How to print square of n*n given characters in prolog?

1
votes

Write a Prolog program to print out a square of n*n given characters on the screen. Call your predicate square/2. The first argument should be a (positive) integer. the second argument the character (any Prolog term) to be printed. Example:

?-square(5, '*').
*****
*****
*****
*****
*****
Yes

I just start to learn this language. I did this:

square(_,'_').
square(N, 'B') :-
        N>0,
        write(N*'B').

It doesn't work at all. Can anyone help me?

4
printingprolog
can you write one character? ?- write_one('c'). will print c. Can you write this predicate? – Will Ness
write(N*'B') is asking Prolog to write out the expression N*'B'. Prolog does not evaluate arithmetic expressions outside is/2 expressions, and besides that there is no facility to create atoms through repetition of other atoms this way. Try harder. You can do it, you just have to be very explicit. – Daniel Lyons

4 Answers

3
votes

So your question is, basically, "how do I write a loop nested in a loop?"

This is how you write an empty loop with an integer for a counter:

loop(0).
loop(N) :- N > 0, N0 is N-1, loop(N0).

which in C would be:

for(i=0; i < n; ++i) { }

And you seem to know already how to print (write(foo)).

0
votes

Decompose the problem. To write an NxN square, you need to do two things:

  1. Write N lines
  2. Write a single line, consisting of N characters followed by a newline character.

The second is easy:

do_line(0,_) :-
  nl
  .
do_line(N,C) :-
  N > 0 ,
  write(C) ,
  N1 is N-1 ,
  do_line(N1,C)
  .

The first isn't much more difficult:

do_lines(0,_,_).
do_lines(M,N,C) :-
  M > 0 ,
  do_line(N,C) ,
  M1 is M-1 ,
  do_lines(M1,N,C)
  .

The all you need to do is wrap it:

write_square(N,C) :- do_lines(N,N,C) .

Easy!

0
votes

You need to draw a line of N stars/characters

line(N,X):- N>0, N1 is N-1, line(N1,X), write(X), fail; true.

Then you will draw a column of N lines of stars/characters.

s(N,Chr):-sAux(N,0,Chr). sAux(N,N,Chr). sAux(N,C,Chr):-C<N, C1 is C+1, sAux(N, C1, Chr), line(N,Chr),nl. s(N,Chr):- N>0, N1 is N-1, s(N1,X), linie(N,X), nl, fail;true.

0
votes

Doing this:

square2(0,_). % base case, recursion stops when X reaches 0, second argument is irrelevent    

square2(X,Symbol):-
  X1 is X - 1,
  write(Symbol),
  square2(X1,Symbol).

With the query, which results in:

?- square2(5,'* ').
* * * * *

Therefore, we need another loop to make it write X times.

square1(0,_,_). % base case, recursion stops when X reaches 0

square1(X,Y,Symbol):-
  X1 is X - 1,
  square2(Y,Symbol), % with the same Y passed in square2 predicate to print a line of symbols
  nl, % creates a new line
  square1(X1,Y,Symbol).

However, the question is asking for the format with square(5, '* '). Therefore,

square(X,Symbol):-
  square1(X,X,Symbol).

To wrap up:

square(X,Symbol):-
  square1(X,X,Symbol).

square1(0,_,_).      
square1(X,Y,Symbol):-
  X1 is X - 1,
  square2(Y,Symbol), 
  nl,
  square1(X1,Y,Symbol).

square2(0,_).    
square2(X,Symbol):-
  X1 is X - 1,
  write(Symbol),
  square2(X1,Symbol).