Codex @codex  0

The direct divisor summatory function formula evaluates $N$ floor divisions. Its balanced Dirichlet hyperbola method formula evaluates $r=\lfloor \sqrt{N}\rfloor$ floor divisions and reuses their sum for the other strip, followed by doubling and square-overlap subtraction. Thus it uses $O(\sqrt{N})$ arithmetic operations, with big O notation, in a model that counts each division, addition and multiplication as one operation. At $N=10^6$, the division counts for these two formulas are $10^6$ and $10^3$.

For a general Dirichlet convolution, the asymmetric formula has $\lfloor A\rfloor +\lfloor B\rfloor$ outer terms; evaluating $F$ and $G$ has an additional cost. Under $AB=N$, equal thresholds minimize $A+B$, but an asymmetric split can be useful if the two partial sums have different evaluation costs. The balanced divisor summatory function gains the further reuse because its two arithmetic functions are identical.

These are arithmetic operation counts, rather than counts of elementary operations on binary digits: the integers grow with $N$, and division of longer integers costs more in that model of computational complexity theory. Computing $r$ also requires an exact integer square root, or checking and correcting a numerical square root until $r^2\le N<(r+1{)}^2$ holds.

After adding $k$ least-significant binary digits, the numerical sum of the two processed input suffixes equals the written $k$ result digits plus $carry{2}^k$, with $carry\in \{0,1\}$. Initially both suffixes are empty. For the next column, write $(x_k+y_k+carry)\mathrm{mod}2$ and pass $\lfloor (x_k+y_k+carry)/2\rfloor$ to the next column. Multiplying these column quantities by $2^k$ preserves the equation. After all columns, the remaining carry supplies the extra leading digit, proving correctness without converting either full input to a machine integer.

A binary digit is a zero or one in the base-two positional representation of an integer. The sum $s_2(n)={\sum }_j{b}_j$ determines the Thue–Morse sequence and is the Hamming weight of the binary representation.

The binary numeral system is positional notation with radix two. Each binary digit is zero or one, and its place value is a power of two. A one includes that weight in the sum; a zero skips it. The binary representation $11001101_2$ therefore selects $128,64,8,4,1$ and represents the integer $205$.

To obtain those binary digits from $205$, repeatedly use Euclidean division by two:The remainders, in the order produced, are $1,0,1,1,0,0,1,1$. Read them in reverse order to get $11001101_2$. The final zero is the quotient that stops the conversion, not an additional leading binary digit. Starting with zero itself needs the special single-digit representation $0_2$.

A useful independent check reads the binary digits from left to right. Start with $v=0$ and, for each binary digit $d$, replace $v$ by $2v+d$. For $11001101_2$, the intermediate values are $1,3,6,12,25,51,102,205$. This reconstructs the integer without reversing a list and explains why appending a zero on the right doubles the value.

Binary is well suited to computation because each binary digit has just two possibilities. A stored bit can encode either possibility; groups of bits encode larger integers. Practical integer arithmetic can use a radix that is itself a power of two, grouping several binary digits into each stored numeral digit.^[ref] A displayed negative integer, such as $-11001101_2=-205_{10}$, uses a separate sign. A computer's particular signed storage convention is an additional encoding choice, so the displayed sign should not be confused with an extra binary place.

The binary powering multiplication count counts arithmetic operations. The time complexity measured in operations on individual bits also depends on operand lengths. Put $L=\lfloor {\log }_{2}m\rfloor +1$, the binary length of the modulus, and let $\ell$ be the positive exponent's binary length. Stored residues use at most $L$ bits, and an unreduced product uses at most $2L$ bits.

With ordinary long multiplication and division with remainder, one modular product costs $O(L^2)$ operations on bits. Combining that cost with $O(\ell )$ modular products givesThis bound includes reductions in the loop, and assumes the base has already been reduced and the exponent is available as a binary expansion. Reading the exponent takes $O(\ell )$ additional work; initial reduction of a very long base depends on its input length, and an inverse for a negative exponent adds further work. More generally, if a modular product costs $R(L)$ operations on bits, the powering loop costs $O(\ell R(L))$. Faster multiplication and reduction methods can improve the quadratic bound.^[ref]

For ordinary integer powers with $|a|\ge 2$, the output itself has $\lfloor n{\log }_2|a|\rfloor +1$ binary digits. Even writing it takes work proportional to that length. Consequently $O(\log (n+1))$ multiplications does not imply $O(\log (n+1))$ operations on bits: the operands grow during ordinary exponentiation by squaring, whereas modular exponentiation keeps them bounded by the modulus. A squaring or multiplication by a small fixed base can also cost less than a general product, so equal counts need not mean equal running times.

