Complete newbie in PROLOG here. I am provided with a list of parts to implement a hypothetical computational system:
%partName(name, type, number, required_types, size, price)
part(a-1, a, 1, 2*b, 0.5, 60 )
part(a-2, a, 2, 3*b, 1.0, 25 )
part(b-1, b, 1, 2*c, 0.5, 25 )
part(b-2, b, 2, -, 1, 45 )
part(c-1, c, 1, -, 0.5, 5 )
part(c-2, c, 2, -, 0.5, 10 )
part(d-1, d, 1, 1*b && 2*c, 0.5, 45 )
part(d-2, d, 2, 1*c-1, 0.5, 25 )
For example: part a-1, of type a and number 1, needs two (2) b-type parts to function (either two b-1s or two b-2s of one of each), and then b-1 requires another two (2) c-type parts to function, etc. Parts also include a size (half size is 0.5, full size is 1) and a price.
(NOTE: parts with "-" as needed type work on their own, part d-1 requires one b-type AND two c types, and d-2 requires one c-1 in particular rather than generally a c type.)
The input is to be a setup of part types, like {a, b}, {c, d}, {a, b, d} etc.
The output is to be the cheapest possible configuration/combination of them in terms of space and money.
Remember: for example, in {a, b}, "a" can either be a-1 or a-2 (same with b) and each part needs the parts listed above in order to function and so on, so I presume it all works in a tree like, somewhat "recursive" manner.
I guess I first have to find all possible combinations/configurations and then somehow find the cheapest/most compact.
Can anyone provide a working solution or basic insight on how to set up the problem?
Please be a lifesaver.
Related
I have some sets of combinations and I want to find out the intersection function between say two of them. Then I want to represent the intersected results in ZDD.
I am thinking about using the CUDD package to do this.
An example:
All the 4-bit strings having hamming distance >= 2 with 1100 =
{ 0001, 0010, 0011,0101, 0110, 0111, 1001, 1010, 1011 }
All the 4-bit strings having hamming distance >= 2 with 0000 =
{ 0011, 0101, 0110, 1001, 1010, 0111, 1011, 1101, 1110 }
Intersected elements of the set (what I want):
{0011, 0101, 0110, 1010, 1001 }
From what I understand, I need to be able to express those sets of combinations first, with boolean functions, e.g. ( f = a b c d ) to represent their corresponding BDDs, convert them to ZDDs and then find out the intersection? Someone experienced with the CUDD package please help.
Your reasoning is correct. You can first build BDDs corresponding to the two string sets, convert them to ZDDs, and then build the intersection (logical AND).
However, you can also first compute the intersection (logical AND) and then translate the result to a ZDD.
It is however not clear what you mean by "find out the intersection" - what do you want to do with it? Print out all the elements? Count the number of elements? Depending what what is your aim, the translation to ZDDs may be unnecessary (or the use of CUDD altogether).
I am following a MOOC, but I don't understand the correct answer nor the other answers.
The MOOC closed and I cannot ask any questions on the forum.
This is the question:
Considering the following relation R:
A B C D
1 0 2 2
4 1 2 2
6 0 6 3
7 1 2 3
1 0 6 1
1 1 2 1
Between all these requests, which one return the same relation R?
ΠA,B,C,D(R⋈δA→D,D→F(R))
R⋈δA→D,D→A(R)
R⋈δB→C,C→B(R)
ΠA,B,C,D(R⋈δB→G,C→F(R)) (note: this is the correct answer)
The only given explanation is :
The first 3 answers loose the tuple(4,1,2,2). In the last joint, no tuple is lost.
Could you details please whats does the answers do?
Thank you very much for your attention!
This is a question about the Relational Algebra's Natural Join, and attribute naming. I presume the squiggly thing in your formulas is for Rename, usually denoted by Greek letter rho ρ (see the wikipedia link).
For Natural Join see the wikipedia example and note
The result of the natural join is the set of all combinations of tuples in R and S that are equal on their common attribute names.
