Recursive Definitions and Structural Induction

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Recursive Definitions and Structural Induction

• CS 202
• Epp section ???
• Aaron Bloomfield

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Recursion

• Recursion means defining something, such as a
  function, in terms of itself
• For example, let f(x) x!
• We can define f(x) as f(x) x f(x-1)

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Recursion example

• Find f(1), f(2), f(3), and f(4), where f(0) 1
• Let f(n1) f(n) 2
• f(1) f(0) 2 1 2 3
• f(2) f(1) 2 3 2 5
• f(3) f(2) 2 5 2 7
• f(4) f(3) 2 7 2 9
• Let f(n1) 3f(n)
• f(1) 3 f(0) 31 3
• f(2) 3 f(1) 33 9
• f(3) 3 f(2) 39 27
• f(4) 3 f(3) 327 81

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Recursion example

• Find f(1), f(2), f(3), and f(4), where f(0) 1
• Let f(n1) 2f(n)
• f(1) 2f(0) 21 2
• f(2) 2f(1) 22 4
• f(3) 2f(2) 24 16
• f(4) 2f(3) 216 65536
• Let f(n1) f(n)2 f(n) 1
• f(1) f(0)2 f(0) 1 12 1 1 3
• f(2) f(1)2 f(0) 1 32 3 1 13
• f(3) f(2)2 f(0) 1 132 13 1 183
• f(4) f(3)2 f(0) 1 1832 183 1 33673

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Fractals

• A fractal is a pattern that uses recursion
• The pattern itself repeats indefinitely

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Fractals
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Fibonacci sequence

• Definition of the Fibonacci sequence
• Non-recursive
• Recursive F(n) F(n-1) F(n-2)
• or F(n1) F(n) F(n-1)
• Must always specify base case(s)!
• F(1) 1, F(2) 1
• Note that some will use F(0) 1, F(1) 1

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Fibonacci sequence in Java

• long Fibonacci (int n)
• if ( (n 1) (n 2) )
• return 1
• else
• return Fibonacci (n-1) Fibonacci (n-2)
• long Fibonacci2 (int n)
• return (long) ((Math.pow((1.0Math.sqrt(5.0)),n)
  -
• Math.pow((1.0-Math.sqrt(5.0)),n)) /
• (Math.sqrt(5) Math.pow(2,n)))

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Recursion definition

• From The Hackers Dictionary
• recursion n. See recursion. See also tail
  recursion.

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Bad recursive definitions

• Consider
• f(0) 1
• f(n) 1 f(n-2)
• What is f(1)?
• Consider
• f(0) 1
• f(n) 1f(-n)
• What is f(1)?

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Defining sets via recursion

• Same two parts
• Base case (or basis step)
• Recursive step
• Example the set of positive integers
• Basis step 1 ? S
• Recursive step if x ? S, then x1 ? S

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Defining sets via recursion

• Give recursive definitions for
• The set of odd positive integers
• 1 ? S
• If x ? S, then x2 ? S
• The set of positive integer powers of 3
• 3 ? S
• If x ? S, then 3x ? S
• The set of polynomials with integer coefficients
• 0 ? S
• If p(x) ? S, then p(x) cxn ? S
• c ? Z, n ? Z and n 0

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Defining strings via recursion

• Terminology
• ? is the empty string
• ? is the set of all letters a, b, c, , z
• The set of letters can change depending on the
  problem
• We can define a set of strings ? as follows
• Base step ? ? ?
• If w ? ? and x ? ?, then wx ? ?
• Thus, ? s the set of all the possible strings
  that can be generated with the alphabet
• Is this countably infinite or uncountably
  infinite?

