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An odd composite number $N$ is a Fermat pseudoprime to a coprime base $b$ when

A composite number is an integer greater than one that is not a prime number. Equivalently, it has a factorization $n=ab$ with integers $a,b>1$. Thus $4=2\cdot 2$, $9=3\cdot 3$, and $12=3\cdot 4$ are composite numbers; one is neither a composite number nor a prime number.

Every composite number has a prime factor at most its square root. In $91=7\cdot 13$, seven is the smaller factor and lies below $\sqrt{91}$; in $49=7^2$, equality occurs. This square-root bound makes trial division finite and lets the sieve of Eratosthenes stop marking with prime numbers beyond $\sqrt{N}$ when listing prime numbers up to $N$.

A prime number is a positive integer $p>1$ whose only positive divisors are $1$ and $p$. The first ten prime numbers are $2,3,5,7,11,13,17,19,23,29$. In contrast, $12=3\cdot 4$ is a composite number because it has a positive divisor strictly between one and itself. The integer one is neither a prime number nor a composite number: the lower bound $p>1$ is part of the definition. Equivalently, the divisor function satisfies $\tau (p)=2$ for a prime number, whereas $\tau (1)=1$ and $\tau (n)\ge 3$ for every composite number $n$.

Prime numbers are the building blocks of positive integers under multiplication. The fundamental theorem of arithmetic says that every integer greater than one is a product of prime numbers, with a unique collection of factors when their order is ignored. For example, $12=2\cdot 2\cdot 3$ and $18=2\cdot 3\cdot 3$. Their common factors reveal the greatest common divisor $6$, and the combined maximum powers reveal the least common multiple $36$. A complicated divisibility question becomes a comparison of prime exponents.

There is also a useful finite test. If a composite number $n$ equals $ab$ with $1<a\le b$, then $a^2\le n$. Therefore some prime factor is at most $\sqrt{n}$, and trial division only needs to check prime numbers up to that square root. For $97$, checking $2,3,5,7$ suffices; none divides $97$, so it is a prime number. For $91$, the same test finds $7$, giving $91=7\cdot 13$. To find many prime numbers together, the sieve of Eratosthenes reuses the work by marking multiples.

The division with remainder calculations for $97$ make the test explicit:Since $9^2<97<10^2$, these are all the prime numbers at most its square root, and every remainder is nonzero. Thus the calculation certifies a prime number, without testing the integers $8,9,10,\dots ,96$ individually. In contrast, $91=7\cdot 13+0$ gives a prime factor and proves that $91$ is a composite number.

The definition has a long history. Book VII of Euclid's Elements distinguishes prime numbers from composite numbers through their possible measures; Book IX, Proposition 20 establishes that no finite list exhausts the prime numbers. The modern argument under Euclid's theorem needs only divisibility, so it proves an infinite conclusion without estimating how frequently prime numbers occur. ^{[ref] [ref]}.

For integers $a,b$ and a positive integer modulus $n$, modular congruence meansThus $a-b=kn$ for some integer $k$: the two integers differ by complete cycles of length $n$. For example, $14\equiv 2(\mathrm{mod}12)$ because $14-2=12$, and $-1\equiv 11(\mathrm{mod}12)$ because $-1-11=-12$. A modular congruence need not be an equality of integers.

On a twelve-position clock labelled $0,1,\dots ,11$, adding five to nine follows $9\to 10\to 11\to 0\to 1\to 2$. Hence $9+5=14\equiv 2(\mathrm{mod}12)$: the final position remembers the remainder while forgetting a complete revolution.

The same idea tracks any count repeating every $n$ steps. It also lets us compute sums and products with small residue classes, reducing intermediate results without changing the final class. The justification comes from addition and multiplication being well-defined on residue classes.

