CSCI 2210 - Programming in Lisp; Instructor: Alok Mehta; 4.ppt1 CSCI 2210: Programming in Lisp ANSI Common Lisp, Chapters 5-10

CSCI 2210 - Programming in Lisp; Instructor: Alok Mehta; 4.ppt

1

CSCI 2210: Programming in Lisp

ANSI Common Lisp, Chapters 5-10


2

Progn

Progn Creates a block of code Expressions in body are evaluated Value of last is returned

(if (< x 0) (progn (format t "X is less than zero ") (format t "and more than one statement ") (format t "needs to be executed in the

IF") (- x) ))

Prog1 Same as progn, except value of first expression is returned


3

Block

Like a progn with A name An "emergency exit"

– return-from - Returns from a named block– return - Returns from a block named NIL

Examples> (block head (format t "Here we go") (return-from head 'idea) (format t "We'll never see this"))

Here we go.IDEA

> (block nil (return 27))27


4

Implicit use of Blocks

Some Lisp constructs implicitly use blocks All iteration constructs use a block named NIL (note return)

> (dolist (x '(a b c d e)) (format t "~A " x) (if (eql x 'c) (return 'done)))

A B CDONE

Defun uses a block with same name as the function> (defun foo () (return-from foo 27))


5

Iteration

DOTIMES (Review)(dotimes (<counter> <upper-bound> <final-result>) <body>)

– Example(dotimes (i 5) (print i)) ;; prints 0 1 2 3 4

DOLIST(dolist (<element> <list-of-elements> <final-

result>) <body>)

– Example(dolist (elem '(a b c d)) (print elem)) ;; prints

a b c d


6

Example of DOLIST

Given a list of ages of people, how many adults?– List of ages

> (setf ages '(3 4 17 21 22 34 2 7))– Adult defined as: >= 21 years old

> (defun adultp (age) (>= age 21))– Using Count-if

(defun count-adult (ages) (count-if #'adultp ages))

– Using dolist(defun count-adult (ages &aux (nadult 0)) (dolist (age ages nadult) (if (adultp age) (setf nadult (+ 1

nadult)))))


7

Example (cont.)

Get the ages of the first two adults(defun first-two-adults (ages &aux (nadult 0) (adults nil)) (dolist (age ages) (if (adultp age) (progn (setf nadult (+ nadult 1)) (push age adults) (if (= nadult 2) (return adults))))))

Notes PROGN (and PROG1) are like C/C++ Blocks { … }

> (prog1 (setf a 'x) (setf b 'y) (setf c 'z))X

> (progn (setf a 'x) (setf b 'y) (setf c 'z))Z

RETURN exits the DOLIST block– Note: does not necessarily return from the procedure!– Takes an optional return value


8

DO

DO is more general than DOLIST or DOTIMES Example

(defun do-expt (m n) ;; Return M^N (do ((result 1) ;; Bind variable ‘Result’ to 1 (exponent n)) ;; Bind variable ‘Exponent’ to N ((zerop exponent) result) ;; test and return

value (setf result (* m result)) ;; Body (setf exponent (- exponent 1)) ;; Body

Equivalent C/C++ definitionint do_expt (int m, int n) { int result, exponent; for (result=1,exponent=n; (exponent != 0); ) { result = m * result; exponent = exponent - 1; } return result;}


9

DO Template

Full DO Template (There is also a DO*)(DO ( (<p1> <i1> <u1>) (<p2> <i2> <u2>) … (<pN> <iN> <uN>) ) ( <term-test> <a1> <a2> … <aN> <result> ) <body> )

Rough equivalent in C/C++for ( <p1>=<i1>, <p2>=<i2>,…,<pN>=<iN>; //Note: Lisp=parallel !(<term-test>); // C/C++ has a “continuation-test” <p1>=<u1>, <p2>=<u2>,…,<pN>=<uN>) { <body>}<a1>; <a2>; … ; <aN>;<result>; // Note: (DO…) in Lisp evaluates to <result>// Note: <p1>,<p2>,…,<pN> are now restored to original values


10

Do-expt, Loop

Here is another (equivalent) definition of do-expt(defun do-expt (m n) (do ((result 1 (* m result)) (exponent n (- exponent 1))) ((zerop exponent) result) ;; Note that there is no body! ))

