A printable PDF version is at www.cs.bu.edu/fac/lnd/toc/z.pdf
The notes are skeletal: they do have (terse) proofs, but exercises, references, intuitive comments, examples are missing or inadequate. The notes can be used for designing a course or by students who want to refresh the known material or are bright and have access to an instructor for questions. Each subsection takes about a week of the course. Versions of these notes appeared in [1] .
Acknowledgments. I am grateful to the University of California at Berkeley, its MacKey Professorship fund and Manuel Blum who made possible for me to teach this course. The opportunity to attend lectures of M. Blum and Richard Karp and many ideas of my colleagues at BU and MIT were very beneficial for my lectures. I am also grateful to the California Institute of Technology for a semester with light teaching load in a stimulating environment enabling me to rewrite the students' notes. NSF grants DCR-8304498, DCR-8607492, CCR-9015276 also supported the work. And most of all I am grateful to the students (see sec. 8) who not only have originally written these notes, but also influenced the lectures a lot by providing very intelligent reactions and criticism.
Copyright © 2025 by the author.
Sections 1, 2 study deterministic computations. Non-deterministic aspects of computations (interaction, randomization, errors, etc.) are crucial and challenging in advanced theory and practice. Defining them as an extension of deterministic computations is simple. The latter, however, while much simpler conceptually, require elaborate models for definition. These models may be sophisticated if we need to measure all required resources precisely. However, if we only need to define what is computable and get a very rough magnitude of the needed resources, all reasonable models turn out equivalent, even to the simplest ones. We will pay significant attention to this surprising and important fact. The simplest models are most useful for proving negative results and the strongest ones for positive results.
We start with terminology common to all models, later adding specifics of models we actually study. We represent computations as graphs: the edges reflect various relations between nodes (events). Nodes, edges have attributes: labels, states, colors, parameters, etc. (affecting the computation, or only its analysis). Causal edges run from each event to all events immediately essential for its occurrence or attributes. They form a directed acyclic graph (though cycles may be added artificially to mark the external input parts).
We will study only synchronous computations. Their nodes have a time parameter. It reflects logical steps, not necessarily a precise value of any physical clock. Causal edges only span short (typically, $\le 3$ moments) time intervals. One event among the causes of a node is called its parent. Pointer edges connect the parent of each event to all its other possible causes and reflect connections that allow simultaneous events to interact and have a joint effect. Pointers with the same source have different labels (colors). The colored subgraph of events/edges at a given time is an instant memory configuration of the model.
Each non-terminal configuration has active nodes/edges around which it may change. The models with only a small active area at any step of the computation are sequential. Others are called parallel.
Complexity. We use the following measures of computing resources of a machine $A$ on input $x$:
Time: The greatest depth $D_{A(x)}$ of causal chains is the number of computation steps. The volume $V_{A(x)}$ is the combined number of active edges in all steps. Time $T_{A(x)}$, depending on the context, means either depth or volume (close for sequential models). Note that time complexity is robust only up to a constant factor: Machines can be modified for a larger labels alphabet, representing several locations in one. It would produce identical results in a fraction of time and space (provided, the time limits suffice for transforming the input and output into the other alphabet).
Space: $S_{A(x)}$ or $S_A(x)$ of a synchronous computation is the greatest (over time) size of its configurations. Sometimes excluded are nodes/edges unchanged since the input.
Growth Rates (typically expressed as functions of bit length
$n=\|x,y\|$ of input/output $x$/$y$):
$O,\Omega$: $f(n)=O(g(n)) \iff g(n)=\Omega(f(n)) \iff
\sup_n\frac{f(n)}{g(n)} <\infty$.
$o,\omega: f(n)=o(g(n)) \iff g(n)=\omega(f(n)) \iff \lim_{n \to
\infty}\frac{f(n)}{g(n)}{=}0$.
$\Theta: f(n)=\Theta(g(n)) \iff$ ($f(n)=O(g(n))$ and $g(n)=O(f(n))$).
(These are customary but somewhat misleading notations. E.g., $n=O(n^3)$ and $n^2=O(n^3)$ do not imply $n=n^2$. The clear notations would be like $f(n)\in O(g(n))$.)
Here are examples of frequently appearing growth rates: negligible $(\log n)^{O(1)}$; moderate $n^{\Theta(1)}$ (called polynomial or P, like in P-time); infeasible: $2^{n^{\Omega(1)}}$, also $n!=(n/e)^n\sqrt{\pi(2n{+}1/3)+ \e/n}$, $\e{\in}[0,\,.1]$. (A rougher estimate $\ln n!=t\ln(t/e)|_{t=1.5}^{n+.5}+O(1)$ follows by using that $|\sum_{i=2}^ng(i)-\int_{1.5}^{n+.5}g(t)\,{\bf d}t|$ is bounded by the total variation $v$ of $g'/8$. For monotone $g'(t){=}\ln'(t){=}1/t$ the $O(1)$ is ${<}v{<}1/12$.)
The reason for ruling out exponential (and neglecting logarithmic) rates is that the known Universe is too small to accommodate exponents. Its radius is about 46.5 giga-light-years $\sim2^{204}$ Plank units. A system of $\gg R^{1.5}$ atoms packed in $R$ Plank Units radius collapses rapidly, be it Universe-sized or a neutron star. So the number of atoms is $<2^{306}\ll 4^{4^4}\ll 5!!$. (Actually, $<10^{80}$.)
Rigid computations have another node parameter: location (cell). Combined with time, it designates the event uniquely. Cells have structure or proximity edges between them. They (or their short chains) indicate the neighbors to which pointers may be directed.
(CA) are a parallel rigid model. Its sequential restriction is the Turing Machine (TM). The configuration of CA is a (possibly multi-dimensional) grid with a finite, independent of the grid size, alphabet of states to label the events. The states include, among other values, pointers to the grid neighbors. At each computation step, the state of each cell can change as prescribed by a transition function (also called program) applied to the previous states of the cell and its pointed-to neighbors. The initial state of the cells is the input for the CA. All subsequent states are determined by the transition function.
An example of technology CA represents is a VLSI chip (Very Large Scale Integration) represented as a grid of cells connected by wires (chains of cells) of different lengths. The propagation of signals along the wires is simulated by changing the state of the wire cells step by step. The clock interval can be set to the time the signals propagate through the longest wire. This way the delays affect the simulation implicitly.
An example: the Game of Life (GL). GL is a plane grid of cells, each holds a 1-bit state (dead/alive) and pointers to the 8 adjacent cells. A cell remains dead or alive if the number $i$ of its live neighbors is $2$. It becomes (or stays) alive if $i{=}3$. In all other cases it dies (of overcrowding or loneliness) or stays dead.
A simulation of a machine $M_1$ by $M_2$ is a correspondence between their memory configurations which is preserved during the computation (may be with some time dilation). Such constructions show that the computation of $M_1$ on any input $x$ can be performed by $M_2$ as well. GL can simulate any CA in this formal sense: (See a sketch of an ingenious proof in the last section of [8].)
We fix space and time periods $a,b$. Cells $(i,j)$ of GL are mapped to cell $(\lfloor i/a\rfloor,\lfloor j/a\rfloor)$ of CA $M$ (compressing $a{\times}a$ blocks). We represent cell states of $M$ by states of $a{\times}a$ blocks of $GL$. This correspondence is preserved after any number $t$ steps of $M$ and $b t$ steps of $GL$ regardless of the starting configuration.
TM is a minimal CA. Its configuration -- tape -- is an infinite to the right chain of cells. Each state of a cell has a pointer to one of the cell's two adjacent neighbors. No adjacent cells can both point away from each other. Only the two cells pointing at each other are active, i.e. can change state. The cell that just turned its pointer is the TM's moving head working on the linked-to tape symbol - its target. The head's state is often set in a separate alphabet. The input is an array of non-blank cells (only the first one is rightward) followed by all blanks at the right.
Another type of CA is a TM $A$ with several non-communicating heads. At most $O(1)$ heads fit in a cell. They can vanish, split, or merge only in the first cell (which, thus, controls the number of active cells). The input $x$ makes an unchangeable "ink" part of each cell's state. The rest of the cell's state is in "pencil" and can be changed by $A$. The computation halts when all heads drop off. The output $A(x)$ is the pencil part of the tape's state. This model has convenient features. E.g. with linear (in $T$) number ($\|p\|^2T$) of state changes (volume) one can solve the Bounded Halting Problem $H(p,x,T)$: find out if the machine with a program $p$ stops on an input $x$ within volume $T$ of computation (see Sec.2.3).
Exercise: Find a method to transform any given multi-head TM $A$ into another one $B$ such that the value of the output of $B(x)$ (as a binary integer) and the volumes of computation of $A(x)$ and of $B(x)$ are all equal within a constant factor (for all inputs $x$).
Hint: Cells of $B$ may have a field to simulate $A$ and maintain (in other fields) two binary counters with their least significant digits at the leftmost cell: $H$ for the number of heads of $A$, and $V$ for $A$'s volume. $H$ adds its highest digit to the same place in $V$ at each step of $A$. $H$ maintains heads in every second cell. To move the carry, $V$ borrows a head from the $H$ segment. The head moves it right, to drop to the leftmost $0$ digit of $V$ or to a next adjacent head freed already of its carry. Heads freed of carry move left toward the $H$ segment which absorbs them. Borrowed or returned heads make sparse or dense heads regions in $H$ (with adjacent heads or adjacent no-heads). Those shift leftward via transitions $|{-}|h|h|\;\to\;|h|{-}|h|$, and $|h|{-}|{-}|\;\to\;|{-}|h|{-}|$, until absorbed at the leftmost cell.
