1. Herbert G. Mayer, PSU CS status 7/29/2013
CS 410 Mastery in Programming Chapter 7 Hints for: Symbolic Differentiation
Herbert G. Mayer, PSU CS
status 7/29/2013

2. Syllabus Problem Assessment Rules of First Derivative
Data Structures and Types
Make Node
Copy Tree
Print Tree
Build Tree
Simplify
Main
References

3. Problem Assessment
Goal of this assignment is to design, code, and execute symbolic differentiation of mathematical formulae or equations
These formulae are simple, meaning:
No partial differentiation
Differentiation only w.r.t a single variable
And that variable is pre-defined to be ‘x’
All numeric constants are integer, and of small value, i.e. single decimal digits; but leave “room” for adding up larger integer values
Goal was not to focus on scanning, i.e. not on lexical analysis
All variable names different from ‘x’ are constant w.r.t. differentiation toward ‘x
If two derivatives are needed, then the output of the first can serve as input to the second differentiation
So all formulae are simple, yet the mathematical problem remains general and highly interesting

4. Problem Assessment
When applying the rules of differentiation, the result can be a formula with ample redundancies, e.g. +0, *1 etc.
In such cases it is advised to simplify the result
A careful programmer will actually submit the originally input formula f(x) to the same simplification algorithm
Each input formula is terminated by a special symbol, the ‘$’ character
For example:
Input f(x) = 2*x+x*3+5+6*x$
Normalized input f(x) = ((((2*x)+(x*3))+5)+(6*x)) $
First derivative output f’(x) = (((((0*x)+(2*1))+((1*3)+(x*0)))+0)+((0*x)+(6*1)))
Simplified output: 11

5. Rules of First Derivative Toward x Function references u and v below are functions of x, i.e. u(x) and v(x) the ‘ operator refers to the first derivative
f(x)
f’(x)
f(x) = some variable != x
f’(x) = 0
f(x) = integer constant
f(x) = x
f’(x) = 1
f(x) = u + v
f’(x) = u’ + v’
f(x) = u - v
f’(x) = u’ - v’
f(x) = u * v
f’(x) = u’*v + u*v’
f(x) = u / v
f’(x)= (u’*v -u*v’) / v2
f(x) = ln(u) = & u
f’(x) = u’ / u
f(x) = u ^ v
f’(x) = u' * v * u ^ ( v - 1 ) + & u * v' * u ^ v
Example: f(x) = x ^ x
f’(x) = 1 * x * x^(x–1) + & x * 1 * x^x
= x^x + & x * x^x
= (& x + 1) * x^x

6. Data Structures and Types
Element of symbolic-differentiation is a node
Each mathematical function f(x) is represented internally as a tree, pointed to by “root”
Root’s type is pointer to structure of node type
Any node is either:
A literal with a stored integer value - single decimal digit for now!
A variable, which could either be the select variable ‘x’ or some other
An operator, with the actual operator itself being stored, like ‘+’ ‘*’ ‘/’ ...
Each node has all the following fields
Class, specifying enumeration type { literal, variable, or operator }
The single character Symbol, e.g. ‘x’ for the special variable
The integer literal value LitVal to remember the integer value
And node pointers Left and Right to the respective subtrees, if any
It will be handy to store special nodes just once; resulting in a DAG

7. Data Structures and Types
// each node has 1 of these class states:
// a Literal, an Identifier (for variable), or an Operator.
// Parenthesized expressions have been reduced
typedef enum { Literal, Identifier, Operator } NodeClass;
typedef struct NodeType * NodePtr; // forward announcement
// now comes the actual node-type structure,
// using the forward declared pointer type: NodePtr
typedef struct NodeType {
NodeClass Class; // 1 of the 3 classes.
char Symbol; // store: Identifier, Operator
int LitVal; // if Literal, this is its value
NodePtr Left; // subtree
NodePtr Right; // subtree
} s_node_tp;

8. Make Node // malloc() new node from heap. All fields are passed in;
// return the pointer to the new node to caller
NodePtr Make( NodeClass Class, char Symbol, int value,
NodePtr Left, NodePtr Right )
{ // Make
NodePtr Node = (NodePtr)malloc( sizeof( struct NodeType ) );
ASSERT( ... node’s space is really there ... );
Node->Class = Class;
Node->Symbol = Symbol;
Node->LitVal = value;
Node->Left = Left;
Node->Right = Right;
return Node;
} //end Make

9. Copy Tree // recursively copy tree pointed to by Root.
// Pass pointer to the copy to its caller
NodePtr Copy( NodePtr Root )
{ // Copy
if ( Root == NULL ) {
return NULL;
}else{
return Make( Root->Class, Root->Symbol,
Root->LitVal,
Copy( Root->Left ),
Copy( Root->Right ) );
} //end if
} //end Copy