Loop An infinite loop, terminated only by a (return)

(loop (print '(Say uncle)) (if (equal (read) 'uncle) (return)))


11

Multiple Values

We said each lisp expression returns a value Actually, can return zero or more values (i.e. multiple values)

> (round 7.6) ; Round returns two values8-0.4

> (setf a (round 7.6)) 8 ; Setf expects only one value, so it takes

the first (8)> a

8

How can you get all values?> (multiple-value-list (round 7.6))

(8 -0.4)> (multiple-value-setq (intpart dif) (round 7.6))

8> intpart

8> dif

-0.4


12

Generating Multiple Values

How? Use values> (values 1 2 3) ; Generates 3 return values

123

> (defun uncons (aList) (values (first aList) (rest aList)))

UNCONS> (uncons '(A B C))

A(B C)

> (format t "Hello World")"Hello World"NIL

> (defun format-without-return (out str &rest args) (apply #'format out str args) (values)) ; A function with no return value!

FORMAT-WITHOUT-RETURN> (format-without-return t "Hello")

Hello>


13

Functions about functions

FBOUNDP - Is a symbol the name of a function?> (fboundp '+)

T

SYMBOL-FUNCTION - returns function> (symbol-function '+)

#<Compiled-Function + 17B4E>> (setf (symbol-function 'sqr) #'(lambda (x) (* x x)))

#<Interpreted-Function SQR>> (defun sqr (x) (* x x)) ; this is equivalent to

aboveSQR

Defun and symbol-function define global functions Local functions can be defined using LABELS

> (defun add3 (x) (labels ((add1 (x) (+ x 1)) (add2 (x) (+ x 2))) (add1 (add2 x))))


14

Closures

Want a function that can save some values for future access/update Can use global variables for this

> (setf *number-of-hits* 0) ; Global variable0

> (defun increment-hits () (incf *number-of-hits*))INCREMENT-HITS

> (increment-hits)1

> (increment-hits)2

But, what if someone clobbers the variable?> (setf *number-of-hits* "This variable is now a string")

"This variable is now a string"> (increment-hits)

Error: '*number-of-hits*' is not of the expected type: NUMBER

Want something like "static" variables in C/C++.


15

Closures (continued)

One use of closures is to implement 'static' variables safely Closure -- Combination of a function and environment

– Environment can include values of variables> (let ((number-of-hits 0)) (defun increment-hits () (incf number-of-

hits)))INCREMENT-HITS

–number-of-hits is a free variable– It is a lexical variable (has scope)– A function that refers to a free lexical variable is called a closure

Multiple functions can share the same environment> (let ((number-of-hits 0)) (defun increment-hits() (incf number-of-hits)) (defun reset-hit-counter() (setf number-of-

hits 0))


16

Closures (cont)

Example 1> (defun make-adder (n) ; returns function to add n to x #'(lambda (x) (+ x n)))> (setf add3 (make-adder 3)) ; returns function to add 3

#<Interpreted function C0EBF6>> (funcall add3 2) ; Call function add3 with argument 2

5> (setf (symbol-function 'add3-b) (make-adder 3))> (add3-b 5)

8

Example 2 - Returns an "opposite" function > (defun our-complement (f) #(lambda (&rest args) (not (apply f args))))> (mapcar (our-complement #'oddp) '(1 2 3 4))

(NIL T NIL T)

Tip of the iceberg in terms of possibilities


17

Trace

Format(trace <procedure-name>)

Example> (defun factorial (x) (if (<= x 1) 1 (factorial (-

x 1))))> (trace factorial)

Causes entry, exit, parameter, and return values to be printed– For EACH procedure being traced

> (factorial 3); 1> FACTORIAL called with arg: 3; | 2> FACTORIAL called with arg: | 2; | | 3> FACTORIAL called with arg: | | 1; | |3< FACTORIAL returns value: | | 1; | 2< FACTORIAL returns value: | 2; 1< FACTORIAL returns value: 66

Untrace stops tracing a procedure: (untrace <procedure-name>)


18

Print, Read

Print Example

> (print '(A B))

