I have to solve a cryptogram which looks like this:
ABC / DEF = 0.GHGHGH...
where length of variables may be different. Also repetition may be different (like 0.XYZXYZ...). I've written a piece of code which I thought will work but it doesn't:
cryp(A,B,C) :-
mn(A, X),
mn(B, Y),
mn(C, Z),
Z = X/Y.
mn([], 0).
mn([H|T], W) :- D is 10, mn(T, W1), length(T, D), P is 10^D, W is W1 + H*P.
I execute it as crypt([A,B,C], [D,E,F], [G,H]). I thought it will at least solve ABC / DEF = GH just to have any part of working solution but it doesn't work.
I don't have any clue how to do it even for one example input. I don't know how to represent 0.GHGHGH....
EDIT:
mn/2 is for converting list of digits to number ([1,2,3] -> 123).
Here is clp(fd)-way
?- use_module(library(clpfd)).
true.
?- Vars=[A,B,C,D,E,F,G,H], Vars ins 0..9, A#\=0, D#\=0, ((10*A + B)*10+C)*99 #= ((10*D+E)*10+F)*(10*G+H),all_distinct(Vars),forall(label(Vars),format('~w~n',[Vars])).
[1,0,8,2,9,7,3,6]
[1,0,8,3,9,6,2,7]
[1,3,0,2,8,6,4,5]
[1,3,0,4,9,5,2,6]
[1,3,8,2,9,7,4,6]
[1,3,8,5,0,6,2,7]
[1,6,0,4,9,5,3,2]
[1,8,0,3,9,6,4,5]
[1,8,0,4,9,5,3,6]
[2,0,4,3,9,6,5,1]
[2,5,9,4,0,7,6,3]
[2,8,4,3,9,6,7,1]
[2,8,7,4,5,1,6,3]
[2,8,7,6,9,3,4,1]
[2,9,0,6,3,8,4,5]
[3,1,0,4,9,5,6,2]
[3,1,0,6,8,2,4,5]
[3,6,0,4,9,5,7,2]
[3,6,0,7,9,2,4,5]
[3,8,0,4,9,5,7,6]
[4,0,8,5,6,1,7,2]
[4,0,8,7,9,2,5,1]
[4,9,3,5,6,1,8,7]
[5,0,4,6,9,3,7,2]
[5,0,4,7,9,2,6,3]
[5,1,8,6,9,3,7,4]
[5,7,4,6,9,3,8,2]
[5,7,4,9,0,2,6,3]
[5,9,4,7,2,6,8,1]
[6,8,0,9,3,5,7,2]
[7,5,6,9,2,4,8,1]
You might do the following identifications: let a=0.GHGHGHGH.... Then
a = 0.GHGH....
GH + a = GH.GHGHGHGH.... = 100 a
GH = 99 a
=> a = GH/99
So you can replace it by GH/99 in your formulas.
Similarly 0.XYZXYZXYZ = XYZ/999.
Related
I am a beginner in Prolog and I have a task to do.
I need to check if the graph is connected.
For now I have that...
graph(
[arc(a,b)],
[arc(a,f)],
[arc(b,c)],
[arc(c,d)],
[arc(c,e)],
[arc(e,d)],
[arc(f,c)],
[arc(f,e)],
[arc(f,g)],
[arc(g,c)],
[arc(c,a)]).
edge(X,Y):-arc(X,Y);arc(Y,X).
path(X,Y):-edge(X,Y).
path(X,Y):-edge(X,Z),path(Z,Y).
triangle(X,Y,Z):-arc(X,Y),arc(Y,Z),arc(Z,X).
cycle(X):-arc(X,Y),path(Y,X).
connectivity([]):-forall(member(edge(X,Y)),path(X,Y)).
Check:
connectivity(graph).
upper I have arc(x,y) and I need check if every pair is connected.
Could u help me ?
