one is designing a robot path.The problem of drawing constrained curves has beendiscussed previously. In Ref. 2, a G2continuous, shape-preserving curve made of rational cubics that interpolates togiven points and that lies on one side of a line, or severallines, is described. In Ref. 4, fair Be zier curves with fixedcross-sectional area are produced. In Ref. 5, a G2continuous curve made of rational cubics that interpolatesto given points inside an arbitrary polygon is presented. InRef. 1, a C1continuous non-parametric interpolatingrational (cubic numerator and linear denominator) that liesabove, below, or between polylines is discussed. In Ref. 6,a G2continuous curve made of non-parametric rationalcubics that lies on one side of a line, or one side of aquadratic curve is found. In Ref. 7, a C1continuous shape-preserving interpolating cubic spline with minimum curva-ture is produced. The curve is constrained to lie inside anarbitrary polygon.In this paper, the boundary consists of straight linesegments and circular arcs, which is different thanboundaries considered by other papers. The curve here isquadratic, most other results use cubics, and the quadraticgenerally results in simpler theory and simpler algorithms.Finally, the curve here is guided by control points ratherthan by the more commonly used interpolation points.TheuserguidesthecurvewithcontrolpointsofaquadraticB-spline with uniform knots. In the simplest case, Section 3,the B-spline control polyline does not intersect the boundary.A B-spline can be expressed as a set of Be zier curves. If aparticularquadraticBe ziercurvedoesnotintersectanypartofthe boundary, it is accepted. If it does intersect any part of theboundary, then it is replaced by a rational quadratic Be ziercurve that has the same Be zier control points, but which doesnotintersecttheboundary.TheuseofthesameBe ziercontrolpoints ensures that the collection of Be zier curves forms a G1continuous curve. The method described in Section 3 canalways determine a rational quadratic curve that does notintersect the boundary (see Theorem 1). In Section 4, the B-spline control polyline is allowed to intersect the boundaryat some points, and a curve that does not cross the boundarycan often be produced.A user could manipulate the B-spline control pointsinteractively to cause the curve to fall within the boundaryrather than use the method presented in this paper. However,the method here attempts to aid the user by partlyautomating the process. The method here can also give a0010-4485/03/$ - see front matter q 2002 Published by Elsevier Science Ltd.PII: S0010-4485(02)00086-6Computer-Aided Design 35 (2003) 591599/locate/cad1Visiting from Universiti Sains Malaysia, 11800 Penang, Malaysia.*Corresponding author. Tel.: 1-204-474-8681; fax: 1-204-474-7609.E-mail address: dereck_meekumanitoba.ca (D.S. Meek).curve that passes exactly (within computational accuracy)through a corner of the boundary, or that is exactly tangentto the boundary. Those results could be difficult to achieveinteractively. The method used here also gives morelocalized control as only one Be zier segment may bechanged to avoid crossing the boundary, while adjusting aB-spline control point will cause three Be zier segments tochange.Special requirements such as having the curve tangent toa line segment of the boundary, having the curve run along aline segment of the boundary, and having a cusp on theboundary are discussed at the end of Section 3.2. Some results on rational quadratic Be zier curvesConsider the family (parameter w) of rational quadraticBe zier curves B(t,w) with non-zero area Be zier controltriangle B0B1B2,Bt;w Qt;wqt;w1 2 t2B0 2w1 2 ttB1 t2B21 2 t2 2w1 2 tt t2;0 , t , 1; 0 , w , 1;1The members of this family of curves are segments ofconics: w , 1 gives a segment of an ellipse, w 1 gives asegment of a parabola, and w . 