i have a quad (2D array) which is composed of numbers which ranges from 0 to 255
and the most frequent value of the array ( in my case 2) is the background value
i have to put all the values of the array except the background value ( i have to ignore all cases that contains a 2) in a 1D array with this arrangement
the line,the column,the value,the line,the next column,the next value
for example i have
{2,3,2,2},
{2,2,2,2},
in the 1D array it will be like { 1,2,3,}
i added spaces to make it more readable
here is my array
int image1[MAXL][MAXC]=
{
{2,3,2,2},
{2,2,2,2},
{2,255,2,2},
{255,2,2,2},
{2,255,2,2}
};
and the loop
for (int i = 0; i<nbc;i++)
{
for (int j=0; j<nbl;j++)
{
if (image1[j][i]==BackColor)
{
}
else
}
}
nbc and nbl are respectively the number of columns and lines
thanks you for your help
EDIT : i completely failed my example i didn't ignored the 2, should be fine now
A simple approach using std::vector
std::vector<int> result;
for (int i = 0; i<MAXL;i++)
{
for (int j=0; j<MAXC;j++)
{
if (image1[i][j] != BackColor) // Notice !=
{
result.push_back(i+1);
result.push_back(j+1);
result.push_back(image1[i][j]);
}
}
}
for (auto x : result)
{
std::cout << x << " ";
}
std::cout << endl;
Output:
1 2 3 3 2 255 4 1 255 5 2 255
http://ideone.com/yJawu5
Related
Recently I have been trying to do a problem that requires me to find all the different combinations with selecting only 1 element from each row. For example, I'm inputting n rows with 2 strings per row. However, I only want to find all the different combinations where I choose 1 string from each row.
Example:
Input:
3
alex bob
straw mat
eat drink
Example combination:
alex straw drink
This results in 2^n combinations, which in this case would be 2^3 = 8 combinations. However, if I was to use n for loops to find the combinations
e.g.
#include <iostream>
#include <cstring>
#include <string>
using namespace std;
int n;
int main(int argc, char ** argv) {
cin >> n; //rows of words
string words[n][2]; //the words with 2 words per row
for (int i = 0; i < n; i++) {
for (int j = 0; j < 2; j++) {
cin >> words[i][j]; //input of words
}
}
//finding all possible combinations
for (int i =0; i<n; i++){
for (int j=0; j<2; j++){
for (int x=0; x<2; x++){
//and so on per n
}
}
}
return 0;
}
this would take n for loops to find out all the combinations of the array with only taking one item from each row. What would be the best and simplest approach to finding all different combinations with size n as I would take 1 string out of the two in each row? Thanks.
You can do recursion.
Assuming C++11, something like this maybe (didn't try to compile this though):
// finding all possible combinations
std::vector<std::vector<std::string>> combinations;
const auto processLine = [&](const std::vector<std::string>& currentCombination, int line) {
std::vector<std::string> combination0 = currentCombination;
std::vector<std::string> combination1 = currentCombination;
combination0.push_back(words[line][0]);
combination1.push_back(words[line][1]);
if (line + 1 < n) {
// process next line
processLine(combination0, line + 1);
processLine(combination1, line + 1);
}
else {
// last line - keep the result
combinations.push_back(combination0);
combinations.push_back(combination1);
}
};
std::vector<std::string> empty;
processLine(empty, 0);
// print results
for (const auto& combination : combinations) {
for (const auto& word : combination) {
std::cout << word << " ";
}
std::cout << std::endl;
}
A very simple solution for a setting where you have always 2 elements per row would be to use datatype integer and interpret each bit as a decision for the first or the second column in the respective row; then simply count from 0 to 2^n - 1 in order to get all combinations.