The memory configuration of a Pointer Machine (PM), called pointer graph, is a finite directed labeled multigraph. One node $R$ is marked as root and has directed paths to all nodes. Edges (pointers) are labeled with colors from a finite alphabet common to all graphs handled by a given program. The pointers coming out of a node must have different colors (which bounds the outdegree).
Some colors are designated as working and not used in inputs/outputs. One of them is called active, as also are pointers carrying it and nodes seeing them. Active pointers must have inverses, form a tree to the root, and can be dropped only in leaves. Nodes can see and change the configuration of their out-neighborhood of constant (2 suffices) depth. All active nodes each step execute an identical program.
At program's first pulling stage, node $A$ acquires copies of all pointers of its children using "composite" colors: e.g., for a two-pointer path $(A,B,C)$ colored $x,y$, the new pointer $(A,C)$ is colored $xy$, or an existing $z$-colored pointer $(A,C)$ is recolored $\{z,xy\}$. $A$ also spawns a new node with pointers to and from it.
Next, $A$ transforms the colors of its set of pointers, drops the pointers left with composite colors, and vanishes if no pointers are left. Nodes with no path from the root are forever invisible and considered dropped, too. The computation is initiated by inserting an active loop-edge into the root. When no active pointers remain, the graph, with all working colors dropped, is the output.
Exercise: Design a PM transforming the input graph into the same one with two extra pointers from each node: to its parent in a BFS spanning tree and to the root. Hint: Nodes with no path to the root can never be activated. but can be copied with pointers, copies connected to the root, the original then dropped.
PM can be parallel, PPM [7] or sequential, SPM: In SPM only pointers to the root, their sources, and nodes connected by two-way pointers with these sources can be active.
A Kolmogorov or Kolmogorov-Uspenskii Machine (KM) [9], is a special case of Pointer Machine [10] with all pointers having inverses. This implies the graph's $O(1)$ in/out-degree which we further assume constant.
Fixed Connection Machine (FCM) is a variant of the PKM with the restriction that pointers, once created, cannot be removed, only re-colored. So when the memory limits are reached, the pointer structure freezes, and the computation can be continued only by changing the colors of the pointers.
PPM is the most powerful model we consider: it can simulate the others in the same space/time. E.g., cellular automata are a simple special case of a PPM, restricting the Pointer Graph to be a grid.
Exercise. Design a machine of each model (TM, CA, KM, PPM) which determines if an input string $x$ has a form $ww$, $w\in\{a,b\}^*$. Analyze time (depth) and space. KM/PPM takes input $x$ in the form of colors of edges in a chain of nodes, with root linked to both ends. The PPM nodes also have pointers to the root. Below are hints for TM,SPM,CA. The space is $O(\|x\|)$ in all three cases.
Turing and Pointer Machines. TM first finds the middle of $ww$ by capitalizing the letters at both ends one by one. Then it compares letter by letter the two halves, lowering their case. The complexity is: $T(x)=O(\|x\|^2)$. SPM acts similarly, except that the root keeps and updates the pointers to the borders between the upper and lower case substrings. This allows constant time access to these borders. So, $T(x){=}O(\|x\|)$.
Cellular Automata. At the start, the leftmost cell sends right two signals. Reaching the end the first signal turns back. The second signal propagates three times slower, so they meet in the middle of $w'w$ and disappear. While alive, the second signal copies the input field $i$ of each cell into a special field $c$. The $c$ symbols will try to move right whenever the next cell's $c$ field is blank.
So the chain of these symbols alternating with blanks will start growing both ways from the middle of $w'w$. Upon reaching the ends they will push the blanks left and pack themselves into a copy of the $w'$ right of the middle. Having no blank at the right to move to, a $c$ symbol compares itself with the $i$ field of the same cell. If they differ, it generates a signal to halt all activity and reject $x$. If all comparisons succeed, the last $c$ generates an accepting signal. The depth is: $T(x)=O(\|x\|)$.
We looked at several models of computation. We will see now how the simplest of them -- Turing Machine -- can simulate all others: these powerful machines can compute no more functions than TM.
Church-Turing Thesis is a generalization of this conclusion: TMs can compute every function computable in any thinkable realistic model of computation. This is not a math theorem because the notion of model is not formally specified. Yet, the long history of studying ways to design real and ideal computing devices makes it very convincing. Moreover, this Thesis has a stronger Polynomial Time version which bounds the volume of computation required by that TM simulation by a polynomial of the volume used by the other models. Both forms of the Thesis play a significant role in foundations of Computer Science.
PKM Simulation of PPM. Assume all PPM nodes have pointers to root, as generated in the above HW. PKM represents PPM configuration with extra colors $l,r,u$ used in a $u$-colored binary tree added to each node $X$. All (unlimited in number) PPM pointers to $X$ are reconnected to its leaves. The inverses, colored $l,r$, added to all $u$-pointers. The number of pointers increases at most 4 times.
To simulate PPM, $X$ gets a binary name formed by the $l,r$ colors on its path through the root tree, and broadcasts it down its own tree. For pulling stage $X$ extends its tree to double depth and merges (with combined colors) its own 2-pointer paths with the same targets (seeing identical names). Then $X$ re-colors its pointers as PPM program requires and rebalances its tree. This simulation of a PPM step takes polylogarithmic parallel time.
TM Simulation of PKM.. Assume the PKM keeps a roughly balanced root tree. TM tape reflects PKM state as the list of all pointers sorted by (source $(l,r)$-name, color). TM's transition table reflects the PKM program. To simulate PKM's pulling stage TM creates a copy of each pointer and sorts copies by their sinks.
Now each pointer, located at source, has its copy near its sink. So both components of 2-pointer paths are nearby, stored as double-colored special pointers, and moved to their sources by resorting on the source names. The re-coloring stage is local: all relevant pointers with the same source are nearby.
Once the root has no active pointers, the TM stops. If a PPM computes $f(x)$ in $t(x)$ steps, using $s(x)$ nodes, the simulating TM uses space $S = O(s\log s)$, ($O(\log s)$ bits for each of $O(s)$ pointers) and time $T= O(S^2)t$, as TM sorting takes quadratic time.
Squaring matters ! TM cannot outperform Bubble Sort. Is its quadratic overhead a big deal? In a short time all silicon gates on your PC run, say, Avogadro order clock cycles $X{=} 6\cdot10^{23}\sim2^{2^{6.3}}$ combined. Silicon parameters double almost annually. Decades may bring clouds of $\mu m$-thin sunlight sails in space in of great computing and physical (light beam) power. Centuries may turn them into a Dyson Sphere enveloping the solar system. Still, the power of such an ultimate computer is limited by the number of photons the Sun emits per second: $Y\sim 2^{2^{7.3}}{=}X^2$. Giga-years may turn much of the known universe into a computer, but its might is still limited by its total entropy $2^{2^{8.3}}{=}Y^2$.
Faster PPM Simulations. Parallel Bubble-Sort on CA or Merge-Sort on sequential FCM take nearly linear time. Parallel FCM can do much better [11]. It represents and updates pointer graphs as the above TM. All steps are straightforward to do locally in parallel polylog time except sorting of pointers. Sophisticated sorting networks sort arbitrary arrays of $n$ integers in $O(\log n)$ parallel steps. We need only a simpler polylog method. Merge-Sort splits an array of two or more entries in two halves and sorts each recursively. Batcher-Merge combines two sorted lists in $O(\log n)$ steps as follows.
Batcher-Merge. A bitonic cycle is the combination of two sorted arrays (one may be shorter), connected by max-to-max and min-to-min entries. Entries in a contiguous half (high-half) of the cycle are $\ge$ than all entries in the other (low) half. Each half (with its ends connected) forms a bitonic cycle itself.
Shuffle Exchange graph links nodes in a $2^k$-nodes array to their flips and shits. The flip flips the highest bit of a node's address; the shift cycle-shifts that bit to the end.
We merge-sort two sorted arrays given as a bitonic cycle on such a graph as follows. Comparing each entry with its flip (half-a-cycle away), and switching if wrongly ordered, fits the high and low halves into respectively the first and last halves of the array. (This rotates the array, and then its left and right halves.) We do so for each half recursively (decrementing $k$ via shift edges).
The first computers were hardware-programmable. To change the function computed, one had to reconnect the wires or even build a new computer. John von Neumann suggested using Turing's Universal Algorithm. The function computed is then specified by giving its description (program) as part of the input instead of changing the hardware. This was a radical idea, since universal functions do not exist in classical math (as we will see in Sec. 2.2).
Let $R$ be the class of all TM-computable functions: total (defined for all inputs) and partial (which may diverge). Surprisingly, there is a universal function $u$ in $R$. It simulates every Turing Machine $M$ with a given tape alphabet (but different head alphabets). Let $M$ compute $f{\in}R$ in time $T$ and space $S$. $u$ uses a program $m$ of length $c$ listing the commands and initial head state of $M$.
Then $u(m x)$ simulates $M(x)$. It operates in cycles, each simulating one step of $M$ in time $c^2$ and space $S{+}c$. Let the tape of $M$ after $i$ steps be $t_i=l_is_ir_i$, where $s_i$ is the state of the pair of active cells. After $i$ cycles $u$ has $t_i=l_ims_ir_i$ on its tape. It looks up $m$ to find the command corresponding to $s_i$, and modifies $t_i$ accordingly. When $M(x)$ halts, $u(m x)$ erases the (penciled) $m$ and halts too. Universal Multi-head TM works similarly but can also determine in time $O(t(x))$ whether it halts in $t$ steps (given $x,t(x)$).
Exercise. Design a universal multi-head TM with a constant factor volume overhead $c^2$.
Hint: When heads split or merge in the first cell, the room $u$ needs for their programs creates sparse or dense content regions that propagate right (sparse faster).
