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If the eigenvalues of $A$ are $\ell /2^n$ with $0\le \ell <2^n$, exact quantum phase estimation on $U=e^{2\pi iA}$ uses $n$ controlled powers $U^{2^j}$. Their Hamiltonian simulation times reach $2\pi 2^{n-1}$. Counting each supplied controlled power as one oracle query is valid in that oracle model, but does not prove a polynomial time implementation from generic sparse-matrix access. In the ordinary HHL algorithm, one instead chooses finite precision appropriate to the output error and the condition number, or exploits additional structure giving efficient access to the needed powers.

Solve the maximum-weight assignment problem forbidding fixed points, obtaining a cycle cover of a directed graph whose weight bounds every Hamiltonian cycle above. Remove a lightest edge from each graph cycle. Each graph cycle has at least two edges, so at least half its nonnegative weight survives. Connect the resulting disjoint graph paths cyclically; completeness supplies the connectors and their nonnegative weights cannot reduce the retained sum. The Hungarian algorithm and linear-time patching give a polynomial time approximation algorithm. Negative weights invalidate a universal multiplicative half guarantee: when every weight is $-1$, every tour has weight $-n<(-n)/2$.

Dynamic programming solves a mathematical optimization problem by choosing reusable subproblems, recording their value functions, and combining already solved cases through a recurrence relation. Its attraction is that many possible decision histories can share one continuation calculation. A dynamic programming state must retain everything needed to determine the allowed future choices and their rewards. The 0-1 knapsack problem, for example, needs both the available capacity and which items remain available; capacity alone would lose the restriction that each item can be used at most once.

Two ideas explain when this works. Optimal substructure makes an optimal continuation valid inside a larger optimal solution. When subproblems overlap, storing each answer avoids repeated work: memoization computes entries as needed, while a table fills them in dependency order. Start from known base cases. For a problem indexed by increasing size, compute smaller cases first; for a finite sequence of future decisions, work backwards from the terminal reward, as in the Bellman equation. See Bertsekas, MIT lecture 2: optimality and state augmentation.

Consider a 0-1 knapsack problem with item weights $(w_1,w_2,w_3)=(2,3,4)$, item values $(v_1,v_2,v_3)=(3,4,5)$, and capacity $5$. Define the value function $V_i(W)$ to be the largest total item value obtainable from the first $i$ items with total weight at most $W$. Here $i\in \{0,1,2,3\}$ and $W\in \{0,1,2,3,4,5\}$ specify the dynamic programming state. All weights and capacities are nonnegative integers. The empty selection supplies the base case $V_0(W)=0$. The recurrence relation isThe two candidates skip or take item $i$. Both use the previous row, so taking item $i$ cannot take that same item again. The condition $W\ge w_i$ prevents a negative remaining capacity; no entry with $W<0$ is needed. The optimal substructure is concrete: after taking item $i$, its companions must solve the smaller 0-1 knapsack problem on the first $i-1$ items with capacity $W-w_i$. Otherwise replacing them by a better feasible selection would improve the whole selection. Compare ^[ref]; those notes index the remaining items in the opposite direction.

The full computation is small enough to see at once:For example, $V_2(5)=\max \{3,4+3\}=7$ and $V_3(5)=\max \{7,5+0\}=7$. Choosing the feasible item with largest individual value takes weight $4$, earns $5$, and leaves capacity $1$, which cannot fit another item. This greedy algorithm misses the two-item selection of weights $2$ and $3$, whose value is $7$. This example tests the largest-individual-value rule; choosing by value divided by weight happens to find the optimum here.

The number $V_3(5)=7$ is the optimal value, while the choices attaining it form a control policy. To reconstruct one optimal selection, start at $(3,5)$. Equality with $V_2(5)$ allows skipping item $3$. At $(2,5)$, taking item $2$ gives $4+V_1(2)=7$, so continue to $(1,2)$. Take item $1$ and finish at $(0,0)$. In a tie, either maximizing choice suffices to recover one optimum. Recording choices during the calculation or comparing stored entries afterwards recovers the control policy; a final optimal value alone does not describe the selection.