10. Print Tree void PrintTree( NodePtr Root ) { // PrintTree
if ( Root != NULL ) {
if ( Root->Class == Operator ) {
printf( "(" );
} //end if
PrintTree( Root->Left );
if ( Root->Class == Literal ) {
printf( "%d", Root->LitVal ); // prints ints > 9
}else{
printf( "%c", Root->Symbol );
PrintTree( Root->Right );
printf( ")" );
} //end PrintTree

11. Build Tree for Expression()
Expression : Term { plus_op Term }
plus_op : ‘+’ | ‘-’
Term : Factor { mult_op Factor }
mult_op : ‘*’ | ‘/’
Factor : Primary { ‘^’ Primary }
Primary : IDENT | LITERAL
| ‘(‘ Expression ‘)’
| ‘&’ Primary // for ln()

12. Build Tree Expression()
// parse expression and build tree
// using Term() and higher priority functions/ops
// all returning pointers to nodes
// in Expression() handle ‘+’ and ‘-’ operators
NodePtr Expression()
{ // Expression
char Op; // remember ‘+’ or ‘-’
NodePtr Left = Term(); // handle all higher prior.
while ( NextChar == ‘+’ || ( NextChar == ‘-’ ) {
Op = NextChar; // remember ‘+’ or ‘-’
GetNextChar(); // skip Op
// note 0 below for LitVal is just a dummy
Left = Make( Operator, Op, 0, Left, Term() );
} //end while
return Left;
} //end Expression

13. Build Tree Term() // multiply operators ‘*’ and ‘/’, later add ‘%’
NodePtr Term( )
{ // Term
char Op; // remember ‘*’ or ‘/’
NodePtr Left = Factor();
while ( NextChar == ‘*' || NextChar == ‘/' ) {
Op = NextChar; // remember ‘*’ or ‘/’
GetNextChar(); // skip over Op
// note 0 below for LitVal is just a dummy
Left = Make( Operator, Op, 0, Left, Factor() );
} //end while
return Left;
} //end Term

14. Build Tree Factor() Left-Assoc.
// exponentiation operator ‘^’ left-associatively
NodePtr Factor()
{ // Factor
NodePtr Left = Primary();
while ( NextChar == ‘^’ ) {
GetNextChar(); // skip over ‘^’
Left = Make( Operator, ‘^’, 0, Left, Primary() );
} //end while
return Left;
} //end Factor
// Think about left- versus right-associativity!!!
// How would you change the code –-and grammar—
// if indeed you make ‘^’ right associative?

15. Build Tree Factor () Right-Assoc.
// exponentiation operator ‘^’ right-associative
NodePtr Factor()
{ // Factor
NodePtr Left = Primary();
if ( NextChar == ‘^’ ) {
GetNextChar(); // skip over ‘^’
Left = Make( Operator, ‘^’, 0, Left, Factor() );
} //end if
return Left;
} //end Factor
// now multiple ^ operators are handled right-to-left
// in line with common precedence of exponentiation

16. Build Tree NodePtr primary( ) { // primary
char Symbol = NextChar; // first_set = { ‘(‘, ‘&’, IDENT, LIT }
NodePtr Temp;
GetNextChar(); // skip over current Symbol
if ( IsDigit( Symbol ) ) {
// end node: don’t recurse
return Make( Literal, Symbol, (int)(Symbol-'0’), NULL, NULL );
}else if ( IsLetter( Symbol ) ) {
// also end node: don’t recurse
return Make( Identifier, tolower( Symbol ), 0, NULL, NULL );
}else if ( ‘(‘ == Symbol ) {
Temp = Expression();
Must_Be( ‘)’ );
return Temp;
}else if ( Symbol == '&' ) {
return Make( Operator, '&', 0, NULL, primary() );
}else{
printf( "Illegal character '%c'.\n", Symbol );
return NULL;
} //end if
// impossible to reach! Hence check Herb!!
} //end primary

17. Derive, 1 NodePtr Derive( NodePtr Root ) { // Derive
if ( NULL == Root ) {
return NULL;
}else{
switch ( Root->Class ) {
case Literal:
return Make( Literal, '0', 0, NULL, NULL );
case Identifier:
if ( ( Root->Symbol == 'x' ) || ( Root->Symbol == 'X' ) ) {
return Make( Literal, '1', 1, NULL, NULL );
} //end if
case Operator:
switch ( Root->Symbol ) {
case '+': case '-':
return Make( Operator, Root->Symbol, 0,
Derive( Root->Left ), Derive( Root->Right ) );
. . .