Evaluates a single argument– Can be a constant (ex. 100), a variable (x), or any single expression

Prints its value on a new line Returns the value printed

Read - reads a single expression Example:

> (read)20 23 (A B) C20 ;; Note: Only the 20 is read; rest is ignored> (progn (print 'Enter-temperature) (read))Enter-temperature 2020


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Format

Format Allows more elegant printing

> (progn (print "Enter temperature: ") (read))"Enter temperature: " 3232

> (progn (format t "~%Enter temperature: ") (read))Enter temperature: 3232

– The second parameter (t) is the output buffer (T=stdout)– The character “~” signifies that a control character follows– The character “%” signifies a newline (Lisp: “~%” C: “\n”)– The characters “~a” tells Lisp to substitute the next value

printf ("The value is ( %d, %d )", x, y); /* A C stmt */> (format t "The value is ( ~a, ~a )" x y) ;; Lisp way> (format t "The value is ( ~10a, ~a )" x y) ; Can get

fancy


20

Streams

Write output to a file (e.g. knowledge base) Prototype of with-open-file

(with-open-file (<stream name> <file specification> :direction <:input or :output>) …)

Example> (setf fact-database '((It is raining) (It is pouring) (The old man is snoring)))> (with-open-file (my-file ”myfile.lsp” :direction

:output) (print fact-database my-file))


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My-Trace-Load

Redefine the built-in Lisp “Load” function Example of Built In function

> (load "a.lsp")

Additional requirements– Want to print each expression that is read in.– Want to print the value returned by the expression

Definition(defun my-trace-load (filename &aux next-expr next-

result) (with-open-file (f filename :direction :input) (do ((next-expr (read f nil) (read f nil))) ((not next-expr)) (format t "~%~%Evaluating '~a'" next-expr) (setf next-result (eval next-expr)) (format t "~% Returned: '~a'" next-

result))))


22

Read-Line, Read-Char

Read-Line Reads an individual line (terminated by a carriage return) Returns it as a character string

> (read-line)Hello World“Hello World”

Read-Char Reads an individual character Returns it as a character

> (read-char)x#\x ;; This is Lisp notation for the character ‘x’


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Symbols

We've used symbols as names for things A symbol is much more than just a name

Includes: name, package, value, function, plist (Property List) A symbol is a "substantial object"

Some functions for manipulating symbols symbol-name, symbol-plist, intern,…

Details are in Chapter 8 of textbook


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Symbol Names

By default Lisp symbol names are upper case> (symbol-name 'abc)

"ABC"

Can use "|" to delimit symbol names> (list '|abc|) ; no conversion

"abc"> (list '|Lisp 1.5| '|| '|abc| '|ABC|)

(|Lisp 1.5| || |abc| ABC) ; Note that only ABC doesn't have delimiter (default)

Intern creates new symbols> (set (intern '|5.1/5)|) 999)

999> |5.1/5)|

999


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Packages

Package A name space for symbols

Large programs use multiple packages> (defpackage "MY-APPLICATION" ;; Creates new package (:use "COMMON-LISP" "MY-UTILITIES") ;; Other packages (:nicknames "APP") (:export "WIN" "LOSE" "DRAW")) ;; Symbols exported

#<The MY-APPLICATION package>> (in-package my-application) ;; Sets this to be default

New symbols are (by default) created in default package The default package, when Lisp is started is COMMON-LISP-USER The COMMON-LISP packages is automatically used


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Numbers

Number crunching is one of Lisp's strengths Many data types, automatically converted from one to another Many numeric functions Example: Factorial program was easier in Lisp (no overflow!) Four distinct types

– Types: Integer, Floating Point Number, Ratio, and Complex Number– Examples: 100, 123.45, 3/2, #c(a b) ; #c(a b) = a+bi– Predicates: integerp, floatp, ratiop, complexp

Basic rules for automatic conversion– If function receives floating point #'s, generally returns floating point #'s– If a ratio divides evenly (for example, 4/2), it will be converted to integer– If complex # has an imaginary part of zero, converted to a real


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Subset of Type HierarchyS u b se t o f Typ e H ie ra rch y

