UNIVERSITY OF SOUTH CAROLINA Department of Computer Science and Engineering Using Structures: Example Programs Notes for Ch.4 of Bratko For CSCE 580 Sp03 Marco Valtorta

UNIVERSITY OF SOUTH CAROLINA Department of Computer Science and Engineering Retrieving Structured Information From a Database Each family has three components: husband, wife, and children, e.g.: family( person( tom, fox, date( 7, may, 1960), works( bbc, 15200)), person( ann, fox, date( 9, may, 1961), unemployed), [ person( pat, fox, date( 5, may, 1983), unemployed), person( jim, fox, data( 5, may, 1983), unemployed) ] ). Database and relations in ch4_1.pl 3 ?- family( H,W,C), total( [H,W|C],I), length([H,W|C],N), I / N <

UNIVERSITY OF SOUTH CAROLINA Department of Computer Science and Engineering Datalog and Relational Operations See UseDefK2.ppt

UNIVERSITY OF SOUTH CAROLINA Department of Computer Science and Engineering Doing Data Abstraction Selectors can be used to hide details of representation.

UNIVERSITY OF SOUTH CAROLINA Department of Computer Science and Engineering Simulating a Non-deterministic Automaton Non-deterministic finite automaton (NFA) –accepts or rejects a string of symbols –has states –has transitions (possibly silent) –represented by a directed graph with self loops with distinguished final states and initial state (e.g. fig. 4.3) –string is accepted if there is a transition path s.t. it starts with the initial state it ends with a final state the arc labels along the path correspond to the complete input string

UNIVERSITY OF SOUTH CAROLINA Department of Computer Science and Engineering Simulation of NFA (II) To simulate an NFA in Prolog, we need: final/1, which defines the finals states trans/3, s.t. trans(S1,X,S2) means that a transition from S1 to S2 is possible when X is read silent/2 s.t. silent( S1,S2) means that a silent move is possible from S1 to S2.

UNIVERSITY OF SOUTH CAROLINA Department of Computer Science and Engineering nfa.pl 6 ?- length(Str,7), accepts(S,Str). Str = [a, a, a, a, a, a, b] S = s1 ; 5 ?- accepts(S, String), length(String, 7). ERROR: Out of local stack 12 ?- accepts(s1,S). ERROR: Out of local stack Why? –If the length of the input string is not limited, the recursion will never hit the empty list: the first clause will be missed. When the input string is limited, after exhausting all possible transitions consistent with it, the input will become empty and the first clause will be used.

UNIVERSITY OF SOUTH CAROLINA Department of Computer Science and Engineering Travel Agent Program to give advice about air travel timetable(Place1,Place2,ListOfFlights) ListOfFlights holds terms of the form DepartureTime/ArrivalTime/FlightNumber/ListOfDays, where / is an infix operator, e.g., timetable( london, edinburgh, [9:40 / 10:50 / ba4733 / alldays, 19:40 / 20:50 / ba4833 / [mo,tu,we,th,fr,su] ] ).

UNIVERSITY OF SOUTH CAROLINA Department of Computer Science and Engineering Finding Routes route( Place1, Place2, Day, Route), where Route is a sequence of flights that satisfy: the start point of the route is Place1 the end point is Place2 all the flights are on the same Day of the week all the flights in Route are in the timetable relation there is enough time for transfer between flights

UNIVERSITY OF SOUTH CAROLINA Department of Computer Science and Engineering Auxiliary Predicates flight( Place1,Place2,Day,FlightNum,DepTime,ArrTime). deptime( Route, Time). transfer( Time1, Time2). Note the similarity between the NFA simulation and the route finding problem: –states cities –transitions flights –path between initial and final state route between start and end city

UNIVERSITY OF SOUTH CAROLINA Department of Computer Science and Engineering Route Relation Direct connection: route( Place1, Place2, Day, [Place1/Place2/Fnum/Dep]) :- flight( Place1, Place2, Day, Fnum, Dep, Arr). Indirect connection with sufficient transfer time: route( P1,P2,Day,[P1/P3/Fnum1/Dep1 | RestRoute]) :- route( P3,P2,Day,RestRoute), flight( P1,P3,Day,Fnum1,Dep1,Arr1), deptime( RestRoute, Dep2), transfer( Arr1,Dep2).

