I am solving this homework of clean programming language;
The problem is we have a number of five digits and we want to check whether it's an odd palindrome or not.
I am stuck at the stage of dividing the number to five separate digits and perform a comparison with the original number, for the palindrome check. With Clean I can't loop over the number and check if it remains the same from the both sides, so I am looking for an alternative solution (Some mathematical operations).
Code block:
isOddPalindrome :: Int -> Bool
isOddPalindrome a
| isFive a <> 5 = abort("The number should be exactly five digits...")
| (/*==> Here should be the palindrome check <==*/) && (a rem 2 <> 0) = True
| otherwise = False
isFive :: Int -> Int
isFive n
| n / 10 == 0 = 1
= 1 + isFive(n / 10)
My idea is to take the number, append it's digits one by one to an empty list, then perform the reverse method on the list and check if it's the same number or not (Palindrome)
Your answer above doesn't have a stopping condition so it will result in stack overflow.
You could try this
numToList :: Int -> [Int]
numToList n
| n < 10 = [n]
= numToList (n/10) ++ [n rem 10]
Start = numToList 12345
and then like you mentioned in the answer, you can reverse it with the 'reverse' function and check if they are equal.
After hours of trying to figure out how to recursively add the digits of our number to an empty list, I did the following:
sepDigits :: Int [Int] -> [Int]
sepDigits n x = sepDigits (n/10) [n rem 10 : x]
Now I can easily check whether the reverse is equal to the initial list :) then the number is palindrome.
I am trying to code a Periodic Vehicle Routing Problem with some inventory constraints in AMPL. I would like to add the subtour constraints dynamically. In order to do this i was inspired by this formulation for a TSP:
https://groups.google.com/d/msg/ampl/mVsFg4mAI1c/ZdfRHHRijfUJ
However, I can not get it to eliminate subtours in my model. I used the following in my model file.
param T; # Number of time-periods
param V; # Number of vehicles
param F; # Number of fuel types
set P ordered; # Number of gas stations
param hpos {P} >= 0;
param vpos {P} >= 0;
set PAIRS := {p in P, j in P};
param dist {(p,j) in PAIRS}
:= sqrt((hpos[j]-hpos[p])**2 + (vpos[j]-vpos[p])**2);
# A binary variable to determine if an arc is traversed.
var H{(p,j) in PAIRS, v in 1..V, t in 1..T} binary;
# A binary variable to determine if a delivery of fuel is made to a station in a given time period.
var StationUsed{p in P, f in 1..F, v in 1..V, t in 1..T} binary;
minimize TransportationCost:
sum {(p,j) in PAIRS} sum {v in 1..V, t in 1..T} dist[p,j] * H[p,j,v,t];
param nSubtours >= 0 integer;
set SUB {1..nSubtours} within P;
subject to Subtour_Elimination {k in 1..nSubtours, m in SUB[k], v in 1..V, t in 1..T, f in 1..F}:
sum {p in SUB[k], j in P diff SUB[k]}
if (p,j) in PAIRS then H[p,j,v,t] else H[j,p,v,t] >=2 * StationUsed[m,f,v,t] ;
I added the StationUsed variable, as my problem unlike TSP does not have to visit all nodes in every timeperiod. H is my binary decision variable declaring if vehicle travels the arc (p,j) in a time period.
Then I used a formulation similar to the TSP in my run file:
set NEWSUB;
set EXTEND;
let nSubtours := 0;
repeat {
solve;
let NEWSUB := {};
let EXTEND := {member(ceil(Uniform(0,card(P))),P)};
repeat {
let NEWSUB := NEWSUB union EXTEND;
let EXTEND := {j in P diff NEWSUB: exists {p in NEWSUB, v in 1..V, t in 1..T}
((p,j) in PAIRS and H[p,j,v,t] = 1 or (j,p) in PAIRS and H[j,p,v,t] = 1)};
} until card(EXTEND) = 0;
if card(NEWSUB) < card(P) then {
let nSubtours := nSubtours + 1;
let SUB[nSubtours] := NEWSUB;
display SUB;
} else break;
};
# Display the routes
display {t in 1..T, v in 1..V}: {(p,j) in PAIRS} H[p,j,v,t];
I am not sure if the above is applicable to my problem with multiple vehicles and multiple time periods. I have tried defining v and t in let EXTEND, at it is needed to use H, but I am not sure if this is a correct method. My models runs, when formulated as above, however it does not eliminate the subtours. Do you guys have any suggestions in this regard?
