**Tags:** math ppc

Rating: 5.0

# Programming01, PPC

We are given a IP and port to nc to 'nc 15.164.75.32 1999':

![alt text](https://raw.githubusercontent.com/GabiCtrlZ/Whitehat-Quals/master/pictures/terminal.png)

At first, this looks quite confusing. What does the `CREATED BY N` mean? but we can use the example to understand.

The challenge refers to how many triangles and you create with sides length less or equal to n.

So to create a triangle you must follow a + b > c, a + c > b and b + c > a.

We may attempt to brute-force but the challenge specifies N can get as high as `10**6` and brute-forcing is about `O(n**3)` so we must improve the efficiency.

First, let's write all the outputs (for now we're using brute-force) for n 4-14 just to see if we can notice something we can work with.

![alt text](https://raw.githubusercontent.com/GabiCtrlZ/Whitehat-Quals/master/pictures/outputs.png)

Now let's see if we can find something in the output, so we'll writing them and let's write the differences also.

```

+1, +2, +4, +6, +9, +12, +16, +20, +25, +30, +36 <---- Differences

0, 1, 3, 7, 13, 22, 34, 50, 70, 95, 125, 161 <---- Outputs

```

You may already notice there are two series here `k**2` and `k(k+1)`.

```

1, 4, 9, 16, 25, 36 <---- k**2

2, 6, 12, 20, 30 <---- k(k+1)

```

So, to know how much triangles there are for a given N we can sum up all the differences up till that N,

so we need the formula for calculating the sum of both of those series.

The `k**2` sum is

![alt text](https://raw.githubusercontent.com/GabiCtrlZ/Whitehat-Quals/master/pictures/n-squared.png)

And we can view `k(k+1)` as `k**2 + k` so the sum of `k(k+1)` is

![alt text](https://raw.githubusercontent.com/GabiCtrlZ/Whitehat-Quals/master/pictures/n-plus-one.png)

Now all that's left is to know for each N what k are we solving for.

Let's take N=10 for example, there are 50 different triangles and if we look again here

```

+1, +2, +4, +6, +9, +12, +16, +20, +25, +30, +36 <---- Differences

0, 1, 3, 7, 13, 22, 34, 50, 70, 95, 125, 161 <---- Outputs

```

It can be represented as the sum of `n**2` series up to 16 plus the `n(n+1)` series up to 12, so that's `k = 4` and `k = 3`.

If we go up to 11 we will get `k = 4` and `k = 4`.

So we can understand if we're dealing with an even N, k would be `n/2 - 1` for the `n**2` series and `n/2 - 2` for the `n(n+1)` series.

And if it's an odd N, k would be `((n - 1) / 2) - 1` for both.

Now with that information let's implement it all in code:

```python

def n_squared_sum(num):

return int(((num)*(num + 1)*((num * 2) + 1)) / 6)

def n_plus_one_sum(num):

return int(n_squared_sum(num) + ((num * (num+1))/2))

def num_of_triangles(num):

if (num % 2) == 0:

n = (num / 2) - 1

return n_plus_one_sum(n - 1) + n_squared_sum(n)

else:

n = ((num - 1) / 2) - 1

return n_plus_one_sum(n) + n_squared_sum(n)

```

Full code can be found in solver.py

![alt text](https://raw.githubusercontent.com/GabiCtrlZ/Whitehat-Quals/master/pictures/solve.png)

Thanks for this writeup. It is very interesting to see how elegantly the challenge can be solved using math.