CS代考计算机代写 prolog algorithm database Prolog

Prolog
Aditya Thakur
ECS 140A Programming Languages – Winter 2019

Outline
• Defining facts, and rules • Solving goals
• Lists in Prolog
• Unification
• Arithmetic, logical operators, and comparison operators
•Control in Prolog (goal and rule ordering)
• Search and backtracking
– Deriving search trees (examples: mylength1, mylength2,
prefix/suffix, etc.)
• Cuts (green and red)
2

Introduction
• What is logic programming?
– Use facts and rules to represent information
– Use logical deduction to answer queries • It is based on first-order logic
– Use logic to represent information
– Use proof rules/techniques to derive more information
• Logic programming view: Algorithm = Logic + Control
– Logic: What needs to be done (provided by the user/programmer)
– Control: How it needs to be done (derived by system based on the logic)
• Programming in terms of deduction rather than evaluation – Functional programming: expression evaluates to a value
– Imperative programming: program runs and updates some store
3

Key to Logic Programming
• You do not need to specify exactly how to compute a result
• But simply need to describe the form of the results
• The language system will determine how to compute the result
4

Building Blocks
• Everything in a Prolog program is built from Prolog terms
1. Constants
Numbers: 1, 2, 1.4
Atoms: a, b, c, parent, append
(Start with lowercase letter)
X, Y, Z
(Any name beginning with an uppercase letter or an underscore)
x(Y,Z), parent(adam,seth)
(An atom followed by a parenthesized, comma-separated list of terms)
2. Variables
3. Compound Terms
5

Prolog Language System
• Collection of facts and rules of inference • Fact: term followed by a period
• Example of facts:
parent(kim,holly).
parent(margaret,kim).
parent(margaret,kent).
parent(esther,margaret).
parent(herbert,margaret).
parent(herbert,jean).
In this example, assume facts are saved in a file family.pl
6

Runtime System
http://www.swi-prolog.org/
https://swish.swi-prolog.org Web-based tool for running Prolog
% swipl
| ?- consult(family).
7

More on Facts
• Facts express unconditional information:
parent(margaret,kent).
parent(fred,john).
• More examples
|- ? parent(margaret,kent). true
|- ? parent(margaret,X).
X = kim ? ;
% margaret is kent’s parent
X = kent ?
true
|- ? parent(X,jean).
X = herbert
true
|- ? parent(margaret,X), parent(X,holly). X = kim ? ;
false
% means “anded” together
8
; means to show more answers

Rules
• A rule is defined in the following way
:- , , …,
HEAD CONDITIONS
– Means if AND AND … AND then
• A fact is just a special case of a rule: with no conditions • Example:
grandparent(GP,GC) :- parent(GP,P), parent(P,GC).
|- ? grandparent(X,Y).
X = margaret,
Y = holly ? ;
…
9

Goals (Queries)
• Questions that we can ask a system • Goal of the form: g1,g2,…,gk
–The gi’s are called subgoals
|- ? parent(X,kim), parent(X,kent).
|- ? parent(_,X). % anybody with a parent • Notice the closed world assumption
– Everybody has a parent, but we only get whatever is represented in the database
– Plus what can be deduced from the facts and rules – Nothing more
10

Lists in Prolog
• Lists:
[] empty list [a,b,c] listwitha,b,&c
• Head/tail notation
[a, b, c] = [a | [b | [c | []]]]
consider the same list in Lisp
(1 2 3) = (1 . (2 . (3 . nil)))
11

Append Predicate (1)
• Let’s look at the append relationship on lists [] empty list
[1,2,3] list with three elements
• append(X,Y,Z) is true if append(X,Y) = Z
append([], [1], [1]).
append([1], [], [1]).
append([1,2], [3], [1,2,3]).
…
…
• This is an infinite set
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Append Predicate (2)
• We can ask some questions about append append([],[],[]). true append([],[1],[1]). true append([1,2],[3], [1,2,3]). true append([1],[],[]). false
• What if we want to ask append(X,[1],[1]).
append([1,2],X,[1,2,3]).
append(X,Y,[1,2]).
append(X,Y,Z).
13