Because of the renaming in the four formulas, in general, the result from renamed R will not have the same attribute names as the original R, or will not be equal on the values in the resulting same-named attributes.
I suggest you go through each four of the renamings, and work out what is the 'heading' of each result -- that is, what are the resulting attribute names.
You'll find in requests 1., 2., 3. there's at least one resulting attribute same-named as the original R but the values for that attribute are not the same.
In request 4., although attributes B, C are renamed, their new names do not clash with any existing attribute in R. So the Natural Join to original R will use attributes A, D. This'll produce an interesting intermediate result: consider the tuples <1, 0, 6, 1>, <1, 1, 2, 1> which each contain equal values in their A attribute and their D attribute.
But then in request 4., the projection will throw away the newly-named attributes G, F and collapse back to the original A, B, C, D. So in general, request 4. always returns exactly the original R.
Requests 1., 2., 3. might sometimes return the original R, depending on the content of R. But with the content you show, there are clashes of newly-same-named attributes with non-equal values, so they do 'lose' tuples.
BTW, although tuple <4, 1, 2, 2> does indeed get 'lost' in those three requests, it's not the only tuple that gets 'lost'. In particular in request 3., note that for the sample data, there are no values in common between B, C, so swapping them round in the rename has the effect of returning an empty result from the Join.
Keep getting the error Arguments are not sufficiently instantiated for the multiplication by addition rule I wrote as shown below.
mult(_, 0, 0). %base case for multiplying by 0
mult(X, 1, X). % another base case
mult(X, Y, Z) :-
Y > 1,
Y1 is Y - 1,
mult(X, Y1, Z1),
Z is X + Z1.
I am new to Prolog and really struggling with even such simple problems.
Any recommendations for books or online tutorials would be great.
I am running it on SWI-Prolog on Ubuntu Linux.
In your definition of mult/3 the first two arguments have to be known. If one of them is still a variable, an instantiation error will occur. Eg. mult(2, X, 6) will yield an instantiation error, although X = 3 is a correct answer ; in fact, the only answer.
There are several options you have:
successor-arithmetics, constraints, or meta-logical predicates.
Here is a starting point with successor arithmetics:
add(0,Y,Y).
add(s(X),Y,s(Z)) :- add(X,Y,Z).
Another approach would be to use constraints over the integers. YAP and SWI have a library(clpfd) that can be used in a very flexible manner: Both for regular integer computations and the more general constraints. Of course, multiplication is already predefined:
?- A * B #= C.
A*B#=C.
?- A * B #= C, C = 6.
C = 6, A in -6.. -1\/1..6, A*B#=6, B in -6.. -1\/1..6.
?- A * B #= C, C = 6, A = 2.
A = 2, B = 3, C = 6.
Meta-logical predicates: I cannot recommend this option in which you would use var/1, nonvar/1, ground/1 to distinguish various cases and handle them differently. This is so error prone that I have rarely seen a correct program using them. In fact, even very well known textbooks contain serious errors!
I think you got the last two calls reversed. Don't you mean:
mult(X,Y,Z):- Y>1,Y1 is Y-1, Z1 is X+Z, mult(X,Y1,Z1).
Edit: nevermind that, looked at the code again and it doesn't make sense. I believe your original code is correct.
As for why that error is occuring, I need to know how you're calling the predicate. Can you give an example input?
The correct way of calling your predicate is mult(+X, +Y, ?Z):
?- mult(5,0,X).
X = 0
?- mult(5,1,X).
X = 5
?- mult(5,5,X).
X = 25
?- mult(4,4,16).
yes
?- mult(3,3,10).
no
etc. Calling it with a free variable in the first two arguments will produce that error, because one of them will be used in the right side of an is or in either side of the <, and those predicates expect ground terms to succeed.
