Recursive Definitions Inductive Proofs Implementing Recursion file- evaluate all basic operations...

RecursionChapter 4 – Self-Reference

Recursive Definitions

Inductive Proofs

Implementing Recursion

© schmiedecke 09 In2-2-Algor.Paradigms 2

Imperative Algorithms

• Based on a basic abstract machine model

- linear execution model

- storage

- control structures

� von Neumann architecture

// Factorial

public int fac1 (int i) {

int result = 1;

for (int j=i; j>1; j--) result *= j;

return result;

}

i

result

j


Applicative Algorithms

• Based on the principle of inductive definition

• Evaluation pattern taken from mathematical function evaluation

• Recursion is a basic step

f(1) = 1

f(n) = n*f(n-1)// Factorial recursive

public int fac1 (int i) {

if (i<2) return 1;

return i*fac1(i-1);

}


Applicative Algorithms

• Idea is to apply a function to an argument

• so you need a function definition like

f(1) = 1

f(n) = n*f(n-1)

• and a method for evaluating that function for a given argument:

- Replace variable(s) by arguments

- Replace left hand side by right hand side

- Repeat until termination

So evaluation is described as a set of rewriting rules.


More Formally

• An applicative algorithm is a list of functions

f0(v01,...,v0n) = t0(v01,...,v0n)

...

fm(vm1,...,vmn) = tm(vm1,...,vmn)

function name function expression

function parameters

• A function evaluation for a given set of arguments

- replace the parameters in the function expression by the corresponding arguments

- evaluate all basic operations and functions in the function expression from left to right according to basic priorities

• The semantics of the applicative algorithm is the evaluation of the first function.


Notation

• To make functions conform the algorithm definition, we introduce the conditional operator:

• Instead of:

f(1) = 1

f(n) = n*f(n-1)

• We write:

f(n) = if n=1 then 1 else n*f(n-1) fi

• In Java, the conditional operator is (?:)

int f(int n) { return n==1 ? 1 : n*f(n-1); }


The Semantics of Recursive Functions

• is the successive application of rewriting rules:

fac(5) = 5 * fac(4)

= 5 * 4 * fac(3)

= 5 * 4 * 3 * fac(2)

= 5 * 4 * 3 * 2 * fac(1)

= 5 * 4 * 3 * 2 * 1


The Execution of Recursive Functions

• uses a stack to save "partial rewritings":

call fac(5) �� push 5, call method




call fac(1)�� push 1, call method

base case

contin.fac(2�� pop, pop, mult(2,1), push 2

contin.fac(3) �� pop, pop, mult(3,2), push 6

contin.fac(4) �� pop, pop, mult(6,4), push 24

contin.fac(5) �� pop, pop, mult(24,5),push 120


Recursion is Natural

• To learn downhill skiing

- climb up a few metres and learn to plow a right bend to the

bottom

- climb up a few metres and learn to plow a left bend to the

bottom

- take the lift all the way up and combine right and left bends until you reach the bottom

• Let's apply this technique to the Towers of Hanoi...


Towers of Hanoi

� For a stack of 1:move disc from start to dest

� For a stack of 2:move disc from start to auxmove disc from start to destmove disc from aux to dest

Call this: move 2 from start to dest using aux

� For a stack of 3:move 2 from start to aux using destmove disc from start to destmove 2 from aux to dest using start

source:http://www.cut-the-knot.org/recurrence/hanoi.shtml

Puzzle invented by Edouard Lucas

in 1883:

• Put an ordered stack of discs

from one peg to another, using athird peg,

• such that all stacks are ordered

at all times.


Inductive Aproach

• A stack of one can be moved directly to the target pole.

• Apparently, we can move a bigger stack of discs correctly to the target pole if we have an auxiliary pole. We have tested it for 2 and 3.

• Use this idea to first move the top of the stack to the auxiliary pole

• leaving the biggest disc behind to be moved directly to the target.

• move it to the target.

• Then move the top of the stack back to the origin, leaving the auxiliary pole empty again for the next step.

• ... until the stack size is 1.

• move the top of the stack is a recursive application of hanoi to a smaller stack!

• Parameters?


