This is my first C++ related question here on StackOverflow
I have a vector of previously initialized 3D points in Openframeworks stored in the Vec3f
vector vertex;
I'm trying to get the middle point of these verticies points. My code at the moment is something like this
*for (int vertOne = 0; vertOne< 39; vertOne++) {
vertex.push_back(vertices[vertOne].getMiddle(vertices[vertOne++]));
}*
I want to have the vertices[vertOne] to move up through the loop like - 0,1,2,3,4,5,6,7,8,9 etc.
But i want .getMiddle(vertices[vertOne++]) to move through the loop starting at 1,2,3,4,5,6,7,8,9 etc.
So i want the first array to start at 0 and the second one to start at 1 and move in pair upwards through the loop
Im new to programming so any help would be great
Thanks!
The postfix ++ operator increases the value stored in the variable and stores the result back in the variable. You want to increment that value but not store the result back in the variable. You can achieve this just by adding 1 (vertOne + 1).
Related
For example if I have a vector of ints
vector<int> my_vector;
my_vector[0] = 6;
my_vector[1] = 3;
So my vector is of size 2 right now.
Now let's say I want to add another integer in my vector. Let's just say this new integer is 10.
I want to be able to set it up 3 (my_vec.size() + 1) ways. In other words, check where placing my new value in my vector of ints would result in the value I'm interested in.
10, 6, 3
6, 10, 3
6, 3, 10
Out of those 3 options, I'll pick the one that best fits my needs. The one I pick, will be my new vector. So if I pick 6, 10, 3, that will be my vector afterwards.
That's the gist of what I want to be able to do.
I have a very inefficient brute force way of creating temp vectors and calculating it that way. I was wondering if there is a simple and optimal way to go about this. I essentially want to be able to compute a new value that I want to add into my vector in all possible areas and look for whatever value I'm interested in.
Just insert new element to the end (most efficient way for the vector) and then move it to the front step by step and test each combination:
vec.push_back( new_value );
test_vector( vec );
for( size_t i = vec.size() - 1; i > 0; --i ) {
std::swap( vec[i], vec[i-1] );
test_vector( vec );
}
live example on ideone for 6,3 + 10. New element will always end at the first position. You need to find best position and then move your element there, that should be pretty obvious.
You really only have 2 cases here. First case: the newly added number is less than the first number in the vector. In this case the number should always be added to the beginning of the vector. If the new number is larger, it should always go in the second spot in the vector (since moving it any farther down the vector will not affect the new total).
I guess I understand your question.
Your computing method is complex. It is a example just now.
Only can you pick a kind of arrangement?
You can implement a function like this:
int GetArrangemnetValue(vector<int>& sourceVec, int nNewPos, int nNewValue);
This function to simulate a whole vetor.
You can do some compute is this function.
Then you can do some select via the return value of funtion.
So you can use this function and not a temporary vector.
I want to remove numbers from a matrix that represents coordinates with the format 'x y z'. One example:
1.211 1.647 1.041
2.144 2.684 1.548
1.657 2.245 1.021
1.657 0.984 2.347
2.154 0.347 2.472
1.211 1.647 1.041
In this example the coordinates 1 and 6 are the same (x, y and z are the same) and I want to remove them but I do not want to remove cases with only one value equal as coordinates 3 and 4 for x-coordinate).
These values are in a text file and I want to print the coordinates without duplication in another file or even in the same one.
A very simple solution would be to treat each line as a string and use a set of strings. As you traverse the file line-wise, you check if the current line exists in the set and if not, you insert and print it.
Complexity: O(nlogn), extra memory needed: almost the same as your input file in the worst case
With the same complexity and exactly the worst case memory consumption as the previous solution, you can load the file in memory, sort it line-wise, and then easily skip duplicates while printing. The same can be done inside the file if you are allowed to re-order it, and this way you need very little extra memory, but be much slower.
If memory and storage is an issue (I'm assuming since you can't duplicate the file), you can use the simple method of comparing the current line with all previous lines before printing, with O(n^2) complexity but no extra memory. This however is a rather bad solution, since you have to read multiple times from the file, which can be really slow compared to the main memory.
How to do this if you want to preserve the order.
Read the coordinates into an array of structures like this
struct Coord
{
double x,y,z;
int pos;
bool deleted;
};
pos is the line number, deleted is set to false.
Sort the structs by whatever axis tends to show the greatest variation.