Here is a function I would like to write but am unable to do so. Even if you
don't / can't give a solution I would be grateful for tips. For example,
I know that there is a correlation between the ordered represantions of the
sum of an integer and ordered set partitions but that alone does not help me in
finding the solution. So here is the description of the function I need:
The Task
Create an efficient* function
List<int[]> createOrderedPartitions(int n_1, int n_2,..., int n_k)
that returns a list of arrays of all set partions of the set
{0,...,n_1+n_2+...+n_k-1} in number of arguments blocks of size (in this
order) n_1,n_2,...,n_k (e.g. n_1=2, n_2=1, n_3=1 -> ({0,1},{3},{2}),...).
Here is a usage example:
int[] partition = createOrderedPartitions(2,1,1).get(0);
partition[0]; // -> 0
partition[1]; // -> 1
partition[2]; // -> 3
partition[3]; // -> 2
Note that the number of elements in the list is
(n_1+n_2+...+n_n choose n_1) * (n_2+n_3+...+n_n choose n_2) * ... *
(n_k choose n_k). Also, createOrderedPartitions(1,1,1) would create the
permutations of {0,1,2} and thus there would be 3! = 6 elements in the
list.
* by efficient I mean that you should not initially create a bigger list
like all partitions and then filter out results. You should do it directly.
Extra Requirements
If an argument is 0 treat it as if it was not there, e.g.
createOrderedPartitions(2,0,1,1) should yield the same result as
createOrderedPartitions(2,1,1). But at least one argument must not be 0.
Of course all arguments must be >= 0.
Remarks
The provided pseudo code is quasi Java but the language of the solution
doesn't matter. In fact, as long as the solution is fairly general and can
be reproduced in other languages it is ideal.
Actually, even better would be a return type of List<Tuple<Set>> (e.g. when
creating such a function in Python). However, then the arguments wich have
a value of 0 must not be ignored. createOrderedPartitions(2,0,2) would then
create
[({0,1},{},{2,3}),({0,2},{},{1,3}),({0,3},{},{1,2}),({1,2},{},{0,3}),...]
Background
I need this function to make my mastermind-variation bot more efficient and
most of all the code more "beautiful". Take a look at the filterCandidates
function in my source code. There are unnecessary
/ duplicate queries because I'm simply using permutations instead of
specifically ordered partitions. Also, I'm just interested in how to write
this function.
My ideas for (ugly) "solutions"
Create the powerset of {0,...,n_1+...+n_k}, filter out the subsets of size
n_1, n_2 etc. and create the cartesian product of the n subsets. However
this won't actually work because there would be duplicates, e.g.
({1,2},{1})...
First choose n_1 of x = {0,...,n_1+n_2+...+n_n-1} and put them in the
first set. Then choose n_2 of x without the n_1 chosen elements
beforehand and so on. You then get for example ({0,2},{},{1,3},{4}). Of
course, every possible combination must be created so ({0,4},{},{1,3},{2}),
too, and so on. Seems rather hard to implement but might be possible.
Research
I guess this
goes in the direction I want however I don't see how I can utilize it for my
specific scenario.
http://rosettacode.org/wiki/Combinations
You know, it often helps to phrase your thoughts in order to come up with a solution. It seems that then the subconscious just starts working on the task and notifies you when it found the solution. So here is the solution to my problem in Python:
from itertools import combinations
def partitions(*args):
def helper(s, *args):
if not args: return [[]]
res = []
for c in combinations(s, args[0]):
s0 = [x for x in s if x not in c]
for r in helper(s0, *args[1:]):
res.append([c] + r)
return res
s = range(sum(args))
return helper(s, *args)
print partitions(2, 0, 2)
The output is:
[[(0, 1), (), (2, 3)], [(0, 2), (), (1, 3)], [(0, 3), (), (1, 2)], [(1, 2), (), (0, 3)], [(1, 3), (), (0, 2)], [(2, 3), (), (0, 1)]]
It is adequate for translating the algorithm to Lua/Java. It is basically the second idea I had.
The Algorithm
As I already mentionend in the question the basic idea is as follows:
First choose n_1 elements of the set s := {0,...,n_1+n_2+...+n_n-1} and put them in the
first set of the first tuple in the resulting list (e.g. [({0,1,2},... if the chosen elements are 0,1,2). Then choose n_2 elements of the set s_0 := s without the n_1 chosen elements beforehand and so on. One such a tuple might be ({0,2},{},{1,3},{4}). Of
course, every possible combination is created so ({0,4},{},{1,3},{2}) is another such tuple and so on.