C om p lex

B it

F ixn u m B ig n u m

In teg er R a tio

R a tion a l

L on g -floa tD ou b le -floa tS in g le -floa tS h ort-floa t

F loa t

R ea l

N u m b er

T

Lisp Expression Return Value(setf a 12); 12(type a 'bit); NIL(type a 'fixnum); T(type a 'integer); T(integerp a); T(rationalp a); T(realp a); T(numberp a); T


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More on Numbers

Conversion– (float) (truncate) (floor) (ceiling) (round) …

> (mapcar #'float '(1 2/3 .5))(1.0 0.6667 0.5)

Comparison– Use = to compare for equality (also, <,<=, >, >=, /=)

> (= 4 4.0)T

Other Misc. functions– Max– Min– (expt x n) = X^n– (exp x) = E^x– (log x n) = logn X; N is optional (default = natural log)– sqrt, sin, cos, tan


29

Eval

Eval - Evaluates an expression and returns its value> (eval '(+ 1 2 3))

6

Top-level is also called the read-eval-print loop Reads expression, evaluates it, prints value, loops back

> (defun our-toplevel () (loop (format t "%~> ") (print (eval (read))))

Eval Inefficient - slower than running compiled code Expression is evaluated without a lexical context


30

Macros

Macros A common way to write programs that write programs Defmacro is used to define macros Example: A macro to set its argument to NIL

> (defmacro nil! (x) (list 'setf x nil))> (nil! A) ;; Expands to (setf A NIL), is then

evaluatedNIL

macro-expand-1 Generates macro expansion (not eval'ed)> (macroexpand-1 '(nil! A))

(SETF A NIL)T


31

Macros

Typical procedure call (defun) Evaluate arguments Call procedure Bind arguments to variables inside the procedure

Macro procedure (defmacro) Macros do not evaluate their arguments When a macro is evaluated, an intermediate form is produced The intermediate form is evaluated, producing a value


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Example: Manually Define Pop

Review of the built in operation “Pop”: Implements Stack data structure “Pop” operation

> (setf a '(1 2 3 4 5))> (pop a)

1> a

(2 3 4 5)

How would you emulate this using other functions?– Attempt 1: Remove the element “1” from A

> (setf a (rest a))(2 3 4 5) ;; A is set correctly to (2 3 4 5), but we want “1” to

be returned– Attempt 2: Remove first element AND return it

> (prog1 (first a) (setf a (rest a)))– Attempt 3: Write a Lisp expression that generates above expression

> (list 'prog1 (list 'first a) (list 'setf a (list 'rest a)))


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Our-Pop, using Macro Convert Lisp expression into a macro (Our-Pop)

> (defmacro our-pop (stack) (list 'prog1 (list 'first stack) (list 'setf stack (list 'rest stack))))

– Note similarity to Defun

Example Call> (OUR-POP a)

Notes– The parameter “A” is NOT evaluated– “A” is substituted for stack wherever the variable stack appears

(list 'prog1 (list 'first a) (list 'setf a (list 'rest a)))

– Intermediate form is generated(prog1 (first a) (setf a (rest a)))

– Intermediate form is evaluatedA is set to (rest A); the first element is returned


34

Our-Pop using Defun

Why doesn’t this (defun) work the same way?> (defun our-pop (stack) (prog1 (first stack) (setf stack (rest

stack))))> (setf a '(1 2 3 4 5))> (our-pop a)

1> a

(1 2 3 4 5)

Reason: Lisp passes parameters “by-value”– The value of A is COPIED into the variable “stack”– Any changes to the variable “stack” are done to the COPY, and NOT the

original variable A– When the function returns, the original value of A is unchanged


35

Significance of Eval Steps

Macro evaluation has several steps (as noted)– The parameter “A” is NOT evaluated– “A” is substituted for stack wherever the variable stack appears– Intermediate form is generated– Intermediate form is evaluated

Note that A is evaluated at step 4 above (not step 1) Why does this matter?

