I am working on the second part of an assignment which asks me to reorder a matrix such that each row is in monotonically increasing order and so that the first element of each row is monotonically increasing. If two rows have the same initial value, the rows should be ordered by the second element in the row. If those are both the same, it should be the third element, continuing through the last element.
I have written a bubble sort that works fine for the first part (reordering each row). I have written a bubble sort for the second part (making sure that the first element of each row is monotonically increasing). However, I am running into an infinite loop and I do not understand why.
I do understand that the issue is that my "inorder" variable is not eventually getting set to true (which would end the while loop). However, I do not understand why inorder is not getting set to true. My logic is the following: once the following code has swapped rows to the point that the rows are all in order, we will pass through the while loop one more time (and inorder will get set to true), which will cause the while loop to end. I am stumped as to why this isn't happening.
inorder = .false.
loopA: do while ( .not. inorder ) !While the rows are not ordered
inorder = .true.
loopB: do i = 1, rows-1 !Iterate through the first column of the array
if (arr(i,1)>arr(i+1,1)) then !If we find a row that is out of order
inorder = .false.
tempArr = arr(i+1,:) !Swap the corresponding rows
arr(i+1,:) = arr(i,:)
arr(i,:) = tempArr
end if
if (arr(i,1)==arr(i+1,1)) then !The first elements of the rows are the same
loopC: do j=2, cols !Iterate through the rest of the row to find the first element that is not the same
if (arr(i,j)>arr(i+1,j)) then !Found elements that are not the same and that are out of order
inorder = .false.
tempArr = arr(i+1,:) !Swap the corresponding rows
arr(i+1,:) = arr(i,:)
arr(i,:) = tempArr
end if
end do loopC
end if
end do loopB
end do loopA
Example input:
6 3 9 23 80
7 54 78 87 87
83 5 67 8 23
102 1 67 54 34
78 3 45 67 28
14 33 24 34 9
Example (correct) output (that my code is not generating):
1 34 54 67 102
3 6 9 23 80
3 28 45 67 78
5 8 23 67 83
7 54 78 87 87
9 14 24 33 34
It is also possible that staring at this for hours has made me miss something stupid, so I appreciate any pointers.
When you get to compare rows where the first element is identical, you then go through the whole array and compare every single item.
So if you have two arrays like this:
1 5 3
1 2 4
Then the first element is the same, it enters the second part of your code.
In second place, 5>2, so it swaps it:
1 2 4
1 5 3
But then it doesn't stop. In third place, 4>3, so it swaps it back
1 5 3
1 2 4
And now you're back to where you were.
Cheers
Related
Some information about the overall project:
I have to find if specific nodes remain connected if i start removing the lowest width edges from a graph. I have a struct solve, which has a member array called connected. In a method of this struct , FindConnections I go over some of the edges, from the Kth till the last and see which nodes are connected. The way I keep track of the connected nodes is to have an array that for each node points to the lowest id node it is connected, with the lowest pointing to itself
for example
if nodes 2 5 6 12 are directly connected
connected[2] =connected[5] =connected[6] =connected[12] = 2
so now if 12 and 23 are connected (and 12 is the lowest connection of 23)
connected [23] = 12 and connected[connected[23]] = 2 so i can reach 2 from 23 with recursion
My problem is that after finishing modifying the connected array inside FindConnections, some of the changes are preserved while other not
Here is the code:
void FindConnections(int index)
{
for (int temp, i = index; i < NumberOfPortals; i++)
{
temp = min(first[i], second[i]); // the nodes which edge i connects
connected[first[i]] = temp;
connected[second[i]] = temp;
}
}
which is called by
void seeAllConnections() // this function is for visualization it will not be included
{
for (int i = NumberOfPortals - 1; i >= 0; --i)
{
printf("Only using %d Portals:\n", NumberOfPortals - i);
FindConnections(i);
seeconnected(); // prints connected array
for (int i = 0; i < NumberOfUniverses; i++) //resets connected array
{
connected[i] = i;
}
}
}
