I picked up prolog a couple of days ago and I 'm kind of stuck to this question. I want to subtract a number recursively until that number becomes less than 0. In pseudocode that would be like:
N:=0
while(Y>=X)
{
Y := Y-X
N := N+1
Y := Y+2
}
So for example if I have Y=20 and X=10 then we would get N=2 and Y=4.
Any ideas? Thanks in advance. Any help appreciated. I'm using SWI Prolog.
EDIT 1
What I've accomplished so far is(although I'm not sure even if its correct):
sufficient(X, Y, M, N, F) :-
F is Y-X,
Y>=X,
plus(M, 1, N),
sufficient(X, F, N, N, F).
I have problem finding my base case, I'm confused on how to implement it. Also, in the sufficient I have implemented, obviously when Y<X it terminates returning false. Is there a way to get the N and F before terminating? I am feeling that I am not thinking the "prolog" way, since I am mostly used on C and that vagues my thinking. Thanks.
EDIT 2
I have found my base case and I can stop recursion however, I can't manage to ge the correct values. My code:
sufficient(X, Y, M, N, F) :- Y<X.
sufficient(X, Y, M, N, F) :-
F is Y-X,
plus(M, 1, N),
sufficient(X, F, N, D, E).
Thing is after the first recursion, if for example I call sufficient as sufficient(10,21,0,N,F). from the swi prolog command prompt, I 'll get N=1 and F=11. That happens because I make 2 new variables D and E. If I don't make those 2 new variables(D and E), and at the 3rd sufficient in the code I call N and F instead of D and E at the line F is Y-X, I get a false, because F is 11 and Y-X is 1. Do I have to set the a subtraction function myself, since F is Y-X is not exactly a subtraction? Any ideas on how to do it?
All recursive functions need at least one base case. In what circumstance should your function say, OK, I have the answer, no need to recurse?
It would be the case in which your pseudocode loop is done, right?
Usually we write it in this format:
factorial(0,1). % The factorial of 0 is 1.
factorial(N,Factorial) :-
N>0, % You may need to test applicability
% of this recursive clause
NMinus1 is N-1, % maybe some setup
factorial(NMinus1,FactorialOfNMinus1), %recursive call
Factorial is N*FactorialOfNMinus1). %and maybe some code after
I wouldn't want to do your homework for you, but this should get you going:
sufficient(X,Y,M,N,F) :- %whatever condition means you're done,
% and results = whatever they should
sufficient(X,Y,M,N,F) :- %whatever condition means you aren't done
% and setting up results w/ a recursive call
One more hint: looks like M is a temporary variable and need not be a parameter.
The distance between a long sequence and a short sequence, is the minimum distance between the short sequence and any subsequence of the long sequence that is the same length as the short sequence.
The distance I am using is I think the Manhattan distance. (But this should be unimportant as I would like to be able to change distance functions).
This first version shows a naive implementation without early abandon. I generate all subsequence of the same length, map these to find the distance between them and the short sequence and then use aggregate/3 to find the min.
abs(X,Y,Z):-
Z is abs(X-Y).
seq_seq_absdis(Seq1,Seq2,Dis):-
same_length(Seq1,Seq2),
maplist(abs,Seq1,Seq2,Dislist),
sumlist(Dislist,Dis).
seq_subseq(List1,List2):-
append(List2,_,List1),
dif(List2,[]).
seq_subseq([_|T],Subseq):-
seq_subseq(T,Subseq).
smallseq_largeseq_dis(Sseq,Lseq,Dis):-
findall(Subseq, (same_length(Subseq,Sseq),seq_subseq(Lseq,Subseq)),Subseqs),
maplist(seq_seq_absdis(Sseq),Subseqs,Distances),
aggregate(min(D),member(D,Distances),Dis).
Example query:
?-smallseq_largeseq_dis([1,2,4],[1,2,3,1,2,5],Dis).
Dis = 1
This next version should be more efficient, as it will abandon calculating the distance between a subsequence of the long sequence and the short sequence once the distance is over the minimum already found.
ea_smallseq_largeseq_dis(Sseq,Lseq,Subseq,Dis):-
retractall(best(_,_)),
assert(best(initial,10000)),
findall(Subseq-Dis, ea_smallseq_largeseq_dis_h(Sseq,Lseq,10000,Subseq,Dis),Pairs),
append(_,[Subseq-Dis|[]],Pairs).