UNIVERSITY OF SOUTH CAROLINA Department of Computer Science and Engineering Full Program fig4_5.pl As for the NFA simulation program, to avoid infinite loops, make sure to limit the length of the route. E.g., 3 ?- route( rome,edinburgh,mo,R). gives an infinite loop, while the following does not: 6 ?- conc(R,_,[_,_,_,_]), route(rome, edinburgh,mo,R). No conc generates routes in order of increasing length, so that shorter routes are tried first

UNIVERSITY OF SOUTH CAROLINA Department of Computer Science and Engineering More Infinite Looping Trouble ?- route( ljubljana, edinburgh, th, R). R = [ljubljana/zurich/jp322/11:30, zurich/london/sr806/16:10, london/edinburgh/ba4822/18:40] ; Action (h for help) ? abort % Execution Aborted ?- conc( R,_,[_,_,_,_]), route( ljubljana, edinburgh, th, R). R = [ljubljana/zurich/jp322/11:30, zurich/london/sr806/16:10, london/edinburgh/ba4822/18:40] ; No

UNIVERSITY OF SOUTH CAROLINA Department of Computer Science and Engineering Eight Queens: Program 1 How to place 8 queens on a cheesboard, so that they do not attack each other solution(Pos) if Pos is a solution to the problem Positions are represented by a list of squares where the queen is sitting, e.g: [1/4,2/2,3/7,4/3,5/6,6/8,7/5,8/1] (fig.4.6, a solution) We generalize to square boards of any size, so that we can use induction

UNIVERSITY OF SOUTH CAROLINA Department of Computer Science and Engineering Program 1, ctd. solution( [ ]). solution( [X/Y | Others]) :- solution( Others), member( Y, [1,2,3,4,5,6,7,8]), noattack( X/Y, Others). This shows how to add a queen to extend a partial solution

UNIVERSITY OF SOUTH CAROLINA Department of Computer Science and Engineering Program 1, ctd. noattack( _,[ ]). noattack( X/Y, [X1/Y1 | Others] ) :- Y =\= Y1, % Different Y-coordinates Y1-Y =\= X1-X, % Different diagonals Y1-Y =\= X-X1, noattack( X/Y, Others). The full program is in fig4_7.pl Finds all (92) solutions upon backtracking and stops

UNIVERSITY OF SOUTH CAROLINA Department of Computer Science and Engineering Eight Queens: Program 2 Represent X queen position by their position in the position list: –[1/Y1, 2/Y2, …., 8/Y8] is replaced by –[Y1, Y2, …., Y8] Generate an ordering of the Y positions, then test that position: solution( S) :- permutation( [1,2,3,4,5,6,7,8], S), %generate safe(S). %test

UNIVERSITY OF SOUTH CAROLINA Department of Computer Science and Engineering Program 2, ctd. noattack/2 is generalized to noattack/3, where the third argument represents the X distance (column distance) of two queens Full program in fig4_9.pl To get all solutions, do: ?- setof( S, solution(S), L), length( L,N). S = _G405 L = [[1, 5, 8, 6, 3, 7, 2, 4], [1, 6, 8, 3, 7, 4, 2|...], [1, 7, 4, 6, 8, 2|...], [1, 7, 5, 8, 2|...], [2, 4, 6, 8|...], [2, 5, 7|...], [2, 5|...], [2|...], [...|...]|...] N = 92

UNIVERSITY OF SOUTH CAROLINA Department of Computer Science and Engineering Eight Queens: Program 3 Program 3 uses a redundant representation of the board, with –columns, x, 1 through 8 –rows, y, 1 through 8 –upward diagonals, u = x – y, -7 through 7 –downward diagonals, v = x + y, 2 through 16 When a queen is placed, its column, row, and diagonals are removed from consideration pl4_11.pl

UNIVERSITY OF SOUTH CAROLINA Department of Computer Science and Engineering Eight Queens: Efficiency The second program is the least efficient: ?- time( setof( S, solution(S), L)). % 1,139,743 inferences in 1.10 seconds ( Lips) In general, generate-and-test is inefficient: it is best to introduce constraints as early as possible in the solution design process First program: ?- time( setof( S, solution1(S), L)). % 171,051 inferences in 0.22 seconds ( Lips) Third program: 1 ?- time( setof( S, solution(S), L)). % 120,544 inferences in 0.15 seconds ( Lips)