In Disquisitiones arithmeticae (1801), Carl Friedrich Gauss defines modular congruence by divisibility in article 1; article 2 describes the integers $a+kn$ and adopts $\equiv$. These passages document his presentation and notation, without claiming that repeating remainders began in 1801. ^[ref]

For integers $d,n$, $d$ divides $n$ when multiplication by some integer $k$ takes $d$ to $n$:Then $d$ is a divisor of $n$, and $n$ is an integer multiple of $d$. The notation $d\nmid n$ means that no such integer $k$ exists. For $d\ne 0$, this is equivalent to saying that the quotient $n/d$ is an integer, or that Euclidean division by $|d|$ leaves remainder zero.

For example, $3\mid 12$ because $12=3\cdot 4$, whereas $5\nmid 12$ because $12=5\cdot 2+2$. Signs do not change divisibility: $-3\mid 12$ because $12=(-3)(-4)$, and $3\mid -12$ because $-12=3(-4)$. In general, changing the sign of either input preserves the relation.

The definition here permits zero as either input. Every integer divides zero, since $0=d\cdot 0$. In particular, $0\mid 0$; however, $0\nmid n$ whenever $n\ne 0$, since $0k$ is always zero. This is a statement about multiplication, and assigns no meaning to division by zero: $0/0$ remains undefined.

Two useful properties come directly from the defining multiplication. Divisibility is a reflexive relation, since $n=n\cdot 1$, and a transitive relation: if $b=ar$ and $c=bs$, then $c=a(rs)$. Moreover, a shared divisor divides every integer linear combination:Indeed, writing $a=du$ and $b=dv$ gives $ax+by=d(ux+vy)$. For example, $6\mid 12$ and $6\mid 18$ imply $6\mid (18-12)$. This small observation explains why the Euclidean algorithm preserves common divisors, and why Bézout's identity can certify a greatest common divisor.

On the positive integers, divisibility is a partial order: if $a\mid b$ and $b\mid a$, then $a=b$. The restriction matters, since $3\mid -3$ and $-3\mid 3$ although $3\ne -3$. The prime factorization viewpoint makes this order concrete: $d\mid n$ exactly when every prime valuation of $d$ is at most the corresponding prime valuation of $n$. Thus one relation connects equal grouping, prime factorization, and modular arithmetic.

Book VII, definitions 3 and 5 of Euclid's Elements describe the corresponding positive-number relation through one number measuring another and through integer multiples. The sign and zero conventions above belong to the present integer formulation, rather than to those ancient definitions. ^[ref]

An integer is a whole number, positive, negative, or zero. The set of all integers isThe positive integers are $1,2,3,\dots$; the negative integers are their opposites $-1,-2,-3,\dots$. Zero is neither positive nor negative. The nonnegative integers include zero and all the positive integers. On a number line, consecutive integers are one unit apart, negative integers lie to the left of zero, and positive integers lie to its right. ^[ref]

The integers let counting describe changes in both directions. If a counter starts at $-3$ and increases by five, it reaches $2$: $-3+5=2$. Decreasing that result by four gives $2+(-4)=-2$. In addition on the number line, the sign of the added integer chooses the direction and its absolute value gives the distance. The additive inverse $-a$ reverses the change made by $a$.

The integers have closure under a binary operation for addition, subtraction and multiplication: whenever $a,b\in \mathbb{Z}$, the results $a+b$, $a-b$ and $ab$ are still integers. Together these operations give the integers their ring structure.^[ref] For instance, $-3+5=2$, $-3-5=-8$, and $(-3)\cdot 5=-15$. General division has no such closure: $1/2$ is a rational number but not an integer. Some divisions are exact, such as $(-6)/3=-2$; division by zero is undefined. Euclidean division instead produces an integer quotient and a remainder, so $7=3\cdot 2+1$ retains the information that ordinary division puts into the fraction $7/3$.

An integer is the value, whereas its written numeral is an encoding. The decimal representation $205_{10}$ and the binary representation $11001101_2$ name the same integer. Changing positional notation changes how that value is displayed and how calculations are organized; it does not change the value itself.

Analytic number theory studies integers, prime numbers, and arithmetic functions using tools from real analysis and complex analysis, especially Dirichlet series and their singularities.

