I have a linear programming model, for the problem of minimum path. This is the model:
/* Min path problem
file: minPath.mod */
set V;
set E within V cross V;
param cost{E};
param S symbolic;
param T symbolic;
var flow{E} integer, >= 0;
minimize min_path: sum{(a,b) in E} cost[a,b] * flow[a,b];
s.t. conservazione{v in V: v != S and v != T}:
sum{(a,b) in E: a == v} flow[a,b] ==
sum{(a,b) in E: b == v} flow[a,b];
s.t. sorgente: sum{(a,b) in E: a == S} flow[a,b] == 1;
s.t. destinazione: sum{(a,b) in E: b == T} flow[a,b] == 1;
display {(a,b) in E} flow[a,b];
data;
set V := A B C D E;
set E := (A,B) (A,C) (B,D) (B,E) (C,D) (D,E);
param S := A;
param T := D;
param cost := [A,B] 2 [A,C] 1 [B,D] 3 [B,E] 1 [C,D] 1 [D,E] 1;
end;
The objective value is 3 for my example, and the minimum path is:
A -> C -> D -> E
For this reason, the vector flow has to be 1 on the edges, that i written here above. By the way, when i display the vector flow with the statement:
display {(a,b) in E} flow[a,b];
the vector is 0 in all the position.
flow[A,B].val = 0
flow[A,C].val = 0
flow[B,D].val = 0
flow[B,E].val = 0
flow[C,D].val = 0
flow[D,E].val = 0
I tried to change the syntax, but i couldn't force glpsol to print the real value.
Am I missing something?
you have to put the statement:
solve;
before the display statement!
Related
I have defined the following module to implement a matrix type:
module MatrixImplementation: MatrixADT.MatrixInterface =
struct
type 'a matrix = {n: int; m: int; c: 'a array array};;
let zeroes n m = {n= n; m= m; c= Array.make_matrix n m 0};;
let identity n =
let m = zeroes n n
in for i = 0 to (n-1) do
m.c.(i).(i) <- 1
done;
(m);
;;
let init n =
let m = zeroes n n
in for i = 0 to (n-1)do
for j = 0 to (n-1) do
m.c.(i).(j) <- (n-1) * i + j;
done;
done;
(m);
;;
(* . . . *)
let rec print_row rl =
match rl with
| [] -> print_string("");
| v::cl -> Format.printf "%2d " v; print_row cl;
;;
let rec print_matrix m =
match Array.to_list m.c with
| [] -> print_string("");
| r::rl ->
print_string "[ ";
print_row (Array.to_list r);
print_string "]\n";
print_matrix {n= ((m.n)-1); m= (m.m); c= Array.of_list rl};
;;
end;;
However when I declare matrices using the module functions (zeroes, identity, init) and try to print them, only part of the lines that compose them are formatted, the first lines are not correctly formatted.
For instance i tried with:
let empty = zeroes 3 5;;
let id = identity 4;;
let mat = init 5;;
print_matrix mat;;
print_matrix empty;;
print_matrix id;;
And I got it as a result:
[ ]
[ ]
[ ]
[ ]
[ ]
[ 0 1 2 3 4 4 5 6 7 8 8 9 10 11 12 12 13 14 15 16 16 17 18 19 20 0 0 0 0 0 ]
[ 0 0 0 0 0 ]
[ 0 0 0 0 0 ]
[ 1 0 0 0 ]
[ 0 1 0 0 ]
[ 0 0 1 0 ]
[ 0 0 0 1 ]
Stripping your code of unnecessary ; and ;; tokens, we can also implement your print_matrix function imperatively. Given that you're using arrays, this simply makes more sense.
Converting from arrays to lists and then back so that you can use recursion and pattern matching is both inefficient and makes your code more difficult to understand.
module MatrixImplementation =
struct
type 'a matrix = {n : int; m : int; c : 'a array array}
let zeroes n m = {n = n; m = m; c = Array.make_matrix n m 0}
let identity n =
let m = zeroes n n in
for i = 0 to (n-1) do
m.c.(i).(i) <- 1
done;
m
let init n =
let m = zeroes n n in
for i = 0 to (n-1) do
for j = 0 to (n-1) do
m.c.(i).(j) <- (n-1) * i + j
done;
done;
m
let print_matrix {n; m; c} =
for i = 0 to n - 1 do
print_string "[";
for j = 0 to m - 1 do
Printf.printf " %d " c.(i).(j)
done;
print_string "]\n"
done
end
utop # MatrixImplementation.(init 3 |> print_matrix);;
[ 0 1 2 ]
[ 2 3 4 ]
[ 4 5 6 ]
- : unit = ()
A note: print_string "" effectively does nothing. If you need to actually do nothing, you can simply return () (unit).