Applied to your example this would look as follows:
int bits meaning
0 000 alex,straw,eat
1 001 alex,straw,drink
2 010 alex,mat,eat
3 011 alex,mat,dring
4 100 bob,straw,eat
5 101 bob,straw,drink
6 110 bob,mat,eat
7 111 bob,mat,drink
For any of the given integer values 0..7, use bit shift operators or &-bitmask to map each bit to a column index:
void getCombinationRepresentedByIntValue(vector<string>& combination, int value) {
int mask = 1;
for (int i=n-1; i>=0; i--) {
if (value & mask)
combination.push_back(words[i][1]);
else
combination.push_back(words[i][0]);
mask = mask << 1;
}
}
That seems to answer your question :
int ct[n]; // count of the number of pass
int current = 0; // index of the current word (n)
/* while not all combinaison have been exploited */
while (current >= 0)
{
cout << words[current][ct[current]]; /* <<<<< can be used another way*/
/* down to the next word */
current ++; // to get the next word
if (current >=n) { // at the end of the list
cout << " ";
current--; // restore last
ct[current]++; // increment number of time we passed
/* find the previous not completely exploited */
while (current >= 0 && ct[current]> 1) /* <<< change 1 to any number of words per line */
{
ct[current] = 0;
current--;
if (current >= 0) ct[current]++;
}
if (current > 0 ) current = 0;
}
}
With your example :
Input :
3
alex bob
straw mat
eat drink
output :
alexstraweat
alexstrawdrink
alexmateat
alexmatdrink
bobstraweat
bobstrawdrink
bobmateat
bobmatdrink
hope it helps !
I need a different version of permutations for my code. I could able to achieve what I want but it is not generic enough. my algorithm keeps going bigger along with my requirements. But that should not be.
This is not a home work for any one, I need it for one my critical projects, wondering if any pre-defined algorithms available from boost or any other.
Below is the standard version of next_permutation using c++.
// next_permutation example
#include <iostream> // std::cout
#include <algorithm> // std::next_permutation
int main ()
{
int myints[] = {1,2,3};
do
{
std::cout << myints[0] << ' ' << myints[1] << ' ' << myints[2] << '\n';
} while ( std::next_permutation(myints,myints+3) );
return 0;
}
That gives below output :
1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1
But my requirement is :- Let's say I have 1 to 9 numbers :
1,2,3,4,5,6,7,8,9
And I need a variable length of permutations and in only ASCENDING order and with out DUPLICATES.
Let's say i need 3 digit length of permutations then i need output as below.
123
124
125
.
.
.
128
129
134 // After 129 next one should be exactly 134
135 // ascending order mandatory
136
.
.
.
148
149
156 // exactly 156 after 149, ascending order mandatory
.
.
.
489 // exactly 567 after 489, because after 3rd digit 9, 2nd digit
567 // will be increased to 49? , so there is no possibility for
. // 3rd digit, so first digit gets incremented to 5 then 6 then
. // 7, in ascending order.
.
.
.
789 // and this should be the last set I need.
My list may contain upto couple of hundred's of numbers and variable length can be 1 to up to Size of the list.
My own algorithm is working for specific variable length, and a specific size, when they both changes, i need to write huge code. so, looking for a generic one.
I am not even sure if this is called as Permutations or there is a different name available for this kind of math/logic.
Thanks in advance.
musk's
Formally, you want to generate all m-combinations of the set [0;n-1].
#include <iostream>
#include <vector>
bool first_combination (std::vector<int> &v, int m, int n)
{
if ((m < 0) || (m > n)) {
return false;
}
v.clear ();
v.resize (m);
for (int i = 0; i < m; i++) {
v[i] = i;
}
return true;
}
bool next_combination (std::vector<int> &v, int m, int n)
{
for (int i = m - 1; i >= 0; i--) {
if (v[i] + m - i < n) {
v[i]++;
for (int j = i + 1; j < m; j++) {
v[j] = v[j - 1] + 1;
}
return true;
}
}
return false;
}
void print_combination (const std::vector<int> &v)
{
for (size_t i = 0; i < v.size(); i++) {
std::cout << v[i] << ' ';
}
std::cout << '\n';
}
int main ()
{
const int m = 3;
const int n = 5;
std::vector<int> v;
if (first_combination (v, m, n)) {
do {
print_combination (v);
} while (next_combination (v, m, n));
}
}
You can use this code and the linked article as inspiration.