Defining our own Append
% swipl
| ?- [user].
myappend([],L,L).
myappend([X|L1],L2,[X|L3]) :- myappend(L1,L2,L3).
user compiled, 3 lines read – 476 bytes written, 25598 ms
true
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myappend
| ?- myappend([1],[2,3],X). X = [1,2,3]
true
| ?- myappend(X,[2,3],[1,2,3]).
X = [1] ? ;
false
| ?- myappend(X,[2,3],[1,3,2]).
false
| ?-
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myappend
| ?- myappend(X,Y,[1,2,3]).
X = []
Y = [1,2,3] ? ;
X = [1]
Y = [2,3] ? ;
X = [1,2]
Y = [3] ? ;
X = [1,2,3]
Y = [] ? ;
false
| ?-
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Unification (1)
• How does prolog do deductions? – Based on a process called unification
• Unification is a process to match two or more terms
– Similar to pattern matching
– But pattern-matching goes only one way: variable to a pattern – Unification can go both ways
• Example
– foo(X,b) and foo(a,Y) matches if X = a, Y = b – f(a,b) is an instance of both
• Unification attempts to find if there is a common instance of two terms
• Definition:Two terms t1 and t2 unify, if they have a common instance
17

Unification (2)
• When does unification happen?
– When two terms are checked for equality |- ? foo(X,b) = foo(a,Y).
• Example:
| ?- foo(X,b)=foo(a,Y).
X=a Y=b true
– When a rule is applied | ?- [user].
same(X,X).
true
| ?- same(foo(X,b), foo(a,Y)). X=a
Y=b
true
18

Control in Prolog (1)
• How is a prolog program evaluated?
– Goal order: subgoals are processed left-to-right – Rule order: rules are applied top-to-bottom
• Answer to a query is affected by
– Goal order in the query
– Rule order in the database of facts and rules
20

Control in Prolog (2)
1 current_goal := query
2 3
4 5 6 7 8 9
while current_goal is nonempty do
assume current_goal = g1,…,gk
choose the leftmost goal g1
if there is a rule that applies to g1 then
select first such rule H :- C1,…,Cn
let s be the most general unifier of H and g1
current_goal := s(C1), …, s(Cn), s(g2), …, s(gk)
else
backtrack
10
11 end-while
12 success
Ignore 􏱚most general unifier􏱛
Just think that it is a solution to a unification equation
21

Arithmetic in Prolog
• Prolog has the usual +, -, *, etc. operators for arithmetic
– And each implementation may have many more – Use infix notation: 3 * 4
• Arithmetic expressions not evaluated unless forced to | ?- X=2+3.
X = 2+3 NOT X = 5
• Use “is” to force their evaluations
| ?- X is 2+3. X =5
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Numeric Comparisons
• Six special operators for comparing numeric values <, >, =<, >=, =:=
• What is the difference between the following:
X = Y.
X is Y.
X =:= Y.
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More Unification Examples
| ?- X = 2 + 3.
X = 2+3
| ?- 5 = 2 + 3.
false
| ?- 2 + 3 = 2 + X. X=3
| ?- 2 * X = Y * (3 + Y). X = 3+2
Y=2
| ?- X = X + 2.
X=X+2.
• How unification is done?
– Unification algorithm (Robinson’s) – Union-find
– We will not cover here
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Logical Operators (1)
• true
– Goal always succeeds
• fail
– Always fails
• = (equality)
t1 = t2 succeeds if t1 and t2 unifies
– Examples
| ?- 5 = 5 succeeds |?-5=2+3 fails
•= (inequality) – Negation of =
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Logical Operators (2)
• not (negation)
– not(X) succeeds if fails to satisfy X
– order matters
• X=2, not(X=1) succeeds because
– To satisfy X=2, we can unify X with 2
– And then 2=1 fails, thus not(X=1) is satisfied • not(X=1), X=2 fails because
– X=1 succeeds, thus not(X=1) fails
– Then the whole goal fails
• Different from the logical interpretation where both are equivalent
• g1; g2: disjunction of goals
– Its meaning is: g1 OR g2
– The goal succeeds if g1 succeeds or g2 succeeds
26

