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.
Related
As is shown below, the function "search" is recursive. I want to transform it into a form without recursion, but have trouble in dealing with the popping and pushing of the variable "path". What's its non-recursive form and the method of converting it?
path current search path (acts like a stack)
node current node (last node in current path)
g the cost to reach current node
f estimated cost of the cheapest path (root..node..goal)
h(node) estimated cost of the cheapest path (node..goal)
cost(node, succ) step cost function
is_goal(node) goal test
successors(node) node expanding function, expand nodes ordered by g + h(node)
ida_star(root) return either NOT_FOUND or a pair with the best path and its cost
procedure ida_star(root)
bound := h(root)
path := [root]
loop
t := search(path, 0, bound)
if t = FOUND then return (path, bound)
if t = ∞ then return NOT_FOUND
bound := t
end loop
end procedure
function search(path, g, bound)
node := path.last
f := g + h(node)
if f > bound then return f
if is_goal(node) then return FOUND
min := ∞
for succ in successors(node) do
if succ not in path then
path.push(succ)
t := search(path, g + cost(node, succ), bound)
if t = FOUND then return FOUND
if t < min then min := t
path.pop()
end if
end for
return min
end function
Iterative deepening A* pseudocode - Wikipedia
Having a hard time trying to figure out how can I count the number of apps or words on a slice using the go-lang code I made.
Hoping someone could help me figure out how to count the number of occurence?
https://play.golang.org/p/KvgI-lCz_c6
package main
import (
"fmt"
)
func main() {
apps := []string{"one", "two", "three", "one", "four"}
fmt.Println("apps:", apps)
o := CountOccurence(apps)
fmt.Println("=== o: ", o)
}
func CountOccurence(apps []string) map[string]int {
dict := make(map[string]int)
for k, v := range apps {
fmt.Println(k, v)
dict[v] = k
}
// fmt.Println("=== dict: ", dict)
return dict
}
Outputs the following
apps: [one two three one four]
0 one
1 two
2 three
3 one
4 four
=== o: map[four:4 one:3 three:2 two:1]
PS: go strings.Count only counts a string, not a []string.
What you currently do is you gather the different elements and you assign their index to them. If a word occurs multiple times, the highest index will be assigned to it.
As you stated, you want to count the words. So instead of the index, assign 1 for new words (first occurrence), and if it's already in the map, increment its value by 1.
Since you can index a map with a non-existing key, in which case the result is the zero value of the value type of the map, which is 0 for int, it will tell you it was found 0 times (so far), so you don't even have to check if a key is already in there, just go ahead and increment it:
dict[v]++
So CountOccurrences() may look like this:
func CountOccurence(apps []string) map[string]int {
dict := make(map[string]int)
for _, v := range apps {
fmt.Println(v)
dict[v]++
}
return dict
}
Which will output (try it on the Go Playground):
apps: [one two three one four]
one
two
three
one
four
=== o: map[four:1 one:2 three:1 two:1]
I have a quick question. I am fairly new to golang. Say I have a map like so:
map[int]string
How could I randomly split it into two maps or arrays and as close to even as possible? So for example, if there are 15 items, it would be split 7 - 8.
For example:
func split(m map[int]string) (odds map[int]string, evens map[int]string) {
n := 1
odds = make(map[int]string)
evens = make(map[int]string)
for key, value := range m {
if n % 2 == 0 {
evens[key] = value
} else {
odds[key] = value
}
n++
}
return odds, evens
}
It is actually an interesting example, because it shows a few aspects of Go that are not obvious for beginners:
range m iterates in a random order, unlike in any other language as far as I know,
the modulo operator % returns the remainder of the integer division,
a function can return several values.
You could do something like this:
myStrings := make(map[int]string)
// Values are added to myStrings
myStrings2 := make(map[int]string)
// Seed system time for random numbers
rand.Seed(time.Now().UTC().UnixNano())
for k, v := range myStrings {
if rand.Float32() < 0.5 {
myStrings2[k] = v
delete(myStrings, k)
}
}
https://play.golang.org/p/6OnH1k4FMu
Given this algorithm, I would like to know if there exists an iterative version. Also, I want to know if the iterative version can be faster.
