The problem is to find a schedule for some people to play golf (or whatever) in groups of a fixed size.
We have to guarantee that every player is only in one group at a time.
Here is my code:
int: gr; % number of groups
int: sz; % size of groups
int: we; % number of weeks
int: n=gr*sz; % number of players
set of int: G=1..gr; % set of group indices
set of int: P=1..n; % set of players
set of int: W=1..we; % set of weeks
% test instance
gr = 2;
sz = 2;
we = 2;
array[G,W] of var set of P: X;
%X[g,w] is the set of people that form group with index g in week w
% forall group x, |x| = sz
constraint forall (g in G, w in W)
(card (X[g,w]) = sz);
% one person cannot play in two groups simultaneously
constraint forall (g in G, w in W, p in X[g,w], g2 in (g+1..gr))
(not(p in X[g2,w]));
solve satisfy;
My problem now is that if I use the G12 lazyfd solver, i.e.
$ minizinc -b lazy this.mzn
I get
X = array2d(1..2 ,1..2 ,[1..2, 1..2, 1..2, 1..2]);
----------
which seems to ignore my second constraint.
On the other hand, using G12 without the lazy option, i.e.
$ minizinc this.mzn
yields
X = array2d(1..2 ,1..2 ,[1..2, 1..2, 3..4, 3..4]);
----------
which is correct.
G12 MIP and Gecode also give a correct result.
How is this possible? And how can I use the lazy solver such that I can rely on it? Or is it just my installation that is messed up somehow?
G12/lazyFD is known to be broken in various places. The problem is that the G12 solvers are no longer being developed and will most likely be removed from the distributions soon.
I would offer Chuffed as an alternative. Chuffed is FD solver written in C++ with lazy clause generation. It should be correct and will perform better then the G12 solver (at least when the solutions are correct).
Chuffed and other MiniZinc solvers can be found on the software page of the MiniZinc website.
Related
I'm trying to solve a Constraint Satisfaction Optimisation Problem that assigns agents to tasks. However, different then the basic Assignment Problem, a agent can be assigned to many tasks if the tasks do not overlap.
Each task has a fixed start_time and end_time.
The agents are assigned to the tasks according to some unary&binary constraints.
Variables = set of tasks
Domain = set of compatible agents (for each variable)
Constraints = unary&binary
Optimisation fct = some liniar function
An example of the problem: the allocation of parking space (or teams) for trucks for which we know the arrival and departure time.
I'm interested if there is in the literature a precise name for these type of problems. I presume it is some kind of assignment problem. Also, if you ever approach the problem, how do you solve it?
Thank you.
I would interpret this as: rectangular assignment-problem with conflicts
which is arguably much more hard (NP-hard in general) than the polynomially-solvable assignment-problem.
The demo shown in the other answer might work and ortools' cp-sat is great, but i don't see a good reason to use discrete-time based reasoning here like it's done: interval-variables, edge-finding and co. based scheduling constraints (+ conflict-analysis / explanations). This stuff is total overkill and the overhead will be noticable. I don't see any need to reason about time, but just about time-induced conflicts.
Edit: One could label those two approaches (linked + proposed) as compact formulation and extended formulation. Extended formulations usually show stronger relaxations and better (solving) results as long as scalability is not an issue. Compact approaches might become more viable again with bigger data (bit it's hard to guess here as scheduling-propagators are not that cheap).
What i would propose:
(1) Formulate an integer-programming model following the basic assignment-problem formulation + adaptions to make it rectangular -> a worker is allowed to tackle multiple tasks while all tasks are tackled (one sum-equality dropped)
see wiki
(2) Add integrality = mark variables as binary -> because the problem is not satisfying total unimodularity anymore
(3) Add constraints to forbid conflicts
(4) Add constraints: remaining stuff (e.g. compatibility)
Now this is all straightforward, but i would propose one non-naive improvement in regards to (3):
The conflicts can be interpreted as stable-set polytope
Your conflicts are induced by a-priori defined time-windows and their overlappings (as i interpret it; this is the core assumption behind this whole answer)
This is an interval graph (because of time-windows)
All interval graphs are chordal
Chordal graphs allow enumeration of all max-cliques in poly-time (implying there are only polynomial many)
python: networkx.algorithms.chordal.chordal_graph_cliques
The set (enumeration) of all maximal cliques define the facets of the stable-set polytope
Those (a constraint for each element in the set) we add as constraints!
