- (2i:0^ in .NET

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5 - (2i:0^
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1 3: Iteration This is clearly true (since k ^ N and k ^ N together imply that k = N). The requirement (13.11) checks the initialization. We have to verify that { (KAf } fe,5 := 0,0 { O^k^N A s = Using the assignment axiom, this follows from A 0= ( Z i : 0 ^ i < which is true by the empty-range rule for summations. The requirement (13.12) checks the loop body. We have to verify that A 5 = ( Z i : 0 < i < f e : a [ i ] ) A k<N k,s := k+l,s+a[k] { O ^ f c ^ J V A 5 = ( I i : 0 ^ i < k : a [ i ] > A N~k<C } . Again, use of the assignment axiom is called for. We get A 5 = <Ii:(Ki<fc:a[i]) A k<N A N-k = C A s+a[k] = ( S i : 0 ^ i < f e + l : a[i]> A N-(k+l)<C . This is true by virtue of the splitting and one-point rules for summation (and simple arithmetic). The final condition (13.10) is clearly satisfied: A 5 = (Si:0^i< A N-k = C }
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1 3.3.2 Evaluating a Polynomial
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Evaluating a polynomial involves a more complicated summation and is the basis of several other algorithms. Suppose we are required to evaluate
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for given real number X and array a. It is possible, of course, to regard this problem as a specific instance of the summation problem just discussed. Doing so means the introduction of variables 5 and k satisfying the invariant property A 5 = <Zi:0^i< The problem is that each iteration of the loop body involves executing the assignment k,5 := fc+1 ,s + a[k]xXk and thus evaluating Xk.
1. 3.3 Basic Arithmetic Operations
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An alternative method, called Homer's rule, is preferable because it uses fewer multiplications. Homer's rule involves computing the values
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a[N-l]xX + a[N-2] , (a[N-l]xX + a[N-2])xX
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+ a[N-3] ,
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and so on. Functionally, we can describe Horner's rule as maintaining invariant the property
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A sxXk = (Z.i:k^ This property is established, initially, by the assignment
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k,s := JV.O ,
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and the required postcondition is satisfied when
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We, therefore, consider an algorithm of the form
k,s := AT.O ;
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{ Invariant: ( K f c ^ N A sxXk = Bound function: k }
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do k > 0 k,s := k-I,S od
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{ 5 = (Zi:Q^ where S, the value to be assigned to 5 in the body of the loop, is the only missing element. We calculate the appropriate value of S using the assignment axiom. The specification of the assignment in the body of the loop is given by the Hoare triple:
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A sxXk = ( I . i : k ^ i < N : a [ i ] x X i )
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A k>0
} . A k>0
{ O^k^N A sxXk = (Zi:k^i<N:a[i]xXi) A sxXk = (Zi:k^i<N:a[i]xXi) A SxX ~ = (Zi:k-l
Applying the assignment axiom, the requirement reduces to
1 90 Clearly,
A k > 0 => (Kk-
1 3: Iteration
So, it is indeed only the appropriate value of 5 that needs to be determined. We now use the summation rules to calculate the appropriate value of the unknowns: SxXk~l = (Zi:k-l^ = { SxXk~l { SxX ~ <=
splitting the range on i = k- 1 , assuming 0 < k ^ N } = (Zi:k^i<N:a[i]xXi) = sxX
+ a[k-l]xXk~l }
assume sxXk = (Zi:k^i<N:a[i]xXi) + a[k-l]xX ~ }
factor out Xk~l
S = sxX + a[k-l] . We have thus determined that A sxXk = (Ei:k^i<N:a[i]xXi) A SxX ~ = (Zi So the loop body is k,s := k-l ,sxX + a[k-l] and the complete algorithm is as follows: k,s := N,0 ; { Invariant: O^k^N k } A sxXk = Bound function:
A 5 = sxX + a[k-l]
do k > 0 k,s := k-l ,
{ s = (Zi:0^ Note that this derivation of the algorithm constitutes a formal proof of Homer's rule. Note also how the calculation of S (as opposed to guessing what it should be) avoids making a 'one-off error. With problems like this one, it is very easy for array indices to be 'one-off. For example, we might have guessed the value sxX + a[k] for S. The consequences are often noticed immediately, but not always. And, in the digital world, a small error of this nature can be disastrous! Exercise 13.13. Verify the correctness of the initialization and the termination condition. D