I have modified the question on n-puzzle. In this scenario, the puzzle has two blanks instead of one blank.
Initial State
3 5 1
4 6 -
7 2 -
Goal State
- 1 7
3 2 -
5 6 4
Is there any algorithm that I can use for this?
All existing algorithms that solve the regular sliding tile puzzle (such as A* or IDA*) can solve this variant as well. The puzzle with multiple blanks is equivalent to a pattern database for the sliding-tile puzzle - the exact solution to the puzzle with some pieces replaced with blanks can be used as a heuristic for the original puzzle with only a single blank.
(To be precise they are equivalent to additive pattern databases. You can combine several together and add their heuristic values as long as the action cost of swapping two blanks is 0 and none of the tiles are duplicated.)
Related
Given a non-negative integer array of length n and m queries consisting of two integers a and b, it is expected to find the maximum in the range of index [a,b] of the array. Note that a can be greater than b, in which case the desired range is from a to n and then from 1 to b. And an input k is also given that signifies that the length of the range to be considered is also constant that is constant
Example:
INPUT:
6 3 5 ---> n,m,k
7 6 2 6 1 5 ---> integer array
1 5 ---> query 1
2 6 ---> query 2
4 2 ---> query 3
OUTPUT:
7
6
7
I referred this article but am not able to get how to take care of the cases where a>b. Is there any other approach for this problem
Sliding window approach:
To solve the problem using approach mentioned i.e. Sliding Window Maximum, Just append the input array to itself like as shown below:
7 6 2 6 1 5 7 6 2 6 1 5
For a<=b case work as normal.
For a>bcase: Consider b = a + k. So your new range is [a,a+k] which you can happily solve without any changes to algorithm.
To optimize the above approach a bit, you can just append first k elements.
If you slide over every time a query arrives, it takes O(n) per query. k being very close or equal to n is the worst case.
Alternative Approach: Use the following approach in case of heavy querying and flexible ranges.
You are looking for range queries and this is what Segment Trees are popular for.
This tutorial finds the minimum in given range. I know you have asked for maximum, which is just a trivial change you have to make in code.
For a>b case, query two times once for [1,b] & then for [a,n] and report the maximum out of the two.
Preprocessing time: O(n)
Extra Space: O(n)
This approach is very efficient as it will answer every query in O(logn) which is quite helpful in case you are querying too much.
Sliding Window is going to output maximum element in all the ranges, but you need the maximum element only in given range. So instead of going with Sliding Window approach go with Segment Trees or Binary Indexed Trees. You'll feel the fun of truly querying within a range and not sliding over. (Just sliding over every time a query arrives won't scale if the range is flexible.)
I think this could be done by using divide and conquer approach, so let's take a look at the above example.
So for the case a>b
find max for range (1,b), say max_b = max_in_range(1,b).
find max for range (a,n), say max_a = max_in_range(a,n).
Now you can easily take up max between two numbers using a in built max method in any language as
ans = max(max_a, max_b)
But problems like this which involes ranges, you can solve it using segment trees, here is the link to start with - https://en.wikipedia.org/wiki/Segment_tree
Hope this helps!
Given a goal state
int final[3][3]={{1,2,3},
{4,5,6},
{7,8,9}};
and a random initial state, I want to sort my array as final only by shifting rows (right or left) and columns (up and down) of my table
7 8 4 by shifting to the right the first row it will become 4 7 8
2 1 9 2 1 9
6 5 3 6 5 3
So I want to use a* search and I'm trying to find a good heuristic.
I've already tried with misplaced array elements.
Any suggestions?
I view this as an algebraic problem. You are given a group of permutation which is generated by 6 cycles (3 rows and 3 columns) and you want to find some more moves which help you to get to any permutation.
First advice: not all permutations are possible! Since every shift is an even permutation (a 3-cycle is the composition of two transpositions) only even permutations are possible. Hence you will not find any solution to a configuration where all is in place but two swapped numbers as in (2,1,3),(4,5,6),(7,8,9).
Second advice. If r is a row shift and c is a coumn shift, compute the action of rcr'c' where r' and c' are the inverse shifts. This "commutator" is again a cycle of 3 elements but this time they are not in a row or column. By choosing different r and c you get a lot of 3-cycles which can be used in the third advice.
Third advice. Consider the region of numbers which are already in their final position. Apply 3-cycles to the complement of this set to reduce it, until you get to a solution.