1 gives a segment of ahyperbola 3. Although the curves as defined in Eq. (1) donot allow w 0 or w tending to infinity, these two extremew-values are of interest. If these values were allowed, B(t,0)would be the line segment B0B2, while B(t,1) would be thepoint B1(see Fig. 1).Result 1: Location of curves (1). The points on a curve(1) are a weighted average of the control points B0, B1, andB2. With the restrictions on t and w in Eq. (1), all of theweights are positive, so all points B(t, w) are strictly insidethe control triangle B0B1B23 (see Fig. 2).AResult 2: The curve (1) that passes through a given point.The w-value of the curve (1) that passes through a givenpoint P can be found as described later. Since all points onthe curves (1) are strictly inside the control triangle B0B1B2(Result 1), P must be strictly inside as well. Thedevelopment below finds the t in (0,1) and w in (0,1)such that Bt;w P:Since P B1, consider the intersection of the linethrough B0and B2, 1 2 rB0 rB2; with the line throughB1and P, 1 2 sB1 sP (see Fig. 1). The r and s values atthe intersection point arer B12 P B12 B0B12 P B22 B0s B12 B0 B22 B0B12 P B22 B0;2wherestands for the scalar cross product of two-dimensional vectors. Since P is inside control triangleB0B1B2, 0 , r , 1 and 1 , s. The point of intersection fora given t is also B(t,0), and this allows the solution of tindependently of w 3. The line B0B2is from Eq. (1)Bt;0 1 2 t2B0 t2B21 2 t2 t2:3so a quadratic equation for t ist21 2 t2 t2 r:4Since 0 , r , 1, there is exactly one root t in (0,1). Thisroot can be found by writing Eq. (4) as1 2 tt?21 2 rr;solving for 1 2 t=t; which must be positive, and finallysolving for t to givet ffiffirpffiffirpffiffiffiffiffiffiffi1 2 rp;5which is in (0,1) for 0 , r , 1. The vectors B(t,0) 2 P andP 2 B1are parallel and, since 1 , s, are pointing in theFig. 1. Curve (1) passing through a given point P.Fig. 2. Curves (1) filling the control triangle and nested.D.S. Meek et al. / Computer-Aided Design 35 (2003) 591599592same direction, orBt;0 2 PP 2 B1 . 0:6Substitution of formula (3) in Eq. (6) shows that1 2 t2B02 PP 2 B1 t2B22 PP 2 B1 . 0: 7Once the t value in Eq. (5) is found, the uniquecorresponding positive (see Eq. (7) value for w is, fromEq. (1),w 1 2 t2B02 PP 2 B1 t2B22 PP 2 B121 2 ttP 2 B1P 2 B1:8AThe earlier two results are summarized in lemma 1.Lemma 1. If P is strictly inside the Be zier control triangleB0B1B2, then there is a unique curve (1) that passes throughP. If P is not inside the control triangle, then there is nocurve (1) that passes through P. The t- and w-values suchthat Bt;w P are given by Eqs. (5), (2) and (8).Result 3: Control triangle filling and nesting property.Result 2 shows that there is a unique curve (1) through anypoint P strictly inside the control triangle B0B1B2. This factmeans the curves (1) completely fill the interior of thetriangle B0B1B2. The fact that the curve (1) through anypoint P is unique means two curves (1) with different w-values cannot intersect inside the control triangle as anintersection point would be a point through which more thanone curve passes (see Fig. 2).The partial derivative of Eq. (1) with respect to w is thevectorBwt;w 21 2 tt1 2 t2 t2?1 2 t2 2w1 2 tt t2?2B12 Bt;0:The direction of this vector is constant with respect to w, sothe locus of points B(t,w) with fixed t and increasing w is astraight line from B(t,0) to B1. Thus, the family of curves (1)forms a nested set of curves as w moves from 0 to 1; thecurve (1) as w tends to 0 is the line segment B0B2, while thecurve (1) as w tends to 1 is the polyline with segments B0B1and B1B2. If a curve (1) with a certain w is drawn, then allcurves (1) with higher w-values are between that curve andthe polyline B0B1B2.AResult 4: The curve (1) that is tangent to a given linesegment. The w-value of the curve (1) that is tangent to agiven line segment S0S1(S0and S1are distinct) can befound as described later in this Result. This result is used inthe method described in Section 4. The procedure is first tofind a curve (1) that is tangent to the infinite line L throughthe points S0and S1, and then to determine if the point oftangency is in the segment S0S1. The line L must passthrough the control triangle for a tangent to be possible (seeFig. 