Basic building blocks for a triple-double intermediate format

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1 Basic building blocks for a triple-double intermediate format corrected version) Christoph Quirin Lauter February 26, 2009 Abstract The implementation of correctly rounded elementary functions needs high intermediate accuracy before final rounding. This accuracy can be provided by pseudo-) expansions of size three, i.e. a triple-double format. The report presents all basic operators for such a format. Triple-double numbers can be redundant. A renormalization procedure is presented and proven. Elementary functions implementations need addition and multiplication sequences. These operators must take operands in double, double-double and triple-double format. The results must be accordingly in one of the formats. Several procedures are presented. Proofs are given for their accuracy bounds. Intermediate triple-double results must finally be correctly rounded to double precision. Two effective rounding sequences are presented, one for round-to-nearest mode, one for the directed rounding modes. Their complete proofs constitute half of the report. Introduction The implementation of correctly rounded double precision elementary functions needs high accuracy intermediate formats [7, 4,, 3]. To give an order of magnitude, in most cases 20 bits of intermediate accuracy are needed for assuring the correct rounding property [3]. In contrast, the native IEEE 754 double precision format offers 53 bits of accuracy. Well-know techniques allow to double this accuracy [5]. Nevertheless the resulting accuracy, 06 bits, is not sufficient. Tripling it would be enough. Using expansions of floating point numbers allows to expand still more the accuracy of a native floating point type. An expansion is a non-evaluated sum of some floating point numbers in a floating point format [6, 9, 0]. However the techniques for manipulating general expansions presented in literature are too costly for the implementation of elementary functions. In this report, we are going to consider expansions of three double precision floating point numbers, i.e. we are going to investigate on a triple double format. In our case, we will hence manipulate floating point numbers x h + x m + x l ) with x h, x m, x l F. After making some definitions and an analysis of a procedure that we call renormalization, we will consider addition and multiplication procedures for triple-double numbers. These will also comprise operations linking the triple double format with the native double and a double-double format. In the end, we will present and prove in particular two final rounding sequences for correctly rounding a triple-double number to the base double format.

2 2 Definitions It will be necessary to define a normal form of a triple-double number because it is clear that the triple-double floating point format - or that of pseudo-)expansions in general - is redundant: there exist numbers ˆx such that there are different triplets x h, x m, x l ) F 3 such that ˆx = x h + x m + x l. It is easy to see that a representation in such a format is unique if the numbers forming the expansion are ordered by decreasing 0 and if the latter are such that there is no binary) digit represented by two bits of the significands of two different numbers of the expansion. The sign bit has in this case the same role as an additional bit of the significant [6]. As we will see, the need for a normal form for each triple-double number is not directly motivated by needs of the addition and multiplication operators we will have to define. As long as the latter operate correctly, i.e. between known error bounds, on numbers whose representation in a triple-double format is not unique, we are not obliged to recompute normal forms. On the other hand, when we want to round down such a higher precision number to a native double in one of the different rounding modes), a normal form will be needed. If we could not provide one, we would assist to a explosion of different cases to be handled by the rounding sequence. Let us first define some notations that are needed for the analysis of numerical algorithms like the ones that we are going to consider. Definition 2. Predecessor and successor of a floating point number) Let be x F a floating point number. Let be < the total ordering on F. If x is positive or zero we will design by x + the direct successor of x in F with regard to < and we will notate x its predecessor. If x is negative we will design by x the successor and by x + the predecessor of x. In any case, we will design by succ x) the successor of x in F with regard to < and pred x) its predecessor. This definition 2. is inspired by [4]. Definition 2.2 Unit on the last place the ulp function) Let be x F a double precision number and let be x + its successor resp. negative). So x + x if x 0 and x + + ulp x) = 2 ulp ) x 2 if x + but x + = + ulp x) if x < 0 This definition is inspired by [4], too. ulp function. Concerning the overlap we define finally: Definition 2.3 Overlap) Let x h, x l F be two non-subnormal double precision numbers. We will say that x h and x l overlap iff predecessor if x is Compare [8] for further research on the subject of the x l ulp x h ) Let be x h, x m, x l F the components of a triple-double number. We will suppose that they are not subnormals. So, we will say that there is overlap iff x h and x m or x m and x l or x h and x l overlap. Definition 2.4 Normal form) Let be x h, x m, x l F three non-subnormal floating point numbers forming the triple-double number x h + x m + x l. We will say that x h + x m + x l are in normal form iff there is no overlap between its components. Further, we will say that x h + x m + x l is normalised. 2

3 Having made this definition, let us remark that it is deeply inspired by our direct needs and not by abstract analysises on how to represent real numbers. Anyway, the fact that an expansion is not overlapping is not a sufficient condition for its representing an unique floating point number. The implementation of addition and multiplication operators will be finally be based on computations on the components of the operands or partial products in the case of a multiplication) followed by a final summing up for regaining the triple-double base format. As we do not statically know the magnitudes of the triple-double operands and its components, we will not be able to guarantee that in any case there will not be any overlap in the result. On the other hand, we strive to develop a renormalization sequence for recomputing normal forms. These will be handy, as we already said, for the final rounding and, if needed, i.e. if sufficiently good static bounds can not be given, before handing over a result as an operand to a following operation. Such a renormalization sequence must guarantee that for each triple-double in argument if needed verifying some preconditions on an initial overlap), the triple-double number returned will be normalised and that the sum of the components for the first number is exactly the same as the one of the components of the latter. Definition 2.5 The Add2 algorithm) Let A be an algorithm taking as arguments two double precision numbers x, y F and returning two double precision floating point numbers r h, r l F. We will call A the Add2 algorithm iff it verifies that x, y F.r h + r l = x + y x, y F. r l 2 53 r h r h = x + y) r l = x + y r h i.e. iff it makes an exact addition of two floating point numbers such that the components of the double-double expansion in result are non-overlapping and if the most significant one is the floating point number nearest to the sum of the numbers in argument. Definition 2.6 The Mul2 algorithm) Let A be an algorithm taking as arguments two double precision numbers x, y F and returning two double precision floating point numbers r h, r l F. We will call A the Mul2 algorithm iff it verifies that x, y F.r h + r l = x y x, y F. r l 2 53 r h r h = x y) r l = x y r h i.e. iff it makes an exact multiplication of two floating point numbers such that the components of the double-double expansion in result are non-overlapping and if the most significant one is the floating point number nearest to the sum of the numbers in argument. We will pass over the existence proof of such algorithms. Consult [5] on this subject. Let us now give some main lemmas that can be deduced from the definition 2.2 of the ulp function. 3