ADDED QUESTION:
I found some inspiration in this model formulated in SAS/OR:
(A bit extensive to read and not necessary for my questions)
http://support.sas.com/documentation/cdl/en/ormpex/67518/HTML/default/viewer.htm#ormpex_ex23_sect009.htm
It eliminates subtours dynamically over d days and I figured it could be translated to my problem with multiple vehicles and multiple periods (days).
To specify my problem a little. A node can only be visited by one vehicle once within a time period. All nodes does not have to be visited in every time period, which is a major difference from the TSP formulation, where all nodes are in the cycle.
I tried with the following approach:
The constraint in the model file is the same as before.
set P ordered; # Number of nodes
set PAIRS := {p in P, j in P: ord(p) != ord(j)};
param nSubtours >= 0 integer;
param iter >= 0 integer;
set SUB {1..nSubtours} within P;
subject to Subtour_Elimination {s in 1..nSubtours, k in SUB[s], f in F, v in V, t in T}:
sum {p in SUB[s], j in P diff SUB[s]}
if (p,j) in PAIRS then H[p,j,v,t] else H[j,p,v,t] >= 2 * StationUsed[k,f,v,t];
My run file looks like this:
let nSubtours := 0;
let iter := 0;
param num_components {V, T};
set P_TEMP;
set PAIRS_SOL {1..iter, V, T} within PAIRS;
param component_id {P_TEMP};
set COMPONENT_IDS;
set COMPONENT {COMPONENT_IDS};
param cp;
param cj;
# loop until each day and each vehicles support graph is connected
repeat {
let iter := iter + 1;
solve;
# Find connected components for each day
for {v in V, t in T} {
let P_TEMP := {p in P: exists {f in F} StationUsed[p,f,v,t] > 0.5};
let PAIRS_SOL[iter, v, t] := {(p,j) in PAIRS: H[p, j, v, t] > 0.5};
# Set each node to its own component
let COMPONENT_IDS := P_TEMP;
let num_components[v, t] := card(P_TEMP);
for {p in P_TEMP} {
let component_id[p] := p;
let COMPONENT[p] := {p};
};
# If p and j are in different components, merge the two component
for {(p,j) in PAIRS_SOL[iter, v, t]} {
let cp := component_id[p];
let cj := component_id[j];
if cp != cj then {
# update smaller component
if card(COMPONENT[cp]) < card(COMPONENT[cj]) then {
for {k in COMPONENT[cp]} let component_id[k] := cj;
let COMPONENT[cj] := COMPONENT[cj] union COMPONENT[cp];
let COMPONENT_IDS := COMPONENT_IDS diff {cp};
} else {
for {k in COMPONENT[cj]} let component_id[k] := cp;
let COMPONENT[cp] := COMPONENT[cp] union COMPONENT[cj];
let COMPONENT_IDS := COMPONENT_IDS diff {cj};
};
};
};
let num_components[v, t] := card(COMPONENT_IDS);
display num_components[v, t];
# create subtour from each component not containing depot node
for {k in COMPONENT_IDS: 1 not in COMPONENT[k]} { . #***
let nSubtours := nSubtours + 1;
let SUB[nSubtours] := COMPONENT[k];
display SUB[nSubtours];
};
};
display num_components;
} until (forall {v in V, t in T} num_components[v,t] = 1);
I get a lot of "invalid subscript discarded", when running the model:
Error at _cmdno 43 executing "if" command
(file amplin, line 229, offset 5372):
error processing set COMPONENT:
invalid subscript COMPONENT[4] discarded.
Error at _cmdno 63 executing "for" command
(file amplin, line 245, offset 5951):
error processing set COMPONENT:
invalid subscript COMPONENT[3] discarded.
(...)
Bailing out after 10 warnings.
I think the script is doing what I am looking for, but it stops, when it has discarded 10 invalid subscripts.
When trying to debug I tested the second for loop.
for {p in P_TEMP} {
let component_id[p] := p;
let COMPONENT[p] := {p};
display component_id[p];
display COMPONENT[p];
};
This is displaying correct, but not before a few errors with "invalid subscript discarded". It seems that p runs through some p not in P_TEMP. For example when P_TEMP is a set consisting of nodes "1 3 4 5", then I get "invalid subscript discarded" for component_id[2] and COMPONENT[2]. My guess is that something similar happens again later on in the IF-ELSE statement.
How do I avoid this?
