For loops in ocaml - for-loop

I want to do something like
let switchgraph cases =
let g = Graph.makeGraph() in
let g = (Graph.addNode g 1) in
for i = 2 to cases do
let g = (Graph.addNode g i) in
done
g
But apparently, this is not possible. How else can i achieve this.

There are two things you need to fix:
you need to use references (see ref, := and !) for this, since let bindings are immutable
to sequence two expressions, you need to use ;
Something like this should work:
let switchgraph cases =
let g = ref (Graph.makeGraph()) in
g := Graph.addNode (!g) 1;
for i = 2 to cases do
g := Graph.addNode (!g) i
done;
!g
Note that g is the reference, and !g the value.

Related

Question: how to implement A=a/1, B = b/3, C=A+B that results in C = a/1 + b/3?

I am very new to Prolog, and would appreciate any help.
I want to implement the following in Prolog:
combine (A,B,C)
D = A/1,
E = B/3,
and C = A + B , in which given expression of A,B, C will return as A/1 + E/3, rather than calculating its value.
Many thanks!

Is these algorithms equivalent?

E is a logic variable (T/F), P and Q are programs
(P)
If E then R
Else
S
(Q)
bool c = E
bool d = not E
While c do
Begin
R
c = d
End
While d do
Begin
S
d = c
End
We knew that, the same input mean the same output, so they are weak-equivalency, but what about the execute time numbers (R)? I am not sure R is for (R,S) or E?
As it is now, both variants are equivalent in the sense that R and S are fulfilled or not in both variants according to the same start conditions.
But the second variant also sets two variables c and d, and obviously they will be used somehow later, for otherwards there is no use in their setting inside while cycles. So, the second variant has additional and independent consequences (c and d are defined and set to false).
If the S and R can cancel the whole algorithm, that additional part becomes NOT independent.

Which way of these two pattern matching is more preferred?

I'm just curious, these two functions would do the same thing. But which one should I use?
let f a =
match a with
b -> a;;
let f a =
match a with
b -> b;;
Or it just depends on your preference?
I feel the second one would be better but I'm not sure.
Performance wise there is no difference. Style-wise b -> a is a bit problematic because you have an unused variable b. _ -> a would make more sense. Other than that, it's just preference.
Personally I would prefer _ -> a over b -> b because it doesn't introduce an extra variable.
PS: I assume in your real code there are more cases than just b - otherwise you could just write let f a = a.
Also, in your particular example I would rewrite using function
let f = function
| b -> b