(The stable-set polytope on the graph in use here would also allow very very powerful semidefinite-relaxations but it's hard to foresee in which cases this would actually help due to SDPs being much more hard to work with: warmstart within tree-search; scalability; ...)
This will lead to a poly-size integer-programming problem which should be very very good when using a good IP-solver (commercials or if open-source needed: Cbc > GLPK).
Small demo about (3)
import itertools
import networkx as nx
# data: inclusive, exclusive
# --------------------------
time_windows = [
(2, 7),
(0, 10),
(6, 12),
(12, 20),
(8, 12),
(16, 20)
]
# helper
# ------
def is_overlapping(a, b):
return (b[1] > a[0] and b[0] < a[1])
# raw conflicts
# -------------
binary_conflicts = []
for a, b in itertools.combinations(range(len(time_windows)), 2):
if is_overlapping(time_windows[a], time_windows[b]):
binary_conflicts.append( (a, b) )
# conflict graph
# --------------
G = nx.Graph()
G.add_edges_from(binary_conflicts)
# maximal cliques
# ---------------
max_cliques = nx.chordal_graph_cliques(G)
print('naive constraints: raw binary conflicts')
for i in binary_conflicts:
print('sum({}) <= 1'.format(i))
print('improved constraints: clique-constraints')
for i in max_cliques:
print('sum({}) <= 1'.format(list(i)))
Output:
naive constraints: raw binary conflicts
sum((0, 1)) <= 1
sum((0, 2)) <= 1
sum((1, 2)) <= 1
sum((1, 4)) <= 1
sum((2, 4)) <= 1
sum((3, 5)) <= 1
improved constraints: clique-constraints
sum([1, 2, 4]) <= 1
sum([0, 1, 2]) <= 1
sum([3, 5]) <= 1
Fun facts:
Commercial integer-programming solvers and maybe even Cbc might even try to do the same reasoning about clique-constraints to some degree although without the assumption of chordality where it's an NP-hard problem
ortools' cp-sat solver has also a code-path for this (again: general NP-hard case)
Should trigger when expressing the conflict-based model (much harder to decide on this exploitation on general discrete-time based scheduling models)
Caveats
Implementation / Scalability
There are still open questions like:
duplicating max-clique constraints over each worker vs. merging them somehow
be more efficient/clever in finding conflicts (sorting)
will it scale to the data: how big will the graph be / how many conflicts and constraints from those do we need
But those things usually follow instance-statistics (aka "don't decide blindly").
I don't know a name for the specific variant you're describing - maybe others would. However, this indeed seems a good fit for a CP/MIP solver; I would go with the OR-Tools CP-SAT solver, which is free, flexible and usually works well.
Here's a reference implementation with Python, assuming each vehicle requires a team assigned to it with no overlaps, and that the goal is to minimize the number of teams in use.
The framework allows to directly model allowed / forbidden assignments (check out the docs)
from ortools.sat.python import cp_model
model = cp_model.CpModel()
## Data
num_vehicles = 20
max_teams = 10
# Generate some (semi-)interesting data
interval_starts = [i % 9 for i in range(num_vehicles)]
interval_len = [ (num_vehicles - i) % 6 for i in range(num_vehicles)]
interval_ends = [ interval_starts[i] + interval_len[i] for i in range(num_vehicles)]
### variables
# t, v is true iff vehicle v is served by team t
team_assignments = {(t, v): model.NewBoolVar("team_assignments_%i_%i" % (t, v)) for t in range(max_teams) for v in range(num_vehicles)}
#intervals for vehicles. Each interval can be active or non active, according to team_assignments
vehicle_intervals = {(t, v): model.NewOptionalIntervalVar(interval_starts[v], interval_len[v], interval_ends[v], team_assignments[t, v], 'vehicle_intervals_%i_%i' % (t, v))
for t in range(max_teams) for v in range(num_vehicles)}
team_in_use = [model.NewBoolVar('team_in_use_%i' % (t)) for t in range(max_teams)]
## constraints
# non overlap for each team
for t in range(max_teams):
model.AddNoOverlap([vehicle_intervals[t, v] for v in range(num_vehicles)])
# each vehicle must be served by exactly one team
for v in range(num_vehicles):
model.Add(sum(team_assignments[t, v] for t in range(max_teams)) == 1)