– Answer: For the same reason that it matters in C/C++ macros– You may not want arguments evaluated at all– Or, you may want them evaluated multiple times– Macros give this flexibility


36

Backquotes

Significance of Evaluation Steps (cont) Consider

> (defmacro our-if-macro (conditional then-part else-part) (list 'if conditional then-part else-part))> (defun our-if-fun (conditional then-part else-part) (if conditional then-part else-part))> (if (= 1 2) (print "Equal") (print "Not Equal"))

– Lisp evaluates all parameters of OUR-IF-FUN before function is called

Backquote Mechanism Forward quotes: Entire next expression is not evaluated

> (defun temp () (setf a '(a b c d e)))

Backquote: Next expression is not evaluated (with exceptions)> (defun temp () (setf a `(a b c d e)))> (defun temp (x) (setf a `(a b c d e ,x))

– The “,x” expression is evaluated; the value of X is used.


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Backquotes (cont.)

Exceptions - Backquote evaluates the following ,variable - Evaluates the value of the variable

> (setf x '(h i j))> (setf a `(a b c ,x e f))

(A B C (H I J) E F)

,@variable - Splices the elements of a list> (setf a `(a b c ,@x e f))

(A B C H I J E F)

Backquotes simplify macro development> (defmacro our-if-macro (conditional then-part

else-part) (list 'if conditional then-part else-part)) ;;

old way> (defmacro our-if-macro (conditional then-part

else-part) `(if ,conditional ,then-part ,else-part))


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Backquotes simplify macros

Original version of our-pop> (defmacro our-pop (stack) (list 'prog1 (list 'first stack) (list 'setf stack (list 'rest stack))))

Our-pop redefined using backquotes> (defmacro our-pop (stack) `(prog1 (first ,stack) (setf ,stack

(rest ,stack))))– Syntax is much closer to the intermediate form

Macros can be defined with following parameters Optional (&optional) Rest (&rest) Key (&key)


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Macro Design

Macro: take a number n, evaluate its body n times Example call

> (ntimes 10 (princ "."))……….NIL

First attempt (incorrect)> (defmacro ntimes (n &rest body) `(do ((x 0 (+ x 1))) ((>= x ,n)) ,@body))

For example call, within body of macro– n bound to 10– body bound to ((princ "."))

(do ((x 0 (+ x 1))) ((>= x 10)) (princ "."))

– Macro works for example. Why is it incorrect?


40

Inadvertent Variable Capture

Consider Initialize x to 10, increment it 5 times

> (let ((x 10)) (ntimes 5 (setf x (+ x 1))) x)

10

Expected value: 15 Why? Look at expansion

> (let ((x 10)) (do ((x 0 (+ x 1))) ((>= x 5)) (setf x (+ x 1))) x)

X is used in let and in the iteration– setf increments iteration variable!


41

Gensym Gives New Symbol

Gensym Generates a new (uninterned) symbol

> (defmacro ntimes (n &rest body) ; Still incorrect though (let ((g (gensym))) `(do ((,g 0 (+ ,g 1))) ((>= ,g ,n)) ,@body)))

The value of symbol G is a newly generated symbol– How does this avoid the problem? – What if the call has a variable G?

Look at the expansion of(let ((x 10)) (ntimes 5 (setf x (+ x 1))) x)


42

Expansion of ntimes (2)

Substitute for N and BODY> (let ((x 10)) (let ((g (gensym))) `(do ((,g 0 (+ ,g 1))) ((>= ,g ,5)) ,@((setf x (+ x 1))))) x)

Generate intermediate form> (let ((x 10)) (do ((#:G34 0 (+ #:G34 1))) ((>= #:G34 5)) (setf x (+ x 1))) x)

Evaluate15

This works for our example … but ...


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Multiple Evaluation What happens when we want to debug…

> (let ((x 10) (niteration 3)) (ntimes niteration (setf x (+ x 1))) x)

13

To debug, insert a print expression> (let ((x 10) (niteration 3)) (ntimes (print niteration) (setf x (+ x 1))) x)

3 3 3 3 13

The argument is evaluated multiple times– Apparent when argument causes side-effects– What if the argument was a (setf …)

Non-intuitive: Expect argument to be evaluated only once.


