So the aim is to take two arrays as shown below
int x[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
int k[4] = {1, 2, 3, 4};
and add each element of k to each element of x in a loop as shown
1 2 3 4 5 6 7 8 9 10
+1 +2 +3 +4 +1 +2 +3 +4 +1 +2
This should give us a final array [2, 4, 6, 8, 6, 8, 10, 12, 10, 12].
Any suggestions as to how I could achieve this in C++
Loop through the indexes of the larger array, using the modulus (%) operator to wrap-around the indexes when accessing the smaller array.
int x[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
int k[4] = {1, 2, 3, 4};
int res[10];
for (int i = 0; i < 10; ++i) {
res[i] = x[i] + k[i % 4];
}
Online Demo
With % you can have the wrap-around behavior and with std::size(from C++17 onwards) the size of the array.
#include <algorithm>
#include <iostream>
int main()
{
int x[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
int k[4] = {1, 2, 3, 4};
for(int i = 0; i < std::size(x); ++i)
{
x[i] = x[i] + k[i%std::size(k)];
}
//lets confirm if x has the right elmennts
for(const int& element: x)
{
std::cout<< element<<std::endl;
}
}
Note that here i have not used a separate array to store the resulting array. Instead the elements are added into the original array x. Storing the result in a new array is trivial.
Given an arbitrary set of numbers S = {1..n} and size k, I need to print all possible combinations of size k ordered by the sum of the combinations.
Let's say S = {1, 2, 3, 4, 5} and k = 3, a sample output would be:
1 2 3 = 6
1 2 4 = 7
1 2 5 = 8
1 3 4 = 8
1 3 5 = 9
2 3 4 = 9
2 3 5 = 10
1 4 5 = 10
2 4 5 = 11
3 4 5 = 12
The first idea that came to my mind is to sort S (if not already sorted) and then perform a BFS search, having a queue of combinations sorted by sum. This is because I want to generate the next combination only if requested (iterator like). But this seems like an overkill and I think that there should be an easier solution. How can this be solved?
Edit:
This is my original idea:
1. Sort S
2. Pick the first k numbers and add them to the queue as a single node (because this is the combination with the smallest sum)
3. While the queue is not empty - pop the first node, output it, and generate successors.
Generating successors:
1. Start with the first number in the node. Make a copy of a node. Find the smallest unselected value that is > than the number, and assign this value.
2. Verify that this node hasn't been seen and calculate the sum.
3. Add the node to the queue.
4. Repeat 1-3 for all of the remaining numbers in the node.
Example:
Queue q = { {1, 2, 3} = 6 }
Pop front, output
1 2 3 = 6
Generate {4, 2, 3} = 9, {1, 4, 3} = 8, {1, 2, 4} = 7
Seen { {1, 2, 3}, {4, 2, 3}, {1, 4, 3}, {1, 2, 4} }
Queue q = { {1, 2, 4} = 7, {1, 4, 3} = 8, {4, 2, 3} = 9 }
Pop front, output
1 2 3 = 6
1 2 4 = 7
Generate {3, 2, 4} = 9 (seen - discard), {1, 3, 4} = 8 (seen - discard), {1, 2, 5} = 8 }
Seen { {1, 2, 3}, {4, 2, 3}, {1, 4, 3}, {1, 2, 4}, {1, 2, 5} }
Queue q = { {1, 2, 5} = 8, {1, 4, 3} = 8, {4, 2, 3} = 9 }
Pop front, output
1 2 3 = 6
1 2 4 = 7
1 2 5 = 8
Generate {3, 2, 5} = 10, {1, 3, 5} = 9
Seen { {1, 2, 3}, {4, 2, 3}, {1, 4, 3}, {1, 2, 4}, {1, 2, 5}, {3, 2, 5}, {1, 3, 5} }
Queue q = { {1, 4, 3} = 8, {1, 5, 3} = 9, {1, 3, 5} = 9, {3, 2, 5} = 10, {1, 4, 5} = 10, {4, 2, 5} = 11 }
...