This task can be done with a simple iterative algorithm. Just increment the first element that can be incremented and rescale the elements before it until there is no element to be incremented.
int a[] = {0,1,2,3,4,5,6,7,8,9}; // elements: must be ascending in this case
int n = sizeof(a)/sizeof(int);
int digits = 7; // number of elements you want to choose
vector<int> indexes; // creating the first combination
for ( int i=digits-1;i>=0;--i ){
indexes.push_back(i);
}
while (1){
/// printing the current combination
for ( int i=indexes.size()-1;i>=0;--i ){
cout << a[indexes[i]] ;
} cout << endl;
///
int i = 0;
while ( i < indexes.size() && indexes[i] == n-1-i ) // finding the first element
++i; // that can be incremented
if ( i==indexes.size() ) // if no element can be incremented, we are done
break;
indexes[i]++; // increment the first element
for ( int j=0;j<i;++j ){ // rescale elements before it to first combination
indexes[j] = indexes[i]+(i-j);
}
}
Output:
0123456
0123457
0123458
0123459
0123467
0123468
0123469
0123478
0123479
0123489
0123567
0123568
0123569
0123578
0123579
0123589
0123678
0123679
0123689
0123789
0124567
0124568
0124569
0124578
0124579
0124589
0124678
0124679
0124689
0124789
0125678
0125679
0125689
0125789
0126789
0134567
0134568
0134569
0134578
0134579
0134589
0134678
0134679
0134689
0134789
0135678
0135679
0135689
0135789
0136789
0145678
0145679
0145689
0145789
0146789
0156789
0234567
0234568
0234569
0234578
0234579
0234589
0234678
0234679
0234689
0234789
0235678
0235679
0235689
0235789
0236789
0245678
0245679
0245689
0245789
0246789
0256789
0345678
0345679
0345689
0345789
0346789
0356789
0456789
1234567
1234568
1234569
1234578
1234579
1234589
1234678
1234679
1234689
1234789
1235678
1235679
1235689
1235789
1236789
1245678
1245679
1245689
1245789
1246789
1256789
1345678
1345679
1345689
1345789
1346789
1356789
1456789
2345678
2345679
2345689
2345789
2346789
2356789
2456789
3456789
Geeting!
I'm working on a console application which read in numbers put them in an array and the output tells you those numbers position in the array where the number is larger then 1000.
so if you have an array like this :
int be[] = { 1001, 1001, 1001, 101, 1001 };
The output should tell you that from the first to the third position you have 1000+ numbers and from the fifth to the last position you have 1000+ numbers. So the outcome would be like this:
1 3 5 6
I solved it with for and if like this:
for (int i=1;i<=5;i++)
{
if (be[i]>=1000)
{
cout<<i<<" ";
for (int j=1;j<5-1;j++)
{
if (be[i+j]<=1000)
{
cout<<i+j-1<<" ";
break;
}
}
}
}
And the main problem is that the outcome will be this:
1 3 2 3 3 3 5 6
So after the break i want the first for loop to continue where the secont one ended but I don't exactly know how to do it. If there's anyone who got the basic idea from this vague explanation and can give me some hint or tips I would be most grateful!
You only need one loop and a bunch of checks.
Here's some pseudo code:
loop through elements
if number > 1000 and not started
print index
else if number <=1000
started = false
print index-1
if started = true
print index - 1 // last element of array
The only comment: you have no 6th position in your array, that is why I output 5 5 lik from 5th element to 5th element
http://ideone.com/k3gF03
int main(void) {
int be[] = { 1001, 1001, 1001, 101, 1001 };
int beSize = (sizeof(be)/sizeof(*be));
bool fl = false;
int i=0;
for (i;i<beSize;i++)
{
if (be[i]>=1000)
{
if (!fl) {
fl=true;
cout<<i+1<<" ";
}
} else {
if (fl) {
fl=false;
cout<<i<<" ";
}
}
}
if (fl) { cout<<i<<" "; }
return 0;
}
If I have understood you correctly you need something like the following
#include <iostream>
int main()
{
const size_t low_bound = 1000;
int be[] = { 1001, 1001, 1001, 101, 1001 };
const size_t N = sizeof( be ) / sizeof( *be );
for ( int x : be ) std::cout << x << ' ';
std::cout << std::endl;
for ( size_t i = 0; i < N; i++ )
{
if ( be[i] >= low_bound )
{
size_t j = i + 1;
while ( j < N && be[j] >= low_bound ) j++;
--j;
std::cout << i + 1 << "-" << j + 1 << " ";
i = j;
}
}
std::cout << std::endl;
}
The program output is
1001 1001 1001 101 1001
1-3 5-5
Take into account that array indices start at 0.