44

Avoiding Multiple Evaluation

Solution is to copy value when you get into macro Use copy when needed within macro

> (defmacro ntimes (n &rest body) (let ((g (gensym)) (h (gensym))) `(let ((,h ,n)) (do ((,g 0 (+ ,g 1))) ((>= ,g ,h)) ,@body))))

This is correct> (let ((x 10) (niteration 3)) (ntimes (print niteration) (setf x (+ x 1))) x)

313


45

Expansion of Ntimes (Final)

Pprint is a useful function for Pretty PRINTing expressions

> (pprint (macroexpand-1 '(ntimes (print niteration) (setf x (+ x

1)))))(LET ((#:G93 (PRINT NITERATION))) (DO ((#:G92 0 (+ #:G92 1))) ((>= #:G92 #:G93)) (SETF X (+

X 1))))

Problems of Multiple Evaluation and Inadvertent Variable Capture Examples of errors that can occur when working with macros Errors are common in Lisp as well as languages like C/C++


46

Multiple Evaluation in Pop

Looking back at our Pop macro It suffers from multiple evaluation We can't use same technique, though

– Need to get the first AND do a setf to change the value> (defmacro our-pop (stack) `(prog1 (first ,stack) (setf ,stack

(rest ,stack))))

Textbook solution avoids multiple evaluation Page 301 Uses a function get-setf-expansion to get to inner

details of setf


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Case Study: Expert Systems

Overview of using Lisp for Symbolic Pattern Matching Rule Based Expert Systems and Forward Chaining Backward Chaining and PROLOG

Motivational example Given:

– A set of Facts– A set of Rules

Desired result– Answer complex questions and queries


48

Smart Animal Guessing Facts about an animal named “Joey”

– F1. (Joey’s mother has feathers)– F2. (Joey does not fly)– F3. (Joey swims)– F4. (Joey is black and white)– F5. (Joey lives in Antarctica)

Rules about animals in general– R1. If (animal X has feathers) THEN (animal X is a bird)– R2. If (animal X is a bird) and (animal X swims) and (animal X does not fly) and

(animal X is black and white) THEN (animal is a penguin)– R3. If (animal X’s mother Z) THEN (animal X Z)

Example: if (animal X’s mother has feathers) then (animal X has feathers)

– R4. If (animal X Z) THEN (animal’s mother Z)

Notes– By combining the facts and rules, we can deduce that Joey is a penguin, and that

the Joey’s mother is a penguin.


49

Symbolic Pattern Matching

Symbolic pattern matching example Match F1 with the IF part of R1

– F1. (Joey’s mother has feathers)– R1. If (animal X has feathers) THEN (animal X is a bird)

The expression (Joey’s mother has feathers) matches the pattern (animal X has feathers).

The association (animal X = Joey’s mother) is implied

In general Symbolic pattern matching

– matching an ordinary expression (e.g. fact) to a pattern expression

Unification: more advanced version of pattern matching– match two pattern expressions to see if they can be made identical– Find all substitutions that lead to this


50

Rule Based Expert System

Rule Based Expert Systems Once the pattern matching step is done, then we know that Rule

R1 can be combined with fact F1– F1. (Joey’s mother has feathers)– R1. If (animal X has feathers) THEN (animal X is a bird)

The association (animal X = Joey’s mother), along with the second part of the rule (animal X is a bird) leads to a derived fact:– (Joey’s mother is a bird)


51

Forward Chaining

Basic philosophy: Given a set of rules R and a set of facts F, what new facts (DF)

can be derived?– DF1: Joey has feathers (R3,F1)– DF2: Joey’s mother is a bird (R1, F1)– DF3: Joey is a bird (R1,DF1) [or, (R3,DF2)]– DF4: Joey’s mother does not fly (R4, F2)– DF5: Joey’s mother swims (R4, F3)– DF6: Joey’s mother is black and white (R4, F4)– DF7: Joey’s mother lives in Antarctica (R4, F5)– DF8: Joey is a penguin (R2, DF3, F2, F3, F4)– DF9: Joey’s mother is a penguin (R4, DF8) or (R2, DF2, DF5, DF4,

DF6)


52

Backward Chaining

Basic philosophy Can a statement (e.g. Joey is a penguin) be proven given the

current set of facts and rules? Work backwards, to determine what facts, if true, can prove that