# what teams are in use?
for t in range(max_teams):
model.AddMaxEquality(team_in_use[t], [team_assignments[t, v] for v in range(num_vehicles)])
#symmetry breaking - use teams in-order
for t in range(max_teams-1):
model.AddImplication(team_in_use[t].Not(), team_in_use[t+1].Not())
# let's say that the goal is to minimize the number of teams required
model.Minimize(sum(team_in_use))
solver = cp_model.CpSolver()
# optional
# solver.parameters.log_search_progress = True
# solver.parameters.num_search_workers = 8
# solver.parameters.max_time_in_seconds = 5
result_status = solver.Solve(model)
if (result_status == cp_model.INFEASIBLE):
print('No feasible solution under constraints')
elif (result_status == cp_model.OPTIMAL):
print('Optimal result found, required teams=%i' % (solver.ObjectiveValue()))
elif (result_status == cp_model.FEASIBLE):
print('Feasible (non optimal) result found')
else:
print('No feasible solution found under constraints within time')
# Output:
#
# Optimal result found, required teams=7
EDIT:
#sascha suggested a beautiful approach for analyzing the (known in advance) time window overlaps, which would make this solvable as an assignment problem.
So while the formulation above might not be the optimal one for this (although it could be, depending on how the solver works), I've tried to replace the no-overlap conditions with the max-clique approach suggested - full code below.
I did some experiments with moderately large problems (100 and 300 vehicles), and it seems empirically that on smaller problems (~100) this does improve by some - about 15% on average on the time to optimal solution; but I could not find a significant improvement on the larger (~300) problems. This might be either because my formulation is not optimal; because the CP-SAT solver (which is also a good IP solver) is smart enough; or because there's something I've missed :)
Code:
(this is basically the same code from above, with the logic to support using the network approach instead of the no-overlap one copied from #sascha's answer):
from timeit import default_timer as timer
from ortools.sat.python import cp_model
model = cp_model.CpModel()
run_start_time = timer()
## Data
num_vehicles = 300
max_teams = 300
USE_MAX_CLIQUES = True
# Generate some (semi-)interesting data
interval_starts = [i % 9 for i in range(num_vehicles)]
interval_len = [ (num_vehicles - i) % 6 for i in range(num_vehicles)]
interval_ends = [ interval_starts[i] + interval_len[i] for i in range(num_vehicles)]
if (USE_MAX_CLIQUES):
## Max-cliques analysis
# for the max-clique approach
time_windows = [(interval_starts[i], interval_ends[i]) for i in range(num_vehicles)]
def is_overlapping(a, b):
return (b[1] > a[0] and b[0] < a[1])
# raw conflicts
# -------------
binary_conflicts = []
for a, b in itertools.combinations(range(len(time_windows)), 2):
if is_overlapping(time_windows[a], time_windows[b]):
binary_conflicts.append( (a, b) )
# conflict graph
# --------------
G = nx.Graph()
G.add_edges_from(binary_conflicts)
# maximal cliques
# ---------------
max_cliques = nx.chordal_graph_cliques(G)
##
### variables
# t, v is true iff point vehicle v is served by team t
team_assignments = {(t, v): model.NewBoolVar("team_assignments_%i_%i" % (t, v)) for t in range(max_teams) for v in range(num_vehicles)}
#intervals for vehicles. Each interval can be active or non active, according to team_assignments
vehicle_intervals = {(t, v): model.NewOptionalIntervalVar(interval_starts[v], interval_len[v], interval_ends[v], team_assignments[t, v], 'vehicle_intervals_%i_%i' % (t, v))
for t in range(max_teams) for v in range(num_vehicles)}
team_in_use = [model.NewBoolVar('team_in_use_%i' % (t)) for t in range(max_teams)]
## constraints
# non overlap for each team
if (USE_MAX_CLIQUES):
overlap_constraints = [list(l) for l in max_cliques]
for t in range(max_teams):
for l in overlap_constraints:
model.Add(sum(team_assignments[t, v] for v in l) <= 1)
else:
for t in range(max_teams):
model.AddNoOverlap([vehicle_intervals[t, v] for v in range(num_vehicles)])
# each vehicle must be served by exactly one team
for v in range(num_vehicles):
model.Add(sum(team_assignments[t, v] for t in range(max_teams)) == 1)
