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How to count possible combination for coin problem
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I have given an array.And i want to find the all permutation of an array so it sum to a specific numbers.ExampleArray a =[2,3,5 ,1] Target = 8`Solution: [2,2,2,2] ,[5,3] ,[3,3,2] ,[5,2,1] and all possible combinationPlease provide me a approach to solve this the problem , the problem i am facing how to handle the repetition of the elements.Target is a large number of 10^6.
I think it is same asThis theory
You are facing a typical Subset Problem. The worst case complexity of this problem is exponential no matter how you put it. You might find good polynomial-time approximations that work wonders for average case though.
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Algorithm for solving Flow Free Game
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I'm looking to create a program to perform a puzzle for me in a similar fashion as this one. I have completed the piece searching portion to identify each of the pieces but I'm stuck at where to being to start solving the puzzle.
Backtracking is an option but for very large board sizes it will have high complexity and take time.
Is there any suitable algorithm to solve the game efficiently. Can using heuristics to solve the board help?
Looking for a direction to start.
It's hard to understand the precise logic of the game you posted. However, if you want something more efficient than DFS/BFS, then you should take a look at A* star: http://theory.stanford.edu/~amitp/GameProgramming/
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I am going through a list of algorithm that I found and try to implement them for learning purpose. Right now I am coding K mean and is confused in the following.
How do you know how many cluster there is in the original data set
Is there any particular format that I have follow in choosing the initial cluster centroid besides all centroid have to be different? For example does the algorithm converge if I choose cluster centroids that are different but close together?
Any advice would be appreciated
Thanks
With k-means you are minimizing a sum of squared distances. One approach is to try all plausible values of k. As k increases the sum of squared distances should decrease, but if you plot the result you may see that the sum of squared distances decreases quite sharply up to some value of k, and then much more slowly after that. The last value that gave you a sharp decrease is then the most plausible value of k.
k-means isn't guaranteed to find the best possible answer each run, and it is sensitive to the starting values you give it. One way to reduce problems from this is to start it many times, with different starting values, and pick the best answer. It looks a bit odd if an answer for larger k is actually larger than an answer for smaller k. One way to avoid this is to use the best answer found for k clusters as the basis (with slight modifications) for one of the starting points for k+1 clusters.
In the standard K-Means the K value is chosen by you, sometimes based on the problem itself ( when you know how many classes exists OR how many classes you want to exists) other times a "more or less" random value. Typically the first iteration consists of randomly selecting K points from the dataset to serve as centroids. In the following iterations the centroids are adjusted.
After check the K-Means algorithm, I suggest you also see the K-means++, which is an improvement of the first version, as it tries to find the best K for each problem, avoiding the sometimes poor clusterings found by the standard k-means algorithm.
If you need more specific details on implementation of some machine learning algorithm, please let me know.
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Call every subunitary ratio with its denominator a power of 2 a perplex.
Number 1 can be written in many ways as a sum of perplexes.
Call every sum of perplexes a zeta.
Two zetas are distinct if and only if one of the zeta has as least one perplex that the other does not have. In the image shown above, the last two zetas are considered to be the same.
Find all the numbers of ways 1 can be written as a zeta with N perplexes. Because this number can be big, calculate it modulo 100003.
Please don't post the code, but rather the algorithm. Be as precise as you can.
This problem was given at a contest and the official solution, written in the Romanian language, has been uploaded at https://www.dropbox.com/s/ulvp9of5b3bfgm0/1112_descr_P2_fractii2.docx?dl=0 , as a docx file. (you can use google translate)
I do not understand what the author of the solution meant to say there.
Well, this reminds me of BFS algorithms(Breadth first search), where you radiate out from a single point to find multiple solutions w/ different permutations.
Here you can use recursion, and set the base case as when N perplexes have been reached in that 1 call stack of the recursive function.
