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### Prolog Programming(Volume 3)

Dr W.F. Clocksin

Mapping: The Full Map

sqlist( , )

sqlist([], []).

sqlist([X|T], [Y|L]) :- Y is X * X, sqlist(T, L).

List of numbers

List of squares of numbers

Mapping: The Full Map (cont’d)

Here is an exercise in compound terms. Map each list element (a number) to a term s(A,B) where A is the number and B is its square.

Input: [1, 9, 15]

Output: [s(1,1), s(9, 81), s(15, 225)]

sqterm([], []).

sqterm([X|T], [s(X,Y)|L]) :-

Y is X * X,

sqterm(T, L).

General Scheme for Full Map

/* fullmap(In, Out) */

fullmap([], []).

fullmap([X|T], [Y|L]) :-

transform(X,Y), fullmap(T, L).

Here is a typical transformation table…

Simple Transformation

transform(cat, gatto).

transform(dog, cane).

transform(hamster, criceto).

transform(X, X).

?- fullmap([cat, dog, goat], Z).

Z = [gatto, cane, goat]

A ‘catchall’ rule

Worksheet 11: Multiple Choices

Sometimes the map needs to be sensitive to the input data:

Input: [1, 3, w, 5, goat]

Output: [1, 9, w, 25, goat]

squint([], []).

squint([X|T], [Y|L]) :-

integer(X),

Y is X * X, squint(T, L).

squint([X|T], [X|L]) :- squint(T,L).

Just use a separate clause for each choice

Multiple Choices

Using the infix binary compound term *, it is easy enough to give some mathematical reality to the map:

Input: [1, 3, w, 5, goat]

Output: [1, 9, w*w, 25, goat*goat]

squint([], []).

squint([X|T], [Y|L]) :-

integer(X),

Y is X * X, squint(T, L).

squint([X|T], [X*X|L]) :- squint(T,L).

Worksheet 12: Partial Maps

Given an input list, partially map it to an output list.

evens([], []).

evens([X|T], [X|L]) :- 0 is X mod 2, evens(T, L)

evens([X|T], L) :- 1 is X mod 2, evens(T, L).

?- evens([1, 2, 3, 4, 5, 6], Q),

Q = [2, 4, 6].

General Scheme for Partial Maps

partial([], []).

partial([X|T], [X|L]) :- include(X), partial(T, L)

partial([X|T], L) :- partial(T, L).

For example,

include(X) :- X >= 0.

?- partial([-1, 0, 1, -2, 2], X),

X = [0, 1, 2].

Partial Maps

Exercise: Write a program that ‘censors’ and input list, by making a new list in which certain prohibited words do not appear. To do this, define a predicate prohibit such that prohibit(X) succeeds if X is a censored word. For example,

prohibit(bother).

prohibit(blast).

prohibit(drat).

prohibit(fiddlesticks).

Worked example

censor([], []).

censor([H|T], T1) :- prohibit(H), censor(T, T1).

censor([H|T], [H|T1]) :- censor(T, T1).

Worksheet 13: Removing Duplicates

setify([], []).

setify([X|T], L) :- member(X,T), setify(T, L).

setify([X|T], [X|L]) :- setify(T, L).

WS14: Partial Maps with a Parameter

Before, we saw how to prevent loops (when searching a graph) by keeping a ‘trail’ of the nodes visited so far. Here is another way to think about the problem.

Keep a list of all the visitable nodes. As each node is visited, strike it off the list and pass the reduced list to the recursive call. Backtracking restores the old list, so alternative paths can be searched.

WS14: Partial Maps with a Parameter

First, a predicate to reduce the list.

Goal reduce(L,X,M) succeeds for input list L, term X and output list M, when M contains the elements of L except for the first occurrence of X. Thus, X is a parameter that controls which element will be omitted from the output list.

reduce([X|T], X, T).

reduce([H|T], X, [H|L]) :- reduce(T, X, L).

