Update 16.1.md

docs/Chap16/Problems/16-1.md

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 > **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.** **Algorithm:** At each step, select the largest coin denomination that does not exceed the remaining amount of change, take as many as possible (i.e., $\lfloor n / C \rfloor$ coins, where $C$ is the denomination), and recursively solve for the remaining amount.
+
+**Proof of Correctness:** Let the coin denominations be $25$ (quarters), $10$ (dimes), $5$ (nickels), and $1$ (pennies). Suppose that in an optimal solution, the numbers of coins used for these denominations are $a, b, c, d$ respectively. First, we prove the upper bounds on the number of coins of each denomination in an optimal solution:
+* $d \le 4$: Otherwise, $5$ pennies ($5$ cents) can be replaced by $1$ nickel ($5$ cents), yielding a solution with $4$ fewer coins.
+* $c \le 1$: Otherwise, $2$ nickels ($10$ cents) can be replaced by $1$ dime ($10$ cents), yielding $1$ fewer coin.
+* $b \le 2$: Otherwise, $3$ dimes ($30$ cents) can be replaced by $1$ quarter and $1$ nickel ($30$ cents), yielding $1$ fewer coin.
+* $b = 2$ and $c = 1$ cannot hold simultaneously: Otherwise, $2$ dimes and $1$ nickel ($25$ cents) can be replaced by $1$ quarter ($25$ cents), yielding $2$ fewer coins.
+
+From these bounds, we can establish upper bounds on the total value that can be made by combinations of lower denominations in an optimal solution:
+* The maximum value made by dimes, nickels, and pennies is:
+  $$10b + 5c + d \le 2 \times 10 + 4 = 24 < 25$$
+  (since $b=2$ and $c=1$ cannot both hold, the max value is either $2 \times 10 + 0 \times 5 + 4 = 24$ or $1 \times 10 + 1 \times 5 + 4 = 19$).
+  Thus, dimes, nickels, and pennies can make up at most $24$ cents in an optimal solution.
+* The maximum value made by nickels and pennies is:
+  $$5c + d \le 1 \times 5 + 4 = 9 < 10$$
+  Thus, nickels and pennies can make up at most $9$ cents in an optimal solution.
+* The maximum value made by pennies is:
+  $$d \le 4 < 5$$
+  Thus, pennies can make up at most $4$ cents in an optimal solution.
+
+Now we show that the greedy choice is always optimal by induction or case analysis on $n$:
+* When $n \ge 25$, if the optimal solution uses $a < \lfloor n/25 \rfloor$ quarters, the remaining amount of change to be made by dimes, nickels, and pennies must be $n - 25a \ge 25$. However, dimes, nickels, and pennies can make up at most $24$ cents in an optimal solution, which is a contradiction. Thus, we must have $a = \lfloor n/25 \rfloor$ quarters, which is exactly the greedy choice.
+* When $10 \le n < 25$, if the optimal solution uses $b < \lfloor n/10 \rfloor$ dimes, the remaining amount of change to be made by nickels and pennies must be $n - 10b \ge 10$. However, nickels and pennies can make up at most $9$ cents in an optimal solution, which is a contradiction. Thus, $b = \lfloor n/10 \rfloor$ dimes, matching the greedy choice.
+* When $5 \le n < 10$, if the optimal solution uses $c < \lfloor n/5 \rfloor = 1$ nickels (i.e., $c = 0$), the remaining amount of change to be made by pennies must be $n - 5c = n \ge 5$. However, pennies can make up at most $4$ cents in an optimal solution, which is a contradiction. Thus, $c = \lfloor n/5 \rfloor = 1$, matching the greedy choice.
+* When $n < 5$, only pennies can be used, which also matches the greedy choice.
+
+By induction, the greedy strategy of choosing the largest denomination at each step yields an optimal solution.
 
 **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 $V\mod c^k$.