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By Ian Parberry
Too usually the matter units in typical set of rules texts are composed of small, idiosyncratic devices of busy-work and beside the point questions - forcing teachers into the time-consuming job of discovering or composing extra difficulties. Designed to fill that hole, this complement presents an intensive and sundry choice of helpful, useful difficulties at the layout, research, and verification of algorithms.
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Extra resources for Problems on algorithms
Example text
8. Comments verify this pattern by induction on i. ) Then choose i to be whatever value it takes to make all of the T (·)s in the pattern turn into the base case for the recurrence. You are usually left with some algebra to do, such as a summation or two. Once you have used repeated substitution to get an answer, it is prudent to check your answer by using induction. ) Let’s do it for this problem. We are required to prove that the solution to the recurrence T (1) = 1, and for all n ≥ 2, T (n) = 3T (n − 1) + 2, is T (n) = 2 · 3n−1 − 1.
1. 2. 3. 4. 268. Prove that the following recursive algorithm for the addition of natural numbers is correct. 1. 2. 3. 269. function increment(y) comment Return y + 1. if y = 0 then return(1) else if y mod 2 = 1 then return(2 · increment( y/2 )) else return(y + 1) function add(y, z, c) comment Return the sum y + z + c, where c ∈ {0, 1}. if y = z = 0 then return(c) else a := y mod 2; b := z mod 2; return(2 · add( y/2 , z/2 , (a + b + c)/2 ) + (a ⊕ b ⊕ c)) Prove that the following recursive algorithm for the multiplication of natural numbers is correct.
7. 252. procedure swap(x, y) comment Swap x and y. x := x + y y := x − y x := x − y function increment(y) comment Return y + 1, where y ∈ IN x := 0; c := 1; d := 1; while (y > 0) ∨ (c > 0) do a := y mod 2; if a ⊕ c then x := x + d; c := a ∧ c; d := 2d; y := y/2 ; return(x) Prove that the following algorithm for the multiplication of natural numbers is correct. 48 Chap. 5. Correctness Proofs 1. 2. 3. 4. 5. 253. Prove that the following algorithm for the multiplication of natural numbers is correct.