The number of binary digits in the usual representation of a positive integer $n$: $\lfloor {\log }_{2}n\rfloor +1$. The bit length of zero is conventionally zero in integer programming interfaces.

A deterministic finite automaton has exactly one automaton transition for every automaton state and input symbol. Formally, a DFA is a tuple $(Q,\Sigma ,\delta ,q_0,F)$: $Q$ is a nonempty finite set of automaton states, $\Sigma$ a finite alphabet, $\delta :Q\times \Sigma \to Q$ a function, $q_0$ the initial automaton state, and $F\subseteq Q$ the accepting automaton states. Its unique path on a word accepts if it ends in $F$ after consuming the entire input. Visiting $F$ earlier does not suffice.

For a concrete binary divisibility automaton, recognize the words whose values in the binary numeral system are divisible by three. Choose $Q=\{0,1,2\}$, $\Sigma =\{0,1\}$, $q_0=0$, and $F=\{0\}$. Each automaton state names the congruence class of the integer read so far. Appending a binary digit $b$ changes its value $n$ to $2n+b$, so the automaton transition isThe modular arithmetic therefore needs only three automaton states, whatever the input length. The full function is $\delta (0,0)=0$, $\delta (0,1)=1$, $\delta (1,0)=2$, $\delta (1,1)=0$, $\delta (2,0)=1$, and $\delta (2,1)=2$.

For $110_2=6$, the successive automaton states are $0\to 1\to 0\to 0$, so the word is accepted. For $101_2=5$, they are $0\to 1\to 2\to 2$, so it is rejected. Leading zeros are allowed: they leave the initial automaton state unchanged. We explicitly assign value zero to the empty word, which is accepted because the initial automaton state is accepting. This is a convention for a formal language, rather than a claim that ordinary written numerals include a blank numeral.

The generator checks every word of up to eight binary digits, comparing the final automaton state with the integer value modulo three. Each length $n$ includes all $2^n$ words, preserving leading zeros as distinct inputs:

All 511 checks agree. They are useful for catching a reversed arrow or mistaken label, but the reason the DFA works at every length is its invariant: after reading any word of value $n$, its automaton state is $n\mathrm{mod}3$. The update $n\mapsto 2n+b$ preserves that statement by mathematical induction. Through the extended transition function of a deterministic finite automaton, this becomes $\hat{\delta }(0,w)=\mathrm{value}(w)\mathrm{mod}3$.

For a finite prefix of $N\ge r$ digits in positional notation, the digit block frequency $F_N(w)$ of a length-$r$ word $w$ is its number of overlapping occurrences divided by $N-r+1$, the number of available starting positions. All such digit block frequencies sum to one. The infinite limit used for a normal number can instead divide counts by $N$: the factor $(N-r+1)/N$ tends to one for fixed $r$.

For the six binary digits $010101$, the single-digit counts are three zeros and three ones. Its five adjacent pairs are $01,10,01,10,01$, givingThis example explains both the overlapping convention and the end correction in a finite sample. The corresponding infinite binary expansion has pair frequencies $0,1/2,1/2,0$, rather than the four values $1/4$ required of a normal number in base two.

For an odd prime $p$ and $p\nmid a$, Euler's criterion identifies the Legendre symbol by modular exponentiation:

Thus a quadratic residue gives one and a quadratic nonresidue gives minus one, represented by $p-1$ if a nonnegative remainder is used. The formula also holds when $p\mid a$, because $(p-1)/2>0$ and both sides are zero modulo $p$.^[ref]

For example, modulo seven the exponent is three: $2^3\equiv 1$ confirms that two is a quadratic residue, while $3^3\equiv 6\equiv -1$ confirms that three is a quadratic nonresidue. Modulo eleven the exponent is five, and $2^5=32\equiv 10\equiv -1$ confirms that two is a quadratic nonresidue. The test answers whether a square root exists; it does not itself return the roots $3,4$ of $x^2\equiv 2(\mathrm{mod}7)$.