This some kind of pseudo-python...
the algorithm returns a reference to root of the tree
make_tree(array a)
if len(a) == 0
return None;
node = pick a random point from the array
calculate distances of the point against the others
calculate median of such distances
node.left = make_tree(subset of the array, such that the distance of points is lower to the median of distances)
node.right = make_tree(subset, such the distance is greater or equal to the median)
return node
A recursive function with only one recursive call can usually be turned into a tail-recursive function without too much effort, and then it's trivial to convert it into an iterative function. The canonical example here is factorial:
# naïve recursion
def fac(n):
if n <= 1:
return 1
else:
return n * fac(n - 1)
# tail-recursive with accumulator
def fac(n):
def fac_helper(m, k):
if m <= 1:
return k
else:
return fac_helper(m - 1, m * k)
return fac_helper(n, 1)
# iterative with accumulator
def fac(n):
k = 1
while n > 1:
n, k = n - 1, n * k
return k
However, your case here involves two recursive calls, and unless you significantly rework your algorithm, you need to keep a stack. Managing your own stack may be a little faster than using Python's function call stack, but the added speed and depth will probably not be worth the complexity. The canonical example here would be the Fibonacci sequence:
# naïve recursion
def fib(n):
if n <= 1:
return 1
else:
return fib(n - 1) + fib(n - 2)
# tail-recursive with accumulator and stack
def fib(n):
def fib_helper(m, k, stack):
if m <= 1:
if stack:
m = stack.pop()
return fib_helper(m, k + 1, stack)
else:
return k + 1
else:
stack.append(m - 2)
return fib_helper(m - 1, k, stack)
return fib_helper(n, 0, [])
# iterative with accumulator and stack
def fib(n):
k, stack = 0, []
while 1:
if n <= 1:
k = k + 1
if stack:
n = stack.pop()
else:
break
else:
stack.append(n - 2)
n = n - 1
return k
Now, your case is a lot tougher than this: a simple accumulator will have difficulties expressing a partly-built tree with a pointer to where a subtree needs to be generated. You'll want a zipper -- not easy to implement in a not-really-functional language like Python.
Making an iterative version is simply a matter of using your own stack instead of the normal language call stack. I doubt the iterative version would be faster, as the normal call stack is optimized for this purpose.
The data you're getting is random so the tree can be an arbitrary binary tree. For this case, you can use a threaded binary tree, which can be traversed and built w/o recursion and no stack. The nodes have a flag that indicate if the link is a link to another node or how to get to the "next node".
From http://en.wikipedia.org/wiki/Threaded_binary_tree
Depending on how you define "iterative", there is another solution not mentioned by the previous answers. If "iterative" just means "not subject to a stack overflow exception" (but "allowed to use 'let rec'"), then in a language that supports tail calls, you can write a version using continuations (rather than an "explicit stack"). The F# code below illustrates this. It is similar to your original problem, in that it builds a BST out of an array. If the array is shuffled randomly, the tree is relatively balanced and the recursive version does not create too deep a stack. But turn off shuffling, and the tree gets unbalanced, and the recursive version stack-overflows whereas the iterative-with-continuations version continues along happily.
#light
open System
let printResults = false
let MAX = 20000
let shuffleIt = true
// handy helper function
let rng = new Random(0)
let shuffle (arr : array<'a>) = // '
let n = arr.Length
for x in 1..n do
let i = n-x
let j = rng.Next(i+1)
let tmp = arr.[i]
arr.[i] <- arr.[j]
arr.[j] <- tmp
// Same random array
let sampleArray = Array.init MAX (fun x -> x)
if shuffleIt then
shuffle sampleArray
if printResults then
printfn "Sample array is %A" sampleArray
// Tree type
type Tree =
| Node of int * Tree * Tree
| Leaf
// MakeTree1 is recursive
let rec MakeTree1 (arr : array<int>) lo hi = // [lo,hi)
if lo = hi then
Leaf
else
let pivot = arr.[lo]
// partition
let mutable storeIndex = lo + 1
for i in lo + 1 .. hi - 1 do
if arr.[i] < pivot then
let tmp = arr.[i]
arr.[i] <- arr.[storeIndex]
arr.[storeIndex] <- tmp
storeIndex <- storeIndex + 1
Node(pivot, MakeTree1 arr (lo+1) storeIndex, MakeTree1 arr storeIndex hi)
// MakeTree2 has all tail calls (uses continuations rather than a stack, see
// http://lorgonblog.spaces.live.com/blog/cns!701679AD17B6D310!171.entry
// for more explanation)
let MakeTree2 (arr : array<int>) lo hi = // [lo,hi)
let rec MakeTree2Helper (arr : array<int>) lo hi k =
if lo = hi then
k Leaf
else
let pivot = arr.[lo]
// partition
let storeIndex = ref(lo + 1)
for i in lo + 1 .. hi - 1 do
if arr.[i] < pivot then
let tmp = arr.[i]
arr.[i] <- arr.[!storeIndex]
arr.[!storeIndex] <- tmp
storeIndex := !storeIndex + 1
MakeTree2Helper arr (lo+1) !storeIndex (fun lacc ->
MakeTree2Helper arr !storeIndex hi (fun racc ->
k (Node(pivot,lacc,racc))))
MakeTree2Helper arr lo hi (fun x -> x)
// MakeTree2 never stack overflows
printfn "calling MakeTree2..."