Since you changed the question after I was almost done I will post what would solve the question before the change and you can figure out how to change it to meet your update.
arc(a,b).
arc(a,f).
arc(b,c).
arc(c,d).
arc(c,e).
arc(e,d).
arc(f,c).
arc(f,e).
arc(f,g).
arc(g,c).
arc(c,a).
edge(X,Y) :-
arc(X,Y), !.
edge(X,Y) :-
arc(Y,X).
path_prime(Visited,X,Y) :-
\+ member(X,Visited),
edge(X,Y), !.
path_prime(Visited,X,Y) :-
\+ member(X,Visited),
edge(X,Z),
path_prime([X|Visited],Z,Y).
path(X,X) :-
ground(X), !.
path(X,Y) :-
path_prime([],X,Y).
nodes(Nodes) :-
setof(A,B^arc(A,B),Starts),
setof(B,A^arc(A,B),Ends),
union(Starts,Ends,Nodes).
connected(X,Y) :-
nodes(Nodes),
member(X,Nodes),
member(Y,Nodes),
path(X,Y).
The first thing that has to be done is to get a list of the unique nodes which will be a set.
This can be done using
nodes(Nodes) :-
setof(A,B^arc(A,B),Starts),
setof(B,A^arc(A,B),Ends),
union(Starts,Ends,Nodes).
Notice that both the start and the end node of an arc are done separately. In particular notice that the node d is only in the destination of an arc.
Since you included edge(X,Y):-arc(X,Y);arc(Y,X). in your question, this indicated that the arcs should not be directional and so it is possible to get cycles. To avoid the cycles the list of visited nodes is added to the argument list and checked before proceeding.
As no test cases or examples of a correct solution were given, some times a node connected to itself is valid and so the clause
path(X,X) :-
ground(X), !.
was added.
This is by no means an optimal or best way to do this, just to give you something that works.
Partial run
?- connected(X,Y).
X = Y, Y = a ;
X = a,
Y = b ;
X = a,
Y = c ;
X = a,
Y = d ;
X = a,
Y = e ;
X = a,
Y = f ;
X = a,
Y = g ;
X = b,
Y = a ;
X = Y, Y = b ;
X = b,
Y = c ;
...
As I often comment, you should do problems with pen an paper first before writing code. If you don't know exactly what the code will be before you start typing the first line of code then why are you typing in code?
Questions from comments:
And setof ,union ,whats mean? Im rly beigneer and I don't understand that language and predicates.
setof/3 collects all of the values from arc/2. Since only one of the two values is needed, ^ tells setup/3 not to bind the variable in the Goal, or in beginner terms to just ignore the values from the variable.
union/3 just combines the to sets into one set; remember that a set will only have unique values.
so I just got started with Prolog this semester, and got the homework to implement a pretty basic d(function, variable, derivative) which I did like this:
d(X,X,1) :- !.
d(C,X,0) :- atomic(C). %, (C \= X).
d(X**E,X,E*X**(E-1)).
d(U+V,X,A+B) :- d(U,X,A), d(V,X,B).
d(U-V,X,A-B) :- d(U,X,A), d(V,X,B).
d(U*V,X,DU*V+U*DV) :- d(U,X,DU), d(V,X,DV).
d(U/V,X,(DU*V-U*DV)/(V*V)) :- d(U,X,DU), d(V,X,DV).
I know this is not complete, but it covers all the tasks required in the exercise.
However,
?- d((x*x+2*x+3)/(3*x),x,R).
leads to
R = ((1*x+x*1+ (0*x+2*1)+0)* (3*x)- (x*x+2*x+3)* (0*x+3*1))/ (3*x* (3*x)).
which doesn't look pretty at all. is/2 unfortunately doesn't like my x as it is not a number...
Is there a simple solution to achieve a cleaner result?
I would rather see this as two separate problems:
First, get derivation right (you're probably getting close, depending on your concrete requirements).
Then, work on simplifying expressions on an algebraic level. Exploit algebraic identities, see if applying the laws of commutativity / associativity / distributivity on some subexpressions enable their rewriting into something equivalent (but simpler / more compact).
As a starting point, you may want to look at the somewhat related question "Replacing parts of expression in prolog".