ea_smallseq_largeseq_dis_h(Sseq,Lseq,BestSofar1,Subseq,Dis):-
same_length(Sseq,Subseq),
seq_subseq(Lseq,Subseq),
best(_,BestSofar2),
( ( BestSofar2 < BestSofar1) ->
accumulate_dis(Sseq,Subseq,BestSofar2,Dis),
retractall(best(_,_)),
assert(best(Subseq,Dis))
;(
accumulate_dis(Sseq,Subseq,BestSofar1,Dis),
retractall(best(_,_)),
assert(best(Subseq,Dis))
)
).
accumulate_dis(Seq1,Seq2,Best,Dis):-
accumulate_dis(Seq1,Seq2,Best,Dis,0).
accumulate_dis([],[],_Best,Dis,Dis).
accumulate_dis(Seq1,Seq2,Best,Dis,Ac):-
Seq1=[H1|T1],
Seq2=[H2|T2],
abs(H1,H2,Dis1),
Ac1 is Dis1 + Ac,
Ac1 <Best,
accumulate_dis(T1,T2,Best,Dis,Ac1).
Query:
?-ea_smallseq_largeseq_dis([1,2,3],[1,2,4,5,6,7,8,1,2,3],Subseq,Dis).
Dis = 0,
Subseq = [1, 2, 3]
But in this I have used assert and retract so I want to have a version which does the same algorithm but with out these. I think I should be able to do this with a dcg with semicontext notation but find it hard to grasp... how do I keep track of the subsequences I am generating by backtracking and at the same time the 'state' of the minimum distance found so far?
The problem I have..
seq_subseq/2 is generating the sub-sequences by back tracking.
The first subseq tested needs to be set to the min distance.
I then want to loop, so back track to generate another sequence. But to back track I have to fail. But then I cant bring back the min distance so far to check on the next sequence.
If I don't want to use backtracking, I think I need to define a state transition predicate for generating the sub-sequences in order.
At the moment
? seq_subseq([1,2,3,4],X).
X = [1]
X = [1, 2]
X = [1, 2, 3]
X = [1, 2, 3, 4]
X = [2]
X = [2, 3]
X = [2, 3, 4]
X = [3]
X = [3, 4]
X = [4]
So I think I need to define a relation:
subseq0_seq_subseq1(Subseq0,Seq,Subseq1)
That would work like:
?-subseq0_seq_subseq1([1,2,3,4],[1,2,3,4],Subseq1).
Subseq1 = [2].
and
?-subseq0_seq_subseq1([1,2,3],[1,2,3,4],Subseq1).
Subseq1 = [1,2,3,4].
But I need to do this in an efficient way.
Update- Thanks to the answer from Mat I now have this, which is a great improvement I think. Can anyone see any further improvements to this? I have a double nested -> structure and a ! in the accumulate_dis/4 definition both of which seem ugly. I have also made it return the sub-sequence of the long-sequence which is the shortest distance away from the short sequence.
It needs to work with non integers so clpfd is not appropriate I think.
abs(X,Y,Z):-
Z is abs(X-Y).
list_subseq_min(Ls, Subs, Min,BestSeq1) :-
prefix_dist(Ls, Subs, 1000, Front, D0),
BestSeq0=Front,
min_sublist(Ls, Subs,BestSeq0,BestSeq1, D0, Min).
prefix_dist(Ls, Ps, Best,Front,D) :-
same_length(Front, Ps),
append(Front, _, Ls),
accumulate_dis(Front, Ps, Best, D).
min_sublist(Ls0, Subs, BestSeq0,BestSeq2, D0, D) :-
( prefix_dist(Ls0, Subs, D0, Front,D1) ->
min_list([D0,D1],D2),
Ls0 = [_|Ls],
( D0 < D1 ->
BestSeq1 =BestSeq0
;
BestSeq1 =Front
),
min_sublist(Ls, Subs, BestSeq1,BestSeq2, D2, D)
; D = D0,BestSeq0 =BestSeq2
).
accumulate_dis(Seq1,Seq2,Best,Dis):-
accumulate_dis(Seq1,Seq2,Best,Dis,0),!.
accumulate_dis([],[],_Best,Dis,Dis).
accumulate_dis(Seq1,Seq2,Best,Dis,Ac):-
Seq1=[H1|T1],
Seq2=[H2|T2],
abs(H1,H2,Dis1),
Ac1 is Dis1 + Ac,
Ac1 <Best,
accumulate_dis(T1,T2,Best,Dis,Ac1).
accumulate_dis(Seq1,Seq2,Best,Dis):-Dis is Best+1.
query:
?- list_subseq_min([2.1,3.4,4,1.1,2,4,10,12,15],[1,2,3],D,B).
D = 1.1,
B = [1.1, 2, 4].