18. Derive, 2 . . . case '*': return Make( Operator, '+', 0,
Make( Operator, '*', 0, Derive( Root->Left ),
Copy( Root->Right ) ),
Make( Operator, '*', 0, Copy( Root->Left ),
Derive( Root->Right ) ) );
case '/':
return Make( Operator, '/', 0,
Make( Operator, '-', 0,
Derive( Root->Right ) ) ),
Make( Operator, '*', 0,
Copy( Root->Right ), Copy( Root->Right ) ) );
} //end switch

19. Derive, 3
Students write code for derive() of ‘^’

20. Derive, 3 case '^': return Make( Operator, '+', 0,
Derive( Root->Left ),
Copy( Root->Right ),
Make( Operator, '^', 0,
Copy( Root->Left ),
Make( Operator, '-', 0,
Copy( & OneNode ) ) ) ) ),
Make( Operator, '&', 0,
NULL,
Copy( Root->Left ) ),
Derive( Root->Right ) ),
Copy( Root->Right ) ) ) );
case '&':
if ( Root->Left != NULL ) {
printf( "ln has only one operand.\n" );
} //end if
// now do the work ...

21. Opportunities for Simplification#
Original expression
Simplified expression 1
x + 0 x 2
0 + x 3
x - 0 4
x - x 5
x * 0 6
0 * x 7
x * 1 8
1 * x 9
x / x 10
x / 1 11
x ^ 1 12
x ^ 0 13
1 ^ x 14
& 1

22. Simplify, 1 NodePtr Simplify( NodePtr Root ) { // Simplify
int val = 0; // accumulate integer values from + - * etc.
if ( !Root ) {
return Root;
}else{
switch ( Root->Class ) {
case Literal:
case Identifier:
case Operator:
Root->Left = Simplify( Root->Left );
Root->Right = Simplify( Root->Right );
switch ( Root->Symbol ) {
case '+':
if ( IsLit( '0', Root->Left ) ) {
return Root->Right;
}else if ( IsLit( '0', Root->Right ) ) {
return Root->Left;
}else if ( BothLit( Root->Left, Root->Right ) ) {
val = Root->Left->LitVal + Root->Right->LitVal;
return Make( Literal, (char)( val + '0' ), val,
NULL, NULL );
return Root; // no other simplifiction for ‘+’
} //end if
. . .

23. Simplify, 2 case '-': if ( IsLit( '0', Root->Right ) ) {
return Root->Left;
}else if ( BothLit( Root->Left, Root->Right ) ) {
val = Root->Left->LitVal - Root->Right->LitVal;
return Make( Literal, (char)( val + '0' ), val, NULL, NULL );
}else if ( IsEqual( Root->Left, Root->Right ) ) {
return & NullNode;
}else{
return Root;
} //end if
case '*':
if ( IsLit( '1', Root->Left ) ) {
return Root->Right;
}else if ( IsLit( '1', Root->Right ) ) {
}else if ( IsLit( '0', Root->Left ) || IsLit( '0', Root->Right ) ) {
}//end if
case '/':
if ( IsLit( '1', Root->Right ) ) {
}else if ( IsLit( '0', Root->Left ) ) {
return & OneNode;
case '^':
if ( IsLit( '0', Root->Right ) ) { // x^0 = 1
}else if ( IsLit( '1', Root->Right ) ) { // x^1 = x
}else if ( IsLit( '1', Root->Left ) ) { // 1^x = 1
. . .

24. Two Equal Trees Students write code for:
bool IsEqual( NodePtr Left, NodePtr Right )

25. Two Equal Trees
// return true only if both subtrees left and right are equal
bool IsEqual( NodePtr Left, NodePtr Right )
{ // IsEqual
if ( ( !Left ) && ( !Right ) ) {
return TRUE;
}else if ( NULL == Left ) {
// Right is known to be not NULL
return FALSE;
}else if ( NULL == Right ) {
// Left is known to be NOT NULL
}else if ( ( Left->Class == Literal ) && ( Right->Class == Literal ) ) {
return ( Left->LitVal ) == ( Right->LitVal );
}else if ( ( Left->Class == Identifier ) && ( Right->Class == Identifier )){
return ( Left->Symbol ) == ( Right->Symbol );
}else{
// must be Operator; same?
if ( ( Left->Symbol ) == ( Right->Symbol ) ) {
// IsEqual yields true, only if both subtrees are equal
return ( IsEqual( Left->Left, Right->Left ) &&
IsEqual( Left->Right, Right->Right ) ) ||
( is_associative( Left->Symbol ) &&
IsEqual( Left->Left, Right->Right ) &&
IsEqual( Left->Right, Right->Left ) );
} //end if
printf( "Impossible to reach in IsEqual.\n" );
} //end IsEqual

26. main() int main () { // main: Differentiation NodePtr root = NULL;
Initialize();
root = Expression();
VERIFY( ( NextChar == '$' ), "$ expected, not found\n" );
SHOW( " original f(x) = ", root );
root = Simplify( root );
SHOW( " Simplified f(x) = ", root );
root = Derive( root );
SHOW( " derived f'(x) = ", root );
SHOW( " reduced f'(x) = ", root );
Or else:
print_tree( simplify( derive( simplify( expression( root )))));
return 0;
} //end main: Differentiation

27. References
Differentiation rules, implementation code samples: on.aspx
More code samples in Lisp: text/sicp/book/node39.html