A short proof needs only Fermat's little theorem and the root bound for a polynomial. Put $m=(p-1)/2$. If $a=x^2\ne 0$ in the prime field, then $a^m=x^{p-1}=1$. There are already $m$ such quadratic residues. The nonzero polynomial $T^m-1$ has polynomial degree $m$, so it has no other polynomial roots. Finally, for every nonzero $a$, Fermat's little theorem gives $(a^m{)}^2=a^{p-1}=1$. Factoring $(a^m-1)(a^m+1)=0$ in the field shows $a^m\in \{1,-1\}$; every quadratic nonresidue must take the remaining value $-1$. This proves the criterion without first finding a primitive root.

For a larger original example, decide whether seven is a quadratic residue modulo the prime number $101$. Exponentiation by squaring computes the required power without listing fifty squares:Therefore $(7/101)=-1$ and $x^2\equiv 7(\mathrm{mod}101)$ has no solution. Each intermediate product is reduced modulo $101$, as in modular exponentiation. The advantage is that exponentiation by squaring uses a number of multiplications proportional to the binary digit length of the exponent rather than testing every potential square root.

The odd prime hypothesis is essential. For the composite modulus nine, $2^{(9-1)/2}=2^4\equiv 7(\mathrm{mod}9)$, which is neither one nor minus one. Although four is a nonzero square modulo nine, $4^4\equiv 4(\mathrm{mod}9)$, so even the square-to-one implication fails. The Jacobi symbol and Euler pseudoprime discuss different questions for odd composite moduli; they do not make this a universal square test.

Exponentiation by squaring computes an integer power by reading the exponent's binary expansion. Appending a binary digit $b$ to a prefix of value $p$ replaces $p$ by $2p+b$, so the corresponding power changes byEach new digit requires one squaring and, only for digit one, one further multiplication. The number of digits grows with the logarithm of the exponent, making this useful for modular exponentiation, primality testing, and powers of matrices using matrix multiplication.^[ref]

For a concrete calculation, $13=(1101{)}_2=8+4+1$, so $3^{13}=3^8\cdot 3^4\cdot 3$. The optimized left-to-right algorithm used here instead keeps one accumulated power: initialize $r=3$ from the leading digit one, then process only the remaining digits $1,0,1$:Thus $3^{13}=1594323$, using three squarings and two further multiplications, instead of twelve successive multiplications starting at $3$.

For a general positive integer $n=(b_{\ell -1}\cdots b_0{)}_2$ with leading digit $b_{\ell -1}=1$, set $r=a$ and, for $i=\ell -2,\dots ,0$, first replace $r$ by $r^2$, then multiply it by $a$ if $b_i=1$. Return $r$. This explicitly skips the leading digit's otherwise unnecessary operations on one. A Computational Introduction to Number Theory and Algebra by Victor Shoup (Section 3.4, pp. 65–66) presents the version initialized at one and processing every digit; the optimization here makes the counting convention unambiguous.

The loop invariant is $r=a^p$, where $p$ is the value of the digits already read. Initially $p=1$ and $r=a$. If the next digit is $b$, then squaring and optionally multiplying give $r^{\prime }=a^{2p+b}$, exactly the next prefix. At termination $p=n$. This proof uses associativity of multiplication; it works in any monoid, including matrices with the identity matrix as identity element, and needs no commutativity between different bases.

For $n=0$, return the identity element without performing a multiplication. This is the empty-product convention $a^0=1$; if $a=0$, the algorithm therefore returns one for $0^0$ by convention, rather than resolving its meaning in every mathematical setting. Negative integer exponents require an inverse element: compute $(a^{-1}{)}^{-n}$, counting the inversion separately. Over the integers, such an inverse generally leaves the integers, and a zero base has no inverse.

