Enter the atomic number Z (2 to 6 or 0 to quit): 2
2 1 1 + a(1)b(2)
1 0 0 - a(2)b(1)
# Even Permutations = 1
Enter the atomic number Z (2 to 6 or 0 to quit): 3
6 3 1 + a(1)b(2)c(3)
5 2 0 - a(1)b(3)c(2)
4 2 0 - a(2)b(1)c(3)
3 1 1 + a(2)b(3)c(1)
2 1 1 + a(3)b(1)c(2)
1 0 0 - a(3)b(2)c(1)
# Even Permutations = 3
Enter the atomic number Z (2 to 6 or 0 to quit): 4
24 12 0 + a(1)b(2)c(3)d(4)
23 11 1 - a(1)b(2)c(4)d(3)
22 11 1 - a(1)b(3)c(2)d(4)
21 10 0 + a(1)b(3)c(4)d(2)
20 10 0 + a(1)b(4)c(2)d(3)
19 9 1 - a(1)b(4)c(3)d(2)
18 9 1 - a(2)b(1)c(3)d(4)
17 8 0 + a(2)b(1)c(4)d(3)
16 8 0 + a(2)b(3)c(1)d(4)
15 7 1 - a(2)b(3)c(4)d(1)
14 7 1 - a(2)b(4)c(1)d(3)
13 6 0 + a(2)b(4)c(3)d(1)
12 6 0 + a(3)b(1)c(2)d(4)
11 5 1 - a(3)b(1)c(4)d(2)
10 5 1 - a(3)b(2)c(1)d(4)
9 4 0 + a(3)b(2)c(4)d(1)
8 4 0 + a(3)b(4)c(1)d(2)
7 3 1 - a(3)b(4)c(2)d(1)
6 3 1 - a(4)b(1)c(2)d(3)
5 2 0 + a(4)b(1)c(3)d(2)
4 2 0 + a(4)b(2)c(1)d(3)
3 1 1 - a(4)b(2)c(3)d(1)
2 1 1 - a(4)b(3)c(1)d(2)
1 0 0 + a(4)b(3)c(2)d(1)
# Even Permutations = 12
Enter the atomic number Z (2 to 6 or 0 to quit):
// AOPermutations.cpp : This file contains the 'main' function.
// Program execution begins and ends there.
// Copyright (c) Saturday, March 29, 2025
// by James Pate Williams, Jr., BA, BS, MSwE, PhD
// Signs of the atomic orbitals in a Slater Determinant
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
int main()
{
char alpha[] = { 'a', 'b', 'c', 'd', 'e', 'f' }, line[128] = {};
int factorial[7] = { 1, 1, 2, 6, 24, 120, 720 };
while (true)
{
int col = 0, counter = 0, row = 0, sign = 1, t = 0, Z = 0, zfact = 0;
int numberEven = 0;
std::cout << "Enter the atomic number Z (2 to 6 or 0 to quit): ";
std::cin.getline(line, 127);
std::string str(line);
Z = std::stoi(str);
if (Z == 0)
{
break;
}
if (Z < 2 || Z > 6)
{
std::cout << "Illegal Z, please try again" << std::endl;
continue;
}
zfact = factorial[Z];
std::vector<char> orb(Z);
std::vector<int> tmp(Z), vec(Z);
for (int i = 0; i < Z; i++)
{
orb[i] = alpha[i];
vec[i] = i + 1;
}
do
{
for (int i = 0; i < (int)vec.size(); i++)
{
tmp[i] = vec[i];
}
t = 0;
do
{
t++;
} while (std::next_permutation(tmp.begin(), tmp.end()));
std::cout << t << '\t' << t / 2 << '\t';
std::cout << (t / 2 & 1) << '\t';
if (Z == 2 || Z == 3)
{
if ((t / 2 & 1) == 0)
{
std::cout << "-\t";
}
else
{
std::cout << "+\t";
numberEven++;
}
}
else
{
if ((t / 2 & 1) == 1)
{
std::cout << "-\t";
}
else
{
std::cout << "+\t";
numberEven++;
}
}
for (int i = 0; i < Z; i++)
{
std::cout << orb[i] << '(' << vec[i] << ')';
}
row++;
std::cout << std::endl;
if (zfact != 2 && row == zfact)
{
std::cout << std::endl;
break;
}
row %= Z;
} while (std::next_permutation(vec.begin(), vec.end()));
std::cout << "# Even Permutations = ";
std::cout << numberEven << std::endl;
}
return 0;
}
Numerical Explorations 