4 Lemma 2.7 The ulp functions with regard to upper bounds) Let be x h and x l two non-subnormal floating point numbers. So x l < ulp x h ) x l 2 52 x h Proof Let us suppose that x l < ulp x h ) and that x l > 2 52 x h. So we get the following inequality 2 52 x h < ulp x h ) Without loss of generality, let us suppose now that x h > 0 and that x + h + where x+ h is the successor of x h in the ordered set of floating point numbers. So we know by the definition of non-subnormal floating point number in double precision that there exists m N and e Z such that x h = 2 e m with 2 52 m < Anyway, one can check that { x + h = 2e m + ) if m + < e otherwise So 2 cases must be treated separately: st case: x + h = 2e m + ) So we get 2 e m + ) 2 e m > e m > 2 52 m In contrast, m 2 52, so we obtain the strict inequality > which contradicts the hypotheses. 2nd case: x + h = 2e In this case, we know that m = We can deduce that 2 e e 2 53 ) > e 2 53 ) > This last inequality is a direct contradiction with the hypotheses. Lemma 2.8 Commutativity of the x + and x operators with unary ) Let be x F a positive floating point number. So, x) + = x +) and x) = x ) Proof Since the set of the floating point numbers is symmetrical around 0, we get x) + = pred x) = succ x) = x +) and x) = succ x) = pred x) = x ) 4

5 Lemma 2.9 x + and x for an integer power of 2) Let be x F a non-subnormal floating point number such that it exists e Z such that where p 2 is the format s precision. So, Proof If x > 0, we get and If x is negative it suffices to apply lemma 2.8. x = ±2 e 2 p x x = 2 x + x ) x x = 2 e 2 p 2 e 2 p+ ) = 2 e x + x = 2 e 2 p + ) 2 e 2 p = 2 e Lemma 2.0 x + and x for a float different from an integer power of 2) Let be x F a non-subnormal floating point number such that it does not exist any e Z such that x = ±2 e 2 p where p 2 is the format s precision. So, x x = x + x Proof If x > 0 we know that there exist e Z and m N such that with x = 2 e m 2 p < m < 2 p because x is not exactly an integer power of 2. Further, one checks that x + = 2 e m + ) even if m = 2 p and that because the lower bound given for m is strict. So one gets If x is negative it suffices to apply lemma 2.8. x = 2 e m ) x x = 2 e m 2 e m ) = 2 e = 2 e m + ) 2 e m = x + x Lemma 2. Factorized integer powers of 2 and the operators x + and x ) Let be x F a non-subnormal floating point number such that 2 x is still not subnormal. So, ) + 2 x = 2 x+ and ) 2 x = 2 x 5

6 Proof Without loss of generality let us suppose that x is positive. Otherwise we easily apply lemma 2.8. So, if x can be written x = 2 e m with m + 2 p+ where p is the precision then Otherwise, ) + 2 x = 2 e m ) + = 2 e m + ) = 2 x+ ) + 2 x = 2 e 2 p+ )) + = 2 e + 2 p = 2 x+ One can check that one obtains a completely analogous result for x. Lemma 2.2 Factor 3 of an integer power of 2 in argument of the x + operator) Let be x F a positive floating point number such that x is not subnormal, x + and 3 x) + are different from +, and that e Z. x = 2 e 2 p where p 3 is the precision of the format F. So the following equation holds 3 x) + + ulp x) = 3 x + Proof We can easily check the following 3 x) + + ulp x) = 3 x) + x + x ) = 3 2 e 2 p ) e 2 p ) + 2 e 2 p = 2 + ) 2 e 2 p ) e 2 p + ) 2 e 2 p = 2 2 e 2 p + 2 ) + 2 2e 2 p + 2 e = 2 e+ 2 p + 2 p )) e = 2 e+ 2 p + 2 p + ) + 2 e = 2 e+ 2 p + 2 p ) + 2 e+ + 2 e = 2 e+ 2 p + 2 p ) e = 3 2 e 2 p e = 3 2 e 2 p + ) = 3 2 e 2 p ) + = 3 x + Lemma 2.3 Monotony of the ulp function) The ulp function is monotonic for non-subnormal positive floating point numbers and it is monotonic for non-subnormal negative floating point numbers, i.e. x, y F. denorm < x y ulp x) ulp y) x, y F. x y < denorm ulp x) ulp y) where denorm is the greatest positive subnormal. Proof As a matter of fact it suffices to show that the ulp function is monotonic for non-subnormal floating point numbers and to apply its definition 2.2 for the negative case. Let us suppose so that we have two floating point numbers x, y F such that denorm < x < y. 6

7 Without loss of generality we suppose that x + + and that y + +. Otherwise we apply definition 2.2 of the ulp function and lemma 2.. So we get ulp y) ulp x) = y + y x + + x It suffices now to show that y + x + y + x 0 We can suppose that we would have x = 2 ex m x y = 2 ey m y with e x, e y Z, 2 p m x, m y < 2 p+. Since y > x, we clearly see that e y, m y ) lex e x, m x ) So four different cases are possible:. x + = 2 ex m x + ) y + = 2 ey m y + ) Hence y + x + y + x = 2 ey m y + ) 2 ex m x + ) 2 ey m y + 2 ex m x = 2 ey 2 ex 0 2. which yields to x + = 2 ex m x + ) y + = 2 ey+ 2 p y + x + y + x = 2 ey+ 2 p 2 ex m x + ) 2 ey 2 p+ ) + 2 ex m x = 2 ey 2 ex 0 3. x + = 2 ex+ 2 p y + = 2 ey m y + ) One checks that y + x + y + x = 2 ey m y + ) 2 ex+ 2 p 2 ey m y + 2 ex 2 p+ ) = 2 ey 2 ex 0 7

8 4. x + = 2 ex+ 2 p y + = 2 ey+ 2 p Thus y + x + y + x = 2 ey+ 2 p 2 ex+ 2 p 2 ey 2 p+ ) + 2 ex 2 p ) = 2 ey 2 ex 0 This finishes the proof. 3 A normal form and renormalization procedures Let us now analyse a renormalization algorithm. We will prove its correctness be a series of lemmas and a final theorem. Let be the following procedure: Algorithm 3. Renormalization) In: a h, a m, a l F verifying the following preconditions: Preconditions: None of the numbers a h, a m, a l is subnormal a h and a m do not overlap in more than 5 bits a m and a l do not overlap in more than 5 bits which means formally: Out: r h, r m, r l F a m 2 2 a h a l 2 2 a m a l 2 4 a h t h, t l ) Add2 a m, a l ) r h, t 2l ) Add2 a h, t h ) r m, r l ) Add2 t 2l, t l ) Consult also [6] on the subject of this algorithm. Let us give now some lemmas on the properties of the values returned by algorithm 3. and on the intermediate ones. Lemma 3.2 Exact sum) For each triple-double number a h +a m +a l, algorithm 3. returns a triple-double number r h +r m +r l such that a h + a m + a l = r h + r m + r l Proof This fact is a trivial consequence of the properties of the Add2 algorithm. 8