Thank you,
Kristian
(previous answer text deleted because I misunderstood the implementation)
I'm not sure if this fully explains your issue, but I think there are a couple of problems with how you're identifying subtours.
repeat {
solve;
let NEWSUB := {};
let EXTEND := {member(ceil(Uniform(0,card(P))),P)};
repeat {
let NEWSUB := NEWSUB union EXTEND;
let EXTEND := {j in P diff NEWSUB: exists {p in NEWSUB, v in 1..V, t in 1..T}
((p,j) in PAIRS and H[p,j,v,t] = 1 or (j,p) in PAIRS and H[j,p,v,t] = 1)};
} until card(EXTEND) = 0;
if card(NEWSUB) < card(P) then {
let nSubtours := nSubtours + 1;
let SUB[nSubtours] := NEWSUB;
display SUB;
} else break;
};
What this does:
solves the problem
sets NEWSUB as empty
randomly picks one node from P as the starting point for EXTEND and adds this to NEWSUB
looks for any nodes not currently in NEWSUB which are connected to a node within NEWSUB by any vehicle journey on any day, and adds them to NEWSUB
repeats this process until there are no more to add (i.e. either NEWSUB equals P, the entire set of nodes, or until there are no journeys between NEWSUB and non-NEWSUB notedes)
checks whether NEWSUB is smaller than P (in which case it identifies NEWSUB as a new subtour, appends it to SUB, and goes back to the start).
if NEWSUB has the same size as P (i.e. is equal to P) then it stops.
This should work for a single-vehicle problem with only a single day, but I don't think it's going to work for your problem. There are two reasons for this:
If your solution has different subtours on different days, it may not recognise them as subtours.
For example, consider a single-vehicle problem with two days, where your cities are A, B, C, D, E, F.
Suppose that the day 1 solution selects AB, BC, CD, DE, EF, FA, and the day 2 solution selects AB, BC, CA, DE, EF, FD. Day 1 has no subtour, but day 2 has two length-3 subtours, so this should not be a legal solution.
However, your implementation won't identify this. No matter which node you select as the starting point for NEWSUB, the day-1 routes connect it to all other nodes, so you end up with card(NEWSUB) = card(P). It doesn't notice that Day 2 has a subtour so it will accept this solution.
I'm not sure whether your problem allows for multiple vehicles to visit the same node on the same day. If it does, then you're going to run into the same sort of problem there, where a subtour for vehicle 1 isn't identified because vehicle 2 links that subtour to the rest of P.
Some of this could be fixed by doing subtour checking separately for each day and for each vehicle. But for the problem as you've described it, there's another issue...
Once the program has identified a closed route (i.e. a set of nodes that are all linked to one another, and not to any other nodes) then it needs to figure out whether this subtour should be prohibited.
For the basic TSP, this is straightforward. We have one vehicle that needs to visit every node - hence, if the cardinality of the subtour is smaller than the cardinality of all nodes, then we have an illegal subtour. This is handled by if card(NEWSUB) < card(P).
However, you state:
my problem unlike TSP does not have to visit all nodes in every timeperiod
Suppose Vehicle 1 travels A-B-C-A and Vehicle 2 travels D-E-F-D. In this case, these routes will look like illegal subtours because ABC and DEF are each smaller than ABCDEF and there are no routes that link them. If you use if card(NEWSUB) < card(P) as your criterion for a subloop that should be forbidden, you'll end up forcing every vehicle to visit all nodes, which is fine for basic TSP but not what you want here.
This one can be fixed by identifying how many nodes vehicle v visits on day t, and then comparing the length of the subtour to that total: e.g. if there are 10 cities total, vehicle 1 only visits 6 of them on day 1, and a "subtour" for vehicle 1 visits 6 cities, then that's fine, but if it visits 8 and has a subtour that visits 6, that implies it's travelling two disjoint subloops, which is bad.
One trap to watch out for here:
Suppose Day 1 requires vehicle 1 to visit ABCDEF. If we get a "solution" that has vehicle 1 ABCA and DEFD on one day, we might identify ABCA as a subtour that should be prevented.
However, if Day 2 has different requirements, it might be that having vehicle 1 travel ABCA (and no other nodes) is a legitimate solution for day 2. In this case, you don't want to forbid it on day 2 just because it was part of an illegal solution for day 1.
Similarly, you might have a "subroute" that is a legal solution for one vehicle but illegal for another.
