I would like to "complete" an array in MiniZinc as follows.
Say I have an array a=[A,H,J,D,C] which is associated to certain integers between 1 and 6 (injectively), e.g. n=[4,2,1,6,5] so here the 3 is missing. I would like to output the array a corresponding to an increasing ordering of n and filling in the missing element (3) by a fixed content, say XXX. In the example it would be
[A,H,XXX,J,D,C]
The way I considered for now is to create a new array of 6 variables, and checking for all of them if they are in a and assigning them at the right position, and putting XXX if the position do not appear. This makes my program terrible slower (about five times slower!).
Are there more efficient options to do so?
You could move the selection logic to functions solely called from within the output section:
int: not_found = 0;
array[1..5] of string: a = ["A", "H", "J", "D", "C"];
array[1..5] of var int: n = [4, 2, 1, 6, 5];
% inspired by https://stackoverflow.com/a/44846993/1911064
% This assumes that the value does not occur more than once.
% Constant not_found has to be 0!
function int: index_of(int: item, array[int] of var int: items) =
sum( [ if item == fix(items[i]) then i else not_found endif | i in index_set(items) ]
);
function string: str_select(array[int] of string: a, array[int] of var int: n, int: i) =
let {
int: index = index_of(i, n);
int: gaps = sum([ index_of(j, n) == not_found | j in 1..i-1 ]);
} in
" " ++ if index != not_found then show(a[i - gaps]) else "xxx" endif;
output [ str_select(a, n, j) | j in 1..6];
This approach does not lead to additional constraints. Therefore, it should not cause extra solution time.
Related
Currently i'm trying
I have a function to calculate the inverse sum of a number
let inverseSum n =
let rec sI n acc =
match n with
| 1 -> acc
| _ -> sI (n - 1) ((1.0 /. float n) +. acc)
in sI n 1.0;;
For example, inverseSum 2 -> 1/2 + 1 = 3/2 = 1.5
I try to test the function with 2 and 5, it's okay:
inverseSum 2;;
inverseSum 5;;
inverseSum 2;;
- : float = 1.5
inverseSum 5;;
- : float = 2.28333333333333321
For the moment, no problem.
After that, I initialize a list which contains all numbers between 1 and 10000 ([1;…;10000])
let initList = List.init 10000 (fun n -> n + 1);;
no problem.
I code a function so that an element of the list becomes the inverse sum of the element
(e.g. [1;2;3] -> [inverseSum 1; inverseSum 2; inverseSum 3])
let rec invSumLst lst =
match lst with
| [] -> []
| h::t -> (inverseSum h) :: invSumLst t;;
and I use it on the list initList:
let invInit = invSumLst initList;;
So far so good, but I start to have doubts from this stage:
I select the elements of invList strictly inferior to 5.0
let listLess5 = List.filter (fun n -> n < 5.0) invInit;;
And I realize the sum of these elements using fold_left:
let foldLess5 = List.fold_left (+.) 0.0 listLess5;;
I redo the last two steps with floats greater than or equal to 5.0
let moreEg5 = List.filter (fun n -> n >= 5.0) invInit;;
let foldMore5 = List.fold_left (+.) 0.0 moreEg5;;
Finally, I sum all the numbers of the list:
let foldInvInit = List.fold_left (+.) 0.0 invInit;;
but at the end when I try to calculate the absolute error between the numbers less than 5, those greater than 5 and all the elements of the list, the result is surprising:
Float.abs ((foldLess5 +. foldMore5) -. foldInvInit);;
Printf.printf "%f\n" (Float.abs ((foldLess5 +. foldMore5) -. foldInvInit));;
Printf.printf "%b\n" ((foldLess5+.foldMore5) = foldInvInit);;
returns:
let foldMore5 = List.fold_left (+.) 0.0 moreEg5;;
val foldMore5 : float = 87553.6762998474733
let foldInvInit = List.fold_left (+.) 0.0 invInit;;
val foldInvInit : float = 87885.8479664799379
Float.abs ((foldLess5 +. foldMore5) -. foldInvInit);;
- : float = 1.45519152283668518e-11
Printf.printf "%f\n" (Float.abs ((foldLess5 +. foldMore5) -. foldInvInit));;
0.000000
- : unit = ()
Printf.printf "%b\n" ((foldLess5+.foldMore5) = foldInvInit);;
false
- : unit = ()
it's probably a rounding problem, but I would like to know where exactly the error comes from?