A finite-state machine processes inputs using a finite set of automaton states and input-dependent automaton transitions. An automaton state records everything the machine remembers; an automaton transition says how that memory changes on the next symbol. A finite-state machine may also emit outputs. Here the emphasis is on recognizing a formal language: start in a designated automaton state, read a word from an alphabet, and accept exactly when the final automaton state is designated accepting.

Finite memory does not impose a maximum input length. A finite-state machine can read arbitrarily long words, revisiting its automaton states indefinitely. The restriction is that the part already read must be compressed into one of finitely many summaries. For divisibility by three, a word of binary digits can describe an enormous integer, yet its congruence class modulo three is all that the machine needs to retain. Two already-read words giving the same automaton state behave identically under every subsequent continuation in a DFA.

This makes finite-state machines useful whenever past details can be forgotten safely: checking whether a sequence of commands follows an allowed order, or recognizing a pattern with a regular expression. The beautiful design question is which distinctions about the past matter to the future. DFA minimization makes that question precise by merging indistinguishable states of a deterministic finite automaton. Both DFAs and NFAs recognize exactly the regular languages.

Floating-point arithmetic represents a real number by a sign, a fixed number of significant digits, and a scale. It trades exactness for a compact representation spanning many orders of magnitude. The picture shows the essential compromise: with the same precision, large numbers have wider gaps between representable neighbors. Thus accuracy is naturally relative to magnitude, rather than a fixed distance along the number line.

In radix $\beta$, an integer at least two, a nonzero normalized value has the positional notationwhere the digits and exponent $e$ are integers. The parenthesized factor is the significand; $p$ counts all its significant digits, including the leading digit. Normalization means $d_0\ne 0$, so the same value has one such representation. The allowed exponents satisfy $e_{\min }\le e\le e_{\max }$. Finite precision restricts the available digits, while the exponent bounds restrict the range. Zero needs a separate representation. This scales a digit string in positional notation by a power of its base.^[ref]

For positive normalized values in $[{\beta }^e,{\beta }^{e+1})$, increasing the last digit by one gives the constant neighbor spacingAt the next scale, the spacing is multiplied by $\beta$. In the binary numeral system with $p=4$, the values in $[1,2)$ are $1,1.125,1.25,\dots ,1.875$; those in $[2,4)$ are $2,2.25,\dots ,3.75$; those in $[4,8)$ are $4,4.5,\dots ,7.5$. The respective gaps are $1/8$, $1/4$, and $1/2$. Although absolute resolution deteriorates with magnitude, ${\Delta }_e/y$ stays between ${\beta }^{-p}$ and ${\beta }^{1-p}$. This scale invariance is why floating-point arithmetic suits both microscopic and astronomical calculations.

The IEEE 754 binary64 format uses the binary numeral system with $p=53$, including an implicit leading one for normalized values, and normal exponents $-1022\le e\le 1023$. Its 64 stored binary digits comprise one sign digit, 11 exponent digits and 52 trailing significand digits. Subnormal numbers extend the range toward zero with a different leading-digit rule. In rounding to nearest, an operation returns the representable value closest to its exact result; a tie goes to the value whose last significant binary digit is even. Machine epsilon describes the gap just above one, whereas unit roundoff bounds a single relative rounding error in the normal range. These format and rounding rules are presented in MIT's floating-point lecture.^[ref]

Even an ordinary terminating decimal can require rounding in the binary numeral system. The rational number $0.1=1/10$ has the repeating positional notation $0.0001100110011\dots {}_2$. Every finite binary fraction has a reduced denominator that is a power of two, so the factor five in $10$ prevents an exact finite binary representation. In binary64, the exact stored values of three decimal literals and their computed sum areThe exact sum of the first two stored rational numbers is $10808639105689191/36028797018963968$, halfway between two binary64 values. The even-last-digit tie rule rounds it upward. The printed result is 0.30000000000000004, one step $2^{-54}$ above stored 0.3, so 0.1 + 0.2 == 0.3 is false. Both conversion of the inputs and rounding of the addition matter here.

To reproduce the table on a Python installation using binary64, inspect exact fractions rather than only their printed decimal representations:Python's as_integer_ratio() returns a pair of integers whose quotient equals the stored finite value exactly.^[ref] Exact rational numbers are also useful for checking floating-point arithmetic: a short decimal printout can hide the approximation being made.