9 Lemma 3.3 Rounding of the middle component) For each triple-double number a h +a m +a l, algorithm 3. returns a triple-double number r h +r m +r l such that r m = r m + r l ) The same way, the intermediate and final value will verify the following properties: t h = a m + a l ) r h = a h + t h ) t l 2 53 t h t 2l 2 53 r h In particular, r m will not be equal to 0 if r l is not equal to 0. Proof This fact is a trivial consequence of the properties of the Add2 algorithm. Lemma 3.4 Upper bounds) For all arguments of algorithm 3., the intermediate and final values t h, t l, t 2l and r m can be bounded upwards as follows: Proof t h 2 a h t l 2 54 a h t 2l 2 52 a h t m 2 5 a h. Upper bound for t h : We have supposed that So we can check that a m 2 2 a h a l 2 4 a h t h a m + a l a h 2 2 a h a h a h 2 a h 2. Upper bound for t l : Using the properties of the Add2 algorithm we can get to know that t l 2 53 t h which yields finally to t l 2 54 a h 3. Upper bound for t 2l : t 2l 2 53 r h 2 53 a h + t h ) 2 53 a h t h a h t h 2 53 a h a h a h a h 2 52 a h 9

10 4. Upper bound for r m : r m t 2l t l t 2l + t l t 2l + t l 2 52 a h a h a h a h 2 5 a h Lemma 3.5 Special case for r h = 0) For all arguments verifying the preconditions of algorithm 3., r h will not be equal to 0 if r m is not equal to 0. Formally: r h = 0 r m = 0 Proof Let us suppose that r h = 0 and that r m 0. So we get because we have already shown that r h = a h + t h ) ) 2 a h = 2 a h t h 2 a h So for r h being equal to 0, a h must be equal to 0. In contrast, this yields to t h = 0 because r h = 0 + t h ) = t h This implies that t l = 0 because of the properties of the Add2 procedure. The same way, we get t 2l = 0. We can deduce from this that 0 r m = 0 + 0) = 0 which is a contradiction. Lemma 3.6 Lower bound for r h ) For all arguments verifying the preconditions of algorithm 3., the final result r h can be bounded in magnitude as follows: r h 2 a h Proof We have that r h = a h t h Clearly, if a h and t h have the same sign, we get r h a h 2 a h Otherwise - a h and t h are now of the opposed sign - we have already seen that t h 2 a h So, in this case, too, we get r h 2 a h 0

11 Corollary 3.7 Additional property on the Add2 procedure) Let be r h and r l two double precision floating point numbers returned by the Add2 procedure. So r l 2 ulp r h) Proof By the definition 2.5 of the Add2 procedure, we have r h + r l = x + y and r h = x + y) and r l = x + y r l. Thus the number r l = x + y) x + y) = x + y) x y) is the absolute error of correctly rounded IEEE 754 addition. It is bounded by 2 ulp r h) which gives us the desired result. Theorem 3.8 Correctness of the renormalization algorithm 3.) For all arguments verifying the preconditions of procedure 3., the values returned r h, r m and r l will not overlap unless they are all equal to 0 and their sum will be exactly the sum of the values in argument a h, a m and a l. Proof The fact that the sum of the values returned is exactly equal to the sum of the values in argument has already been proven by lemma 3.2. Without loss of generality, we will now suppose that neither r h nor r m will be 0 in which case all values returned would be equal to 0 as we have shown it by lemmas 3.5 and 3.3. Using lemma 3.3, we know already that r m and r l do not overlap. Let us show now that r h and r m do not overlap by proving that the following inequality is true There are two different cases to be treated. st case: t 2l = 0 We know that r m 3 4 ulp r h) < ulp r h ) r m = t 2l + t l ) = 0 + t l ) = t l When showing lemma 3.4, we have already proven that Using lemma 3.6, we therefore know that which is the result we wanted to prove. 2nd case: t 2l 0 Still using lemma 3.4, we have shown that t l 2 54 a h r m 2 53 r h t h 2 a h In consequence, when the IEEE 754 [2] addition r h = a h t h is ported out, the rounding will be done at a bit of weight heigher than one ulp t h ) because t 2l is strictly greater than 0 and because a h and t h do not completely overlap. Therefore we can check that t 2l ulp t h ) With the result of lemma 3.4 we already mentioned, we can deduce that t 2l 2 ulp t h) 2 t 2l

12 So one can verify the following upper bound using among others lemma 3.7: r m = t 2l + t l ) ) 3 2 t 2l ) 3 4 ulp r h) = 3 4 ulp r h) One remarks that the last simplification is correct here because ulp is always equal to an integer power of 2 and because the precision of a double is greater than 4 bits. 4 Operators on double-double numbers Since we dispose now of a renormalization procedure which is effective and proven, we can now consider the different addition and multiplication operators we need. They will surely work finally on expansions of size 3, but the double-double format [5] must be analysed, too, because it is at the base of the triple-double format. We already mentioned that on definition and analysis of this operators, we need not care such a lot of the overlap in the components of a triple-double number any more: as long as the overlap does not make us loose a too much of the final accuracy because several bits of the significand are represented twice, overlap is not of an issue for intermediate values. At the end of triple-double computations, it will be sufficient to apply once the renormalization procedure. In order to measure the consequences of an overlap in the operands on final accuracy and in order to be able to follow the increase of the overlap during computations in triple-double, we will indicate for each operator which produces a triple double result or which takes a triple-double operand not only a bound for relative and absolute rounding errors but also a bound for the maximal overlap of the values returned. All this bounds will be parameterised by a variable representing the maximal overlap of the triple-double arguments. 4. The addition operator Add22 Let us analyse first the following addition procedure: Algorithm 4. Add22) In: two double-double numbers, a h + a l and b h + b l Out: a double-double number r h + r l Preconditions on the arguments: a l 2 53 a h b l 2 53 b h Algorithm: 2