Minimizing chunks in a matrix

Suppose I have the following matrix:
The matrix can be broken down into chunks such that each chunk must, for all rows, have the same number of columns where the value is marked true for that row.
For example, the following chunk is valid:
This means that rows do not have to be contiguous.
Columns do not have to be contiguous either, as the following is a valid chunk:
However, the following is invalid:
That said, what is an algorithm that can be used to select chunks such that the minimal number of chunks will be used when finding all the chunks?
Given the example, above, the proper solution is (items with the same color represent a valid chunk):
In the above example, three is the minimal number of chunks that this can be broken down into.
Note that the following is also a valid solution:
There's not a preference to the solutions, really, just to get the least number of chunks.
I thought of counting using adjacent cells, but that doesn't account for the fact that the column values don't have to be contiguous.
I believe the key lies in finding the chunks with the largest area given the constraints, removing those items, and then repeating.
Taking that approach, the solution is:
But how to traverse the matrix and find the largest area is eluding me.
Also note, that if you want to reshuffle the rows and/or columns during the operations, that's a valid operation (in order to find the largest area), but I'd imagine you can only do it after you remove the largest areas from the matrix (after one area is found and moving onto the next).
You are doing circuit minimization on a truth table. For 4x4 truth tables, you can use a K map. The Quine-McCluskey algorithm is a generalization that can handle larger truth tables.
Keep in mind the problem is NP-Hard, so depending on the size of your truth tables, this problem can quickly grow to a size that is intractable.
This problem is strongly related to Biclustering, for which there are many efficient algorithms (and freely available implementations). Usually you will have to specify the number K of clusters you expect to find; if you don't have a good idea what K should be, you can proceed by binary search on K.
In case the biclusters don't overlap, you are done, otherwise you need to do some geometry to cut them into "blocks".
The solution I propose is fairly straightforward, but very time consuming.
It can be decomposed in 4 major steps:
find all the existing patterns in the matrix,
find all the possible combinations of these patterns,
remove all the incomplete pattern sets,
scan the remaining list to get the set with the minimum number of elements
First of, the algorithm below works on either column or row major matrices. I chose column for the explanations, but you may swap it for rows at your convenience, as long as it remains consistent accross the whole process.
The sample code accompanying the answer is in OCaml, but doesn't use any specific feature of the language, so it should be easy to port to other ML dialects.
Step 1:
Each column can be seen as a bit vector. Observe that a pattern (what you call chunk in your question) can be constructed by intersecting (ie. and ing) all the columns, or all the rows composing it, or even a combinations. So the first step is really about producing all the combinations of rows and columns (the powerset of the matrix' rows and columns if you will), intersecting them at the same time, and filter out the duplicates.
We consider the following interface for a matrix datatype:
module type MATRIX = sig
type t
val w : int (* the width of the matrix *)
val h : int (* the height ........ *)
val get : t -> int -> int -> bool (* cell value getter *)
end
Now let's have a look at this step's code:
let clength = M.h
let rlength = M.w
(* the vector datatype used throughought the algorithm
operator on this type are in the module V *)
type vector = V.t
(* a pattern description and comparison operators *)
module Pattern = struct
type t = {
w : int; (* width of thd pattern *)
h : int; (* height of the pattern *)
rows : vector; (* which rows of the matrix are used *)
cols : vector; (* which columns... *)
}
let compare a b = Pervasives.compare a b
let equal a b = compare a b = 0
end
(* pattern set : let us store patterns without duplicates *)
module PS = Set.Make(Pattern)
(* a simple recursive loop on #f #k times *)
let rec fold f acc k =
if k < 0
then acc
else fold f (f acc k) (pred k)
(* extract a column/row of the given matrix *)
let cr_extract mget len =
fold (fun v j -> if mget j then V.set v j else v) (V.null len) (pred len)
let col_extract m i = cr_extract (fun j -> M.get m i j) clength
let row_extract m i = cr_extract (fun j -> M.get m j i) rlength
(* encode a single column as a pattern *)
let col_encode c i =
{ w = 1; h = count c; rows = V.set (V.null clength) i; cols = c }
let row_encode r i =
{ h = 1; w = count r; cols = V.set (V.null rlength) i; rows = r }
(* try to add a column to a pattern *)
let col_intersect p c i =
let col = V.l_and p.cols c in
let h = V.count col in
if h > 0
then
let row = V.set (V.copy p.rows) i in
Some {w = V.count row; h = h; rows = row; clos = col}
else None
let row_intersect p r i =
let row = V.l_and p.rows r in
let w = V.count row in
if w > 0
then
let col = V.set (V.copy p.cols) i in
Some { w = w; h = V.count col; rows = row; cols = col }
else None
let build_patterns m =
let bp k ps extract encode intersect =
let build (l,k) =
let c = extract m k in
let u = encode c k in
let fld p ps =
match intersect p c k with
None -> l
| Some npc -> PS.add npc ps
in
PS.fold fld (PS.add u q) q, succ k
in
fst (fold (fun res _ -> build res) (ps, 0) k)
in
let ps = bp (pred rlength) PS.empty col_extract col_encode col_intersect in
let ps = bp (pred clength) ps row_extract row_encode row_intersect in
PS.elements ps
The V module must comply with the following signature for the whole algorithm:
module type V = sig
type t
val null : int -> t (* the null vector, ie. with all entries equal to false *)
val copy : t -> t (* copy operator *)
val get : t -> int -> bool (* get the nth element *)
val set : t -> int -> t (* set the nth element to true *)
val l_and : t -> t -> t (* intersection operator, ie. logical and *)
val l_or : t -> t -> t (* logical or *)
val count : t -> int (* number of elements set to true *)
val equal : t -> t -> bool (* equality predicate *)
end
Step 2:
Combining the patterns can also be seen as a powerset construction, with some restrictions: A valid pattern set may only contain patterns which don't overlap. The later can be defined as true for two patterns if both contain at least one common matrix cell.
With the pattern data structure used above, the overlap predicate is quite simple:
let overlap p1 p2 =
let nullc = V.null h
and nullr = V.null w in
let o v1 v2 n = not (V.equal (V.l_and v1 v2) n) in
o p1.rows p2.rows nullr && o p1.cols p2.cols nullc
The cols and rows of the pattern record indicate which coordinates in the matrix are included in the pattern. Thus a logical and on both fields will tell us if the patterns overlap.
For including a pattern in a pattern set, we must ensure that it does not overlap with any pattern of the set.
type pset = {
n : int; (* number of patterns in the set *)
pats : pattern list;
}
let overlap sp p =
List.exists (fun x -> overlap x p) sp.pats
let scombine sp p =
if overlap sp p
then None
else Some {
n = sp.n + 1;
pats = p::sp.pats;
}
let build_pattern_sets l =
let pset l p =
let sp = { n = 1; pats = [p] } in
List.fold_left (fun l spx ->
match scombine spx p with
None -> l
| Some nsp -> nsp::l
) (sp::l) l
in List.fold_left pset [] l
This step produces a lot of sets, and thus is very memory and computation intensive. It's certainly the weak point of this solution, but I don't see yet how to reduce the fold.
Step 3:
A pattern set is incomplete if when rebuilding the matrix with it, we do not obtain the original one. So the process is rather simple.
let build_matrix ps w =
let add m p =
let rec add_col p i = function
| [] -> []
| c::cs ->
let c =
if V.get p.rows i
then V.l_or c p.cols
else c
in c::(add_col p (succ i) cs)
in add_col p 0 m
in
(* null matrix as a list of null vectors *)
let m = fold (fun l _ -> V.null clength::l) [] (pred rlength) in
List.fold_left add m ps.pats
let drop_incomplete_sets m l =
(* convert the matrix to a list of columns *)
let m' = fold (fun l k -> col_extract m k ::l) [] (pred rlength) in
let complete m sp =
let m' = build_matrix sp in
m = m'
in List.filter (fun x -> complete m' x) l
Step 4:
The last step is just selecting the set with the smallest number of elements:
let smallest_set l =
let smallest ps1 ps2 = if ps1.n < ps2.n then ps1 else ps2 in
match l with
| [] -> assert false (* there should be at least 1 solution *)
| h::t -> List.fold_left smallest h t
The whole computation is then just the chaining of each steps:
let compute m =
let (|>) f g = g f in
build_patterns m |> build_pattern_sets |> drop_incomplete_sets m |> smallest_set
Notes
The algorithm above constructs a powerset of a powerset, with some limited filtering. There isn't as far as I know a way to reduce the search (as mentioned in a comment, if this is a NP hard problem, there isn't any).
This algorithm checks all the possible solutions, and correctly returns an optimal one (tested with many matrices, including the one given in the problem description.
One quick remark regarding the heuristic you propose in your question:
it could be easily implemented using the first step, removing the largest pattern found, and recursing. That would yeld a solution much more rapidly than my algorithm. However, the solution found may not be optimal.
For instance, consider the following matrix:
.x...
.xxx
xxx.
...x.
The central 4 cell chunck is the largest which may be found, but the set using it would comprise 5 patterns in total.
.1...
.223
422.
...5.
Yet this solution uses only 4:
.1...
.122
334.
...4.
Update:
Link to the full code I wrote for this answer.

