Some implicit constraint - prolog

When i execute this code(shown below) , it always sets implicit kind of constraint.
As you can see below , it always says that D1 = D2 but there is no such explicit constraints nor any pattern matching which forces this.
Or in otherwords there is some reference between D1 and D2 such that whenever D1 gets initialized, D2 gets automatically initialized. I can't see how this is happening. Can someone explain me, i tried ot figure it out with debugger but it did not help.
It's a puzzle "GERALD + DONALD = ROBERT", initially three lists contains these variables.
I add code below if anyone wants to test it:
sum(N1,N2,N):-
sum1(N1,N2,N,0,0,[0,1,2,3,4,5,6,7,8,9],_).
sum1([],[],[],0,0,Digits,Digits).
sum1([D1|N1],[D2|N2],[D|N],C1,C,Digs1,Digs2):-
sum1(N1,N2,N,C1,C2,Digs1,Digs2),
digitSum(D1,D2,C2,D,C,Digs2,Digs).
digitSum(D1,D2,C1,D,C,Digs1,Digs):-
del(D1,Digs1,Digs2),
del(D2,Digs2,Digs3),
del(D,Digs3,Digs),
S is D1 + D2 + C1,
D is S mod 10,
C is D div 10.
del(A,L,L):-
nonvar(A),!.
del(A,[A|L],L).
del(A,[B|L],[B|L1]):-
del(A,L,L1).
Query:
?- sum( [D,O,N,A,L,D], [G,E,R,A,L,D], [R,O,B,E,R,T] ).

When it says D1 = D2 then you should see that both D1 and D2 have same variable from the list and List is a functor and one variable is visible in whole functor.
You should see that when the recursion ends, then you have have D's(last element) from GERALD and DONALD , since this D is visible in whole functor, D1 and D2 both refer to same variable.

Related

Prolog - create simple function

I have a function that uses 4 parameters, called tile . It is designed to work the following way :
tile(?E, ?S, ?W, ?N, ?ID)
I would like a getter function that given an ID, it returns the first 4 parameters: E, S, W and N.
I have tried something like:
coordonates(tile(E,S,W,N,L), (E,S,W,N)).
But it does not return the actual values, only true.
If I type tile(E, S, W, N, #1) in the terminal I get the desired result but I do not know what exactly is returned (a list maybe?).
Let's suppose our facts describing tile looks as follows:
tile(p1,p2,p3,p4,id1).
tile(q1,q2,q3,q4,id2).
tile(r1,r2,r3,r4,id3).
In this we have a finite number of facts. That can be checked by the most general query for tile:
?- tile(E,S,W,N,I).
E = p1,
S = p2,
W = p3,
N = p4,
I = id1 ; % <---- user input ; to continue
E = q1,
S = q2,
W = q3,
N = q4,
I = id2 ; % <---- user input ; to continue
E = r1,
S = r2,
W = r3,
N = r4,
I = id3. % <---- toplevel outputs . -- we're done
So in theory, we could define coordonates as follows:
coordonates(id1, t(p1, p2, p3, p4)).
coordonates(id2, t(q1, q2, q3, q4)).
coordonates(id3, t(r1, r2, r3, r4)).
which could be queried for id2 as follows:
?- coordonates(id2,X).
X = t(q1, q2, q3, q4).
I used the functor t to group the solution, to make clear that it is not the predicate tile we defined earlier. There's also a lot of repetition in this definition which is already a hint, that we can do better. What we are looking for is a rule which tells us how, given we have a answer for tile, we can describe coordonates. In logical terms, this is written as an implication of the form: goal1 ∧ ... ∧ goalN → head. which means "Suppose I know that goal1 to goalN is true, then I also know that head is true." In Prolog, this is written backwards:
head :-
goal1,
% ...
goalN.
Let's go back to our task: we know something about a tile and we want to describe how the projection looks like. This means, our code looks as follows:
coordonates( ... ) :-
% ...
tile(E,S,W,N,I).
The body tile(E,S,W,N,I) is the most general form we can write (see our query above) and can be read as "suppose I have any tile at coordinates E S W N with id I". Now we only need to fill in, how coordonates should look like. We know it has two arguments, because it relates the id with the four other elements. Lets give them names, say Id and Coords:
coordonates(Id, Coords) :-
% ...
tile(E,S,W,N,I).
Now we only need to find out how to relate E,S,E,N and I with Id and Coords. One is easy: Id is just I. The other one is also not too hard, we just need to group the coordinates into one term. We can pick an arbitrary one, but already above decided to take t, so we will stick with it:
coordonates(Id, Coords) :-
Id = I,
Coords = t(E,S,W,N),
tile(E,S,W,N,I).
This already works as we expect:
?- coordonates(X,Y).
X = id1,
Y = t(p1, p2, p3, p4) ;
X = id2,
Y = t(q1, q2, q3, q4) ;
X = id3,
Y = t(r1, r2, r3, r4).
Now we can make one observation: if two terms are equal, we can use one instead of the other. So instead of writing Id = I, we can just reuse Id. The same goes for Coords and t(E,S,W,N):
coordonates(I, t(E,S,W,N)) :-
tile(E,S,W,N,I).
It couldn't be much shorter :-)
You have to declare 'E, S, W and N' so that prolog can unify those parameters with the input when you make the query. Something like (In the most basic case):
tile(['cordE1','cordS1','cordW1','cordN1'],1).
tile(['cordE2','cordS2','cordW2','cordN2'],2).
tile(['cordE3','cordS3','cordW3','cordN3'],3).
Query:
?- tile(C,2).
C = [cordE2, cordS2, cordW2, cordN2].
?- tile(C,1).
C = [cordE1, cordS1, cordW1, cordN1].
?- tile(C,3).
C = [cordE3, cordS3, cordW3, cordN3].