13 t a h b h if a h b h then else Compare [] concerning algorithm 4.. t 2 a h t t 3 t 2 b h t 4 t 3 b l t 5 t 4 a l t 2 b h t t 3 t 2 a h t 4 t 3 a l t 5 t 4 b l end if r h, r l ) Add2 t, t 5 ) Theorem 4.2 Relative error of algorithm 4. Add22 without occurring of cancellation) Let be a h + a l and b h + b l the double-double arguments of algorithm 4. Add22. If a h and b h have the same sign, so we know that r h + r l = a h + a l ) + b h + b l )) + ε) where ε is bounded as follows: ε 2 03,5 Proof Since the algorithm Add22 ends by a call to the Add2 procedure, it suffices to show that t + t 5 = a h + a l ) + b h + b l )) + ε) Further, since the two branches of the algorithm are symmetrical, we can suppose that a h b h and consider only one branch without loss of generality. Finally, we remark that the following lines of the Add22 procedure constitute a non-conditional Add2 with arguments a h and b h and the result t + t 3 : Thus, we get with t = a h b h t 2 = a h t t 3 = t 2 b h t 5 = t 4 a l = t 4 + a l ) + ε ) = t 3 + bl) + ε 2 ) + a l ) + ε ) = t 3 + a l + b l + δ δ = t 3 ε 2 + b l ε 2 + t 3 ε + b l ε + t 3 ε 2 ε 2 + b l ε 2 ε 2 + a l ε For giving an upper bound for δ, let us first give an upper bound for t 3, a l and b l as function of a h + b h using the following bounds that we know already: a l 2 53 a h b l 2 53 b h t t 3

14 We get therefore and than t t = 2 53 a h b h 2 53 a h + b h a h + b l a l 2 53 a h 2 53 a h + b h The last bound is verified because we suppose that a h and b h have the same sign. Finally, since a h b h, Thus we get for δ : b l 2 53 b h 2 53 a h 2 53 a h + b h δ a h + b h ) a h + b h ) Let us now give a lower bound for a h + a l + b h + b l as a function of a h + b h in order to be able to give a relative error bound for the procedure Add22. We have that So we can check that Concerning δ, this yields to δ a h + a l + b h + b l One easily checks that a l + b l a l + b l 2 53 a h b h 2 52 a h 2 52 a h + b h a h + a l + b h + b l 2 52) a h + b h ) ) 2 03,5 from which one trivially deduces the affirmation. Theorem 4.3 Relative error of algorithm 4. Add22 with a bounded cancellation) Let be a h + a l and b h + b l the double-double arguments of 4. Add22. If a h and b h are of different sign and if one can check that for µ so the returned result will verify where ε is bounded as follows: b h 2 µ a h r h + r l = a h + a l ) + b h + b l )) + ε) ε µ 2 µ

15 Proof Let us reuse the results obtained at the proof 4. and let us start by giving an upper bound for b l as a function of a h : b l 2 53 b h 2 53 µ a h Let us now continue with a lower bound for a h + b h still as a function of a h : We get in consequence and Thus we can check that a h + b h a h 2 µ) a l µ a h + b h b l 2 53 µ 2 µ a h + b h δ a h + b h µ 2 µ Once again, we must give a lower bound for a h + a l + b h + b l with regard to a h + b h : We know that So Thus we get for δ a l + b l a l + b l 2 52 a h µ a h + b h a h + a l + b h + b l a h + b h 2 µ µ δ 2 µ a h + a l + b h + b l 2 µ µ 2 µ = a h + a l + b h + b l µ 2 µ 2 52 So finally the following inequality is verified for the relative error ε: ε µ 2 µ 2 52 We can still give a less exact upper bound for this term by one that does not depend on µ because µ : 2 µ 3 2 µ so ε 2 02 Theorem 4.4 Absolute error of algorithm 4. Add22 general case)) Let be a h + a l and b h + b l the double-double arguments of algorithm 4. Add22. The result r h + r l returned by the algorithm verifies where δ is bounded as follows: r h + r l = a h + a l ) + b h + b l ) + δ δ max 2 53 a l + b l, 2 02 a h + a l + b h + b l ) 5

16 Proof Without loss of generality, we can now suppose that 2 a h b h a h and that a h and b h have different signs because for all other cases, the properties we have to show are a direct consequence of theorems 4.2 and 4.3. So we have 2 a h b h 2 a h and So the floating point operation 2 b h a h 2 b h t = a h b h is exact by Sterbenz lemma []. In consequence, t 3 will be equal to 0 because, as we have already mentioned, the operations computing t and t 3 out of a h and b h constitute a Add2 whose properties assure that We can deduce that t 3 = a h + b h t t 4 = t 3 b l = b l and can finally check that with So we get with t 5 = t 4 a l = t 4 + a l ) + ε ) ε 2 53 r h + r l = t + t 5 = a h + a l + b h + b l ) + δ δ 2 53 a l + b l which yields to the bound to be proven δ = max 2 53 a l + b l, 2 02 a h + a l + b h + b l ) Theorem 4.5 Output overlap of algorithm 4. Add22) Let be a h + a l and b h + b l the double-double arguments of algorithm 4. Add22. So the values r h and r l returned by the algorithm will not overlap at all and will verify r l 2 53 r h Proof The proof of the affirmed property is trivial because the procedure Add22 sequence Add2 which assures it. ends by a call to 6

17 4.2 The multiplication operator Mul22 Let us now consider the multiplication operator Mul22: Algorithm 4.6 Mul22) In: two double-double numbers, a h + a l and b h + b l Out: a double-double number r h + r l Preconditions on the arguments: Algorithm: a l 2 53 a h b l 2 53 b h t, t 2 ) Mul2 a h, b h ) t 3 a h b l t 4 a l b h t 5 t 3 t 4 t 6 t 2 t 5 r h, r l ) Add2 t, t 6 ) Compare also to [] concerning algorithm 4.6. Theorem 4.7 Relative error of algorithm 4.6 Mul22) Let be a h + a l and b h + b l the double-double arguments of algorithm 4.6 Mul22. So the values returned r h and r l verify r h + r l = a h + a l ) b h + b l )) + ε) where ε is bounded as follows: ε 2 02 Further r h and r l will not overlap at all and verify r l 2 53 r h Proof Since algorithm 4.6 ends by a call to the Add2 procedure, the properties of the latter yield to the fact that r h and r l do not overlap at all and that r l 2 53 r h. In order to give upper bounds for the relative and absolute error of the algorithm, let us express t 6 as a function of t 2, a h, a l, b h and b l joined by the error term δ. We get t 6 = t 2 a h b l a l b h ) = t 2 + a h b l + ε ) + a l b h + ε 2 )) + ε 3 )) + ε 4 ) where ε i 2 53, i =, 2, 3, 4. Simplifying this expression, we van verify that with t 6 = t 2 + a h b l + a l b h + δ δ a l b l + a h b l ε + a l b h ε 2 + a h b l ε 3 + a h b l ε ε 3 + a l b h ε 3 + a l b h ε ε 3 +a h b l ε 4 + a l b h ε 4 + a h b l ε ε 4 + a l b h ε 2 ε 4 + a h b l ε 3 ε 4 +a h b l ε ε 3 ε 4 + a l b h ε 3 ε 4 + a l b h ε 2 ε 3 ε 4 a h b h ) a h b h