I have a large vector of numbers, say 500 numbers. I would like a program to detect patterns (reoccurrence in this case) in such vector based on following rules:
A sequence of numbers is a pattern if:
The size of the sequence is between 3 and 20 numbers.
The RELATIVE positions of the numbers in sequence is repeated at
least one other time in a vector. So let's say if I have a sequence
(1,4,3) and then (3,6,5) somewhere else in the vector then (1,4,3) is
a pattern. (as well as (2,5,4), (3,6,5) etc.)
The sequences can't intersect. So, a vector (1,2,3,4,5) does not
contain patterns (1,2,3) and (3,4,5)(we can't use the same number for
both sequences). However, (1,2,3,3,4,5) does contain a pattern
(1,2,3) (or (3,4,5))
A subset A of a pattern B is a pattern ONLY IF A appears somewhere
else outside B. So, a vector (1,2,3,4,7,8,9,2,3,4,5) would contain
patterns (1,2,3,4) and (1,2,3), because (1,2,3,4) is repeated (in a
form of (2,3,4,5)) and (1,2,3) is repeated (in a form (7,8,9)).
However, if the vector was (1,2,3,4,2,3,4,5) the only pattern will
be (1,2,3,4), because (1,2,3) appeares only in context of (1,2,3,4).
I'd like to know several things:
First of all I hope the rules don't go against each other. I made them myself so there might be a clash somewhere that I didn't notice, please let me know if you do notice it.
Secondly, how would one implement such system in the most efficient way? Maybe someone can point out towards some particular literature on the subject? I could go number by number starting with searching a sequence repetition for all subsets of 3, then 4,5 and till 20. But that seems to be not very efficient..
I am interested in implementation of such system in C, but any general guidance is very welcome.
Thank you in advance!
Just a couple of observations:
If you're interested in relative values, then your first step should be to calculate the differences between adjacent elements of the vector, e.g.:
Original numbers:
1 4 3 2 5 1 1 3 6 5 6 2 5 4 4 4 1 4 3 2
********* ********* ********* *********
Difference values:
3 -1 -1 3 -4 0 2 3 -1 1 4 3 -1 -3 0 -3 3 -1 -1
****** ****** ****** ******
Once you've done that, you could use an autocorrelation method to look for repeated patterns in the data. This can be computed in O(n log n) time, and possibly even faster if you're only concerned with exact matches.
I'm having trouble with my n-puzzle solver. Thought it was working, but it turns out it is solving insoluble puzzles. I've tried to trace it, but that's a lot of tracing and so far I see no cheating. I think I understand the algorithm for determining solubility, and my implementation agrees with the odd/even parity of some examples from the web... that is to say, when I count up the number of tiles after a given tile that are smaller than it, for every tile, and then add the row index of the blank tile, I get the same odd or even number as others have gotten.
So a thought that has occurred to me. In my model of, say, the 8-puzzle, my solution state is:
_ 1 2
3 4 5
6 7 8
Rather than
1 2 3
8 _ 4
7 6 5
Or
1 2 3
4 5 6
7 8 _
As it is in some other formulations. Could this be affecting which puzzles are soluble and which are not?
Thanks!
z.
In general, yes: If a configuration is solvable to the standard solution, it will not be solvable to an unsolvable configuration.
In particular, it depends on the exact configuration you're using as a solution. You will need to check to see if you can solve from that configuration to the standard one.
EDIT: This of it this way:
Let A be the standard solution.
Let B be your preferred solution.
Let C be your starting configuration.
If you can get from A to B, and you can get from C to A, then you can get from C to B.
But if you can't get from A to B, and you can get from C to A, then you can't get from C to B.
I came across an algorithmic problem to find out the number of inversion pairs in an array in O(nlogn) time. I got the solution to this. But, my question is that what is the real-life application of this problem? Like I want to know some applications where we need to know the inversion pairs.
One example is the fifteen puzzle. If you want to randomly shuffle a grid of numbers, can you tell at a glance if
1 14 5 _
7 3 2 12
6 9 13 15
4 10 8 11
can be solved by sliding moves or not? The parity of the permutation will tell you that it is not.
Here is the use of inversion count in real life..
suppose you want to know how similar two list are..based on ranking..
on any movie site..two wishlist of movies are compared and few of them who are similar , are shown to users who have same choice.
Same logic applies to shopping list on any shopping website.. for recommending shopping items based on his activity..