An arithmetic function is a function whose domain of a function is the positive integers, usually taking values in the complex numbers. Thus $f(1),f(2),\dots$ are its values; $f(0)$ is outside this domain of a function. Examples include the divisor function, with $\tau (12)=6$, and the sum-of-divisors function, with ${\sigma }_1(12)=28$. Studying an arithmetic function packages a question about each positive integer into one object whose values can be combined using Dirichlet convolution. ^[ref]

A linear functional minus one half of a positive-definite quadratic has the unique maximizer obtained by solving its linear first-order equation.

Mathematical finance uses probability theory, linear algebra and mathematical optimization to study financial markets, portfolio decisions and contingent claim prices. One question asks which portfolio best serves a specified objective; another asks what price a contingent claim must have to exclude arbitrage. The latter can have an exact answer even when future stock prices are uncertain.

The central idea is claim replication: construct a self-financing portfolio delivering the same financial payoff as a contract in every possible state. The law of one price then identifies the contract's price with the portfolio's initial cost. The binomial market makes this beautiful connection between uncertainty and linear algebra visible using just two equations. Risk-neutral pricing expresses the same cost as a discounted expectation.

The one-step example below fixes today's stock price at 100, terminal prices at 120 or 80, and the bank account's gross growth factor at 1.05. A European call option with strike price 100 pays 20 or zero. Its replicating portfolio in a binomial market costs $250/21$, independently of the physical up probability, provided both states remain possible. These are conclusions within the specified financial market model.

Game theory studies decisions whose consequences depend on what other decision makers choose. A bidder's gain depends on competing bids in an auction; a driver's travel time depends on other drivers in a congestion game. This strategic interdependence makes choosing well a different problem from optimization against a fixed environment: each participant's best response depends on the others' choices.

A normal-form game specifies the players, their available pure strategies, and their payoffs for each strategy profile. Larger payoffs mean more preferred outcomes for that player. Unless stated otherwise, a two-player table below lists $(u_1,u_2)$: the row player's payoff first, the column player's second. Each player maximizes their own expected utility; the numbers need not represent money, and comparing different players' numerical scales is unnecessary.

The central question is whether everyone's choices fit together. A Nash equilibrium consists of mutual best responses: no unilateral deviation improves a player's expected payoff. It can involve definite choices, or mixed strategies that deliberately randomize. Matching pennies shows why randomization can be essential; the coordination game shows how players can share an interest in coordinating while disagreeing over which outcome to choose. Such shared gains are absent from a two-player zero-sum game, where one player's gain is exactly the other's loss.

These ideas are useful both for analyzing incentives and for designing them. Mechanism design asks how changing the rules changes participants' best responses; evolutionary game theory studies how successful strategies spread. The mathematical appeal is that local comparisons of payoffs can reveal a globally consistent pattern of choices. Consistency alone does not imply Pareto efficiency, uniqueness, or that actual play will converge to a Nash equilibrium.

John Nash's ^[ref] established existence of a mixed strategy Nash equilibrium for every finite game. His ^[ref] developed the theory beyond two-player zero-sum games. The distinction between conflicting and shared interests is fundamental to the subject.

A convex set permits every intermediate mixture of two of its points. If two choices meet its constraints, continuously averaging them also meets those constraints. This geometric property makes many feasible regions easier to explore: there is no forbidden gap along a straight move between feasible choices.

Formally, a subset $C$ of a real vector space is convex whenThe parameter $t$ traverses the entire line segment, including both endpoints. The definition does not require a closed subset, a bounded set, or a nonempty interior. The empty set, the whole vector space, and every affine subspace are convex; an open unit disk and a closed unit disk are both convex.^[ref]

The triangle inequality verifies the disk example directly: if $\Vert x\Vert ,\Vert y\Vert \le 1$, then $\Vert (1-t)x+ty\Vert \le (1-t)\Vert x\Vert +t\Vert y\Vert \le 1$. For the open disk, the last inequality is strict. The unit circle fails the same test: $(1,0)$ and $(-1,0)$ belong to it, but their midpoint $(0,0)$ does not. A curved route along the unit circle connects the endpoints, so being path-connected is weaker than being a convex set.