18 For checking the given bound, we have supposed the following inequalities: and a l 2 53 a h a l 2 53 a h Let us now give a lower bound for a h + a l ) b h + b l ) as a function of a h b h. For doing so, we give an upper bound for a h b l + a l b h + a l b l. We verify since: a h b l + a l b h + a l b l a h b l + a l b h + a l b l 2 53 a h b h a h b h a h b h 2 5 a h b h This yields to a h + a l ) b h + b l ) a h b h 2 5) 2 a h b h from which we deduce that δ 2 02 a h + a l ) b h + b l ) which gives us as an bound for the relative error r h + r l = a h + a l ) b h + b l ) + ε) with ε Addition operators for triple-double numbers 5. The addition operator Add33 We are going to consider now the addition operator Add33. We will only analyse a simplified case where the arguments values verify some bounds statically known. Algorithm 5. Add33) In: two triple-double numbers, a h + a m + a l and b h + b m + b l Out: a triple-double number r h + r m + r l Preconditions on the arguments: b h 3 4 a h a m 2 αo a h a l 2 αu a m b m 2 βo b h b l 2 βu b m α o 4 α u β o 4 β u Algorithm: 8

19 r h, t ) Add2 a h, b h ) t 2, t 3 ) Add2 a m, b m ) t 7, t 4 ) Add2 t, t 2 ) t 6 a l b l t 5 t 3 t 4 t 8 t 5 t 6 r m, r l ) Add2 t 7, t 8 ) Theorem 5.2 Relative error of algorithm 5. Add33) Let be a h +a m +a l and b h +b m +b l the triple-double arguments of algorithm 5. Add33 verifying the given preconditions. So the following egality will hold for the returned values r h, r m and r l where ε is bounded by: r h + r m + r l = a h + a m + a l ) + b h + b m + b l )) + ε) ε 2 minαo+αu,βo+βu) minαo,βo) 98 The returned values r m and r l will not overlap at all and the overlap of r h and r m will be bounded by the following expression: r m 2 minαo,βo)+5 r h Proof The procedure 5. ends by a call to the Add2 sequence. One can trivially deduce that r m and r l do not overlap at all and verify r l 2 53 r m Further, it suffices that the bounds given at theorem 5.2 hold for t 7 and t 8 because the last addition computing r m and r l will be exact. The same way, one can deduce the following inequalities out of the properties of the Add2 procedure. They will become useful during this proof. t 2 53 r h t t 2 t t 7 Let us start the proof by giving bounds for the magnitude of r h with regard to a h : We have on the one hand and on the other r h = a h + b h ) = a h + b h ) a h + b h ) a h + 3 ) 4 a h 2 a h ) = 2 a h r h = a h + b h ) ) 4 a h = 4 a h 9

20 So we know that 4 a h r h 2 a h. It is now possible to give the following upper bounds for t, t 2, t 3, t 7, t 4, t 6 and t 6 : t 2 53 r h a h = 2 5 a h t 2 a m + b m ) a m + b m ) 2 αo a h + 2 βo b h ) 2 αo a h + 2 βo 3 ) 4 a h ) 2 minαo,βo)+ a h = 2 minαo,βo)+ a h t t minαo,βo)+ a h = 2 minαo,βo) 52 a h t 7 t + t 2 ) t + t 2 ) ) 2 5 a h + 2 minαo,βo)+ a h ) 2 minαo,βo)+2 a h = 2 minαo,βo)+2 a h t t minαo,βo)+2 a h = 2 minαo,βo) 5 a h and finally t 6 a l + b l ) 2 αo 2 αu a h + 2 βo 2 βu 3 ) 4 a h ) 2 minαo+αu,βo+βu)+ a h = 2 minαo+αu,βo+βu)+ a h t 5 t 3 + t 4 ) ) 2 minαo,βo) 52 a h + 2 minαo,βo) 5 a h ) 2 minαo,βo) 50 a h = 2 minαo,βo) 50 a h 20

21 Using the fact that the addition Add2 is exact, it is easy to show that we have exactly r h + t 7 + t 3 + t 4 = a h + a m + b h + b m Further, we can check t 8 = t 3 2 t 4 ) a l 3 b l ) = t 3 + t 4 + a l + b l + δ with δ t 3 ε 3 + t 4 ε 2 + a l ε 3 + b l ε 3 + t 3 ε + t 4 ε + t 3 ε ε 2 + t 4 ε ε 2 + a l ε + b l ε 2 + a l ε ε 3 + b l ε ε 3 where for i {, 2, 3}, ε i is the relative error bound of the floating point addition i and verifies So we get immediately ε i 2 53 r h + r m + r l = r h + t 7 + t 8 = a h + a m + a l ) + b h + b m + b l ) + δ Let us now express a h + a m + a l ) + b h + b m + b l ) as a function of a h : and, the same way round, which allows for noting a h + a m + a l a h + a m + a l a h + 2 αo a h + 2 αo αu a h 2 a h b h + b m + b l 2 b h 3 2 a h a h + a m + a l ) + b h + b m + b l ) 2 2 a h In order to give a lower bound for this term, let us prove an upper bound for b h + a m + b m + a l + b l as follows So we get b h + a m + b m + a l + b l b h + a m + b m + a l + b l 3 4 a h + 2 αo a h + 2 βo 3 4 a h + 2 αo αu a h +2 βo βu 3 4 a h 7 8 a h a h + a m + a l ) + b h + b m + b l ) 8 a h Using this bounds, we can give upper bounds for the absolute error δ first as a function of a h and than as a function of a h + a m + a l ) + b h + b m + b l ) for deducing finally a bound for the relative error. So we get δ t 3 ε 3 + t 4 ε 2 + a l ε 3 + b l ε 3 + t 3 ε + t 4 ε + t 3 ε ε 2 + t 4 ε ε 2 + a l ε + b l ε 2 + a l ε ε 3 + b l ε ε 3 a h ε 2

22 with This yields to with ε minαo,βo) minαo,βo) αo αu βo βu minαo,βo) minαo,βo) minαo,βo) minαo,βo) αo αu βo βu αo αu βo βu 2 minαo,βo) minαo+αu,βo+βu) 50 r h + r m + r l = a h + a m + a l ) + b h + b m + b l )) + ε) ε 2 minαo,βo) minαo+αu,βo+βu) 47 In order to finish the prove, it suffices now to give an upper bound for the maximal overlap between r h and r m because we have already shown that r m and r l do not overlap at all. So we can check r 8 t 5 + t 6 ) ) 2 minαo,βo) 50 a h + 2 minαo+αu,βo+βu)+ a h ) 2 minαo,βo) 48 r h + 2 minαo+αu,βo+βu)+3 r h and continue by giving the following upper bound r m = t 7 + t 8 ) t 7 + t 8 ) 2 minαo,βo)+4 r h + r h )) 2 minαo,βo) 48 r h + 2 minαo+αu,βo+βu)+3 r h 2 minαo,βu) minαo,βo) minα+αu,βo+βu) minαo,βo) minαo+αu,βo+βu) 50)) 2 minαo,βo)+5 r h This is the maximal overlap bound we were looking for; the proof is therefore finished. Theorem 5.3 Special case of algorithm 5. Add33) Let be a h + a m + a l and b h + b m + b l the triple-double arguments of algorithm 5. Add33 that a h = a m = a l = 0 So the values r h, r m and r l returned will be exactly equal to such r h + r m + r l = b h + b m + b l The values r m and r l will not overlap at all. The overlap of r h and r m must still be evaluated. 22

