Question

You have some LEGO plastic bricks, all bricks are 1x1x1. Also you have one tile, 1xN (N <= 80), on which you should place the LEGO bricks. You can order them in sequences (one sequence is correct if it has 3 or more consecutive bricks), also you must have at least 1 empty space between 2 sequences. You should calculate the number of different combination you can place the bricks on the tiles.

Here’s an example:

If the tile is 1×7, there are 17 different combinations.

input : 7
output : 17

_{(source: mendo.mk)}

Also if you have no bricks it’s counted as 1 combination.

I have worked on this problem and i found the way to calculate possible combinations of if the max length of the tile is 14 (3 sequences). I found it using for loops.

My biggest problem is the huge number of for loops that i need to run. For example, for 1 sequence i use 1 for loops, for 2 sequences 2 loops + 1 for 1 sequnce…so if i use all 80 bricks, i can create 20 sequence and i will have to use over 210 for loops which is huge number. So it’ll be good if i can nest them in one. I tried it and it get messy and it doesn’t give correct answers.

New code:

#include <iostream>
using namespace std;
int main()
{
    long long int places, combinations = 1;
    cin >> places;
    long long int f[80], g[80];
    f[0] = 0;
    f[1] = 0;
    f[2] = 0;
    g[0] = 1;
    g[1] = 1;
    g[2] = 1;
    for(int i = 3; i<=places; i++)
    {
        f[i] = f[i-1] + g[i-3];
        g[i] = f[i-1] + g[i-1];
    }
    combinations = f[places] + g[places];
    cout << combinations;
    return 0;
}

Answer 1

If this is a counting problem (not outputting combination, rather just counting them) it’s an easy one. Assume we solved it for n ≥ 3 now to solve it for n+1, we solve it by induction:

Assume f is a function that shows the number of possible ways such that the last item is a brick.
Analogously g is a function that shows the number of possible ways such that the last item is not brick.
Let define h = f+g, to be the number of all possible ways.

So we have:

f(n+1) = f(n) + g(n-2)
g(n+1) = g(n) + f(n)

With initial condition:

for n=0,1,2: g=1, f= 0.
for n = 3: g=1,f=1

Samples:

n=4: g=2,f=2 ==> h=4
n=5: g=4, f= 3 ==> h=7
n=6: g=7, f= 4 ==> h=11
n=7: g=11,f=6 ==> h=17

We can solve it with one for loop in O(n).

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Why:

f(n+1) = f(n) + g(n-2)
g(n+1) = g(n) + f(n)

First, let prove first part:

Remember that we assumed f(n) is a working solution which has a plastic brick in the last item, and g(n) is a working solution which doesn’t have a brick in the last item.

f(n+1) can be obtained from f(n) by adding one brick at the last place.
Also f(n+1) can be obtained by adding three brick after g(n-2), it means cells of n-1,n,n+1.

Note that we can’t add brick after g(n-1) or g(n) to create a valid solution for f(n+1) because they are not valid solutions (number of consecutive bricks is less than 3). Also, note that we don’t need to count the number of ways which arises by adding bricks after g(n-3) because they are previously enumerated by f(n). So we have f(n+1) = f(n) + g(n-2).

In the same way we can prove g(n+1) = f(n)+g(n) this case is easier, because g(n+1) simply can be made from any valid solution up to n, as there is no 3 consecutive brick barrier here, they can come after any valid solution.