1 / 20

Using Structures: Example Programs

Using Structures: Example Programs. Notes for Ch.4 of Bratko For CSCE 580 Sp03 Marco Valtorta. 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)),

faris
Télécharger la présentation

Using Structures: Example Programs

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Using Structures: Example Programs Notes for Ch.4 of Bratko For CSCE 580 Sp03 Marco Valtorta

  2. 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 < 10000.

  3. Datalog and Relational Operations • See UseDefK2.ppt

  4. Doing Data Abstraction • Selectors can be used to hide details of representation.

  5. 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

  6. 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.

  7. 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.

  8. 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] ] ).

  9. 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

  10. 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

  11. 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).

  12. 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

  13. 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

  14. 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

  15. 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

  16. 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

  17. 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

  18. 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

  19. 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

  20. Eight Queens: Efficiency • The second program is the least efficient: ?- time( setof( S, solution(S), L)). % 1,139,743 inferences in 1.10 seconds (1034640 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 (776387 Lips) • Third program: 1 ?- time( setof( S, solution(S), L)). % 120,544 inferences in 0.15 seconds (802471 Lips)

More Related