23 Proof We will suppose that the Add2 procedure is exact for a h = a m = a l = 0 if even we are using its unconditional version. Under this hypothesis, we get thus: r h = 0 + b h ) = b h t = 0 + b h b h = 0 t 2 = b m t 3 = 0 t 7 = 0 + b m ) = b m t 4 = 0 t 6 = 0 b l = b l t 5 = 0 0 = 0 t 8 = 0 b l = b l r m + r l = b m + b l In consequence, the following holds for the values returned: r h + r m + r l = b h + b m + b l Clearly r m and r l do not overlap because the Add2 line assures this property. procedure the algorithm calls at its last 5.2 The addition operator Add233 Let us consider now the addition operator Add233. We will only analyse a simplified case where the arguments of the algorithm verify statically known bounds. Algorithm 5.4 Add233) In: a double-double number a h + a l and a triple-double number b h + b m + b l Out: a triple-double number r h + r m + r l Preconditions on the arguments: Algorithm: b h 2 2 a h a l 2 53 a h b m 2 βo b h b l 2 βu b m r h, t ) Add2 a h, b h ) t 2, t 3 ) Add2 a l, b m ) t 4, t 5 ) Add2 t, t 2 ) t 6 t 3 b l t 7 t 6 t 5 r m, r l ) Add2 t 4, t 7 ) Theorem 5.5 Relative error of algorithm 5.4 Add233) Let be a h + a l and b h + b m + b l the values taken in argument of algorithm 5.4 Add233. Let the preconditions hold for this values. So the following holds for the values returned by the algorithm r h, r m and r l r h + r m + r l = a h + a m + a l ) + b h + b m + b l )) + ε) 23

24 where ε is bounded by ε 2 βo βu βo The values r m and r l will not overlap at all and the overlap of r h and r m will be bounded by: r m 2 γ r h with γ min 45, β o 4, β o + β u 2) Proof We know using the properties of the Add2 procedure that r h + t = a h + b h t 2 + t 3 = a l + b m t 4 + t 5 = t + t 2 r m + r l = t 4 + t 7 Supposing that we dispose already of a term of the following form t 7 = t 5 + t 3 + b l + δ with a bounded δ, we can note that r h + r m + r l = a h + a l ) + b h + b m + b l ) + δ Let us now express t 7 by t 5, t 3 and b l : with ε 2 53 and ε We get in consequence t 7 = t 5 t 6 = t 5 t 3 b l ) = t 5 + t 3 + b l ) + ε )) + ε 2 ) t 7 = t 5 + t 3 + b l + t 3 ε + b l ε + t 5 ε 2 + t 3 ε 2 + b l ε 2 + t 3 ε ε 2 + b l ε ε 2 and we can verify that the following upper bound holds for the absolute error δ: δ = t 3 ε + b l ε + t 5 ε 2 + t 3 ε 2 + b l ε 2 + t 3 ε ε 2 + b l ε ε t b l t b l t b l 2 52 t b l t 5 Let us get now some bounds for t 3, b l and t 5, all as a function of a h : b l 2 βo βu b h 2 βo βu 2 a h which can be obtained using the preconditions hypotheses. Further t t 2 = 2 53 a l + b m ) 2 52 a l + b m 2 52 a l b m 2 05 a h + 2 βo 52 b h 2 05 a h + 2 βo 54 a h 24

25 and finally t t 4 = 2 53 t + t 2 ) 2 52 t + t t t r h a l + b m ) 2 05 a h + b h ) a l + b m 2 04 a h + b h a l b m 2 04 a h a h a h + 2 βo 53 a h a h βo 53) So we have δ a h βo βo βu βo 06) a h 2 βo βu βo ) Let us now give a lower bound for a h + a l ) + b h + b m + b l ) as a function of a h by getting out an upper bound for a l + b h + b m + b l as such a function: a l + b h + b m + b l a l + b h + b m + b l a h βo βo βu 2) Since β o, β u we can check that a l + b h + b m + b l 2 a h In consequence a h + a l + b h + b m + b l ) 2 a h Using this lower bound, we can finally give an upper bound for the relative error ε of the considered procedure: r h + r m + r l = a h + a l ) + b h + b m + b l )) + ε) with ε 2 βo βu βo Last but not least, let us now analyse the additional overlaps generated by the procedure. It is clear that r m and r l do not overlap at all because they are computed by the Add2 procedure. Let us merely examine now the overlap of r h and r m. We begin by giving a lower bound for r h as a function of a h : r h = a h + b h ) a h + b h ) ) 3 4 a h ) 2 a h = 2 a h Let us then find an upper bound for r m using also here a term which is a function of a h : r m = r m + r l ) 25

26 2 r m + r l = 2 t 4 + t 7 2 t t 5 + t 3 + b l + δ 2 t t t b l + 2 δ 2 t 4 + a h βo 52) + a h βo 53) + a h 2 βo βu + a h 2 βo βu βo ) By bounding finally still t 4 by a term that is function of a h we obtain We finally check that we have t 4 = t + t 2 ) 2 t + 2 t r h + 2 a l + b m ) 2 5 a h + b h + 4 a l + b m a h 2 βo ) 2 48 r h a h +2 βo βo βo βo βu +2 βo βu βo ) from which we can deduce the following bound a h βo βo βu) r m r h βo βo βu+) r m 2 γ r h with This finishes the proof. γ min 45, β o 4, β o + β u 2) 6 Triple-double multiplication operators 6. The multiplication procedure Mul23 Let us go on with an analysis of the multiplication procedure Mul23. Algorithm 6. Mul23) In: two double-double numbers a h + a l and b h + b l 26

