CSE240 Lecture 26

Introduction to Programming Languages
Recursion in Prolog
(202204)

Transcript

1. jgs
   CSE 240
   Introduction to Programming Languages
   Lecture 26: Recursion in Prolog
   Dr. Javier Gonzalez-Sanchez
   javiergs@asu.edu
   javiergs.engineering.asu.edu | javiergs.com
   PERALTA 230U
   Office Hours: By appointment
   

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2. jgs
   Previously
   

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3. Javier Gonzalez-Sanchez | CSE240 | Spring 2020 | 3
   jgs
   Arithmetic Queries
   ?- X is 10+5.
   X = 15
   ?- 6 is 10+5.
   false
   ?- 15 is 10+5.
   true
   ?- 10+5 is X.
   Error: Arguments are not sufficiently instantiated
   ?- 10+5 is 10+5.
   false
   

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4. Javier Gonzalez-Sanchez | CSE240 | Spring 2020 | 4
   jgs
   Arithmetic Queries
   § X is 5 /3, X = 1.
   false
   § X is 5 /3, Y = 1.
   X=1.666666667,
   Y=1
   § X is 5 /3; X = 1.
   X = 1.6666666667
   X = 1
   

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5. jgs
   Rules and Recursion
   

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6. Javier Gonzalez-Sanchez | CSE240 | Spring 2020 | 6
   jgs
   Rules as functions
   § doSomething(A,B,P) :- P is A + B.
   ?-doSomething(3,2,P).
   ?-doSomething(3,2,10).
   

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7. Javier Gonzalez-Sanchez | CSE240 | Spring 2020 | 7
   jgs
   Recursion | Factorial
   % this is fact
   factorial(0, 1).
   % this is a rule
   % the factorial of F is N*F1
   % if N>0 and
   % N1 is N-1 and
   % the factorial of N1 is F1
   factorial(N, F) :-
   N>0,
   N1 is N - 1,
   factorial(N1, F1),
   F is N * F1.
   ?- factorial (3, W).
   W=6
   1. factorial(3, W)
   apply rule 2
   2. 3>0, N1 is 3-1, factorial (N1, F1), F is 3 *F1
   solve the individual parts
   true, 2 is 3-1, factorial (2, F1), F is 3*F1
   apply rule 2 again
   3. 2>0, N2 is 2-1, factorial (N2, F2), F1 is 2 * F2
   solve the individual parts
   true, 1 is 2-1, factorial (1, F1), F1 is 2*F2
   apply rule 2 again
   4. 1>0, N3 is 1-1, factorial (N3, F3), F2 is 2 * F3
   solve the individual parts
   true, 0 is 1-1, factorial (0, F3), F2 is 1*F3
   apply rule 1
   5. factorial (0, 1)
   F3 is 1
   6. Go back, replace F3 to get F2, then F2 to get F1,
   then F1 to get F.
   

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8. Javier Gonzalez-Sanchez | CSE240 | Spring 2020 | 8
   jgs
   Recursion | Factorial
   

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9. Javier Gonzalez-Sanchez | CSE240 | Spring 2020 | 9
   jgs
   Recursion | Fibonacci
   fib(0, 0).
   fib(1, 1).
   fib(X, Y) :- X > 1,
   X2 is X – 2, fib(X2, Y2),
   X1 is X – 1, fib(X1, Y1),
   Y is Y1 + Y2.
   

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10. Javier Gonzalez-Sanchez | CSE240 | Spring 2020 | 10
    jgs
    Recursion | Fibonacci
    

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11. jgs
    Test Yourselves
    C/C++
    

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12. Javier Gonzalez-Sanchez | CSE240 | Spring 2020 | 12
    jgs
    Test Yourselves
    § A recursive function rec is defined as follows
    Write a set of rules in Prolog to define the solution to this function.
    ï
    î
    ï
    í
    ì
    ³
    -
    -
    +
    -
    =
    £
    =
    3
    )
    3
    (
    *
    )
    2
    (
    )
    1
    (
    2
    1
    1
    0
    )
    (
    n
    n
    rec
    n
    rec
    n
    rec
    n
    n
    n
    rec
    

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13. jgs
    CSE 240 Introduction to Programming Languages
    Javier Gonzalez-Sanchez, Ph.D.
    javiergs@asu.edu
    Fall 2021
    Copyright. These slides can only be used as study material for the class CSE240 at Arizona State University.
    They cannot be distributed or used for another purpose.
    

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