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Dynamic programming specifies a collection of overlapping subproblems, expresses each answer using smaller subproblem answers, and stores answers so each subproblem is solved once. One may fill a table in dependency order or use memoization to compute only requested entries. The key design choice is a state that retains every feature needed by later decisions.

The principle of optimality says that, after a decision in an optimal solution, the remaining decisions must solve the corresponding residual subproblem optimally. Otherwise, replacing that residual solution by a better one would improve the entire solution. This replacement argument requires the state to describe the residual constraints completely. Its useful form here isThe maximization applies to these value-maximization problems; minimization uses the analogous minimum recurrence.

In the 0-1 knapsack problem, item $j$ has weight $w_j$ and value $v_j$. Choose a subset of the $n$ items, using each at most once, whose total weight is at most $W$ and whose total value is as large as possible. For the capacity-indexed dynamic program, take $W$ and the weights to be nonnegative integers. The empty choice is allowed.

Let $K[j,c]$ be the greatest value attainable using only items $1,\dots ,j$ with capacity $c$. The base case is $K[0,c]=0$. With item $j$ unavailable when it is too heavy, the recurrence isEvery feasible optimum either excludes item $j$, giving the first candidate, or includes it exactly once, leaving capacity $c-w_j$ for earlier items. The principle of optimality proves that the remaining earlier items must attain the indicated residual optimum. Conversely, either candidate constructs a feasible solution, so taking the larger is both necessary and sufficient. Mathematical induction on $j$ proves the table's meaning and the final answer $\boxed{K[n,W]}$.

Store item weights and values in arrays and the values in a two-dimensional array with $(n+1)(W+1)$ entries. Fill rows in increasing $j$; each entry uses at most two values from the previous row. There are $n(W+1)$ constant-work updates, givingFor positive growing capacities this is the familiar $\Theta (nW)$ bound. The $+1$ terms also cover capacity zero and zero-weight items: one must not initialize every $K[j,0]$ to zero if zero-weight, positive-value items are allowed.

To recover an actual subset, store the chosen alternative at each entry or compare the two candidates while tracing backward from $(n,W)$. A selected item changes $(j,c)$ to $(j-1,c-w_j)$; an excluded item changes it to $(j-1,c)$. This takes $O(n)$ extra time.

If only the optimal value is needed, keep two rows, using $\Theta (W+1)$ space. With strictly positive weights, a single capacity row updated in decreasing capacity also prevents item reuse. The two-row version handles zero weights without special in-place reasoning. The running time is pseudopolynomial, since the binary input length of $W$ is only logarithmic in its numerical value.

For the two-bag 0-1 knapsack problem, let $D[j,a,b]$ be the maximum total value using the first $j$ items, with remaining capacity limits $a$ in the first bag and $b$ in the second. Initialize $D[0,a,b]=0$. Item $j$ has exactly three possible decisions: omit it, place it in bag one if it fits, or place it in bag two if it fits. HenceA braced candidate with a false condition is omitted. All candidates refer to layer $j-1$, which prevents the same item from entering both bags. Combining two independently optimized single-bag knapsacks would not enforce that restriction.

The principle of optimality proves correctness exactly as in part (b): condition on the location of the final item and replace the earlier items by the optimal feasible residual assignment. Every candidate is realizable, and every feasible assignment belongs to one of the three cases. The final answer is $\boxed{D[n,W_1,W_2]}$.

Fill the three-dimensional array in increasing item index. Each of its $n(W_1+1)(W_2+1)$ updates inspects at most three candidates:For positive capacities the customary expression is $\Theta (n{W}_1{W}_2)$. Keeping only two item layers reduces space to $\Theta ((W_1+1)(W_2+1))$ for the optimal value. Full decision storage also permits backward reconstruction of each item's bag assignment. Nonnegative integer weights and capacities are the assumptions behind this capacity-indexed dynamic programming table.

