Simple Java programs I found on an old hard drive.
Language: Java
Start: 2013
A few old simple Java programs I found on an old hard drive. Among them, two sorting algorithms and two intriguing calculations whose original source I have forgotten.
The following calculation:
$$a_i = i \cdot i \cdot i$$
is performed in two different ways:
private static void test1()
{
for (int i = 0; i < a.length; i++)
a[i] = i*i*i;
}
private static void test2()
{
int i = 0;
int c = 0;
int k = 1;
int m = 6;
while (i < a.length)
{
a[i] = c;
c = c + k;
k = k + m;
m = m + 6;
i = i + 1;
}
}The second one is not straightforward but it can be proven to be the same:
$$a_i = c_{i - 1}$$
$$c_i = \sum_{j = 0}^i k_j$$
$$k_i = 1 + \sum_{j = 1}^i m_{j - 1}$$
$$m_i = \sum_{j = 0}^i 6 = (i + 1) \cdot 6$$
where c1, k1 and m1 are the values at the end of the first loop where i is equal to 1. We can now write:
$$a_i = \sum_{j = 0}^{i - 1} k_j = \sum_{j = 0}^{i - 1} (1 + \sum_{l = 1}^j m_{l-1}) = \sum_{j = 0}^{i - 1} (1 + \sum_{l = 1}^j l \cdot 6) = \sum_{j = 0}^{i - 1} 1 + \sum_{j = 0}^{i - 1} \sum_{l = 1}^j l \cdot 6 $$
and since:
$$\sum_{j = 0}^{i - 1} 1 = i$$
$$\sum_{l = 1}^j l \cdot 6 = 6 \cdot \sum_{l = 1}^j l = 6 \cdot \frac{j^2 + j}{2}$$
we can write:
$$a_i = i + 6 \cdot \sum_{j = 0}^{i - 1} \frac{j^2 + j}{2} = i + 6 \cdot \sum_{j = 1}^{i - 1} \frac{j^2 + j}{2} = i + 3 \cdot \sum_{j = 1}^{i - 1} (j^2 + j)$$
we can simplify further the previous summation as follows:
$$\sum_{j = 1}^{i - 1} (j^2 + j) = \sum_{j = 1}^{i - 1} j^2 + \sum_{j = 1}^{i - 1} j =$$
$$= \frac{(i - 1)(i - 1 + 1)(2 (i - 1) + 1)}{6} + \frac{(i - 1)(i - 1 + 1)}{2} =$$
$$= \frac{i(i - 1)(2i - 1) + 3i(i - 1)}{6} = \frac{1}{3} (i^3 + i^2 - i^2 - i) = \frac{1}{3} (i^3 - i)$$
we can plug the simplified summation in the previous formula:
$$a_i = \bcancel{i} + \cancel{3} \frac{1}{\cancel{3}}(i^3 - \bcancel{i}) = i^3$$