So you can say:
function(int N <-- perplexes, ArrayList<Double> currentNumbers, double dividedNum)
if N == 0, then you're done - enter the currentNumbers array into a hashtable
clone the currentNumbers ArrayList as cloneNumbers
remove dividedNum from cloneNumbers and add 2 dividedNum/2
iterate through index of cloneNumbers
for every number x in cloneNumbers, call function(N--, cloneNumbers, x)
This is a rough, very inefficient but short way to do it. There's obviously a lot of ways you can prune the algorithm(reduce the amount of duplicates going into the hashtable, prevent cloning as much as possible, etc), but because this shows the absolute permutation of every number, and then enters that sequence into a hashtable, the hashtable will use its equals() comparison to see that the sequence already exists(such as your last 2 zetas), and reject the duplicate. That way, you'll be left with the answer you want.
The efficiency of the current algorithm: O(|E|^(N)), where |E| is the absolute number of numbers you can have inside of the array at the end of all insertions, and N is the number of insertions(or as you said, # of perplexes). Obviously this isn't the most optimal speed, but it does definitely work.
Hope this helps!
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Suppose you have a quantum computer that can run Shor's algorithm for factorization of integers.
Is it then possible to produce an oracle that determines if no solution exists for an instance of the Subset Product problem, with 100% confidence, in sub-exponential time?
So, the oracle is given a sequence x1, ... xn, as the description of a subset product problem.
It responds either Yes, a solution to this instance does not exist, or No, a solution to this instance may or may not exist.
If we take he prime factors of all elements in the sequence and then check to see if all of them are present in the target product's factors, this should tell us if a solution is not at all possible. A solution exist may exist if and only if all the prime factors are accounted for. On quantum computers, prime factorization is sub-exponential.
Would like some feedback on if this is correct logic- if it works- and if the complexity is indeed different between classical and quantum systems for this oracle/algorithm. Would also appreciate an explanation on reductions - can Subset Product be reduced to 3SAT without consequence?
Your algorithm, if I understood it correctly, will fail for the elements [6, 15] and the target 10. It will determine that 6*15 = 2*3*3*5, which has all of the factors used in 10=2*5, and incorrectly assert that this means you can make 10 by multiplying 6 and 15.
There are two reasons that it's unlikely you'll be able to fix this algorithm:
Subset Product is NP-Complete. Finding a polynomial time quantum algorithm for it, or showing that no such algorithm exists, is probably as hard as determining if P=NP. That is to say, very very hard.
You don't want the prime factors, you want the "no-need-to-reduce" factors. For example, if every time a number in the problem has a prime factor of 13 it's accompanied by a factor of 17 then there's no need to break 221 into 13*17. You can apply Euclid's gcd algorithm to various combinations of elements to find these no-need-to-reduce factors, no quantum-ness required.
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The problem is here
Say I have just 25-cent, 10-cent and 4-cent coins and my total amount is 41. Using greedy, I'll pick 25-cent and then 10-cent, and then the remaining 6 cents can not be made.
So my question is, does greedy in this case will tell me that there is no solution?
It looks like your problem was answered right in the the Greedy algorithm wiki: http://en.wikipedia.org/wiki/Greedy_algorithm#Cases_of_failure
Imagine the coin example with only 25-cent, 10-cent, and 4-cent coins. The greedy algorithm would not be able to make change for 41 cents, since after committing to use one 25-cent coin and one 10-cent coin it would be impossible to use 4-cent coins for the balance of 6 cents, whereas a person or a more sophisticated algorithm could make change for 41 cents with one 25-cent coin and four 4-cent coins.
The greedy algorithm mentioned in your link assumes the existence of a unit coin. Otherwise there are some integer amounts it can't handle at all.
Regarding optimality - as stated there, it depends on the available coins. For {10,5,1} for example the greedy algorithm is always optimal (i.e. returns the minimum number of coins to use). For {1,3,4} the greedy algorithm is not guaranteed to be optimal (it returns 6=4+1+1 instead of 6=3+3).
It seems that greedy algorithm is not always the best and this case is used as example to illustrate when it doesn't work
See example in Wikipedia