All s Well That Ends Well: Supplementary Proofs

Transcription

1 All s Well That Ens Well: Guarantee Resolution of Simultaneous Rigi Boy Impact 1:1 All s Well That Ens Well: Supplementary Proofs This ocument complements the paper All s Well That Ens Well: Guarantee Resolution of Simultaneous Rigi Boy Impact an provies etaile proofs of several claims therein: that pairwise Gauss-Seiel-like algorithms an Generalize Reflections, when moifie accoring to the template shown in Algorithm 3, satisfy all the inexact impact operator esierata, an hence are guarantee to terminate, just as are their exact arithmetic counterparts A DETAILED INEXAT ARITHETI PROOFS Here we will prove the claims in 7: that both the inexact pairwise Gauss-Seiel metho escribe in Algorithm 3, as well as the Smith et al s Generalize Reflections algorithm 01], satisfy the inexact impact operator axioms ϵnor ϵod We will assume the following computation moel: real numbers are approximate using floating-point arithmetic, with machine epsilon ε < 1 an minimum representable magnitue η < ε We assume that no intermeiate calculation overflows; we then have an associate rouning operator flx], so that every exact quantity x, x x ε η flx] x + x ε + η For calculations we will make use of the weaker, more convenient boun x x ε ε flx] x + x ε + ε Arithmetic operations an square roots are assume to take place in infinite precision, an then roune; we will write fle] to enote that every operation in the expression E is perme in this way, eg flx + y] = flflx] + fly]] Finally, we will assume that q i an small integer constants are represente exactly, but that, 1, an N must be roune If ε is too large, the properties ϵnor, ϵdrift, an ϵod cannot be guarantee We will prove that both pairwise Gauss-Seiel an Generalize Reflections satisfy these properties ε sufficiently small, an give a constructive boun ε in terms of the magnitues of input quantities like q 0,, N, etc For both algorithms, we will first look at rift, an construct a which is use in the efinition of ϵdrift as a certificate that energy cannot grow unboune over the course of several iterations The proof of no rift will alreay impose a boun on ε; intuitively, if the machine precision is too large, the renormalization of the velocity after every iteration in Algorithms 3 an 4 itself introuces so much error into the computation of q i+1 that espite the renomalization, its magnitue cannot be boune Once we have constructe a, we also nee an ϵ We will show that ϵnor imposes a lower boun of ϵ, an that this lower boun ecreases to zero as ε ecreases We en by proving ϵod hol, provie that ϵ is not too large The upper boun is constant, an the lower boun shrinks as ε shrinks, so that it is always possible to fin an ϵ if ε is sufficiently small A1 Pairwise Gauss-Seiel In this section, we erive an ϵ an which the moifie pairwise GS algorithm escribe in section 7 satisfies the six criteria ϵnor ϵod Three of these, ϵkin, ONE an ϵvio, are obvious from the construction of the algorithm We first prove ϵdrift by inuction on the iteration i: suppose it hols the first i iterations of Algorithm 3 Then 1 q i 1 q 0 + q i q 0 + q i λ max q 0 + q i α 1 + β 1, α 1 = β 1 = λ max q 0 an λ max are the minimum an maximum eigenvalue of, respectively Since q i q i we also have that q i α 1 + β 1 We now boun p = flq i q i, n 1 n], n is some constraint graient selecte by Algorithm 3 The following fact will be useful: a sequence of numbers x 1,, x, it can be shown by inuction on that fl x i flx i ] j=1 j=1 + flx i ] ε1 + ε 1 j=1 We now procee to boun p First, flq j i nj ] q j i nj q j i fln j ] + 1 ε n j enotes the jth coorinates of the vector n We can write these bouns as flq j i nj ] q j i nj ε α + β α = α 1 n 1 + ε + ε + 1 β = β 1 n 1 + ε + ε Since q j i nj qi n n α 1 + β 1, summing over j gives fl q i, n ] q i, n ε α 3 + β 3 α 3 = 1 + n α 1 + εα 1 + ε 1 β 3 = n β 1 + εβ 1 + ε 1 A Transactions on Graphics, Vol 36, No 4, Article 1 Publication ate: July 017