3).It will now be argued that a tangent line L cannot cutB0B2and cannot pass through B1. Curve (1) is a segment ofa conic and conics have the property that they do not crossany of their tangent lines. The curve (1) joins B0to B2, so itwould cross any line that cuts B0B2; thus, a tangent line Lcannot cut B0B2.In particular, L cannot pass through B1and cross B0B2. IfL passes through B1and does not cross B0B2, then curves (1)cannot be tangent to L because L does not enter the controltriangle B0B1B2. Thus, a tangent line L cannot pass throughB1.In the calculations that follow, first check that L is atangent to the curve B(t,w) and then check that the point oftangency is in the segment S0S1. If a curve (1) is tangent toline L, then the point of tangency B(t,w) must be on L, orBt;w 2 S0 S12 S0 0;9and the partial derivative of B(t,w) with respect to t must beparallel to L, orBtt;w S12 S0 0:10Using Eq. (1), Eq. (9) can be rewritten,Qt;w 2 qt;wS0 S12 S0 0:11Differentiation of the relation qt;wBt;w Qt;w fromEq. (1) with respect to t givesqtt;wBt;w qt;wBtt;w Qtt;w:12Taking the cross product of both sides of Eq. (12) on theright by S12 S0and using Eqs. (9) and (10) givesQtt;w 2 qtt;wS0 S12 S0 0:13The left side of Eq. (11) is the quadratic in t1 2 t2B02 S0 S12 S0 2w1 2 ttB12 S0 S12 S0 t2B22 S0 S12 S0:14Eqs. (11) and (13) show that this quadratic has a double zeroint.SinceB1cannotbeonlineL,(B12 S0) (S12 S0) 0;Fig. 3. Curve (1) tangent to the line segment S0S1at P.D.S. Meek et al. / Computer-Aided Design 35 (2003) 591599593dividing Eq. (14) by this non-zero quantity gives1 2 t2A 2w1 2 tt t2C 0;15whereA B02 S0 S12 S0B12 S0 S12 S0;C B22 S0 S12 S0B12 S0 S12 S0:16Thinking of A, w, and C as the y-values of a quadratic Be ziercurve, and recalling that a positive value for w is required andthatadoublerootofEq.(15)isrequiredin(0,1),itmustbethatA , 0 and C , 0. The condition for a double root isw2 AC or w ffiffiffiffiACp:17The double root occurs att w 2 A2w 2 A 2 C;18wherein the denominator is non-zero because w . 0, A , 0,and C , 0. Although this t-value is in (0,1), the point oftangencyP Bt;w;maynotbeinthelinesegmentS0S1,soa final check is needed, namely,0 ,P 2 S0S12 S0S12 S0S12 S0, 1:19AThe above result is summarized in lemma 2.Lemma 2. If the Be zier control triangle B0B1B2has non-zero area, the infinite line L that passes through the distinctpoints S0and S1does not cross B0B2, B1isnot on L, A and Cin Eq. (16) are both negative, and Eq. (19) is satisfied, thenthere is a unique curve (1) tangent to the line segment S0S1.Otherwise, there is no curve (1) that is tangent to linesegment S0S1. The w-, t-values at the point of tangency aregivenin Eqs. (17) and (18),where Aand Care defined in Eq.(16).Result 5: The curve (1) that is tangent to a given circulararc. The w-value of the curve (1) that is tangent to a givenarc can be found as described later. An arc will be specifiedby two distinct endpoints S0and S1, and by a non-zerosigned sweep angleufrom S0and S1(counterclockwise ispositive). Let C be the circle of which the arc is a part. Theprocedure is first to find a curve (1) that is tangent to C, thentest if the point of tangency is in the given arc (see Fig. 4).Ifuis positive, let U be the unit vector that is a rotation ofp/2 from the direction S12 S0, while ifuis negative, let Ube the unit vector that is a rotation of 2p/2 from thedirection S12 S0. The centre E and radius r of circle C areE S0 S12kS12 S0k2 tanu2?U;r kS12 S0k2 sinu2?:By analogy with Result 4, if a curve (1) is tangent to a circlecenter E and radius r, then the equationBt;w 2 E?Bt;w 2 E? 2 r2 020has a double root in t. Using Eq. (1), Eq. (20) becomesQt;w 2 qt;wE?Qt;w 2 qt;wE? 2 r2qt;w?2 0:21The left hand side of Eq. (21) in Bernstein form isc01 2 t4 4c1w1 2 t3t 2c2 2c3w2?1 2 t2t2 4c4w1 2 tt3 c5t4;22wherec0 B02 EB02 E 2 r2;c1 B02 EB12 E 2 r2;c2 B02 EB22 E 2 r2;c3 B12 EB12 E 2 r2;c4 B12 EB22 E 2 r2;c5 B22 EB22 E 2 r2:Converting to power representation, polynomial (22)becomesa4t4 4a3t3 6a2t2 4a1t a023wherea4 4c3w22 4c1 c4w c0 2c2 c5;a3 22c3w2 3c1 c4w 2 c02 c2;a223c3w22 2c1w c013c2;a1 c1w 2 c0;a0 c0:Tangents occur when Eq. (23) has a double zero in t; amethod for checking for a double zero is described inAppendix A. The condition for Eq. (23) to have a doubleFig. 