I am in need of help proving that an algorithm has greedy choice property and optimal substructure.
Context of the problem:
Consider a problem where a company owns n gas stations that are connected by a highway.
Each gas station has a limited supply g_i of gas-cans. Since the company don't know which gas station is most visited they want all of them to have the same amount of gas.
So they hire a fueling-truck to haul gas between the stations in their truck. However, truck also consumes 1 gas-can per kilometer driven.
Your task will be to help the chain calculate the largest amount of gas-cans g_bar
they can have at all Stations.
Consider the example: Here we have g = (20, 40, 80, 10, 20) and
p = (0, 5, 13, 33, 36) (in kilometers). In order to send one gas-can from station 3 to
station 4 we need to put 41 gas-cans in the truck, as the fueling-truck will consume 40 before reaching their destination (to send two gas-cans we need to put 42 in the truck).
The optimal g_bar for the example is 21 and can be achieved as follows:
Station 2 sends 11 gas-cans towards Station 1. One gas-can arrives while ten are consumed on the way.
Station 3 sends 59 gas-cans towards Station 4. 19 arrive while 40 are consumed on the way.
Station 4 now has 29 gas-cans and send eight towards Station 5. Two of these arrive
and six are consumed on the way.
The final distribution of gas-cans is: (21, 29, 21, 21, 22).
Given an integer g_bar. Determine whether it is possible to get at least g_bar gas-cans in every Gas Station.
in order for the greedy choice property and optimal substructure to make sense for a decision problem, you can define an optimal solution to be a solution with at least g_bar gas-cans in every gas station if such a solution exists; otherwise, any solution is an optimal solution.
Input: The position p_i and gas-can supply g_i of each bar. Here g_i is the supply for the bar at position p_i. You may assume that the positions are in sorted order – i.e. p_1 < p_2 < . . . < p_n.
Output: The largest amount g_bar, such that each gas-station can have a gas-can supply of at least g_bar after the truck have transferred gas-cans between the stations.
How can i prove Greedy Choice and Optimal Substructure for this?
Let's define an optimal solution: Each station has at least X gas cans in each station (X = g_bar).
Proving greedy property
Let us assume our solution is sub-optimal. There must exist a station i such that gas[i] < X. Based on our algorithm, we borrow X - gas[i] from station i+1 (which is a valid move, since we had already found a solution). Now station i has gas = X. This contradicts the original assumption that there must exist a station i such that gas[i] < X, which means our solution isn't suboptimal. Hence, we prove the optimality.
Proving optimal substructure
Assume we have a subset of the stations of size N', and our solution is suboptimal. Again, if the solution is suboptimal, there must exist a station i such that gas[i] < X. You can use the greedy proof again to prove that our solution isn't suboptimal. Since we have optimal solution for each arbitrary subset, we prove that we have optimal substructure.
Related
The Problem is making n cents change with quarters, dimes, nickels, and pennies, and using the least total number of coins. In the particular case where the four denominations are quarters,dimes, nickels, and pennies, we have c1 = 25, c2 = 10, c3 = 5, and c4 = 1.
If we have only quarters, dimes, and pennies (and no nickels) to use,
the greedy algorithm would make change for 30 cents using six coins—a quarter and five pennies—whereas we could have used three coins, namely, three dimes.
Given a set of denominations, how can we say whether greedy approach creates an optimal solution?
What you are asking is how to decide whether a given system of coins is canonical for the change-making problem. A system is canonical if the greedy algorithm always gives an optimal solution. You can decide whether a system of coins which includes a 1-cent piece is canonical or not in a finite number of steps. Details, and more efficient algorithms in certain cases, can be found in http://arxiv.org/pdf/0809.0400.pdf.
How would you reach a given sum in the most optimal manner possible given a set of coins ?
Let's say that in this case we have random numbers of 1, 5, 10, 20 and 50 cent coins with the biggest coins getting the priority.
My first intuition would be to use all the biggest coins possible to fit and then use up the next smallest coin in value if the sum is exceeded.
Would this do or are there any shortfalls to this approach ? Are there any more efficient approaches ?
There are shortfalls to simply giving out the largest coins first.
Let's say your vending machine is out of every coin except twenty each of 50c, 20c and 1c coins and you have to deliver 60c in change.
A "prioritise-largest" (or greedy) scheme will give you eleven coins, one 50c coin and ten 1c coins.
The better solution is three 20c coins.
Greedy schemes only give you local optimum solutions. For global optima, you generally need to examine all possibilities (though there may be minimax-type algorithms to reduce the search space) to be certain which, for delivering change, is usually quite within the limits of computability.
Greedy Algorithms (what you are doing right now) are usually chosen for this type of things and implemented as Final State Machines to be used in vending machines (for this particular case).