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Defining strings via recursion

• Let ? 0, 1
• Thus, ? is the set of all binary numbers
• Or all binary strings
• Or all possible computer files

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String length via recursion

• How to define string length recursively?
• Basis step l(?) 0
• Recursive step l(wx) l(w) 1 if w ? ? and x
  ? ?
• Example l(aaa)
• l(aaa) l(aa) 1
• l(aa) l(a) 1
• l(a) l() 1
• l() 0
• Result 3

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Strings via recursion example

• Give a recursive definition for the set of string
  that are palindromes
• We will define set P, which is the set of all
  palindromes
• Basis step ? ? P
• Second basis step x ? P when x ? ?
• Recursive step xpx ? P if x ? ? and p ? P

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Recursion pros

• Easy to program
• Easy to understand

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Recursion cons

• Consider the recursive Fibonacci generator
• How many recursive calls does it make?
• F(1) 1
• F(2) 1
• F(3) 3
• F(4) 5
• F(5) 9
• F(10) 109
• F(20) 13,529
• F(30) 1,664,079
• F(40) 204,668,309
• F(50) 25,172,538,049
• F(100) 708,449,696,358,523,830,149 ? 7 1020
• At 1 billion recursive calls per second
  (generous), this would take over 22,000 years
• But that would also take well over 1012 Gb of
  memory!

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Trees

• Rooted trees
• A graph containing nodes and edges
• Cannot contain a cycle!

Cycle not allowed in a tree
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Rooted trees

• Recursive definition
• Basis step A single vertex r is a rooted tree
• Recursive step
• Let T1, T2, , Tn be rooted trees
• Form a new tree with a new root r that contains
  an edge to the root of each of the trees T1, T2,
  , Tn

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(Extended) Binary trees

• Recursive definition
• Basis step The empty set is an extended binary
  tree
• Recursive step
• Let T1, and T2 be extended binary trees
• Form a new tree with a new root r
• Form a new tree such that T1 is the left
  subtree,and T2 is the right subtree

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Full binary trees

• Recursive definition
• Basis step A full binary tree consisting only of
  the vertex r
• Recursive step
• Let T1, and T2 be extended binary trees
• Form a new tree with a new root r
• Form a new tree T such that T1 is the left
  subtree, and T2 is the right subtree (denoted by
  T T1T2)
• Note the only difference between a regular binary
  tree and a full one is the basis step

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Binary tree height

• h(T) denotes the height of tree T
• Recursive definition
• Basis step The height of a tree with only one
  node r is 0
• Recursive step
• Let T1 and T2 be binary trees
• The binary tree T T1T2 has heighth(T) 1
  max ( h(T1), h(T2) )
• This definition can be generalized to non-binary
  trees

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Binary tree size

• n(T) denotes the number of vertices in tree T
• Recursive definition
• Basis step The number of vertices of an empty
  tree is 0
• Basis step The number of vertices of a tree
  with only one node r is 1
• Recursive step
• Let T1 and T2 be binary trees
• The number of vertices in binary tree T T1T2
  isn(T) 1 n(T1) n(T2)
• This definition can be generalized to non-binary
  trees

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End of lecture on 11 April 2007
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Recursion vs. induction

• Consider the recursive definition for factorial
• f(0) 1
• f(n) n f(n-1)
• Sort of like induction

Base case
The step
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Recursion vs. induction

• Consider the set of all integers that are
  multiples of 3
• 3, 6, 9, 12, 15,
• x x 3k and k ? Z
• Recursive definition
• Basis step 3 ? S
• Recursive step If x ? S and y ? S, then xy ? S

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Recursion vs. induction

• Proof via induction prove that S contains all
  the integers that are divisible by 3
• Let A be the set of all ints divisible by 3
• Show that S A
• Two parts
• Show that S ? A
• Let P(n) 3n ? S
• Base case P(1) 31 ? S
• By the basis step of the recursive definition
• Inductive hypothesis assume P(k) 3k ? S is
  true
• Inductive step show that P(k1) 3(k1) is
  true
• 3(k1) 3k3
• 3k ? S by the inductive hypothesis
• 3 ? S by the base case
• Thus, 3k3 ? S by the recursive definition
• Show that A ? S
• Similar steps (not reproduced here)

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What did we just do?

