Programming Languages

Transcription

1 Programming Languages Jörg Kreiker Chair for Theoretical Computer Science Prof. Esparza TU München winter term 2010/2011

2 Lecture 13 Databases

3 3 Organizational Issues assignments corrected today: final mini test results of assignment and mini test bonus available on campus.tum.de next week

4 4 Exam Thursday, Feb 17; we meet at am! seating plan will be at door large lecture hall in physics building: PH1 bring only a pen, no auxiliary means allowed ten minutes to read exam 60 minutes to answer questions 40 points, app. 2/3 Haskell, 1/3 Prolog

5 5 A Love Story Luca software developer at JuventusFC die-hard Java enthusiast takes the programming language class at TUM in love with Monica

6 6 A Love Story Monica mathematician at insurance company passionate Prolog hacker despises Java undecided about Luca

7 A Task Here is your task write a Prolog predicate allperm(l, P) such that P is the list of all permutations of L until the end of the lecture then...

8 ??????????? 8

9 9 The Enemy Sébastien Monica s ex boy-friend Luca s manager at JuventusFC part-time rugby player database nerd

10 10 The Enemy country Fix this!!! (spain, capital(madrid, ), languages([spanish]), cities([barcelone, valencia, bilbao]), success([football(ec,2008)])).

11 Facts as data/knowledge-base We can use facts to represent knowledge. For instance: an almanac. country (spain, capital(madrid, ), languages([spanish]), cities([barcelone, valencia, bilbao]), success([football(ec,2008)])).

12 Queries Define the following predicates! country/1 yields country names capital/1 yields capitals bigcapital/1 yields capitals with population > 10 6 sport(x,y) yields countries with at least one success in sport Y huge(x) yields countries with more than one language, success, and city define a predicate to find all countries (ring a bell?)

13 13 Changing information Spain won the football WC 2010 (unfortunately) need to replace the success subterm of country(spain,..) this is a special case of substitution (known from type inference) in general, replace each occurrence subterm S1 by subterm S2 in term T1 to obtain T2 subst(s1,s2,t1,t2)

14 Term manipulations In order to do this, we need some ways of manipulating terms. =.. is a binary predicate (in infix notation) that decomposes terms in functor and arguments X =.. [F Args] succeeds, if X = f(t 1,..., t k ) and F = f and Args = [t 1,..., t k ] examples:?- f(a,b) =.. [F Args].?- X+a =.. [F Args].?- [1,2,3] =.. [F Args].

15 15 Types Terms can be tested for their types by predicates. atom(x) succeeds if X is bound to an atom. integer(x) succeeds if X is bound to an integer. atomic(x) :- atom(x) ; integer(x). var(x) succeeds if X is bound to a variable. nonvar(x) analogously.

16 16 Substitution subst(s1,s2,t1,t2) replace S1 by S2 in T1 to obtain T2 solution in program file on web site observation 1: what if T1=S1? observation 2: what if T1=T2? observation 3: what if T1=f(A1,..,Ak), T2=f(B1,..,Bk)? how do we make Spain WC 2010 winner?

17 17 All right then... Get lost!

18 18 Good friends always help Fabio Fibonacci Luca s best friend nerd found famous sequence 1, 1, 2, 3, 5, 8,... and its relation to rabbit population

19 19 Luca, how do you implement Fibonacci? write a predicate fib/2

20 20 Easy, eh? fib(1,1). fib(2,1). fib(n,x) :- N>2, N1 is N-1, N2 is N-2, fib(n1,x2), fib(n2,x2), X is X1+X2.

21 Not quite... re-computation of results solution (in Haskell): dynamic programming solution in Prolog: self-modifying code

22 Adding/deleting information self-modifying code: add and remove clauses assert/1: add a clause to the program can be done in the interpreter and in programs asserta/1 adds clause at beginning assertz/1 adds clause at end predicates need to be declared dynamic:?- dynamic(p). retract/1: removes matching clause from program

23 23 What about Fibonacci then? We can store intermediate results using assert! fib(1,1). fib(2,1). fib(n,x) :- N>2, N1 is N-1, N2 is N-2, fib(n1,x2), fib(n2,x2), X is X1+X2, asserta( fib(n,x) ). This is indeed the most common use of assert.

24 24 Caution self-modifying code is error-prone! assert and retract are impure Prolog just like cuts, negation by failure, arithmetic

25 Aaah, I see... 25

26 26 You again?!?!? Write this mini test!

27 27 The Key Idea find(x,g, ) :- call( G ), assertz( stack(x) ), fail. find(,,xs) :- assertz( stack(end) ), c(xs). c(l) :- retract( stack(x) ),!, ( X == end,!, L=[] ; L = [X Rest], c(rest) ).

28 28 This is how it works call( G ) calls goal G, succeeds if G succeeds, fails otherwise example:?- f(x, perm([1,2,3],x), Ps). call yields a solution, binding it to X solution is saved on a stack fail makes find fail: backtracking! the second clause for find is only called, when call( G ) fails then the end of the stack is written to the program c(l) collects everything stored on stack so we can find all permutations

29 29 allperm(l,ps) :- f(x, perm(l,x), Ps). So, Monica Do you accept this solution?

30 True Love 30

31 31 But, Luca... Your f predicate already exists in Prolog It s called findall/3 Examples?- findall(pair(x,y), take(x,y,[1,2,3]), Z). allperm(x,y) :- findall(z, perm(x,z), Y).?- findall(x, (retract(fib(x,1)), assert(a(x))), Y). next week: Sudoku