Presentation Number 6

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Presentation Number 6

• C Programming Course

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Last Time in Class

• float mat33
• mat02 5.0f
• mat00 mat10 4.0f

Multi-dimensional arrays
mat
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Functions

• Function definition
• return-type name(arg_type1 arg_name1, arg_type2
  arg_name2, )
• function body
• return value
• Return statement.
• Void type.
• Call by-value.

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Functions

• Function declaration
• return-type name(arg_type1 arg_name1, arg_type2
  arg_name2, )

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Questions?
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Assertions

• include ltassert.hgt
• include ltstdio.hgt
• int f( int a, int b )
• int g( int c )
• int main( void )
• int a 0, b 0, c 0
• ....
• scanf( dd, a, b )
• .....
• c f( a,b )
• assert( c gt 0 )
• .....

An assertion makes sure that the statement
evaluates to true.
If the assertion evaluates to false it aborts the
run of the program and prints a message to the
screen.
an assertion
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Assertions

• Assertions are useful when debugging an
  application.
• They require the inclusion of ltassert.hgt
• Assertions work only in debug mode.
• They will not have no effect in an optimized
  version of the application.
• You can use them when debugging your homework ...
  However do not leave them in the submitted code.

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Thinking About Algorithms
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Advanced Calculator Explained

• Input
• Output 8
• Assuming that we solved the problem for part of
  the array and we are now working on the ith
  element ... which part of the array do we want to
  have solved?
• Do we want to have the length of the sequence for
  elements i1 to n-1 or elements 0 to i-1?

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Elements i1 to n-1?

• Elements
• Lengths
• Can I use information about the next elements to
  calculate the length of the sequence for the
  current element?
• Lengths stores the length of the sequence that
  begins at element j.
• If (element j gt element i) then length i 1
  length j

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Solution

• Define an additional array that will store the
  sequence lengths for each element (initialized to
  1).
• Solve the problem for element i (with i running
  from n-1 to 0).
• Compare element i to element j (with j running
  from i1 to n-1).
• Length i equals maxlength i, 1 length j.

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Elements 0 to i-1?

• Elements
• Lengths
• Can I use information about the previous elements
  to calculate the length of the sequence for the
  current element?
• Lengths stores the length of the sequence that
  ends at element j.
• If (element j lt element i) then length i 1
  length j

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Solution

• Define an additional array that will store for
  each element (initialized to 1) the length of the
  sequence that ends at that element.
• Solve the problem for element i (with i running
  from 0 to n-1).
• Compare element i to element j (with j running
  from 0 to i-1).
• Length i equals maxlength i, 1 length j.

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Recursions
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Recursion

• Definition 1
• Recursion - a method for defining functions in
  which the function being defined is used within
  its own definition.
• Definition 2
• Recursion - see recursion.
• Definition 3
• Recursion If you still dont get it See
  Recursion.

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Recursion

• n! 1 2 3 ... (n-1) n
• This is equivalent to
• n! (n-1)! n
• Thus, we can also define factorial to be
• 0! 1 n 0
• n! (n-1)! n n gt 0

This is a recursive definition
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A recursive definition

• C functions can call themselves!
• When a function calls itself we call this a
  recursive calls.
• Two things you must remember about recursions
• Change the parameters from call to call!!!
• Make sure you defined a boundary condition!!!

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Factorial Example
reucrsive version

• int factorial ( int n )
• if ( n lt 1 )
• return 1
• else
• return ( n factorial( n 1 ) )
• int factorial( int n )
• int product 1
• for( n gt 1 --n )
• product n
• return product

boundary condition
recursive call
iterative version
RecFactorial.c
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Another Way To Think About It

• int factorial ( int n )
• if ( n lt 1 )
• return 1
• else
• return ( n factorial( n 1 ) )
• In other words ... I know how to solve a simple
  problems.
• Let the function deal with the complicated
  problems.

I know how to solve this factorial
I know that n! n (n-1)! Let the function
solve the (n-1)!
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Examples
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Write Backwards

• include ltstdio.hgt
• void wrt_back()
• int main(void)
• printf( Input a line )
• wrt_back()
• printf( \n\n )
• return 0
• void wrt_back()
• int c 0
• if ( ( c getchar() ) ! \n )
• wrt_back()
• putchar( c )

Call order
Print order
RecReverse.c
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Recursive Power

• double power( double val, unsigned int pow )
• if ( pow 0 )
• return( 1.0 )
• else
• return( power( val, pow 1 ) val )

pow( X, 0 ) returns 1
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Fibonacci Number

• int fib( int num )
• switch( num )
• case 0
• return(0)
• break
• case 1
• return(1)
• break
• default
• return( fib( num 1 ) fib( num 2 ) )
• break

the recursive calls
Fibonacci.c
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Trace Fibonacci
In this implementation the function calls itself
twice
Recursion can consume a lot of memory (possible
stack overflow) and are therefore considered
inefficient.
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Iterative Solution

• int fib( int num )
• int i 0, fn 0, fn_1 1, fn_2 2
• if ( num 0 )
• return 0
• if (num 1)
• return 1
• for ( i 2 i lt num i )
• fn fn_2 fn_1
• fn_2 fn_1
• fn_1 fn
• return fn

Recursive functions are sometimes the simplest
and shortest answers
There is always an alternative non-recursive
solution available
Fibonacci.c
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Hanoi Towers
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Mutually Recursive Functions(Hanoi Towers)

• include ltstdio.hgt
• void move_right(int id)
• void move_left(int id)
• void move_right(int id)
• if ( 0 id ) return
• move_left( id 1 )
• printf( Shift disc d one peg to right.\n, id
  )
• move_left(id - 1)

move id 1 discs one peg to the left
move id 1 discs one peg to the left
Hanoi.c
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Mutually Recursive Functions(Hanoi Towers)

• void move_left( int id )
• if ( 0 id ) return
• move_right( id 1 )
• printf( Shift disc d one peg to left.\n, id
  )
• move_right( id 1 )
• int main( void )
• move_left( 3 )
• return 0

move id 1 discs one peg to the right
move id 1 discs one peg to the right
solve the 3 disc problem
Hanoi.c