Joey is a penguin (or prove that Joey is not a penguin).– B1. R2: (Joey is a penguin) IF (a) Joey is a bird; (b) Joey swims; (c) Joey does

not fly; and (d) Joey is black and white– B2. R1: (Joey is a bird) IF (Joey has feathers)– B4. R3: (Joey has feathers) IF (Joey’s mother has feathers)– DF1. (Joey has feathers), since we know (Joey’s mother has feathers (F1)– DF2. (Joey is a bird), since we know (Joey has feathers) (DF1) – DF3. (Joey is a penguin), since (a), (b), (c), and (d) are known to be true

(DF2, F3, F2, F4 respectively)

The fact (Joey is a penguin) can be derived


53

Does this apply in the real world? Given a clinical specimen, need to know what tests to perform?

– Example facts about a specific patient specimen (to test whether a person has Syphilis):

F1. Automated Reagin Test result is Reactive, Titer=8F2. Microheme Agglutination Test is Non-ReactiveF3. Specimen is from a pregnant womanF4. Prior history indicates a result of Reactive, Titer=2F5. Prior test was performed 12 months ago

– Example rulesR1. IF (Automated Reagin Test is Reactive, Titer >= 4) AND (Microheme Agglutination Test is Non-Reactive) THEN (Rapid Plasma Reagin test must be performed)R2. IF (Specimen is from a pregnant woman) THEN (Microheme Agglutination Test must be performed)R3. IF (Specimen is from a pregnant woman) THEN (Automated Reagin Test must be performed twice)

– Sample questions:What tests need to be performed? (Forward chaining)Should we do the RPR test? (Backward chaining)Is the specimen considered abnormal? (Backward chaining)


54

Lisp

Lisp is a good language for implementing expert systems. Concise programs Flexible processing of lists Basic implementations are shown in chapters 24-27

Other applications of expert systems Mathematics: Calculus, geometry Computer configuration, electronic circuits Evaluate geological formations, planned investments Diagnosis of infections


55

Building an Expert System

Knowledge representation how to represent facts, patterns, rules how to represent sets of these

Build a pattern matcher Build the inference engine

Forward Chaining and/or Backward Chaining


56

Implementing Pattern Matching

Want a procedure to match patterns Input

– Animal X has feathers (pattern)((? X) has feathers) ; Uses (? Var) for pattern

variable– Joey’s mother has feathers (regular expression or Fact)

((mother Joey) has feathers)

Returns – Mapping between pattern variables

( (X (mother Joey)) )

Example call> (match '((? X) has feathers) '((mother Joey) has feathers) )

((X (mother Joey)))> (match '((? X) has (? Z)) '(Joey has feathers)))

((X Joey) (Z feathers))


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Return values of Match

Return value is a set of variable bindings– Example

((X Joey) (Z feathers))– General Form

( (var1 value1) (var2 value2) … (varN valueN) )

What if the two patterns don’t match?– First attempt: On failure, return NIL

> (match '((? X) has feathers)) '(Joey does not fly))NIL

But consider...> (match '(Joey has feathers) '(Joey has feathers)))

– This matches, but there is no need to bind any variables.– So, need to return SUCCESS with a list of zero variable bindings: ( )

NIL <-- NIL = ( )

Need to differentiate between failed match and match with no variable bindings


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Return values of Match (cont.)

Approach taken by Winston/Horn Note: This is NOT the only way to do it

– NIL = Success, empty list of pattern variables– FAIL = Symbol returned when the pattern and datum don’t match

Examples> (match '((? X) has feathers) '((mother Joey) has

feathers))((X (mother Joey)))

> (match '((? X) has (? Z)) '(Joey has feathers)))((X Joey) (Z feathers))

> (match '((? X) has feathers)) '(Joey does not fly))

FAIL> (match '(Joey has feathers) '(Joey has feathers)))