2 1: E Vouga et al Switching gears, fl 1 k j ] an so that 1 k j 1 ε + ε fl n j ] n j n + 1ε, fl 1 ] k j n j 1 k j n j εα 4 α 4 = 7 1 n n + 3 Summing again over j we can boun ] k k fl 1 n n 1 εα 5 α 5 = Now since we have that q fl i, n n + εα ε 1 fl 1 n k ] 1 n + εα 5 ] j j 1 n q i, n n 1 εα 6 + β 6 α 6 = α 3 1 n + α 5 n α 1 + α 3 α 5 β 6 = β 3 1 n + α 5 n β 1 + α 5 β 3, we have mae liberal use of the fact that ε < ε to simplify the above expressions Then ] j j q fl i, n 1 n + q i, n n 1 εα 7 + β 7 α 7 = 1 + 4α n α 1 β 7 = 4β n β 1 Finally, we boun p in terms of p = q i q i, n 1 n We have that p j p j εα8 + β 8 4 α 8 = 1 + α 1 + α n α 1 β 8 = β 1 + β n β 1 Next, we nee to boun the norm fl p ] in the enominator of the coefficient of the velocity upate step We can use the fact that p j p p = q i q 0 + α 9 = α 8 + q 0 β 9 = + 3 β 8 + The summation mula then gives ] fl p k p k εα10 + β 10 α 10 = α 9 + β 10 = β 9 + q εα ε 1 + εβ ε 1 Next, combining the last several bouns, fl p j p j ] p j p j εα11 + β 11 + γ11 α 11 = 1 + q 0 + α 8 α 10 + α 10 + α 8 q 0 β 11 = q 0 + α 8 β 10 + α 10 β 8 + β 10 + β 8 q 0 + α 10 + α 8 γ 11 = + β 8 β 10 + β 10 + β 8 We apply the summation mula a secon time to get the square norm, ] fl p T p p T p ε α 1 + β 1 + γ1, α 1 = α q 0 + εα ε 1 β 1 = β εβ ε 1 γ 1 = γ 11 + εγ ε 1 We can rewrite this boun in more convenient m, by completing the square, in anticipation of taking the square root: ] fl p T p p T p β1 ε + γ 1 + ε α 1 β 1 γ 1 4γ 1 to get ] fl k j p j k j p j εα9 + β 9 Finally, we have a boun on the norm of p: fl p ] p εα 13 + β 13, 5 A Transactions on Graphics, Vol 36, No 4, Article 1 Publication ate: July 017

3 All s Well That Ens Well: Guarantee Resolution of Simultaneous Rigi Boy Impact 1:3 α 13 = 1 + q ε ε α 1 β 1 4γ 1 + β 1 γ 1 β 13 = 1 + ε ε γ1 α 0 = q 0 α 13 + α 19 + q 0 + q 0 β 0 = q 0 β 13 + β 19 + q 0 + Notice that since, γ 1 > 1 an so the enominators in α 13 are boune well away from zero The last piece we nee computing q i+1 is the norm of the initial velocity, q 0 To begin with, fl k j q j 0 ] k j q j 0 εα 14 α 14 = q 0 + q Since k j q j 0 q 0, applying the summation mula yiels fl q0 j ] q 0 j εα15 α 15 = α q 0 + εα ε 1 Then fl q j 0 q 0 j ] q j 0 q 0 j εα16 α 16 = q 0 + q j 0 α Applying the summation mula a secon time gives fl q0 ] q0 εα17 Finally with α 17 = α q 0 + εα ε 1 fl q 0 ] q 0 εα 18 6 α 18 = 1 + q ε ε α17 ombining equations 4 an 6 gives fl q0 p j ] q 0 p j εα19 + β 19 α 19 = 1 + q 0 + α 18 q 0 + q 0 α 8 + α 18 α 8 β 19 = q 0 + α 18 + q 0 β 8 + α 18 β 8 Now, we are at last prepare to boun the next velocity iterate q j q0 i+1 = fl p j ] p Suppose that p > εα 13 + β 13 Then by the previous boun, an equation 5, qj i+1 q 0 p j p ε α 0 + β 0 7 p εα 13 + β 13 Theree Let qi+1 q 0 α 0 + β 0 q 0 λ max ε p εα 13 + β 13 α 0 + β 0 + ελ max p εα 13 + β 13 1 α 1 = 4λ max α0 + q 0 4λ max α0 1 β 1 = 4λ max β0 + q 0 8λ max α 0 β 0 1 γ 1 = q 0 4λ max β0 Lemma A1 If ε < q 0 α 13, ε < γ 1, an εβ 1 + ε β1 + 4εα 1 εγ 1 q 0 εα 13 4 εγ εβ 13, then pairwise Gauss-Seiel satisfies ϵdrift Notice that these conitions are satisfie if ε is sufficiently small Proof Take = 1 q 0 εα εβ 13 εβ 1 + ε β1 + 4εα 1 εγ εγ 1 Since ε < q 0 α 13, q 0 1 εα 13 + β 13 q 0 an 1 p εα 13 + β 13 q 0, hence the boun in equation 7 is vali oreover we can substitute this inequality into the boun on q i+1 to get qi+1 q 0 εα1 + β 1 + γ1 Then q i+1 satisfies ϵdrift whenever εγ 1 εβ 1 εα 1 0, an in particular, whenever εβ 1 + ε β1 + 4εα 1 εγ 1 4 εγ 1 A Transactions on Graphics, Vol 36, No 4, Article 1 Publication ate: July 017