...
javascript sample like below;
<!DOCTYPE html>
<html>
<head>
<meta charset = utf-8>
<title>s{1..n} by k</title>
</head>
<body>
<header>s{1..n} by k</header>
<form action="#" onsubmit="return false;">
s:<input type="text" id="sArrayObj" name="sArrayObj" value="4 5 1 8 9 3 6 7"/><br/>
n:<input type="text" id="kVarObj" name="kVarObj" value="3"/><br/>
<button id="scan" name="scan" onclick="scanX()"> CALCULATE </button><br/>
<textarea id="scanned" name="scanned" cols=40 rows=20>
</textarea>
</form>
<script>
const headFactorial = n => {
if ( n > 1 ) return n * headFactorial( n - 1 );
else return 1;
}
const tailFactorial = n => {
if ( n === 1 ) return 1;
else return n * tailFactorial( n - 1 );
}
const combinations = ( collection, combinationLength ) => {
let head, tail, result = [];
if ( combinationLength > collection.length || combinationLength < 1 ) { return []; }
if ( combinationLength === collection.length ) { return [ collection ]; }
if ( combinationLength === 1 ) { return collection.map( element => [ element ] ); }
for ( let i = 0; i < collection.length - combinationLength + 1; i++ ) {
head = collection.slice( i, i + 1 );
tail = combinations( collection.slice( i + 1 ), combinationLength - 1 );
for ( let j = 0; j < tail.length; j++ ) { result.push( head.concat( tail[ j ] ) ); }
}
return result;
}
const sumArr = (anArr) => {
var s=0;
for (j=0; j < anArr.length; j++) s += 1*anArr[j];
return s;
}
function scanX(){
sArray = document.getElementById("sArrayObj").value.split(" ");
kVar = document.getElementById("kVarObj").value;
sArray.sort();
resultObj= document.getElementById("scanned");
resultObj.value = "";
//for (i=0;kVar > i;i++){
// resultObj.value+=sArray[i]+' ';
//}
resultArr = combinations(sArray, kVar);
for (i=0;i<resultArr.length;i++){
resultObj.value += resultArr[i]+"="+sumArr(resultArr[i])+"\n";
}
return false;
}
</script>
</body>
</html>
Im using VS2013 along with the SystemC library from Allegro. I was trying to initialize two arrays as follows:
int pathObs1[19] = {10,9,8,7,6,5,4,3,2,1,2,3,4,5,6,7,8,9,10};
int Map[10][4] = {
{ 0, 3, 1, 4 }, //Grid 1
{ 1, 3, 2, 4 }, //Grid 2
{ 2, 3, 3, 4 }, //Grid 3
{ 3, 3, 4, 4 }, //Grid 4
{ 4, 3, 5, 4 }, //Grid 5
{ 5, 3, 6, 4 }, //Grid 6
{ 6, 3, 7, 4 }, //Grid 7
{ 6, 2, 7, 3 }, //Grid 8
{ 6, 1, 7, 2 }, //Grid 9
{ 6, 0, 7, 1 } //Grid 10
};
However i received the error the above error. I saw some questions on SO which had the same issue, however I dont think they were dealing with SystemC. Any easy workaround for this in SystemC since im trying to initialize inside my SC_MODULE header/constructor?
Edit: I had a typo in my array initialization. Still get the same error.
2dArray[m][n] means m rows n columns so you can keep n values in each row but in your code you defined matrix which had 3 columns but still you are assigning 4 values.
You can use a loop for filling the array:
#include <iostream>
#include <stdlib>
int main()
{
srand(time(null));
int map[10][4];
for (int i = 0; i < 10; i++)
{
for (int j = 0; j < 4; j++)
{
map[i][j] = rand(); // you can write smth like rand() % 5 to make a limit of the values
}
}
return 0;
}
Here is my array:
int grid[gridsize+1] = { 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 3, 3, 3, 2, 2, 4, 1, 5, 3, 3, 6, 2, 6, 4, 5, 5, 5, 3, 6, 2, 6, 4, 4, 5, 5, 5, 6, 6, 6, 4, 7, 7, 8, 5, 8, 8, 8, 4, 7, 7, 8, 8, 8, 8, 8, 4, 7, 7, 7, 7, 8, 8, 8 };
Each number represents a colour, I would like to create multiple arrays for each unique number. The created arrays will store the locations of that number from the original array.
e.g
colour1[5]
[0]=0 //because the number 1 is stored in element 0.
[1]=1
[2]=8
[3]=9
The numbers in grid will change every time I run, so things need to be dynamic?
I can write inefficient code that accomplishes this, but it's just repetitive and I can't comprehend a way to turn this into something I can put in a function.
Here is what I have;
int target_number = 1
grid_size = 64;
int counter = -1;
int counter_2 = -1;
int colour_1;
while (counter < grid_size + 1){
counter = counter + 1;
if (grid[counter] == target)
counter_2 = counter_2 + 1;
colour_1[counter_2] = counter;
}
}
I have to do this for each colour, when I try to make a function, it cannot access the main array in main so is useless.
You can just use vector<vector<int>> to represent your counters. No maps or sorting are needed.
EDIT: added additional pass to determine maximum color, so no run-time resize is needed.