Run through the array comparing the value of the axis you were using in the sort from the previous item to the value in the current item. If the difference is less than a certain preset delta (.i.e. if you care about three digits after the decimal point you would look for a difference of 0.000999999 or so) you compare the remaining values and set deleted for any line where x,y,z are close enough.
for(int i=1;i<count;i++)
{
if(fabs(arr[i].x-arr[i-1].x)<0.001)
if(fabs(arr[i].y-arr[i-1].y)<0.001)
if(fabs(arr[i].z-arr[i-1].z)<0.001)
arr[i].deleted=true;
}
sort the array again, this time ascending by pos to restore the order.
Go through the array and output all items where deleted is false.
In c++, you can use the the power of STL to solve this problem. Use the map and store the three coordinates x, y and z as a key in the map. The mapped value to the key will store the count of that key.
Key_type = pair<pair<float,float>,float>
mapped_type = int
Create a map m with the above given key_type and mapped_type and insert all the rows into the map updating the count for each row. Let's assume n is the total number of rows.
for(i = 0; i < n; i++) {
m[make_pair(make_pair(x,y),z)]++;
}
Each insertion takes O(logn) and you have to insert n times. So,overall time complexity will be O(nlogn). Now, loop over all the rows of the matrix again and if the mapped_value of that row is 1, then it is unique.
I am working on a binary linear program problem.
I am not really familiar with any computer language(just learned Java and C++ for a few months), but I may have to use computer anyway since the problem is quite complicated.
The first step is to declare variables m_ij for every entry in (at least 8 X 8) a matrix M.
Then I assign corresponding values of each element of a matrix to each of these variables.
The next is to generate other sets of variables, x_ij1, x_ij2, x_ij3, x_ij4, and x_ij5, whenever the value of m_ij is not 0.
The value of x_ijk variable is either 0 or 1, and I do not have to assign values for x_ijk variables.
Probably the simplest way to do it is to declare and assign a value to each variable, e.g.
int* m_11 = 5, int* m_12 = 2, int* m_13 = 0, ... int* m_1n = 1
int* m_21 = 3, int* m_12 = 1, int* m_13 = 2, ... int* m_2n = 3
and then pick variables, the value of which is not 0, and declare x_ij1 ~ x_ij5 accordingly.
But this might be too much work, especially since I am going to consider many different matrices for this problem.
Is there any way to do this automatically?
I know a little bit of Java and C++, and I am considering using lp_solve package in C++(to solve binary integer linear program problem), but I am willing to use any other language or program if I could do this easily.
I am sure there must be some way to do this(probably using loops, I guess?), and this is a very simple task, but I just don't know about it because I do not have much programming language.
One of my cohort wrote a program for generating a random matrix satisfying some condition we need, so if I could use that matrix as my input, it might be ideal, but just any way to do this would be okay as of now.
Say, if there is a way to do it with MS excel, like putting matrix entries to the cells in an excel file, and import it to C++ and automatically generate variables and assign values to them, then this would simplify the task by a great deal!
Matlab indeed seems very suitable for the task. Though the example offered by #Dr_Sam will indeed create the matrices on the fly, I would recommend you to initialize them before you assign the values. This way your code still ends up with the right variable if something with the same name already existed in the workspace and also your variable will always have the expected size.
Assuming you want to define a square 8x8 matrix:
m = zeros(8)
Now in general, if you want to initialize a three dimensional matrixh of size imax,jmax,kmax:
imax = 8;
jmax = 8;
kmax = 5;
x = zeros(imax,jmax,kmax);
Now assigning to or reading from these matrices is very easy, note that length and with of m have been chosen the same as the first dimensions of x:
m(3,4) = 4; %Assign a value
myvalue = m(3,4) %read the value
m(:,1) = 1:8 *Assign the values 1 through 8 to the first column
x(2,4,5) = 12; %Assign a single value to the three dimensional matrix
x(:,:,2) = m+1; Assign the entire matrix plus one to one of the planes in x.
In C++ you could use a std::vector of vectors, like
std::vector<std::vector<int>> matrix;
You don't need to use separate variables for the matrix values, why would you when you have the matrix?
I don't understand the reason you need to get all values where you evaluate true or false. Instead just put directly into a std::vector the coordinates where your condition evaluates to true:
std::vector<std::pair<int, int> true_values;
for (int i = 0; i < matrix.size(); i++)
{
for (int j = 0; j < matrix[i].size(); j++)
{
if (some_condition_for_this_matrix_value(matrix[i][j], i, j) == true)
true_values.emplace_back(std::make_pair(i, j));
}
}
Now you have a vector of all matrix coordinates where your condition is true.
