1. LING 388: Language and Computers Sandiway Fong Lecture 5: 9/5

2. 2 Administrivia Homework 1 –reviewed today This Thursday –Computer Lab class –meet in Social Sciences 224 –Homework 2

3. 3 Homework 1 Review Use database bird(tweety). “tweety is a bird” bird(woody). “woody is a bird” (2pts) What do the following queries return as an answer? The queries ?- \+ \+ bird(tweety). ?- \+ bird(X). (6pts) Give the logical reasoning for each answer ?- \+ \+ bird(tweety). Yes logic ?- trace. Yes [trace] ?- \+ \+ bird(tweety). Call: (9) bird(tweety) ? creep Exit: (9) bird(tweety) ? creep Yes (... doesn’t really give enough information) \+ \+ bird(tweety) is true if \+ bird(tweety) is false if bird(tweety) is true but bird(tweety) is true since it’s in the database so \+ bird(tweety) is false and so \+ \+ bird(tweety) is true that’s why Prolog says Yes

4. 4 Homework 1 Review Use database bird(tweety). “tweety is a bird” bird(woody). “woody is a bird” (2pts) What do the following queries return as an answer? The queries ?- \+ \+ bird(tweety). ?- \+ bird(X). (6pts) Give the logical reasoning for each answer ?- \+ bird(X). No logic [trace] ?- \+ bird(X). Call: (8) bird(_G312) ? creep Exit: (8) bird(tweety) ? creep No (... shows us X gets instantiated as tweety ) \+ bird(X) is true if bird(X) is false but bird(X) is true when X=tweety so \+ bird(X) is false that’s why Prolog says No

5. 5 Today’s Topics More on recursion Last time –we defined last/2 and len/2 to pick the last element of a list and compute the length of a list, respectively. last([X],X). (base case) last([X|L],Y) :- last(L,Y).(recursive case) len([],0). (base case) len([X|L],N) :- len(L,M), N is M+1. (recursive case)

6. 6 Today’s Topics More on recursion –Look at taking apart and putting lists together –And symbols (atoms)

7. 7 Building and Taking Atoms Apart Built-in predicate atom_chars/2 has two modes of usage –atom_chars(symbol,list) 1.takes names apart ?- atom_chars(will,X). – X is the list of characters representing will 2.builds names from a list of characters ?- atom_chars(X,[ ’ J’,o,h,n]). –use quotes around capitalized J to avoid J being interpreted as a variable ?- atom_chars(X,[w,o,n,’’’’,t]). – ’’’’ denotes the single quote (or ’\’’ )

8. 8 Building and Taking Atoms Apart more example queries ?- atom_chars(has,[h,a,s]). –Yes ?- atom_chars(will,[w,X,l,l]). –X = i more example queries ?- atom_chars(X,Y). what happens here? ?- atom_chars(X,[J,o,h,n]).

9. 9 Building and Taking Lists Apart append/3 is a built-in predicate in SWI-Prolog defined as follows: –append(L1,L2,L3) is true –if list L3 is the linear concatenation of lists L1 and L2 [1,2] append [3,4] = [1,2,3,4] Example –?- append([1,2],[3,4],[1,2,3,4 ]). Note –append/3 has multiple modes of usage –i.e. it can be used to take apart as well as concatenate lists let’s run example queries –?- append([1],[2,3],X). –?- append(X,Y,[1,2]). –?- append(_,[X],[1,2,3]). –?- append(X,Y,Z). Note: –the underscore character ‘_’ is a variable with no name –it’s a variable but we don’t care about its final value –no binding will be reported by the interpreter for underscores

10. 10 Building and Taking Lists Apart append/3 can be defined recursively as follows: –app([],L,L). Base case –app([X|L1],L2,[X|L3]) :- app(L1,L2,L3). Recursive case we use the predicate name app/3 here to avoid a naming clash append/3 is already defined (built-in)

11. 11 Building and Taking Lists Apart app([],L,L). Base case app([X|L1],L2,[X|L3]) :- app(L1,L2,L3). Recursive case Base case says: –if we concatenate [] with an arbitrary list L, we get L Recursive case says: –if I have a list headed by X and the tail is L1, i.e. [X|L1], and I want to concatenate that to another list L2 –the answer is going to be a list headed by X –the tail is some other list L3 –where L3 is defined as –the concatenation of L1 and L2 Example ?- app([1,2,3],[4,5],L). L = [1|L3] where L3 is given by ?- app([2,3],[4,5],L3).

12. 12 Building and Taking Lists Apart app([],L,L). Base case app([X|L1],L2,[X|L3]) :- app(L1,L2,L3). Recursive case Base case says: –if we concatenate [] with an arbitrary list L, we get L Recursive case says: –if I have a list headed by X and the tail is L1, i.e. [X|L1], and I want to concatenate that to another list L2 –the answer is going to be a list headed by X –the tail is some other list L3 –where L3 is defined as –the concatenation of L1 and L2 Example ?- app([1,2,3],[4,5],L). L = [1|L3] where L3 is given by ?- app([2,3],[4,5],L3). L3 = [2|L3’] where L3’ is given by ?- app([3],[4,5],L3’).