27 Out: a triple-double number r h + r m + r l Preconditions on the arguments: a l 2 53 a h b l 2 53 b h Algorithm: r h, t ) Mul2 a h, b h ) t 2, t 3 ) Mul2 a h, b l ) t 4, t 5 ) Mul2 a l, b h ) t 6 a l b l t 7, t 8 ) Add22 t 2, t 3, t 4, t 5 ) t 9, t 0 ) Add2 t, t 6 ) r m, r l ) Add22 t 7, t 8, t 9, t 0 ) Theorem 6.2 Relative error of algorithm 6. Mul23) Let be a h + a l and b h + b l the values taken by arguments of algorithm 6. Mul23 So the following holds for the values returned r h, r m and r l : where ε is bounded as follows: r h + r m + r l = a h + a l ) b h + b l )) + ε) ε 2 49 The values returned r m and r l will not overlap at all and the overlap of r h and r m will be bounded as follows: r m 2 48 r h Proof Since algorithm 6. is relatively long, we will proceed by analysing sub-sequences. consider first the following sequence: t 2, t 3 ) Mul2 a h, b l ) t 4, t 5 ) Mul2 a l, b h ) t 7, t 8 ) Add22 t 2, t 3, t 4, t 5 ) So let us Clearly t 7 and t 8 will not overlap. The same way t 2 and t 3 and t 4 and t 5 will not overlap and we know that we have exactly the following egalities Further we can check that t 2 + t 3 = a h b l t 4 + t 5 = b h a l t a h b l ) 2 52 a h b l and similarly So we have on the one side t b h a l 2 53 t 3 + t t t a h b l b h a l 27

28 and on the other 2 02 t 2 + t 3 + t 4 + t b h a l + a h b l Using theorem 4.4 it is possible to note with 2 02 b h a l a h b l t 7 + t 8 = a h b l + a l b h + δ δ 2 02 a h b l a l b h Let us now consider the following sub-sequence of algorithm 6.: r h, t ) Mul2 a h, b h ) t 6 a l b l t 9, t 0 ) Add2 t, t 6 ) Trivially t 9 and t 0 do not overlap. Additionally, one sees that we have exactly r h + t = a h b h and, exactly too, So using where ε 2 53 we get t 9 + t 0 = t + t 6 t 6 = a l b l = a l b l + ε) with r h + t 9 + t 0 = r h + t + t 6 Let us now bound t 9 with regard to a h b h : We have t 9 t + t 6 ) = a h b h + t 6 = a h b h + a l b l + δ 2 δ a l b l t + a l b l )) t a h b h )) 2 53 a h b h a h b h ) 2 5 a h b h With inequalities given, we can bound now the absolute and relative error of algorithm Mul We know already that t 7 + t 8 = a h b l + a l b h + δ where and where δ 2 02 a h b l b h a l r h + t 9 + t 0 = a h b h + a l b l + δ 2 δ a l b l 28

29 One remarks that t t 7 t t 9 and easily checks that and that 2 53 t 0 + t t t t 9 + t 0 + t 7 + t t t 7 So by means of the theorem 4.4, we obtain that r m + r l = t 9 + t 0 + t 7 + t 8 + δ 3 where So finally we get δ t t 7 r h + r m + r l = a h b h + a h b l + a l b h + a l b l + δ where δ = δ + δ 2 + δ 3 And for t 7 we obtain the following inequalities t 7 t 7 + t 8 ) δ + δ 2 + δ a l b h a h b l a l b l a h b h +2 0 t 7 2 t 7 + t 8 2 a h b l + a l b h a l b h a h b l ) 8 a h b l In consequence we can check δ 2 50 a h b h Let us give now an upper bound for a h b l + a l b h + a l b l as a function of a h b h : We have from which we deduce that a h b l + a l b h + a l b l a h b l + a l b h + a l b l 2 5 a h b h a h b h + a h b l + a l b h + a l b l ) a h b h 2 5) 2 a h b h Thus δ 2 49 a h b h + a h b l + a l b h + a l b l 29

30 So we can finally give an upper bound for the relative error ε of the multiplication procedure Mul23 defined by algorithm 6.: r h + r m + r l = a h + a l ) b h + b l ) + ε) with ε 2 49 Before concluding, we must still analyse the overlap of the different components of the tripledouble number returned by the algorithm. It is clear that r m and r l do not overlap because the Add22 brick ensures this. Let us now consider the magnitude of r m with regard to the one of r h. We give first a lower bound for r h : and then an upper bound for r m : r m r m + r l ) t 7 + t 8 + t 9 + t 0 + δ 3 ) r h = a h b h ) a h b h ) 2 a h b h a h b l + a l b h + δ + t 9 + t 0 + δ 3 ) a h b h )) 2 49 a h b h From this we can deduce the final bound r m 2 48 r h 6.2 The multiplication procedure Mul233 Let us concentrate now on the multiplication sequence Mul233. Algorithm 6.3 Mul233) In: a double-double number a h + a l and a triple-double number b h + b m + b l Out: a triple-double number r h + r m + r l Preconditions on the arguments: with Algorithm: a l 2 53 a h b m 2 βo b h b l 2 βu b m β o 2 β u 30

31 r h, t ) Mul2 a h, b h ) t 2, t 3 ) Mul2 a h, b m ) t 4, t 5 ) Mul2 a h, b l ) t 6, t 7 ) Mul2 a l, b h ) t 8, t 9 ) Mul2 a l, b m ) t 0 a l b l t, t 2 ) Add22 t 2, t 3, t 4, t 5 ) t 3, t 4 ) Add22 t 6, t 7, t 8, t 9 ) t 5, t 6 ) Add22 t, t 2, t 3, t 4 ) t 7, t 8 ) Add2 t, t 0 ) r m, r l ) Add22 t 7, t 8, t 5, t 6 ) Theorem 6.4 Relative error of algorithm 6.3 Mul233) Let be a h + a l and b h + b m + b l the values in argument of algorithm 6.3 Mul233 given preconditions hold. So the following will hold for the values r h, r m and r l returned such that the where ε is bounded as follows: r h + r m + r l = a h + a l ) b h + b m + b l )) + ε) ε 2 99 βo βo βu βo+ 2 βo βu βo βo βu The values r m and r l will not overlap at all and the following bound will be verified for the overlap of r h and r m : r m 2 γ r h where γ min 48, β o 4, β o + β u 4) Proof During this proof we will once again proceed by basic bricks that we will assemble in the end. Let us therefore start by the following one: Since we have the exact egalities and t 2, t 3 ) Mul2 a h, b m ) t 4, t 5 ) Mul2 a h, b l ) t, t 2 ) Add22 t 2, t 3, t 4, t 5 ) t 2 + t 3 = a h b m t 4 + t 5 = a h b l and since we know that t 2 and t 3 and t 4 and 5 do not overlap, it suffices to apply the bound proven at theorem 4.4. So we can check on the one hand and on the other 2 53 t 3 + t t t t t a h b m a h b l 2 05 βo a h b m βo βu a h b l 2 02 t 2 + t 3 + t 4 + t 5 = 2 02 a h b m + a h b l 2 02 a h b m a h b l 2 02 βo a h b h βo βu a h b l 3

