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Induction
Lecture 5 CS2110 – Summer 2008
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Overview
y Recursion
A programming strategy that solves a problem by reducing it to simpler or smaller instance(s) of the same problem
y Induction
A mathematical strategy for proving statements about natural numbers 0,1,2,... (or more generally, about inductively defined objects)
y Induction and recursion are very closely related
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Defining Functions
y 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
Recursive definition S(0) = 0 S(n) = n + S(n-1) for n > 0
Another characterization: closed form S(n) = n(n+1)/
n i=
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Sum of Squares
y 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
y Definition (iterative form): SQ(n) = 0^2 + 1^2 + … + n^2
y Is there an equivalent closed-form expression?
Closed-Form Expression for SQ(n)
y Sum of integers between 0 through n was n(n+1)/
which is a quadratic in n
y Inspired guess: perhaps sum of squares of
integers between 0 through n is a cubic in n
y Conjecture: SQ(n) = an^3 +bn^2 +cn+d
where a, b, c, d are unknown coefficients
y How can we find the values of the four unknowns?
Idea: Use any 4 values of n to generate 4 linear equations, and then solve
Finding Coefficients
y 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 +1^2 +…+n^2 = an^3 +bn^2 +cn+d
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Is the Formula Correct?
y This suggests
SQ(n) = 0^2 + 1^2 + … + n^2
= n^3 /3 + n^2 /2 + n/
= n(n+1)(2n+1)/
y 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
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One Approach
y Try a few other values of n to see if they work.
Try n = 5: SQ(n) = 0+1+4+9+16+25 = 55 Closed-form expression: 5·6·11/6 = 55 Works!
y Try some more values…
y We can never prove validity of the closed-form
solution for all values of n this way, since there are an
infinite number of values of n
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A Recursive Definition
y 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)
y This leads to the following recursive definition
SQ(0) = 0
SQ(n) = SQ(n-1) + n^2 , n > 0
y 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 Case
Recursive Case
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Are These Two Functions Equal?
y SQ r (r = recursive)
SQ r(0) = 0
SQ r(n) = SQ r(n-1) + n^2 , n > 0
y SQ c (c = closed-form)
SQ c (n) = n(n+1)(2n+1)/
Induction over Integers
y To prove that some property P(n) holds for all
integers n ≥ 0,
Basis: Show that P(0) is true Induction Step: Assuming that P(k) is true for an unspecified integer k, show that P(k+1) is true
y Conclusion: Because we could have picked any k, we
conclude that P(n) holds for all integers n ≥ 0
Dominos
y Assume equally spaced dominos, and assume that spacing between dominos is less than domino length y How would you argue that all dominos would fall? y 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 ... y Is there a more compact argument we can make?
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What are the “Dominos”?
y In some problems, it can be tricky to determine how to
set up the induction
y This is particularly true for geometric problems that
can be attacked using induction
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A Tiling Problem
y A chessboard has one square cut out of it y Can the remaining board be tiled using tiles of the shape shown in the picture (rotation allowed)? y Not obvious that we can use induction!
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Proof Outline
Consider boards of size 2n^ x 2n^ for n = 1,2,…
y Basis: Show that tiling is possible for 2 x 2 board
y Induction Hypothesis: Assume the 2k^ x 2 k^ board can
be tiled
y Inductive Step: Using I.H. show that the 2k+1^ x 2k+
board can be tiled
y Conclusion: Any 2 n^ x 2n^ board can be tiled, n = 1,2,…
Our chessboard (8 x 8) is a special case of this argument We will have proven the 8 x 8 special case by solving a more general problem!
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Basis
y The 2 x 2 board can be tiled regardless of which one of the four pieces has been omitted
2 x 2 board
4 x 4 Case
y Divide the 4 x 4 board into four 2 x 2 sub-boards
y One of the four sub-boards has the missing piece By the I.H., that sub-board can be tiled since it is a 2 x 2 board with a missing piece
y Tile the center squares of the three remaining sub-boards as shown This leaves three 2 x 2 boards, each with a missing piece We know these can be tiled by the Induction Hypothesis
2 k+1^ x 2k+1^ case
y Divide board into four sub-boards and tile the center squares of the three complete sub-boards y The remaining portions of the sub-boards can be tiled by the I.H. (which assumes we can tile 2k^ x 2k^ boards)
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When Induction Fails
y Sometimes an inductive proof strategy for some
proposition may fail
y This does not necessarily mean that the proposition is
wrong
It may just mean that the particular inductive strategy you are using is the wrong choice
y A different induction hypothesis (or a different proof
strategy altogether) may succeed
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Tiling Example (Poor Strategy)
Let’s try a different induction strategy
y Proposition
Any n x n board with one missing square can be tiled
y Problem
A 3 x 3 board with one missing square has 8 remaining squares, but our tile has 3 squares; tiling is impossible
y Thus, any attempt to give an inductive proof of this
proposition must fail
y Note that this failed proof does not tell us anything
about the 8x8 case
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A Seemingly Similar Tiling Problem
y A chessboard has opposite corners cut out of it. Can the remaining board be tiled using tiles of the shape shown in the picture (rotation allowed)?
y Induction fails here. Why? (Well…for one thing, this board can’t be tiled with dominos.)
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Strong Induction
y We want to prove that some property P holds for all n
y Weak induction
P(0): Show that property P is true for 0 P(k) ⇒ P(k+1): Show that if property P is true for k, it is true for k+ Conclude that P(n) holds for all n
y Strong induction
P(0): Show that property P is true for 0 P(0) and P(1) and … and P(k) ⇒ P(k+1): show that if P is true for numbers less than or equal to k, it is true for k+ Conclude that P(n) holds for all n
y Both proof techniques are equally powerful
Conclusion
y Induction is a powerful proof technique
y Recursion is a powerful programming technique
y Induction and recursion are closely related
We can use induction to prove correctness and complexity results about recursive programs