13. 13 Building and Taking Lists Apart app([],L,L). Base case app([X|L1],L2,[X|L3]) :- app(L1,L2,L3). Recursive case Base case says: –if we concatenate [] with an arbitrary list L, we get L Recursive case says: –if I have a list headed by X and the tail is L1, i.e. [X|L1], and I want to concatenate that to another list L2 –the answer is going to be a list headed by X –the tail is some other list L3 –where L3 is defined as –the concatenation of L1 and L2 Example ?- app([1,2,3],[4,5],L). L = [1|L3] where L3 is given by ?- app([2,3],[4,5],L3). L3 = [2|L3’] where L3’ is given by ?- app([3],[4,5],L3’). L3’ = [3|L3”] where L3” is given by ?- app([],[4,5],L3”).

14. 14 Building and Taking Lists Apart app([],L,L). Base case app([X|L1],L2,[X|L3]) :- app(L1,L2,L3). Recursive case Base case says: –if we concatenate [] with an arbitrary list L, we get L Recursive case says: –if I have a list headed by X and the tail is L1, i.e. [X|L1], and I want to concatenate that to another list L2 –the answer is going to be a list headed by X –the tail is some other list L3 –where L3 is defined as –the concatenation of L1 and L2 Example ?- app([1,2,3],[4,5],L). L = [1|L3] where L3 is given by ?- app([2,3],[4,5],L3). L3 = [2|L3’] where L3’ is given by ?- app([3],[4,5],L3’). L3’ = [3|L3”] where L3” is given by ?- app([],[4,5],L3”). L3” = [4,5] ( Base case)

15. 15 Building and Taking Lists Apart app([],L,L). Base case app([X|L1],L2,[X|L3]) :- app(L1,L2,L3). Recursive case Example ?- app(L1,L2,[1,2]). matches app([],L,L). Base case L1 = [], L2 = [1,2] 1st answer

16. 16 Building and Taking Lists Apart app([],L,L). Base case app([X|L1],L2,[X|L3]) :- app(L1,L2,L3). Recursive case Example ?- app(L1,L2,[1,2]). matches app([],L,L). Base case L1 = [], L2 = [1,2] head of recursive case app([X|L1’],L2’,[X|L3’]) L1 = [1|L1’] L2 = L2’ [1,2] = [X|L3’] L3’=[2] ?- app(L1’,L2’,L3’). ?- app(L1’,L2’,[2]).

17. 17 Example ?- app(L1,L2,[1,2]). matches app([],L,L). Base case L1 = [], L2 = [1,2] head of recursive case app([X|L1’],L2’,[X|L3’]) L1 = [1|L1’] L2 = L2’ [1,2] = [X|L3’] L3’=[2] ?- app(L1’,L2’,L3’). ?- app(L1’,L2’,[2]). matches app([],L,L). Base case L1’ = [] L2’ = L = [2] Building and Taking Lists Apart app([],L,L). Base case app([X|L1],L2,[X|L3]) :- app(L1,L2,L3). Recursive case 2nd answer L1 = [1] L2 = [2]

18. 18 Building and Taking Lists Apart app([],L,L). Base case app([X|L1],L2,[X|L3]) :- app(L1,L2,L3). Recursive case Example ?- app(L1,L2,[1,2]). matches app([],L,L). Base case L1 = [], L2 = [1,2] head of recursive case app([X|L1’],L2’,[X|L3’]) L1 = [1|L1’] L2 = L2’ [1,2] = [X|L3’] L3’=[2] ?- app(L1’,L2’,L3’). ?- app(L1’,L2’,[2]). matches app([],L,L). Base case L1’ = [] L2’ = L = [2] Example contd. ?- app(L1’,L2’,[2]). matches head of recursive case app([X|L1”],L2”,[X|L3”]) L1’ = [2|L1”] L2’ = L2” [2] = [X|L3”] L3” =[] ?- app(L1”,L2”,L3”). ?- app(L1”,L2”,[]).

19. 19 Building and Taking Lists Apart app([],L,L). Base case app([X|L1],L2,[X|L3]) :- app(L1,L2,L3). Recursive case Example ?- app(L1,L2,[1,2]). matches app([],L,L). Base case L1 = [], L2 = [1,2] head of recursive case app([X|L1’],L2’,[X|L3’]) L1 = [1|L1’] L2 = L2’ [1,2] = [X|L3’] L3’=[2] ?- app(L1’,L2’,L3’). ?- app(L1’,L2’,[2]). matches app([],L,L). Base case L1’ = [] L2’ = L = [2] Example contd. ?- app(L1’,L2’,[2]). matches head of recursive case app([X|L1”],L2”,[X|L3”]) L1’ = [2|L1”] L2’ = L2” [2] = [X|L3”] L3” =[] ?- app(L1”,L2”,L3”). ?- app(L1”,L2”,[]). matches app([],L,L). Base case L1” = [] L2”=L=[] 3rd answer L1 = [1,2] L2 = []

20. 20 Building and Taking Lists Apart app([],L,L). Base case app([X|L1],L2,[X|L3]) :- app(L1,L2,L3). Recursive case Example ?- app(L1,L2,[1,2]). has 3 answers L1 = [], L2 = [1,2] L1 = [1], L2 = [2] L1 = [1,2], L2 = [] in each case it is true that L1 concatenated with L2 is [1,2]