If you really want to have both true and false values, you could use a std::unordered_map with a std::pair containing the matrix coordinates as key and bool as value:
// Create a type alias, as this type will be used multiple times
typedef std::map<std::pair<int, int>, bool> bool_map_type;
bool_map_type bool_map;
Insert into this map all values from the matrix, with the coordinates of the matrix as the key, and the map value as true or false depending on whatever condition you have.
To get a list of all entries from the bool_map you can remove any false entries with std::remove_if:
std::remove_if(bool_map.begin(), bool_map.end(),
[](const bool_map_type::value_type& value) {
return value.second == false;
};
Now you have a map containing only entries with their value as true. Iterate over this map to get the coordinates to the matrix
Of course, I may totally have misunderstood your problem, in which case you of course are free to disregard this answer. :)
I know both C++ and Matlab (not Python) and in your case, I would really go for Matlab because it's way easier to use when you start programming (but don't forget to come back to C++ when you will find the limitations to Matlab).
In Matlab, you can define matrices very easily: just type the name of the matrix and the index you want to set:
m(1,1) = 1
m(2,2) = 1
gives you a 2x2 identity matrix (indices start with 1 in Matlab and entries are 0 by default). You can also define 3d matrices the same way:
x(1,2,3) = 2
For the import from Excel, it is possible if you save your excel file in CSV format, you can use the function dlmread to read it in Matlab. You could also try later to implement your algorithm directly in Matlab.
Finally, if you want to solve your binary integer programm, there is already a built-in function in Matlab, called bintprog which can solve it for you.
Hope it helps!
I have searched Google and Stackoverflow for this question, but I still don't understand how a minimax function works.
I found the wikipedia entry has a pseudocode version of the function:
function integer minimax(node, depth)
if node is a terminal node or depth <= 0:
return the heuristic value of node
α = -∞
for child in node: # evaluation is identical for both players
α = max(α, -minimax(child, depth-1))
return α
Several other minimax functions I found with Google are basically the same thing; I'm trying to implement this in C++, and this is what I have come up with so far:
double miniMax(Board eval, int iterations)
{
//I evaluate the board from both players' point of view and subtract the difference
if(iterations == 0)
return boardEval(eval, playerNumber) - boardEval(eval, opponentSide());
/*Here, playerTurn tells the findPossibleMoves function whose turn it is;
I mean, how do you generate a list of possible moves if you don't even know
whose turn it's supposed to be? But the problem is, I don't see where I can
get playerTurn from, as there are only 2 parameters in all the examples of
minimax I've seen*/
vector<int> moves = eval.findPossibleMoves(playerTurn);
//I'm assuming -∞ in the wikipedia article means a very low number?
int result = -999999999;
//Now I run this loop to evaluate each possible move
/*Also, the Lua example in the wiki article has
alpha = node.player==1 and math.max(alpha,score) or math.min(alpha,score)
Is alpha a boolean there?!*/
for(int i = 0; i * 2 < moves.size(); i++)
{
//I make a copy of the board...
Board temp = eval;
/*and make the next possible move... once again playerTurn crops up, and I
don't know where I can get that variable from*/
temp.putPiece(moves[i * 2], moves[i * 2 + 1], playerTurn);
/*So do I create a function max that returns the bigger of two doubles?*/
result = max(result, -miniMax(temp, iterations - 1));
}
return result;
/*So now I've returned the maximum score from all possible moves within a certain
# of moves; so how do I know which move to make? I have the score; how do I know
which sequence of moves that score belongs to?*/
}
As you can see, I'm pretty confused about this minimax function. Please at the very least give me some hints to help me with this.
Thanks! :)
That sample from Wikipedia is doing NegaMax with Alpha/Beta pruning.
You may be helped by getting the naming straight:
The basis is MiniMax, a literal implementation would involve 2 methods that take turns (mutually recursive), 1 for each side.
Lazy programmers turn this into NegaMax, one method with a strategically placed - operator.
Alpha/Beta pruning is keeping track of a Window of best moves (over multiple depths) to detect dead branches.
Your playerTurn is used to determine whose turn it is . In NegaMax you can derive this from the depth (iterations) being odd or even. But it would be easier to use 2 parameters (myColor, otherColor) and switch them at each level.
