"Unique Object Codes for Inter-Group Connections"

Language: C++
Author: Jerry Obrian
Comments: 0
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Solution:

To ensure the uniqueness of every code representing an object as explained, you need to think of the connections forming a bipartite graph where you have group 1 nodes connected to group 2 nodes. In this scenario, you can think of the objects within the same group as being equivalent, and you need to guarantee that by assigning unique codes to these objects.

Here's a suggested approach to guarantee uniqueness:

1. **Create a Unique Character Mapping**: For each object `x` in group 1, you can assign a unique character `C_x` to it. Similarly, for each object `y` in group 2, you can assign a unique character `C_y` to it.

2. **Character Building Logic**:
- For each connection [a, b]:
- Set `codes[a][b] = C_b` and `codes[b][b] = C_b`.

3. **Filling up the Rest of the Code**:
- For each position `pos` greater than the number of connections, assign a unique character to ensure uniqueness within each group.

Here is an updated version of your code reflecting these changes:

cpp
#include
#include
#include

using namespace std;

int main()
{
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);

int n, p, M;
cin >> n >> p >> M;

// Create a unique character mapping for each object
vector group1Char(n + 1);
vector group2Char(n + 1);
for (int i = 1; i <= n; ++i) {
group1Char[i] = 'A' + (i - 1);
group2Char[i] = 'A' + (i - 1);
}

set> connections;
for (int i = 0; i < p; ++i)
{
int a, b;
cin >> a >> b;

connections.insert({a, b});
}

// Initialize codes with a unique character for each object
vector codes(2 * n + 1, string(M, ' '));
for (int a = 1; a <= n; ++a)
{
codes[a][b] = group1Char[a]; // Assign unique character for the object in group 1
codes[a + n][b] = group2Char[a]; // Assign unique character for the object in group 2
}

// Set characters in codes to create required connections
int pos = n + 1;
for (auto [a, b] : connections)
{
codes[a][pos] = group2Char[b]; // Connect object a with b in group 2
codes[b + n][pos] = group2Char[b]; // Connect object b with a in group 1
}

// Assign unique characters for the rest of the positions
for (int i = pos + 1; i <= M; ++i)
{
codes[i][i] = 'A' + n + 1 + (i - pos); // Assign unique characters for unmatched positions
}

// Output the codes
for (const string& code : codes)
{
cout << code << endl;
}

return 0;
}
This logic should ensure the uniqueness of every code while minimizing the number of characters needed to create the required connections.


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