Update 16-1.md

docs/Chap16/Problems/16-1.md

@@ -8,7 +8,22 @@
 >
 > **d.** Give an $O(nk)$-time algorithm that makes change for any set of $k$ different coin denominations, assuming that one of the coins is a penny.
 
-**a.** Always give the highest denomination coin that you can without going over. Then, repeat this process until the amount of remaining change drops to $0$.
+**a.** Always give the highest denomination coin that you can without going over, that is, $\lfloor n / C \rfloor$ coins of the largest denomination $C \le n$. Then, repeat this process on the remaining $n \bmod C$ cents until it drops to $0$.
+
+Let $a, b, c, d$ be the numbers of quarters, dimes, nickels, and pennies in an optimal solution. Each of the following would contradict optimality:
+
+- $d \ge 5$: $5$ pennies can be replaced by $1$ nickel, saving $4$ coins.
+- $c \ge 2$: $2$ nickels can be replaced by $1$ dime, saving $1$ coin.
+- $b \ge 3$: $3$ dimes can be replaced by $1$ quarter and $1$ nickel, saving $1$ coin.
+- $b = 2$ and $c = 1$: $2$ dimes and $1$ nickel can be replaced by $1$ quarter, saving $2$ coins.
+
+Hence, in an optimal solution,
+
+- pennies make up $d \le 4 < 5$ cents,
+- nickels and pennies make up $5c + d \le 5 + 4 = 9 < 10$ cents, and
+- dimes, nickels, and pennies make up $10b + 5c + d \le \max(2 \cdot 10 + 4, 10 + 5 + 4) = 24 < 25$ cents.
+
+Since dimes, nickels, and pennies make up less than $25$ cents, $n - 25a < 25$, so $a = \lfloor n / 25 \rfloor$, which is the greedy choice. Let $n_1 = n - 25a$. Since nickels and pennies make up less than $10$ cents, $b = \lfloor n_1 / 10 \rfloor$. Let $n_2 = n_1 - 10b$. Since pennies make up less than $5$ cents, $c = \lfloor n_2 / 5 \rfloor$, and finally $d = n_2 - 5c$. Every count in the optimal solution is exactly the one the greedy algorithm picks, so the greedy solution is optimal.
 
 **b.** Given an optimal solution $(x_0, x_1, \dots, x_k)$ where $x_i$ indicates the number of coins of denomination $c_i$ . We will first show that we must have $x_i < c$ for every $i < k$. Suppose that we had some $x_i \ge c$, then, we could decrease $x_i$ by $c$ and increase $x_{i + 1}$ by $1$. This collection of coins has the same value and has $c − 1$ fewer coins, so the original solution must of been non-optimal. This configuration of coins is exactly the same as you would get if you kept greedily picking the largest coin possible. This is because to get a total value of $V$, you would pick $x_k = \lfloor V c^{−k} \rfloor$ and for $i < k$, $x_i\lfloor (V\mod c^{i + 1})c^{-i} \rfloor$. This is the only solution that satisfies the property that there aren't more than $c$ of any but the largest denomination because the coin amounts are a base $c$ representation of $V\mod c^k$.