Your miniMax() function should remember the best move it found so far. So instead of this code:
/*So do I create a function max that returns the bigger of two doubles?*/
result = max(result, -miniMax(temp, iterations - 1));
You should do something like this:
/*So do I create a function max that returns the bigger of two doubles?*/
double score = -miniMax(temp, iterations - 1);
if (score > result)
{
result = score;
bestMove = i;
}
Of course, you need a variable "bestMove" and a way to return the best move found to the caller.
Add the playerTurn variable as an argument to miniMax, and call miniMax which the current player's move initially and recursively.
Also, opponentSide needs to be a function of playerTurn.
A good place to start with game tree searching is the chess programming wiki. For your question about the move: I think it is most common to have two max-functions. The difference between the two max functions is that one returns only the score and the other returns the score and the best move. A recursive call order would be like following:
maxWithBestMoveReturn(...) --> min(...) --> max(...) --> min(...)
There are some good papers with pseudocode for the Alpha Beta algorithm:
TA Marsland - Computer Chess and Search
J Schaeffer - The games Computers (and People) Play
To your question in the comment: and math.max(alpha,score) or math.min(alpha,score) Is alpha a boolean there?!
No alpha is a window bound in a alpha beta algorithm. The alpha value gets updated with a new value. Because alpha and beta are swapped with the recursive call of the negamax-Function the alpha variable refers to the beta variable in the next recursive call.
One note to the playerTurn variable: The minimax or alpha-beta algorithm doesn't need this information. So i would give the information -- who's next --, into the Board-Structure. The functions findPossibleMoves and boardEval get all information they need from the Board-Structure.
One note to the recursive break condition: If i understand your code right, then you only have the one with iterations == o. I think this means the algorithm has reached the desired depth. But what if there are no possible moves left befor the algorithm reaches this depth. Maybe you should write following:
vector<int> moves = findPossibleMoves(...);
if (!moves.size())
return boardEval(...);
In your pseudocode, the node variable has to contain all the information about the current board position (or whatever). This information would include whose turn it is to move.
I'm working on a program to solve the n queens problem (the problem of putting n chess queens on an n x n chessboard such that none of them is able to capture any other using the standard chess queen's moves). I am using a heuristic algorithm, and it starts by placing one queen in each row and picking a column randomly out of the columns that are not already occupied. I feel that this step is an opportunity for optimization. Here is the code (in C++):
vector<int> colsleft;
//fills the vector sequentially with integer values
for (int c=0; c < size; c++)
colsleft.push_back(c);
for (int i=0; i < size; i++)
{
vector<int>::iterator randplace = colsleft.begin() + rand()%colsleft.size();
/* chboard is an integer array, with each entry representing a row
and holding the column position of the queen in that row */
chboard[i] = *randplace;
colsleft.erase(randplace);
}
If it is not clear from the code: I start by building a vector containing an integer for each column. Then, for each row, I pick a random entry in the vector, assign its value to that row's entry in chboard[]. I then remove that entry from the vector so it is not available for any other queens.
I'm curious about methods that could use arrays and pointers instead of a vector. Or <list>s? Is there a better way of filling the vector sequentially, other than the for loop? I would love to hear some suggestions!
The following should fulfill your needs:
#include <algorithm>
...
int randplace[size];
for (int i = 0; i < size; i ++)
randplace[i] = i;
random_shuffle(randplace, randplace + size);
You can do the same stuff with vectors, too, if you wish.
Source: http://gethelp.devx.com/techtips/cpp_pro/10min/10min1299.asp
Couple of random answers to some of your questions :):
As far as I know, there's no way to fill an array with consecutive values without iterating over it first. HOWEVER, if you really just need consecutive values, you do not need to fill the array - just use the cell indices as the values: a[0] is 0 and a[100] is 100 - when you get a random number, treat the number as the value.
You can implement the same with a list<> and remove cells you already hit, or...
For better performance, rather than removing cells, why not put an "already used" value in them (like -1) and check for that. Say you get a random number like 73, and a[73] contains -1, you just get a new random number.
Finally, describing item 3 reminded me of a re-hashing function. Perhaps you can implement your algorithm as a hash-table?
Your colsleft.erase(randplace); line is really inefficient, because erasing an element in the middle of the vector requires shifting all the ones after it. A more efficient approach that will satisfy your needs in this case is to simply swap the element with the one at index (size - i - 1) (the element whose index will be outside the range in the next iteration, so we "bring" that element into the middle, and swap the used one out).
And then we don't even need to bother deleting that element -- the end of the array will accumulate the "chosen" elements. And now we've basically implemented an in-place Knuth shuffle.