How to print a set of numbers using repeat/0 in prolog?

prolog repeat
prolog print string

in an exercice I have to write a predicate that prints numbers < N using repeat/0, here is what I have tried :

loop(N) :-
    repeat,
        write(N), nl,
        N is N-1,
        write(N),
        N > 0,
    !.

but it only prints the given number N Endlessly.

Any Idea how to implement this using repeat/0 ?? thank you


What about:

loop(N) :-
        between(1, N, X),
        writeln(X),
        false.

SWI-Prolog -- repeat/0, Numbers: Strings of digits with or without a decimal A Prolog program is a sequence of statements, called clauses, of some group of actions. Prolog implements repetition using recursion. sum := sum+odd end while; write sqrt end. Trace n sqrt odd sum. 28. 0. 1. 1. 1. 3. 4. 2 value of N is printed on the current output. LOOPING using PROLOG December 7, 2009 Posted by integereuler in TUTORIAL. trackback. Now, we will discuss about Looping in PROLOG. As we know, most of conventional programming languages have a looping facility that enables a set of instructions to b executed for a several times, or until a condition is met.


you can use assert/retract eg

:- dynamic value/1.
loop(N) :-
    retractall(value(_)),
    assert(value(N)),
    repeat,
            retract(value(V)),
            writeln(V),
            V1 is V - 1,
            assert(value(V1)),
            V = 0,
    !.

[PDF] Logic Programming with Prolog, Although, PROLOG has no looping facilities, similar looping effects can a set of instructions to be executed a fixed numbers of times, PROLOG is testloop(0). The testloop predicate is defined as 'loop from N, write the value of N, and output them to the screen, until a word end is encountered, using  Prolog variables are similar to "unknowns'' in algebra: Prolog tries to find values for the variables such that the entire clause can succeed. Once a value has been chosen for a variable, it cannot be altered by subsequent code; however, if the remainder of the clause cannot be satisfied, Prolog may backtrack and try another value for that


Your original one doesn't work for various reasons, but in particular

loop(N) :-
    repeat,
        write(N), nl,
        N is N-1,                <--- no!
        write(N),
        N > 0,
    !.

"is" unifies if the left side is the mathematical result of the right. the left side will be bound to the right. you can't change the value of a var once you've set it in Prolog.

LOOPING using PROLOG, If a list contains repeated elements they should be replaced with a single copy of the element. P24 (*) Lotto: Draw N different random numbers from the set 1. Given a range of integers by its lower and upper limit, print a list of all the result could be Hs = [hc(a,'0'), hc(b,'101'), hc(c,'100'), hc(d,'111'), hc(e  This is the same as ~:d, but using the fixed English locale. e Output next argument as a floating point number in exponential notation. The numeric argument specifies the precision. Default is 6 digits. Exact representation depends on the C library function printf(). This function is invoked with the format %.<precision>e. E


Abderrahmane asked about how to decrement.

Couple ways to do this. Most people would do

dec(0).
dec(N) :-
   write(N), nl,
   NewN is N - 1,
   dec(NewN).

if you insist on a failure driven loop, how about

dec(N) :-
    between(1,N,X),
    Dec is 11 - X,
    write(Dec), nl,
    Dec = 1.

or, and this is ugly and bad style (and slow as molasses):

dec(N) :-
      assert(current_n(N)),
      repeat,
      current_n(Y),
      write(Y), nl,
      NewY is Y - 1,
      retractall(current_n(_)).
      assert(current_n(NewY)),
      Y > 0.

P-99: Ninety-Nine Prolog Problems, Most of the I/O predicates have two forms, one with a stream ID, the other without. for example to read all the clauses in a file, you can use the repeat/0 predicate before the I/O predicate. They will be set up as the current input/output streams. Stream Properties. Each stream has a number of properties associated with it. tail is bound to the non-list term and len is set to the number of list-cells skipped. It is allowed to pass 0 for tail and NULL for len.


I finally got to make it work using the predicate between/3 this is my final code :

loop(N) :-
    repeat,
        between(1, N, X),
        writeln(X),
        X = N,
    !.

Edit : as @CookieMonster said, repeat/0 is not needed, so here is the last version of the code :

loop(N) :-
        between(1, N, X),
        writeln(X),
        false.

Input and Output, Prolog, which stands for PROgramming in LOGic, is the most widely uniform data structures, and input and output are interchangeable. Constants: numbers​, begin with lowercase letter, or enclosed in single quotes. An alternate representation is to represent an array as a set of facts in a the data base  I wrote the following program based on the logic that a prime number is only divisible by 1 and itself. So I just go through the process of dividing it to all numbers that are greater than one and less than itself, but I seem to have a problem since I get all entered numbers as true. Here's my code


Prolog Tutorial, Prolog uses the first food fact and responds with Edible = apple, using the variable a semicolon we see the full set of values for which that can be proven. The collection In fact, the sequence abc \== xyz causes Prolog to create a structure. display/1 can Prolog tried the two clauses for the predicate showfoods/0 in turn. Then, to help you with your original problem of incrementing a variable in-place: you can't do that in Prolog. Once a variable is bound, you can't change its binding. When programming in Prolog, you have to think in terms of relations and recursion, not mutable state. Here's how to increment in Prolog: incr(X, X1) :- X1 is X+1.


[PDF] Prolog basics, set array to {1,2,3,4,5} repeat with i in array display dialog i end repeat 10 0 DO I . LOOP \ Prints the numbers from 0 to 9 10 0 DO I . 2 +LOOP There are three primitive methods of looping in Prolog: recursion, fail-driven  The answer is false, instead of the count of number elements – Alexandre Rocha Jun 12 '17 at 23:33 Yes the answer that prolog gives you is right!! – coder Jun 12 '17 at 23:38 count([X], 0)) . succeeds, but X = 1, count([X], 0) fails. – false Jun 13 '17 at 8:04


Loop Structures, Take a string and repeat it some number of times. 124 Python; 125 Prolog; 126 Pure; 127 PureBasic; 128 Python; 129 R; 130 Racket; 131 Raku; 132 RapidQ; 133 REALbasic; 134 REBOL; 135 Red 165.1 Using printf; 165.2 Using repeat; 165.3 Using head -c output += string; if (n mod 2) > 0 then set out to out & dbl As format/2, but write the output on the given Output. The de-facto standard only allows Output to be a stream. The SWI-Prolog implementation allows all valid arguments for with_output_to/2. 136Earlier versions defined sformat/3 .