One important note: You should have clearified that you are talking about the Manhatten distance between lists. This was only clear from your code, whereas your wording can easily lead readers to assume you are talking about the edit distance.
Here is a solution that simply walks the list, keeps track of the minimum, and eventually yields the found minimum.
list_subseq_min(Ls, Subs, Min) :-
prefix_dist(Ls, Subs, D0),
min_sublist(Ls, Subs, D0, Min).
absdiff(X, Y, Z):- Z #= abs(X-Y).
lists_dist(Ls1, Ls2, D) :-
maplist(absdiff, Ls1, Ls2, Ds),
sum(Ds, #=, D).
prefix_dist(Ls, Ps, D) :-
same_length(Front, Ps),
append(Front, _, Ls),
lists_dist(Front, Ps, D).
min_sublist(Ls0, Subs, D0, D) :-
( prefix_dist(Ls0, Subs, D1) ->
D2 #= min(D0,D1),
Ls0 = [_|Ls],
min_sublist(Ls, Subs, D2, D)
; D #= D0
).
Example query and its result:
?- list_subseq_min([1,2,3,1,2,5], [1,2,4], D).
D = 1.
It's quite straight-forward, and since the bookkeeping is limited to only one predicate, using semicontext notations does not really pay off. It is especially useful to use semicontext notation—and DCGs in general—when what is being described spans different rules and communication between them would otherwise be harder.
The running time is in O(N×M).
And now the point, which I leave as an exercise: Modify this solution to prune earlier, if the previously found minimum is already exceeded. Do so in a pure way, or at least as pure as possible: assertz/1 and friends are definitely out of the question, since they prevent your testing these predicates in isolation. Pass around arguments and build the distance more incrementally! This may help you to improve the average running time, though of course not the worst case complexity.
It is for this passing around states between different clauses that semicontext notation may become useful at last.
EDIT: Very well, you have implemented a solution that does the pruning. I will now also show mine. I will reuse the auxiliary predicates absdiff/3 and lists_dist/3 from above, and the following additional auxiliary predicate:
same_length_prefix(Ls, Ps, Front) :-
same_length(Front, Ps),
append(Front, _, Ls).
list_subseq_min/3 is now slightly different:
list_subseq_min(Ls, Subs, Min) :-
same_length_prefix(Ls, Subs, Front),
lists_dist(Front, Subs, D0),
phrase(min_sublist(Ls, Subs), [D0-Front], [Min-_]).
And now the point: min_sublist//2 is a DCG nonterminal that concisely describes the main idea of the algorithm:
min_sublist(Ls0, Subs) -->
( front(Ls0, Subs) ->
{ Ls0 = [_|Ls] },
min_sublist(Ls, Subs)
; []
).
From this description, it is very clear that we are considering the list element by element. It uses fewer (explicit) arguments than previously. The additional two arguments are implicitly passed around as a pair D-Front, which keeps track of the best distance and subsequence found so far. Note how DCG notation exposes the core of the computation, and hides what is not relevant at this place.
The rest is quite self-explanatory, and is analogous to the pruning you have implemented. I highlight the single use of semicontext notation in this program, which lets us express any change of the optimal sequence found so far.
front(Ls, Subs), [D-Front] -->
[Current],
{ same_length_prefix(Ls, Subs, Front1),
capped_dist(Front1, Subs, Current, 0-Front1, D-Front) }.
capped_dist([], [], _, DF, DF).
capped_dist([L|Ls], [P|Ps], Current, D1-Front1, DF) :-
absdiff(L, P, D2),
D3 #= D1 + D2,
Current = D0-_,
( D3 #> D0 -> DF = Current
; capped_dist(Ls, Ps, Current, D3-Front1, DF)
).
I can't bring myself to accept the nastiness and primitiveness of contemporary floating-point numbers, so I have retained the integer arithmetic and simply multiply all numbers you show so that they become integers:
?- list_subseq_min([21,34,40,11,20,40,100,120,150], [10,20,30], D).
D = 11.
I leave extending this so that it also shows the found subsequence as an easy exercise.
One important note: The capping only affects the calculation of the distance; note in particular that the running time is Θ(N×M) due to the way same_length_prefix/3 is used in front//2! I leave improving this as a slightly harder exercise.
Kindly, could you help me in the following:
I am writing a Prolog program that takes two numbers digits then combine them as one number, for example:
Num1: 5
Num2: 1
Then the new number is 51.
Assume V1 is the first number digit and V2 is the second number digit. I want to combine V1 and V2 then multiply the new number with V3, so my question is how I can do it?
calculateR(R, E, V1, V2, V3, V4):-
R is V1 V2 * V3,
E is R * V4.