32 In consequence, using the mentioned theorem, we obtain with t + t 2 = a h b m + a h b l + δ δ 2 02 βo a h b h βo βu a h b l Let us continue with the following part of the algorithm: We have and So we get with t 6, t 7 ) Mul2 a l, b h ) t 8, t 9 ) Mul2 a l, b m ) t 3, t 4 ) Add22 t 6, t 7, t 8, t 9 ) 2 53 t 7 + t t t t t a l b h a l b m 2 58 a h b h βo a h b h 2 02 t 6 + t 7 + t 8 + t 9 = 2 02 a l b h + a l b m 2 02 a l b h a l b m 2 55 a h b h βo a h b h t 3 + t 4 = a l b h + a l b m + δ 2 δ a h b h βo a h b h Let us now consider the brick that produces t 5 and t 6 out of the values in argument. By the properties of the Add22 procedure, t and t 2 and t 3 and t 4 do not overlap at all and verify thus the preconditions of the next Add22 brick that will compute t 5 and t 6. So it suffices to apply once again the absolute error bound of this procedure for obtaining with δ 3 which remains to be estimated. So we have on the one hand t 5 + t 6 = t + t 2 + t 3 + t 4 + δ t 2 + t t t 4 which is an upper bound that can still be estimated by t = t + t 2 ) t + t 2 ) ) 2 t + t 2 2 a h b m + a h b l + δ 2 06 t t 3 2 a h b m + a h b l + δ ) 2 2 βo a h b h + 2 βo βu a h b h + 2 βo 02 a h b h + 2 βo βu 02 a h b h ) a h b h 2 βo βo βu+2) 32

33 which means, using also the following inequalities that t 3 = t 3 + t 4 ) 2 t 3 + t 4 2 a l b h + a l b m + δ 2 2 a l b h + a l b m + δ 2 ) a h b h βo a h b h a h b h βo a h b h ) 2 50 a h b h we can finally check that we have on the one side And on the other 2 53 t 2 + t 4 a h b h 2 βo βo βu ) 2 02 t + t 2 + t 3 + t a h b m + a h b l + a l b h + a l b m + δ + δ a h b h 2 β o So we know that with With this result we can note now with +2 βo βu βo βo βo βu βo 55) a h b h 2 βo βo βu ) t 5 + t 6 = t + t 2 + t 3 + t 4 + δ 3 δ 3 a h b h 2 βo βo βu ) t 5 + t 6 = a h b m + a h b l + a l b h + a l b m + δ 4 δ 4 a h b h 2 β o βo βu βo βo 0 +2 βo βu 0) a h b h 2 βo βo βu ) Let us give now an upper bound for δ 5 defined by the following expression: r m + r l = t + a l b l + t 5 + t 6 + δ 5 33

34 It is clear that t 7 and t 8 do not overlap. In contrast the Add2 operation which adds t to t 0 is necessary because t and t 0 can overlap and even overtake each other: and t 2 06 a h b h t = 0 t 0 2 βo βu 52 a h b h The same argument tells us that the Add2 must be conditional. So we have t 7 + t 8 = t + t 0 and with t 0 = a l b l + δ δ 2 06 βo βu a h b h Let us apply once again the bound for the absolute error of the Add22 procedure: So we have on the one hand We can estimate this by and by t 5 = t 5 + t 6 ) 2 t 5 + t t 8 + t t t 8 t 7 t + t 0 ) 2 t + t 0 2 t + 2 t t t r h + 2 a l b l ) 2 52 a h b h ) a l b l 2 5 a h b h βo βu a h b h 2 a h b m + a h b l + a l b h + a l b m + δ 4 a h b h 2 βo+ + 2 βo βu βo βo βo βu ) a h b h 2 βo βo βu ) So finally, we have on the one hand 2 53 t 8 + t 6 a h b h βo βu βo βo βu ) And on the other a h b h 2 βo βo βu ) 2 02 t 7 + t 8 + t 5 + t t + a l b l + δ + a h b m + a h b l + a l b h +a l b m + δ r h βo βu a h b h +2 βo a h b h +2 βo βu a h b h 34

35 +2 53 a h b h which means that we finally obtain the following βo a h b h βo βu a h b h + a h b h 2 00 βo βo βu )) a h b h 2 0 βo βo βu ) with where Thus we can check that r m + r l = t + a l b l + a h b m + a h b l + a l b h + a l b m + δ 6 δ 6 δ 4 + δ 5 δ 5 a h b h 2 0 βo βo βu ) δ 6 a h b h 2 00 βo βo βu βo βo βu ) a h b h 2 99 βo βo βu ) Let us now integrate the different intermediate results: Since we know that the following egality is exact we can check that r h + t = a h b h r h + r m + r l = a h + a l ) b h + b m + b l ) + δ 6 We continue by giving a lower bound for a h + a l ) b h + b m + b l ) using a term which is a function of a h b h. We do so for being able to give a relative error bound. We first bound by such a term. We have a h b m + a h b l + a l b h + a l b m + a l b l a h b m + a h b l + a l b h + a l b m + a l b l a h b m + a h b l + a l b h + a l b m + a l b l a h b h 2 βo + 2 βo βu βo βo 53) and we get 2 βo+ a h b h + 2 βo βu+ a h b h a h b h a h + a l ) b h + b m + b l ) a h b h βo+ 2 βo βu+) from which we deduce since β o 2, β u ) a h b h βo+ 2 βo βu+ a h + a l ) b h + b m + b l ) Using this inequality we can finally give a bound for the relative error ε as follows: r h + r m + r l = a h + a l ) b h + b m + b l ) + ε) 35

36 with ε 2 99 βo βo βu βo+ 2 βo βu+ Let us recall that for this inequality, β o 2, β u must hold which is the case. It is certainly possible to estimate ε by a term which is slightly less exact: because for β o 2, β u. ε 2 97 βo βo βu βo+ 2 βo βu+ 4 In order to finish this proof, we must still give an upper bound for the maximal overlap generated by the algorithm 6.3 Mul233. Clearly r m and r l do not overlap because of the properties of the basic brick Add22. Let us give an upper bound for the overlap between r h and r m giving a term of the following form: r m 2 γ r h where we constate a lower bound for γ using a term in β o and β u. Let us start by giving a lower bound for r h as a function of a h b h. We have r h = a h b h ) 2 a h b h Then, let us give an upper bound for r m using a function of a h b h : This implies that r m r m + r l ) 2 r m + r l 2 t + a l b l + a h b m + a h b l + a l b h + a l b m + δ 6 2 a h b h βo βu βo + 2 βo βu βo βo βo βu ) a h b h βo βo βu+2) r m 2 r h βo βo βu+2) r h βo βo βu+3) From which we can deduce with r m 2 γ r h γ min 48, β o 4, β o β u 4) This result finishes the proof. 36