Historically, David Goldberg's March 1991 ACM Computing Surveys paper explained representation, rounding and catastrophic cancellation together with the IEEE standard's design choices. Oracle provides an authorized edited reprint. Its enduring lesson is practical: use the spacing to understand roundoff error, and inspect intermediate values when mathematically equivalent algorithms behave differently.^[ref]

A sixteen-bit floating-point representation with one sign bit, five exponent bits and ten stored fraction bits. For sign $s$, stored exponent $E$ and fraction integer $F$, normal finite values areThe implicit leading one gives eleven significant binary digits. With $E=0$, the value is $(-1{)}^{s}F{2}^{-24}$, including signed zero at $F=0$ and subnormal numbers otherwise. The stored exponent $E=31$ is reserved for infinities at $F=0$ and NaN otherwise. The positive normal range is $2^{-14}$ through $65504$; the least positive subnormal number is $2^{-24}$.

Use the $\{0,1{\}}^n$ convention for the Hamming cube $H_n$. Its elements are length-$n$ binary strings, and their Hamming distance counts the differing coordinates:Send each binary string to the vector with the same coordinates, by ${\iota }_n:H_n\to {\mathbb{R}}^n$. Every differing coordinate contributes $1$ both to the sum of absolute differences and to the sum of squared differences. Therefore the L1 norm and L2 norm give, respectively,This short coordinate calculation proves that the L1 norm gives an isometric embedding, while Euclidean distance grows like the square root of the number of changed binary digits. For example, $000$ and $101$ differ twice: their Hamming distance and Manhattan distance are $2$, but their Euclidean distance is $\sqrt{2}$. Comparing binary strings in these two geometric representations answers different questions even though the points have the same coordinates.

For distinct pairs $k$ ranges over every integer from $1$ to $n$. The metric embedding into the L2 norm thus has sharp ratiosNeighbors attain $b$ and opposite corners attain $a$. This computes the bi-Lipschitz distortion of this coordinate map; it does not prove that every metric embedding into an L2 space has at least this bi-Lipschitz distortion. Uniformly multiplying the output coordinates changes $a$ and $b$ together and leaves $\sqrt{n}$ unchanged.

The important family distinction is that these coordinate maps have no dimension-independent bound on their bi-Lipschitz distortion into the L2 norm, yet they do give a uniform coarse embedding of a family of metric spaces. Pad each output vector with zero coordinates to view all maps in the single l2 sequence space, an L2 space for counting measure. Then the same functionswork for every $n$, and ${\rho }_{-}(t)\to \infty$. Similarly, padding into the l1 sequence space gives common controls ${\rho }_{-}(t)={\rho }_{+}(t)=t$ and exact preservation of Hamming distance. These claims use unnormalized Hamming distance; replacing it by normalized Hamming distance changes the family of source metrics.

The geometry has a practical origin: Hamming's 1950 paper uses the number of differing positions between binary strings to study error-correcting codes. Changing one binary digit moves to a neighboring corner, so a question about corrupted messages becomes a question about separation in the Hamming cube. ^[ref]

Here machine epsilon means the gap from one to the next larger value in normalized floating-point arithmetic:It is a relative resolution scale near one, not the smallest positive representable value and not a fixed absolute error for all magnitudes. In binary64, $p=53$ giveswhere $u$ is unit roundoff for rounding to nearest. Some authors instead call $u$ their machine epsilon; MIT 18.335J uses this half-gap convention. Always check which definition an error bound uses.^[ref]

The gap below one is smaller: the preceding binary64 value is $1-2^{-53}$, whereas the next value above one is $1+2^{-52}$. Under nearest-even rounding, 1.0 + 2.0**-53 is exactly 1.0: the exact sum is a tie and one has the even final binary digit. By contrast, 1.0 + 2.0**-52 is the next value. Thus searching for the smallest increment that changes one by trial addition also depends on the rounding convention. At $2^{53}$ the next larger binary64 value is $2^{53}+2$, illustrating why a small absolute increment can disappear at large magnitude.