The Realization
At first the set to work with is created (s = range(sum(args))). Then this set and the arguments are passed to the recursive helper function helper.
helper does one of the following things: If all the arguments are processed return "some kind of empty value" to stop the recursion. Otherwise iterate through all the combinations of the passed set s of the length args[0] (the first argument after s in helper). In each iteration create the set s0 := s without the elements in c (the elements in c are the chosen elements from s), which is then used for the recursive call of helper.
So what happens with the arguments in helper is that they are processed one by one. helper may first start with helper([0,1,2,3], 2, 1, 1) and in the next invocation it is for example helper([2,3], 1, 1) and then helper([3], 1) and lastly helper([]). Of course another "tree-path" would be helper([0,1,2,3], 2, 1, 1), helper([1,2], 1, 1), helper([2], 1), helper([]). All these "tree-paths" are created and thus the required solution is generated.
Given a set** S containing duplicate elements, how can one determine the total number all the possible subsets of S, where each subset is unique.
For example, say S = {A, B, B} and let K be the set of all subsets, then K = {{}, {A}, {B}, {A, B}, {B, B}, {A, B, B}} and therefore |K| = 6.
Another example would be if S = {A, A, B, B}, then K = {{}, {A}, {B}, {A, B}, {A, A}, {B, B}, {A, B, B}, {A, A, B}, {A, A, B, B}} and therefor |K| = 9
It is easy to see that if S is a real set, having only unique elements, then |K| = 2^|S|.
What is a formula to calculate this value |K| given a "set" S (with duplicates), without generating all the subsets?
** Not technically a set.
Take the product of all the (frequencies + 1).
For example, in {A,B,B}, the answer is (1+1) [the number of As] * (2+1) [the number of Bs] = 6.
In the second example, count(A) = 2 and count(B) = 2. Thus the answer is (2+1) * (2+1) = 9.
The reason this works is that you can define any subset as a vector of counts - for {A,B,B}, the subsets can be described as {A=0,B=0}, {A=0,B=1}, {0,2}, {1,0}, {1,1}, {1,2}.
For each number in counts[] there are (frequencies of that object + 1) possible values. (0..frequencies)
Therefore, the total number of possiblities is the product of all (frequencies+1).
The "all unique" case can also be explained this way - there is one occurence of each object, so the answer is (1+1)^|S| = 2^|S|.
I'll argue that this problem is simple to solve, when viewed in the proper way. You don't care about order of the elements, only whether they appear in a subset of not.
Count the number of times each element appears in the set. For the one element set {A}, how many subsets are there? Clearly there are only two sets. Now suppose we added another element, B, that is distinct from A, to form the set {A,B}. We can form the list of all sets very easily. Take all the sets that we formed using only A, and add in zero or one copy of B. In effect, we double the number of sets. Clearly we can use induction to show that for N distinct elements, the total number of sets is just 2^N.
Suppose that some elements appear multiple times? Consider the set with three copies of A. Thus {A,A,A}. How many subsets can you form? Again, this is simple. We can have 0, 1, 2, or 3 copies of A, so the total number of subsets is 4 since order does not matter.
In general, for N copies of the element A, we will end up with N+1 possible subsets. Now, expand this by adding in some number, M, of copies of B. So we have N copies of A and M copies of B. How many total subsets are there? Yes, this seems clear too. To every possible subset with only A in it (there were N+1 of them) we can add between 0 and M copies of B.
So the total number of subsets when we have N copies of A and M copies of B is simple. It must be (N+1)*(M+1). Again, we can use an inductive argument to show that the total number of subsets is the product of such terms. Merely count up the total number of replicates for each distinct element, add 1, and take the product.
See what happens with the set {A,B,B}. We get 2*3 = 6.
For the set {A,A,B,B}, we get 3*3 = 9.