I am having a problem with what I think should be an easy piece of code. I have a 2D array that is N x M, currently stored in a boost multi_array. The N columns represent spatial dimensions e.g. x,y,z and the M rows are points along each dimension.
What I would like to do is print all possible combinations of points along each dimension
For example, if my array is:
-1 -1
1 1
I want to print:
-1 -1
1 -1
-1 1
1 1
I just cant make it work and I always go out of range on the array. I have tried using iterators and accessing the elements as A[i][j], but with no luck.
Anyone have any suggestions or thoughts?
You're likely not resetting the iterators by setting them back to begin(). A better strategy is not reusing the iterators at all. Make them local to the loop that you're using them in.
Related
The problem is:
You have to sort an array in ascending order(permutation: numbers from 1 to N in a random order) using series of swaps. Every swap has a price and there are 5 types of prices. Write a program that sorts the given array for the smallest price.
There are two kinds of prices: priceByValue and priceByIndex. All of the prices of a kind are given in 2 two-dimensional arrays N*N. Example how to access prices:
You want to swap the 2nd and the 5th elements from the permutation with values of 4 and 7. The price for this swap will be priceByValue[4][7] + priceByIndex[2][5].
Indexes of all arrays are counted from 1 (, not from 0) in order to have access to all of the prices (the permutation elements’ values start from 1): priceByIndex[2][5] would actually be priceByIndex[1][4] in code. Moreover, the order of the indexes by which you access prices from the two-dimensional arrays doesn’t matter: priceByIndex[i][j] = priceByIndex[j][i] and priceByIndex[i][i] is always equal to 0. (priceByValue is the same)
Types of prices:
Price[i][j] = 0;
Price[i][j] = random number between 1 and 4*N;
Price[i][j] = |i-j|*6;
Price[i][j] = sqrt(|i-j|) *sqrt(N)*15/4;
Price[i][j] = max(i,j)*3;
When you access prices by index i and j are the indexes of the elements you want to swap from the original array; when you access prices by value i and j are the values of the elements you want to swap from the original array. (And they are always counted from 1)
Things given:
N - an integer from 1 to 400, Mixed array, Type of priceByIndex, priceByIndex matrix, Type of priceByValue, priceByValue matrix. (all elements of a matrix are from the given type)
Things that should 'appear on the screen': number of swaps, all swaps (only by index - 2 5 means that you have swapped 2nd and 3rd elements) and the price.
As I am still learning C++, I was wondering what is the most effective way to sort the array in order to try to find the sort with the smallest cost.
There might be a way how to access series of swaps that result a sorted array and see which one is with the smallest price and I need to sort the array by swapping the elements which are close by both value and index, but I don’t know how to do this. I would be very grateful if someone can give me a solution how to find the cheapest sort in code. Thank you in advance!
More: this problem might have no solution, I am just trying to get a result close to the ideal.
Dynamic Programming!
Think of the problem as a graph. Each of the N-factorial permutations represents a graph vertex, and the allowed swaps are just arcs between vertices. The price-tag of a swap is just the weight on the arc.
When you look at the problem this way, it can be easily solved with Dijkstra's algortihm for finding the lowest cost path through a graph from one vertex to another.
This is also called Single Pair Shortest Path
you can use an algorithm for sorting an array in lexicographical order and modify it so that it fits your needs ( you did not mention the sorting criteria like the desired result i.e. least value first, ... ) there are multiple algorithms available for this, i.e. quick sort,...
a code example is in https://www.geeksforgeeks.org/lexicographic-permutations-of-string/
My question is similar to this one.
I have two lists: X with n elements and Y with m elements - let's say they hold row and column indices for a n x m matrix A. Now, I'd like to write something to k random places in matrix A.
I thought of two solutions:
Get a random element x from X and a random element y from Y. Check if something is already written to A[x][y] and if not - write. But if k is close to m*n I can shoot like this for ever.
Create an m*n array with all possible combinations of indices, shuffle it, draw first k elements and write there. But the problem I see here is that if both n and m are very big, the newly created n*m array may be huge (and shuffling may take some time too).
Karoly Horvath suggested to combine the two. I guess I'd have to pick threshold t and:
.
if( k/(m*n) > t ){
use option 2.
}else{
use option 1.
}
Any advice on how to pick t then?
Are there any other (better) approaches I missed?
