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95 lines
3.1 KiB
Prolog
95 lines
3.1 KiB
Prolog
% Lab 2 - Recursive Arithmetic in Prolog
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% LABSHEET:
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% https://teaching.csse.uwa.edu.au/units/CITS3005/labs/lab02-arithmetic.pdf
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% Build a Prolog program to do recursive arithmetic. It should have
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% 1. a constant, zero, to represent zero.
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zero.
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% 2. a function, next(X), that represents the next number (so next(zero) represents one).
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next(X) :- X.
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% 3. a predicate, sum(X,Y,Z), which is true if X + Y = Z.
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% sum(zero,zero,Z) :- Z is zero.
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% sum(zero,Y,Z) :- Z is Y.
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% sum(next(zero),Y,Z) :- Z is next(Y).
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sum(zero,zero,zero). % CASE 1 : 0 + 0 == 0
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sum(zero,X,X). % CASE 2 : 0 + X == X
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sum(next(X),Y,next(Z)) :- sum(X,Y,Z). % CASE 3 : (X+1) + Y == (Z+1) (Works based on previous two cases???)
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% TEST 1:
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% sum(next(zero),next(zero),next(next(zero)))
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%=sum(zero,next(zero),next(zero)) [ TRUE ]
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% TEST 2:
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% sum(next(next(zero)),zero,next(zero))
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%=sum(next(zero),zero,zero) [ FALSE ]
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% 4. a predicate, mult(X,Y,Z) which is true if X × Y = Z.
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mult(zero,zero,zero).
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mult(zero,_,zero).
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mult(next(zero),X,X).
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mult(next(X),Y,Z) :- mult(X,Y,Zp), sum(Y,Zp,Z). % (x+1)*y == x*y+y
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% 5. a predicate, equals(X,Y) which is true if X = Y .
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equals(X,X).
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% 6. a predicate, lessThan(X,Y) which is true if X < Y ¿
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lessThan(zero,next(_)).
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lessThan(next(X),next(Y)) :- lessThan(X,Y). % X+1 < Y+1 => X < Y. As X->0, Y->N => 0 < N.
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% binary_(zero,0).
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% binary_(next(X),IntValue) :- binary_(X,PrevIntValue), write(PrevIntValue), IntValue is PrevIntValue+1.
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% binary(X) :- binary_(X,Buffer), write(Buffer).
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% Test your program and consider the efficinecy of the implementations. Is there a way to get faster annswers?
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% Next, add additional functions:
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% 1. Write a program binary(X) that will print the binary representation of X out, so for example, binary(next(next(zero)
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% will print out 10.
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% NOTE: Binary array is in reverse!
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increment_b([],[]).
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increment_b([ODigit | ORest],[Digit | Rest]) :-
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ODigit =:= 0 -> Digit is 1,
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equals(Rest,ORest);
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ODigit =:= 1 -> Digit is 0,
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(
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equals(ORest,[]) -> equals(Rest,[1]);
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increment_b(ORest,Rest)
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).
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% NOTE: Binary array is in reverse!
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binary_arr(zero,[0]).
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binary_arr(next(X),NextBinary) :-
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binary_arr(X,Binary),
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increment_b(Binary,NextBinary).
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binary_arr_print([]).
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binary_arr_print([Digit|Rest]) :- write(Digit), binary_arr_print(Rest).
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binary(X) :-
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binary_arr(X,BinaryArrRev),
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reverse(BinaryArrRev, BinaryArr),
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binary_arr_print(BinaryArr).
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% 2. Implement predicates for odd, even and prime numbers.
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even(zero).
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even(next(X)) :- \+even(X).
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odd(X) :- \+even(X).
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% Cheating a bit here ... would be harder doing the whole primality test in the above notation, since I don't have a modulo predicate defined
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to_integer(zero,0).
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to_integer(next(X),Integer) :- to_integer(X,OInteger), Integer is OInteger + 1.
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not_prime_i(I,Comp) :-
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Comp*Comp < I+1,
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(I mod Comp =:= 0; I mod (Comp + 2) =:= 0;
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not_prime_i(I,Comp+6)).
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prime_i_entry(I) :-
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I > 1, (I =:= 2; I =:= 3;
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I mod 2 =\= 0, I mod 3 =\= 0,
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\+not_prime_i(I, 5)).
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prime(X) :- to_integer(X,I), prime_i_entry(I). |