4 1:4 E Vouga et al We now prove the remaining properties, which are relatively straightwar First, we have that Lemma A Let be as in the previous lemma, an suppose ϵ > ε λ max α0 + β 0 q 0 q 0 + an α 0 + β 0 ϵ ελ max q 0 q 0 Then pairwise Gauss-Seiel satisfies ϵnor Notice that both righthan sies vanish as ε ecreases Proof Let be as in the previous lemma By construction of the algorithm an ϵvio we know that the value of λ is λ = q i, n > ϵ q i ϵ q 0 + the last inequality follows from ϵdrift From the boun 7 on the components of c, we have that c ελ max α 0 + β 0 q 0 an this is less than ϵλ when ελ max α 0 + β 0 q 0 ϵ q 0 + Lastly since q i q 0, we have that c ϵ q i whenever α 0 + β 0 ϵ ελ max q 0 q 0 Lemma A3 Pairwise Gauss-Seiel satisfies ϵod when ϵ < 1 Proof At every iteration a constraint with graient n is violate, A Generalize Reflections q i+1 q i = λ 1 n + c λ c 1 ϵ λ > 0 The generalize reflection operator of Smith et al 01] improves on pairwise Gauss-Seiel by guaranteeing preservation of symmetries an more accurately moeling shock propagations, at the cost of an R that is more expensive to compute Algorithm 4 shows how to moify it so that it satisfies all the inexact esierata require guarantee termination Notice that these moifications mirror those of Gauss-Seiel: constraints whose violation oes not excee a threshol are prune from consieration every time a reflection is applie, an the velocity is renormalize every step to prevent energy rift omputing λ at each iteration of Algorithm 4 requires solving a quaratic program QP Let λ be the exact solution to this QP, ξ the Algorithm 4 Inexact Generalize Reflections 1: function ResolveImpactsApproxq, q, ϵ : N ActiveonstraintGraientsq 3: q 0 q 4: i := 1, o 5: N V ViolateNq i // q T i N V < ϵ q i 1 6: if N V = then 7: return q i 8: en if 9: λ argmin λ 1 N V λ + q i st λ 0 q 10: q i+1 0 q i + 1 qi + N V λ 1 N V λ 11: en 1: en function corresponing positivity constraint Lagrange multipliers, an λ, ξ the compute solution We assume that λ approximately satisfies the KKT conitions of the QP, NV T 1 N V λ + N T V q i ξ ε κ 1 q i λ 0 ξ 0 λ ξ, κ 1 is an accuracy parameter inepenent of q i ; notice that this conition is a stanar relative error termination criterion in numerical QP coes The goal now will be to boun the intermeiate step p = fl q i + 1 N V λ] in terms of the true step p = q i + 1 N V λ; the proof of ϵdrift will then follow irectly from ientical calculations to that in pairwise Gauss-Seiel Once we have a value of, we will prove that inexact GR satisfies ϵnor an ϵod As in the case of Gauss-Seiel, ϵkin, ONE, an ϵvio all hol by construction of Algortihm 4 Let N A N V be the set of constraints that are active in the inexact QP solution, an λ A the corresponing parts of λ The matrix N T V 1 N V has ones along the iagonal, an off-iagonal entries of magnitue at most one; theree by the Gershgorin ircle Theorem its maximum eigenvalue is at most m, the number of total constraints in N Then we have the following useful boun on λ: λ = λ A NT A 1 N A λa m N T A 1 N A λa m m NT A q i + εκ 1 q i m q i + εκ 1 q i m λ max κ + µ, A Transactions on Graphics, Vol 36, No 4, Article 1 Publication ate: July 017