Here's a simplistic sketch how you could do the simplification—using iwhen/2 to safeguard against insufficient instantiation:
expr_simplified(A, B) :-
iwhen(ground(A), xpr_simplr(A,B)).
xpr_simplr(A, B) :-
( atomic(A)
-> A = B
; ( A = X+0 ; A = 0+X ; A = 1*X ; A = X*1 )
-> xpr_simplr(X, B)
; ( A = 0*_ ; A = _*0 )
-> B = 0
; A = X+X
-> B = X*2
; A = X*X
-> B = X**2
; A = X**1
-> B = X
; A =.. [F|Xs0], % defaulty catch-all
maplist(xpr_simplr, Xs0, Xs),
B =.. [F|Xs]
).
Let's see what it does with the expression you gave. We apply expr_simplified/2 until we reach a fixed point:
?- A = ((1*x+x*1+(0*x+2*1)+0)*(3*x)-(x*x+2*x+3)*(0*x+3*1))/(3*x*(3*x)),
expr_simplified(A,B),
expr_simplified(B,C),
expr_simplified(C,D).
A = ((1*x+x*1+(0*x+2*1)+0)*(3*x)-(x*x+2*x+3)*(0*x+3*1))/(3*x*(3*x)),
B = ((x+x+(0+2))*(3*x)-(x**2+2*x+3)*(0+3))/(3*x)**2,
C = ((x*2+2)*(3*x)-(x**2+2*x+3)*3)/(3*x)**2,
D = C. % fixed point reached
As imperfect as the simplifier is, the expression got a lot more readable.
a possibility to get a number is to replace each instance of variable x with a value, visiting the derived tree. You should do writing a clause to match each binary operator, or use a generic visit, like
set_vars(E, Vs, Ev) :-
E =.. [F,L,R],
set_vars(L, Vs, Lv),
set_vars(R, Vs, Rv),
Ev =.. [F,Lv,Rv].
set_vars(V, Vs, N) :- memberchk(V=N, Vs).
set_vars(V, _, V).
that yields
?- d((x*x+2*x+3)/(3*x),x,R), set_vars(R,[x=5],E), T is E.
R = ((1*x+x*1+ (0*x+2*1)+0)* (3*x)- (x*x+2*x+3)* (0*x+3*1))/ (3*x* (3*x)),
E = ((1*5+5*1+ (0*5+2*1)+0)* (3*5)- (5*5+2*5+3)* (0*5+3*1))/ (3*5* (3*5)),
T = 0.29333333333333333
but, there is a bug in your first clause, that once corrected, will allow to evaluate directly the derived expression:
d(X,V,1) :- X == V, !.
...
now, we can throw away the utility set_vars/3, so
?- d((T*T+2*T+3)/(3*T),T,R), T=8, V is R.
T = 8,
R = ((1*8+8*1+ (0*8+2*1)+0)* (3*8)- (8*8+2*8+3)* (0*8+3*1))/ (3*8* (3*8)),
V = 0.3177083333333333.
I'm new to Prolog as I'm just starting to learn and write up my own small set of database rules. Using my own .pl file of database rules, I'm having a small problem with a query that I enter in Prolog, using these rules. Below shows my small database of rules:
staff(andy,18235,3).
staff(beth,19874,4).
staff(andy,18235,5).
staff(carl,16789,2).
staff(earl,34567,9).
sum([], 0).
sum([H|T], X) :-
sum(T, X1),
X is X1 + H.
getincome(Name, Income) :-
findall(Income,staff(Name,_,Income),Member),
sum(Member, Income).
As you can see, I have written a rule that finds the total income for a particular member of staff. This works very fine, as when I input:
?- getincome(andy, X).
The program always returns:
X = 8
As it should do, however whenever I instead input:
?- getincome(andy, 8).
This always returns false, when it should be true.
However when I also input:
?- getincome(andy, 3).
This returns true, due to already being in the database.
I'm just wondering, how could I modify this rule so that this could output true for the correct summation value, entered for any given staff (most particularly Andy), as opposed to the value already in the given database?