We now describe in detail a simpler but slower universal [14] TM $U$. It simulates any other TM $M$ that uses only $0,1$ bits on the tape, lacking even the blank that usually marks the end of input. (Larger alphabets can be encoded as bit strings.) $U$ has 6 symbols + 11 head states. Its transition table at the right shows the states and tape symbols only when changed, except that primes always show. The head is on the tape: lower case states look left, upper case -- right. Input calls, halt, etc. are special states for $M$; for $U$ they are shown as "A/B" or "=". $U$ works in cycles, each simulating one transition of $M$. The tape consist of $0/1$ segments, each preceded with a *. Some symbols are primed.
| 1 | 0 | * | 1 ' | 0 ' | * ' | ||
|---|---|---|---|---|---|---|---|
| A | f | f | e0 | ||||
| B | F | F | e1 | ||||
| f,F | b* | a* | F | c | |||
| D | d ' | -- | e ' | ||||
| E | = | -- | ' | ' | ' | e ' | |
| d | ' | ' | ' | ' | D | ||
| b | ' | ' | ' | a ' | D | ||
| a | ' | ' | ' | b ' | F | E ' | |
| c | ' | ' | = | F | E ' | ||
| e | ' | ' | ' | B | A | A/B |
The rightmost part of one or two segments is always a copy of $M$'s (infinite to the right) tape. The other segments describe one transition command each: $(s,b)\to (s',b',d)$ for $M$ to change head state $s$ to $s'$, tape bit $b$ to $b'$ and turn left or right. The commands segments are sorted in order of $(s,b)$ and never change, except for priming. Each transition is represented as $*S d b$, where $b$ is the bit to write, $d$ the direction $R{=}0/L{=}1$ to turn. $S$ points to the next state represented as $1^k$, if it is $k$ segments to the left, or $0^k$ (if to the right). Each cycle starts with $U$'s head in state $F$ or $f$, located at the site of $M$'s head. Primed are the digits of $S$ in the prior command and all digits to their left. An example of the configuration: $ *'0'0'0'1'0'*'0'0'0'0'01*011*\ldots *00\ \fbox{head}\ 00\ldots$ The leftward head in the leftmost cell halts.
$U$ first reads the bit of an $M$'s cell changing the state from $F$ or $f$ to $a/b$, puts a * there, moves left to the primed state segment $S$, finds from it the command segment and moves there. It does this by repeatedly priming nearest unprimed * and 1s of $S$ (or unpriming 0s) while alternating the states $c/F$ or $d/D$. When $S$ is exhausted, the target segment $\|S\|+b$ stars away is reached. Then $U$ reads (changing state from $e$ to $A$ or $B$) the rightmost symbol $b'$ of the command, copies it at the * in the $M$ area, goes back, reads the next symbol $d$, returns to the just overwritten (and first unprimed) cell of $M$ area and turns left or right.
In CA format, $M$ and $U$ have in each cell three standard bits: present $d$ and previous $d'$ pointer directions and a "content" bit to store M's symbol. In addition $U$ needs just one "trit" of its own!
Notations: Let us choose a special mark and after its $k$-th occurrence, break any string $x$ into Prefix$_k(x)$ and Suffix$_k(x)$. Let $f^+(x)$ be $f($Prefix$_k(x)\ x)$ and $f^-(x)$ be $f($Suffix$_k(x))$. We say $u$ $k$-simulates $f$ iff for some $p=$Prefix$_k(p)$ and all $s$, $u(ps)=f(s)$. The prefix can be intuitively viewed as a program which simulating function $u$ applies to the suffix (input). We also use a symmetric variant of relation "$k$-simulate" which makes some proofs easier. Namely, $u$ $k$-intersects $f$ iff $u(ps)=f(ps)$ for some prefix$_k$ $p$ and all $s$. E.g., length preserving functions can intersect but cannot simulate one another.
We call universal for a class $F$, any $u$ which $k$-simulates all functions in $F$ for a fixed $k$. When $F$ contains $f^-,f^+$ for each $f\in F$, universality is equivalent to (or implies, if only $f^+\in F$) completeness: $u$ $k$-intersects all $f\in F$. Indeed, $u$ $k$-simulates $f$ iff it $k$-intersects $f^-$; $u$ $2k$-intersects $f$ if it $k$-simulates $f^+$.
Exercise: Describe explicitly a function, complete for the class of all linear (e.g., $5x$ or $23x$) functions.
A negation of a (partial or total) function $f$ is the total predicate $\neg f$ yielding $1$ iff $f(x){=}0$ and $0$ otherwise. Obviously, no closed under $\neg$ class of functions contains a complete one. So, the class of all (computable or not) predicates contains no universal one. This is the prominent Cantor Theorem that the set of all sets of strings (as well as the sets of all functions, reals etc.) is not countable.
Formal proof systems are computable functions $A(P)$ which check if $P$ is an acceptable proof and output the proven statement. $\vdash s$ means $s=A(P)$ for some $P$. $A$ is rich iff it allows computable translations $s_x$ of statements "$u_2(x)=0$", provable whenever true, and refutable ($\vdash\neg s_x$), whenever $u_2(x)=1$.
$A$ is consistent iff at most one of any such pair $s_x,\neg s_x$ is provable, and complete iff at least one of them always (even when $u(x)$ diverges) is. There is no complete function among the total computable ones, as this class is closed under negation. So the universal in $R$ function $u$ (and $u_2=(u\bmod2)$) has no total computable extensions. Thus, rich consistent and complete formal systems cannot exist, since they would provide an obvious total extension $u_A$ of $u_2$ (by exhaustive search for $P$ to prove or refute $s_x$). This is the famous Goedel's Theorem -- one of the shocking surprises of the 20th century science. (Here $A$ is any extension of the formal Peano Arithmetic; we skip the details of its formalization and proof of richness.)
A closer look at this proof reveals another famous Goedel theorem: Consistency $C$ of $A$ (expressible in rich $A$, as divergence of the search for contradictions) is itself an unprovable example. Indeed, $u_2$ intersects $1{-}u_A$ for some prefix $a$. $C$ implies that $u_A$ extends $u_2$ and, thus, $u_2(a),u_A(a)$ both diverge. So, $C{\Rightarrow\neg} s_a$. Peano Arithmetic can formalize this proof, thus ${\vdash}C\Rightarrow {\vdash\neg}s_a$. But $\vdash\neg s_a$ implies $u_A(a)$ converges, so $\vdash C$ contradicts $C$. So, consistency of $A$ is provable in $A$ if and only if false !
Computable Functions. Another byproduct is that the Halting (of $u$) Problem would yield a total extension of $u$ and, thus, is not computable. This leads to many other uncomputability results. Another source is an elegant Fixed Point Theorem by S. Kleene: any total computable $A$ transforming programs (prefixes) $p$ maps some $p$ into equivalent ones (computing the same functions). Indeed, the complete $u(ps)$ intersects computable $u(A(p)s)$. This implies Rice theorem: the only computable invariant (i.e. equal on equivalent programs) property of programs is constant (exercise).
Computable (partial and total) functions are also called recursive (due to an alternative definition). Their ranges (and, equivalently, domains) are called (computably) enumerable or c.e. sets. A c.e. set with a c.e. complement is called computable or decidable. A function $f$ is computable iff its graph $G$ is c.e. If $f$ is also total, $G$ is decidable. Each infinite c.e. set is the range of an injective total computable function ("enumerating" it, hence the name c.e.).
We can reduce membership problem of a set $A$ to one of a set $B$ by finding a computable function $f$ s.t. $x\in A \iff f(x)\in B$. Then $A$ is called m- (or many-to-1-) reducible to $B$. More complex Turing reduction are by algorithms which, starting from input $x$, interact with $B$ by generating strings $s$ and receiving answers to $s\in?B$ questions. Eventually this stops and tells if $x{\in}A$. C.e. sets (like Halting Problem) to which all c.e. sets m-reduce are called c.e.-complete. This (and, thus, undecidability) can be shown by reducing the Halting Problem to them. Ju.Matijasevich so proved completeness of Diophantine Equations Problem: find if a given multivariate integer polynomial of degree 4 has integer roots. Such ideas are broadly used in Theory of Algorithms and should be learned from any standard text, e.g., [6].
The $t$-restriction $u_t$ of $u$ aborts and outputs $1$ if $u(x)$ does not halt within $t(x)$ steps, i.e. $u_t$ computes the $t$-Bounded Halting Problem ($t$-BHP). It remains complete for the $\neg\,$-closed class of functions computable in time $o(t(x))$. ($o(1)$ and padding of $p$ absorb $O(\|p\|^2)$ overhead.) So, $u_t$ is not in the class, i.e. cannot be computed in time $o(t(x))$ [17]. (And neither can be any function agreeing with $t$-BHP on a dense (i.e. having strings with each prefix) subset.) E.g. $2^{\|x\|}$-BHP requires exponential time. However for some trivial input programs the BHP can obviously be answered by a fast algorithm. The following theorem provides another function $P_f(x)$ (which can be made a predicate) which has only a finite number of such trivial inputs.
We state the theorem for the Multi-Head Turing Machine volume of computation. It can be reformulated in terms of Pointer Machine time or regular Turing Machine space (or, with smaller accuracy, time).
Definition: A function $f(x)$ is called constructible iff it can be computed in volume $V(x)=O(f(x))$.
Examples: $f(x){=}2^{\|x\|}$ is constructible, as $V_f(x)=O(\|x\|)\ll2^{\|x\|}$. Yet, $2^{\|x\|}{+}h(x)$, where $h(x){\in}\{0,1\}$ is $0$ iff $V_{u(x)}<3^{\|x\|}$, is not.