For $n$ items and integer capacity $W\ge 1$, the table uses time complexity $O(nW)$ and space complexity $O(nW)$, counting arithmetic operations and stored numbers using Big O notation. This is pseudo-polynomial time: when weights and capacity use the binary numeral system, the capacity needs only $O(\log (W+1))$ digits. A capacity near $2^b$ creates about $2^b$ columns despite needing only about $b$ digits. Thus this algorithm is practical for moderate capacities but does not establish polynomial time in the encoded input length. See MIT 6.006 recitation 19: knapsack and input size. For the optimal value alone, one array reduces space complexity to $O(W)$: for each item, update capacities in descending order, from $W$ down to $w_i$. Ascending order would read entries already updated for that item and could reuse it. Reconstruction then requires extra information or recomputation.

The link between the knapsack problem and dynamic programming is already explicit in a 1957 paper by Richard Bellman at RAND: its abstract discusses an exact knapsack problem solution using functional equations and cites his 1957 book. See Bellman, Comment on Dantzig's Paper on Discrete Variable Extremum Problems (1957).

The ellipsoid method repeatedly asks a separation oracle about the center of a containing ellipsoid and replaces that ellipsoid by a smaller one containing its intersection with a separating half-space. The volume decreases by a dimension-dependent factor at each iteration. For rational polyhedra, the exact feasibility theorem has polynomial bit complexity in the dimension and coefficient encoding bounds when separation is polynomial time. It includes lower-dimensional feasible sets through rational precision bounds or suitable relaxations; one must not declare infeasibility just because the original feasible set has zero volume.

An existential second-order logic sentence has the formwhere each $R_i$ is a second-order relation variable of fixed finite arity $k_i$ and $\phi$ is a first-order sentence over the input first-order signature expanded by the $R_i$. The first-order sentence may contain both universal quantification and existential quantification over individuals; only the outer quantification over second-order relation variables is required to be existential. Under full semantics for second-order logic, truth means that some expansion by witness relations satisfies $\phi$.

For a fixed existential second-order logic sentence on a domain of cardinality $n$, the witnesses need ${\sum }_{i=1}^m{n}^{k_i}$ membership bits. The arities and the number of witnesses are constants because the sentence is fixed. After guessing those bits, a verifier evaluates the fixed first-order sentence by enumerating assignments to its finitely many individual variables, taking polynomial time. In the second-order encoding of three-colorability, the three witnesses require $3n$ bits, and the edge constraints can be checked over at most $n^2$ ordered vertex pairs. This is the logical form of the NP principle that a positive answer has a short complexity certificate.

The fixed-sentence qualification matters: it is the input first-order structure that varies, rather than an arbitrarily large formula supplied with each input. Fagin's theorem states the converse as well: every isomorphism-invariant NP property of finite relational first-order structures has such a description. The coloring example uses only unary second-order relation variables; the general theorem allows any fixed finite arities.

Fix a finite relational first-order signature $\tau$, and let $\mathcal{K}$ be a class of finite nonempty $\tau$-first-order structures closed under isomorphism. Fagin's theorem states that $\mathcal{K}$ is definable by one existential second-order logic sentence under full semantics for second-order logic if and only if its language of standard finite encodings belongs to NP. A standard encoding numbers the domain $\{0,\dots ,n-1\}$ and lists membership bits for each input relation, together with $n$; take an explicit encoding whose length is at least $n$. The encoding language contains all numbered copies of members of $\mathcal{K}$, with malformed encodings rejected. The isomorphism condition makes membership a property of the structure rather than an arbitrary choice of labels. Changing between the usual explicit encodings preserves the NP characterization by polynomial time translations.