Ignore my question above!
Thanks for the help 'false'. I'm also having another issue, this time to do with working out and displaying the sum of the income for each member. I have modified my rules, in order to display this, as follows:
getincome(Name, I) :- staff(Name, _, _ ), findall(Income,staff(Name,_,Income),Member), sum(Member, I).
Whenever I enter the query:
?- getincome(X, Y).
I keep getting duplicate results of staff (most notably Andy, of course), as show below:
X = andy,
Y = 8 ;
X = beth,
Y = 4 ;
X = andy,
Y = 8 ;
X = carl,
Y = 2 ;
X = earl,
Y = 9.
What changes can I make to avoid these duplicates?
library(aggregate) offers a clean interface to solve such kind of problems:
?- aggregate(sum(S), K^staff(E,K,S), I).
E = andy,
I = 8 ;
E = beth,
I = 4 ;
E = carl,
I = 2 ;
E = earl,
I = 9.
One way to do this is to use bagof to collect each set of incomes:
person_income(Name, Income) :-
bagof(I, X^staff(Name,X,I), Incomes), % Incomes for a given name
sumlist(Incomes, Income). % Sum the incomes
With results:
| ?- person_income(Name, Income).
Income = 8
Name = andy ? a
Income = 4
Name = beth
Income = 2
Name = carl
Income = 9
Name = earl
yes
| ?- person_income(andy, Income).
Income = 8
yes
| ?- person_income(Name, 8).
Name = andy ? a
no
| ?-
I named this person_income to emphasize that it's a relation between a person and their income, rather than getincome which is more of an imperative notion, and doesn't really reflect that you can do more with the relation than just "get the income". Also, I'm using SWI Prolog's sumlist/2 here. GNU Prolog has sum_list/2. And as #CappeliC indicates in his answer, SWI Prolog has a handy aggregate predicate for operations like this.
I try to check the correctness of student mathematical expression using Prolog (SWI-Prolog). So, for example if the student were asked to add three variable x, y, and z, and there's a rule that the first two variable that must be added are: x and y (in any order), and the last variable that must be added is z then I expect that prolog can give me true value if the student's answer is any of these:
x+y+z
(x+y)+ z
z+(x+y)
z+x+y
y+x+z
and many other possibilities.
I use the following rule for this checking:
addData :-
assert(variable(v1)),
assert(variable(v2)),
assert(variable(v3)),
assert(varName(v1,x)),
assert(varName(v2,y)),
assert(varName(v3,z)),
assert(varExpr(v1,x)),
assert(varExpr(v2,y)),
assert(varExpr(v3,z)).
add(A,B,R) :- R = A + B.
removeAll :- retractall(variable(X)),
retractall(varName(X,_)),
retractall(varExpr(X,_)).
checkExpr :-
% The first two variable must be x and y, in any combination
( (varExpr(v1,AExpr), varExpr(v2,BExpr));
(varExpr(v2,AExpr), varExpr(v1,BExpr))
),
add(AExpr, BExpr, R1),
% store the expression result as another variable, say v4
retractall(variable(v4)),
retractall(varName(v4, _)),
retractall(varExpr(v4, _)),
assert(variable(v4)),
assert(varName(v4, result)),
assert(varExpr(v4, R1)),
% add the result from prev addition with Z (in any combination)
( (varExpr(v3,CExpr), varExpr(v4,DExpr));
(varExpr(v4,CExpr), varExpr(v3,DExpr))
),
add(CExpr, DExpr, R2),
R2 = z + x + y. % will give me false
% R2 = z + (x + y). % will give me true
% Expected: both should give me true
checkCorrect :- removeAll,
addData,
checkExpr.
You should try to specify a grammar and write a parser for your expressions.
Avoid assert/retract, that make the program much more difficult to understand, and attempt instead to master the declarative model of Prolog.
Expressions are recursive data structures, using operators with known precedence and associativity to compose, and parenthesis to change specified precedence where required.