The distinction between a convex set and a convex function matters. The former concerns where one may choose a point; the latter concerns how a value changes along line segments. Convex optimization combines both: averaging preserves feasibility, and a convex function obeys an upper bound by the corresponding average of endpoint values. A convex set alone does not make an arbitrary objective a convex function or eliminate its inferior local minima.

Linear programming optimizes a linear function subject to finitely many linear equalities and inequalities.

For nonnegative $a_1,\dots ,a_n$,with equality exactly when all the $a_i$ are equal.

For continuously differentiable functions $f,g$ on an open subset of ${\mathbb{R}}^n$, suppose $z_{\ast }$ is a local extremum of $f$ subject to $g(z)=b$, with $\nabla g(z_{\ast })\ne 0$. Then a real number $\lambda$, the Lagrange multiplier, satisfiesThe nonzero gradient assumption excludes singular points of the optimization constraint. With several equality optimization constraints, require their gradients to be linearly independent; the gradient of $f$ is then a linear combination of them. These are necessary conditions, not automatic certificates of a global maximum or global minimum.

Why can the gradient be nonzero at an optimum? In unconstrained optimization, every direction is available at an interior local extremum, so all directional derivatives vanish and $\nabla f=0$. On a regular equality feasible set, only directions in its tangent space are available to first order. The gradient of $f$ must be perpendicular to those directions, so it is parallel to the gradient of $g$. In two dimensions this explains the tangency of their level sets. See MIT, Lagrange multipliers and contour geometry.

For the product example, set $g(x,y)=x+y$ and use the optimization LagrangianIts partial derivatives give $y-\lambda =0$, $x-\lambda =0$, and $x+y=10$. Hence $x=y=\lambda =5$. At $(5,5)$, $\nabla f=(5,5)=5(1,1)=\lambda \nabla g$: increasing both inputs would help, but is infeasible. Along the allowed direction $(1,-1)$, the directional derivative is zero. The endpoints must still be checked because of the nonnegativity optimization constraints; completing the square establishes the unique global maximum.

The sign convention matters: writing $L=xy+\mu (x+y-10)$ instead gives $\mu =-5$. Neither equality Lagrange multiplier has a universal nonnegativity requirement. Also, the displayed $L$ is unbounded above along $x=y=t$, so stationarity here does not meet the global hypothesis of the Lagrangian sufficiency theorem.

The Lagrange multiplier also measures the value of extra resources. For a positive budget $b$, the optimizer is $(b/2,b/2)$, the optimum value is $V(b)=b^2/4$, and $V^{\prime }(b)=b/2=\lambda$. At $b=10$, an extra $0.1$ units predicts a gain of $0.5$; the exact gain is $V(10.1)-V(10)=0.5025$. This is the envelope theorem in a computed sensitivity analysis: the multiplier translates a binding budget constraint into a marginal benefit.

Convex optimization is a minimization problem with a convex function as its objective function and a convex set of feasible points. For a finite-valued $f:C\to \mathbb{R}$, the problem is to find $z_{\ast }\in C$ with $f(z_{\ast })\le f(z)$ for every $z\in C$. The displayed infimum describes its value even when no minimizing point exists. The two convexity requirements concern different objects: the allowed decisions and the scores assigned to them.

A convex set contains the whole line segment between any two of its points:A convex function lies below the interpolation of its endpoint values:Thus mixing two allowed decisions preserves feasibility, and the mixed score is no worse than the same mixture of endpoint scores. The objective function may have flat regions or corners. For a strictly convex function, the second inequality is strict when $u\ne v$ and $0<t<1$.

A useful constraint representation iswhere $f,g_i$ are convex functions on a common convex set $D$, and the equalities are affine functions. Each inequality defines a convex sublevel set; their intersection with $D$ and the equality set is a convex set. An equality involving a general convex function need not preserve this geometry: $x^2=1$ allows $-1$ and $1$ but excludes their midpoint $0$. See Boyd and Vandenberghe, Convex Optimization (2004), §4.2.1 for the standard problem form.