The forward direction is the witness-and-verifier calculation in existential second-order logic. For the reverse direction, the guiding idea is to encode an accepting nondeterministic computation using second-order relation variables: fixed-length tuples of domain elements index enough time steps and storage positions for a polynomial time computation. A first-order sentence checks the initial input, legal transitions, and acceptance. The number of tuples grows with the input, although the checking sentence stays fixed. No total order must be supplied as part of the input first-order signature: an auxiliary total order can itself be existentially guessed and constrained by a first-order sentence. Isomorphism invariance ensures that a change of this auxiliary numbering does not change membership in $\mathcal{K}$.

The second-order encoding of three-colorability is a small instance of the forward direction. The beauty of Fagin's theorem is the complete correspondence: the computational class NP can be characterized by the shape of a logical description, without mentioning a particular machine in the statement. The scope is finite relational first-order structures and fixed sentences; it is not a claim about deciding arbitrary second-order logic sentences or about arbitrary infinite domains.

Ronald Fagin published the result in Generalized First-Order Spectra and Polynomial-Time Recognizable Sets, SIAM-AMS Proceedings 7 (1974), pp. 43–73, Section 4. The original paper frames the correspondence using classes of finite expansions and their encodings. Fagin's author-maintained publication list explains that the abstract in the available copy was added in 2019; its modern summary should therefore not be mistaken for part of the 1974 printing.

Given finitely encoded linear inequalities, a maximum feasible subsystem is a largest collection that can be simultaneously satisfied. Its decision version asks whether some $k$ inequalities admit a common solution. This is NP-complete, despite polynomial time feasibility testing for each fixed collection. For a graph, use nonnegative variables $x_v$, constraints $x_v\ge 1$ for vertices, and $M>|V|$ copies of $x_u+x_v\le 1$ for each edge. A subsystem of size $M|E|+k$ must retain a copy of every edge constraint and at least $k$ vertex constraints; those vertices form an independent set. Conversely an independent set of size $k$ supplies such a feasible subsystem by its indicator vector.

$\mathbf{NP}$ is the class of decision problems whose positive instances have polynomial-length certificates verifiable in polynomial time. Equivalently, it is polynomial time on a nondeterministic computation.

Fagin's theorem states that, for a finite relational signature, an isomorphism-invariant class of finite structures has its standard encoding language in NP if and only if it is definable by an existential second-order logic sentencewhere $\psi$ is first-order and the relation variables have fixed finite arities. The forward implication includes the substantive encoding of a polynomially bounded accepting computation by relations. In the reverse implication the relations are polynomial-length certificates and the fixed first-order sentence can be checked in polynomial time. This is a finite-structure characterization; no order need be named in the input signature.

For the metric travelling salesman problem, a tour is a polynomial-size complexity certificate: list the vertices in cyclic visiting order. In polynomial time verify that every vertex occurs exactly once, add the encoded integer edge costs and compare the sum with the threshold. The certificate has $O(n\log n)$ bits, and addition uses a number of bits polynomial in the input length. If validity of the metric input is not treated as a promise, nonnegativity, symmetry and all triangle inequalities can also be checked in polynomial time. Therefore the decision problem belongs to NP.

An evaluation algorithm immediately decides a threshold instance by computing the optimum and comparing it with $L$. Conversely, suppose there is a polynomial-time decision algorithm. The optimum is an integer in $[0,nM]$, where $M$ is the largest edge cost; any fixed cyclic ordering gives the upper bound. Binary search on this interval using threshold queries finds the exact minimum. It makes $O(\log (nM+1))$ queries, and every queried threshold has polynomial encoding length. Thus the time is polynomial in the bit length, even when $M$ is numerically large. Zero-cost and trivial small instances can be handled directly. ConsequentlyThis concerns the optimum value; constructing an optimal tour is the separate optimization task.

For the hardness reduction, start with an undirected graph $G$ on $n\ge 3$ vertices and form the complete graph with costsThese costs satisfy the triangle inequality, since every direct cost is at most two and every two-edge path costs at least two. Set the threshold to $n$. Every tour has exactly $n$ edges, each costing at least one. Hence it has cost at most $n$ precisely when all its edges came from $G$, which is precisely the existence of a Hamiltonian cycle in $G$. The construction and threshold encoding are polynomial-time. Thus the assumed NP-completeness of the Hamiltonian circuit problem gives NP-hardness of metric TSP; together with its NP membership this proves NP-completeness of metric TSP decision.