# what teams are in use?
for t in range(max_teams):
model.AddMaxEquality(team_in_use[t], [team_assignments[t, v] for v in range(num_vehicles)])
#symmetry breaking - use teams in-order
for t in range(max_teams-1):
model.AddImplication(team_in_use[t].Not(), team_in_use[t+1].Not())
# let's say that the goal is to minimize the number of teams required
model.Minimize(sum(team_in_use))
solver = cp_model.CpSolver()
# optional
solver.parameters.log_search_progress = True
solver.parameters.num_search_workers = 8
solver.parameters.max_time_in_seconds = 120
result_status = solver.Solve(model)
if (result_status == cp_model.INFEASIBLE):
print('No feasible solution under constraints')
elif (result_status == cp_model.OPTIMAL):
print('Optimal result found, required teams=%i' % (solver.ObjectiveValue()))
elif (result_status == cp_model.FEASIBLE):
print('Feasible (non optimal) result found, required teams=%i' % (solver.ObjectiveValue()))
else:
print('No feasible solution found under constraints within time')
print('run time: %.2f sec ' % (timer() - run_start_time))
I am trying to write some codes to find the global maximum of an equation, e.g. f = -x**4.
Here is what I have got at the moment.
import sympy
x = sympy.symbols('x')
f = -x**4
df = sympy.diff(f,x)
ans = sympy.solve(df,x)
Then I am stuck. How should I substitute ans back into f, and how would I know if that would be the maximum, but not the minimum or a saddle point?
If you are just looking for the global maximum and nothing else, then there is already a function for that. See the following:
from sympy import *
x = symbols('x')
f = -x**4
print(maximum(f, x)) # 0
If you want more information such as the x value that gives that max or maybe local maxima, you'll have to do more manual work. In the following, I find the critical values as you have done above and then I show the values as those critical points.
diff_f = diff(f, x)
critical_points = solve(diff_f, x)
print(critical_points) # x values
for point in critical_points:
print(f.subs(x, point)) # f(x) values
This can be extended to include the second derivative test as follows:
d_f = diff(f, x)
dd_f = diff(f, x, 2)
critical_points = solve(d_f, x)
for point in critical_points:
if dd_f.subs(x, point) < 0:
print(f"Local maximum at x={point} with f({point})={f.subs(x, point)}")
elif dd_f.subs(x, point) > 0:
print(f"Local minimum at x={point} with f({point})={f.subs(x, point)}")
else:
print(f"Inconclusive at x={point} with f({point})={f.subs(x, point)}")
To find the global max, you would need to take all your critical points and evaluate the function at those points. Then pick the max from those.
outputs = [f.subs(x, point) for point in critical_points]
optimal_x = [point for point in critical_points if f.subs(x, point) == max(outputs)]
print(f"The values x={optimal_x} all produce a global max at f(x)={max(outputs)}")
The above should work for most elementary functions. Apologies for the inconsistent naming of variables.
If you are struggling with simple things like substitution, I suggest going through the docs for an hour or two.
I have a custom (discrete) probability distribution defined somewhat in the form: f(x)/(sum(f(x')) for x' in a given discrete set X). Also, 0<=x<=1.
So I have been trying to implement it in python 3.8.2, and the problem is that the numerator and denominator both come out to be really small and python's floating point representation just takes them as 0.0.
