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Induction
A well-known scientist (some say it was Bertrand Russell) once gave a public lecture on astronomy. He described how the earth orbits around the sun and how the sun, in turn, orbits around the center of a vast collection of stars called our galaxy. At the end of the lecture, a little old lady at the back of the room got up and said: "What you have told us is rubbish. The world is really a flat plate supported on the back of a giant tortoise." The scientist gave a superior smile before replying, "What is the tortoise standing on?" "You're very clever, young man, very clever", said the old lady. "But it's turtles all the way 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 function where S(n) is the sum of
the integers from 0 to n. For example,
S(0) = 0 S(3) = 0+1+2+3 = 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 function that gives the sum of the
squares of integers from 0 to n:
SQ(0) = 0 SQ(3) = 0 2
- 1 2
- 2 2
- 3 2 = 14 Definition (iterative form): SQ(n) = 0 2
- 1 2
- … + n 2 Is there an equivalent closed-form expression?
Finding Coefficients
Use n = 0, 1, 2, 3
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 = 1/3 b = 1/2 c = 1/6 d = 0
SQ(n) = 0
2
2
+…+n
2
= an
3
+bn
2
+cn+d
Is the Formula Correct?
This suggests SQ(n) = 0 2
- 1 2
- … + n 2 = n 3 /3 + n 2 /2 + n/ = n(n+1)(2n+1)/ Question: Is this closed-form solution true for all n?
Remember, we only used n = 0,1,2,3 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-1) 2
- n 2 The part in the box is just SQ(n-1)
This leads to the following recursive definition
SQ(0) = 0 SQ(n) = SQ(n-1) + n 2 , n > 0
Thus,
SQ(4) = SQ(3) + 4 2 = SQ(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 Case Recursive Case
Are These Two Functions Equal?
SQ
r
(r = recursive)
SQ
r
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?
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+1 (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: SQr(0) = 0 SQr(n) = SQr(n-1) + n 2 , n > 0 SQc(n) = n(n+1)(2n+1)/ Let P(n) be the proposition that SQr(n) = SQc(n) Basis: P(0) holds because SQr(0) = 0 and SQc(0) = 0 by definition Induction Hypothesis: Assume SQr(k) = SQc(k) Inductive Step: SQr(k+1) = SQr(k) + (k+1) 2 by definition of SQr(k+1) = SQc(k) + (k+1) 2 by the Induction Hypothesis = k(k+1)(2k+1)/6 + (k+1) 2 by definition of SQc(k) = (k+1)(k+2)(2k+3)/6 algebra = SQc(k+1) by definition of SQc(k+1) Conclusion: SQr(n) = SQc(n) for all n 0
Another Example
Prove that 0+1+...+n = n(n+1)/ Basis: Obviously holds for n = 0 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)/2 + (k+1) by I.H. = (k+1)(k+2)/2 algebra Conclusion: 0+1+…+n = n(n+1)/2 for all n ≥ 0
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