To avoid this, you might need to maintain a different list of prohibited subroutes for each vehicle x day, instead of using one list for all. Unfortunately this is going to make your implementation a bit more complex.
1.we have cycling table
2.a man must be sit next to a woman and a woman next to a man
3. guests must share hobbies (at least one common between their hobbies)
4. there are couples of furious guests. they must not sit near to next to each other
5. nobody of list o furious guests must sit at start(seat 1) or end (seat N)
-pR is the number of furious couples
my model:
int :N;
set of int: GUESTS = 1..N;
set of int: POSITIONS = 1..N;
array[GUESTS] of 1..2 : gender;
array[GUESTS] of set of int: hobbies;
enum PAIR = {first,second};
int : pR;
set of int: LIST = 1..pR;
array[LIST,PAIR] of GUESTS : furious;
array[POSITIONS] of var GUESTS : guest_at;
array[POSITIONS] of var 1..2: table_gender;
constraint forall(i in 1..length(table_gender)-1)(
table_gender[i]!=table_gender[i+1]
/\
table_gender[1]!=table_gender[length(table_gender)]
)
;
include "alldifferent.mzn";
constraint alldifferent(guest_at);
constraint forall(i in 2..N-1)(card(hobbies[guest_at[i+1]] intersect hobbies[guest_at[i]]) >0);
constraint card(hobbies[guest_at[N]] intersect hobbies[guest_at[1]]) >0;
constraint forall(i in 2..N-1,l in LIST, p in PAIR)(if guest_at[i]=furious[i,first] then guest_at[i+1]!=furious[i,second] /\ guest_at[i-1]!=furious[i,second] else true endif);
constraint forall(l in LIST, p in PAIR)(guest_at[1]!=furious[l,p] /\ guest_at[N]!=furious[l,p]);
solve satisfy;
output
["guest_at = \(guest_at);"]
++ ["\ntable_gender = \(table_gender); \n" ]
++ ["Furious Placement\n"]
++ [show_int(4,furious[i,j]) | i in LIST, j in PAIR] ++["\n"]
++ [if fix(guest_at[p]) = furious[i,j] then show_int(4,p) else "" endif | i in LIST, j in PAIR, p in POSITIONS]
;
my model's bugs:
C:/Users/�������/Documents/������/����������/Gala/gala.mzn:36:
in call 'forall'
in array comprehension expression
with i = 4
with l = 3
with p = 1
in if-then-else expression
in binary '=' operator expression
in array access
WARNING: Further warnings have been suppressed.
This constraint, where there errors are referring to, contains a couple of strange things:
constraint
forall(i in 2..N-1,l in LIST, p in PAIR) (
if guest_at[i]=furious[i,first] then
guest_at[i+1]!=furious[i,second] /\
guest_at[i-1]!=furious[i,second]
else
true
endif
);
1) The second and third loop parameters l in List and p in PAIR is never used, so they are meaningless.
2) The main reason for the warning is that the furious matrix is just two rows, but in the loop variable i goes from 2 to 16. The error (array access out of bounds) indicates that when i is larger than 2 it's out of bound of the furious matrix.
New MiniZinc user here ... I'm having a problem understanding the syntax of the counting constraint:
predicate exactly(int: n, array[int] of var int: x, int: v)
"Requires exactly n variables in x to take the value v."
I want to make sure each column in my 10r x 30c array has at least one each of 1,2 and 3, with the remaining 7 rows equal to zero.
If i declare my array as
array[1..10,1..30] of var 0..3: s;
how can I use predicate exactly to populate it as I need? Thanks!
Well, the "exactly" constraint is not so useful here since you want at least one occurrence of 1, 2, and 3. It's better to use for example the count function:
include "globals.mzn";
array[1..10,1..30] of var 0..3: s;
solve satisfy;
constraint
forall(j in 1..30) (
forall(c in 1..3) (
count([s[i,j] | i in 1..10],c) >= 1
)
)
;
output [
if j = 1 then "\n" else " " endif ++
show(s[i,j])
| i in 1..10, j in 1..30
];
You don't have do to anything about 0 since the domain is 0..3 and all values that are not 1, 2, or 3 must be 0.
Another constraint is "at_least", see https://www.minizinc.org/2.0/doc-lib/doc-globals-counting.html .
If you don't have read the MiniZinc tutorial (https://www.minizinc.org/downloads/doc-latest/minizinc-tute.pdf), I strongly advice you to. The tutorial teaches you how to think Constraint Programming and - of course - MiniZinc.