The 15 Puzzle is a classical problem for modelling algorithms involving heuristics. Commonly used heuristics for this problem include counting the number of misplaced tiles and finding the sum of the Manhattan distances between each block and its position in the goal configuration. Note that both are admissible, i.e., they never overestimate the number of moves left, which ensures optimality for certain search algorithms such as A*.
What Heuristic do you think is proper, A* seems to work nice, do you have an example, maybe in c or java?
Heuristic
My heuristic of choice is to find if the sum of all inversions in a permutation is odd or even - if it is even, then the 15Puzzle is solvable.
The number of inversions in a permutation is equal to that of its inverse permutation (Skiena 1990, p. 29; Knuth 1998).
Only if I know it can be solved does it make sense to solve it. The task then is to reduce inverses and - viola problem solved. To find a solution should be no more then 80 moves.
Even more help
The equation for permutation is:
Factorials in range of 0 to 16 are {1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600, 6227020800, 87178291200, 1307674368000, 20922789888000}. If you need more of them, search WolframAlpha for Range[1,20]!
If you want to learn more about it read: 15Puzzle.
Fifteen puzzle implemetation in C++ using A* algorihtm https://gist.github.com/sunloverz/7338003
Related
I came across the following question in this course:
Consider a variation of the Knapsack problem where we have two
knapsacks, with integer capacities ๐1 and ๐2. As usual, we are given
๐ items with positive values and positive integer weights. We want to
pick subsets ๐1,๐2 with maximum total value such that the total weights of ๐1 and ๐1 are at most ๐1 and ๐2, respectively. Assume that every item fits in either knapsack. Consider the following two algorithmic approaches.
(1) Use the algorithm from lecture to pick a max-value feasible solution ๐1 for the first knapsack, and then run it again on the remaining items to pick a max-value feasible solution ๐2 for the second knapsack.
(2) Use the algorithm from lecture to pick a max-value feasible solution for a knapsack with capacity ๐1+๐2, and then split the chosen items into
two sets ๐1+๐2 that have size at most ๐1 and ๐2, respectively.
Which of the following statements is true?
Algorithm (1) is guaranteed to produce an optimal feasible solution to the original problem provided ๐1=๐2.
Algorithm (1) is guaranteed to
produce an optimal feasible solution to the original problem but
algorithm (2) is not.
Algorithm (2) is guaranteed to produce an
optimal feasible solution to the original problem but algorithm (1) is
not.
Neither algorithm is guaranteed to produce an optimal feasible
solution to the original problem.
The "algorithm from lecture" is on YouTube. https://www.youtube.com/watch?v=KX_6OF8X6HQ, which is 0-1 knapsack problem for one bag.
The correct answer to this question is option 4. This, this and this post present solutions to the problem. However, I'm having a hard time finding counterexamples showing that options 1 through 3 are incorrect. Can you cite any?
Edit:
The accepted answer doesn't provide a counterexample for option 1; see 2 knapsacks with same capacity - Why can't we just find the max-value twice for that.
(Weight; Value): (3;10), (3;10), (4;2)
capacities 7, 3
The first method chooses 3+3 into the first sack, remaining items does not fit into the second one
(Weight; Value): (4;10), (4;10), (4;10), (2:1)
capacities 6, 6
The second method chooses (4+4+4) but this set cannot fit into two sacks without loss, while (4+2) and (4) is better
my professor gave me a simple task to program - at least i thought so. The basics are simple: I have a randomized maze out of n > 20.000 different rooms. I thought "hey great, let's use A* to calculate the optimal route". Now i am encountering the following problem:
The java-program from my professor randomizes the rooms into a large array saving all successor rooms from room X. Example:
Rooms[2427][{42},{289},{2833}] // Room 2427 is next to the rooms 42, 289 and 2833 - those 3 rooms can be entered from room 2427.
A* needs a good heuristic to measure the following costs for each room. Example:
TotalCosts for Room Successor = CompleteCosts for CurrentRoom + Costs from CurrentRoom to Successor + estimated rest costs.
The "estimated rest costs" are the problem. If i had a 2D array i could easily use the air distance from a to b as a basic heuristic cause there will be at least X costs (air distance from a to b). However i can't use that now, because i don't know where the rooms are! They are completely randomized and when i don't know where Room 23878 will be.
Are there any other kinds of heuristics i can use? I cannot just set the estimated rest costs for every open room to 0, can i? Are there better ways?
Thanks!
Using 0 for the heuristic should just work. The 0-heuristic is actually a special case of the A* algorithm, also called the Dijkstra algorithm.
I believe there is no admissible heuristic estimate in this case, i.e. h(x) = 0 in all cases.
However, it seems that the edges of the graph are not weighted. You can simply code BFS, which will run optimally fast (and easy to implement) in O(V+E).
I'm currently trying to solve the following problem, but am unsure which algorithm I should be using. Its in the area of mass identification.
I have a series of "weights", *w_i*, which can sum up to a total weight. The as-measured total weight has an error associated with it, so is thus inexact.
I need to find, given the total weight T, the closest k possible combinations of weights that can sum up to the total, where k is an input from the user. Each weight can be used multiple times.
Now, this sounds suspiciously like the bounded-integer multiple knapsack problem, however
it is possible to go over the weight, and
I also want all of the ranked solutions in terms of error
I can probably solve it using multiple sweeps of the knapsack problem, from weight-error->weight+error, by stepping in small enough increments, however it is possible if the increment is too large to miss certain weight combinations that could be used.
The number of weights is usually small (4 ->10 weights) and the ratio of the total weight to the mean weight is usually around 2 or 3
Does anyone know the names of an algorithm that might be suitable here?