Because here I am using an interpreter, so I see the error "1.45519152283668518e-11"
But if I used a compiler like ocamlpro, I would just get 0.000000 and false on the terminal and I wouldn't understand anything.
So I would just like to know if the problem comes from one of the functions of the code, or from a rounding made by the Printf.printf function which wrote the result with a "non scientific" notation.
OCaml is showing you the actual results of the operations you performed. The difference between the two sums is caused by the finite precision of floating values.
When adding up a list of large-ish numbers, by the time you reach the end of the list the number is large enough that the lowest-order bits of the new values simply can't be represented. But when adding a list of small-ish numbers, fewer bits are lost.
A system that shows foldLess5 +. foldMore5 as equal to foldInvInit is most likely lying to you for your convenience.
I have a system of n equations and n unknown variables under symbol sum. I want to create a loop to solve this system of equations when inputting n.
y := s -> 1/6cos(3s);
A := (k, s) -> piecewise(k <> 0, 1/2exp(ksI)/abs(k), k = 0, ln(2)exp(s0I) - sin(s));
s := (j, n) -> 2jPi/(2*n + 1);
n := 1;
for j from -n to n do
eqn[j] := sum((A(k, s(j, n))) . (a[k]), k = -n .. n) = y(s(j, n));
end do;
eqs := seq(eqn[i], i = -n .. n);
solve({eqs}, {a[i]});
enter image description here
Please help me out!
I added some missing multiplication symbols to your plaintext code, to reproduce it.
restart;
y:=s->1/6*cos(3*s):
A:=(k,s)->piecewise(k<>0,1/2*exp(k*s*I)/abs(k),
k=0,ln(2)*exp(s*I*0)-sin(s)):
s:=(j,n)->2*j*Pi/(2*n+1):
n:=1:
for j from -n to n do
eqn[j]:=add((A(k,s(j,n)))*a[k],k=-n..n)=y(s(j,n));
end do:
eqs:=seq(eqn[i],i=-n..n);
(-1/4+1/4*I*3^(1/2))*a[-1]+(ln(2)+1/2*3^(1/2))*a[0]+(-1/4-1/4*I*3^(1/2))*a[1] = 1/6,
1/2*a[-1]+ln(2)*a[0]+1/2*a[1] = 1/6,
(-1/4-1/4*I*3^(1/2))*a[-1]+(ln(2)-1/2*3^(1/2))*a[0]+(-1/4+1/4*I*3^(1/2))*a[1] = 1/6
You can pass the set of names (for which to solve) as an optional argument. But that has to contain the actual names, and not just the abstract placeholder a[i] as you tried it.
solve({eqs},{seq(a[i],i=-n..n)});
{a[-1] = 1/6*I/ln(2),
a[0] = 1/6/ln(2),
a[1] = -1/6*I/ln(2)}
You could also omit the indeterminate names here, as optional argument to solve (since you wish to solve for all of them, and no other names are present).
solve({eqs});
{a[-1] = 1/6*I/ln(2),
a[0] = 1/6/ln(2),
a[1] = -1/6*I/ln(2)}
For n:=3 and n:=4 it helps solve to get a result quicker here if exp calls are turned into trig calls. Ie,
solve(evalc({eqs}),{seq(a[i],i=-n..n)});
If n is higher than 4 you might have to wait long for an exact (symbolic) result. But even at n:=10 a floating-point result was fast for me. That is, calling fsolve instead of solve.
fsolve({eqs},{seq(a[i],i=-n..n)});
But even that might be unnecessary, as it seems that the following is a solution for n>=3. Here all the variables are set to zero, except a[-3] and a[3] which are both set to 1/2.
cand:={seq(a[i]=0,i=-n..-4),seq(a[i]=0,i=-2..2),
seq(a[i]=0,i=4..n),seq(a[i]=1/2,i=[-3,3])}:
simplify(eval((rhs-lhs)~({eqs}),cand));
{0}
I am new to SML. I am trying to create a function that outputs a list of all prime numbers which are smaller than or equal to a given integer n > 2 using the Sieve of Eratosthenes. I am running into a problem however where the output is only showing as [1]. I would like to be able to have an input such as 5 and get [1,3,5] as a result.
