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Fix the first tape of the group multiplier automaton $M_x$ to be $u=a_1\cdots a_m$. Construct its possible runs by dynamic programming, one column at a time. A record consists of a state and a flag indicating whether the output tape has ended. At column $i\le m$, the first symbol is $a_i$ and the second symbol may be any letter of the alphabet, or $\#$; after the second tape first uses $\#$ it must continue to use $\#$. At later columns the first symbol is $\#$, and an additional column must have a genuine letter on the second tape. Never add a $(\#,\#)$ column.

Keep one predecessor record and its output symbol for each reachable state-and-flag pair in each layer. Two runs reaching the same record have exactly the same possible future completions, so this merging loses no accepting output. The merged dead state of a finite automaton can be discarded. At each layer $i\ge m$, test whether a reached state is accepting; the initial layer is included when $m=0$. Upon success, follow predecessor records backwards and omit padding symbols to recover $v$.

The automatic structure for a group supplies a representative in $L$ of $\overline{u}x$, so an accepted padded convolution of words exists and this search terminates. The output is the required representative:The search uses only the finite transition table; it does not presuppose a solution to the word problem for a group.

Part (a)(i), applied to $w=u$ and $g=\overline{u}x$, guarantees an accepted output of length at most $m+N$. Its padded convolution of words has at most $m+N$ columns. Thus the dynamic programming search in part (b) reaches an accepting layer by this bound.

Each layer has at most twice the fixed number of states of $M_x$, and each record has at most $|A|+1$ outgoing choices. These are constants for the fixed automatic structure for a group. Storing a pointer and a symbol takes constant time; copying whole partial output strings at every step would unnecessarily spoil this bound. Reconstruction also takes at most $m+N$ steps. Therefore automatic multiplication has linear time and bounded length increase:For nonempty inputs this is the requested $O(|u|)$ time complexity. The additive $1$ merely includes the constant work on the empty word.

For the intended random walk model, take $S_0$ to be deterministic, or independent of all the independent and identically distributed random variables $Y_j=S_j-S_{j-1}$. If $\nu$ is their common probability distribution, then $Y_{t+1}$ is independent of ${\mathcal{F}}_t=\sigma (S_0,Y_1,\dots ,Y_t)$. This supplies the Markov property that the backward dynamic programming argument needs. Define the transition operatorand the deterministic optimal stopping value function recursively byAt any state where the conditional law is evaluated, independence givesBackward induction, starting with $U_T=f(S_T)$, proves the intended representationIntegrability along the actual random walk makes the recursion finite at the states used by that process, up to null sets. If finite real-valued value functions on all of $\mathbb{R}$ are intended, a convenient sufficient convention is statewise integrability: $\mathbb{E}|f(s+Y_1+\cdots +Y_k)|<\infty$ for every $s$ and $0\le k\le T$. Indeed each stopping value is bounded in absolute value by the expectation of the sum of the absolute rewards. This convention makes the displayed recursion finite and measurable everywhere; the source only assumes integrability from the given starting process.

The printed independence of the increments alone does not suffice if $S_0$ can reveal future increments. Here is a bounded finite-state counterexample, also with a convex reward. Let $T=2$, let ${\epsilon }_1,{\epsilon }_2$ be independent fair signs, and setThe two increments are exactly ${\epsilon }_1,{\epsilon }_2$, hence independent and identically distributed. But ${\mathcal{F}}_1$ already knows both signs, so $U_1=\max \{S_1^{+},S_2^{+}\}$. On the two positive-probability histories with $S_1=0$, its values are respectively $0$ and $1$. Thus no deterministic $V(1,0)$ can work. Adding independence of $S_0$ from the increments repairs this missing Markov property hypothesis.

For the plug-in Neyman allocation, the estimated Bernoulli distribution variances are $6/25$ and $3/16$, soThis uses the target ratio directly as the allocation odds; a procedure that actively corrects previous allocation imbalances would need an additional rule.

For the randomized play-the-winner rule RPW$(1,1)$, return the drawn ball and add one ball of the same treatment after a success or of the opposite treatment after a failure. Treatment 0 has three successes and two failures, and treatment 1 has one success and three failures. The urn therefore has $1+3+3=7$ treatment-0 balls and $1+1+2=4$ treatment-1 balls. HenceFor dynamic programming with only the final patient left, there is no future value from learning. Independent uniform prior distributions and Beta-binomial conjugacy give posteriors $p_0\mid \mathcal{D}\sim \mathrm{Beta}(4,3)$ and $p_1\mid \mathcal{D}\sim \mathrm{Beta}(2,4)$. Their posterior predictive probabilities of success are $4/7$ and $2/6=1/3$. The optimal terminal action therefore givesThis is optimal for expected successes with no imposed lower bound on the randomization probabilities.

The dynamic programming principle over a short interval $[t,t+\Delta t]$ givesA first-order Taylor expansion of the value function givesAfter subtracting $F(x,t)$, dividing by $\Delta t$ and taking the limit, the Hamilton-Jacobi-Bellman equation is

A divide-and-conquer algorithm splits an instance into smaller instances of the same kind, solves them recursively, and combines their answers. The smaller instances should be sufficiently smaller to ensure algorithm termination; the combination cost determines whether splitting saves work. The two halves of merge sort illustrate this separation.

Dynamic programming identifies a collection of subproblems and dependencies among them, solves each distinct subproblem once, and stores its answer. A table in dependency order or memoization prevents repeated evaluation of overlapping subproblems. For optimization, an optimal substructure argument justifies constructing an optimal answer from optimal smaller answers.