After calculating these probabilities, I need to sample a random element from an array, whose each index may be selected with the corresponding probability in the distribution. So if my distribution is [p1,p2,p3,p4], and my array is [a1,a2,a3,a4], then probability of selecting a2 is p2 and so on.
So how can I implement this in an elegant and efficient way?
Is there any way I could use the np.random.beta() in this case? Since the difference between the beta distribution and my actual distribution is only that the normalization constant differs and the domain is restricted to a few points.
Note: The Probability Mass function defined above is actually in the form given by the Bayes theorem and f(x)=x^s*(1-x)^f, where s and f are fixed numbers for a given iteration. So the exact problem is that, when s or f become really large, this thing goes to 0.
You could well compute things by working with logs. The point is that while both the numerator and denominator might underflow to 0, their logs won't unless your numbers are really astonishingly small.
You say
f(x) = x^s*(1-x)^t
so
logf (x) = s*log(x) + t*log(1-x)
and you want to compute, say
p = f(x) / Sum{ y in X | f(y)}
so
p = exp( logf(x) - log sum { y in X | f(y)}
= exp( logf(x) - log sum { y in X | exp( logf( y))}
The only difficulty is in computing the second term, but this is a common problem, for example here
On the other hand computing logsumexp is easy enough to to by hand.
We want
S = log( sum{ i | exp(l[i])})
if L is the maximum of the l[i] then
S = log( exp(L)*sum{ i | exp(l[i]-L)})
= L + log( sum{ i | exp( l[i]-L)})
The last sum can be computed as written, because each term is now between 0 and 1 so there is no danger of overflow, and one of the terms (the one for which l[i]==L) is 1, and so if other terms underflow, that is harmless.
This may however lose a little accuracy. A refinement would be to recognize the set A of indices where
l[i]>=L-eps (eps a user set parameter, eg 1)
And then compute
N = Sum{ i in A | exp(l[i]-L)}
B = log1p( Sum{ i not in A | exp(l[i]-L)}/N)
S = L + log( N) + B
A poster asked how to compare functions in Alloy. While testing a small example (comparing predicates instead of functions) to answer the question with, I've noticed the following behavior, which puzzles me.
The analyzer finds no counterexamples whenever the boundary of the check command is higher than 3 and the fact 'f1' is active. Inactivating the fact, the analyzer works as expected. Why does the redundant fact 'f1' modify the analyzer's operation so and why just in the case the boundary is higher than 3?
open util/ordering [V]
sig V {}
fact f1 {
# V > 0
}
pred p1 [x: V] {
x = last
}
pred p2 [x: V] {
x = first
}
assert a1 {
all x: V | p1[x] <=> p2[x]
}
check a1 for 3
It appears that whenever the check boundary is 4 or higher and 'f1' is active the analyzer reports '0 vars. 0 primary vars. 0 clauses.'
I am unable at the moment to look into the details, but it seems likely that you're seeing overflow behavior, based in part on the fact that Alloy's integers are very narrow (4 bits by default, I believe?) twos-complement integers, so overflow happens regularly.
Several changes might be instructive, separately or together, to see if they affect the behavior.
replace fact f1 with some V
turn on the "Forbid Overflow" option
provide an explicit bit width for Int using the scope command (for other signatures, the scope number specifies a maximum number of instances; for Int it specifies a bit width)
As Loïc Gammaitoni has put it in another question here "You should always be careful when playing with numbers in Alloy".
Local alignment between X and Y, with at least one column aligning a C
to a W.
Given two sequences X of length n and Y of length m, we
are looking for a highest-scoring local alignment (i.e., an alignment
between a substring X' of X and a substring Y' of Y) that has at least
one column in which a C from X' is aligned to a W from Y' (if such an
alignment exists). As scoring model, we use a substitution matrix s
and linear gap penalties with parameter d.
Write a code in order to solve the problem efficiently. If you use dynamic
programming, it suffices to give the equations for computing the
entries in the dynamic programming matrices, and to specify where
traceback starts and ends.
My Solution:
I've taken 2 sequences namely, "HCEA" and "HWEA" and tried to solve the question.