How to understand `u=r÷s`, the division operator, in relational algebra?

let be a database having the following relational-schemes: R(A,B,D) and S(A,B) with the attributes of same name in the same domain and with the instances r and s respectively.
An instance of r
An instance of s
What is the scheme and what are the tuples of u=r÷s? How to define them in English with r and s?
My attempt
I know that
u=r÷s=
Which leads me to think that it would only be an array of one column A, but I'm not sure enough to know what will be ther result within the array.
Can you help me understand u=r÷s?
An intuitive property of the division operator of the relational algebra is simply that it is the inverse of the cartesian product. For example, if you have two relations R and S, then, if U is a relation defined as the cartesian product of them:
U = R x S
the division is the operator such that:
U ÷ R = S
and:
U ÷ S = R
So, you can think of the result of U ÷ R as: “the projection of U that, multiplied by R, produces U”, and of the operation ÷, as the operation that finds all the “parts” of U that are combined with all the tuples of R.
However, in order to be useful, we want that this operation can be applied to any couple of relations, that is, we want to divide a relation which is not the result of a cartesian product. For this, the formal definition is more complex.
So, supposing that we have two relations R and S with attributes respectively A and B, their division can be defined as:
R ÷ S = πA-B(R) - πA-B((πA-B(R) x S) - R)
that can be read in this way:
πA-B(R) x S: project R over the attributes of R which are not in S, and multiply (cartesian product) this relation with S. This produces a relation with the attributes A of R and with rows all the possible combinations of rows of S and the projection of R;
From the previous result subtract all the tuples originally in R, that is, perform (πA-B(R) x S) - R. In this way we obtain the “extra” tuples, that is the tuples in the cartesian product that were not present in the original relation.
Finally, subtract from the original relation those extra tuples (but, again, perform this operation only on the attributes of R which are not present in S). So, the final operation is: πA-B(R) - πA-B(the result of step 2).
So, coming to your example, the projection of r on D is equal to:
(D)
d1
d2
d3
d4
and the cartesian product with s is:
(A, B, D)
a1 b1 d1
a1 b1 d2
a1 b1 d3
a1 b1 d4
Now we can remove from this set the tuples that were also in the original relation r, i.e. the first two tuples and the last one, so that we obtain the following result:
(A, B, D)
a1 b1 d3
And finally, we can remove the previous tuples (projected on D), from the original relation (again projected on D), that is, we remove:
(D)
d3
from:
(D)
d1
d2
d3
d4
and we obtain the following result, which is the final result of the division:
(D)
d1
d2
d4
Finally, we could double check the result by multiplying it with the original relation s (which is composed only by the tuple (a1, b1)):
(A B D)
a1 b1 d1
a1 b1 d2
a1 b1 d4
And looking at the rows of the original relation r, you can see this fact, that should give you an important insight on the meaning of the division operator:
the only values of the column D in r that are present together with (a1, b1) (the only tuple of s), are d1, d2 and d4.
You can also see another example in Wikipedia, and for a detailed explanation of the division, together with its transformation is SQL, you could look at these slides.