4. Curve (1) tangent to an arc from S0to S1at P.D.S. Meek et al. / Computer-Aided Design 35 (2003) 591599594zero in t, D 0 where D is defined in the Appendix A, leadsto an eighth degree polynomial equation of even powers inw. Here only positive w-values whose corresponding t-values are in (0,1) are acceptable. For each point oftangency to the circle, a check is made to see if the point isin the arc being considered. The result of this process is a setof possible w-values for curves (1) (the set may be empty).Note that a curve (1) may be tangent to the arc in twoplaces and there may be several curves (1) tangent to the arc.Since the largest w-value will be taken later, thismultiplicity of curves (1) does not cause any difficulty. AResult 5 can be restated in the following lemma 3.Lemma 3. If the area of the Be zier control triangle B0B1B2is non-zero, S0and S1are distinct endpoints of an arc, thesweep angle of the arcuis non-zero, there are positive w-values such that Eq. (21) has a double zero in t for t in (0,1),the points of tangency are in the arc from S0to S1,then thereexist one or more curves (1) that are tangent to the arc fromS0to S1. Otherwise, no curve (1) exists that is tangent to thearc from S0to S1.3. A solution of the constrained curve drawing problemA solution to the constrained curve drawing problempresented in this section is based on a quadratic B-splinewith uniform knots. In this section, assume that the B-splinecontrol polyline does not intersect the constraining bound-ary. A typical segment of the quadratic B-spline controlpolyline has control points P0, P1, P2. The correspondingquadratic Be zier control points are:B0P0 P12;B1 P1;B2P1 P22:If the control polyline of a quadratic B-spline is closed, thenthe control polyline of the corresponding quadratic Be ziercurves follows the same path; if the control polyline of aquadratic B-spline is open, then the control polyline of thecorresponding quadratic Be zier curves follows the samepath, but is slightly shorter (at both ends). Thus, if the B-spline control polyline does not cut the boundary, then thecorresponding Be zier control polyline does not cut theboundary. Henceforth, references will be to the Be ziercontrol polyline and its corresponding Be zier curves ratherthan to the B-spline control polyline and curve. Further-more, each Be zier segment can be considered separatelybecause overall G1continuity is guaranteed from the Be ziercontrol polyline.Before discussing the algorithm, it is helpful to considera default weight w, and to examine two simple cases.Whenever possible, a quadratic curve (parabola, w 1) willbe used rather than a rational quadratic (conic, w 1). Thisdefault choice seems to give nice curves, although it issomewhat arbitrary and could be chosen differently. Thechoice w 1 does give the algebraically simplest curves.Even if a different choice were made, it would probably notbe too far from 1. Fig. 5 shows the control polyline in grayand quadratic B-splines with three different weight values;the weight w 1 gives the nicest curve of the three.Case 1: (boundary does not enter the control triangleB0B1B2). Here, the control triangle is entirely on one side ofthe boundary (see Fig. 6). Since all the curves (1) areentirely inside the triangle B0B1B2, none of them intersectthe boundary. The quadratic Be zier segment, curve (1) withw 1; will be used for part of the final curve.ACase 2: (boundary enters the control triangle B0B1B2bycrossing line segment B0B2). Since the Be zier controlpolyline does not intersect the boundary, the only way theBe zier curve segment can intersect the boundary is if theboundary cuts line segment B0B2(see Fig. 7). For allendpoints of straight-line boundary segments inside theFig. 