Here is my code. Have I fulfilled what is asked in the question? If am wrong kindly tell me where I've gone wrong so that I will modify my code.
Also is there any other way to solve the question? If its available can anyone post a pseudo code or algorithm, so that I'll be able to code for it.
public class Q1 {
public static void main(String[] args) {
// Input Protein Sequences
String seq1 = "HCEA";
String seq2 = "HWEA";
// Array to store the score
int[][] T = new int[seq1.length() + 1][seq2.length() + 1];
// initialize seq1
for (int i = 0; i <= seq1.length(); i++) {
T[i][0] = i;
}
// Initialize seq2
for (int i = 0; i <= seq2.length(); i++) {
T[0][i] = i;
}
// Compute the matrix score
for (int i = 1; i <= seq1.length(); i++) {
for (int j = 1; j <= seq2.length(); j++) {
if ((seq1.charAt(i - 1) == seq2.charAt(j - 1))
|| (seq1.charAt(i - 1) == 'C') && (seq2.charAt(j - 1) == 'W')) {
T[i][j] = T[i - 1][j - 1];
} else {
T[i][j] = Math.min(T[i - 1][j], T[i][j - 1]) + 1;
}
}
}
// Strings to store the aligned sequences
StringBuilder alignedSeq1 = new StringBuilder();
StringBuilder alignedSeq2 = new StringBuilder();
// Build for sequences 1 & 2 from the matrix score
for (int i = seq1.length(), j = seq2.length(); i > 0 || j > 0;) {
if (i > 0 && T[i][j] == T[i - 1][j] + 1) {
alignedSeq1.append(seq1.charAt(--i));
alignedSeq2.append("-");
} else if (j > 0 && T[i][j] == T[i][j - 1] + 1) {
alignedSeq2.append(seq2.charAt(--j));
alignedSeq1.append("-");
} else if (i > 0 && j > 0 && T[i][j] == T[i - 1][j - 1]) {
alignedSeq1.append(seq1.charAt(--i));
alignedSeq2.append(seq2.charAt(--j));
}
}
// Display the aligned sequence
System.out.println(alignedSeq1.reverse().toString());
System.out.println(alignedSeq2.reverse().toString());
}
}
#Shole
The following are the two question and answers provided in my solved worksheet.
Aligning a suffix of X to a prefix of Y
Given two sequences X and Y, we are looking for a highest-scoring alignment between any suffix of X and any prefix of Y. As a scoring model, we use a substitution matrix s and linear gap penalties with parameter d.
Give an efficient algorithm to solve this problem optimally in time O(nm), where n is the length of X and m is the length of Y. If you use a dynamic programming approach, it suffices to give the equations that are needed to compute the dynamic programming matrix, to explain what information is stored for the traceback, and to state where the traceback starts and ends.
Solution:
Let X_i be the prefix of X of length i, and let Y_j denote the prefix of Y of length j. We compute a matrix F such that F[i][j] is the best score of an alignment of any suffix of X_i and the string Y_j. We also compute a traceback matrix P. The computation of F and P can be done in O(nm) time using the following equations:
F[0][0]=0
for i = 1..n: F[i][0]=0
for j = 1..m: F[0][j]=-j*d, P[0][j]=L
for i = 1..n, j = 1..m:
F[i][j] = max{ F[i-1][j-1]+s(X[i-1],Y[j-1]), F[i-1][j]-d, F[i][j-1]-d }
P[i][j] = D, T or L according to which of the three expressions above is the maximum
Once we have computed F and P, we find the largest value in the bottom row of the matrix F. Let F[n][j0] be that largest value. We start traceback at F[n][j0] and continue traceback until we hit the first column of the matrix. The alignment constructed in this way is the solution.
Aligning Y to a substring of X, without gaps in Y
Given a string X of length n and a string Y of length m, we want to compute a highest-scoring alignment of Y to any substring of X, with the extra constraint that we are not allowed to insert any gaps into Y. In other words, the output is an alignment of a substring X' of X with the string Y, such that the score of the alignment is the largest possible (among all choices of X') and such that the alignment does not introduce any gaps into Y (but may introduce gaps into X'). As a scoring model, we use again a substitution matrix s and linear gap penalties with parameter d.
