[PDF] - F unction Symb ols in Prolog F rom Deductive Databases to PDF Document

SLIDE 1 F unction Symb

ls

in Prolog F rom Deductive Databases to Logic Programs In logic there a re t w

kinds
f
bjects

p redicates and functions

Predicates

rep resent statements ab

ut

the w

rld

John hates Mary hatesjohnmary John is short shortjohn hates is a p redicate symb

l

shortjohn is an atomic fo rmula

F

unction terms rep resent

bjects

in the w

rld

the mother

f

Mary motherofmary a rectangle

f

length

and

width

rectangle

motherofmary is a function term rectangle is a function symb

l

SLIDE 2 F unction terms do not have values In Pro log they act as data structures let pXY denote a p

int

in dim space let pXYZ denote a p

int

in dim space W rite a Prolog p rogram SQDISTPointPointD that returns the squa re

f

the distance b e t w een t w

p
ints

The p rogram should w

rk

fo r

and

dim p

ints

W ant SQDISTp p D returns D

and

SQDISTp p D returns D

and

SQDISTp p D is undefined

SLIDE 3 Prolog Program

SQDISTpXY

pXY D

XD

is XX YD is YY D is XDXD

YDYD
SQDISTpXYZ

pXYZ D

XD

is XX YD is YY ZD is ZZ D is XDXD

YDYD
ZDZD

Query

SQDISTp

p D This query unies with the head

f

rule

with

fXn Yn Xn Yng so XD is XX

YD

is YY

D

is

So

D is returned Note the query do es not unify with the head

f

rule

so
nly

rule

is

used

SLIDE 4 Prolog Program

SQDISTpXY

pXY D

XD

is XX YD is YY D is XDXD

YDYD
SQDISTpXYZ

pXYZ D

XD

is XX YD is YY ZD is ZZ D is XDXD

YDYD
ZDZD

Query

SQDISTppD

This query unies with the head

f

rule

with

fXn Yn Zn Xn Yn Zng so XD is

YD

is

ZD

is

D

is

So

D is returned Note the query do es not unify with the head

f

rule

so
nly

rule

is

used

SLIDE 5 Prolog Program

SQDISTpXY

pXY D

XD

is XX YD is YY D is XDXD

YDYD
SQDISTpXYZ

pXYZ D

XD

is XX YD is YY ZD is ZZ D is XDXD

YDYD
ZDZD

Query

SQDISTp

p D Note this query do es not unify with any rule so Prolog simply returns no ie no answ ers fo r D

SLIDE 6 Returning F unction T erms as Answ ers eg

given

a p

int

pXY return a new p

int

with double the co

rdinates

eg

Query
doublepP

Answ er

P
p

Prolog Program doublepXY pXY

X

is X Y is Y In Plain English if X

X

and Y

Y

then the double

f

pXY is pXY An equivalent p rogram using

doublepXY

P

X

is X Y is Y P

pXY

Here

is

b eing used to assign a value to va riable P T ry to avoid this It reects p ro cedural thinking

SLIDE 7 Sample Execution Prolog Program doublepXY pXY

X

is X Y is Y Query

doublepP

The query unies with the head

f

the rule where the mgu is fXn Yn PnpXYg The b

dy
f

the rule then evaluates X is X ie

Y

is Y ie

The

mgu b ecomes fXn Yn Pnpg So the answ er is P

p

SLIDE 8 Recursion with F unction Symb

ls

Example Electrical circuits

Tw
resisto

rs in series

with

resistances R

and

R

resp

ectively

T
tal

resistance

f

the circuit is

Can

rep resent the circuit as a function term series

Tw
resisto

rs in pa rallel

T
tal

resistance

f

the circuit is

Rep

resent the circuit as a function term par

SLIDE 9 Mo re Complex Circuits

par

series

seriespar

par

SLIDE 10 Problem W rite a Prolog p rogram that computes the total resistance

f

any circuit F

r

example Query

resistanceseries

R Answ er

R
Query
resistancepar

R Answ er

R
Query
resistanceseriespar

R Answ er

R
Query
resistance

R Answ er

R

SLIDE 11 Solution

resistanceRR
numberR
resistanceseriesCC

R

resistanceC

R resistanceC R R is RR

resistanceparCC

R

resistanceCR

resistanceCR R is RRRR Sample Query resistanceseriespar TR ie compute the total resistance TR

f

the follo wing circuit

SLIDE 12

rule (1) R1 = 3

UNWINDING PHASE RECURSIVE (WINDING) PHASE

rule (2) rule (1) rule (1) R21 = 6 R22 = 3 resistance(series(3,par(6,3)),R), R = 5 resistance(3,R1), resistance(par(6,3),R2), R is R1+R2 is 5 resistance(6,R21), resistance(3,R22), R2 is (R21R22)/(R21+R22) is (63)/(6+3) is 2

a

SLIDE 13

rule (1) R1 = 3

UNWINDING PHASE RECURSIVE (WINDING) PHASE

rule (2) rule (1) rule (1) R21 = 6 R22 = 3 resistance(series(3,par(6,3)),R), R = 5 resistance(3,R1), resistance(par(6,3),R2), R is R1+R2 is 5 resistance(6,R21), resistance(3,R22), R2 is (R21*R22)/(R21+R22) is (6*3)/(6+3) is 2

a

rule (2) rule (1) rule (1) R21 = 6 R22 = 3 resistance(series(3,par(6,3)),R), R = 5 resistance(3,R1), resistance(par(6,3),R2), R is R1+R2 is 5 resistance(6,R21), resistance(3,R22), R2 is (R21R22)/(R21+R22) is (63)/(6+3) is 2