Other Predicates for Lists
• Provable if the list Y contains the element X member(X,Y).
• Provable if X is a list of length Y length(X,Y).
• Provable if Y is a list that contains the elements of list X in reverse order
•
reverse(X,Y).
Provable when List1, with Elem removed, results in
List2.
select(Elem,List1,List2).
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Example: Length of a List
• length of a list (correct and buggy versions): mylength1([],0).
mylength1([_|T],L) :- mylength1(T, K), L is K+1.
mylength2([],0).
mylength2([_|T],L) :- K is L-1, mylength2(T, K).
What happens when we pose the following queries?
mylength1([1,2,3],3). ✓ mylength2([1,2,3],3). ✓
How about the following queries?
Show the search tree
Show the search tree
mylength1([1,2,3],X). ✓ mylength2([1,2,3],X). ✗
WHY?
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Trace for mylength1
| ?- mylength1([1,2,3],3).
1 1 Call: mylength1([1,2,3],3) ?
2 2 Call: mylength1([2,3],_348) ?
3 3 Call: mylength1([3],_372) ?
4 4 Call: mylength1([],_396) ?
4 4 Exit: mylength1([],0) ?
5 4 Call: _424 is 0+1 ?
5 4 Exit:1is0+1?
3 3 Exit: mylength1([3],1) ?
6 3 Call: _453 is 1+1 ?
6 3 Exit:2is1+1?
2 2 Exit: mylength1([2,3],2) ? 7 2 Call:3is2+1?
7 2 Exit:3is2+1?
1 1 Exit: mylength1([1,2,3],3) ?
(1 ms) true
29