This is my code so far, I am very new so I know it is most likely not written correctly.
fun createList(ending) =
let
fun createListX(start, ending) =
if start = ending then []
else start :: createListX(start + 1, ending)
in
createListX(1, ending + 1)
end;
fun removeMult ([], n) = []
| removeMult (x::xs, n) =
if x mod n = 0 then
removeMult(xs, n)
else
x :: removeMult(xs, n);
fun sieve([], primes) = primes
| sieve(n::ns, primes) = sieve(removeMult(ns, n), n :: primes);
fun dosieve(n) = sieve(createList(n-1), []);
Your removeMult function works nicely.
Your sieve function works perfectly too. Too perfectly.
Consider what happens when you call dosieve(10) for instance:
dosieve(10)
sieve(createList(9), [])
sieve([1,2,3,4,5,6,7,8,9], [])
From there:
sieve(removeMult([2, 3, 4, 5, 6, 7, 8, 9], 1), 1 :: [])
sieve([], [1])
[1]
Oops. You removed all multiples of 1, but of course they're all multiples of 1.
Perhaps something like:
fun sieve([], primes) = primes
| sieve(1::ns, primes) = sieve(ns, 1 :: primes)
| sieve(n::ns, primes) = sieve(removeMult(ns, n), n :: primes);
I've got a function that needs to work out the minimum change required to break down a certain amount of money.
A user would enter a value of money and the possible denominations, and it would output how many of each denomination would be needed to make up the money eg.
> coinChange 34 [1, 5, 10, 25, 50, 100]
[4,1,0,1,0,0]
Here is what I have so far:
coinChange :: Integer -> [Integer] -> [Integer]
coinChange _ [] = []
coinChange 0 xs = []
coinChange v xs = do
dropWhile (>v) (reverse xs)
createNewList :: Integer -> [Integer]
createNewList 0 = []
createNewList x = 0 : createNewList (x - 1)
I have a separate function called createNewList which I want to call to set up a new list of 0s at the beginning.
However, because I plan on making coinChange a recursive function when working out how much money is left, if I just call makeNewList in coinChange, at the moment it would reset the list because it would be called every recursion.
So my question is, is there a way to make a function call another function only once, before proceeding with the recursion.
Hope I've made it clear, thanks
In general you are allowed to create local helper functions in Haskell. Maybe this example will inspire you:
prepareData x = if x < 0 then -x else x
mainFunction n =
let recursiveLocalFunction x =
if x < 2 then 1 else x * recursiveLocalFunction (x - 1)
in recursiveLocalFunction (prepareData n)
I would like to define a list using a for loop and I need to do it using a function of the n-iterate.
I have:
Initialization
In[176]: Subscript[y, 0] = {1, 2, 3}
Out[180]: {1,2,3}
The function:
In[181]: F[n_] := For[l = 1, l++, l <= 3, Subscript[y, n + 1][[l]] :=Subscript[y, n][[l]]+ n]
I call the function
F[0]
and I get:
In[183]: Subscript[y, 1]
Out[183]: Subscript[0, 1]
I should have {1,2,3}.
Anyone know why it isn't working as it should?
I have troubles recreating your error, problem.
I understand you want to add n to your vector, where n is the number of the subscript.
Here's another way to have a go at your question, avoiding the loop and the subscripts:
Clear#y;
y[0] = {1, 2, 3};
y[n_Integer] : =y[n - 1] + n
(as Plus is Listable, you can just add n to the vector, avoiding the For)
and then call it using, e.g.
y[0]
{1,2,3}
or
y[5]
{16,17,18}
Alternatively, using memoization, you could define y as follows:
y[n_Integer] := y[n] = y[n - 1] + n
This will then store already calculated values (check ?y after executing e.g. y[5]). Don't forget to Clear y, if y changes.
Obviously, for a function as this one, you might want to consider:
y[n_Integer] := y[0] + Total[Range[n]]