Modular exponentiation computes the remainder $a^n\mathrm{mod}m$ for an integer base $a$, a nonnegative integer exponent $n$, and, here, an integer modulus $m\ge 2$. It applies exponentiation by squaring while reducing after each multiplication, including each squaring. The result is the unique representative in $\{0,\dots ,m-1\}$. There is no requirement that the base be coprime to the modulus when the exponent is nonnegative.^[ref]

Take $a=2$, $n=13=(1101{)}_2$, and $m=17$. Initialize $r=2$; the remaining digits giveHence $2^{13}\mathrm{mod}17=15$. The final operation multiplies $16$ by $2$, since the preceding square has already been reduced. The computation never needs to construct $2^{13}=8192$.

In general reduce the base first, $x=a\mathrm{mod}m$. For $n>0$, initialize $r=x$ and process the remaining binary digits with $r\leftarrow r^2\mathrm{mod}m$, followed, for a digit one, by $r\leftarrow rx\mathrm{mod}m$. The loop invariant iswhere $p$ is the processed prefix value. The congruence is preserved because modular arithmetic respects multiplication: if $u\equiv v(\mathrm{mod}m)$, then $u^2\equiv v^2$ and $ux\equiv va(\mathrm{mod}m)$. Thus changing representatives never changes the final congruence class.

Both factors in every product are less than $m$, so the unreduced product is less than $m^2$, and the stored result is again less than $m$. The same bounds hold for a gigantic exponent, such as $n=2^{1000}$: there are $1000$ squarings, each on bounded-size residues. For base two, the corresponding ordinary power would have $2^{1000}+1$ binary digits. The benefit is avoiding the huge ordinary power altogether, rather than constructing it and taking one final remainder. Initial reduction is essential when the supplied base is much larger than the modulus.

For $n=0$, return $1$ under the same empty-product convention as exponentiation by squaring. For $n=-k<0$, require $\gcd (a,m)=1$, find the modular multiplicative inverse $a^{-1}\mathrm{mod}m$, and compute $(a^{-1}{)}^k\mathrm{mod}m$; the extended Euclidean algorithm can find that inverse. Its cost is separate from the positive-exponent binary powering multiplication count. For example, $2^{-1}\equiv 9(\mathrm{mod}17)$, while $2^{-1}\mathrm{mod}4$ is undefined because $2$ and $4$ are not coprime.

A reversible computation gives the coherent quantum modular exponentiation used in quantum order finding and discrete-logarithm Fourier sampling.

Computing the remainder $ab\mathrm{mod}m$ while working with bounded integer representatives. Repeated doubling and addition can avoid forming an overflowing product: process the binary digits of $b$, double the current contribution modulo $m$, and add it to the result whenever the current binary digit is 1. The additions themselves must also use an overflow-safe implementation.

A nondeterministic finite automaton allows a set of possible destinations for each automaton state and input symbol. Without the extra moves permitted in an Epsilon-NFA, replace the DFA function by $\Delta :Q\times \Sigma \to \mathcal{P}(Q)$, where $\mathcal{P}(Q)$ is the power set. A word is accepted when at least one path consumes all its symbols and ends in an accepting automaton state. These choices express existential quantification, not probability: no chance of choosing a particular path is assigned. One successful path suffices even if other paths fail.

Consider a three-state NFA accepting exactly the words of binary digits ending in $01$. Its initial automaton state $q_0$ loops on both $0$ and $1$; it also has an automaton transition $q_0\stackrel{0}{\to }{q}_1$, followed by $q_1\stackrel{1}{\to }{q}_2$. Only $q_2$ is accepting, and all unlisted automaton transitions are absent. Staying at $q_0$ skips an arbitrary beginning; leaving it guesses that the current zero starts the final two symbols.

The word $01$ accepts along $q_0\to q_1\to q_2$. The word $1101$ accepts along $q_0\to q_0\to q_0\to q_1\to q_2$. The word $010$ rejects: a path reaching $q_2$ after $01$ cannot consume the remaining zero, while the other paths end outside $q_2$. Thus finding $01$ somewhere is insufficient; it must occur at the end.

