cs774 (prasad)l5arithstruc1 arithmetic ; structures [email protected]

cs774 (Prasad) L5ArithStruc 1

Arithmetic ; Structures

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http://www.knoesis.org/tkprasad/

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Arithmetic?- X = 1 + 2.

• Succeeds with X bound to the term 1 + 2.

• To make the system evaluate the expression, we use “is” construct.

?- X is 1 + 2.• Succeeds with X bound to the term 3.

• If variables occur in the RHS expression, they must be bound to values (i.e., must be ground) before the expression can be evaluated.

?- Y = 1, X is Y + 2.


(cont’d)?- Y is X+2, X=1.

• ERROR: Arguments are not sufficiently instantiated.

?- X = 1, Y = 2, Z is X * Y + 3.

• Succeeds with Z bound to the term 5.

• Re-ordering of (atomic) goals in the query can cause the query to fail.

• A different control strategy that delays evaluation of goals containing non-ground expressions may succeed in certain cases.



“Equalities” in Prolog• We have three different but related “equality”

operators:– X is Y evaluates Y and unifies the result with X:

3 is 1+2 succeeds, but 1+2 is 3 fails– X = Y unifies X and Y, with no evaluation: both 3 = 1+2 and 1+2 = 3 fail

– X =:= Y evaluates both and compares: both 3 =:= 1+2 and 1+2 =:= 3 succeed

• Any evaluated term must be fully instantiated.

Equalities and their negations

• Matching

T1 = T2 T1 \= T2

• Literal Equality

T1 == T2 T1 \== T2

5 \== 3+2

• Arithmetic Equality

5 =:= 3+2 5 =\= 2+2

• Arithmetic Relations

5 =< 8 5 >= 2

Sum of Squares


b

ai

i2

sum(A,B,0) :- A > B.sum(A,A,Res) :- Res is A*A.sum(A,B,Res) :-

A < B, T is A + 1,sum(T, B, R),Res is A*A + R.

Factorial Function : Recursive definition


!or 1

n

ini

rfact(0,1).rfact(N,R) :-

N > 0, M is N-1,rfact(M,S),R is N * S.

Alternative: Tail recursive definition


!or 1

n

ini

fact(N,R) :- trfact(N,1,R).trfact(0,R,R). trfact(N,T,R) :-

N > 0, M is N-1,S is N * T,trfact(M,S,R).

Matching vs Arithmetic Comparison

?- 1 + 2 =:= 2 + 1.

YES

?- 1 + 2 = 2 + 1.

NO

?- A + 1 = 2 + B.

A = 2

B = 1

•


Permutation Sort


psort(L,R) :- permutation(L,R),sorted(R).

sorted([]). sorted([_]). sorted([X,Y|R]) :-

X =< Y, sorted([Y|R]).

Obviously correctbut prohibitively inefficient

Selection Sort


ssort([],[]). ssort(L,[Min|Srest]) :-

least(L,Min,Rest),ssort(Rest,Srest).

least([Min],Min,[]). least([H|T],Min,Rest) :-

least(T,M,L), (M > H -> (Min, Rest) = (H,T);

(Min, Rest) = (M,[H|L])).

Quadratic

Structures via Travel Agent Example


Travel Planning• Querying a database of flight information

timeTable(Place1, Place2, ListOfFlights).

• Form of flight itemdepTime / arrTime / fltNo / days

• One day travels

?- route(Place1, Place2, Day, Route).


operator

Examples of Fact and Query

timetable(edinburgh, london,

[ 9:40 / 10:50 / ba4733 / alldays,

13:40 / 14:50 / ba4773 / alldays,

19:40 / 20:50 / ba4833 / [mo,we,fr]]).

• Note that ‘:’ should bind stronger than ‘/’.

?- route(london, paris, tu, Route)• Form of items in the route

from / to / fltNo / depTime


Prolog Code% A Flight Route Planner

:- op(50, xfy, :).

% route(Place1, Place2, Day, Route).

% Route is a sequence of flights on Day, starting at Place1 and ending at Place2

route(P1, P2, Day, [P1 / P2 / Fnum / Deptime]) :-

flight(P1, P2, Day, Fnum, Deptime, _).

% direct flight

route(P1, P2, Day, [(P1 / P3 / Fnum1 / Dep1) | RestRoute]) :-

flight(P1, P3, Day, Fnum1, Dep1, Arr1),

route(P3, P2, Day, RestRoute),

deptime(RestRoute, Dep2),

transfer(Arr1, Dep2).

% Indirect flight - ensure enough transfer time

% optimize by propagating that constraint


% decoding timetable

flight(Place1, Place2, Day, Fnum, Deptime, Arrtime) :-

timetable(Place1, Place2, Flightlist),

member(Deptime / Arrtime / Fnum / Daylist, Flightlist),

flyday(Day, Daylist).

flyday(Day, Daylist) :- member(Day, Daylist).

flyday(Day, alldays) :- member(Day, [mo,tu,we,th,fr,sa,su]).

deptime([( _ / _ / _ / Dep) | _], Dep).

transfer(Hours1:Mins1, Hours2:Mins2) :-

60 * (Hours2 - Hours1) + Mins2 - Mins1 >= 40.