In the two first iterations of the for loop in seeAllConnections, everything is good, the edges that are examined are
first second width(irrelevant for now)
6 7 255
26 2 111
11 7 36
in the beginning everyone is connected with himself
in the first one we get the output
(I am placing ** around the values that are changed and !! around the one that was supposed to change but didn't , just so you can see it better, the program prints just the numbers)
Only using 1 Portals:
connected are:
0 1 2 3 4 5 6 7 8 9 10 *7* 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
and we can see that connected[11] = 7 just like we wanted to
in the second one
Only using 2 Portals:
connected are:
0 1 2 3 4 5 6 7 8 9 10 *7* 12 13 14 15 16 17 18 19 20 21 22 23 24 25 *2* 27 28 29
connected[11] =7 just like before
connected[26] = 2 just like we wanted
in the third one
Only using 3 Portals:
connected are:
0 1 2 3 4 5 6 !7! 8 9 10 *7* 12 13 14 15 16 17 18 19 20 21 22 23 24 25 *2* 27 28 29
connected [7] = 7 , not 6
moreover, when i use gdb, inside the FindConnections loop, connected[7] = 6 like we wanted
(gdb) print first[i]
$10 = 6
(gdb) print second[i]
$11 = 7
(gdb) print connected[first[i]]
$12 = 6
(gdb) print connected[second[i]]
$13 = 6
but when it exits the function and returns to seeAllConnected
connected[7] = 7
What Am I doing wrong? how can the first two changes be preserved form the same function in the same struct in the same loop, while the second one isn't?
Also after every time I call FindConnections I reset the array to it's original values, so the changes couldn't have been preserved from before
Thank you in advance
I found out what was wrong, as it was a reverse iteration connected[7] got overwritten.
I have a 2d Array of Vectors declared as
Vector<int> v1[10][11];
and I want to accessing the elements inside the array. Would I treat the 2d array of vectors as a 3d array and access elements like this:
v1[9][10][0];
to access the first element of the last member of the array?
std::vector allows the [] operator, but I'm pretty sure your [9][10][0] is going to be out of bounds - you don't have anything there.
Assuming that you're opting for an array something like
[0 1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17 18 19
...
100 101 102 103 104 105 106 107 108 109]
then the last element is going to be v1[9][10], corresponding to 109 in the picture.
If you want each element to be an array, I think you'll need
std::vector<int[sizeofarray]> v1[10][11];
//or just vector<int[size]> v1[10][11]; if using namespace std;
I've made a program to create the pascal's triangle. the program takes number of rows as input and displays the triangle on the console. I've used the setw() function to set the distance between numbers. it's of for unit single digits but when the numbers get greater than 10,the width is not being adjusted properly,right now I've :
if(P<10){
std::cout << P ;
std::cout <<std::setw(2);
}
if(P>=10){
std::cout<<std::setw(3) << P ;
std::cout<<std::setw(2);
}
here's the ouput from the console:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84126126 84 36 9 110
I want it to appear like a proper triangle,Could someone help me out please???
If you read e.g. this reference of std::setw you will see
The width property of the stream will be reset to zero (meaning "unspecified") if any of the following functions are called
And then goes on to list basically all output operators.
This means that when you do
std::cout <<std::setw(2);
the width will only be set for the next output operation. If you do any kind of output after that the width will be reset to zero.
You get 10 numbers that you have to split into two lists where the sum of numbers in the lists have the smallest difference possible.
so let's say you get:
10 29 59 39 20 17 29 48 33 45
how would you sort this into two lists where the difference in the sum of the lists is as small as possible
so in this case, the answer (i think) would be:
59 48 29 17 10 = 163
45 39 33 29 20 = 166
I'm using mIRC script as the language but perl or C++ is just as good for me.
edit: actually there can be multiple answers such as in this scenario, it could also be:
59 48 29 20 10 = 166
45 39 33 29 17 = 163
to me, it doesn't matter so long as the end result is that the difference of the sum of the lists is as small as possible
edit 2: each list must contain 5 numbers.