Note also that in print_matrix I've used pattern matching right in the function arguments (let print_matrix {n; m; c} =) to save myself some time. This could have been written as:
let print_matrix m =
for i = 0 to m.n - 1 do
print_string "[";
for j = 0 to m.m - 1 do
Printf.printf " %d " m.c.(i).(j)
done;
print_string "]\n"
done
We could take advantage of the Array.iter function.
let print_matrix {n; m; c} =
Array.(
let print_row arr =
print_string "[";
arr |> iter (Printf.printf " %d ");
print_string "]\n"
in
c |> iter print_row
)
If we consider your original approach, the following will fix it.
let rec print_row rl =
List.iter (Printf.printf " %2d ") rl
Reviewing the code of my module and its incorrect behavior, the problem is not the way the print functions are defined (leaving out the tricks suggested by #Chris in his answer), but the module used: the Format module should be used to make more complex printouts (API reference), the formatting error is probably due to the arguments of the fmt string; for simpler printing, the Printf module and its printf function (same arguments) are more suitable.
Therefore a possible change to the printing functions is as follows:
let rec print_row = function
| [] -> ()
| c::cl -> Printf.printf "%2d " c; print_row cl;
and print_matrix m =
match m.c with
| [] -> ()
| rows ->
for i= 0 to (List.length rows)-1 do
print_string("[ ");
print_row (List.nth rows i);
print_string("]\n");
done;
;;
(In this case the field c (content) of type definition matrix is int list list to avoid conversions)
I'm learning SageMath (uses Python 3)
and playing with the Goldbach conjecture.
I wrote this function (it works!):
def Goldbach(n):
if n % 2 != 0 or n <= 2:
show("No és parell")
else:
for i in srange(n):
if is_prime(i):
for j in srange(n):
if is_prime(j) and i + j == n:
a = [i, j]
show(a)
return
Now I'm trying (no idea) to do the following plot:
Denoting by r(2k) the number of Goldbach partitions of 2k,
the conjecture affirms that r(2k) > 0 if k > 1.
I have to do a graph of points (k, r(2k)), k > 2.
How could I do it?
First of all, let us get some better implementation in Sage
of the routine counting the number r(K) (for K > 2 some even integer) of the solutions for p + q = 2k, p, q prime numbers.
We count both solutions (p, q) and (q, p) when they differ.
def r(K):
if K not in ZZ or K <= 2 or K % 2:
return None
if K == 4:
return 1
count = 0
for p in primes(3, K):
for q in primes(3, K + 1 - p):
if p + q == K:
count += 1
return count
goldbach_points = [(K, r(K)) for K in range(4, 100,2)]
show(points(goldbach_points))
This gives:
Is it possible to have nested if without else statements. I wrote the following useless program to demonstrate nested ifs. How do I fix this so it's correct in terms of syntax. lines 5 and 6 gives errors.
let rec move_helper b sz r = match b with
[] -> r
|(h :: t) ->
if h = 0 then
if h - 1 = sz then h - 1 ::r
if h + 1 = sz then h + 1 ::r
else move_helper t sz r
;;
let move_pos b =
move_helper b 3 r
;;
let g = move_pos [0;8;7;6;5;4;3;2;1]
You can't have if without else unless the result of the expression is of type unit. This isn't the case for your code, so it's not possible.
Here's an example where the result is unit:
let f x =
if x land 1 <> 0 then print_string "1";
if x land 2 <> 0 then print_string "2";
if x land 4 <> 0 then print_string "4"
You must understand that if ... then is an expression like any other. If no else is present, it must be understood as if ... then ... else () and thus has type unit. To emphasize the fact that it is an expression, suppose you have two functions f and g of type, say, int → int. You can write
(if test then f else g) 1
You must also understand that x :: r does not change r at all, it constructs a new list putting x in front of r (the tail of this list is shared with the list r). In your case, the logic is not clear: what is the result when h=0 but the two if fail?
let rec move_helper b sz r = match b with
| [] -> r
| h :: t ->
if h = 0 then
if h - 1 = sz then (h - 1) :: r
else if h + 1 = sz then (h + 1) :: r
else (* What do you want to return here? *)
else move_helper t sz r
When you have a if, always put an else. Because when you don't put an else, Java will not know if the case is true or false.