Trace for mylength2
| ?- mylength2([1,2,3],3).
1 1 Call: mylength2([1,2,3],3) ?
2 2 Call: _351 is 3-1 ?
2 2 Exit:2is3-1?
3 2 Call: mylength2([2,3],2) ?
4 3 Call: _403 is 2-1 ?
4 3 Exit:1is2-1?
5 3 Call: mylength2([3],1) ?
6 4 Call: _455 is 1-1 ?
6 4 Exit:0is1-1?
7 4 Call: mylength2([],0) ?
7 4 Exit: mylength2([],0) ?
5 3 Exit: mylength2([3],1) ?
3 2 Exit: mylength2([2,3],2) ?
1 1 Exit: mylength2([1,2,3],3) ?
(1 ms) true
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Practice Exercise
• reversing a list: myreverse([], []).
myreverse([H|T], L1) :- myreverse(T,L2),myappend(L2,[H],L1).
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• Example :
Another Example
append([],Y,Y).
append([H|X],Y,[H|Z]) :- append(X,Y,Z). prefix(X,Z) :- append(X,Y,Z).
suffix(Y,Z) :- append(X,Y,Z). sublist1(S,L) :- prefix(X,L), suffix(S,X). sublist2(S,L) :- suffix(S,X), prefix(X,L).
• A graphical representation of the logic of sublist1 and sublist2
|<-- X -->| +——-+-+—–+ | |S| | +——-+-+—–+ |<-- L -->|
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How Does Search Work – Example 1
| ?- suffix([a],L), prefix(L,[a,b,c]).
L = [a].
• Search tree with no backtracking (_i fresh variables)
suffix([a],L), prefix(L,[a,b,c])
pick this subgoal and match rule: suffix(Y,Z) :- append(X,Y,Z). unifier: Y = [a], Z = L, X = _1 (X can be anything, let’s use _1)
append(_1,[a],L), prefix(L,[a,b,c])
pick this subgoal and match the fact: append([], Y, Y). unifier: Y = [a], L = [a]
prefix([a], [a,b,c])
pick this subgoal and match the rule: prefix(X, Z) :- append(X, Y, Z). unifier: X = [a], Z = [a,b,c], Y = _2 (call Y _2)
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How Does Search Work – Example 1
append([a], _2, [a,b,c])
pick this subgoal, match rule:
append([H|X], Y, [H|Z]) :- append(X,Y,Z).
unifier: H = a, X = [], Z = [b,c], Y = _2 (still anything)
append([], _2, [b,c])
pick this subgoal and match the fact: append([], Y, Y). unifier: Y = [b,c]
yes
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How Does Search Work – Example 2
| ?- suffix([b],L), prefix(L,[a,b,c]).
L = [a,b].
• Search tree with backtracking
suffix([b],L), prefix(L,[a,b,c]).
matches rule suffix(Y, Z) :- append(X, Y, Z). unifier: Y = [b], Z = L, X = _1
append(_1, [b], L), prefix(L, [a,b,c])
matches rule append([], Y, Y) unifier: Y = L = [b]
prefix([b], [a,b,c])
matches prefix(X, Z) :- append(X, Y, Z). unifier: X = [b], Z = [a,b,c], Y = _2
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How Does Search Work – Example 2
append([b], _2, [a,b,c])
matches no rule or fact, backtrack TO:
append(_1, [b], L), prefix(L, [a,b,c])
match append([H|X], Y, [H|Z]) :- append(X,Y,Z) unifier: _1 = [_3|_4], Y = [b], L =[_3|_5]
append(_4, [b], _5), prefix([_3|_5], [a,b,c])
match append([], Y, Y) unifier: _4 = [], Y = [b] = _5
prefix([_3, b], [a,b,c])
match prefix(X, Z) :- append(X, Y, Z) unifier: X = [_3,b], Y = _6, Z = [a,b,c]
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How Does Search Work – Example 2
append([_3,b], _6, [a,b,c])
match append([H|X], Y, [H|Z]) :- append(X, Y, Z). unifier: H = a = _3, Z = [b,c], Y = _6, X = [b]
append([b], _6, [b,c])
match append([H|X], Y, [H|Z]) :- append(X, Y, Z). unifier: H = b, Z = [c], Y = _6, X = []
append([], _6, [c])
match append([], Y, Y). unifier: Y = _6 = [c]
yes
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Order in the Goals and Rules (1)
• What’s the difference between the the sublist predicates? sublist2(S,L) :- suffix(S,X), prefix(X,L).
sublist1(S,L) :- prefix(X,L), suffix(S,X).
Hint: use same example where suffix is [b] and the list is [a,b,c].
suffix([b],L), prefix(L,[a,b,c]). ✗ prefix(L,[a,b,c],suffix([b],L). ✓
WHY?
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Order in the Goals and Rules (2)
• What’s the difference between the following definitions? ancestor1(X,Y) :- parent(X,Y).
ancestor1(X,Y) :- parent(Z,Y),ancestor1(X,Z). ancestor2(X,Y) :- parent(Z,Y),ancestor2(X,Z).
ancestor2(X,Y) :- parent(X,Y).
Minor difference: same solutions but in different order Sample query:
-? ancestor1(X,holly).
39

Order in the Goals and Rules (3)
• What’s the difference between the following definitions?
ancestor3(X,Y) :- ancestor3(X,Z),parent(Z,Y).
ancestor3(X,Y) :- parent(X,Y).
ancestor4(X,Y) :- parent(X,Y).
ancestor4(X,Y) :- ancestor4(X,Z),parent(Z,Y).
Very different to ancestor1 and ancestor2 Non-terminating in all or some cases (left recursive!) Sample query:
-? ancestor3(kim,holly).
40

Cuts
• The cut, in Prolog, is a goal, written as !, which always succeeds, but cannot be backtracked past.
–Cuts allow you to prune out or “cut out” an unexplored part of a prolog search tree
•Cuts are written as
C :- A_1, A_2, …, A_m, !, B_1, B_2, …, B_n.
– Semantics: tells control to backtrack past A_1, …, A_m without considering any of the rules for them
41

Cuts
• Rules for max: max(A,B,B) :- A =< B. max(A,B,A) :- A > B.
• Optimize the code:
max(A,B,B) :- A =< B,!. % no need to check other rule max(A,B,A) :- A > B. % if backtracking
-? max(1,3,M). M=3
The meaning of the program remains unchanged, but the program is more efficient
What happens if we remove the cut?
42