Here is the code:
#include <iostream>
#include <vector>
#include <algorithm>
int main() {
int grid[] = { 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 3, /*...*/};
const size_t gridSize = std::end(grid) - std::begin(grid);
int maxColor = *std::max_element(std::begin(grid), std::end(grid));
std::vector<std::vector<int>> colorPos(maxColor);
for (size_t i = 0; i < gridSize; ++i)
colorPos[grid[i] - 1].push_back(i);
for (size_t i = 0; i < colorPos.size(); ++i) {
std::cout << (i + 1) << ": ";
for (int p : colorPos[i])
std::cout << p << ' ';
std::cout << std::endl;
}
return 0;
}
The output:
1: 0 1 8 9 10 17
2: 2 3 4 5 6 7 14 15 22 30
3: 11 12 13 19 20 28
4: 16 24 32 33 40 48 56
5: 18 25 26 27 34 35 36 44
6: 21 23 29 31 37 38 39
7: 41 42 49 50 57 58 59 60
8: 43 45 46 47 51 52 53 54 55 61 62 63
I think you'd be best off using a counting sort, which is a sorting algorithm that works very well for sorting large groups of simple types with many duplicate values in better than O(n log n) time. Here's some sample code, annotated for clarity:
// set up our grid
int grid_raw[] = { 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 3, 3, 3, 2, 2, 4, 1, 5, 3, 3, 6, 2, 6, 4, 5, 5, 5, 3, 6, 2, 6, 4, 4, 5, 5, 5, 6, 6, 6, 4, 7, 7, 8, 5, 8, 8, 8, 4, 7, 7, 8, 8, 8, 8, 8, 4, 7, 7, 7, 7, 8, 8, 8};
// build a vector using our raw list of numbers. This calls the range constructor:
// (number 3) http://www.cplusplus.com/reference/vector/vector/vector/
// The trick to using sizeof is that I don't have to change anything if my grid's
// size changes (sizeof grid_raw gives the number of bytes in the whole array, and
// sizeof *grid_raw gives the number of bytes in one element, so dividing yields
// the number of elements.
std::vector<int> grid(grid_raw, grid_raw + sizeof grid_raw / sizeof *grid_raw);
// count the number of each color. std::map is an associative, key --> value
// container that's good for doing this even if you don't know how many colors
// you have, or what the possible values are. Think of the values in grid as being
// colors, not numbers, i.e. ++buckets[RED], ++buckets[GREEN], etc...
// if no bucket exists for a particular color yet, then it starts at zero (i.e,
// the first access of buckets[MAUVE] will be 0, but it remembers each increment)
std::map<int, int> buckets;
for (vector<int>::iterator i = grid.begin(); i != grid.end(); ++i)
++buckets[*i];
// build a new sorted vector from buckets, which now contains a count of the number
// of occurrences of each color. The list will be built in the order of elements
// in buckets, which will default to the numerical order of the colors (but can
// be customized if desired).
vector<int> sorted;
for (map<int, int>::iterator b = buckets.begin(); b != buckets.end(); ++b)
sorted.insert(sorted.end(), b->second, b->first);
// at this point, sorted = {1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, ...}
Read more about the Counting Sort (includes example python code)
Here's an ideone that demonstrates sorting your grid.
I'm not 100% sure this answers your question... but you included sorting in the title, even though you didn't say anything about it in the body of your question.
Maybe it would be better to use some associative container as for example std::unordered_map or std::multimap.