Express Hanoi in Java

public class Hanoi extends Applet {

private Tower start, aux, dest;

private int size;

public Hanoi () {

this.size = 8; // replace by reading from ComboBox..

start = new Tower(size);

aux = new Tower(0);

dest = new Tower(0);

}

public void start() { move(size, start, dest, aux); }

private void move(int number, Tower s, Tower d, Tower a) {

if (number==1) { d.add(s.remove()); repaint(); }

else { move(n-1,s,a,d); move(1,s,d,a); move(n-1,a,d,s); }

}

public void paint(Graphics g) {

start.display(g, size, 50, 10);

aux.display(g, size, 100, 10);

dest.display(g, size, 150, 10);

}

}


Properties of Hanoi

• Is it correct?

Theorem: if a disc is moved to a stack, it is smaller that ist base

Inductive proof:

� Any movement of a partial stack ends with one stack empty (Lemma1)

� if a disc is moved which is bigger than the top of a stack, it is moved to an empty stack (Lemma2)

- true for first step.

- true for second step.

- if a partial stack of n-1 discs has been moved properly, a biggerdisc is moved to an empty stack in the nth step.


Towers of Hanoi - Complexity


Complexity Argument


Recursion is Powerful

• It can be shown that every algorithm that can be specified

can be specified recursively!

• So, recursion has maximum computation potential.

• In most situations, you can replace recursion by iteration.

• Can you do so for Hanoi?

• Later we will learn about a certain type of complex

recursion which cannot be expressed iteratively (so-called

R Gramars).


Replacing Recursion by Iteration

• Lets consider two "famous" recusive algorithms:

• the factorial n!

fac(n) = if n=1 then 1 else n*f(n-1) fi

• and the Fibonacci numbers (rabbit generations)

fib(n) = if n<=1 then 1 else fib(n-1)+fib(n-2)



• the factorial n!

fac(n) = if n=1 then 1 else n*f(n-1) fi

int fac(int n) {

int result = 1;

for (int i=2; i<=n; i++) result *= i;

return result;

}

Solution is straightforward:

use local variable to store intermediate results

otherwise returned by recursive calls



• and the Fibonacci numbers (rabbit generations)

fib(n) = if n<=1 then 1 else fib(n-1)+fib(n-2)

int fib(int n) {int current = 1, prev = 0, temp;

for (int i=1; i<n; i++)

{ temp = current;

current = current + prev;

prev = temp;

}

return current;

}

Same idea,

use local variables to store values that

would result from recursive calls


The Termination Problem

• What is the domain of the factorial function?

fac(n) = if n=1 then 1 else n*fac(n-1) fi

int fac(int n) {int result = 1;

for (int i=2; i<=n; i++) result *= i;

return result;

}

• Oh dear, the applicative agorithm will run indefinitely for negative n.....

• Generally, algorithms must be suspected of being partial functions,

• i.e they may not terminate on certain arguments.

• It may be hard to find out, or even impossible....

• There are algorithms for which we do not know whether they terminate for all valid input data. (There is one in your lab assignments ☺)


Tail Recursion

• There is a "nice" type of recursion, called tail recursion:

• Such functions have a direct solution for a fixed set of arguments.

• For every recursive call, a monotonic modification is made to the parameters moving them "towards" the direct solution arguments.

• Tail recursion is easily translated into iteration.

• The termination of tail recursive functions can generally be determined easily (but not always).

• So, in programming, try to use tail recursion and make itobvious.


Recursion and Complexity

• How do you determine the complexity of a recursive function?

• Use recursive call as basic step.How much does it cost except for the rec. call?How many recursions occur?

the number of calls depends on the parameter value.

• Multipy step cost by recursion depth.

• Fac is called n times, each step has constant complexity, so the complexity is O(n):fac(n) = if n=1 then 1 else n*f(n-1) fi

• Fib is called 2*n times, so the complexity is O(n) too:fic(n) = if (n=1|n=2) then 1 else fib(n-1)+fib(n-2) fi

• Those were easy to assess, because the parameter is decreasing steadily. But what about this function:f(x) = if x>100 then x-10 else f(f(x+11)) fi


Incredible Growth:The Ackermann Function

• Here is the worst example ever known:

ack(x,y) = if x<=0 then y+1

else if y<=0 then f(x-1,1)

else f(x-1,f(x, y-1))

• Some function results

f(0,0) = 1f(1,0) = 2

f(1,1) = 3

f(2,2) = 7

f(3,3) = 61

f(4,4) = 2**2**2**2**2**2**2 –3 = 2**2**2**65536 –3

• As this growth is achieved by adding 1, the time consumption, i.e. the functions comlexity, is just as incredible


Mathematics of Applicative Programming:The Lambda Calculus

• Inventend by Church and Kleene in the 1930ies.