A greedy algorithm repeatedly makes a locally best feasible choice and commits to it. It avoids exploring alternative continuations. Algorithm correctness therefore needs a greedy-choice property, often proved by replacing the first choice of an optimal solution with the greedy choice without worsening it.

Divide-and-conquer algorithms suit problems with largely independent, balanced subproblems and an inexpensive combination step. In merge sort, the two halves are independently sorted and merged in linear time. Its recurrence relation is $T(n)=T(\lfloor n/2\rfloor )+T(\lceil n/2\rceil )+\Theta (n)$, giving $\Theta (n\log n)$ time complexity.

Dynamic programming suits overlapping subproblems with a compact state description. For the coin change problem, let $C(v)$ be the minimum number of coins needed to make amount $v$. With positive denominations $D$ containing $1$,Every optimal nonempty answer has a last coin $d$ and an optimal answer for $v-d$; otherwise replacing the remainder improves it. Filling $C(0),\dots ,C(V)$ takes $O(V|D|)$ time complexity and $O(V)$ space complexity. Repeated amounts are computed just once. This bound is in the numeric amount $V$, not its binary input length.

A greedy algorithm suits problems where one can prove a greedy-choice property. For unweighted interval scheduling, choose an interval with earliest finishing time, discard overlapping intervals, and repeat. Among optimal schedules, replace the first interval by the earliest-finishing one. The replacement finishes no later, so every later interval remains feasible. Applying the same argument to the remaining intervals proves optimality. After sorting by finish time, a single scan suffices.

These examples separate the relevant structural reasons: independent halves, repeated amount states, and an exchange argument certifying each irreversible choice.

In the coin change problem, take denominations $\{1,3,4\}$ and target amount $6$. The natural greedy algorithm repeatedly chooses the largest denomination not exceeding the remaining amount. It chooses $4$, then $1$, then $1$, for three coins. Yet $3+3=6$ uses only two.

Two is optimal: one coin cannot make $6$ because denomination $6$ is absent, and the displayed pair attains the lower bound. ThereforeThe locally largest first coin loses the possibility of using the better pair of threes. This supplies a concrete failure of the greedy-choice property; optimal substructure alone does not justify the greedy choice. The dynamic programming recurrence from part (b) would compare all feasible first or last denominations and find the pair.

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.

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.

Store pred[j] = i whenever a strict improvement sets D[j] = D[i] + 1. Initialize predecessors to NONE. After the dynamic programming pass, return NONE if the final position is unreachable; otherwise follow predecessors backwards:

route = []
j = n - 1
while j != 0:
    route.append(j)
    j = pred[j]
route.append(0)
reverse(route)
return route

Every stored predecessor is smaller than the current index, so reconstruction terminates. Its stored distance is exactly one smaller, so the output has precisely $D[n-1]$ jumps and each jump satisfies its input bound. Therefore $\boxed{\text{store a minimizing predecessor; reverse the predecessor chain from }n-1\text{ to }0}$. Reconstruction takes $O(n)$ additional time complexity and $O(n)$ output space; the total dynamic programming bounds remain $\Theta (n^2)$ time and $\Theta (n)$ space complexity. Keeping the first predecessor on ties selects one optimal route deterministically.

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 a problem by storing solutions to overlapping subproblems and combining them according to a recurrence. Each dynamic programming state describes all information needed by subsequent choices. Evaluating states in an order where their dependencies are already available avoids recomputing the same subproblem. A typical optimization recurrence chooses the best transition; a decision recurrence instead combines feasible transitions with Boolean disjunction.

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.

Use dynamic programming in increasing order of the maximum permitted part and then of the remaining sum. Let Count be a sufficiently wide integer type and let table have dimensions (MaxN+1) by (MaxN+1):Require $0\le n\le \text{MaxN}$. The outer loop invariant says all rows below $i$ equal $p(\cdot ,\cdot )$. The inner loop invariant additionally says entries before column $j$ in the current row are correct. Its dependencies are the previous row and the earlier column $j-i$, so no future entry is read. Each state is evaluated once, giving $\Theta (n^2)$ arithmetic operations and $\Theta (n^2)$ stored counts. A memoization table around part (a)'s recursion is another valid way to share states; its entries need a separate “computed” flag because zero is a legitimate answer. Growing count sizes affect bit complexity and can cause integer overflow in a fixed-width Count.

Dynamic programming stores answers to overlapping subproblems so they can be reused. Its optimal substructure requirement, the principle of optimality, says that an optimal answer's remaining subproblem solution must itself be optimal once the final decision and state are fixed.

For global sequence alignment without mismatches, use the two prefix lengths as the state. A final column is either a letter of the first string opposite a gap, a gap opposite a letter of the second, or two equal letters. Removing that column leaves one of three smaller prefix problems. If its alignment were suboptimal, replacing it by a better alignment would improve the whole answer. Many different alignments reach the same prefix state, making reuse valuable. Unequal-letter columns and double-gap columns are forbidden, rather than assigned ordinary finite penalties.

Dynamic programming defines subproblem optima and combines their stored answers through a recurrence. It is useful when optimal substructure and overlapping subproblems allow repeated work to be shared. Several alternative choices can be compared before an optimum is selected.

A greedy algorithm instead commits to a locally selected choice and does not later reconsider it. To be correct, its choice needs a justification such as an exchange argument or a safe-cut property, beyond merely having optimal substructure. Neither method is automatically applicable to every optimization problem. Prim's algorithm illustrates a justified greedy choice; the maximum-segment recurrence below illustrates dynamic programming.