% must have 40 min gap


% A Flight Database

timetable(edinburgh, london,

[ 9:40 / 10:50 / ba4733 / alldays,

13:40 / 14:50 / ba4773 / alldays,

19:40 / 20:50 / ba4833 / [mo,tu,we,th,fr,su]]).

timetable(london, edinburgh,

[ 9:40 / 10:50 / ba4732 / alldays,

11:40 / 12:50 / ba4752 / alldays,

18:40 / 19:50 / ba4822 / [mo,tu,we,th,fr]]).

timetable(london, ljubljana,

[ 13:20 / 16:20 / jp212 / [mo,tu,we,fr,su],

16:30 / 19:30 / ba473 / [mo,we,th,sa]]).


timetable(london, zurich,

[ 9:10 / 11:45 / ba614 / alldays,

14:45 / 17:20 / sr805 / alldays]).

timetable(london, milan,

[ 8:30 / 11:20 / ba510 / alldays,

11:00 / 13:50 / az459 / alldays]).

timetable(ljubljana, zurich,

[ 11:30 / 12:40 / jp322 / [tu,th]]).

timetable(ljubljana, london,

[ 11:10 / 12:20 / jp211 / [mo,tu,we,fr,su],

20:30 / 21:30 / ba472 / [mo,we,th,sa]]).


timetable(milan, london,

[ 9:10 / 10:00 / az458 / alldays,

12:20 / 13:10 / ba511 / alldays]).

timetable(milan, zurich,

[ 9:25 / 10:15 / sr621 / alldays,

12:45 / 13:35 / sr623 / alldays]).

timetable(zurich, ljubljana,

[ 13:30 / 14:40 / jp323 / [tu,th]]).

timetable(zurich, london,

[ 9:00 / 9:40 / ba613 / [mo,tu,we,th,fr,sa],

16:10 / 16:55 / sr806 / [mo,tu,we,th,fr,su]]).

timetable(zurich, milan,

[ 7:55 / 8:45 / sr620 / alldays]).


Queries• What days of the week is there is a direct

evening flight from Ljubljana to London?

?- flight(ljubljana, london, Day, _, DepHour:_,_), DepHour >= 18.

• How can I get from Ljubljana to London on Thursday?

?- route(ljubljana, london, th, R).

• Queries can result in infinite loops or stack overflow on Prolog.


SWI-Prolog vs XSBXSB: tabletable route/4.

• How can I get from Ljubljana to Edinburgh on Thursday?

?- route(ljubljana, edinburgh, th, R).

SWI-Prolog: Out of local stack

XSB: [ ljubljana / zurich / jp322 / 11 : 30,

zurich / london / sr806 / 16 : 10,

london / edinburgh / ba4822 / 18 : 40];


Chess : Eight Queens Problems


Encoding of a solution : list(Column/Row)[1/1, 2/5, 3/8, 4/6, 5/3, 6/7, 7/2, 8/4]

solution([]).

solution([C/R | Others]) :-

solution(Others),

member(R,[1,2,3,4,5,6,7,8]),

noAttack(C/R, Others).

noAttack(_,[]).

noAttack(C/R, [C1/R1 | Others]) :-

R =\= R1, R1 – R =\= C1 – C,

R1 – R =\= C – C1,

noAttack(C/R, Others ).

template([1/_,2/_,3/_,4/_,5/_,6/_,7/_,8/_]).

?- template(S), solution(S).


Eight Queens Solutions (92 in all)S = [1/4, 2/2, 3/7, 4/3, 5/6, 6/8, 7/5, 8/1] ;

S = [1/5, 2/2, 3/4, 4/7, 5/3, 6/8, 7/6, 8/1] ;

S = [1/3, 2/5, 3/2, 4/8, 5/6, 6/4, 7/7, 8/1] ;

S = [1/3, 2/6, 3/4, 4/2, 5/8, 6/5, 7/7, 8/1] ;

…

S = [1/5, 2/7, 3/2, 4/6, 5/3, 6/1, 7/8, 8/4] ;

S = [1/3, 2/6, 3/2, 4/7, 5/5, 6/1, 7/8, 8/4] ;

S = [1/6, 2/2, 3/7, 4/1, 5/3, 6/5, 7/8, 8/4] ;

S = [1/3, 2/7, 3/2, 4/8, 5/6, 6/4, 7/1, 8/5] ;

S = [1/6, 2/3, 3/7, 4/2, 5/4, 6/8, 7/1, 8/5] ;

S = [1/4, 2/2, 3/7, 4/3, 5/6, 6/8, 7/1, 8/5] ;

… S = [1/8, 2/2, 3/4, 4/1, 5/7, 6/5, 7/3, 8/6] ;

S = [1/7, 2/2, 3/4, 4/1, 5/8, 6/5, 7/3, 8/6] .


cs774 (prasad)l5arithstruc1 arithmetic ; structures [email protected]

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