NIL ; Treated as an empty list


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Match Function Definition

Function definition for MATCH: 4 basic branches;; Calculates and returns bindings (if successful);; Or, returns 'FAIL(defun match (p d &optional bindings) ;; 1. If P and D are both atoms, ;; If they’re equal, it’s a match, otherwise FAIL ;; e.g. (match 'FEATHERS 'FEATHERS) ;; 2. If P is a pattern variable ;; Assign the value of D to the pattern variable in P ;; e.g. (match '(? X) 'JOEY) ;; should assign the value JOEY to the variable X ;; 3. If P and D are both Lists ;; Recursively solve for matches ;; e.g. (match '(A B (? X) (D E)) '(A B C (D E))) ;; should recursively call match on (A vs. A) ;; (B vs. B) ((? X) vs. C) ((D E) vs. (D E)) ;; 4. Any other case (e.g. atom vs. list, etc.) ;; FAIL)


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Match: Case 1, Case 4

Part 1: Build up Cond infrastructure Implement case 1 (P and D are both atoms) Implement case 4 (Everything Else)

(defun match (p d &optional bindings) (cond ((and (atom p) (atom d)) ;; Case 1: Both p and d are atoms ;; If P and D are equal: Match. Return

bindings ;; Otherwise, return FAIL (if (eql p d) bindings 'FAIL)) (…Case 2…) (…Case 3…) (t ;; Case 4: Any other case. Return FAIL 'FAIL) ))


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Match: Case 3

Case 3 (P and D are lists and need to be solved recursively)– Algorithm

;; A) Match the first pair, to get new bindings;; B) If first pair failed, ;; C) return fail;; D) Otherwise, using the bindings returned;; by step (A), match remaining pairs

– Code(defun match (p d &optional bindings) (cond (…Case 1…) (…Case 2…) ((and (listp p) (listp d)) ; P and D are both Lists ;; Recursively solve for matches (let ((result (match (first p) (first d) bindings))) ;(A) (if (eq 'fail result) ;(B) 'FAIL ;(C) (match (rest p) (rest d) result)))) ;(D) (…Case 4…) ))


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Match, Case 2

Case 2: P is a variable, D is a piece of data Example: P=(? X); D=Joey Make the binding (X Joey) Add it to the bindings already defined

– Old Bindings: ( (A apple) (B banana) )– After adding: ( (X Joey) (A apple) (B banana) )

Code for adding a new binding to a list of bindings(defun add-binding (p d bindings) (cons (list (second p) d) bindings))


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Match, Case 2 (continued)

Note:– Before adding a binding, need to check if the binding already exists– If binding exists, it should match previous binding

Example 1> (match '((? X) (? Y) implies (? X) (? Z)) (Joey smokes implies Joey (will get cancer)))

( (X Joey) (Y smokes) (Z (will get cancer)))NOT((X Joey) (Y smokes) (X Joey) (Z (will get cancer)))

– When the algorithm finds (X Joey), no new binding should be created

Example 2> (match '((? X) (? Y) implies (? X) (? Z)) (Joey smokes implies Mary (will get cancer)))

FAIL– This should fail because X cannot be bound to both Joey and Mary– When the algorithm finds (X Joey) while trying to bind (X Mary), the routine

should return FAIL


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– Example 1(match '(? X) 'Joey '((X Joey) (Y smokes) (Z (will get cancer)))

– Example 2(match '(? X) 'Mary '((X Joey) (Y smokes) (Z (will get cancer)))

– Example 3(match '(? X) 'Joey '((Y smokes) (Z (will get cancer)))

– Algorithm;; 1. Check if variable is already bound;; 2. If bound, try to match the value of the variable;; to the datum;; 3. If bound and the binding matches, return

binding;; (nothing new needs to be done). (Example 1);; 4. If bound and the binding doesn’t match,

return FAIL;; (Example 2);; 5. If not bound, add a binding (Example 3)


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– Find-binding: Uses Assoc> (find-binding '(? X) '((X Joey) (Y smokes) (Z (will

get…)))(X Joey)

– Match(defun match (p d &optional bindings) (cond (…Case 1…) ((and (listp p) (eq (second p) '?)) ; Is p=~ (?

X) (let (binding (find-binding p bindings)) ; (1) (if binding ; (2) (match (second binding) d bindings) ;

(3,4) (add-binding p d bindings) ; (5) ))) (…Case 3…) (…Case 4…) ))

Documents

CSCI 2210 - Programming in Lisp; Instructor: Alok Mehta; 4.ppt1 CSCI 2210: Programming in Lisp ANSI Common Lisp, Chapters 5-10