Induction and Recursion (Sections 4.1-4.3) [Section 4.4 optional] - PDF document

Induction and Recursion (Sections 4.1-4.3) [Section 4.4 optional] Based on Rosen and slides by K. Busch 1 Induction Induction is a very useful proof technique In computer science, induction is used to prove properties of algorithms Induction and recursion are closely related • Recursion is a description method for algorithms • Induction is a proof method suitable for recursive algorithms 2 1

Use induction to prove that a proposition is true: ( n ) P Inductive Basis: Prove that is true ( 1 ) P Inductive Hypothesis: Assume is true ( k ) P (for any positive integer k) (  Inductive Step: Prove that is true 1 ) P k 3 Inductive Hypothesis: Assume is true ( k ) P (for any positive integer k) (  Inductive Step: Prove that is true 1 ) P k In other words in inductive step we prove:   ( ) ( 1 ) P k P k for every positive integer k 4 2

Inductive basis Inductive Step   ( ) ( 1 ) P k P k ( 1 ) P True True Proposition true for all positive integers     ( 1 ) ( 2 ) ( 3 ) ( 4 )  P P P P 5 Induction as a rule of inference:       [ ( 1 ) ( ( ) ( 1 ))] ( ) P k P k P k nP n 6 3

 ( 1 ) n n      Theorem:  ( ) : 1 2 3 P n n 2 Proof:  1 ( 1 1 ) Inductive Basis:  ( 1 ) : 1 P 2 Inductive Hypothesis: assume that it holds  ( 1 ) k k     ( ) : 1 2 P k k 2 Inductive Step: We will prove    ( 1 )(( 1 ) 1 ) k k        ( 1 ) : 1 2 ( 1 ) P k k k 2 7 Inductive Step: (inductive       ( 1 ) : 1 2 ( 1 ) P k k k hypothesis)  ( 1 ) k k    ( 1 ) k 2    ( 1 ) 2 ( 1 ) k k k  2    ( 1 )(( 1 ) 1 ) k k  2 End of Proof 8 4

Harmonic numbers 1 1 1 1        1 H j 2 3 4 j  1 , 2 , 3 ,  j 1 1 1 25      Example: 1 H 4 2 3 4 12 9 n Theorem:    1 H 0 n n 2 2 Proof:  Inductive Basis: 0 n 0 n        1 1 1 H H H 0 1 n 2 2 2 2 10 5

n  Inductive Hypothesis: k k   Suppose it holds: 1 H k 2 2  k  Inductive Step: 1 n  1 k   We will show: 1 H  1 k 2 2 11 1 1 1 1 1           1 H    k 1 1 2 k k k 2 3 2 2 1 2 1 1      H   k 1 2 k k 2 1 2    from inductive 1 1 k        1   hypothesis   1 k k 2 2 1 2    1 k      k 1 2    1 k 2 2    1 k     1   2 2  1 k   1 End of Proof 2 12 6

Theorem:  1   H n 0 n n 2 Proof:  Inductive Basis: 0 n        1 1 0 1 H H H n n 0 1 2 2 13 n  Inductive Hypothesis: k  1  Suppose it holds: H k k 2  k  Inductive Step: 1 n We will show:    1 ( 1 ) H k  1 k 2 14 7

1 1 1 1 1           1 H   1  k 1 2 k k k 2 3 2 2 1 2 1 1      H   k 2 k k 1 2 1 2 from inductive 1 1         1 k hypothesis   1 k k 2 1 2 1       k 1 2 k  k 2 1      1 1 k    1 ( 1 ) k End of Proof 15 n     We have shown: 1 1 H n n 2 2   log k       1 1 log H k k 2  log H k k (for large k ) 16 8

Triominos 2  3 3 2 2  2 2 2 2  2 hole hole hole 17   Theorem: Every checkerboard n n 2 2 , 1 n with one square removed can be tiled with triominoes Proof:  Inductive Basis: 1 n 2  1 1 2 hole 18 9

n  Inductive Hypothesis: k 2  Assume that a checkerboard k k 2 can be tiled with the hole anywhere 2  k k 2 Hole can be anywhere 19  k  Inductive Step: 1 n    1 1 k k 2 2 20 10