Your help is appreciated.
Here is another solution that is based on the idea of #aBathologist and that relies on ISO predicates only, and does not dependent on SWI's idiosyncratic modifications and extensions. Nor does it have most probably unwanted solutions like calculateR('0x1',1,1,17). nor calculateR(1.0e+30,0,1,1.0e+300). Nor does it create unnecessary temporary atoms.
So the idea is to restrict the definition to decimal numbers:
digit_digit_number(D1, D2, N) :-
number_chars(D1, [Ch1]),
number_chars(D2, [Ch2]),
number_chars(N, [Ch1,Ch2]).
Here is a version which better clarifies the relational nature of Prolog - using library(clpfd) which is available in many Prolog systems (SICStus, SWI, B, GNU, YAP). It is essentially the same program as the one with (is)/2 except that I added further redundant constraints that permit the system to ensure termination in more general cases, too:
:- use_module(library(clpfd)).
digits_radix_number(Ds, R, N) :-
digits_radix_numberd(Ds, R, 0,N).
digits_radix_numberd([], _, N,N).
digits_radix_numberd([D|Ds], R, N0,N) :-
D #>= 0, D #< R,
R #> 0,
N0 #=< N,
N1 #= D+N0*R,
digits_radix_numberd(Ds, R, N1,N).
Here are some uses:
?- digits_radix_number([1,4,2],10,N).
N = 142.
?- digits_radix_number([1,4,2],R,142).
R = 10.
?- digits_radix_number([1,4,2],R,N).
R in 5..sup, 4+R#=_A, _A*R#=_B, _A in 9..sup, N#>=_A,
N in 47..sup, 2+_B#=N, _B in 45..sup.
That last query asks for all possible radices that represent [1,4,2] as a number. As you can see, not anything can be represented that way. The radix has to be 5 or larger which is not surprising given the digit 4, and the number itself has to be at least 47.
Let's say we want to get a value between 1450..1500, what radix do we need to do that?
?- digits_radix_number([1,4,2],R,N), N in 1450..1500.
R in 33..40, 4+R#=_A, _A*R#=_B, _A in 37..44,
N in 1450..1500, 2+_B#=N, _B in 1448..1498.
Gnah, again gibberish. This answer contains many extra equations that have to hold. Prolog essentially says: Oh yes, there is a solution, provided all this fine print is true. Do the math yourself!
But let's face it: It is better if Prolog gives such hard-to-swallow answer than if it would say Yes.
Fortunately there are ways to remove such extra conditions. One of the simplest is called "labeling", where Prolog will "try out" value after value:
?- digits_radix_number([1,4,2],R,N), N in 1450..1500, labeling([],[N]).
false.
That is clear response now! There is no solution. All these extra conditions where essentially false, like all that fine print in your insurance policy...
Here's another question: Given the radix and the value, what are the required digits?
?- digits_radix_number(D,10,142).
D = [1,4,2]
; D = [0,1,4,2]
; D = [0,0,1,4,2]
; D = [0,0,0,1,4,2]
; D = [0,0,0,0,1,4,2]
; ... .
So that query can never terminate, because 00142 is the same number as 142. Just as 007 is agent number 7.
Here is a straight-forward solution that should work in any Prolog close to ISO:
digits_radix_to_number(Ds, R, N) :-
digits_radix_to_number(Ds, R, 0,N).
digits_radix_to_number([], _, N,N).
digits_radix_to_number([D|Ds], R, N0,N) :-
N1 is D+N0*R,
digits_radix_to_number(Ds, R, N1,N).
?- digits_radix_to_number([1,4,2],10,R).
R = 142.
Edit: In a comment, #false pointed out that this answer is SWI-Prolog specific.
You can achieve your desired goal by treating the numerals as atoms and concatenating them, and then converting the resultant atom into a number.
I'll use atom_concat/3 to combine the two numerals. In this predicate, the third argument with be the combination of atoms in its first and second arguments. E.g.,
?- atom_concat(blingo, dingo, X).
X = blingodingo.
Note that, when you do this with two numerals, the result is an atom not a number. This is indicated by the single quotes enclosing the the result:
?- atom_concat(5, 1, X).
X = '51'.
But 51 \= '51' and we cannot multiply an atom by number. We can use atom_number/2 to convert this atom into a number:
?- atom_number('51', X).
X = 51.
That's all there is to it! Your predicate might look like this:
calculateR(No1, No2, Multiplier, Result) :-
atom_concat(No1, No2, NewNoAtom),
atom_number(NewNoAtom, NewNo),
Result is NewNo * Multiplier.