Use this for searching as follows

path(X, X, L).

path(X, Y, L) :-

a(X, Z),

reduce(L, Z, L1),

path(Z, Y, L1).

Using the arc relation on Worksheet 9,

?- path(a, b, [a, b, c, d, e, f, g, h]).

yes.

WS 15: Multiple Disjoint Partial Maps

Maps a list into several disjoint lists.

Separating sheep from goats. Define the predicate herd, such that the goal herd(L, S, G) succeeds if S is a list of all the sheep in L and G is a list of all the goats in L.

herd([], [], []).

herd([sheep|T], [sheep|S], G) :- herd(T, S, G).

herd([goat|T], S, [goat|G]) :- herd(T, S, G).

What do the following goals do?

?- herd([sheep, goat, goat, sheep, goat], X, Y).

?- herd([goat, sheep, stone, goat, tree], X, Y).

?- herd(X, [sheep, sheep], [goat, goat]).

How to deal with other objects?

?- herd([goat, sheep, stone, goat, tree], X, Y).

The above goal fails. Instead, we could ignore other objects by having a catchall:

herd([X|T], S, G) :- herd(T, S, G).

Or, collect them in a list also

input sheep goats extras

herd([], [], [], []).

herd([sheep|T], [sheep|S], G, E) :- herd(T, S, G, E).

herd([goat|T], S, [goat|G], E) :- herd(T, S, G, E).

herd([X|T], S, G, [X|E]) :- herd(T, S, G, E).

Example: Alternating a list

Maps an input list into a pair of lists, alternating the elements.

alternate([a, b, c, d, e, f], [a, c, e], [b, d, f])

Alternating by index (which element of the input list it is):

altx([], [], []).

altx([A, B | T], [A | T1], [B| T2]) :- altx(T, T1, T2).

Or, alternating by value

Maps an input list into a pair of lists, alternating the elements.

altv([1, 2, 3, 4], [2, 4], [1, 3])

altv([], [], []).

altv([A | T], [A | T1], B) :-

0 is A mod 2, altv(T, T1, B).

altv([B | T], A, [B | T2]) :-

1 is B mod 2, altv(T, A, T2).

See how this is the same as sheep and goats?

WS 17: Full maps with state

/* mapsum(list of integers, cumulative sum) */

?- mapsum([1, 3, 2, 5, 4], X).

X = [1, 4, 6, 11, 15].

0

mapsum

Use an accumulator as a state variable that helps to determine the value of each element of the output list.

/* ms(input list, accumulator, output list) */

mapsum(A, B) :- ms(A, 0, B)

ms([], _, []).

ms([H|T], N, [C|L]) :- C is H + N, ms(T, C, L).

initialise accumulator

anonymous variable (don’t care about accumulator if end of list)

WS 18: Sequential Partial Map with State

Example: Run length encoding is a useful data representation and compression technique.

Represent sequential ‘runs’ of N identical terms T as the term N*T, for example

[12, 2, 2, w, 3, 3, s, s, s]

maps to

[1*12, 2 * 2, 1 * w, 2 * 3, 3 * s].

WS 18: Sequential Partial Map with State

[12, 2, 2, w, 3, 3, s, s, s][1*12, 2 * 2, 1 * w, 2 * 3, 3 * s]

runcode(input, current term, current count, output)

runcode([], C, N, [N*C]).

runcode([H|T], H, N, Z) :-

N1 is N+1, runcode(T, H, N1, Z).

runcode([H|T], C, N, [N*C|Z]) :-

H \== C, runcode(T, H, 1, Z).

Sequential Partial Map with State

runcode([], C, N, [N*C]).

runcode([H|T], H, N, Z) :-

N1 is N+1, runcode(T, H, N1, Z).

runcode([H|T], C, N, [N*C|Z]) :-

H \== C, runcode(T, H, 1, Z).

End of list? Deal withleftover accumulator values

Have seen this before...

Otherwise, discharge the current run...

...and start a new run

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