Let and construct versus let in sequence

Consider this OCaml code:
let coupe_inter i j cases =
let lcases = Array.length cases in
let low,_,_ = cases.(i)
and _,high,_ = cases.(j) in
low,high,
Array.sub cases i (j-i+1),
case_append (Array.sub cases 0 i) (Array.sub cases (j+1) (lcases-(j+1)))
Why the expression let ... and ... in is used in place of a let ... in let ... in sequence (like F# force you to do)? This construct seems quite frequent in OCaml code.
Thanks!
let x = a and y = b in c has the effect of defining x and y "simultaneously". This means that the order of evaluation (a after or before b) is unspecified (you must not assume that a will be evaluated before), and that x is not bound in b and y not bound in a, they are only available in c.
I rarely use this construction, because I have been bitten in the past by the evaluation order thing. I often use the recursive variant of it, let rec ... and ... in ... (where all variable bound are available everywhere), however, to define mutually recursive functions.
let rec even n = (n = 0) || odd (n - 1)
and odd n = (n <> 0) && even (n - 1)
In F# let ... and ... is prohibited, but you still can write:
let rec low,_,_ = cases.[i]
and _,high,_ = cases.[j]
As #gasche said, let rec ... and ... is mainly used for defining mutually recursive functions/types. I think using a sequence of let is more intuitive and less error-prone hence should be preferred.

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