let tree2 = MakeTree2 sampleArray 0 MAX
if printResults then
printfn "MakeTree2 yields"
printfn "%A" tree2
// MakeTree1 might stack overflow
printfn "calling MakeTree1..."
let tree1 = MakeTree1 sampleArray 0 MAX
if printResults then
printfn "MakeTree1 yields"
printfn "%A" tree1
printfn "Trees are equal: %A" (tree1 = tree2)
Yes it is possible to make any recursive algorithm iterative. Implicitly, when you create a recursive algorithm each call places the prior call onto the stack. What you want to do is make the implicit call stack into an explicit one. The iterative version won't necessarily be faster, but you won't have to worry about a stack overflow. (do I get a badge for using the name of the site in my answer?
While it is true in the general sense that directly converting a recursive algorithm into an iterative one will require an explicit stack, there is a specific sub-set of algorithms which render directly in iterative form (without the need for a stack). These renderings may not have the same performance guarantees (iterating over a functional list vs recursive deconstruction), but they do often exist.
Here is stack based iterative solution (Java):
public static Tree builtBSTFromSortedArray(int[] inputArray){
Stack toBeDone=new Stack("sub trees to be created under these nodes");
//initialize start and end
int start=0;
int end=inputArray.length-1;
//keep memoy of the position (in the array) of the previously created node
int previous_end=end;
int previous_start=start;
//Create the result tree
Node root=new Node(inputArray[(start+end)/2]);
Tree result=new Tree(root);
while(root!=null){
System.out.println("Current root="+root.data);
//calculate last middle (last node position using the last start and last end)
int last_mid=(previous_start+previous_end)/2;
//*********** add left node to the previously created node ***********
//calculate new start and new end positions
//end is the previous index position minus 1
end=last_mid-1;
//start will not change for left nodes generation
start=previous_start;
//check if the index exists in the array and add the left node
if (end>=start){
root.left=new Node(inputArray[((start+end)/2)]);
System.out.println("\tCurrent root.left="+root.left.data);
}
else
root.left=null;
//save previous_end value (to be used in right node creation)
int previous_end_bck=previous_end;
//update previous end
previous_end=end;
//*********** add right node to the previously created node ***********
//get the initial value (inside the current iteration) of previous end
end=previous_end_bck;
//start is the previous index position plus one
start=last_mid+1;
//check if the index exists in the array and add the right node
if (start<=end){
root.right=new Node(inputArray[((start+end)/2)]);
System.out.println("\tCurrent root.right="+root.right.data);
//save the created node and its index position (start & end) in the array to toBeDone stack
toBeDone.push(root.right);
toBeDone.push(new Node(start));
toBeDone.push(new Node(end));
}
//*********** update the value of root ***********
if (root.left!=null){
root=root.left;
}
else{
if (toBeDone.top!=null) previous_end=toBeDone.pop().data;
if (toBeDone.top!=null) previous_start=toBeDone.pop().data;
root=toBeDone.pop();
}
}
return result;
}
I have a quite simple question, I think.
I've got this problem, which can be solved very easily with a recursive function, but which I wasn't able to solve iteratively.
Suppose you have any boolean matrix, like:
M:
111011111110
110111111100
001111111101
100111111101
110011111001
111111110011
111111100111
111110001111
I know this is not an ordinary boolean matrix, but it is useful for my example.
