All possible ways scores $(1,2,3)$ add up to $n$

Question

Could someone please explain mathematical explanation behind this?

You can win three kinds of basketball points, 1 point, 2 points, and 3 points. Given a total score $n$, print out all the combination to compose $n$.

Examples:
For n = 1, the program should print following:
1

For n = 2, the program should print following:
1 1
2

For n = 3, the program should print following:
1 1 1
1 2
2 1
3

For n = 4, the program should print following:
1 1 1 1
1 1 2
1 2 1
1 3
2 1 1
2 2
3 1

Algorithm:

• At first position we can have three numbers 1 or 2 or 3. First put 1 at first position and recursively call for n-1.
• Then put 2 at first position and recursively call for n-2.
• Then put 3 at first position and recursively call for n-3.
• If n becomes 0 then we have formed a combination that compose n, so print the current combination.

I've solved in using JS as per below. But I don't quite understand the mathematical reasoning behind it.

https://jsfiddle.net/d2Lft7d1/

Answer 1

It is a Fibonacci like sequence.
Let $\,A_{\small n}\,$ be the total number of $\,(1,\,2,\,3)\,$ combinations that compose $\,n\,$, Then: $$ A_{\small 1}=1,\,A_{\small 2}=2,\,A_{\small 3}=4,\quad\color{red}{A_{\small n}=A_{\small n-1}+A_{\small n-2}+A_{\small n-3}} \\[4mm] \Rightarrow\quad \left\{A_{\small n}\right\}=\left\{1,\,2,\,4,\,7,\,13,\,24,\,44,\,\cdots\right\} $$ And the idea behind that for $\,n\gt3\,$, you will have the ability to add a Most Significant Digit (MSD) equals $\,1,\,2,\text{ or } \,3\,$. This should left you with $\,n-1,\,n-2,\text{ and } \,n-3\,$ respectively. For Example: $$ \begin{align} n &=5 \\[2mm] \text{MSD} &=\color{red}{1} \quad\Rightarrow\text{ The comination of }\,(n-1=4)= \begin{cases} \color{red}{1}\,1\,1\,1\,1 \\ \color{red}{1}\,1\,1\,2 \\ \color{red}{1}\,1\,2\,1 \\ \color{red}{1}\,1\,3 \\ \color{red}{1}\,2\,1\,1 \\ \color{red}{1}\,2\,2 \\ \color{red}{1}\,3\,1 \end{cases} \\[2mm] \text{MSD} &=\color{blue}{2} \quad\Rightarrow\text{ The comination of }\,(n-2=3)= \begin{cases} \color{blue}{2}\,1\,1\,1 \\ \color{blue}{2}\,1\,2 \\ \color{blue}{2}\,2\,1 \\ \color{blue}{2}\,3 \end{cases} \\[2mm] \text{MSD} &=\color{Green}{3} \quad\Rightarrow\text{ The comination of }\,(n-3=2)= \begin{cases} \color{Green}{3}\,1\,1 \\ \color{Green}{3}\,2 \end{cases} \\[2mm] A_{\small5} &= \color{red}{A_{\small4}}+\color{blue}{A_{\small3}}+\color{green}{A_{\small2}} = \color{red}{7}+\color{blue}{4}+\color{green}{2} = 13 \end{align} $$

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For other similar combination $\,\left({\small\text{e.g }}\,(1,2)\,,(1,2,4)\,,\cdots\right)\,$, we start by computing the first required terms, then we apply the concept of Fibonacci sequence and Most Significant Digit (MSD).

$\underline{\bf(1,2)}$: $$ \begin{align} (n=1) &\rightarrow \begin{cases} \color{red}{1} \end{cases} \qquad\Rightarrow\, A_{\small 1}=1 \\[2mm] (n=2) &\rightarrow \begin{cases} \color{blue}{1}\,\color{red}{1} \\ \color{blue}{2} \end{cases} \quad\Rightarrow\, A_{\small 2}=2 \\[2mm] A_{\small n} &= A_{\small n-1}+A_{\small n-2} = \left\{1,\,2,\,3,\,5,\,8,\,13,\,21,\,\cdots\right\} \end{align} $$ $\underline{\bf(1,2,4)}$: $$ \begin{align} (n=1) &\rightarrow \begin{cases} \color{red}{1} \end{cases} \qquad\qquad\Rightarrow\, A_{\small 1}=1 \\[2mm] (n=2) &\rightarrow \begin{cases} \color{blue}{1}\,\color{red}{1} \\ \color{blue}{2} \end{cases} \quad\qquad\Rightarrow\, A_{\small 2}=2 \\[2mm] (n=3) &\rightarrow \begin{cases} \color{green}{1}\,\color{blue}{1}\,\color{red}{1} \\ \color{green}{1}\,\color{blue}{2} \\ \color{green}{2}\,\color{red}{1} \end{cases} \qquad\Rightarrow\, A_{\small 3}=3 \\[2mm] (n=4) &\rightarrow \begin{cases} 1\,\color{green}{1}\,\color{blue}{1}\,\color{red}{1} \\ 1\,\color{green}{1}\,\color{blue}{2} \\ 1\,\color{green}{2}\,\color{red}{1} \\ 2\,\color{blue}{1}\,\color{red}{1} \\ 2\,\color{blue}{2} \\ 4 \end{cases} \quad\Rightarrow\, A_{\small 4}=6 \\[2mm] A_{\small n} &= A_{\small n-1}+A_{\small n-2}+A_{\small n-4} = \left\{1,\,2,\,3,\,6,\,10,\,18,\,31,\,\cdots\right\} \end{align} $$

Answer 2

Consider the set of sequences of points that add up to $n$ and let's call it $S(n)$. There are three different cases to consider regarding the type of sequence we might have.

1. The first point value is 1. This can only happen if the sum of the remaining points is $n-1$ so the rest of the subsequence must be an element of $S(n-1)$.

2. The first point value is 2. This can only happen if the sum of the remaining points is $n-1$ so the rest of the subsequence must be an element of $S(n-2)$.

3. The first point value is 3. This can only happen if the sum of the remaining points is $n-3$ so the rest of the subsequence must be an element of $S(n-3)$.

These are the only possible cases, each of which is mutually exclusive so the algorithm will print out every possible sequence exactly once.

EDIT: As an example, look at your $n=4$ case, in other words, sequences in $S(4)$. Notice that every sequence is one of the three forms:

1. The number $1$ followed by some sequence in the $n=3$ case since $S(n-1) = S(3).$
2. The number $2$ followed by some sequence in the $n=2$ case since $S(n-1) = S(2).$
3. The number $3$ followed by some sequence in the $n=1$ case since $S(n-1) = S(1).$