5 All s Well That Ens Well: Guarantee Resolution of Simultaneous Rigi Boy Impact 1:5 with κ = + εκ 1 q 0 λ max m µ = + εκ 1, λ max m as usual we have use ϵ < ϵ to simplify expressions Now n j the jth row of N V, we have the boun fl n k j λ ] k n k j λ k εκ3 + µ 3 κ 3 = 3 N κ + κ + 1 µ 3 = 3 N µ + µ, so applying the summation mula gives NV ] j fl λ j N V λ εκ4 + µ 4 Then κ 4 = κ m N κ 1 + ε 1 µ 4 = µ 3 + m N µ 1 + ε 1 fl 1 ] k j j NV λ εκ 5 + µ 5 1 k j NV λ j κ 5 = κ N κ µ 5 = µ N µ, so that applying the summation mula gives ] j j fl 1 N V λ 1 N V λ εκ 6 + µ 6 κ 6 = κ N κ + κ ε 1 µ 6 = µ 5 + µ N µ 1 + ε 1 Bee we can boun p, we nee to relate the impulse using the approximate multipliers λ to that using the exact multipliers We can o so by making use of the fact that the QP s KKT conitions are nearly satisfie λ: 1 N V λ 1 N V λv = λ λ T N T V 1 N V λ N T V 1 N V λ λ, ξ λ, ξ + λ λ κ1 ε q i λ λ κ1 ε q i λ + λ κ 1 ε q 0 + κ + µ κ1 ε q 0 + ε κ 7 + µ 7 + ν7 with κ 7 = q 0 κ κ 1 µ 7 = κ κ 1 + q 0 κ1 µ ν 7 = κ 1 µ ompleting the square gives 1 N V λ 1 N V λ 1 λ max 1 N V λ 1 N V λ εκ 8 + µ 8 Theree ] j fl 1 N V λ κ 8 = 1 µ 7 λ max + ν 7 κ 7 µ 7 4ν 7 ν7 µ 8 = λ max 1 N V λ j εκ 9 + µ 9 simply κ 9 = κ 6 + κ 8 an µ 9 = µ 6 + µ 8 We then have p j p j εκ10 + µ 10 κ 10 = κ N V κ + α µ 10 = µ N V µ + β 1 The proof of ϵdrift now follows ientically the arguments pairwise Gauss-Seiel, with κ 10 an µ 10 taking the place of α 8 an β 8 As in the pairwise GS case, construction of a certifying ϵdrift requires that ε be sufficiently small We now prove that GR satisfies the remaining properties, ϵnor an ϵod Lemma A4 Let be as in the proof of ϵdrift, an suppose that ϵ a + a + b, a = ϵ κ 1 q 0 + b = 4mελ max α 0 + β 0 q0, an α 0 + β 0 ϵ ελ max q 0 q 0 Then Generalize Reflections satisfies ϵnor Notice that both righthan sies vanish as ε ecreases an Proof Since at least one constraint must be violate, by ϵvio, N T V 1 N V λ ϵ q i ε κ 1 q i λ ϵ ε κ 1 q 0 +, m A Transactions on Graphics, Vol 36, No 4, Article 1 Publication ate: July 017

6 1:6 E Vouga et al we have use ϵdrift an again the fact that the largest eigenvalue of N T V 1 N V is at most m From the boun 7 on the components of c, we have that c ελ max α 0 + β 0 q 0 an this is less than ϵ λ 1 when α 0 + β 0 ελ max ϵ ϵ ε κ 1 q 0 + q 0 m Rearranging gives ϵ ϵ ε κ 1 q 0 + mελ max α 0 + β 0 q0 0 an the first inequality above Lastly since q i q 0, we have that c ϵ q i whenever α 0 + β 0 ϵ ελ max q 0 q 0, as in the case of pairwise Gauss-Seiel At last we en with Lemma A5 If ϵ < 4, then Generalize Reflections satisfies ϵod Proof At every iteration a constraint is violate, q i+1 q i = 1 N V λ + c 1 N v λ c λ 1 ϵ q i c λ 1 ϵ q i ϵ λ 1 q i / λ 1 ϵ q i ϵ λ 1 q i / The right-han sie is positive when ϵ < 4 A Transactions on Graphics, Vol 36, No 4, Article 1 Publication ate: July 017