The greedy algorithm determines the minimum number of coins to give
while making change. These are the steps a human would take to emulate
a greedy algorithm
The assumption to exhaust largest denomination will not be the best solution each time. Example:
Input: coins[] = {25, 10, 5}, V = 30
Output: Minimum 2 coins required
We can use one coin of 25 cents and one of 5 cents
Input: coins[] = {9, 6, 5, 1}, V = 11
Output: Minimum 2 coins required
We can use one coin of 6 cents and 1 coin of 5 cents (min)
As per logic of exhausting largest coins first, we would end up with one
coin of 9 cents and 2 coins of 1 cent
Refer this answer for more clarification.
I have a complex problem and I want to know if an existing and well understood solution model exists or applies, like the Traveling Salesman problem.
Input:
A calendar of N time events, defined by starting and finishing time, and place.
The capacity of each meeting place (maximum amount of people it can simultaneously hold)
A set of pairs (Ai,Aj) which indicates that attendant Ai wishes to meet with attendat Aj, and Aj accepted that invitation.
Output:
For each assistant A, a cronogram of all the events he will attend. The main criteria is that each attendants should meet as many of the attendants who accepted his invites as possible, satisfying the space constraints.
So far, we thought of solving with backtracking (trying out all possible solutions), and using linear programming (i.e. defining a model and solving with the simplex algorithm)
Update: If Ai already met Aj in some event, they don't need to meet anymore (they have already met).
Your problem is as hard as minimum maximal matching problem in interval graphs, w.l.o.g Assume capacity of rooms is 2 means they can handle only one meeting in time. You can model your problem with Interval graphs, each interval (for each people) is one node. Also edges are if A_i & A_j has common time and also they want to see each other, set weight of edges to the amount of time they should see each other, . If you find the minimum maximal matching in this graph, you can find the solution for your restricted case. But notice that this graph is n-partite and also each part is interval graph.
P.S: note that if the amount of time that people should be with each other is fixed this will be more easier than weighted one.
If you have access to a good MIP solver (cplex/gurobi via acedamic initiative, but coin OR and LP_solve are open-source, and not bad either), I would definitely give simplex a try. I took a look at formulating your problem as a mixed integer program, and my feeling is that it will have pretty strong relaxations, so branch and cut and price will go a long way for you. These solvers give remarkably scalable solutions nowadays, especially the commercial ones. Advantage is they also provide an upper bound, so you get an idea of solution quality, which is not the case for heuristics.
Formulation:
Define z(i,j) (binary) as a variable indicating that i and j are together in at least one event n in {1,2,...,N}.
Define z(i,j,n) (binary) to indicate they are together in event n.
Define z(i,n) to indicate that i is attending n.
Z(i,j) and z(i,j,m) only exist if i and j are supposed to meet.
For each t, M^t is a subset of time events that are held simulteneously.
So if event 1 is from 9 to 11, event 2 is from 10 to 12 and event 3 is from 11 to 13, then
M^1 = {event 1, event 2) and M^2 = {event 2, event 3}. I.e. no person can attend both 1 and 2, or 2 and 3, but 1 and 3 is fine.
Max sum Z(i,j)
z(i,j)<= sum_m z(i,j,m)
(every i,j)(i and j can meet if they are in the same location m at least once)
z(i,j,m)<= z(i,m) (for every i,j,m)
(if i and j attend m, then i attends m)
z(i,j,m)<= z(j,m) (for every i,j,m)
(if i and j attend m, then j attends m)
sum_i z(i,m) <= C(m) (for every m)
(only C(m) persons can visit event m)
sum_(m in M^t) z(i,m) <= 1 (for every t and i)
(if m and m' are both overlapping time t, then no person can visit them both. )
As pointed out by #SaeedAmiri, this looks like a complex problem.
My guess would be that the backtracking and linear programming options you are considering will explode as soon as the number of assistants grows a bit (maybe in the order of tens of assistants).
Maybe you should consider a (meta)heuristic approach if optimality is not a requirement, or constraint programming to build an initial model and see how it scales.
To give you a more precise answer, why do you need to solve this problem? what would be the typical number of attendees? number of rooms?
Okay here is a puzzle I come across a lot of times-
Given a set of 12 balls , one of which is defective (it weighs either less or more) . You are allow to weigh 3 times to find the defective and also tell which weighs less or more.
The solution to this problem exists, but I want to know whether we can algorithmically determine if given a set of 'n' balls what is the minimum number of times you would need to use a beam balance to determine which one is defective and how(lighter or heavier).
A wonderful algorithm by Jack Wert can be found here
http://www.cut-the-knot.org/blue/OddCoinProblems.shtml
(as described for the case n is of the form (3^k-3)/2, but it is generalizable to other n, see the writeup below)
A shorter version and probably more readable version of that is here
http://www.cut-the-knot.org/blue/OddCoinProblemsShort.shtml
For n of the form (3^k-3)/2, the above solution applies perfectly and the minimum number of weighings required is k.
In other cases...
Adapting Jack Wert's algorithm for all n.
In order to modify the above algorithm for all n, you can try the following (I haven't tried proving the correctness, though):
First check if n is of the from (3^k-3)/2. If it is, apply above algorithm.