I have written myPercolation.ml.
open MyUnionFind
module type MyPercolationSig = sig
type percolation
val create_percolation : int -> percolation
val open_site : percolation -> int -> int -> unit
val is_open : percolation -> int -> int -> bool
val is_full : percolation -> int -> int -> bool
val can_percolates : percolation -> bool
end
module MyPercolation : MyPercolationSig = struct
exception IndexOutOfBounds;;
type percolation =
{n : int;
sites: bool array;
union : MyUnionFind.union_find};;
let create_percolation n =
{n = n; sites = Array.make (n*n) false; union = MyUnionFind.create_union (n*n)};;
let open_site p i j =
let {n;_;union} = p
in
if not (is_open p i j) then
begin
sites.(index_of n i j) <- true;
if i - 1 >= 1 && i - 1 <= n && is_open n (i-1) j then
MyUnionFind.union union (index_of n i j) (index_of n (i-1) j)
else if i + 1 >= 1 && i + 1 <= n && is_open n (i+1) j then
MyUnionFind.union union (index_of n i j) (index_of n (i+1) j)
else if j - 1 >= 1 && j - 1 <= n && is_open n i (j-1) then
MyUnionFind.union union (index_of n i j) (index_of n i (j-1))
else if j + 1 >= 1 && j + 1 <= n && is_open n i (j+1) then
MyUnionFind.union union (index_of n i j) (index_of n i (j+1))
end;;
let index_of n i j = n * (i - 1) + j;;
let is_open {n;sites;_} i j =
if i < 1 || i > n || j < 1 || j > n then
raise IndexOutOfBounds
else
sites.(index_of n i j);;
let is_full {n;_;union} i j =
let rec is_connected_top j' =
if j = 0 then false
else
if MyUnionFind.is_connected union (index_of n i j) (index_of n 0 j') then true
else is_connected_top (j'-1)
in is_connected_top n;;
let can_percolates p =
let {n;_;_} = p
in
let rec is_full_bottom j =
if j = 0 then false
else
if is_full p n j then true
else is_full_bottom (j-1)
end
Please ignore the package MyUnionFind package. It is just a homemade implementation for union-find algorithm.
when I try to compile the myPercolation.ml, I got such an error:
$ ocamlc -c myPercolation.ml
File "myPercolation.ml", line 25, characters 11-12:
Error: Syntax error: '}' expected
File "myPercolation.ml", line 25, characters 8-9:
Error: This '{' might be unmatched
I think the error is talking about let {n;_;union} = p in function of let open_site p i j.
I have read through that line and all code many times, but I still don't see any mismatched {} in that line.
can anyone help please?
Another possible error: {n;_;_} should be {n;_} Only 1 underscore is necessary. Think of it like the _ wildcard in a match statement.
The expression let {n; _; union} = p is not well formed OCaml. I think what you want is let {n; union} = p. The way to handle fields you don't care about in a record pattern is not to mention them.
Update:
As rgrinberg points out, a much better way to describe the problem is that the _ has to appear as the last field. That's why the compiler is expecting to see } afterward. It might be good style to include the _ as an indicator that you're purposely matching only a subset of the fields of the record. You can, in fact, turn on a compiler option that checks for this.
Update 2:
The warning for incomplete record patterns is warning number 9, and also is associated with the letter R. Here's how to use R:
$ ocaml -w +R
OCaml version 4.00.0
# type r = { a: int; b: char };;
type r = { a : int; b : char; }
# let {a} = {a=3; b='x'} in a;;
Warning 9: the following labels are not bound in this record pattern:
b
Either bind these labels explicitly or add '; _' to the pattern.
- : int = 3
The command-line syntax is the same for the compiler.
I have 3 angles a b c
a=315
b=20
c=45
ok so would like to know giving all three if b is in between a and c
i have the long way of doing this adding and subtracting that's seems to work. I would just like to get something smaller and maybe more efficient.
thanks
EDIT
Here is a picture what i am trying to say.
Ok I have angle L(currently 0) i add 45(or any angle) and subtract 45(or any angle) to get a and b (my view angle).