To have the first loop continue where the second ended just do that
int j = 1;
for (int i=j;i<=5;i++)
{
if (be[i]>=1000)
{
cout<<i<<" ";
for (;j<4 and be[i+j]>1000 ;j++);
if(j < 5 )
{
cout<<i+j-1<<" ";
}
}
}
have not really checked if it does what you want?
Very simple and not polite code. Sorry. Array in c++ begin with index 0. So there is a first dummy element.
int be[] = { 99, 1001, 1001, 1001, 101, 1001 }; // 99 is dummy
bool in=false;
int j;
int i;
for (i=1;i<=5;i++)
{
if (be[i]>=1000 && in==false)
{
in=true;
cout << i << " ";
}
if (be[i]<1000 && in==true)
{
in=false;
cout << i-1 << " ";
}
}
if (be[i]<1000 && in==true)
cout << i << " "; // or i-1 if you prefer the last one
I can't figure out the logic behind this one... here's what I have so far:
#include <iostream>
using namespace std;
int thearray[4][4];
int NbPos = 4;
int main() {
int i2;
int q2;
for(int i = 1; i < 4; i++) {
for(int q = 1; q < 4; q++) {
for(int c = 0; c < NbPos; c++) {
thearray[i][q] = c;
}
}
}
}
This is filling the array up to the end is still:
3 3 3
3 3 3
3 3 3
but it's doing so without hitting anywhere near every possible combination.
Ideally once it gets to:
0 0 0
0 0 0
0 0 3
the next step SHOULD be:
0 0 0
0 0 0
0 1 0
so it hits a TON of combinations. Any ideas on how to make it hit them all? I'm stumped on the logic!
with the way you're iterating over this, a 1-dimensional array would make the looping simpler. you can still mentally treat it to have rows and columns, however they are just layed out end-to-end in the code.
you could try something like this; however if you want it in a 2D format specifically that challenge is left to you ;)
#include <iostream>
using namespace std;
#define rows 4
#define columns 4
int main() {
int thearray[rows * columns] = {0};
int NbPos = 4;
int lastPos = rows * columns - 1;
while (true) {
thearray[lastPos]++;
int pos = lastPos;
while (thearray[pos] == NbPos and pos >= 1) {
thearray[pos - 1]++;
thearray[pos] = 0;
pos--;
}
bool finished = true;
for (int i = 0; i < rows * columns; i++) {
if (thearray[i] != NbPos - 1) {
finished = false;
}
}
if (finished) {
break;
}
}
for (int i = 0; i < rows * columns; i++) {
std::cout << thearray[i] << " ";
if (i % rows == rows - 1) {
cout << endl; // makes it look like a 2D array
}
}
}
It makes sense to have the final form as all 3s , since you loop every element of the array and you assign it at the end with 3 .
So the next element will only take into account the combination with the final value of the previous element (which will be 3).
Thinking in math terms, your complexity is N^3 so to speak (actually is N^2 * 4 , but since your N is 3 ...).
Your approach is wrong, since you want to find permutations, which are defined by a factorial function , not a polinomial function.
The necessary complexity for the output doesn't match the complexity of your algorithm (your algorithm is incredbily fast for the amount of output needed).
What you are looking for is backtracking (backtacking will match the complexity needed for your output).