Generic prolog cryptarithmetic program

I'm trying to write a program to solve general cryptarithmetic puzzles such as AM+PM=DAY, SEND+MORE=MONEY..
The program I've written is:
gsolve([H1,T1],[H2|T2],[H3|T3]):-
gsolvehelper([H1,T1],[H2|T2],[H3|T3],[0,1,2,3,4,5,6,7,8,9],0).
gsolvehelper([H1,T1],[H2|T2],[H3|T3],D,C):-
( var(H1)->select(H1,D,D1);D1=D),
( var(H2)->select(H2,D1,D2);D2=D1),
( X1 is H1+H2+C),
( H3 is mod(X1,10)),
( C1 is X1//10),
gsolvehelper(T1,T2,T3,D2,C1).
The input is of the form:
gsolve([A,M],[P,M],[D,A,Y]).
The head of the first two lists is added to the carry to find the head of the third list and the new carry and so on.
The error I get is:
ERROR: is/2: Type error: `[]' expected, found `[2,_G3922]' ("x" must hold one character)
I cannot figure out what this error is. Can someone please help me with this?
The error you report is due to what is probably a typo: [H1,T1], appearing in several places. Due to this typo H2 gets -- unintentionally -- instantiated to a list, causing problems in the summation statement. The following reproduces the error in SWI7:
?- X is 1 + [1,_] + 3.
ERROR: is/2: Type error: `[]' expected, found `[1,_G11068]' (a list) ("x" must hold one character)
There are other ways in which you can improve your code:
Since you simply pass on the arguments of gsolve/3 you do not need to use the head/tail-notation [H|T] here.
You can simply name your second predicate gsolve/5, as it is already different from gsolve/3 due to its arity.
Several brackets in bsolve/5 are superfluous. Spacing around the if/then operators can be improved as well.
Using the combination of mod/2 and ///2 may introduce problems since the rounding of ///2 (i.e., towards zero, typically at least...) is not consistent with mod/2. Use div/2 (rounding towards negative infinity) instead.
Making alterations based upon the above I get:
gsolve(L1, L2, L3):-
gsolve(L1, L2, L3, [0,1,2,3,4,5,6,7,8,9], 0).
gsolve([H1|T1], [H2|T2], [H3|T3], D, C):-
( var(H1)
-> select(H1, D, D1)
; D1 = D
),
( var(H2)
-> select(H2, D1, D2)
; D2=D1
),
X1 is H1 + H2 + C,
H3 is mod(X1, 10),
C1 is div(X1, 10),
gsolve(T1, T2, T3, D2, C1).
This does not solve the puzzle yet... However, since it solves the error you mentioned -- and some other things as well -- you should be able to take it on from here.

Algorithms to create a tabular representation of a DAG?

Given a DAG, in which each node belongs to a category, how can this graph be transformed into a table with a column for each category? The transformation doesn't have to be reversible, but should preserve useful information about the structure of the graph; and should be a 'natural' transformation, in the sense that a person looking at the graph and the table should not be surprised by any of the rows. It should also be compact, i.e. have few rows.
For example given a graph of nodes a1,b1,b2,c1 with edges a1->b1, a1->b2, b1->c1, b2->c1 (i.e. a diamond-shaped graph) I would expect to see the following table:
a b c
--------
a1 b1 c1
a1 b2 c1
I've thought about this problem quite a bit, but I'm having trouble coming up with an algorithm that gives intuitive results on certain graphs. Consider the graph a1,b1,c1 with edges a1->c1, b1->c1. I'd like the algorithm to produce this table:
a b c
--------
a1 b1 c1
But maybe it should produce this instead:
a b c
--------
a1 c1
a1 b1
I'm looking for creative ideas and insights into the problem. Feel free to vary to simplify or constrain the problem if you think it will help.
Brainstorm away!
Edit:
The transformation should always produce the same set of rows, although the order of rows does not matter.
The table should behave nicely when sorting and filtering using, e.g., Excel. This means that mutliple nodes cannot be packed into a single cell of the table - only one node per cell.
What you need is a variation of topological sorting. This is an algorithm that "sorts" graph vertexes as if a---->b edge meant a > b. Since the graph is a DAG, there is no cycles in it and this > relation is transitive, so at least one sorting order exists.
For your diamond-shaped graph two topological orders exist:
a1 b1 b2 c1
a1 b2 b1 c1
b1 and b2 items are not connected, even indirectly, therefore, they may be placed in any order.
After you sorted the graph, you know an approximation of order. My proposal is to fill the table in a straightforward way (1 vertex per line) and then "compact" the table. Perform sorting and pick the sequence you got as output. Fill the table from top to bottom, assigning a vertex to relevant column:
a b c
--------
a1
b2
b1
c1
Now compact the table by walking from top to bottom (and then make similar pass from bottom to top). On each iteration, you take a closer look to a "current" row (marked as =>) and to the "next" row.
If in a column nodes in current and next node differ, do nothing for this column:
from ----> to
X b c X b c
-------- --------
=> X1 . . X1 . .
X2 . . => X2 . .
If in a column X in the next row there is no vertex (table cell is empty) and in the current row there is vertex X1, then you sometimes should fill this empty cell with a vertex in the current row. But not always: you want your table to be logical, don't you? So copy the vertex if and only if there's no edge b--->X1, c--->X1, etc, for all vertexes in current row.
from ---> to
X b c X b c
-------- --------
=> X1 b c X1 b c
b1 c1 => X1 b1 c1
(Edit:) After first (forward) and second (backward) passes, you'll have such tables:
first second
a b c a b c
-------- --------
a1 a1 b2 c1
a1 b2 a1 b2 c1
a1 b1 a1 b1 c1
a1 b1 c1 a1 b1 c1
Then, just remove equal rows and you're done:
a b c
--------
a1 b2 c1
a1 b1 c1
And you should get a nice table. O(n^2).
How about compacting all reachable nodes from one node together in one cell ? For example, your first DAG should look like:
a b c
---------------
a1 [b1,b2]
b1 c1
b2 c1
It sounds like a train system map with stations within zones (a,b,c).
You could be generating a table of all possible routes in one direction. In which case "a1, b1, c1" would seem to imply a1->b1 so don't format it like that if you have only a1->c1, b1->c1
You could decide to produce a table by listing the longest routes starting in zone a,
using each edge only once, ending with the short leftover routes. Or allow edges to be reused only if they connect unused edges or extend a route.
In other words, do a depth first search, trying not to reuse edges (reject any path that doesn't include unused edges, and optionally trim used edges at the endpoints).
Here's what I ended up doing:
Find all paths emanating from a node without in-edges. (Could be expensive for some graphs, but works for mine)
Traverse each path to collect a row of values
Compact the rows
Compacting the rows is dones as follows.
For each pair of columns x,y
Construct a map of every value of x to it's possible values of y
Create another map For entries that only have one distinct value of y, mapping the value of x to its single value of y.
Fill in the blanks using these maps. When filling in a value, check for related blanks that can be filled.
This gives a very compact output and seems to meet all my requirements.