Why local information can certify a global answer. Every local minimum of a convex optimization problem is a global minimum, including minima on the boundary. If a feasible point $v$ had $f(v)<f(u)$, points $u+t(v-u)$ would remain feasible and have value strictly below $f(u)$ for every $0<t<1$. Taking $t$ arbitrarily small contradicts a local minimum at $u$. This argument explains the special power of convexity: a better distant decision would reveal an improving direction immediately nearby.

If a minimizer exists and the objective function is strictly convex on $C$, it is unique. Two distinct minimizers of value $p_{\ast }$ would have a feasible midpoint with value strictly below $p_{\ast }$. Existence is a separate question: the exponential function $e^x$ is strictly convex on $\mathbb{R}$, but its infimum $0$ is approached only as $x\to -\infty$. Even a bounded convex set need not suffice: $x^2$ on $0<x<1$ also has unattained infimum $0$. A continuous function on a nonempty compact set attains its minimum by the extreme value theorem.

Nor does strict convexity imply differentiability. For $f(x)=x^2+|x|$, the squared term gives the strict inequality while the triangle inequality gives the weak inequality for the absolute value. At zero the left and right derivatives are $-1$ and $1$, so no derivative exists there, although zero is the unique minimizer. Such objectives require subgradients or proximal operators rather than an ordinary gradient at every point.

For a differentiable convex function on an open convex set containing $C$, the constrained first-order condition isIndeed, the subgradient inequality $f(z)\ge f(z_{\ast })+\nabla f(z_{\ast }{)}^T(z-z_{\ast })$ makes this sufficient; taking the one-sided derivative along feasible line segments makes it necessary. At an interior minimizer, both directions are feasible and the gradient must vanish. At a boundary minimizer it can be nonzero: it only needs to make every feasible direction nondecreasing. See Boyd and Vandenberghe, §§3.1.3 and 4.2.3.

A fully computed nearest-point problem. Project $a=(2,1)$ onto the half-space $C=\{(x,y):x+y\le 1\}$ by minimizing half the squared Euclidean distance:The Hessian matrix is the identity matrix, so the objective function is strictly convex. The unconstrained minimizer $(2,1)$ violates the inequality. With the convention $g\le 0$, the optimization Lagrangian is $L=f+\lambda g$, where the Lagrange multiplier satisfies $\lambda \ge 0$. Its Karush-Kuhn-Tucker conditions areThe last two equations give $x=2-\lambda$, $y=1-\lambda$. The choice $\lambda =0$ is infeasible, so complementary slackness forces $x+y=1$. Consequently $3-2\lambda =1$, giving $\lambda =1$, $(x_{\ast },y_{\ast })=(1,0)$ and $f(x_{\ast },y_{\ast })=1$. The Euclidean distance is $\sqrt{2}$, rather than the objective function value $1$. The gradient $(-1,-1)$ at the answer balances $\lambda \nabla g=(1,1)$; the displacement from $(1,0)$ to $(2,1)$ is a normal vector to the boundary.

There is an exact certificate independent of numerical search. Completing squares givesFor every feasible point, $g\le 0$ implies $L(x,y,1)\le f(x,y)$. Thus every feasible objective is at least $1$, and $(1,0)$ attains this bound. More generally the Lagrangian dual problem has objective $q(\lambda )={\inf }_{x,y}L(x,y,\lambda )=2\lambda -{\lambda }^2=1-(\lambda -1{)}^2$ for $\lambda \ge 0$. Its maximum is also $1$, illustrating strong duality through matching computed bounds. This makes the example a model for Euclidean projection onto a convex set and the projection step in projected gradient descent.