5. Quadratic B-splines with various weights.Fig. 6. Case 1: boundary does not enter the Be zier control triangle B0B1B2.D.S. Meek et al. / Computer-Aided Design 35 (2003) 591599595control triangle B0B1B2, find curves (1) that pass throughthose endpoints using Result 2. The curve (1) with thehighest w-value will avoid all the straight-line boundarysegments. The curves (1) that are tangent to line segmentsare not considered here because the curves (1) through theend points of any line segment have higher w-values than acurve (1) that is tangent, and later the curve (1) with thelargest w-value is used.For all circular arc boundary segments inside the controltriangle B0B1B2, find curves (1) that pass through theendpoints of those boundary segments and the curves (1)that are tangent to the boundary segments using Results 2and 5 (see Fig. 7). Again, the curve (1) with the highest w-value will avoid all the boundary segments.AAlgorithm. While the quadratic B-spline is beingdesigned with a B-spline control polyline, generate thecontrol points of the corresponding quadratic Be zier curves.For each of the Be zier curves (1), check all boundarysegments that are partly or entirely inside the Be zier controltriangle to determine the w-value of the curve (1) that avoidsall those boundary segments. Let wLbe the largest of all thew-values of the curves thus produced, then w max1;wLis the w-value of a curve (1) that does not intersect any of theboundary segments. Here the preference is to use w 1 ifthe boundary segments allow that value.In Fig. 8, the curve (1) derived from B-spline controlpoints P1, P2, and P3uses a w-value higher than 1 to avoidcutting a circular part of the boundary by being tangent to it.The curve (1) derived from B-spline control points P3, P4,and P5uses a w-value higher than 1 to avoid cutting theboundary by passing through a boundary corner. The curve(1) with w 1 in these two cases is shown in gray. In all theother segments, w 1 gives a curve that does not cross theboundary. The curve that avoids the boundary is shown inblack.Theorem 1. If the guiding B-spline control polyline does notintersect the boundary, then the G1continuous curveproduced by the earlier algorithm does not intersect theboundary.Proof. The theorem follows from the fact that byincreasing the w-parameter sufficiently the curve (1)moves as close to the Be zier control polyline B0B1B2asdesired. Since the Be zier control polyline does notintersect the boundary, w-values can be found so that nosegment of the whole G1continuous curve intersects theboundary. Lemmas 1 and 3 state that a boundary avoidingcurve (1) can be found with both straight line and circulararc boundary segments.ASpecial behavior. There are several special effects thatcan be achieved by the placement of the B-spline controlpoints.To force the curve to be tangent to a straight line segmentof the boundary, put two consecutive B-spline control pointson the same straight line segment of the boundary. This willproduce a curve that is tangent at the midpoint of the twoFig. 8. A curve constrained by a closed boundary.Fig. 7. Case 2: the boundary crosses the line segment B0B2.D.S. Meek et al. / Computer-Aided Design 35 (2003) 591599596consecutive control points. In Fig. 9, the B-spline controlpoints P1and P2are on the boundary and the curve istangent to the boundary at P1 P2=2: To force the curve tobe tangent to the boundary at a given point, arrange twoconsecutive B-spline control points on the boundary so thatthe given point is midway between them.To run the curve along a straight line segment of theboundary, put more than two consecutive control pointsalong the same line segment of the boundary. The quadraticB-spline will yield some quadratic Be zier segments that arestraight lines while maintaining G1continuity. In Fig. 9, theB-spline control points P4, P5, and P6are on the boundary.There is a straight line segment in the curve from P4P5=2 to P5 P6=2:To force a cusp at a boundary point, put a double B-spline control point on the boundary. Of course, the curvewill not be G1continuous at the cusp. In Fig. 9, B-splinecontrol points P8and P9are the same and there is a cusp atthat point.4. Extensionallow the control polyline to cross theboundaryIt is possible to have some B-spline control pointsoutside of the boundary and still produce a