If not,
If n = 3t (i.e. n is a multiple of 3), you find the least m > n such that m is of the form (3^k-3)/2. The number of weighings required will be k. Now form the groups 1, 3, 3^2, ..., 3^(k-2), Z, where 3^(k-2) < Z < 3^(k-1) and repeat the algorithm from Jack's solution.
Note: We would also need to generalize the method A (the case when we know if the coin is heavier of lighter), for arbitrary Z.
If n = 3t+1, try to solve for 3t (keeping one ball aside). If you don't find the odd ball among 3t, the one you kept aside is defective.
If n = 3t+2, form the groups for 3t+3, but have one group not have the one ball group. If you come to the stage when you have to rotate the one ball group, you know the defective ball is one of two balls and you can then weigh one of those two balls against one of the known good balls (from among the other 3t).
Trichotomy ! :)
Explanation :
Given a set of n balls, subdivide it in 3 sets A, B and C of n/3 balls.
Compare A and B. If equal, then the defective ball is in C.
etc.
So, your minimum number of times is the number of times you can divide n by three (sorry, i do not know the english word for that).
You could use a general planning algorithm: http://www.inf.ed.ac.uk/teaching/courses/plan/
If there is more than one constraint (for example, both a volume limit and a weight limit, where the volume and weight of each item are not related), we get the multiply-constrained knapsack problem, multi-dimensional knapsack problem, or m-dimensional knapsack problem.
How do I code this in the most optimized fashion? Well, one can develop a brute force recursive solution. May be branch and bound.. but essentially its exponential most of the time until you do some sort of memoization or use dynamic programming which again takes a huge amount of memory if not done well.
The problem I am facing is this
I have my knapsack function
KnapSack( Capacity, Value, i) instead of the common
KnapSack ( Capacity , i ) since I have upper limits on both of those. can anyone guide me with this? or provide suitable resources for solving these problems for reasonably large n
or is this NP complete ?
Thanks
Merge the constraints. Look at http://www.diku.dk/~pisinger/95-1.pdf
chapter 1.3.1 called Merging the Constraints.
An example is say you have
variable , constraint1 , constraint2
1 , 43 , 66
2 , 65 , 54
3 , 34 , 49
4 , 99 , 32
5 , 2 , 88
Multiply the first constraint by some big number then add it to the second constraint.
So you have
variable , merged constraint
1 , 430066
2 , 650054
3 , 340049
4 , 990032
5 , 20088
From there do whatever algorithm you wanted to do with one constraint. The main limiter that comes to mind with this how many digits your variable can hold.
As a good example would serve the following problem:
Given an undirected graph G having positive weights and N vertices.
You start with having a sum of M money. For passing through a vertex i, you must pay S[i] money. If you don't have enough money - you can't pass through that vertex. Find the shortest path from vertex 1 to vertex N, respecting the above conditions; or state that such path doesn't exist. If there exist more than one path having the same length, then output the cheapest one. Restrictions: 1
Pseudocode:
Set states(i,j) as unvisited for all (i,j)
Set Min[i][j] to Infinity for all (i,j)
Min[0][M]=0
While(TRUE)
Among all unvisited states(i,j) find the one for which Min[i][j]
is the smallest. Let this state found be (k,l).
If there wasn't found any state (k,l) for which Min[k][l] is
less than Infinity - exit While loop.
Mark state(k,l) as visited
For All Neighbors p of Vertex k.
If (l-S[p]>=0 AND
Min[p][l-S[p]]>Min[k][l]+Dist[k][p])
Then Min[p][l-S[p]]=Min[k][l]+Dist[k][p]
i.e.
If for state(i,j) there are enough money left for
going to vertex p (l-S[p] represents the money that
will remain after passing to vertex p), and the
shortest path found for state(p,l-S[p]) is bigger
than [the shortest path found for
state(k,l)] + [distance from vertex k to vertex p)],
then set the shortest path for state(i,j) to be equal
to this sum.
End For
End While
Find the smallest number among Min[N-1][j] (for all j, 0<=j<=M);
if there are more than one such states, then take the one with greater
j. If there are no states(N-1,j) with value less than Infinity - then
such a path doesn't exist.
Knapsack with multiple constraints is a packing problem. Read up. http://en.wikipedia.org/wiki/Packing_problem
There are greedy like heuristics that calculate an "efficiency" for each item, that run quickly and yield approximate solutions.
You can use a branch and bound algorithm. You can get an initial lower bound using a greedy like heuristic, which can be used to initialize the incumbent solution. You can calculate upper bounds for various sub-problems by considering each of the m constraints one at time (relaxing the other constraints in the problem), then use the lowest of these bounds as an upper bound for the original problem. This technique is due to Shih. However this technique probably won't work well if no particular constraint tends to dominate the solution, or if the initial solution from the greedy like heuristic is not close to the optimum.
There are better more modern algorithms which are harder to implement, see "multidimensional knapsack problem" papers by J Puchinger!
As you said vol and weight both are positive quantities, try to use that fact that weight always decreases:
knap[position][vol][t]
Now t=0 when wt is positive, t=1 when wt is negative.