Compression Theorem [16]. For each constructible function $f$, a function $P_f$ exists such that for all functions $t$, the following two statements are equivalent:
(1) $P_f(x)$ is computable in volume $t(x)$.
(2) $t$ is constructible and $f(x)=O(t(x))$.
Proof. Let $t$-time Kolmogorov Complexity $K_t(i|x)$ of $i$ given $x$ be the least $\|p\|$ of programs $p$ such that $u(p,x)=i$, $V_{u(p,x)}{<}t$. Let $P_f(x)$ be the smallest $i$, with $2K_t(i|x)>\log(f(x)|t)$ for all $t$. $P_f$ is computed in volume $f$ by generating all $i$ of low complexity, sorting them and taking the first missing. It satisfies the Theorem, since computing $i{=}P_f(x)$ faster would violate the complexity bound defining it. (Some extra efforts can make $P$ Boolean.) Q.E.D.
Speed-up Theorem [12]. There exists a total computable predicate $P$ such that for any algorithm computing $P(x)$ in volume $t(x)$, there exists another algorithm doing it in volume $O(\log t(x))$.
Though stated here for exponential speed-up, this theorem holds for any computable unbounded monotone function in place of $\log$. So, not even remotely optimal algorithm computing $P$ exists. This case is in a sharp contrast with Compression Theorem cases.
The general case. So, the complexity of some predicates $P$ cannot be characterized by a single constructible function $f$, as in Compression Theorem. However, the Compression Theorem remains true (with harder proof) if the requirement that $f$ is constructible is dropped (replaced with being computable).
In this form it is fully general so that every computable predicate (or function) $P$ satisfies the statement of the theorem with an appropriate computable function $f$. There is no contradiction with Blum's Speed-up, since the complexity $f$ (not constructible itself) cannot be reached. See a review in [15].
(The Compression theorem proof stands if constructibility of $f$ is weakened to being semi-constructible, i.e. one with an algorithm $A(n,x)$, such that $V_{(n,x)}=O(n)$, and $A(n,x){=}f(x)$ if $n{>}f(x)$. The sets of programs $t$ whose volumes (where finite) satisfy either (1) or (2) of the Theorem (for computable $P,f$) are in $\Sigma^0_2$ (i.e. defined with 2 quantifiers). Both generate monotone classes of constructible functions closed under $\min(t_1,t_2)/2$. Then any such class is shown to be the $\Omega(f)$ for some semi-constructible $f$.)
In this section we consider a more interesting provably intractable problem: playing two players zero sum games with full information. We will see that even some simple games yield to no much faster algorithms than exhaustive search of all possible configurations. The rules of an $n$-player game $G$ are set by families $f,v$ of information and value functions and a transition rule $g$. Players $i{\in}I$ at each step participate in transforming a configuration (game position) $x{\in}C$ into the new configuration $g(x,m)$, $m:I{\to}M$ by choosing moves $m_i{=}m(i)$ based only on their knowledge $f_i(x)$ of $x$. The game ends upon reaching a terminal configurations $t{\in}T{\subset}C$. Then $v_i(t)$ is the loss or gain of the $i$-th player.
Our games will have zero sum $\sum v_i(t)=0$ and full information: $f_i(x){=}x$, $g(x,m)=g'(x,m_{a(x)})$ : $a(x)$ indicates the single active player. We consider binary, two-players, no-draw games, taking $0\notin C\subset\Z$, $M\subset\Z$, $T=I=\{\pm1\}$, $a(x)=x/|x|$ (its sign), $v_i(t)=a(t)i$, $|g(x,m)|<|x|$. (In our examples this "${<}|x|$" is by implicitly prepending configurations with a counter of remaining steps.)
An example of such games is chess. Examples of games without full information are card games, where only a part $f_i(x)$ (player's own hand) of the position $x$ is known. Each player may have a strategy providing a move for each position. A strategy $S$ is winning at $x$ if starting at a position $x$ it guarantees victory whatever the opponent does, even if he knows $S$. We can extend $v_1$ on $T$ to $V$ on all positions with a winning strategy for one side so that $a(x)V(x)=\sup_m\{a(x)V(g(x,m))\}$. ($\sup\{\}$ taken as $-1$.)
Evaluating or solving a game, means computing $V$. This ability is close to the ability to find a good move in a modified game. Indeed, modify a game $G$ into $G'$ by adding a preliminary stage to it. At this stage the player $A$ offers a starting position for $G$ and her opponent $B$ chooses which side to play. Then $A$ may either start playing $G$ or decrement the counter of unused positions and offer another one. Obviously, $B$ wins if he can determine the winning side of every position. If he cannot while $A$ can, $A$ wins.
Also, any game can be modified into one with two moves: $M\subset \{0,1\}$ by breaking a string move into several bit-moves. (Position of the new game consist of a position $x$ of the old one and a prefix $y$ of a move. The active player adds bits to $y$ until $m$ is complete and the next position generated by $g(x,m)$.) Evaluating such games is clearly sufficient for choosing the right move.
Theorem. [18] Each position of any full information game has a winning strategy for one side.
(This theorem fails for partial information games: either player may lose if the adversary knows his strategy. E.g.: 1. Blackjack (21); 2. Each player picks a bit; their equality determines the winner.)
Playing all strategies against each other can solve the game. There are $2^n$ positions of length $n$, $(2^n)^{2^n} = 2^{n\times 2^n}$ strategies and $2^{n\times 2^{n+1}}$ pairs of them. For a 5-bit game that is $2^{320}$. But the proof of this Theorem gives a much faster (yet still exponential time!) strategy.
Proof. Make the graph of all $\le x$ positions and moves; Set $V{=}0$; reset $V{=}v$ on $T$. Repeat until idle: If $V(x){=}0$, set $V(x){=}a(x)\sup_m\{a(x)V(g(x,m))\}$. The cycles end, with empty $V^{-1}(0)$, since $|g(x,m)|{<}|x|$ in our games keep decreasing. Q.E.D.
Games may be categorized by the difficulty to compute $r$. We will consider only $r$ computable in linear space $O(\|x\|)$. Then, the $2^{2\|x\|}$ possible moves can be computed in exponential time, say $2^{3\|x\|}$. The algorithm tries each move in each step. Thus, its total running time is $2^{3\|x\|+1}$: extremely slow ($2^{313}$ for a 13-byte game) but still much faster than the previous (double exponential) algorithm.
Can the remaining exponent be eliminated, too ? See the sec.3.2.
Exercise: the Match Game. Consider 3 boxes, each with 3 matches: $\fbox{! ! !}~\fbox{! ! !}~\fbox{! ! !}$. The players alternate turns taking any positive number of matches from a single box. The taker of the last match from table loses. Use the above algorithm to evaluate all positions and list the evaluations after each its cycle.
Exercise: Modify the chess game by giving one side the option to make an extra move out of turn during the first 10 moves. Prove that this side have a non-loosing strategy.
A simple example of a full information game is Linear Chess, played on a finite linear board. Each piece has a 1-byte type, including loyalty to one of two sides: W (weak) or S (shy), gender M/F and a 6-bit rank. All cells of the board are filled and all W's are always on the left of all S's. Changes occur only at the active border where W and S meet (and fight). The winner of a fight is determined by the following Gender Rules: Shy gets confused and loses if Weak is of different sex. Otherwise, Weak loses. The party of a winning piece A replaces the loser's piece B by its own piece C. The choice of C is restricted by the table of rules listing all allowed triples (ABC). We will see that this game cannot be solved in a subexponential time.
We first prove that (see [20]) for an artificial game. Then we reduce this Halting Game to Linear Chess showing that any fast algorithm to solve Linear Chess, could be used to solve Halting Game, thus requiring exponential time. For Exp-Time Completeness of regular (but $n\times n$) Chess, Go, Checkers see: [22], [21].
We use a universal Turing Machine $u$ (defined as 1-pointer cellular automata) restricted to either diverge, or halt within $2^{\|x\|}$ steps by its head rolling off of the tape's left end, leaving a blank. This cannot be determined in $o(2^{\|x\|})$ steps. We will now convert this problem into the Halting Game:
The players are: $W$ claiming $u(x)$ halts (and has winning strategy iff this is true); His opponent is $S$. The board has four parts: the $C$ diagram, the input $x$ to $u$, positive integers $p$ (position) and $t$ (time in the execution of $u(x)$) The diagram shows the states $C_p^{t+1},C_p^t$ of cell $p$ at times $t{+}1,t$, and $C^t_{p'}$ of cell $p'=p{+}d',\,d'{\in}\{\pm1\}$ at time $t$. $C$ include present $d$ and previous $d'$ pointers direction; $C^t$ may be replaced by "?".
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Some board configurations are illegal: if (1) both $C_p,C_{p'}$ point away from each other, or (2) $C^{t+1}$ differs from the result prescribed by the transition rules for $C^t$, or (3) $t{=}1$, while $C^1_p \ne x_p$. (At $t{=}1$, $u$ has the input $x$ on its tape at the left with the head in the initial state, followed by blanks at the right.)
The Game Rules: The starting configuration of the game is at the right. $W$, in its moves, replaces the ?s with symbols claiming to reflect the state of cells $p',p$ at step $t$ of $u(x)$. $S$ then chooses $s{\in}\{0,1\}$, replaces $p$ with $p{+}sd'$, moves $C_p$ to top $C$ box, fills lower $C$ boxes with ?s, and decrements $t$. The game can only check that compliance with "local" rules, not refer to past moves or to actual computation of $u(x)$.
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Note that $W$ may lie (i.e distort the actual computation of $u(x)$ in filling "?"), as long as he is compliant.
