TITLE: An interesting problem with "decomposing" natural numbers.
QUESTION [35 upvotes]: Consider the following one-person game:

A player starts with score $0$ and writes the
number $20$ on an empty whiteboard. At each step, she may erase any one integer (call it $a$)
and writes two positive integers (call them $b$ and $c$) such that $b + c = a$. The player then
adds $b × c$ to her score. She repeats the step several times until she ends up with all $1$s on
the whiteboard. Then the game is over, and the final score is calculated.

Example: At the first step, a player erases $20$ and writes $14$ and $6$, and gets a score of
$14 × 6 = 84$. In the next step, she erases $14$, writes $9$ and $5$, and adds $9 × 5 = 45$ to her
score. Her score is now $84 + 45 = 129$. In the next step, she may erase any of the remaining
numbers on the whiteboard: $5$, $6$ or $9$. She continues until the game is over.
Alya and Bob play the game separately. Alya manages to get the highest possible final score.
Bob, however, manages to get the lowest possible final score. What is the difference between
Alya’s and Bob’s final scores?
I tried to "decompose" into a few numbers and I get the same scores. I am not sure how to prove the conjecture that any numbers will yield the same score no matter which path is taken.

REPLY [2 votes]: Compared to Misha Larov's answer, my solution has essentially the same idea but a different interpretation.
Let's say the number we start off with is $n$. At any stage of the game, we assign a complete graph $K_i$ to any number $i$ written on the board.
The action of splitting a number $a$ into $b$ and $c$ can be reinterpreted as

Choosing disjoint vertex subsets $B$ and $C$ of $K_a$ with respectively $b$ and $c$ elements
Deleting every edge connecting $i \in B$ and $j \in C$
Obtaining new complete graphs $K_b$ and $K_c$.

The score the player gets after this splitting is the number of edges deleted in step 2. Throughout the game, we actually count the total number of edges deleted.
At the final condition, where all graphs are $K_1$, which are individual vertices, we have eliminated all edges of the initial $K_n$. Hence the final score is always the number of edges in $K_n$, $\frac{n(n-1)}{2}$.