Parsing expressions with an undefined number of arguments

I'm trying to parse a string in a self-made language into a sort of tree, e.g.:
# a * b1 b2 -> c * d1 d2 -> e # f1 f2 * g
should result in:
# a
* b1 b2
-> c
* d1 d2
-> e
# f1 f2
* g
#, * and -> are symbols. a, b1, etc. are texts.
Since the moment I know only rpn method to evaluate expressions, and my current solution is as follows. If I allow only a single text token after each symbol I can easily convert expression first into RPN notation (b = b1 b2; d = d1 d2; f = f1 f2) and parse it from here:
a b c -> * d e -> * # f g * #
However, merging text tokens and whatever else comes seems to be problematic. My idea was to create marker tokens (M), so RPN looks like:
a M b2 b1 M c -> * M d2 d1 M e -> * # f2 f1 M g * #
which is also parseable and seems to solve the problem.
That said:
Does anyone have experience with something like that and can say it is or it is not a viable solution for the future?
Are there better methods for parsing expressions with undefined arity of operators?
Can you point me at some good resources?
Note. Yes, I know this example very much resembles Lisp prefix notation and maybe the way to go would be to add some brackets, but I don't have any experience here. However, the source text must not contain any artificial brackets and also I'm not sure what to do about potential infix mixins like # a * b -> [if value1 = value2] c -> d.
Thanks for any help.
EDIT: It seems that what I'm looking for are sources on postfix notation with a variable number of arguments.
I couldn't fully understand your question, but it seems what you want is a grammar definition and a parser generator. I suggest you take a look at ANTLR, it should be pretty straightforward with it to define a grammar for either your original syntax or the RPN.
Edit: (After exercising self-criticism, and making some effort to understand the question details.) Actually, the language grammar is unclear from your example. However, it seems to me, that the advantages of the prefix/postfix notations (i.e. that you need neither parentheses nor a precedence-aware parser) stem from the fact that you know the number of arguments every time you encounter an operator, therefore you know exactly how many elements to read (for prefix notation) or to pop from the stack (for postfix notation). OTOH, I beleive that having operators which can have variable number of arguments makes prefix/postfix notations not simply difficult to parse but outright ambiguous. Take the following expression for example:
# a * b c d
Which of the following three is the canonical form?
(a, *(b, c, d))
(a, *(b, c), d)
(a, *(b), c, d)
Without knowing more about the operators, it is impossible to tell. Of course you could define some sort of greedyness of the operators, e.g. * is greedier than #, so it gobbles up all the arguments. But this would beat the purpose of a prefix notation, because you simply wouldn't be able to write down the second variant from the above three; not without additinonal syntactic elements.
Now that I think of it, it is probably not by sheer chance that none of the programming languages I know support operators with a variable number of arguments, only functions/procedures.

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