For the approximation, compute a minimum spanning tree $T$. Deleting any one edge of an optimal tour leaves a spanning tree, so nonnegative costs implyDouble every tree edge. The resulting connected multigraph has even degrees and hence an Euler circuit, of cost $2c(T)$. Traverse it and skip previously visited vertices, except for the final return to the start. Each skip replaces a path by its direct edge; repeated use of the triangle inequality shows that no replacement increases cost. The result is a tour visiting every vertex exactly once and satisfyingThe Euler traversal and shortcutting take polynomial time, so the assumed polynomial-time MST routine supplies the requested polynomial-time algorithm. This is the double-tree approximation for metric TSP. In the relative-error convention, $c/\mathrm{OPT}-1\le 1$ when the optimum is positive, explaining the printed term “1-approximation”; in the multiplicative convention it is a factor-two approximation. If the optimum is zero, the same bound forces the returned tour to have zero cost as well.

A decision problem is in NP if every yes-instance has a complexity certificate of length bounded by a polynomial in the input length, and a deterministic verifier can check that certificate in polynomial time. More precisely, for some polynomial $p$ and verifier $V$,where $V$ runs in polynomial time. Equivalently, a nondeterministic machine decides the problem in polynomial time. A problem is NP-complete if it belongs to NP and is NP-hard: every problem in NP has a polynomial-time many-one reduction to it. Such a reduction maps instances $w$ to instances $r(w)$ with $w\in L$ if and only if $r(w)$ is a yes-instance of the target problem.

For 3-SAT, the complexity certificate is one truth value for each Boolean variable. Check every Boolean literal, negate its variable's truth value when appropriate, evaluate the disjunction in each Boolean clause, and check their conjunction. This takes time linear in the encoded Boolean formula, apart from equally polynomial bookkeeping for variable names. A satisfying assignment passes; an unsatisfiable Boolean formula has no passing assignment. Thus 3-SAT belongs to NP.

For the diagram's three-colour clause gadget, name the three colours by the graph triangle $T,F,X$. Each input $a,b,\bar{c}$ is adjacent to $X$, so each has colour $T$ or $F$. Suppose that none has colour $T$. Then all three have colour $F$. Because $x_1$ is adjacent to $a$ and $x_2$ to $b$, neither has colour $F$. The edge $x_1{x}_2$ forces these two colours to be $T$ and $X$ in some order. The graph triangle $x_1,x_2,x_3$ now forces $x_3$ to have colour $F$. Since $x_4$ is adjacent to $x_3$ and $T$, it must have colour $X$. The graph triangle $x_4,x_5,T$ forces $x_5$ to have colour $F$. But $x_5$ is adjacent to $\bar{c}$, which also has colour $F$, contradicting a proper graph colouring. Therefore at least one input has colour $T$.

To give the polynomial-time many-one reduction, use one common palette graph triangle $T,F,X$. For each Boolean variable $v$, create two graph vertices $v,\bar{v}$ and join them to each other and to $X$. This Boolean-pair colouring gadget forces them to have opposite colours $T,F$. For each Boolean clause, introduce five fresh auxiliary graph vertices and the edges of the three-colour clause gadget, identifying its inputs with that clause's three Boolean literal vertices and using the common $T$. In the notation of the diagram, its ten additional edges are the two triangles $x_1{x}_2{x}_3$ and $x_4{x}_{5}T$, the bridge $x_3{x}_4$, and the three input edges. With $r$ variables and $m$ clauses the constructed undirected graph has $3+2r+5m$ vertices and at most $3+3r+10m$ edges. Repeated literals do not affect the argument. Its construction takes polynomial time.