Cuts
• Rules for max: max(A,B,B) :- A =< B. max(A,B,A) :- A > B.
• Can simplify by removing redundant comparison: max(A,B,B) :- A =< B. max(A,B,A). ?- max(1,3,M). M = 3; M = 1; 43 Cuts • Attempt 1: max(A,B,B) :- A =< B,!. max(A,B,A). ?- max(1,3,M). M=3 ?- max(1,3,1). true • Attempt 2: max(A,B,C) :- A =< B,!,B=C. max(A,B,A). ?- max(1,3,1). false What happens if we remove the cut? 44 Cuts - Example • DB and search tree for rule: b :- c. b :- d. d. e. c :- 1 = 2. a(1) :- b. a(2) :- e. ?- a(X). X = 1; X = 2; 45 Cuts – Example • Ifwechangethefirstruleto b:-!,c. b :- !, c. b :- d. d. e. c :- 1 = 2. a(1) :- b. a(2) :- e. ?- a(X). X = 2; 46 Practice Exercise • How to represent binary trees? – Use predicates, e.g., "tree” void: empty tree tree(K,L,R): labeled with K, left subtree L, right subtree R – Examples: tree(2,void,void). tree(4,tree(2,void,void),tree(10,void,void)). 47 Practice Exercise • tree height height(void, 0). height(tree(_, L, R), H) :- height(L, H1), height(R, H2), H1 > H2,
H is H1+1.
height(tree(_, L, R), H) :- height(L, H1),
height(R, H2),
H1 =< H2, H is H2+1. 48 Practice Exercise • Write a predicate insert to insert an element into a binary search tree insert(K,void,tree(K,void,void)). % base case insert(K,tree(K,L,R),tree(K,L,R)). % no duplicates insert(K,tree(N,L,R),tree(N,Lnew,R)) :- K < N, insert(K,L,Lnew). % to the left insert(K,tree(N,L,R),tree(N,L,Rnew)) :- K > N, insert(K,R,Rnew). % to the right
49

Practice Exercise
• Write a predicate member (binary search) member(K, tree(K,_,_)).
member(K, tree(N,L,_)) :- K < N, member(K, L). member(K, tree(N,_,R)) :- K > N, member(K, R).
• How does it work?
• Compare to what you have to do in other languages
• Here, it is nice because we don’t even need to create a new datatype for tree, but simply how you want to traverse the tree
50

Additional Practice Exercises
• factorial: fac(N,N!)
fac(0,1).
fac(N,R) :- N > 0,
M is N-1,
fac(M,S),
R is S*N.
•Fibonacci: fib(N,M) — M is the Nth Fibonacci number
fib(0,0).
fib(1,1).
fib(N,M) :- N > 1,
N1 is N-1,
N2 is N-2,
fib(N1,M1),
fib(N2,M2),
M is M1+M2.
51

Additional Practice Exercises
• quicksort(L,S): L is the input, and S is the sorted list
• split(X,L,K,M): X is the value to be splitted upon L is the input list
K is the smaller list
M is the larger list
quicksort([],[]).
quicksort([X|L],K) :- split(X,L,L1,L2),
quicksort(L1,K1),
quicksort(L2,K2),
append(K1,[X|K2],K).
split(_,[],[],[]).
split(X,[Y|L],K,[Y|M]) :- X < Y, split(X,L,K,M). split(X,[Y|L],[Y|K],M) :- X >= Y, split(X,L,K,M).
52

Summary
• Computation through manipulation of relationships
• A Prolog program consists of
– Facts: unconditional relationships among terms
– Rules: relationships among terms that may be true under certain conditions
– Queries that must be true give a set of facts and rules • A prolog program merely specifies the various
relationships among terms
• Prolog system defines how the computation is carried out
– Unification of variables and terms: assignment of values to variables so that relationships match
– Implement control by searching the different rules: Employ backtracking to search for all solutions
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