Give an efficient dynamic programming algorithm that solves this problem optimally in polynomial time. It suffices to give the equations that are needed to compute the dynamic programming matrix, to explain what information is stored for the traceback, and to state where the traceback starts and ends. What is the running-time of your algorithm?
Solution:
Let X_i be the prefix of X of length i, and let Y_j denote the prefix of Y of length j. We compute a matrix F such that F[i][j] is the best score of an alignment of any suffix of X_i and the string Y_j, such that the alignment does not insert gaps in Y. We also compute a traceback matrix P. The computation of F and P can be done in O(nm) time using the following equations:
F[0][0]=0
for i = 1..n: F[i][0]=0
for j = 1..m: F[0][j]=-j*d, P[0][j]=L
for i = 1..n, j = 1..m:
F[i][j] = max{ F[i-1][j-1]+s(X[i-1],Y[j-1]), F[i][j-1]-d }
P[i][j] = D or L according to which of the two expressions above is the maximum
Once we have computed F and P, we find the largest value in the rightmost column of the matrix F. Let F[i0][m] be that largest value. We start traceback at F[i0][m] and continue traceback until we hit the first column of the matrix. The alignment constructed in this way is the solution.
Hope you get some idea about wot i really need.
I think it's quite easy to find resources or even the answer by google...as the first result of the searching is already a thorough DP solution.
However, I appreciate that you would like to think over the solution by yourself and are requesting some hints.
Before I give out some of the hints, I would like to say something about designing a DP solution
(I assume you know this can be solved by a DP solution)
A dp solution basically consisting of four parts:
1. DP state, you have to self define the physical meaning of one state, eg:
a[i] := the money the i-th person have;
a[i][j] := the number of TV programmes between time i and time j; etc
2. Transition equations
3. Initial state / base case
4. how to query the answer, eg: is the answer a[n]? or is the answer max(a[i])?
Just some 2 cents on a DP solution, let's go back to the question :)
Here's are some hints I am able to think of:
What is the dp state? How many dimensions are enough to define such a state?
Thinking of you are solving problems much alike to common substring problem (on 2 strings),
1-dimension seems too little and 3-dimensions seems too many right?
As mentioned in point 1, this problem is very similar to common substring problem, maybe you should have a look on these problems to get yourself some idea?
LCS, LIS, Edit Distance, etc.
Supplement part: not directly related to the OP
DP is easy to learn, but hard to master. I know a very little about it, really cannot share much. I think "Introduction to algorithm" is a quite standard book to start with, you can find many resources, especially some ppt/ pdf tutorials of some colleges / universities to learn some basic examples of DP.(Learn these examples is useful and I'll explain below)
A problem can be solved by many different DP solutions, some of them are much better (less time / space complexity) due to a well-defined DP state.
So how to design a better DP state or even get the sense that one problem can be solved by DP? I would say it's a matter of experiences and knowledge. There are a set of "well-known" DP problems which I would say many other DP problems can be solved by modifying a bit of them. Here is a post I just got accepted about another DP problem, as stated in that post, that problem is very similar to a "well-known" problem named "matrix chain multiplication". So, you cannot do much about the "experience" part as it has no express way, yet you can work on the "knowledge" part by studying these standard DP problems first maybe?
Lastly, let's go back to your original question to illustrate my point of view:
As I knew LCS problem before, I have a sense that for similar problem, I may be able to solve it by designing similar DP state and transition equation? The state s(i,j):= The optimal cost for A(1..i) and B(1..j), given two strings A & B
What is "optimal" depends on the question, and how to achieve this "optimal" value in each state is done by the transition equation.
With this state defined, it's easy to see the final answer I would like to query is simply s(len(A), len(B)).
Base case? s(0,0) = 0 ! We can't really do much on two empty string right?
So with the knowledge I got, I have a rough thought on the 4 main components of designing a DP solution. I know it's a bit long but I hope it helps, cheers.