Strategy for W: If $u(x)$ does indeed halt, then the initial configuration is true to the computation of $u(x)$. Then $W$ has an obvious (though hard to compute) winning strategy: just tell truly (and thus always consistently) what happens in the computation. W will win when $t{=}1$, so $S$ cannot decrement it further.
Strategy for S: If the initial configuration is a lie, $S$ can force $W$ to lie all the way down to $t{=}1$ (when the lie is exposed). How is this done?' If the upper box $C^{t+1}_p$ of a legal configuration is false then the lower boxes $C^t_{p'}$ $C^t_p$ cannot both be true, since the rules of $u$ determine $C^{t+1}_p$ uniquely from them. If $S$ correctly points the false $C$ and brings it to the top on his move, then $W$ is forced to keep on lying. At time $t{=}1$ the lie is illegal, as the configuration doesn't match the actual input string $x$. So, solving this game is to decide if the initial configuration is correct, i.e. $u(x)$ halts in $2^{\|x\|}$ steps: impossible in time $o(2^{\|x\|})$.
This Halting Game is artificial, still has a Halting Problem flavor, though it does not refer to exponents. We will now reduce it to a sleeker game -- Linear Chess -- to prove it exponentially hard, too.
To reduce (see definition in sec. 2.2) Halting Game to Linear Chess, we introduce a few concepts. A non-deterministic Turing Machine (NTM) is a TM that sometimes offers a (restricted) transition choice, made by a driver, a function (of the TM configuration) of unrestricted complexity. A deterministic (ordinary) TM $M$ accepts a string $x$ if $M(x){=}$yes; an NTM $M$ does if there exists a driver $d$ s.t. $M_d(x){=}$yes.
NTM represent single player games -- puzzles -- with a simple transition rule, e.g., Rubik's Cube. One can compute the winning strategy in exponential time by exhaustive search of all $d$.
Home Work: Prove or refute that all such games have P-time winning strategies. Will get you "A" for the course, $\$1,000,000$ award and a senior faculty rank at a school of your choice.
Alternating TM (ATM) are a variation of the NTM driven by two alternating drivers $l,r$. A string is accepted iff $M_{l,r}(x){=}$ yes for some $l$ and all $r$. Our games could be viewed as ATM returning the result of the game in linear space but possibly exponential time. $M$ prompts players $l$ and $r$ alternatingly to choose their moves (in several steps for multi-bit moves) and computes the resulting positions, until a winner emerges. Accepted strings describe winning (i.e. having a winning strategy) positions.
Linear Chess (LC) Simulation of TM-Games. We first represent the Halting Game as an ATM computation simulated by the Ikeno TM (Sec.2.1) (using the "A/B" command for players' input). It is viewed as an array of 1-pointer cellular automata: Weak cells as rightward, Shy leftward. Upon termination, the TM head is set to move to the end of the tape, eliminating all loser pieces.
This is viewed as a game of 1d-Chess (1dC), a variant of LC, where a table, not the "Gender Rule", determines the victorious piece, and not only the losing piece is replaced, but also the winning piece may change rank. The types are states of Ikeno TM showing Loyalty (pointer direction $d{\in}\{W,S\}$), gender $\delta$ (previous $d$), and 6/6/6/5 ranks (trit $t{\in}\{0,1,*\}$ with $'$ bit $p$).
Exercise: Describe LC simulation of 1dC. Hint: Each transition of 1dC is simulated in several LC stages. Let L,R be the pieces active in 1dC. In odd stages L (in even stages R) turns pointer twice and changes gender. The last stage turns pointer once and possibly changes gender. In the first stage L appends its rank with R's $p$ bit. All other stages replace old 1dC rank with the new one. R appends its old $t$ bit (only if $t\ne*$) to its new rank. Subsequent stages drop both old bits, marking L instead if it is the new 1dC head. Up to 4 more stages are used to fix any mismatch with 1dC new $d,\delta$ bits.
Space-Time Trade-off. Deterministic linear space computations are games where each position has a single (and easy to compute) move. A big open problem: what time is required to compute their outcome.
We know no general subexponential upper or superlinear lower bounds. Recall that on a parallel machine time is the number of steps until the last processor halts, space is the memory size used; volume is the combined number of steps of all processors. "Small" will refer to values bounded by a polynomial of the input length; "large" to exponential. Call computations narrow if either time or space are small, and compact if both (thus, volume too) are.
An open question: Do all exponential volume algorithms (e.g., one solving Linear Chess) allow an equivalent narrow computation? Alternatively, can every narrow computation be converted into a compact one? This is equivalent to the existence of a P-time algorithm for solving any fast game, i.e. a game with a P-time transition rule and a move counter with a small bound.
The sec. 3.1 algorithm can run in parallel P-time for such games. Converse, too, is similar to Halting Game.
[19] solve compact games in P-space: With $M{\subset}\{0,1\}$ run depth-first search on the tree of all games (sequences of moves). On exiting each node it is marked as the active player's win if some move leads to a child so marked; else as his opponent's. Children's marks are then erased.
Conversely, compact games can simulate any P-space algorithms. Player A declares the result of the space-$k$, time-$2^k$ computation. If he lies, player B asks him to declare the memory state in the middle of that time interval, and so by a k-step binary search catches A's lie on a mismatch of states at two adjacent times.
Thus, while narrow alternating computations (general games) correspond to large deterministic ones, the compact alternating computations (fast games) correspond to either type of narrow deterministic ones. This has some flavor of trade-offs such as saving time at the expense of space in dynamic programming.
We now enter Terra Incognita with deterministic computations extended by tools like random choices, non-deterministic guesses, etc., the power of which is completely unknown. Yet fascinating discoveries were made there, a glimpse of which we will get now.
Consider a P-time function $F$. Assume $\|F(x)\|=\|x\|$, for convenience (often $\|F(x)\|=\|x\|^{\Theta(1)}$ suffices). Inverting $F$ means given $y$, finding some $x\in F^{-1}(y)$, i.e. such that $F(x) = y$.
To invert $F$ we may check $F(x)=y$ for all possible $x$. This takes small space but absolutely infeasible $\Omega(2^{\|x\|})$ time. No method is currently proven much better in the worst case. Nor could we prove some inversion problems to require more time than $F$ does. This is the sad present state of Computer Science !
An Example: Factoring. Let $F(x_1,x_2)=x_1x_2$ be the product of integers. For simplicity, assume $x_1,x_2$ are primes. A fast algorithm in sec. 5.1 determines if an integer is prime. If not, no factor is given, only its existence. To invert $F$ means to factor $F(x)$. The density of $n$-bit primes is $\Theta(1)/n$. So, factoring by exhaustive search takes exponential time! In fact, even the best known algorithms for this ancient problem run in time about $2^{\sqrt{\|x\|}}$, despite centuries of efforts by most brilliant people. The task is commonly believed infeasible and the security of many popular cryptographic schemes depends on this unproven faith.
One-Way Functions: $F:x\to y$ are those easy to compute but hard to invert $(y\mapsto x)$ for most $x$. Even their existence is a sort of religious belief in Computer Theory. It is unproven, though many functions seem to be one-way. Some functions, however, are proven to be one-way iff any one-way functions exist. Many theories and applications are based on this hypothetical existence.
Let us compare inversion problems with another type -- search problems -- solving P-time computable relations $F$: given $x$, (a) find $w$ (called witness) s.t. $F(x,w), \|w\|=\|x\|$ (constructive problem) or (b) just decide if such $w$ exist (decision problem). Any inversion problem is a search problem and any search problem can be restated as an inversion problem. E.g., finding Hamiltonian cycles $C$ in a graph $G$, is to invert $f$ where $f(G,C)$ is $(G,0\ldots0)$ if $C$ is in fact a Hamiltonian cycle of $G$, or $(0\ldots0)$ if $C$ is not. Similarly any search problem can be reduced to another one equivalent to its decision version. For instance, factoring $x$ reduces to bounded factoring: given $x,b$ find $p{<}b,q$ such that $pq{=}x$. (Decisions yield construction by binary search.)
Exercise: Generalize the above to reduce any search problem to an inversion and to a decision problems.
(Search problems are games with P-time transition rules and one move duration. A great hierarchy of problems results from allowing more moves and/or other complexity bounds for transition rules.)
The problem's language is the set of instances it accepts: the range of $f$ for inversion problems, $\{x:\exists w\,F(x,w)\}$ for search. An NP language is the set of inputs acceptable by a P-time non-deterministic TM (sec. 3.3). All three language classes -- search, inversion and NP -- coincide. (NP $\iff$ search is trivial.)
Interestingly, P-space bounded deterministic and non-deterministic TMs have equal power [23]. Assume TM has a unique accepting configuration and space bound $s$. Any accepted string is accepted in time $\le2^s$. We need to check $A(x,w,k)$: whether the NTM can be driven from configuration $x$ to $w$ in time $\le2^k$, $k{\le}s$. For this we check $A(x,z,k{-}1)$ and $A(z,w,k{-}1)$ for each $z$. This takes space $t_k\le t_{k{-}1}{+}\|z\|{+}O(1)$. So, $t_k{=}O(sk){=}O(s^2)$.
Nobody knows if all search (or inversion, NP, etc.) problems are solvable in P-time (said being in P). They all yield to exponential time but no better bounds are known. This question (P=?NP) is probably the most famous one in Computer Science. Many problems seem hopeless to solve, while similar or slightly modified problems yielded to (often extraordinary) efforts. Examples:
1. Linear Programming: Given an integer $n{\times}m$
matrix $A$ and vector $b$, find a rational vector $x$ with $Ax 2. Primality test: Determine if a given integer $p$ has a factor?
Solution: Trying all $2^{\|p\|}$ possible integer factors of $p$ is
infeasible, but more sophisticated algorithms, run fast (see
section 5.1).
