pyramid brick pressure function

Question

There is a two-dimensional pyramid of bricks. The weight of the 1st - 1 kg, crushes evenly for 2 below it at 0.5 kg. That is, if the upper one presses +100 kg, then below it each will be (100 + 1) / 2 kg - the lower one presses the weight of each one above it.

1,0 = .5 2,0 = .75 2,1 = 1.5 3,0 = .875 3,1 = .2.125

How to calculate the pressure on any of them, indicating the position, for example, k4-2, k8-4 ... k (row Above, number Left) - as a function with 2 parameters, which returns the weight?

I thought, a number of * .5 + (number of 1) / 2 or 1 * (c + 1) / 2 * p + 1 * c / 2 * p, various other options, but I just can not find the exact one. (preferably on java)

In fact, for the pyramid the loads will not be uniform, since for such structures the load on each element depends on how far it is from the center.
But if we assume that the load is uniform, then this problem is solved easily.
The output format k4-2 is also unclear, ... Here 4 is a row on top, and 2 is a load on a brick?
@CodeGust, maybe this is a false idea, but the digits in your pyramid are quite symmetrical and look like Pascal's triangle, I'm not good at math, but maybe you should try to solve the problem not through recursion, but through binomial coefficients, I'm not strong at this topic , but you can dig deeper

Evgeny Glushkov Evgeny Glushkov 540 one ten · Accepted Answer · 2018-06-28T08:13:15

And why not use recursion? Weight values, row number and brick number in the row, insert what you need:

public static void main(String[] args) { float weight = 1; //вес кирпча int row = 5; //номер ряда начиная с 1 с верху int num = 3; //номер кирпича слева (или с права, как больше нравится) с 1 System.out.println(pressueOnBrick(row, num, weight)); } //давление на один кирпич public static float pressueOnBrick(int row, int num, float weight) { return pressue(row, num, weight) - weight; //давление оказываемое на один кирпич = полное //давление которое оказывает кирпич минус вес кирпича } //полное давление оказываемое кирпичём (включая свой вес) public static float pressue(int row, int num, float weight) { //в ряду не может быть кирпичей с номером меньше 1 и больше чем номер ряда(число кирпичей в ряду = номеру ряда) if (num < 1 || num > row) { return 0; } //давление которое оказывает кирпич равно сумме веса кирпича и половине веса кирпичей которые давят на него //слева и справа return weight + (pressue(row - 1, num - 1, weight) + pressue(row - 1, num, weight)) / 2; }

Validation of input values on your conscience.

At a minimum, you need to save the previously calculated weights into an array, and re-use them when you re-query ... we get O (n ^ 2), although you can also O (1).
reference values: 1.0 = .5 2.0 = .75 2.1 = 1.5 3.0 = .875 3.1 = .2.125
I do not know what is your coordinate system in the pyramid, but if I take the top row of one brick as 1 and further down where 2 is the second row, the 3rd and the third and so on and count in the row with 1 brick, it turns out: 1-1 = 0 (top brick, does not press anything on it), 2-1 = 0.5, 3-1 = 0.75, 4-1 = 0.875, 4-2 = 2.125, 25-14 = 20.660532.
If I understand correctly, then you have the numbering of rows and positions of bricks in a row starting at 0, the given recursive solution is easy to adapt to this.

Answer 2 · 2018-06-28T09:22:52

 #include <iostream> #include <map> int main() { unsigned ranks{}; std::cin >>ranks; // общее количество рядов double weight{}; std::map<unsigned, double> loads; unsigned k{1}; for (; k <= ranks; ++k) { loads.insert(std::make_pair(k, weight)); weight = (weight + 1)/2; } std::cout << std::endl << "вводим нужный ряд: " << std::endl; std::cin >> k; std::cout << "нагрузка для кирпичей " << k << " - го ряда: " << loads[k]; return 0; }

After discussing with goldstar_labs to rewrite, because the first option is true for two rows and for extreme bricks.

 #include <iostream> #include <vector> using std::vector; using std::cin; void setValue(const vector<double>& p1, vector<double>& p2) { size_t k = p2.size(); if (k <= 2) return; for (size_t i = 1; i < k - 1; ++i) p2[i] += (p1[i] + 1)/2; } int main() { unsigned ranks{}, k{}; cin >>ranks; // общее количество рядов double w{}; vector< vector<double> > weights; for (unsigned n = 1; n <= ranks; ++n) { vector<double> v(n, w); weights.emplace_back(v); if (n > 2 ) setValue(weights[n - 2], weights[n - 1]); w = (w + 1)/2; } std::cout << std::endl << "вводим нужный ряд: " << std::endl; std::cin >> k; k %= ranks + 1; std::cout << "нагрузки для кирпичей " << k << " - го ряда:\n"; for (double d : weights[k - 1]) std::cout << d << ' '; return 0; }

Output, I think you can do it yourself in the desired format.

This is a partial solution that will load the outer layer of bricks in the pyramid.
@goldstar_labs, I wrote about this in a comment, but the question says about uniform load, so the answer is appropriate
The question is muddy, I agree) on the one hand, evenly, on the other hand, "1.0 = .5 2,0 = .75 2,1 = 1.5 3,0 = .875 3,1 = .2.125", probably still uneven .
For uneven, it looks like a Pascal triangle, where there is an incomprehensible coefficient on each line, I'll go and think about it)
@goldstar_labs, maybe it was meant that the pressure is evenly distributed to the next two bricks.

pyramid brick pressure function

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