• Mathematical model of function application (to arguments)

• Computability question: Which problems can be solved algorithmically, and can it be decided whether a problem has a solution.

• Abstract machine for applicative algorithms: If it terminates on a given algorithm, it is said to be computable.


Lamda Calculus Simplified

• Lambda stands for argument binding:

• Function f applied to argument a results r:Lambda x f a � r

• If f is x+2, and a is 7 we getLambda x x+2 7� 9

• With nesting:Lambda x x+(Lambda y y+3 4) 5

� Lambda x x+7 5� 12

• So, computation is described as rewriting of one list of terms by another


The LISP language

• Describing computation as a set of rewriting rules has many offsprings

in artificial intelligence.

• Here is a very early rewriting language, invented in 1958 by John Mc

Carthy:

• LISP – List Programming

- The idea is that every program is a list of rewriting rules.

- The result of a rewriting is again a list.

- Terminal symbols can be specified to be excluded from

rewriting.

- Lambda binding is applied to introduce parameters.

- Special list operations cons, car and cdr allow list

manipulation and traversion.

• LISP is still in use – implemented in Unix, control language in the EMACS editior.


Some LISP

(DEFUN THROW-DIE ()

(+ (RANDOM 6) 1) )

(DEFUN THROW-DICE ()

(LIST (THROW-DIE) (THROW-DIE)) )

(DEFUN BOXCARS-P (A B)

(AND (EQUAL '6 A) (EQUAL '6 B) ) )

(DEFUN SNAKE-EYES-P (A B)

(AND (EQUAL '1 A) (EQUAL '1 B) ) )


Factorial in LISP

(DEFUN FAC (N)

(COND ((EQUAL N 1) 1)

((MULT N (FAC (MINUS N 1)) ) ) )

Evaluate Call of FAC (3) by Rewriting:

��FAC (3)

�MULT 3 (FAC (MINUS 3 1))

�MULT 3 (FAC (2))

�MULT 3 (MULT 2 (FAC (MINUS 2 1))

�MULT 3 (MULT 2 (FAC (1)))

�MULT 3 (MULT 2 1)

�MULT 3 2

�6


What's so Charming about LISP?

• Fully declarative – non-operational.

• Completely general.

• Easy to extend programs by adding definitions.

• Results can be added as (redundant) definitions,such as

(DEFUN FAC(3) 6)

• � Supports artifical "learning"


Recursive Fun:Fractals

• If you draw a simple shape repeatedly with decreasing

size and changing orientation, you get a pattern.

• The simplest way to do that is recursion:

- draw a pattern

- spawning a similar pattern of smaller size and modified position and orientation

- which again spawns another smaller pattern

- until the size is below a certain limit

• Such patterns are usually called fractals...

• And they are fun doing ☺


Two Fractal Classics

• Pythagoras Tree:- Draw a square starting from ist base line

- then, for a given angle, draw a pair of

Pythagoras squares opposite to its base line

- continue for each sqare until too small.

• Sierpinski's Triangle:- Take a line between two points

- Replace it by three lines of half that length,

turning -60°, +60°,+60° left

- Replace each of these lines again by three,

turning +60°, -60°, -60° for the first line,

-60°, +60°, +60° for the middle line,

+60°, -60°, -60° for the last line.

- Repeat for each line, changing orientation every time

until too small


Famous simple Fractals

Pythagoras Fractal Szerpinsky Triangle


Pythagoras

public void paint (Graphics g) {

g.clearRect(0,0,500,500);

paintTree(g, double x1, double y1, // tree defined

double x2, double y2); // by base line

}

// paint tree paints the major square

// determines the base lines of the smaller squares

// and calls itself recursively for the two smaller squares


Everybody loves

Mandelbrodt Fractals.

Have Fun...

That's enough for today!!

Next time, we will return to recursion toconsider the most general problem solver:

Recursive Backtracking,

Before Jumping into the Tradition of Sorting and Searching

Recursive Definitions Inductive Proofs Implementing Recursion file- evaluate all basic operations...

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