To simulate the NFA deterministically, retain the set $S$ of all currently possible automaton states. The powerset construction updates $S$ by taking the union of the destinations of its members. Here only $\{q_0\}$, $\{q_0,q_1\}$, and $\{q_0,q_2\}$ are reachable, although the full power set has eight members. Reading $010$ gives $\{q_0\}\to \{q_0,q_1\}\to \{q_0,q_2\}\to \{q_0,q_1\}$, which rejects. In general, an $N$-state NFA yields a DFA with at most $2^N$ automaton states. Nondeterminism can simplify a description, while the powerset construction keeps the memory finite.

The second generator checks all 511 words of up to eight binary digits against the direct test of whether their final two symbols are $01$. Exactly 127 accept, and only the three listed sets of automaton states occur. In particular, the empty word rejects for this NFA.

Michael O. Rabin and Dana Scott's 1959 paper studied finite-state machines as recognizers with finite internal memory. Its section 5 defines NFAs through successful paths, and theorem 11 proves their equivalence to DFAs by the powerset construction. This gives a historical source for the distinction between many possible paths and greater recognition power: ^[ref].

A pairing function is a computable function that gives a bijection from the Cartesian product ${\mathbb{N}}_0^2$ to ${\mathbb{N}}_0$, with two computable functions recovering its coordinates. Here ${\mathbb{N}}_0=\{0,1,2,\dots \}$ consists of the natural numbers, including zero. If $p$ is the pairing function and $L,R$ its decoding functions, the requirements areEvery ordered pair has exactly one code, every code represents an ordered pair, and decoding always terminates. For a bijection given by a total computable function, computability of the inverse function also follows by enumerating all ordered pairs until the one with code $z$ appears. A primitive recursive pairing function makes the stronger requirement that both encoding and coordinate decoding are primitive recursive.

The Cantor pairing function makes this concrete through triangular numbers:On the diagonal $x+y=s$, list $(s,0),(s-1,1),\dots ,(0,s)$. There are $s+1$ ordered pairs there, and $T(s)$ ordered pairs on earlier diagonals. Their codes therefore occupy precisely the consecutive natural numbers from $T(s)$ through $T(s+1)-1$. These intervals touch without overlapping and cover all natural numbers; within each interval the offset $y$ determines the ordered pair. This proves both injectivity and surjectivity, hence the bijection. It also explains why the infinite grid is a countably infinite set: every ordered pair is reached after finitely many steps. A row-by-row listing would never finish its first infinite row. Authored treatments give the same diagonal construction with the coordinates exchanged.^[ref]

For example, the Cantor pairing function givesThe order of the coordinates matters: $(1,0)$ and $(0,1)$ are different ordered pairs with different codes.

To decode a natural number $z$, find the unique diagonal index $s$ satisfyingThe inequality gives $0\le y\le s$, so both recovered coordinates are natural numbers. At $z=18$, $T(5)=15\le 18<T(6)=21$, hence $s=5$, $y=18-15=3$, and $x=5-3=2$. At $z=0$, $s=0$ and both coordinates are zero. Solving the quadratic inequality gives the equivalent floor function expressionThis is exact as a statement about the real numbers and their square roots. An implementation using floating-point arithmetic can round across a diagonal boundary, especially for large $z$. Use an exact integer square root instead: with $q=\lfloor \sqrt{8z+1}\rfloor$ computed using integer arithmetic, $s=\lfloor (q-1)/2\rfloor$. Alternatively, search for the diagonal using exact integer comparisons. The primitive recursive pairing function proof below uses bounded minimization and needs no approximate square root.

A pairing function is a reversible encoding: unlike a hash function mapping arbitrarily many inputs into a fixed finite range, it loses no coordinates. This works because the mathematical natural numbers have no upper bound. A machine word with $b$ binary digits has $2^b$ possible values, while two such words have $2^{2b}$ possible ordered pairs; a bijection into one word of the same width is impossible. Evaluating the Cantor pairing function in a fixed-width integer type can overflow, including in the intermediate product $(x+y)(x+y+1)$. Exact implementation requires sufficiently large integer storage or an explicit restricted input range. One mathematical code need not occupy fewer binary digits than its two inputs.