See this answer for a parser and evaluator, that accepts input from text. In your question you show expressions from code. Then you are using Prolog' parser to do the dirty work, and can simply express your requirements on the resulting syntax tree:
expression(A + B) :-
expression(A),
expression(B).
expression(A * B) :-
expression(A),
expression(B).
expression(V) :-
memberchk(V, [x,y,z]).
?- expression(x+y+(x+z*y)).
true .
edit: we can provide a template of what we want and let Prolog work out the details by means of unification:
% enumerate acceptable expressions
checkExpr(E) :-
member(E, [F = A + D, F = D + A]),
F = f,
A = c * N,
N = 1.8,
D = d.
And so on...
Test:
?- checkExpr(f=(c*1.8)+d).
true.
?- checkExpr(f=(c*1.8)+e).
false.
?- checkExpr(f=d+c*1.8).
true.
I'm trying to solve an exercise in order to become more familiar with prolog.
The task is following:
% Sten wants to send Lisa 100 flowers. He can choose from lilies, roses and tulips.
% One lily costs $50, rose $10 and tulip $1. Find how many flowers of each type he
% must buy, so that he spends exactly $500.
I have solved that exercise, but in somewhat bulky way I guess. My code is:
% numbers 1..100
digit(1). digit(2). digit(3). digit(4). digit(5). digit(6). digit(7). digit(8).
digit(9). digit(10). digit(11). digit(12). digit(13). digit(14). digit(15). digit(16).
digit(17). digit(18). digit(19). digit(20). digit(21). digit(22). digit(23). digit(24).
digit(25). digit(26). digit(27). digit(28). digit(29). digit(30). digit(31). digit(32).
digit(33). digit(34). digit(35). digit(36). digit(37). digit(38). digit(39). digit(40).
digit(41). digit(42). digit(43). digit(44). digit(45). digit(46). digit(47). digit(48).
digit(49). digit(50). digit(51). digit(52). digit(53). digit(54). digit(55). digit(56).
digit(57). digit(58). digit(59). digit(60). digit(61). digit(62). digit(63). digit(64).
digit(65). digit(66). digit(67). digit(68). digit(69). digit(70). digit(71). digit(72).
digit(73). digit(74). digit(75). digit(76). digit(77). digit(78). digit(79). digit(80).
digit(81). digit(82). digit(83). digit(84). digit(85). digit(86). digit(87). digit(88).
digit(89). digit(90). digit(91). digit(92). digit(93). digit(94). digit(95). digit(96).
digit(97). digit(98). digit(99). digit(100).
quantity(A1,A2,A3):-
var(A1), var(A2), var(A3),
digit(A1), digit(A2), digit(A3),
X is A1+A2+A3, X is 100,
Y is (A1*50)+(A2*10)+(A3*1), Y is 500.
Can somebody suggest a better method for initializing those rules? For example in Haskell I could do something like this:
let numbers = [1..100]
Thanks in advance.
Using SWI-Prolog:
:- use_module(library(clpfd)).
flowers(L, R, T) :-
[L,R,T] ins 0..sup,
L+R+T #= 100,
L*50 + R*10 + T*1 #= 500.
Example query:
?- flowers(Lilies, Roses, Tulips), label([Lilies,Roses,Tulips]).
Lilies = 1,
Roses = 39,
Tulips = 60 ;
false.
Some versions of Prolog have the between/3 predicate. You could say
digit(X):-between(1,100,X).
If between is not available, you could say
digit(X):-member(X,[1,2,3,4,5 and so on]).
If you do not want to use member/2, use recursion.
Edit: you can also implement between/3 like this:
my_between(X,Y,Z):-X<Y,(Z=X;X2 is X+1,my_between(X2,Y,Z)).
A robust and efficient implementation of between/3 may be more complicated, but for your purposes, this should be enough.
quantity(lilies(L),roses(R),tulips(T)) :-
between(0,100,L),
between(0,100,R),
between(0,100,T),
L + R + T =:= 100,
L*50 + R*10 + T =:= 500 .