3. Graph Isomorphism: Are two given graphs $G_1,G_2$, isomorphic?
I.e., can the vertices of $G_1$ be re-numbered to make it equal to $G_2$?
Solution: Checking all $n!$ enumerations of vertices is infeasible
(exceeds the number of atoms in the known Universe for $n{=}100$).
[25] found an $O(n^d)$ steps algorithm where
$d$ is the degree.
This is a P-time for $d = O(1)$.
4. Independent Edges (Matching): Find a given number of
independent (i.e., not sharing nodes) edges in a given graph.
Solution: Max flow algorithm solves a bipartite graph case.
A more sophisticated algorithm by J.Edmonds solves the general case.
Many other problems have been battled for decades or centuries and no
P-time solution has been found. Even modifications of the previous four
examples have no known answers:
(1) Linear Programming: All known solutions produce rational $x$.
No reasonable algorithm is known to find integer $x$.
(2) Factoring: Given an integer, find a factor. Centuries of quest for
fast algorithm were unsuccessful. Takes $n^{\sqrt{n}}$ time so far.
(3) Sub-graph isomorphism: find if a graph is isomorphic to a part of
another. No P-time solutions known, even for $O(1)$-degrees.
(4) Independent nodes: Find $k$ independent (i.e., not sharing edges)
nodes in a given graph. No P-time solution is known.
Exercise: Restate the above problems as
inverting easily computable functions.
We proved the exponential complexity of Linear Chess and some other
games. None of the above or any other search/inversion/NP problem,
however, have been proven to require super-polynomial time.
When, therefore, do we stop looking for an efficient solution?
NP-Completeness theory is an attempt to answer this question.
See results by S.Cook, R.Karp, L.Levin, and others surveyed in
[26], [27].
A P-time function $r$ reduces one NP-predicate $p_1(x)$ to $p_2(x)$ iff
$p_1(x) = p_2(r(x))$, for all $x$. $p_2$ is NP-complete if all
NP problems can be so reduced to it. Thus, NP-complete problems are the
hardest in the worst-case in the class NP. Any P-time solution for one of
them would solve all other NP (or inversion, or search) problems. This
may suggest giving up on fast algorithms for them.
Faced with problem's NP-completeness we may try to restate it, find a
similar one which is easier (possibly with additional tools) but still
gives the information we really want. Relaxing factoring to testing
primality (see Sec. 5.1) is such a case. Now we
proceed with an example of NP-completeness.
A tile is one of $26^4$ unit squares with a Latin letter at each corner.
Two tiles may be placed next to each other if the letters on the shared
side match. (See an example at the right.)
Tiling Problem. Find a tiled square, given its first row and the
list of tiles used.
We will now reduce all search problems to that one, starting with some
"standard" NP problem. An obvious candidate is existence of $w$ with
$U(v,w)$, where the universal Turing Machine $U$ simulates $F(x,w)$ for
$v = px$. But $U$ has no P-time limit, so we must impose one on it. How
to make such fixed degree polynomial sufficient to simulate any
polynomial (even of higher degree) time?
Padding Argument. Let $u(v,w)$ for $v{=}00\ldots01px$ simulate
$\|v\|^2$ steps of $U(px,w){=}F(x,w)$. If the $0^*$ padding in $v$
is long enough, $u$ has time to simulate $F$, even though $u$ runs in
quadratic time, while $F$'s time limit may be, say, cube (of a shorter
string $xw$).
So an NP problem $F(x,w)$ is reduced to $u(v,w)$ by mapping instances $x$
to $v{=}0^*1px$, with $\|0^*\|$ determined by $F$'s time limit. So, if
some NP problem cannot be solved in P-time then neither can be
$\exists? w: u(v,w)$. And if the latter HAS a P-time solution
then so have all search problems. (Which of these alternatives
is true we do not know.) It remains to reduce the search problem $u$ to
Tiling.
The Reduction. $u(v,w)$ is run by a TM represented as an array
of 1-pointer cellular automata that runs for $\|v\|^2$ steps and stops if
$w$ does NOT solve the relation $F$. Otherwise it enters an infinite
loop.
An instance $x$ has a solution iff $u(v,w)$ runs forever for some $w$.
Set $n$ to $u$'s time/space $\|v\|^2$. In the space-time diagram of
computation of $u(v,w)$, each row represents configuration of $u$ at the
respective time. The first row is the initial configuration: $\fbox{ v
?...? #...#}$. It has a special symbol "?" as a placeholder for the
solution $w$ which is filled in at the second step below it. A faulty
table may reflect no actual computation. Then it must have 4 cells
sharing a corner impossible to appear in the computation of $u$ on any
input.
Proof. As the input $v$ and the guessed solution $w$ are the
same in both the right and wrong tables, the first 2 rows agree. The
third row starts the actual computation. In the first mismatching row a
transition of some cell from the previous row is wrong. This is clear
from the state in both rows of this cell and the cell it points to,
making an impossible combination of four cells sharing a corner.
For a given $x$, the existence of $w$ satisfying $F(x,w)$ is equivalent
to the existence of a table with the prescribed first row, no halting
state, and permissible patterns of each 4 adjacent squares (cells).
We now convert the table into the Tiling Problem: Cut
each cell into 4 parts by a vertical and a horizontal lines through its
center and copy cell's content in each part.
Combine into a tile the 4 parts sharing a corner of 4 cells. The cells
at the right are separated by "---"; the tiles by dashes; If these cells
are permissible in the table, then so is the respective tile. Thus any
matching tiled square solves the respective NP problem. This reduces
inversion of $u$, and thus every NP problem, to Tiling.
Exercise: Find a P-time algorithm for $n\times\log n$ Tiling
Problem.
The factoring problem seems very hard. But to test if a number
has factors turns out to be much easier than to find them.
This is crucial for many applications, such as, e.g., security tools.
It also helps to supply the computer with a coin-flipping device.
We now consider a Monte Carlo algorithm, i.e. one that with high
probability rejects any composite number, but never a prime.
See: [29], [30],
[28].
Residue Arithmetic. $p|x$ means $p$ divides $x$. $x\equiv
y\pmod p$ means $p|(x{-}y)$. $y=(x \bmod p)$ denotes the residue of $x$
when divided by $p$, i.e. $x\equiv y\in[0,p{-}1]$. Residues can be added,
multiplied and subtracted with the result put back in the range $[0,p
{-}1]$ via shifting by an appropriate multiple of $p$. E.g., $-x$ means
$p{-}x$ for residues $\bmod\ p$. $\pm x$ means either $x$ or $-x$.
The Euclidean Algorithm finds $\gcd\,(x,y)$ -- the greatest
(multiple of any other) common divisor of $x$, $y$: $\gcd\,(x,0){=}x$;
$\gcd\,(x,y)=\gcd\,(y,(x\bmod y))$ for $y{>}0$. By induction,
$d=\gcd\,(x,y)=A{*}x{-}B{*}y$, where integers $A{=}(d/x\bmod y)$ and
$B{=}{-}(d/y\bmod x)$ are produced as a byproduct of that algorithm.
It allows division $(\bmod\ p)$ by any $r$ coprime with $p$ (i.e.
$\gcd\,(r,p)$ $=1$), and operations $+,-,*,/$ obey all usual arithmetical
laws. We also need to compute $(x^q\bmod p)$ in polynomial time.
We cannot do $q{>}2^{\|q\|}$ multiplications. Instead we compute all
$x_i{=}(x_{i{-}1}^2\bmod p){=}(x^{2^i}\bmod p)$, $i{<}\|q\|$.
Then we take $q$ in binary, i.e. as a sum of powers of $2$ and multiply
$\bmod\ p$ the needed $x_i$'s.
Fermat Test. The Little Fermat Theorem for each prime
$p{\not|}\,x$ says: $x^{(p{-}1)}{\equiv}1\pmod p$.
Indeed, $i\to(ix\bmod p)$ is a permutation of $[1,p{-}1]$.
So, $1\equiv(\prod_{i Square Root Test. For each y and prime p, $x^2\equiv y\pmod p$
has at most one pair of solutions $\pm x$.
Proof. Let $x,x'$ be two solutions: $y\equiv x^2\equiv x'^2\pmod
p$. Then $p\,|\,x^2{-}{x'}^2 = (x{-}x')(x{+}x')$. Thus $p|(x{-}x')$ or
$p|(x{+}x')$, making $x\equiv\pm x'$, if $p$ is prime. Otherwise $p$ is
composite, and $\gcd\,(p,x{+}x')$ even gives its factor.
Random Choice. We say $d$ kills $\Z_p^*$ if
$x^d\equiv1\pmod p$ for all $x$ (in $\Z_p^*$, i.e. coprime with $p$). If
$x^d\not\equiv1$, then for all $y$ either $y^d\not\equiv1$ or
$(xy)^d\not\equiv1$. Same with $x^d\not\equiv\pm1\pmod p$. So existence
of a single such $x$ implies the same for most of randomly chosen
$y$.
Miller-Rabin Test $T_x$ factors a composite $p$ given $d$ that
kills $\Z_p^*$. If $d=p{-}1$ does not, then Fermat Test confirms $p$ is
composite. If it does, let $p{-}1{=}2^k q$, with odd $q$. $T_x$ sets
$x_0{=}(x^q\bmod p)$, $x_i=(x_{i{-}1}^2\bmod p)=(x^{2^i q}\bmod p)$,
$i{\le}k$. $x_k{=}1$. If $x_0{=}1$, or one of $x_i$ is $-1$, $T_x$ gives
up for this $x$.
