Prolog examples continued

Transcription

1 Prolog examples continued arithmetic functions translating a simple Lisp function modelling an addition circuit puzzle solving 1 arithmetic functions we keep arithmetic functions as functions for convenience:?- X is exp(1). % exp(1) = e 1 X = How does is compare with =, and with assignment? 2

2 operators some functors are represented by infix or prefix operators. is is an infix operator others? =, +, *, /, etc., (as "and") :- ("if") +, - are both prefix and infix :- as prefix is a command: means "do"; used for declarations 3 Lisp to Prolog (DEFUN 2Tosquare (X) (EXPT 2 (* X X))) square(x, Result) :- Result is X*X. twotosquare(x, Result) :- square(x, Square), Result is 2**Square. a simpler version: twotosquare(x, Result) :- Result is 2**(X*X). 4

3 more translation (DEFUN equalreverse (phr1 phr2) (EQUAL phr1 (REVERSE phr2) ) ) Boolean functions are already relational! equalreverse(phr1, phr2) :- reverse(phr2, phr2reversed), phr1 = phr2reversed. simplification: eliminate the = equalreverse(phr1, phr2) :-reverse(phr2, phr1). 5 reverse is built in?- help(reverse). reverse(+list1, -List2) Reverse the order of the elements in List1 and unify the result with the elements of List2. + = input, - = output but reverse allows either for each parameter! 6

4 reverse examples?- reverse(x, X). X = [] ; X = [_G230] ; X = [_G230, _G230]...?- reverse(x, Y). X = [] Y = [] ; X = [_G245] Y = [_G245] ; X = [_G245, _G248] Y = [_G248, _G245] reversible programming good predicates are steadfast they gives correct answers even if unusual values are supplied e. g. variables for 'inputs', constants for 'outputs'. non-steadfast predicates require specific arguments to be instantiated (input) or variables (output). 8

5 online-help examples?- help(lists). No help available for lists Yes?- apropos(lists). merge/3 append/3 Section 11-1 Section Merge two sorted lists Concatenate lists % try apropos (list) for a long list of predicates "lists: List Manipulation" "lists" Yes?- help(merge/3). merge(+list1, +List2, -List3) List1 and List2 are lists, sorted to the standard order of terms (see section 4.6). List3 will be unified with an ordered list holding both the elements of List1 and List2. Duplicates are not removed. (END) (exit with q) 9 apropos is not steadfast?- apropos(x). ERROR: Arguments are not sufficiently instantiated ^ Exception: (11) findall(_g233-_g234- _G231, apropos_predicate(_g180, _G233, _G234, _G231), _L199)? 10

6 adder example 11 logic facts construct a 2-bit adder in Prolog, using predicates for binary logic. the logic predicates: xor(0,0,0). xor(0,1,1). xor(1,0,1). xor(1,1,0). or(0,0,0). or(0,1,1). or(1,0,1). or(1,1,1). and(0,0,0). and(0,1,0). and(1,0,0). and(1,1,1). 12

7 a half-adder X Y Carry Sum 13 half-adder code halfa([x, Y], [Sum, Carry]) :- xor(x, Y, Sum), and(x,y, Carry). note the "destructuring" 14

8 1 bit adder X Y CarryIn halfa Sum1 halfa Carry1 Carry2 Sum CarryOut 15 1-bit adder code adder1([x,y,carryin], [Sum, CarryOut]) :- halfa([x,y], [Sum1, Carry1]), halfa([sum1,carryin], [Sum, Carry2]), or(carry1, Carry2, CarryOut). arrows show data flow for addition 16

9 subtracting with an adder?-adder1([x,y, 0], [1, 0]). X = 0 Y = 1 ; X = 1 Y = 0 ; no?-adder1([x, 0, Y], [1, 1]). no % no two bits add to bit subtraction adder1([x,y,carryin], [Sum, CarryOut]) :- halfa([x,y], [Sum1, Carry1]), halfa([sum1,carryin], [Sum, Carry2]), or(carry1, Carry2, CarryOut). arrows show data flow for subtraction example 18

10 2-bit adder code 2-bit adder with "ripple" carry: X1 X0 10 Y1 Y0 11 Sum1 Sum bit adder X1 Y1 X0 Y0 0 adder1 adder1 Carry0 CarryOut Sum1 Sum0 20

11 code for 2-bit adder adder2([x1, X0], [Y1, Y0], [Sum1, Sum0, CarryOut]) :- adder1([x0, Y0, 0], [Sum0, Carry0]), adder1([x1, Y1, Carry0], [Sum1, CarryOut]). 21 solving a logic puzzle with Prolog 22

12 the zebra puzzle ************ THE FACTS ************************* 1. There are 5 houses, occupied by politically-incorrect gentlemen of 5 different nationalities, who all have different coloured houses, keep different pets, drink different drinks, and smoke different (now-extinct) brands of cigarettes. */ 2. The Englishman lives in a red house. 3. The Spaniard keeps a dog. 4. The owner of the green house drinks coffee.. 6. The ivory house is just to the left of the green house The Chesterfields smoker lives next to a house with a fox.... Who owns the zebra and who drinks water? 23 prolog implementation represent the houses by a list of 5 terms house(color, Nationality, Pet, Drink, Cigarettes) create a partial structure using variables, to be filled in by the solution process use constraints to bind variables 24

13 house building makehouses(0,[]). makehouses(n, [house(color,nat,pet,drink,cig) List]) :- N>0, N1 is N - 1, makehouses(n1,list). we could express this more cleanly using anonymous variables: makehouses(n, [house(_, _,_, _, _) List]) :- N>0, N1 is N - 1, makehouses(n1,list). Why is this equivalent? (See p. 159.) 25 entering code at the terminal?- consult(user). : makehouses(0, []). : makehouses(n, [house(color,nat,pet,drink,cig) List]) : :- N>0, N1 is N - 1, makehouses(n1,list). Warning: (user://1:12): Singleton variables: [Color, Nat, Pet, Drink, Cig] : end_of_file. % control-d also works % user://1 compiled 0.00 sec, 796 bytes 26

14 the empty houses?- makehouses(5, List). List = [house(_g233, _G234, _G235, _G236, _G237), house(_g245, _G246, _G247, _G248, _G249), house(_g257, _G258, _G259, _G260, _G261), house(_g269, _G270, _G271, _G272, _G273), house(_g281, _G282, _G283, _G284, _G285)] 27 constraints % The Englishman lives in a red house. house( red, englishman, _, _, _) on List, % The Spaniard keeps a dog. house( _, spaniard, dog, _, _) on List, % The owner of the green house drinks coffee. house(green, _, _, coffee, _) on List, % The Ukrainian drinks tea. house( _, ukrainian, _, tea, _) on List, % etc. 28

15 on = member on is a user-defined infix operator version of member/2 see p for definition of member :- op(100,xfy,on). % a bit of grammar X on List :- member(x, List). equivalent to X on [X R]. X on [_ R] :- X on R. 29