There's an elegant algorithm due to Floyd for sampling without replacement from a range of integers. You can map the resulting integers in [0, n*m) to coordinates by the C++ function [m](int i) { return std::make_pair(i / m, i % m); }.
The best approach depends on how full your resulting matrix will be.. If you are going to fill more than half of it your rate of collision (aka getting random spot that is already "written" to) is going to be high and will cause you to loop a lot more than you would want.
I would not generate all possibilities, but instead I would build it as you go using a lists of lists. One for all possible values of X and from that a list of possible values of Y. I would initialize the X list but not the Y ones.
every time you pick a value of x for the first time you create a dictionary or list of m elements, then remove the one you use. then next time you pick x you will have m-1 elements, once a X value runs out of elements then remove it from the list so it does not get picked again.. this way you can guarantee never to pick a occupied space again, and you do not need to generate n*m possible options.
You have n x m elements, e.g. 200 elements for a 10 x 20 matrix. Picking one out of 200 should be easy. Point is, whatever you do, you can flatten the two dimensions into one, reducing that part of the problem.
Notes:
Use floor divide and modulo operations to get row and column out of the index.
Blacklist: Store the picked index in a set to quickly skip those that were already written.
Whitelist: Store the indices that are not yet picked in a set. If this is better than blacklisting depends on how full your set is.
Using the right container type for the set might come important, it doesn't have to be std::set. For the blacklist, you only need fast lookup and fast insertion, a vector<bool> might actually work pretty well. For the whitelist, you need fast random access and fast deletion, a vector<unsigned> with the remaining indices would be a good choice.
Prepare to switch between either method depending on the circumstances.
for a nxm matrix, you can consider [0..n*m-1] the indexes for the matrix elements.
Filling in a random index is rather trivial, just generate a random number between 0 and n*m-1, and that is the position to be filled.
Subsequently doing this operation can be a little more tricky:
you can test weather you have already written something to a position and regenerate the random number; but as you fill the matrix you will have a larger number of index regeneration.
a better solution is to put all the indexes in a vector of n*m elements. As you generate an index, you remove it from the list and next time generate a random index between 0 and N-1
example:
vector<int> indexVec;
for (i=0;i<n*m;i++)
indexVec.push_back(i);
nrOfIndexes = n*m-1;
while (nrOfIndexes>1)
{
index = rand()% nrOfIndexes;
processMatrixLocation(index);
indexVec.erase(indexVec.begin()+index);
nrOfIndexes--;
}
processMatrixLocation(indexVec[0]);
"Write a function which takes as an input an object of vector type
removes an element at the rank k in the constant time, O(1) [constant]. Assume that the order of elements does not matter."
I thought I might have had an idea about this. But, as I started to try by using .erase(), I looked up what the big-O notation was and found out it was O(n),as in linear relation. I can't think of any other way at the moment. I don't want any code, but I think pseudo code will at least point me in the right direction if anyone can help
Assume that the order of elements does not matter.
This is what you need to pay attention to.
Suppose you have a vector
0 1 2 3 4 5 6
and you want to remove the 3. You can turn this into
0 1 2 6 4 5
in O(1) without any issues.
Actually, there is a way to do it. Here is the pseudocode:
If the element you are trying to remove is the last element in the vector, remove it, done.
Read the last element of the vector and write it over the element-to-be-removed.
Remove the last element of the vector.
You can swap and pop_back in constant time.
std::swap(vec.back(), vec[rank]);
vec.pop_back();
My sincere apologies for such a naive question. I know this is simple. But nothing comes to my mind now.
I am using C++. I'm a bit concerned about efficiency since this is targeted for an embedded hardware with very less processing power and RAM.
I have 2 integer arrays with 50 members local to a function. I need to determine what is the corresponding number in the second array when an element in the first array is specified and vice versa. I have the information that the element provided to me for look-up belongs to which array i.e. array 1 or array 2.
Ex : Array1 => 500 200 1000 300 .....
Array2 => 250 170 500 400 .....
Input 500 , Output will be 250
Input 400 , Output will be 300
input 200 , Output will be 170 and so on
I think an array look-up will be least efficient. Is stl::map the best option or do i have to look for any efficient search algorithms? I would like to know if you have to do this, which option you will be choosing.
Any thoughts?
