Here\'s the essence of the problem I want to solve. We have workers taking care of children in a nursery for set times during the weekend. There\'s 16 diff
In case anyone comes across a similar problem, here's the solution I went with:
I ended up using a Backtracking Search Algorithm for Constraint Satisfaction Problems. All it does it do a normal backtracking search but checks that all constraints are satisfied as it traverses the tree. Here is the pseudo-code:
function BACKTRACKING-SEARCH(csp) returns a solution, or failure
return BACKTRACK({ }, csp)
function BACKTRACK(assignment, csp) returns a solution, or failure
if assignment is complete then return assignment
var = SELECT-UNASSIGNED-VARIABLE(csp)
for each value in ORDER-DOMAIN-VALUES(var, assignment, csp) do
if value is consistent with assignment then
add {var = value} to assignment
result = BACKTRACK(assignment, csp)
if result != failure then
return result
remove {var = value}
return failure
The variable is a time slot. The possible values assigned to a variable are the workers. The combination of a variable and its actual assignment is a node in the tree. Thus the search space is all the possible combinations of time slots and workers. The constraints prune nodes from the search space.
A constraint can be a workers availability. So if time slot A is assigned Worker X, but X is not able to work in time slot A, then this node will be ruled inconsistent.
I solved the problem of a particular time slot being assigned multiple workers by considering each time slot / worker combination to be it's OWN time slot. So if the childrens' nursery has 2 workers to fill, I consider it as two individual time slots to fill, each getting assigned its own worker. That way each variable is assigned just one value. It makes for MUCH simpler algorithms.
Thanks for all the help whittling this down to a solvable problem.
The "hospitals/residents problem" could indeed work but it depends of your constraints :
In your case the hospitals are workers and the residents are slots.
If that's ok for you then you have to possibilities :
you want to advantage workers : "hospital oriented case".
You will try to assign workers to their preferred slot(s).
you want to advantage slots : "resident oriented case"
Each slot will have their preferred workers.
I had to code it last year, here is the code.
/*
RO : needed for Resident-Oriented version
HO : needed for Hospital-Oriented version
*/
const int MAX_R = 1000;
const int MAX_H = 1000;
const int INF = 1000*1000*1000;
You need to fill the input variables. Everything is straightforward :
That's all.
// Input data
int R, H; // Number of Residents/Hospitals
int C[MAX_H]; // Capacity of hospitals
vector<int> R_pref[MAX_R], H_pref[MAX_H]; // Preferences : adjency lists
/*RO*/int H_rank[MAX_H][MAX_R]; // Rank : rank of r in H_pref[h]
/*HO*/int R_rank[MAX_R][MAX_H]; // Rank : rank of h in R_pref[r]
No need to touch below.
// Internal data
int RankWorst[MAX_H]; // Rank of the worst r taken by h
/*RO*/int BestH[MAX_R]; // Indice of the best h in R_pref the r can get
/*HO*/int BestR[MAX_H]; // Indice of the best r in H_pref the h can get
int Size[MAX_H]; // Number of residents taken by h
// Output data
int M[MAX_R];
void stable_hospitals_RO()
{
for(int h = 0 ; h < H ; h++)
RankWorst[h] = H_pref[h].size()-1;
fill_n(BestH, R, 0);
fill_n(Size, H,0);
fill_n(M,R,INF);
for (int i = 0; i < R; i++)
for (int r = i; r >= 0;)
{
if(BestH[r] == int(R_pref[r].size()))
break;
const int h = R_pref[r][BestH[r]++];
if(Size[h]++ < C[h])
{
M[r] = h;
break;
}
int WorstR = H_pref[h][RankWorst[h]];
while(WorstR == INF || M[WorstR] != h) // Compute the worst
WorstR = H_pref[h][--RankWorst[h]];
if(H_rank[h][r] < RankWorst[h]) // Ranked better that worst
{
M[r] = h;
M[r = WorstR] = INF; // We have eliminate it, he need to put it somewhere
}
}
}
void stable_hospitals_HO()
{
fill_n(BestR, H, 0);
fill_n(Size, H,0);
fill_n(M,R,INF);
vector<int> SH;
for (int h = 0; h < H; h++)
SH.push_back(h);
while(!SH.empty())
{
int h = SH.back();
if(Size[h] == C[h] || BestR[h] == int(H_pref[h].size())) // Full or no r available
{
SH.pop_back();
break;
}
const int r = H_pref[h][BestR[h]++];
// r is unassigned or prefer h to current hospital
if(M[r] == INF || R_rank[r][h] < R_rank[r][M[r]])
{
if(++Size[h] == C[h]) // Will be full
SH.pop_back();
if(M[r] != INF) // Delete from M[r]
{
Size[M[r]]--;
SH.push_back(M[r]);
}
M[r] = h;
}
}
}
Example of use to show how to build rank from prefs. (In that case the preference lists were on the stdin).
int main()
{
scanf("%d%d",&R,&H);
int num;
// put inf
for(int r = 0 ; r < R ; r++)
{
scanf("%d",&num);
R_pref[r].resize(num);
for(int h = 0 ; h < num ; h++)
{
scanf("%d",&R_pref[r][h]);
R_rank[r][R_pref[r][h]] = h;
}
}
for(int h = 0 ; h < H ; h++)
{
scanf("%d",&C[h]);
scanf("%d",&num);
H_pref[h].resize(num);
for(int r = 0 ; r < num ; r++)
{
scanf("%d",&H_pref[h][r]);
H_rank[h][H_pref[h][r]] = r;
}
}
stable_hospitals_RO();
printf("\n\n\n\n");
stable_hospitals_HO();
return 0;
}
On an example : Hospitals 1 to 3, 6 résidents.
H_pref :
R_pref :
H_rank :
Similar for R_rank.
Hospital don't have to rank everyone et can also rank less people than their capacity.