Otherwise $x_i\not\equiv\pm1$ for some $i{<}k$, while $x_i^2\equiv
x_{i+1}\equiv1$, and the Square Root Test factors $p$. Now, $T$ succeeds
with some (thus most!) $x\in\Z_p^*$. If $p{=}a^{i+1},i{>}0$, then
$(1{+}a^i)^{p-1}{=}1{+}a^i(p{-}1){+}a^{2i}c\equiv$
$1{-}a^i\not\equiv1\pmod p$.
Else, $p{=}ab$, $\gcd(a,b){>}1$: Take the greatest $i$ such that
$2^iq$ does not kill $\Z_p^*$. It exists (as $(-1)^q\equiv-1$
for odd $q$) and has $x_i\not\equiv1\equiv(x_i)^2\pmod p$.
Then $x'{=}1{+}b(1/b\bmod a)(x{-}1)\equiv1{\equiv}x'_i\pmod b$, while
$x'_i{\equiv}x_i\not\equiv1\pmod a$. So, $x'_i\not\equiv\pm1\pmod p$.
Las-Vegas algorithms, unlike Monte-Carlo, never answer wrongly.
Unlucky coin-flips just make them run longer than expected.
Quick-Sort is a simple example. It is about as fast as deterministic
sorters, but popular due to its simplicity. It sorts an array by
choosing a random pivot in it, splitting the rest of it in two by
comparing with the pivot, and calling itself recursively on each half.
For easy reference, rename the entries with their positions $1,\ldots,n$
in the sorted output (no effect on the algorithm). Denote
$t(i)$ the (random) time $i$ is chosen as a pivot. Then $i$ will ever be
compared with $j$ iff either $t(i)$ or $t(j)$ is the smallest among
$t(i),\ldots,t(j)$.
This has $2$ out of $|j{-}i|+1$ chances. So, the expected number of
comparisons is $\sum_{i,j>i} 2/(1{+}j{-}i)= -4n+ (n{+}1)\sum_{k=1}^n
2/k=$ $2n(\ln n-O(1))$. Note, that the expectation of the sum of
variables is the sum of their expectations (not true, say, for product).
The above Monte-Carlo and Las-Vegas algorithms flip random coins
themselves. (Or use pseudo-random generators instead, in hope,
rarely supported by proofs, that their outputs have all the statistical
properties of truly random coin flips used in the analysis.) These
scenaria should not be confused with analysis of algorithms when inputs
are chosen randomly with uniform distribution.
Random Inputs. Consider an example: Someone is interested in
knowing whether or not certain graphs contain Hamiltonian Cycles.
He offers graphs $G$ and pays $\$100$ if we show either that $G$
has or has not one. The problem is NP-Complete, so it
likely is very hard for some, but not necessarily for
most graphs. In fact, if our patron chooses $G$ uniformly, a fast
algorithm can earn us the $\$100$ most of the time ! Let the
graphs have $n$ nodes, $e The probability that a given node $A$ of the graph is isolated is
$(1-1/n)^e>(1-O(1/n))/\sqrt n$. The probability that none
of them are (and we lose our $\$100$ is $O((1-1/\sqrt n)^n)=$
$O(1)/e^{\sqrt n}$, vanishing fast. Similar calculations can be made
whenever $r = \lim e/(n\ln n)<1$. If $r{>}1$, other fast algorithms can
find a Hamiltonian Cycle. See: [32],
[33], [34].
(See also [35] for a proof that another
graph problem is NP-complete even on average.)
How do this HC algorithm and the above primality test differ? The
primality test works for all instances. It tosses its own coin
and can repeat it for a more reliable answer. The HC algorithms only work
for most instances (with isolated nodes or generic HC). In the
HC case, we trust the customer's honest random choices.
If he cheats, producing rare graphs often, the analysis breaks down.
Randomness comes into computing in great many very different ways, we can
only have a glimpse at few examples. Here, one more: Symmetry
Breaking (avoiding conflicts for shared resources).
Dining Philosophers. They sit at a circular table. Between each
pair is either a knife or a fork, alternating. Neighbors must share the
utensils, cannot eat at the same time. How can they complete the dinner
while all following the same procedure with no central organizer? If two
diners grab at the same utensil, neither succeeds.
They could each flip a coin, and sit still if it comes up heads,
otherwise try to grab the utensils. If they repeat this procedure enough
times, most likely each philosopher will soon get a turn for both knife
and fork without interference.
In section 3, to win games, the player
(call him Merlin), must beat another almighty wizard (say, Lady of the
Lake). To make the challenge realistic we replace the Lady with a simple
player, Arthur, who chooses at random the moves for both sides. Merlin,
instead, must evaluate (by a formula) all moves. Any wrong formula of his
will be clearly inconsistent with correct formulas for the results of
most choices of the next move.
So, wrongness will persist This trick, called arithmetization, was
proposed in Noam Nisan's Fall-89 article widely distributed over email,
and quickly used in a flood of follow-ups for matching various high
complexity classes. We follow [36],
[37]. The trick is based on that degree $d$
polynomials agree on either ${<}d$ points or the whole field. We express
boolean functions as low degree polynomials, and apply them to
$\Z_p$-tokens (let us call them bytes) instead of bits. This reduces
generic games to those in which any Merlin's strategy in any losing
configuration has exponentially small chance to win.
The reduction holds for games with any (exponential, polynomial, etc.)
limit on the counter $c$ of the remaining moves.
This $c$ will be included implicitly in games configurations below.
Take the (ATM-complete) game of 1d-Chess
(Sec.3.3), $g(m,x)$ with
$x{=}x_1\ldots x_s$, $m,x_i{\in}\{0,1\}$ being its transition rule.
Configurations include $x$ and a remaining moves counter $c\le2^s$.
They are terminal if $c{=}0$, player $x_1$ winning. Intermediate
configurations $(m,x,y)$ have $y$ claimed as a prefix of $g(m,x)$.
Let $t(m,x,y)$ be $1$ if $y{=}g(m,x)$, else $t{=}0$.
1d-Chess is simple, so $t$ can be expressed as a product of $s$
multilinear $O(1)$-sized terms, any variable shared by $\le2$ terms. Thus
$t$ is a polynomial, quadratic in each $m,x_i,y_i$.
Let $V_c(x)$ indicate if the active player has a strategy to win in $c$
moves, i.e. $V_0(x)\edf x_1$, $V_{c+1}(x)=1{-}V_c(g(0,x))V_c(g(1,x))$,
$V_c(g(m,x))=V'_c(m,x,\{\})$, where $V'_c(m,x,y)\edf V_c(y)t(m,x,y)$ for
$y=y_1\ldots y_s$ and
$V'_c(m,x,y)\edf V'_c(m,x,y{\circ}0){+}V'_c(m,x,y{\circ}1)$
for shorter $y$. ($\circ$ denotes concatenation.)
In our game $G$ Merlin must prove his value $v$ for $V'$. At the start
he chooses a $2s$-bit prime $p$. Configurations $X{=}(m,x,y,c,v)$ of $G$
replace $x_i,m,y_i, v$ bits with $\Z_p$-bytes. The polynomial $V'$
retains the above inductive definition,
thus is quadratic in each $x_i,m$, as $t(m,x,y)$ is.
Then $y_i$ have degree $\le 4$ in $V_c(y)$ and $\le 6$ in $V'$.
At his steps Merlin chooses a degree $6$ polynomial $P(r)$. $v$ must be
$t(m,x,y)(1{-}P(1)P(0))$ for $s$-byte $y$, or $P(0){+}P(1)$ for shorter
$y$. Arthur then takes a random $r{\in}\Z_p$, sets $X$ to
$(r,y,\{\},c{-}1,P(r))$, or $(m,x,y{\circ}r,c,P(r))$, respectively. For
$X$ with a correct $v$ Merlin's obvious winning strategy
is to always provide the correct $P$.
If $v$ is wrong then $P(1)$, $P(0)$ cannot both be right.
So $P$ will differ from $V$, they can agree only on few (up to degree)
points (i.e. on exponentially small fraction of random $r$). The wrong
$v$ will then propagate throughout the game, becoming obvious at $c{=}0$.
Thus any Merlin strategy has exponentially small winning chance.
This reduction of Section 3 games yields a
hierarchy of powers of Arthur-Merlin games and of computations mutually
reducible with $V_c(x)$ of such games. The one-player games with
randomized transition rule $g$ running in space $O(\|x\|)$ are equivalent
to exponential time deterministic computations. If instead the running
time $T$ of $g$ combined for all steps is limited by a polynomial, then
the games (called interactive proofs, IP) are equivalent to P-space
deterministic computations.
An interesting twist comes in 1-move games with polylog $T$, too tiny to
examine the initial configuration $x$ and the move $m$.
Yet, a little care not only removes this obstacle but achieves
equivalence to NP. Namely, $x,m$ are set in an error correcting code, and
$g$ is given $O(\log\|x\|)$ coin-flips and random access to the digits of
$x,m$. Then the membership proof $m$ is reliably verified by the polylog
randomized $g$. See [38] for details and
references.
Now let us look into the concept of Randomness itself. Intuitively,
random sequences are those with the same properties as coin flips.
But what are these properties? Kolmogorov resolved the issue
with a robust definition of random sequences: those with no description
noticeably shorter than their length.
See survey and history in [39]ku87,
[40].
Kolmogorov Complexity $K_A(x|y)$ of the string $x$ given $y$ is
the length of the shortest program $p$ which lets algorithm $A$ transform
$y$ into $x$: $\min\{(\|p\|):A(p,y)=x\}$. A Universal Algorithm $U$
exists s.t, $K_U(x)\le K_A(x)+O(1)$, for every algorithm $A$. This $O(1)$
is bounded by the length of the program $U$ needs to simulate $A$. And as
$|K_U(x|y)-K_U(x|y)|=O(1)$ for any such $U,U'$, we select some $U$ and
abbreviate $K_U(x|y)$ as $K(x|y)$, or $K(x)$ for empty $y$. E.g.: for
$A\,{:}x{\to}x$, $K_A(x){=}\|x\|$, so
$K(x){\le}K_A(x){+}O(1){=}\|x\|{+}O(1)$.