Usage example:
?- calculateR(5, 1, 3, X).
X = 153.
Of course, you'll need more if you want to prompt the user for input.
I expect #Wouter Beek's answer is more efficient, since it doesn't rely on converting the numbers to and from atoms, but just uses the assumption that each numeral is a single digit to determine the resulting number based on their position. E.g., if 5 is in the 10s place and 1 is in the 1s place, then the combination of 5 and 1 will be 5 * 10 + 1 * 1. The answer I suggest here will work with multiple digit numerals, e.g., in calculateR(12, 345, 3, Result), Result is 1234 * 3. Depending on what you're after this may or may not be a desired result.
If you know the radix of the numbers involved (and the radix is the same for all the numbers involved), then you can use the reverse index of the individual numbers in order to calculate their positional summation.
:- use_module(library(aggregate)).
:- use_module(library(lists)).
digits_to_number(Numbers1, Radix, PositionalSummation):-
reverse(Numbers1, Numbers2),
aggregate_all(
sum(PartOfNumber),
(
nth0(Position, Numbers2, Number),
PartOfNumber is Number * Radix ^ Position
),
PositionalSummation
).
Examples of use:
?- digits_to_number([5,1], 10, N).
N = 51.
?- digits_to_number([5,1], 16, N).
N = 81.
(The code sample is mainly intended to bring the idea across. Notice that I use aggregate_all/3 from SWI-Prolog here. The same could be achieved by using ISO predicates exclusively.)
I hope this hasn't been asked before, if so I apologize.
EDIT: For clarity, the following notation will be used: boldface uppercase for matrices, boldface lowercase for vectors, and italics for scalars.
Suppose x0 is a vector, A and B are matrix functions, and f is a vector function.
I'm looking for the best way to do the following iteration scheme in Mathematica:
A0 = A(x0), B0=B(x0), f0 = f(x0)
x1 = Inverse(A0)(B0.x0 + f0)
A1 = A(x1), B1=B(x1), f1 = f(x1)
x2 = Inverse(A1)(B1.x1 + f1)
...
I know that a for-loop can do the trick, but I'm not quite familiar with Mathematica, and I'm concerned that this is the most efficient way to do it. This is a justified concern as I would like to define a function u(N):=xNand use it in further calculations.
I guess my questions are:
What's the most efficient way to program the scheme?
Is RecurrenceTable a way to go?
EDIT
It was a bit more complicated than I tought. I'm providing more details in order to obtain a more thorough response.
Before doing the recurrence, I'm having problems understanding how to program the functions A, B and f.
Matrices A and B are functions of the time step dt = 1/T and the space step dx = 1/M, where T and M are the number of points in the {0 < x < 1, 0 < t} region. This is also true for vector the function f.
The dependance of A, B and f on x is rather tricky:
A and B are upper and lower triangular matrices (like a tridiagonal matrix; I suppose we can call them multidiagonal), with defined constant values on their diagonals.
Given a point 0 < xs < 1, I need to determine it's representative xn in the mesh (the closest), and then substitute the nth row of A and B with the function v( x) (transposed, of course), and the nth row of f with the function w( x).
Summarizing, A = A(dt, dx, xs, x). The same is true for B and f.
Then I need do the loop mentioned above, to define u( x) = step[T].
Hope I've explained myself.
I'm not sure if it's the best method, but I'd just use plain old memoization. You can represent an individual step as
xstep[x_] := Inverse[A[x]](B[x].x + f[x])
and then
u[0] = x0
u[n_] := u[n] = xstep[u[n-1]]
If you know how many values you need in advance, and it's advantageous to precompute them all for some reason (e.g. you want to open a file, use its contents to calculate xN, and then free the memory), you could use NestList. Instead of the previous two lines, you'd do
xlist = NestList[xstep, x0, 10];
u[n_] := xlist[[n]]
This will break if n > 10, of course (obviously, change 10 to suit your actual requirements).
Of course, it may be worth looking at your specific functions to see if you can make some algebraic simplifications.
I would probably write a function that accepts A0, B0, x0, and f0, and then returns A1, B1, x1, and f1 - say
step[A0_?MatrixQ, B0_?MatrixQ, x0_?VectorQ, f0_?VectorQ] := Module[...]
I would then Nest that function. It's hard to be more precise without more precise information.
Also, if your procedure is numerical, then you certainly don't want to compute Inverse[A0], as this is not a numerically stable operation. Rather, you should write
A0.x1 == B0.x0+f0
and then use a numerically stable solver to find x1. Of course, Mathematica's LinearSolve provides such an algorithm.