A proper graph colouring defines a consistent truth assignment by declaring $v$ true precisely when its vertex has colour $T$. The only-if result just proved makes every clause true. Conversely, a satisfying assignment colours every literal vertex consistently, and the supplied if-direction extends each satisfied clause's colours to its fresh auxiliary vertices. These extensions do not conflict, since different clauses share only already-coloured palette and literal vertices. Thus the Boolean formula is satisfiable exactly when the constructed undirected graph is three-colourable. Since 3-SAT is NP-complete, composition of reductions proves NP-hardness of the three-colourability problem.

Finally, a list of three colour labels, one for each graph vertex, is a complexity certificate for the three-colourability problem. Checking all edges takes $O(|V|+|E|)$ time. Hence this problem is in NP as well, and the three-colourability problem is NP-complete.

For a connected undirected weighted graph, a spanning tree connects all vertices without a graph cycle. A minimum spanning tree minimizes the sum of its edge weights. Disconnected input has no spanning tree, although the same methods produce a minimum spanning forest. Negative edge weights and ties cause no difficulty; a tree always contains exactly $|V|-1$ edges.

The central exchange argument is the minimum spanning tree cut property. Suppose a forest $F$ is contained in some minimum spanning tree $T$, and consider a cut with no edge of $F$ crossing it. Let $e$ be a lightest crossing edge. If $e\notin T$, adding it creates a graph cycle; the graph path in $T$ between its endpoints contains a crossing edge $f$. That $f$ is not in $F$, and $w(e)\le w(f)$. Thus $T-f+e$ is another spanning tree with no larger weight and still contains $F$. It is also minimum and contains $F\cup \{e\}$. This proves that the chosen edge is safe, rather than merely asserting a greedy choice.

Kruskal's algorithm sorts all edges by increasing weight, starts with the empty forest, and adds an edge precisely when its endpoints are in different components. Each added edge is lightest across a component cut respected by the current forest. The exchange proof inductively keeps the forest inside a minimum spanning tree; connectedness makes the final forest a spanning tree, hence an optimum. Sorting takes $O(|E|\log |E|)$ comparisons, and even a simple component-merging implementation takes only polynomial additional work. Efficient component tracking improves that overhead further.

Prim's algorithm instead grows a single tree, repeatedly adding a lightest edge from its vertex set to the outside. The same cut argument proves correctness. A dense implementation using an array runs in $O(|V{|}^2)$ comparisons. The dual minimum spanning tree cycle property says a uniquely heaviest edge of a graph cycle belongs to no minimum spanning tree: exchanging it for a lighter crossing edge strictly improves any tree containing it. Ties must be handled carefully; a merely nonunique heaviest edge can occur in an optimum.

Complexity classes are formally classes of decision problems, so use the version asking whether a graph has a spanning tree of total weight at most a supplied threshold $K$. Assume integer or rational weights and threshold are encoded in binary. The input length includes all these bits, not only the numbers of vertices and edges. P consists of decision problems decidable by a deterministic algorithm in time polynomial in that input length. NP consists of those with polynomial-size complexity certificates verified in deterministic polynomial time, equivalently decidable in polynomial time by a nondeterministic machine.

Kruskal's algorithm constructs the optimum and compares its weight with $K$. Binary integer comparisons and sums are polynomial-time operations. Rational comparisons can be made by cross multiplication, and their summed numerator and denominator have polynomially bounded bit length: a product of input denominators has bit length no greater than the sum of their bit lengths. The combinatorial algorithm makes polynomially many such operations. Hence the minimum-spanning-tree decision problem is in P; the optimization version likewise has a polynomial-time algorithm.

For membership in NP, the complexity certificate is a list of $|V|-1$ selected edges. Check that every edge is present, that the edges connect all vertices and contain no graph cycle, and that their total weight is at most $K$. A graph search and binary arithmetic do this in polynomial time. A successful complexity certificate need not establish that its tree is minimum: it proves the threshold question's yes-answer. Conversely, a yes-instance has such a tree as a complexity certificate. ThereforeMembership in NP does not arithmetic mean NP-hardness; here the explicit greedy algorithm provides the stronger tractability result.