1. CS 163 Data Structures Chapter 10 Symbolic Differentiation
Herbert G. Mayer, PSU
Status 6/11/2015

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 redundancies, e.g. +0, *1, /1 etc.
In such cases: advisable to simplify the result
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 symbolizes 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 binary 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, stored as a single character, like ‘+’ ‘*’ ‘/’ ...
Each node has all the following fields, sometimes NOT needed:
Class, specifying enumeration type { Literal, Identifier, Operator }
The single character Symbol, e.g. ‘x’ for the special variable
The integer literal value LitVal to remember the integer value, e.g. 7
And node pointers Left and Right to the respective subtrees

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.
// return a pointer to the copy to caller
NodePtr Copy( NodePtr Root ) // clever code!!!
{ // Copy
if ( NULL == Root ) {
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()
Quick BNF Intro (Backus Naur Form)
Grammar is a set of rules, defining a language
Rules define nonterminals via terminals, nonterminals, or special symbols (like ‘+’ or ‘;’ or ‘/’)
Left side nonterminal : is defined by right side
One nonterminal is the start symbol, here Expression
: operator separates left from right side
| introduces another alternative on r.h.s.
{ and } group 0 or more phrases on r.h.s.
Special symbols are numeric literals (e.g. 6), identifiers (e.g. y), and char and string literals (e.g. ‘#’)

12. Build Tree for Expression()
Make C program from BNF Grammar
For each left nonterminal define a suitable C function by that nonterminal name
For each nonterminal used, call that C function
For each terminal used in a phrase in a non-first position, require that symbol, i.e. error if not found
For each terminal at the start of an alternative, see if this terminal is in the source, then enter that alternative, else find another alternative
Insert needed semantic actions

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

14. 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 ‘+’ or ‘-’
// note 0 below for LitVal is just a dummy
Left = Make( Operator, Op, 0, Left, Term() );
} //end while
return Left;
} //end Expression

15. 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

16. 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?

17. 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

18. Build Tree: Primary() 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! No need to check Herb!!
} //end Primary

19. Derive, 1 // Real action: Derive(Root) derives tree pointed to by Root
// First left, then right subtree
// When done, focus on the Root node
// Both u and v are f(x)
// derive( x ) = derive x -> 1
// derive( a ) = any variable derived except x -> 0
// derive( # ) = any number derived -> 0
// derive( u + v ) = u' + v' where u = f(x) and v = g(x)
// derive( u - v ) = u' - v'
// derive( u * v ) = u' * v + u * v'
// derive( u / v ) = u' * v - u * v' / ( v * v)
// derive( u ^ v ) = u' * v * u ^ ( v - 1 ) + ln u * v' * u ^ v
// derive( ln u ) = derive( & u ) = u' / u

20. Derive, 2 NodePtr Derive( NodePtr Root ) { // Derive
if ( Root == NULL ) {
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 ) );
case '*':
return Make( Operator, '+', 0, . . Next page

21. Derive, 3 case '*': return Make( Operator, '+', 0,
Derive( Root->Left ), Copy( Root->Right ) ),
Copy( Root->Left ), Derive( Root->Right ) ) );
case '/':
return Make( Operator, '/', 0, Make( Operator, '-', 0,
Copy( Root->Left ), Derive( Root->Right ) ) ),
Copy( Root->Right ), Copy( Root->Right ) ) );
case '^':
. . . Next page

22. Derive, 4 case '^': return Make( Operator, '+', 0,
Make( Operator, '*', 0, Derive( Root->Left ),
Make( Operator, '*', 0, Copy( Root->Right ),
Make( Operator, '^', 0, Copy( Root->Left ),
Make( Operator, '-', 0, Copy( Root->Right ),
Copy( & OneNode ) ) ) ),
Make( Operator, '*', 0,
Make( Operator, '&', 0, NULL, Copy( Root->Left ) ),
Derive( Root->Right ) ),
Make( Operator, '^', 0,
Copy( Root->Left ), Copy( Root->Right ) ) );
case '&': Next page

23. Derive, 5 case '&': if ( Root->Left != NULL ) {
printf( "ln has only one operand.\n" );
} //end if
return Make( Operator, '/', 0,
Derive( Root->Right ), Copy( Root->Right ) );
default:
printf( "Impossible operator.\n" );
return NULL;
} //end switch Root->Symbol
printf( "Unknown Root->Class\n" );
} //end switch Root->Class

24. 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

25. 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
. . .

26. 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
. . .

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

28. 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

29. 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

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