Here is a demonstrative program
#include <iostream>
#include <map>
int main()
{
int grid[] =
{
1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 3, 3, 3, 2, 2,
4, 1, 5, 3, 3, 6, 2, 6, 4, 5, 5, 5, 3, 6, 2, 6,
4, 4, 5, 5, 5, 6, 6, 6, 4, 7, 7, 8, 5, 8, 8, 8,
4, 7, 7, 8, 8, 8, 8, 8, 4, 7, 7, 7, 7, 8, 8, 8
};
std::multimap<int, int> m;
int i = 0;
for ( int x : grid )
{
m.insert( { x, i++ } );
}
std::multimap<int, int>::size_type n = m.count( 1 );
std::cout << "There are " << n << " elements of color 1:";
auto p = m.equal_range( 1 );
for ( ; p.first != p.second ; ++p.first )
{
std::cout << ' ' << p.first->second;
}
std::cout << std::endl;
return 0;
}
The output
There are 6 elements of color 1: 0 1 8 9 10 17
Or
#include <iostream>
#include <map>
int main()
{
int grid[] =
{
1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 3, 3, 3, 2, 2,
4, 1, 5, 3, 3, 6, 2, 6, 4, 5, 5, 5, 3, 6, 2, 6,
4, 4, 5, 5, 5, 6, 6, 6, 4, 7, 7, 8, 5, 8, 8, 8,
4, 7, 7, 8, 8, 8, 8, 8, 4, 7, 7, 7, 7, 8, 8, 8
};
std::multimap<int, int> m;
int i = 0;
for ( int x : grid )
{
m.insert( { x, i++ } );
}
for ( auto first = m.begin(); first != m.end(); )
{
auto n = m.count( first->first );
std::cout << "There are " << n
<< " elements of color " << first->first << ":";
auto p = m.equal_range( first->first );
for ( ; p.first != p.second ; ++p.first )
{
std::cout << ' ' << p.first->second;
}
std::cout << std::endl;
first = p.first;
}
return 0;
}
the output is
There are 6 elements of color 1: 0 1 8 9 10 17
There are 10 elements of color 2: 2 3 4 5 6 7 14 15 22 30
There are 6 elements of color 3: 11 12 13 19 20 28
There are 7 elements of color 4: 16 24 32 33 40 48 56
There are 8 elements of color 5: 18 25 26 27 34 35 36 44
There are 7 elements of color 6: 21 23 29 31 37 38 39
There are 8 elements of color 7: 41 42 49 50 57 58 59 60
There are 12 elements of color 8: 43 45 46 47 51 52 53 54 55 61 62 63
If you are not forced to use plain arrays, I can propose a map of colors to a vector of positions:
the map is an associative container, that for any color key returns a reference
the referenced used here will be a vector (a kind of dynamic array) containing all the positions.
Your input grid contains color codes:
typedef int colorcode; // For readability, to make diff between counts, offsets, and colorcodes
colorcode grid[] = { 1, 1, /* .....input data as above.... */ ,8 };
const size_t gridsize = sizeof(grid) / sizeof(int);
You would then define the color map:
map<colorcode, vector<int>> colormap;
// ^^^ key ^^^ value maintained for the key
With this approach, your color1[..] would then be replaced by a more dynamic corlormap[1][..]. And it's very easy to fill:
for (int i = 0; i < gridsize; i++)
colormap[grid[i]].push_back(i); // add the new position to the vector returned for the colormap of the color
To verify the result, you may iterate through the map, and for each existing value iterate through the positions:
for (auto x : colormap) { // for each color in the map
cout << "Color" << x.first << " : "; // display the color (key)
for (auto y : x.second) // and iterate through the vector of position
cout << y << " ";
cout << endl;
}
You don't know for sure how many different color codes you have, but you want to store for accodes
I have for example the following matrix B which is stored in COO and CSR format (retrieved from the non-symetric example here). Could you please suggest an efficient c++ way to apply the matlab sum(B,2) function using the coo or csr(or both) storing format? Because it is quit possible to work with large arrays can we do that using parallel programming (omp or CUDA (e.g, thrust))?
Any algorithmic or library based suggestions are highly appreciated.
Thank you!
PS: Code to construct a sparse matrix and get the CSR coordinates can be found for example in the answer of this post.
COO format: CSR format:
row_index col_index value columns row_index value
1 1 1 0 0 1
1 2 -1 1 3 -1
1 3 -3 3 5 -3
2 1 -2 0 8 -2
2 2 5 1 11 5
3 3 4 2 13 4
3 4 6 3 6
3 5 4 4 4
4 1 -4 0 -4
4 3 2 2 2
4 4 7 3 7
5 2 8 1 8
5 5 -5 4 -5
For COO its pretty simple:
struct MatrixEntry {
size_t row;
size_t col;
int value;
};
std::vector<MatrixEntry> matrix = {
{ 1, 1, 1 },
{ 1, 2, -1 },
{ 1, 3, -3 },
{ 2, 1, -2 },
{ 2, 2, 5 },
{ 3, 3, 4 },
{ 3, 4, 6 },
{ 3, 5, 4 },
{ 4, 1, -4 },
{ 4, 3, 2 },
{ 4, 4, 7 },
{ 5, 2, 8 },
{ 5, 5, -5 },
};
std::vector<int> sum(5);
for (const auto& e : matrix) {
sum[e.row-1] += e.value;
}
and for large matrixes you can just split up the for loop into multiple smaller ranges and add the results at the end.
If you only need the sum of each row (and not columwise) CSR is also straight forward (and even more efficient):
std::vector<int> row_idx = { 0, 3, 5, 8, 11, 13 };
std::vector<int> value = { 1, -1, -3, -2, 5, 4, 6, 4, -4, 2, 7, 8, -5 };
std::vector<int> sum(5);
for(size_t i = 0; i < row_idx.size()-1; ++i) {
sum[i] = std::accumulate(value.begin() + row_idx[i], value.begin() + row_idx[i + 1], 0);
}
Again, for parallelism you can simply split up the loop.