What the multiplier conditions require. For differentiable convex functions $f,g_i$ on an open convex set $D$ and equalities $Az=b$ defined by affine functions, the KKT conditions combine feasibility, ${\lambda }_i\ge 0$, complementary slackness ${\lambda }_i{g}_i(z_{\ast })=0$, and stationaritywith unrestricted equality Lagrange multipliers $\nu$. They suffice to certify a global minimum without an additional regularity assumption: the optimization Lagrangian is convex, stationarity minimizes it, and feasibility with complementary slackness gives the same certificate as above. To guarantee their necessity at an attained finite optimum, a regularity hypothesis suffices: Slater's condition requires some $\bar{z}\in D$ with $g_i(\bar{z})<0$ and $A\bar{z}=b$. The projection example has the strictly feasible point $(0,0)$. See Boyd and Vandenberghe, §§5.2.3 and 5.5.3. This hypothesis guarantees suitable multipliers, not existence of a primal minimizer.

Without regularity, necessity can fail even for convex functions: minimize $f(x)=x$ subject to $g(x)=x^2\le 0$. The sole feasible point $x=0$ is optimal, but stationarity would require $1+\lambda \cdot 0=0$. No Lagrange multiplier works. There is no strictly feasible point, and the gradient of the constraint vanishes at the answer.

The nonconvex contrast. The polynomial $h(x)=(x^2-1{)}^2+0.2x$ has derivative $h^{\prime }(x)=4x^3-4x+0.2$ and second derivative $h^{\prime \prime }(x)=12x^2-4$. Numerically solving the cubic gives stationary points $-1.024120300$, $0.050125947$, and $0.973994353$. Their objective values are respectively $-0.202440434$, $1.005006281$, and $0.197434153$, with second derivatives $8.585869$, $-3.969849$, and $7.383980$. Hence both outer points are local minima, and the middle point is a local maximum. Since $h(x)\to +\infty$ at both ends, its global minimum exists and must be one of the stationary minima: it is the left one. A method stopping at the right local minimum misses a better value by approximately $0.399874587$.

The practical appeal of convex optimization is the ability to check an answer globally using geometry and matching Lagrangian duality bounds. The projection certificate certifies every feasible alternative at once; the tilted quartic shows why a zero gradient alone cannot do that for general nonlinear optimization.

Historically, an early landmark is Kuhn and Tucker's ^[ref]. Its §3 explicitly states a regularity hypothesis for inequality-constrained Lagrange multipliers; its discussion of saddle points explains how concave functions support global conclusions for maximization. This documents an early formulation of the multiplier approach; it does not establish sole discovery of today's KKT conditions or of convex optimization as a whole.

The Karush-Kuhn-Tucker conditions combine membership of the feasible set with a balance of gradients and complementary slackness. For a minimization problem with differentiable functions on an open subset of ${\mathbb{R}}^n$, writeand use the optimization LagrangianA candidate $(x_{\ast },\lambda ,\mu )$ satisfies the KKT conditions exactly when all four requirements hold:

The sign of each inequality Lagrange multiplier follows the chosen convention $g_i\le 0$: on the feasible set, ${\lambda }_i{g}_i\le 0$ makes the optimization Lagrangian a lower bound on the objective function. An equality contributes zero on the feasible set for either sign of its Lagrange multiplier. Complementary slackness says an inequality with $g_i(x_{\ast })<0$ must have ${\lambda }_i=0$, while ${\lambda }_i>0$ implies an active constraint. An active constraint can still have zero Lagrange multiplier; the quadratic minimization with a moving upper bound below gives an explicit example.

The name records William Karush's 1939 master's thesis and the 1951 work of Harold W. Kuhn and Albert W. Tucker.^[ref]

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).

For maximizing $f$ subject to $g_i(x)\ge 0$ and $h_j(x)=0$, suppose a feasible $x^{\ast }$ and multipliers ${\lambda }_i\ge 0,{\mu }_j$ satisfy complementary slackness and $x^{\ast }$ globally maximizesThen $x^{\ast }$ globally maximizes $f$ on the feasible set. Convexity or concavity hypotheses are commonly used to prove the required global extremum of the Lagrangian from first-order conditions.

Biopolymer mechanics studies bending, stretching, thermal fluctuations, and force-induced deformation of filamentous biological macromolecules.