You can use std::map for readability and a little efficiency as well, though in your case efficiency is of small matter
std::map<int,int> mapping;
.... //populate
cout <<mapping[200]; //170
This is only 1 way (Array 1 -> Array 2) though. Im not sure if any easier way to do the other way, but create a second map.
To support reverse lookup, or going from (Array 2 -> Array 1), Reverse map lookup suggests using Boost.Bimap
According to me there are 2 ways of doing it both have already been suggested;
put the both arrays in a map as key pair value and traverse map to find the corresponding value or key.
Traverse the array for which the input is there and calculate the index. Get the value for that index int he other array.
I would go for the second solution as it easier. Moreover with only 50 elements in a static array you don't need to worry about performance.
Say I have a range of number, say {2,3,4,5}, stored in this order in a std::vector v, and that I want to list all possibles subsets which end with 5 using STL... that is :
2 3 4 5
2 3 5
2 4 5
3 4 5
2 5
3 5
4 5
5
( I hope i don't forget any:) )
I tried using while(next_permutation(v.begin(),v.end())) but didn't come up with the wanted result :)
Does anyone have an idea?
PS : those who have done the archives of google code jam 2010 may recognize this :)
Let's focus on the problem of printing all subsets. As you know, if you have vector of n elements, you'll have 2^n possible subsets. It's not coincidence, that if you have n-bit integer, the maximal stored value is 2^n. If you consider each integer as a vector of bits, then iterating over all possible values will give all possible subsets of bits. Well, we have subsets for free by iterating integer!
Assuming vector has not more than 32 elements (over 4 billion possible subsets!), this piece of code will print all subset of vector v (excluding empty one):
for (uint32_t mask =1; mask < (1<<v.size()); ++mask)
{
std::vector<int>::const_iterator it = v.begin();
for (uint32_t m =mask; m; (m>>=1), ++it)
{
if (m&1) std::cout << *it << " ";
}
std::cout << std::endl;
}
I just create all possible bit masks for size of vector, and iterate through every bit; if it's set, I print appropriate element.
Now applying the rule of ending with some specific number is piece of cake (by checking additional condition while looping through masks). Preferably, if there is only one 5 in your vector, you could swap it to the end and print all subsets of vector without last element.
I'm effectively using std::vector, const_iterator and std::cout, so you might think about it as being solved using STL. If I come up with something more STLish, I'll let you know (well, but how, it's just iterating). You can use this function as a benchmark for your STL solutions though ;-)
EDIT: As pointed out by Jørgen Fogh, it doesn't solve your subset blues if you want to operate on large vectors. Actually, if you would like to print all subsets for 32 elements it would generate terabytes of data. You could use 64-bit integer if you feel limited by constant 32, but you wouldn't even end iterating through all the numbers. If your problem is just answering how many are desired subsets, you definitely need another approach. And STL won't be much helpful also ;-)
As you can use any container I would use std::set because it is next to what we want to represent.
Now your task is to find all subsets ending with 5 so we take our initial set and remove 5 from it.
Now we want to have all subsets of this new set and append 5 to them at the end.
void subsets(std::set<std::set<int>> &sets, std::set<int> initial)
{
if(initial.empty())
return;
sets.insert(initial);//save the current set in the set of sets
std::set<int>::iterator i = initial.begin();
for(; i != initial.end(); i++)//for each item in the set
{
std::set<int> new_set(initial);//copy the set
new_set.erase(new_set.find(*i));//remove the current item
subsets(sets, new_set);//recursion ...
}
}
sets is a set that contains all subsets you want.
initial is the set that you want to have the subsets of.
Finally call this with subsets(all_subsets, initial_list_without_5);
This should create the subsets and finally you can append 5 to all of them. Btw don't forget the empty set :)
Also note that creating and erasing all these sets is not very efficient. If you want it faster the final set should get pointers to sets and new_set should be allocated dynamically...
tomasz describes a solution which will be workable as long as n<=32 although it will be take a very long time to print 2^32 different subsets. Since the bounds for the large dataset are 2 <= n <= 500 generating all the subsets is definitely not the way to go. You need to come up with some clever way to avoid having to generate them. In fact, this is the whole point of the problem.
You can probably find solutions by googling the problem if you want. My hint is that you need to look at the structure of the sets and avoid generating them at all. You should only calculate how many there are.
use permutation to create a vector of vectors. Then use std::partition with a function to sort it into the vectors that end with 5 and those that don't.