What you have listed is exactly the partition problem (for more details look at http://en.wikipedia.org/wiki/Partition_problem).
The point is that this is a NP-complete problem, therefore it does not exist a program able to solve any instance of this problem (i.e. with a bigger amount of numbers).
But if your problem is always with only ten numbers to divide into two lists of exactly five items each, then it becomes feasible, also to try naively all possible solutions, since they are only p^N, where p=2 is the number of partitions, and N=10 is the number of integers, thus only 2^10=1024 combinations, and each takes only O(N) to be verified (i.e. compute the difference).
Otherwise you can implement the greedy algorithm described in the Wikipedia page, it is simple to implement but there is no guarantee of optimality, in fact you can see this implementation in Java:
static void partition() {
int[] set = {10, 29, 59, 39, 20, 17, 29, 48, 33, 45}; // array of data
Arrays.sort(set); // sort data in descending order
ArrayList<Integer> A = new ArrayList<Integer>(5); //first list
ArrayList<Integer> B = new ArrayList<Integer>(5); //second list
String stringA=new String(); //only to print result
String stringB=new String(); //only to print result
int sumA = 0; //sum of items in A
int sumB = 0; //sum of items in B
for (int i : set) {
if (sumA <= sumB) {
A.add(i); //add item to first list
sumA+=i; //update sum of first list
stringA+=" "+i;
} else {
B.add(i); //add item to second list
sumB+=i; //update sum of second list
stringB+=" "+i;
}
}
System.out.println("First list:" + stringA + " = " + sumA);
System.out.println("Second list:"+ stringB+ " = " + sumB);
System.out.println("Difference (first-second):" + (sumA-sumB));
}
It does not return a good result:
First list: 10 20 29 39 48 = 146
Second list: 17 29 33 45 59 = 183
Difference (first-second):-37
I'm trying to implement an ACO for 01MKP. My input values are from the OR-Library mknap1.txt. According to my algorithm, first I choose an item randomly. then i calculate the probabilities for all other items on the construction graph. the probability equation depends on pheremon level and the heuristic information.
p[i]=(tau[i]*n[i]/Σ(tau[i]*n[i]).
my pheremon matrix's cells have a constant value at initial (0.2). for this reason when i try to find the next item to go, pheremon matrix is becomes ineffective because of 0.2. so, my probability function determines the next item to go, checking the heuristic information. As you know, the heuristic information equation is
n[i]=profit[i]/Ravg.
(Ravg is the average of the resource constraints). for this reason my prob. functions chooses the item which has biggest profit value. (Lets say at first iteration my algorithm selected an item randomly which has 600 profit. then at the second iteration, chooses the 2400 profit value. But, in OR-Library, the item which has 2400 profit value causes the resource violation. Whatever I do, the second chosen is being the item which has 2400 profit.
is there anything wrong my algorithm? I hope ppl who know somethings about ACO, should help me. Thanks in advance.
Input values:
6 10 3800//no of items (n) / no of resources (m) // the optimal value
100 600 1200 2400 500 2000//profits of items (6)
8 12 13 64 22 41//resource constraints matrix (m*n)
8 12 13 75 22 41
3 6 4 18 6 4
5 10 8 32 6 12
5 13 8 42 6 20
5 13 8 48 6 20
0 0 0 0 8 0
3 0 4 0 8 0
3 2 4 0 8 4
3 2 4 8 8 4
80 96 20 36 44 48 10 18 22 24//resource capacities.
My algorithm:
for i=0 to max_ant
for j=0; to item_number
if j==0
{
item=rand()%n
ant[i].value+=profit[item]
ant[i].visited[j]=item
}
else
{
calculate probabilities for all the other items in P[0..n]
find the biggest P value.
item=biggest P's item.
check if it is in visited list
check if it causes resource constraint.
if everthing is ok:
ant[i].value+=profit[item]
ant[i].visited[j]=item
}//end of else
}//next j
update pheremon matrix => tau[a][b]=rou*tau[a][b]+deltaTou
}//next i