• Notice what we did
• Showed the base case
• Assumed the inductive hypothesis
• For the inductive step, we
• Showed that each of the parts were in S
• The parts being 3k and 3
• Showed that since both parts were in S, by the
  recursive definition, the combination of those
  parts is in S
• i.e., 3k3 ? S
• This is called structural induction

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Structural induction

• A more convenient form of induction for
  recursively defined things
• Used in conjunction with the recursive definition
• Three parts
• Basis step Show the result holds for the
  elements in the basis step of the recursive
  definition
• Inductive hypothesis Assume that the statement
  is true for some existing elements
• Usually, this just means assuming the statement
  is true
• Recursive step Show that the recursive
  definition allows the creation of a new element
  using the existing elements

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Tree structural induction example

• Show that n(T) 2h(T) 1 for full binary trees
• Basis step Let T be the full binary tree of just
  one node r
• h(T) 0
• n(T) 1
• n(T) 2h(T) 1
• 1 20 1
• 1 1

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Tree structural induction example

• Show that n(T) 2h(T) 1
• Inductive hypothesis
• Let T1 and T2 be full binary trees
• Assume that n(T1) 2h(T1) 1 for some tree T1
• Assume that n(T2) 2h(T2) 1 for some tree T2
• Recursive step
• Let T T1 T2
• Here the operator means creating a new tree
  with a root note r and subtrees T1 and T2
• New element is T
• By the definition of height and size, we know
• n(T) 1 n(T1) n(T2)
• h(T) 1 max ( h(T1), h(T2) )
• Therefore
• n(T) 1 n(T1) n(T2)
• 1 2h(T1) 1 2h(T2) 1
• 1 2max ( h(T1), h(T2) ) the sum of two
  non-neg s is at least
• as large as the larger of the two
• 1 2h(T)
• Thus, n(T) 2h(T) 1

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String structural induction example

• Part (a) Give the definition for ones(s), which
  counts the number of ones in a bit string s
• Let ? 0, 1
• Basis step ones(?) 0
• Recursive step ones(wx) ones(w) x
• Where x ? ? and w ? ?
• Note that x is a bit either 0 or 1

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String structural induction example

• Part (b) Use structural induction to prove that
  ones(st) ones(s) ones(t)
• Basis step t ?
• ones (s?) ones(s) ones(s)0 ones(s)
  ones(?)
• Inductive hypothesis Assume ones(st) ones(s)
  ones(t)
• Recursive step Want to show that ones(stx)
  ones(s) ones(tx)
• Where s, t ? ? and x ? ?
• New element is ones(stx)
• ones (stx) ones ((st)x)) by associativity
  of concatenation
• xones(st) by recursive definition
• x ones(s) ones(t) by inductive hypothesis
• ones(s) (x ones(t)) by commutativity and
  assoc. of
• ones(s) ones(tx) by recursive definition
• Proven!

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Induction methods compared
Weak mathematical Strong Mathematical Structural
Used for Usually formulae Usually formulae not provable via mathematical induction Only things defined via recursion
Assumption Assume P(k) Assume P(1), P(2), , P(k) Assume statement is true for some "old" elements
What to prove True for P(k1) True for P(k1) Statement is true for some "new" elements created with "old" elements
Step 1 called Base case Base case Basis step
Step 3 called Inductive step Inductive step Recursive step
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Induction types compared

• Show that F(n) lt 2n
• Where F(n) is the nth Fibonacci number
• Actually F(n) lt 20.7n, but we wont prove that
  here
• Fibonacci definition
• Basis step F(1) 1 and F(2) 1
• Recursive step F(n) F(n-1) F(n-2)
• Base case (or basis step) Show true for F(1) and
  F(2)
• F(1) 1 lt 21 2
• F(2) 1 lt 22 4

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Via weak mathematical induction

• Inductive hypothesis Assume F(k) lt 2k
• Inductive step Prove F(k1) lt 2k1
• F(k1) F(k) F(k-1)
• We know F(k) lt 2k by the inductive hypothesis
• Each term is less than the next, thereforeF(k)
  gt F(k-1)
• Thus, F(k-1) lt F(k) lt 2k
• Therefore, F(k1) F(k) F(k-1) lt 2k 2k
  2k1
• Proven!

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Via strong mathematical induction

• Inductive hypothesis Assume F(1) lt 21, F(2) lt
  22, , F(k-1) lt 2k-1, F(k) lt 2k
• Inductive step Prove F(k1) lt 2k1
• F(k1) F(k) F(k-1)
• We know F(k) lt 2k by the inductive hypothesis
• We know F(k-1) lt 2k-1 by the inductive hypothesis
• Therefore, F(k) F(k-1) lt 2k 2k-1 lt 2k1
• Proven!