You can note there is sort of zero-paths in there...
I want to make a function that receives this matrix and a point where a zero is stored and that transforms every zero in the same area into a 2 (suppose the matrix can store any integer even it is initially boolean)
(just like when you paint a zone in Paint or any image editor)
suppose I call the function with this matrix M and the coordinate of the upper right corner zero, the result would be:
111011111112
110111111122
001111111121
100111111121
110011111221
111111112211
111111122111
111112221111
well, my question is how to do this iteratively...
hope I didn't mess it up too much
Thanks in advance!
Manuel
ps: I'd appreciate if you could show the function in C, S, python, or pseudo-code, please :D
There is a standard technique for converting particular types of recursive algorithms into iterative ones. It is called tail-recursion.
The recursive version of this code would look like (pseudo code - without bounds checking):
paint(cells, i, j) {
if(cells[i][j] == 0) {
cells[i][j] = 2;
paint(cells, i+1, j);
paint(cells, i-1, j);
paint(cells, i, j+1);
paint(cells, i, j-1);
}
}
This is not simple tail recursive (more than one recursive call) so you have to add some sort of stack structure to handle the intermediate memory. One version would look like this (pseudo code, java-esque, again, no bounds checking):
paint(cells, i, j) {
Stack todo = new Stack();
todo.push((i,j))
while(!todo.isEmpty()) {
(r, c) = todo.pop();
if(cells[r][c] == 0) {
cells[r][c] = 2;
todo.push((r+1, c));
todo.push((r-1, c));
todo.push((r, c+1));
todo.push((r, c-1));
}
}
}
Pseudo-code:
Input: Startpoint (x,y), Array[w][h], Fillcolor f
Array[x][y] = f
bool hasChanged = false;
repeat
for every Array[x][y] with value f:
check if the surrounding pixels are 0, if so:
Change them from 0 to f
hasChanged = true
until (not hasChanged)
For this I would use a Stack ou Queue object. This is my pseudo-code (python-like):
stack.push(p0)
while stack.size() > 0:
p = stack.pop()
matrix[p] = 2
for each point in Arround(p):
if matrix[point]==0:
stack.push(point)
The easiest way to convert a recursive function into an iterative function is to utilize the stack data structure to store the data instead of storing it on the call stack by calling recursively.
Pseudo code:
var s = new Stack();
s.Push( /*upper right point*/ );
while not s.Empty:
var p = s.Pop()
m[ p.x ][ p.y ] = 2
s.Push ( /*all surrounding 0 pixels*/ )
Not all recursive algorithms can be translated to an iterative algorithm. Normally only linear algorithms with a single branch can. This means that tree algorithm which have two or more branches and 2d algorithms with more paths are extremely hard to transfer into recursive without using a stack (which is basically cheating).
Example:
Recursive:
listsum: N* -> N
listsum(n) ==
if n=[] then 0
else hd n + listsum(tl n)
Iteration:
listsum: N* -> N
listsum(n) ==
res = 0;
forall i in n do
res = res + i
return res
Recursion:
treesum: Tree -> N
treesum(t) ==
if t=nil then 0
else let (left, node, right) = t in
treesum(left) + node + treesum(right)
Partial iteration (try):
treesum: Tree -> N
treesum(t) ==
res = 0
while t<>nil
let (left, node, right) = t in
res = res + node + treesum(right)
t = left
return res
As you see, there are two paths (left and right). It is possible to turn one of these paths into iteration, but to translate the other into iteration you need to preserve the state which can be done using a stack:
Iteration (with stack):
treesum: Tree -> N
treesum(t) ==
res = 0
stack.push(t)
while not stack.isempty()
t = stack.pop()
while t<>nil
let (left, node, right) = t in
stack.pop(right)
res = res + node + treesum(right)
t = left
return res
This works, but a recursive algorithm is much easier to understand.
If doing it iteratively is more important than performance, I would use the following algorithm:
Set the initial 2
Scan the matrix for finding a 0 near a 2
If such a 0 is found, change it to 2 and restart the scan in step 2.
This is easy to understand and needs no stack, but is very time consuming.
A simple way to do this iteratively is using a queue.
insert starting point into queue
get first element from queue
set to 2
put all neighbors that are still 0 into queue
if queue is not empty jump to 2.