curve that doesnot cross the boundary. The Be zier points B0and B2mustremain inside the boundary while B1can be outside. Inthis situation, part of the boundary can enter the controltriangle B0B1B2by crossing from line segment B0B1toline segment B1B2. The segments of the boundary in sucha part give upper bounds on the w-parameter; let wHbethe smallest of them (see Fig. 10). At the same time, thepart of the boundary can enter the control triangle B0B1B2by crossing in and out over the line segment B0B2asmentioned in Section 3. Segments of such a part put lowerbounds on the w-parameter; let wLbe the largest of them.If wH, wL, it is impossible to satisfy both restrictions, sothere is no curve (1) that matches the G1Hermite data andthat does not cross the boundary. Otherwise, there arethree cases for the choice of the w-parameter. If1 , wL# wH, choose wL; if wL# 1 # wH, choose 1;and if wL# wH, 1, choose wH. In brief, the allowableweight closest to 1 is chosen.An example showing a curve contained entirely inside aboundary appears in Fig. 11. Note that many of the B-splinecontrol points are outside the boundary.Fig. 9. Special featurestangent to boundary, run along boundary, cusp onboundary.Fig. 10. Extensionallow control points outside of the boundary.Fig. 11. Extensionsome control points outside of the boundary.D.S. Meek et al. / Computer-Aided Design 35 (2003) 5915995975. ConclusionsA method for drawing a G1planar curve that avoids aconstraint boundary consisting of straight line segments andcircular arcs has been presented. The curve is guided bycontrol points and consists of parametric quadratics andparametric rational quadratics. A quadratic equation mustbe solved for straight line boundary segments, while a fourthdegree equation must be solved for circular arc boundarysegments.AcknowledgmentsThe authors acknowledge the financial support of theNatural Sciences and Engineering Research Council ofCanada for this research. The authors appreciate the usefulcomments from two anonymous referees; their advicehelped improve the presentation of this paper.Appendix AThe condition that a fourth degree polynomial has a zeroof higher multiplicity than one is developed here. Thecondition is derived from the fact that for multiple zeros,the greatest common divisor of the polynomial and itsderivative is of degree higher than 0. The general case isfollowed by some special cases.General case: Let the fourth degree polynomial be r0tand let its derivative be 4r1(t), wherer0t at4 4bt3 6ct2 4dt er1t at3 3bt2 3ct d;and assume a 0. If r0t is divided by r1t; the result willhave fractions with the variable a as a denominator. Toavoid those fractions, start with a r0t to getar0t q2tr1t r2t;where the quotient q2t and remainder r2t areq2t at br2t 23At2 3ad 2 bct ae 2 bd;andA b22 ac:Assume A 0 (see later in the Appendix for case A 0). Ifr1t is divided by r2t; the result will have fractions withdenominators 3A2. To avoid those fractions, start with3A2r1t to get3A2r1t q3t2r2t ar3t;where the quotient q3t and multiple of the remainder r3tareq3t aAt 3bA aad 2 bc?r3t Bt C;andB ae 8bd 2 9c2A 3ad 2 bc2;C 3be 2 cdA ad 2 bcae 2 bd:Assume B 0 (see later in the Appendix for caseB 0). If r2t is divided by r3t; the result will havefractionswithdenominatorsB2.Toavoidthosefractions, start with B2r2t to get2B2r2t q4tr3t r4t;where the quotient q4t and remainder r4t areq4t 3ABt 2 3AC ad 2 bcB?r4t 2A2D;and whereD 227b2e2 27ace22 64bd3 36c2d22 54c3e2 6ad2e108bcdeAa227ad4 44bcd3218c3d2 a2e3 9ac2e22 12abde2 27c4e 48acd2e2 72bc2de:If D 0, then r0t does not have a double zero; ifD 0, r0t has a double zero at t 2C=B; the zeroof r3t:ACase A 0. If A 0 and ad 2 bc 0; r2t is linear.The remainder after dividing r1t by r2t is a constant. Ifthis constant is non-zero, then r0t does not have a multiplezero; if this constant is zero, then r0t has a double zero att 2ae 2 bd=3ad 2 bc?:If A 0 and ad 2 bc 0; then r2t is the constant ae 2bd: If this constant is non-zero, then r0t does not have amultiple zero; if this constant is zero, then r0t has a zero
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