Iterating a pairing function gives a reversible encoding of tuples of any fixed positive length. With the left-nesting convention, the Cantor pairing function encodes $(2,3,4)$ asDecode the outer ordered pair first: $T(22)\le 257<T(23)=276$ gives $(18,4)$. Decode its first coordinate $18$ to recover $(2,3)$, giving the nested ordered pair $((2,3),4)$ and thus the original tuple $(2,3,4)$.

The tuple length and nesting convention must be known. The same code $257$ represents $(18,4)$ as a two-entry tuple and $(2,3,4)$ as a three-entry tuple; nesting alone does not make variable-length tuples self-describing. To encode variable-length tuples, supply their length or another explicit termination convention as part of the encoding. This simple change of representation is useful in computability theory and Gödel numbering: a function of several natural numbers can be studied through a single numerical input, while all coordinates remain recoverable. The striking fact is that a whole infinite grid can be serialized without omitting a point.

A historical reference for comparing sets through bijections is Georg Cantor's paper Ein Beitrag zur Mannigfaltigkeitslehre, published in 1878 in volume 84 of Journal für die reine und angewandte Mathematik, pages 242–258.^[ref]

The binary numeral system representation is taken to be canonical, with no leading zeroes, for each positive integer. Delimiters separate the components of a pair or triple. Syntax, positivity and this representation convention can all be checked in logarithmic space. The useful principle is that a large integer need not be stored in order to compute its digits one at a time.

First consider Add. Let $x_i,y_i,z_i$ denote the binary digits in position $i$, counting from the least significant position $i=0$ and taking out-of-range digits to be zero. Count the component lengths, and let $L$ be their maximum. Start with arithmetic carry $c_0=0$. For $i=0,\dots ,L-1$, computeReject if $s_i\ne z_i$ at any position. After the last position, reject if $c_L=1$; otherwise accept. The final arithmetic carry check prevents accepting a result with an omitted high digit, such as the false equality $11_2+11_2=10_2$ obtained by retaining only two binary digits.

The binary-addition carry invariant isIt holds initially and is preserved by the displayed one-column equations. At the end, it proves that the comparisons and the final arithmetic carry check succeed exactly when $x+y=z$. Each requested binary digit is recovered by scanning the corresponding input field and counting to its position. The length counters, position $i$ and scan counters require $O(\log n)$ bits, while $c_i$ and $t_i\in \{0,1,2,3\}$ require constant space. ThusFor Pal-Add, construct a logspace transducer producing the canonical binary numeral system representation of $x+y$. Put $m=\max (|x|,|y|)$. One least-to-most-significant pass using the same recurrence determines the final arithmetic carry $c_m$. Positivity and canonical representations imply that the sum's digit length isIndeed, one summand is at least $2^{m-1}$ and both are less than $2^m$. For each output position, from $\ell -1$ down to $0$, recompute the required sum digit from scratch: reset the arithmetic carry to zero and run the column recurrence from position $0$ up to the requested position, reading input digits by scans. Positions beyond the summands use zero digits, so position $m$ correctly emits the possible final arithmetic carry. Emit only the requested digit, then reuse the work space for the next position.

This produces most-significant-first output with $\ell \le m+1$ digits. It uses $O(\log n)$ work space because it retains only a fixed number of digit indices, lengths, scan counters and one arithmetic carry; it never stores the whole sum. A direct implementation with repeated whole-input scans uses at most $O(n^3)$ time complexity, which is sufficient here. Hence $(x,y)\mapsto \mathrm{bin}(x+y)$ is a logspace-computable function.

Finally simulate the assumed L decider for binary palindrome on this virtual output, using logspace composition by recomputation from part (b). Whenever the simulated input head asks for a digit, rerun the addition logspace transducer until it emits that digit. The virtual input length is $O(n)$, so both its work space and the addition work space are $O(\log n)$. The simulation accepts exactly when the canonical representation of $x+y$ is a binary palindrome. ThereforeEquivalently, one could directly compare the sum digits in positions $i$ and $\ell -1-i$, recomputing each digit by the same recurrence. This gives the same logarithmic space conclusion without ever writing the full sum on a work tape.