Now i need to know if the green dot is between a and b
(g> a || g > 0) && (g < b)
so in this picture only the top green dot will be true..
Sorry if I am not making my self clear my first language is not English
I had a similar problem. I got it. All the calculations are in degrees.
I needed to calculate id a gps location is inside a rectangle.
Or, I needed to see if an angle x is between angle check+r and angle check-r.
check-r<x<check+r.
If you need a<x<b, find the angle check in the middle of a and b and then the distance (r) of check from a or b.
The method normalize, changes the angles from -infinity...infinity to -180...180.
The method check, takes the arguments
x: the angle that we need to see if it is between the angles check-r and check+r.
check: the angle to check with.
r: the radius around angle check.
private static double normalize(double x) {
x = x % 360;
if (x>=180) {
return x-360;
}
if (x<-180) {
return x+360;
}
return x;
}
public static boolean check(double x, double check, double r) {
x = x - check;
x = normalize(x);
return x<r && x>-r;
}
1st off, every angle is between 2 other angles, what you're really asking is:
For given angles: a, b, and g, is g outside the reflex angle between a and b?
You can just go ahead and define a as the leftmost angle and b as the rightmost angle or you can solve for that, for example if either of these statements are true a is your leftmost angle:
a ≤ b ∧ b - a ≤ π
a > b ∧ a - b ≥ π
For simplicity let's say that your leftmost angle is l and your rightmost angle is r and you're trying to find if g is between them.
The problem here is the seem. There are essentially 3 positive cases that we're looking for:
l ≤ g ≤ r
l ≤ g ∧ r < l
g ≤ r ∧ r < l
If you're just defining a to be leftmost and b to be rightmost you're done here and your condition will look like:
a <= g && g <= b ||
a <= g && b < a ||
g <= b && b < a
If however you calculated the l and r you'll notice there is an optimization opportunity here in doing both processes at once. Your function will look like:
if(a <= b) {
if(b - a <= PI) {
return a <= g && g <= b;
} else {
return b <= g || g <= a;
}
} else {
if(a - b <= PI) {
return b <= g && g <= a;
} else {
return a <= g || g <= b;
}
}
Or if you need it you could expand into this nightmare condition:
a <= b ?
(b - a <= PI && a <= g && g <= b) || (b - a > PI && (b <= g || g <= a)) :
(a - b <= PI && b <= g && g <= a) || (a - b > PI && (a <= g || g <= b))
Note that all this math presumes that your input is in radians and in the range [0 : 2π].
Live Example
I personally had the same problem recently and found counterexamples for all the answers posted yet, so I will share my own approach.
Let a be the start angle and b the end angle and we are checking whether c is between them clockwise, that means when you go from a to b you must pass c. The approach of checking whether c is in the range from a to b gives you false positives when b is greater than a numerically. For example:
a=80°, b=320° and c=150°: a <= c <= b which would mean that c is between a and b.
But it isn't.
The approach that's working is to subtract 360 from b if it is greater than a and also subtract 360 from c if c is greater than a. Then check whether a <= c <= b.
In Java:
public static boolean isBetween(double c, double a, double b) {
if (b > a) b -= 360;
if (c > a) c -= 360;
return a <= c && c <= b;
}
This assumes that a, b and c are in range 0 to 360.
Some example:
isBetween(150, 80, 320) => false
isBetween(30, 80, 320) => true
isBetween(340, 80, 320) => true
isBetween(140, 0, 160) => true
isBetween(180, 0, 160) => false
There is an issue with the suggested solutions when handling negative angles (e.g. from=30 to=-29)
The suggested (kotlin) fix should be:
fun isBetween(from:Float,to:Float,check:Float,inclusive:Boolean = true):Boolean{
var a1 = to - from
a1 = (a1 + 180f).mod(360f) - 180f
if(a1<0f) a1+=360f
var a2 = check - from
a2 = (a2 + 180f).mod(360f) - 180f
if(a2<0f) a2+=360f
val between = if(inclusive) a2<=a1 else a2<a1 && a2>0f
println("$from->$to, $check, $between ($a1,$a2)")
return between }
Assuming a > c, you would actually use:
( b < a ) && ( b > c )
This is the same as checking if a value is between a lower and upper bound. Them being angles makes no difference, unless you want to take into account the fact that as you go round a circle, an angle of 405 is the same as an angle of 45. In which case you can just use a % 360 to get the angle betweeen 0 and 360.