Dynamic programming solves related subproblems once, stores their results, and combines those results in an order consistent with their dependencies. The state must retain all information that determines the remaining choices and their costs. Here an adequate dynamic programming state is the current array index, since future jumps depend only on that index and the fixed input.

The principle of optimality says that the remaining part of an optimal solution must be optimal for the residual problem defined by its state. Equivalently here, the prefix ending at a particular intermediate position of a shortest route must be a shortest route to that position: replacing it by a shorter prefix would shorten the entire route. This is the optimal substructure needed by the dynamic programming recurrence.

Thus $\boxed{\text{store an optimum for each state; combine previously solved states by an optimality recurrence}}$. Overlapping routes share states, so storing results avoids repeatedly solving the same subproblem.

Let $D[j]$ be the minimum number of jumps from index zero to index $j$, with $\infty$ denoting an unreachable position. The minimum forward-jump dynamic programming recurrence iswhere the minimum of an empty set is $\infty$ and $1+\infty =\infty$. Compute it in increasing index order:

D[0] = 0
for j = 1, ..., n-1:
    D[j] = INFINITY
    for i = 0, ..., j-1:
        if D[i] != INFINITY and A[i] >= j-i:
            D[j] = min(D[j], D[i] + 1)
return D[n-1]

For algorithm correctness, classify any route to $j>0$ by its penultimate index $i$. Every jump advances, so $i<j$, and legality requires $A[i]\ge j-i$. By the principle of optimality, a shortest prefix to $i$ can be used; this gives the lower bound represented by the recurrence. Conversely, a minimizing reachable predecessor supplies an actual route of length $D[i]+1$, proving attainability. Mathematical induction on $j$ therefore proves every stored value correct.

There are $n(n-1)/2$ predecessor tests, and two arrays are unnecessary when only the number is required. ThusFor the source example, the computed array isWhen $n=1$, zero jumps are needed even if the lone entry is zero. An unreachable final position returns $\infty$. Testing A[i] >= j-i also avoids forming a potentially overflowing sum i + A[i] in bounded implementations. A faster frontier-based algorithm is possible, but the displayed method directly supplies the requested dynamic programming solution.

Dynamic programming describes a problem by reusable subproblem states and a recurrence, computing each required state once through memoization or a bottom-up order. Its benefit comes from sharing answers to overlapping subproblems.

The principle of optimality says that, after the information needed by a state has been fixed, the remaining part of an optimal solution is optimal for that state. Otherwise replacing it by a better subsolution would improve the whole solution. A correct state must retain every constraint that can affect later choices.

Both divide-and-conquer algorithms and dynamic programming derive answers from smaller problems. A typical divide-and-conquer algorithm, such as merge sort, splits an instance into separate parts and combines their answers. Dynamic programming additionally identifies repeated states across different decompositions and reuses their answers; matrix-chain multiplication is a standard example.

Thus a recursion tree may repeatedly solve the same state, whereas a dynamic programming table or memoization stores it once. The resulting recurrence and dependency order determine the running time. The principle of optimality supports combining optimal state answers; it does not by itself justify a greedy algorithm.

Dynamic programming solves overlapping subproblems once and stores their values for reuse. The principle of optimality says that the remaining decisions within an optimal solution must solve the corresponding residual subproblem optimally; otherwise replacing them by a better residual solution would improve the original one. Here removing one chosen stamp leaves an optimal way to form the remaining postage.

Let $F[x]$ be the minimum number of stamps for total $x$. Since denomination 1 is available, every nonnegative total is reachable. The dynamic programming recurrence isFill the table in increasing order of $x$:Every representation of $x$ has a last stamp $d_i$, so its remaining count is at least $F[x-d_i]$. Conversely, appending $d_i$ to an optimal residual representation attains each candidate. Thus the recurrence is exact by the principle of optimality. Worst-case time is $O(mn)$, with $O(m)$ storage. This is pseudo-polynomial time in the numerical target, rather than polynomial time in its binary encoding length.