Your problem effectively resembles the knapsack problem which is a NP-complete problem.
For really limited number of weights, you could run over every combinations with repetition followed by a sorting which gives you a quite high number of manipulations; at best: (n + k - 1)! / ((n - 1)! ยท k!) for the combination and nยทlog(n) for the sorting part.
Solving this kind of problem in a reasonable amount of time is best done by evolutionary algorithms nowadays.
If you take the following example from deap, an evolutionary algorithm framework in Python:
ga_knapsack.py, you realise that by modifying lines 58-59 that automatically discards an overweight solution for something smoother (a linear relation, for instance), it will give you solutions close to the optimal one in a shorter time than brute force. Solutions are already sorted for you at the end, as you requested.
As a first attempt I'd go for constraint programming (but then I almost always do, so take the suggestion with a pinch of salt):
Given W=w_1, ..., w_i for weights and E=e_1,.., e_i for the error (you can also make it asymmetric), and T.
Find all sets S (if the weights are unique, or a list) st sum w_1+e_1,..., w_k+e_k (where w_1, .., w_k \elem and e_1, ..., e_k \elem E) \approx T within some delta which you derive from k. Or just set it to some reasonably large value and decrease it as you are solving the constraints.
I just realise that you also want to parametrise the expression w_n op e_m over op \elem +, - (any combination of weights and error terms) and off the top of my head I don't know which constraint solver would allow you to do that. In any case, you can always fall back to prolog. It may not fly, especially if you have a lot of weights, but it will give you solutions quickly.
Let P(x) denote the polynomial in question. The least fixed point (LFP) of P is the lowest value of x such that x=P(x). The polynomial has real coefficients. There is no guarantee in general that an LFP will exist, although one is guaranteed to exist if the degree is odd and โฅ 3. I know of an efficient solution if the degree is 3. x=P(x) thus 0=P(x)-x. There is a closed-form cubic formula, solving for x is somewhat trivial and can be hardcoded. Degrees 2 and 1 are similarly easy. It's the more complicated cases that I'm having trouble with, since I can't seem to come up with a good algorithm for arbitrary degree.
EDIT:
I'm only considering real fixed points and taking the least among them, not necessarily the fixed point with the least absolute value.
Just solve f(x) = P(x) - x using your favorite numerical method. For example, you could iterate
x_{n + 1} = x_n - P(x_n) / (P'(x_n) - 1).
You won't find closed-form formula in general because there aren't any closed-form formula for quintic and higher polynomials. Thus, for quintic and higher degree you have to use a numerical method of some sort.
Since you want the least fixed point, you can't get away without finding all real roots of P(x) - x and selecting the smallest.
Finding all the roots of a polynomial is a tricky subject. If you have a black box routine, then by all means use it. Otherwise, consider the following trick:
Form M the companion matrix of P(x) - x
Find all eigenvalues of M
but this requires you have access to a routine for finding eigenvalues (which is another tricky problem, but there are plenty of good libraries).
Otherwise, you can implement the Jenkins-Traub algorithm, which is a highly non trivial piece of code.
I don't really recommend finding a zero (with eg. Newton's method) and deflating until you reach degree one: it is very unstable if not done properly, and you'll lose a lot of accuracy (and it is very difficult to tackle multiple roots with it). The proper way do do it is in fact the above-mentioned Jenkins-Traub algorithm.
This problem is trying to find the "least" (here I'm not sure if you mean in magnitude or actually the smallest, which could be the most negative) root of a polynomial. There is no closed form solution for polynomials of large degree, but there are myriad numerical approaches to finding roots.
As is often the case, Wikipedia is a good place to begin your search.
If you want to find the smallest root, then you can use the rule of signs to pin down the interval where it exists and then use some numerical method to find roots in that interval.
A and B are sets of N dimensional vectors (N=10), |B|>=|A| (|A|=10^2, |B|=10^5). Similarity measure sim(a,b) is dot product (required). The task is following: for each vector a in A find vector b in B, such that sum of similarities ss of all pairs is maximal.
My first attempt was greedy algorithm:
find the pair with the highest similarity and remove that pair from A,B
repeat (1) until A is empty
But such greedy algorithm is suboptimal in this case:
a_1=[1, 0]
a_2=[.5, .4]
b_1=[1, 1]
b_2=[.9, 0]
sim(a_1,b_1)=1
sim(a_1,b_2)=.9
sim(a_2,b_1)=.9
sim(a_2, b_2)=.45
Algorithm returns [a_1,b_1] and [a_2, b_2], ss=1.45, but optimal solution yields ss=1.8.
Is there efficient algo to solve this problem? Thanks
This is essentially a matching problem in weighted bipartite graph. Just assume that weight function f is a dot product (|ab|).
I don't think the special structure of your weight function will simplify problem a lot, so you're pretty much down to finding a maximum matching.
You can find some basic algorithms for this problem in this wikipedia article. Although at first glance they don't seem viable for your data (V = 10^5, E = 10^7), I would still research them: some of them might allow you to take advantage of your 'lame' set of vertixes, with one part orders of magnitude smaller than the other.
This article also seems relevant, although doesn't list any algorithms.
Not exactly a solution, but hope it helps.
I second Nikita here, it is an assignment (or matching) problem. I'm not sure this is computationally feasible for your problem, but you could use the Hungarian algorithm, also known as Munkres' assignment algorithm, where the cost of assignment (i,j) is the negative of the dot product of ai and bj. Unless you happen to know how the elements of A and B are formed, I think this is the most efficient known algorithm for your problem.