Definition:
public natural value (string w)
{// Returns the natural number value of the given binary string
if (w == emptystring) return 0;
string abl = allButLast (w);
if (last(w) == '0')
return 2 * value (abl);
else return 2 * value (abl) + 1;}
Writing Exercise: Prove by induction on all strings w that
value(w) terminates and returns the correct value according to the
definition.
Solution: The base case is w = λ. Here the
defined value is 0, and the computed value is clearly 0 from line 3 of the code.
For the inductive case, assume that Let v be the defined value of x, which by the IH is equal to the output
of value(x) terminates with
the right answer, and we will prove that each of value(x0) and
value(x1) terminate with the right answer.
value(x). For x0, the defined value is 2v by the definition.
Tracing the code on input x0, we see that it sets abl to x, then
goes down the if branch of the if statement and returns 2 * value (x), which is
2v as desired. We can see that it terminates under the assumption that the
recursive call terminates.
For x1, the defined value is 2v+1 by the definition. The code
sets abl to be x, then goes down the else branch of the if
statement, where it returns 2 * value (x) + 1, which is 2v+1 as desired.
Again we see that it terminates under the assumption that the recursive call
terminates.
Definition:
public string rep (natural n)
{// Returns canonical string (no leading 0's) representing n.
if (n == 0) return "0";
if (n == 1) return "1";
string w = rep (n/2);
if (n%2 == 0)
return append (w, '0');
else return append (w, '1');}
Writing Exercise: Prove by (strong) induction on all naturals n
that
rep(n) terminates and returns the correct value according to the
definition.
Solution: The base case is n = 0. Here the
defined value is "0", and the computed value is clearly "0"
from line 3 of the code.
For the strong inductive case, we assume that If n+1 happens to be 1, the defined value is "1" and we can see from line
4 of the code that the output value is also "1". So now assume that n ≥ 1.
The defined value is the string xa, where x is the defined value for (n+1)/2 and
a is the defined value of (n+1)%2. Let's look at the code on input n+1. The
variable w is set on line 5 to the result of the recursive call to
rep((n+1)/2). But since 1 ≤ n, we know that n+1 ≤ 2n and hence that
(n+1)/2 ≤ n, by halving both sides. Thus by the strong IH, we know that
the recursive call terminates with an output that is equal to the defined value
for (n+1)/2, that is, the string we have called x.
Now the code reaches the if statement and checks the value of (n+1)%2.
If this is 0, it goes down the if branch and terminates with output x0. If
this is 1, it goes down the else branch and terminates with output x1. In each
case this is equal to the defined value for n+1, so we have completed the
strong induction.
rep(j)
terminates with the right answer for all j such that
j ≤ n, and we will prove
that rep(n+1)
terminates with the right answer.
Last modified 24 October 2007