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Induction
Lecture 20 CS
Spring 2013 A well-known scientist (some say it was Bertrand Russell) once gave apublic lecture on astronomy. He described how the earth orbits around thesun and how the sun, in turn, orbits around the center of a vast collection ofstars called our galaxy.At the end of the lecture, a little old lady at the back of the room got up andsaid: "What you have told us is rubbish. The world is really a flat platesupported on the back of a giant tortoise." The scientist gave a superiorsmile before replying, "What is the tortoise standing on?" "You're veryclever, young man, very clever", said the old lady. "But it's turtles all theway down! ”
Overview:
Reasoning
about
Programs
Our
broad
problem:
code
is
unlikely
to
be
correct
if
we
don’t
have
good
reasons
for
believing
it
works
We need clear problem statements - And then a rigorous way to convince ourselves that what we wrote solves the problem
But
reasoning
about
programs
can
be
hard
Especially with recursion, concurrency - Today focus on recursion
Defining
Functions
It is often useful to describe a function in different ways
Let S : int → int be the func Ɵ on where S(n) is the sum of the integers from 0 to n. For example, S(0) = 0 S(3) = 0+1+2+ = 6 - Definition: iterative form - S(n) = 0+1+ … + n = Σ i - Another characterization: closed form - S(n) = n(n+1)/ n i=
Sum
of
Squares
A
more complex example
Let SQ : int → int be the func Ɵ on that gives the sum of the squares of integers from 0 to n: SQ(0)
SQ(3)
2
2
2
2
Definition (iterative form): SQ(n)
2
2
n 2
Is there an equivalent closed
form expression?
Finding
Coefficients
Use n
SQ(0)
0
a ∙ 0
b ∙ 0
c ∙ 0
d - SQ(1) = 1 = a ∙ 1 + b ∙ 1 + c ∙ 1 + d - SQ(2) = 5 = a ∙ 8 + b ∙ 4 + c ∙ 2 + d - SQ(3) = 14 = a ∙ 27 + b ∙ 9 + c ∙ 3 + d - Solve these 4 equations to get - a = 1/ b = 1/ c = 1/ d = 0
SQ(n) = 0
2
2
+…+n
2
= an
3
+bn
2
+cn+d
Is
the
Formula
Correct?
This suggests SQ(n)
2
2
n 2 = n 3
n 2
n/
n(n+1)(2n+1)/
Question: Is this closed
form solution true for all n?
Remember, we only used n = 0,1,2, to determine these coefficients - We do not know that the closed ‐ form expression is valid for other values of n
A
Recursive
Definition
To solve this problem, let’s express SQ(n) in a different way: - SQ(n) = 0 2
1 2
…
(n ‐
2
n 2
The part in the box is just SQ(n ‐
This leads to the following recursive definition
SQ(0)
0
SQ(n)
SQ(n ‐
n 2 ,^ n > 0
Thus,
SQ(4)
SQ(3)
4 2 = SQ(2)
3 2
4 2 = SQ(1)
2 2
3 2
4 2 = SQ(0)
1 2
2 2
3 2
4 2 = 0
1 2
2 2
3 2
4 2 Base CaseRecursive Case
Are
These
Two
Functions
Equal?
SQ r (r = recursive) SQ r (0) = 0 SQ r (n) = SQ r (n ‐ 1) + n 2 , n > 0
SQ c (c = closed ‐ form) SQ c (n) = n(n+1)(2n+1)/
Dominos
Assume equally spaced dominos, and assume that spacing between dominos is less than domino length - How would you argue that all dominos would fall? - Dumb argument: - Domino 0 falls because we push it over - Domino 0 hits domino 1, therefore domino 1 falls - Domino 1 hits domino 2, therefore domino 2 falls - Domino 2 hits domino 3, therefore domino 3 falls - ... - Is there a more compact argument we can make? 0 1 2 3 5 4
Better
Argument
Argument: - Domino 0 falls because we push it over (Base Case or Basis) - Assume that domino k falls over (Induction Hypothesis) - Because domino k’s length is larger than inter ‐ domino spacing, it will knock over domino k+ (Inductive Step) - Because we could have picked any domino to be the k th one, we conclude that all dominos will fall over (Conclusion)
This is an inductive argument
This version is called weak induction
There is also strong induction (later)
Not only is this argument more compact, it works for an arbitrary number of dominoes!
Proof
(by
Induction)
Recall: SQ (0)r = 0 SQ (n)r = SQ (nr ‐ 1) + n 2 ,^ n > 0 SQ (n)c^ = n(n+1)(2n+1)/
Let P(n) be the proposition that SQ (n)r = SQ c (n)
Basis: P(0) holds because SQ (0)r = 0 and SQ c (0) = 0 by definition
Induction Hypothesis: Assume SQ (k)r = SQ (k)c^
Inductive Step: SQ (k+1)r = SQ (k)r
(k+1) 2 by definition of SQ (k+1)r = SQ c (k)
(k+1) 2 by the Induction Hypothesis = k(k+1)(2k+1)/
(k+1) 2 by definition of SQ c (k) = (k+1)(k+2)(2k+3)/ algebra = SQ c (k+1) by definition of SQ c (k+1)
Conclusion: SQ (n)r = SQ (n)c^ for all n 0
Another
Example
Prove that 0+1+...+n
n(n+1)/
Basis: Obviously holds for n
Induction Hypothesis: Assume 0+1+…+k
k(k+1)/
Inductive Step: 0+1+…+(k+1)
[0+1+…+k]
(k+1) by def = k(k+1)/
(k+1) by
I.H.
(k+1)(k+2)/ algebra
Conclusion: 0+1+…+n
n(n+1)/ for all n
Weak
Induction:
Nonzero
Base
Case
Claim: You can make any amount of postage above 8¢ with some combination of 3¢ and 5¢ stamps - Basis: True for 8¢: 8 = 3 + 5 - Induction Hypothesis: Suppose true for some k ≥ 8 - Inductive Step: - If used a 5¢ stamp to make k, replace it by two 3¢ stamps. Get k+1. - If did not use a 5¢ stamp to make k, must have used at least three 3¢ stamps. Replace three 3¢ stamps by two 5¢ stamps. Get k+1. - Conclusion: Any amount of postage above 8¢ can be made with some combination of 3¢ and 5¢ stamps
What
are
the
Dominos
In
some
problems,
it
can
be
tricky
to
determine
how
to
set
up
the
induction
This
is
particularly
true
for
geometric
problems
that
can
be
attacked
using
induction