The recursion function should be something like this (thinking on a 1D array, with 9 elements):
RecursiveGeneratePermutations(int* curArray, int curIdx)
{
if (curIDX==9)
{
for (int i=0; i<9;i++)
{
// write the array
}
} else {
curArray[curIdx]=0;
RecursiveGeneratePermutations(curIdx+1);
curArray[curIdx]=1;
RecursiveGeneratePermutations(curIdx+1);
curArray[curIdx]=2;
RecursiveGeneratePermutations(curIdx+1);
curArray[curIdx]=3;
RecursiveGeneratePermutations(curIdx+1);
}
}
Now you only need to call the function for the index 0 :
RecursiveGeneratePermutations(arrayPtr,0);
Then wait...allot :).
I've been working on a brute force algorithm to generate all permutations of a given set. Eventually, I want to feed each of these permutations into a nxn matrix to test if it is a valid magic square or not.
--I KNOW THAT THERE IS A WAY TO GENERATE A MAGIC SQUARE EASILY--
That is not what I want to do, though. I'm focusing on the brute force aspect of it.
For a set of 3 elements, it works wonderfully. However, once I use 4 or more elements, I lose out on a few permutations. Just from looking at the output of 4, I am missing 7 permutations.
#include <stdio.h>
#include <iostream>
using namespace std;
//ms = magic square
//n = size
void perm(int ms[], int n) {
int pivot = 0;
int index = 0;
int pivBit = 1;
int fin = 0;
int hold = 0;
//While we are not finished
while (fin == 0) {
//Incriment the index
++index;
if (index >= n) {
index = 0;
}
//if index is equal to the pivot
if (index == pivot) {
//Is this the first time visiting the pivot?
if (pivBit == 0) {
//Are we at the beginning again?
if (index == 0 && pivot == 0)
{
fin = 1;
}
pivBit = 1;
++index;
}
//Second time visiting?
else {
pivBit = 0;
++pivot;
if (pivot >= n) {
pivot = 0;
}
}
}
//If we are out of bounds
if (index >= n) {
index = 0;
}
//swap
hold = ms[index];
ms[index] = ms[pivot];
ms[pivot] = hold;
for (int i = 0; i < n; ++i) {
cout << ms[i];
if (i < n - 1) {
cout << ", ";
}
else {
cout << endl;
}
}
}
}
int main() {
cout << "Are you ready to brute force, my brother?" << endl;
//Set
int magicsquare[] = { 1, 2, 3, 4};
int size = 4;
perm(magicsquare, size);
getchar();
return 0;
}
My output is:
2 1 3 4
3 1 2 4
4 1 2 3
1 4 2 3
1 2 4 3
1 3 4 2
3 1 4 2
3 4 1 2
3 4 2 1
2 4 3 1
2 3 4 1
2 3 1 4
4 3 1 2
4 2 1 3
4 2 3 1
1 2 3 4
2 1 3 4
Looking at it, I can already see that I am missing both 1 4 3 2 and 1 3 2 4.
Where've I gone wrong in my algorithm?
The wiki article on permutation includes a common algorithm used to produce all permutations in lexicographic order, starting with an array of sequentially increasing integers, ending with that array reversed. wiki next permutation.
If dealing with an array of objects, you can generate an array of indices 0 through n-1 and use next permutation on the indices to produce all permutations of the array of objects.
You can also do a web search for next permutation to find similar algorithms. The recursive ones produce all permutations, but not in lexicographic order.
The simplest way to generate all permutations is recursive. For each i, swap the i'th element to the 0 position. Then recursively find all permutations of of the remaining array.
int buf[1000], n; // better to wrap these in a class...
void permute(int *a, int a_len) {
if (a_len == 1) {
for (int i = 0; i < n; i++) printf("%d ", buf[i]);
printf("\n");
} else {
for (int i = 0; i < a_len; i++) {
swap(a, 0, i);
permute(a + 1, a_len - 1);
swap(a, 0, i);
}
}
}
void run(int buf_len) {
for (int i = 0; i < buf_len; i++) buf[i] = i + 1;
n = buf_len;
permute(buf, buf_len);
}
This assumes no repeated elements in the original array. It's not to hard to have it take repeated elements into account.