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Via structural induction

• Inductive hypothesis Assume F(n) lt 2n
• Recursive step
• Show true for new element F(n1)
• We know F(n) lt 2n by the inductive hypothesis
• Each term is less than the next, therefore F(n)
  gt F(n-1)
• Thus, F(n-1) lt F(n) lt 2n
• Therefore, F(n) F(n-1) lt 2n 2n 2n1
• Proven!

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Another way via structural induction

• Inductive hypothesis Assume F(n) lt 2n and
  F(n-1) lt 2n-1
• The difference here is we are using two old
  elements versus one, as in the last slide
• Recursive step
• Show true for new element F(n1)
• F(n1) F(n) F(n-1)
• We know F(n) lt 2n by the inductive hypothesis
• We know F(n-1) lt 2n-1 by the inductive hypothesis
• Therefore, F(n) F(n-1) lt 2k 2k-1 lt 2k1
• Proven!

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But wait!

• In this example, the structural induction proof
  was essentially the same as the weak or strong
  mathematical induction proof
• Its hard to find an example that works well for
  all of the induction types
• Structural induction will work on some recursive
  problems which weak or strong mathematical
  induction will not
• Trees, strings, etc.

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Recursive definition examples

• Give the recursive definition of the following
  sequences
• Note that many answers are possible!
• an 4n 2
• Terms 2, 6, 10, 14, 16, etc.
• a1 2
• an an-1 4
• an 1 (-1)n
• Terms 0, 2, 0, 2, 0, 2, etc.
• a1 0, a2 2
• an an-2
• an n(n1)
• Terms 2, 6, 12, 20, 30, 42, etc.
• a1 2
• an an-1 2n
• an n2
• Terms 1, 4, 9, 16, 25, 36, 49, etc.
• a1 1
• an an-1 2n - 1

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More more examples

• Show that f12 f22 f32 fn2 fnfn1
• Base case n 1
• f12 f1f2
• 12 11
• Inductive hypothesis Assume
• f12 f22 f32 fk2 fkfk1
• Inductive step Prove
• f12 f22 f32 fk2 fk12 fk1fk2

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More more more examples

• Inductive hypothesis Assume
• f12 f22 f32 fk2 fkfk1
• Inductive step Prove
• f12 f22 f32 fk2 fk12 fk1fk2
• fkfk1 fk12 fk1fk2
• fkfk1 fk12 fk1 (fk fk1)
• fkfk1 fk12 fkfk1 fk12

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More4 examples

• Show that f1 f2 f3 f2n-1 f2n
• Base case n 1
• f1 f21
• 1 1
• Inductive hypothesis Assume
• f1 f2 f3 f2k-1 f2k
• Inductive step Prove
• f1 f2 f3 f2k-1 f2(k1)-1 f2(k1)
• f1 f2 f3 f2k-1 f2k1 f2k2

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More5 examples

• Inductive hypothesis Assume
• f1 f2 f3 f2k-1 f2k
• Inductive step Prove
• f1 f2 f3 f2k-1 f2k1 f2k2
• f2k f2k1 f2k2
• True by definition of f2k2

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More6 examples

• Show that the set S defined by
• Basis step 1 ? S
• Recursive step s t ? S when s ? S and t ? S
• is the set of positive integers
• Z 1, 2, 3,
• Note the (somewhat recursive) definition of the
  positive integers
• 1 is a positive integer
• For any arbitrary n that is a positive integer,
  n1 is also a positive integer
• Proof by structural induction
• Basis step 1 ? S and 1 ? Z
• Inductive hypothesis Assume k ? S
• Recursive step Show k1 ? S
• k ? S by the inductive hypothesis
• 1 ? S by the base case
• k1 ? S by the recursive step of the recursive
  definition above

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More7 examples

• Give a recursive definition of the reversal of a
  string
• Basis step ?R ?
• Note that the superscripted R means reversal of a
  string
• Recursive step Consider a string w ? ?
• Rewrite w as vy where v ? ? and y ? ?
• v is the first n-1 characters in w
• y is the last character in w
• wR y(vR)
• Parentheses are for our benefit