By inductive hypothesis squares 2  k k 2 with a hole can be tiled 2  2  k k k k 2 2 add three artificial holes 21 2 3 x 2 3 case: 22 11

2  2  k k k k 2 2 Replace the three holes with a triomino Now, the whole area can be tiled 23 2 3 x 2 3 case: End of Proof 24 12

Strong Induction To prove : ( n ) P Inductive Basis: Prove that is true ( 1 ) P Inductive Hypothesis:    Assume is true  ( 1 ) ( 2 ) ( ) P P P k (  Inductive Step: Prove that is true 1 ) P k 25 Theorem: Every integer  2 n is a product of primes (Fundamental Theorem of Arithmetic) Proof: (Strong Induction)  Inductive Basis: 2 n Number 2 is a prime  n  Inductive Hypothesis: 2 k Suppose that every integer between and is a product of primes 2 k 26 13

 k  Inductive Step: 1 n  If is prime then the proof is finished 1 k  If is not a prime then it is composite: 1 k  1    ,  2 k a b a b k 27  1    ,  2 k a b a b k By the inductive hypothesis:  , 1 i j   a p p p     1 2 i   1 k a b p p q q 1 1 i j   b q q q primes 1 2 j primes End of Proof 28 14

Theorem: Every postage amount  12 n can be generated by using 4-cent and 5-cent stamps Proof: (Strong Induction) Inductive Basis: We examine four cases (because of the inductive step)         12 4 4 4 14 5 5 4 n n         13 4 4 5 15 5 5 5 n n 29  n  Inductive Hypothesis: 12 k Assume that every postage amount between and can be generated 12 k by using 4-cent and 5-cent stamps     4 5 n a b  k  Inductive Step: 1 n  k  If then the inductive step 12 14 follows directly from inductive basis 30 15

     Consider: 15 1 ( 3 ) 4 k k k    12 ( 3 ) k k Inductive hypothesis        ( 3 ) 4 5 k a b            1 ( 3 ) 4 ( 1 ) 4 5 k k a b End of Proof 31 Recursion Recursion is used to describe functions, sets, algorithms  Example: Factorial function ( ) ! f n n  Recursive Basis: ( 0 ) 1 f     Recursive Step: ( 1 ) ( 1 ) ( ) f n n f n 32 16

Recursive algorithm for factorial factorial( ) { n  if then //recursive basis 1 n return 1 else //recursive step return n  factorial( n- 1 ) } 33 Fibonacci numbers  , , , , f f f f 0 1 2 3   Recursive Basis: 0 , 1 f f 0 1    Recursive Step: f f f  1 2 n n n   2 , 3 , 4 , n 34 17

 0 f 0  1 f 1      1 0 1 f f f 2 1 0      1 1 2 f f f 3 2 1      2 1 3 f f f 4 3 2      3 2 5 f f f 5 4 3      5 3 8 f f f 6 5 4      8 5 13 f f f 7 6 5  35 Recursive algorithm for Fibonacci function fibonacci( ) { n if then  //recursive basis { 0 , 1 } n return n else //recursive step return n-  fibonacci( 1 ) fibonacci( 2 ) n- } 36 18

Iterative algorithm for Fibonacci function fibonacci( ) { n  if then  0 0 y n else {  0 x  1 y   for to do { 1 1 i n   z x y x  y y  z } return y } 37    Theorem: 2 n for f  3 n n 1  5   (golden ratio) 2 Proof: Proof by (strong) induction   Inductive Basis: 3 4 n n  2   f 3    2 3 f 4 38 19

 n  Inductive Hypothesis: 3 k    Suppose it holds 2 n f n  k  Inductive Step: 1 n      We will prove for ( 1 ) k 4 f k k 1 39   x   is the solution to equation 2 1 0 x     2 1                   1 2 3 3 2 3 k k k k k ( 1 )            2 3 1 k k k f f f   1 1 k k k induction hypothesis End of Proof 40 20

Euclidean Algorithm for Greatest Common Divisor Recursive Basis:  gcd( , 0 ) a a  Recursive Step: gcd( , ) gcd( , mod ) a b b a b a  b 41 Recursive Euclidean algorithm for greatest common divisor gcd( ) { //assume a>b a , b if then  //recursive basis 0 b return a else //recursive step return gcd(b, mod ) a b } 42 21