We cannot compute $K(x)$ by running all programs $p$ of length
${\le}\|x\|{+}O(1)$ to find the shortest one generating $x$, as some $p$
diverge, and the halting problem is unsolvable. In fact, no algorithm can
compute $K$ or even any its lower bounds except $O(1)$.
Consider the Berry Paradox expressed in the phrase:
" The smallest integer that cannot be uniquely and clearly
defined by an English phrase of less than two hundred characters."
There are $<128^{200}$ English phrases of $< 200$ characters.
So there must be integers not expressible by such phrases and the
smallest one among them. But isn't it described by the above phrase?
A similar argument proves that $K$ is not computable. Assume an algorithm
$L(x){\ne}O(1)$ gives a lower bound for $K(x)$. We can use it to compute
$f(n)$ that finds $x$ with $n< L(x)\le K(x)$. But $K(x)\le K_f(x)+ O(1)$
and $K_f(f(n))\le \|n\|$. So $n< K(f(n))\le \|n\|+O(1)= \log O(n)\ll n$:
a contradiction. Thus, $K$ and its non-constant lower bounds are not
computable.
An important application of Kolmogorov Complexity measures the Mutual
Information: $I(x:y)= K(x)+ K(y)- K(x,y)$. It has many uses which we
cannot consider here.
Some upper bounds of $K(x)$ are close in some important cases. One such
case is of $x$ generated at random. Define its rarity for uniform on
$\{0,1\}^n$ distribution as $d(x)=n-K(x|n)\ge-O(1)$.
What is the probability of $d(x){>}i$ (i.e. of $K(x|n){<}n{-}i$), for
uniformly random $n$-bit $x$ ? $n{-}i$-bit programs generate ${<}2^{n-i}$
strings. So, probability of such $x$ is $<2^{n-i}/2^n= 2^{-i}$ for
any $n$. Even for a billion-bit $n$, the probability of
$d(x){>}300$ is really negligible (assuming $x$ was indeed generated by
fair coin flips).
Small rarity implies all other enumerable properties of random strings.
Indeed, let such property "$x{\not\in}P$" have a negligible chance: a
small, say $<2^{-s_n}$, fraction of $n$-bits strings violate $P$. To
generate $x$, we need only the algorithm enumerating $S_n$
and the $n{-}s_n$-bit position of $x$ in that enumeration.
Then the rarity $d(x)> s_n{-}O(1)$ is large. Each $x$ violating $P$ will
thus also violate the "small rarity" requirement. In particular, the
small rarity implies unpredictability of bits of random strings:
Any short algorithm with high prediction rate assures large $d(x)$.
However, the randomness can only be refuted, cannot be confirmed: we saw,
$K$ and its lower bounds are not computable.
Rectification of Distributions. Sources of randomness with
precisely known distribution are rare. (Note: distributions of $10^6$
coin flips with $.49$ and $.51$ head chances are nearly orthogonal.)
But there are very efficient ways to convert "roughly" random sources
into perfect ones. We follow [43], which also uses
some earlier U. and V. Vazirani ideas. Assume a sequence with weird
unknown distribution. We only know that its $m$ bits long segments have
some, say ${>}k{+}i$, min-entropy i.e. probability ${<}1/2^{k+i}$, given
all previous bits. (Without such $m$ arbitrary long segments could be
fully predictable, thus useless.)
No relation is required between $n,m,i,k$, but useful are small $m,i,k$
and huge $n$ (within $o(2^k/i)$). Fold $X$ into an $n{\times}m$ matrix.
We also need a small $m\times i$ matrix $Z$, really
independent of $X$ and uniformly random (or random Toeplitz, i.e. s.t.
$Z_{a+1,b+1}=Z_{a,b}$). The same $Z$ can be reused repeatedly with many
$X$. Then the $n\times i$ product $XZ$ has uniform with accuracy
$O(\sqrt{n i/2^k})$ distribution.
The above definition of randomness is very robust, if not practical. True
random generators are rarely used in computing. The problem is not
in physics of random sources: we just saw efficient ways to perfect the
distributions of messy random noise. The reason lies in many extra
benefits provided by pseudorandom generators. E.g., designing. debugging,
or using programs often requires to repeat the exact same sequence. With
a physical random generator, one would need to record all its bits: slow
and costly.
[41], [42]
justified an alternative: generating pseudo-random strings from a
short seed. First, take any one-way permutation $f_n(x)$ (see
sec. 6.3) with a hard-core bit
$b_t(x)$ (in $\{0,1\}$ or $\{\pm1\}$): easy to compute from $(x,t)$, but
infeasible to guess from $(f_n(x),n,t)$ with any noticeable correlation
(see below). Then take random $k$-bit seeds $x_0,t,n$, and run:
$x_{i+1}\gets f_n(x_i)$, $S_i\gets b_t(x_i)=b_t((f_n)^i(x_0))$.
We will see how distinguishing outputs $S$ of this generator from strings
of coin flips would imply the ability to invert $f$ : infeasible if $f$
is one-way. But if P=NP (a famous open problem), no one-way $f$, and no
pseudorandom generators could exist.
By Kolmogorov standards, pseudo-random strings $S{=}G(s)$, are not
random: Take $\|S\|{\gg}\|s\|$. Then $K(S)\le O(1){+}\|s\|\ll\|S\|$, so
fail Kolmogorov's definition.
We can distinguish between truly random and pseudo-random strings by
simply trying all short seeds. This, however, takes exponential time
$2^{\|s\|}$. Realistically, pseudo-random strings will be as good as
truly random ones if they can't be distinguished in any feasible time.
Such generators we call perfect.
Theorem: [42] Let $G(s)=S\in\{0,1\}^n$
run in time $T_G$. Let an algorithm $A_w$ in expected (over internal coin
flips $w$) time $T_A$ accept $G(s)$ and truly random strings with
probabilities different by $d$. Then, for random $i$, one can use $A$ to
guess $S_i$ from $S_{i+}\,\edf\,S_{i+1},{\ldots},S_n$ in time $T_A{+}T_G$
with correlation $d/n$.
Proof. Let $p_{i,\sigma}$ (or $p_i$ for $\sigma{=}\{\}$) be the
probabilities of $A$ accepting $r\sigma S_{i+}$, for randomly chosen $r$
and $\|r\sigma\|=i$. So $A$ accepts $G(s)$ and random $r$ with respective
probabilities $p_0$, $p_n{=}p_0\pm d$. So, $p_{i-1}{-}p_i{=}d/n$, for
randomly chosen $i$. Then $(p_{i,1}{+}p_{i,0})/2$ averages to $p_i$,
while $p_{i,S_i}{=}p_{i,0}{+}(p_{i,1}{-}p_{i,0})S_i$ averages to
$p_{i-1}$ and $(p_{i,1}{-}p_{i,0}) (S_i{-}1/2)$ to $p_{i-1}{-}p_i=d/n$.
So, (samplable)
$p_{i,1}{-}p_{i,0}$ has the stated correlation with $S_i$ Q.E.D.
If the above generator $S_i{=}b_t(f_n(x_i))$ was not perfect, one could
guess $S_i$ from $S_{i+}$ with correlation $d/n$. But, $S_{i+}$ can be
computed from $(f_n(x_i),n,t)$. So, this employs $A$ to guess $b_t(x_i)$
from $(f_n(x_i),n.t,w)$ with correlation $d/n$ contrary to $b$ being
hard-core.
Hard Core. The key to pseudorandom generation is a one-way $f$
(e.g., Rabin's, sec. 6.3crypt) with a hard core.
And $(x\cdot t)=\sum_ix_ip_i\bmod2$ is hard-core for any one-way
$F(x,n,t){=}(f_n(x),n,t)$. Indeed, [43] converts
any method $a$ of guessing $(x\cdot t)$ from $(f_n(x),n,t)$ with correlation
$\e$ into one for inverting $f$ (slower $\e^2$ times than $a$).
[3] has more details and references.
Proof. (Simplified with some ideas of Charles Rackoff.) Let
$k{=}\|x\|{=}\|f_n(x)\|$, $j{=}\log(2k/\e^2)$, $v_i{=}0^{i-1}10^{k-i}$,
$b_t(x){=}(-1)^{(x\cdot t)}$. Throughout the proof $y{=}(f_n(x),n,w)$ is
fixed. Assume $a_t(y){\in}\{\pm1\}$ guesses $b_t(x)$ with correlation
$\sum_t2^{-\|t\|}b_t(x)a_t(y){>}\e$. $b_t(x)a_t$ averaged over
${>}2k/\e^2$ random pairwise independent $t$ is $\e$-close to its mean
(over all $t$) (and so ${>}0$) with probability $>1{-}1/2k$. Same for
$b_{t+v_i}(x)a_{t+v_i}{=}(-1)^{(x\cdot t)} a_{t+v_i} (-1)^{(x\cdot
v_i)}$.
Take a random $k{\times}j$ binary matrix $P$. The vectors $Pr$,
$0{\ne}r{\in}\{0,1\}^j$, are pairwise independent. So, for a fraction
$\ge1-1/2k$ of $P$, sign $(\sum_r(-1)^{xPr}a_{Pr+v_i})=(-1)^{(x\cdot
v_i)}$. We could thus find all bits $(x\cdot v_i)$ of $x$ with
probability ${>}.5$ if we knew $z{=}xP$. But $z$ is short: we can try all
its $2^j$ possible values and check $f_n(x)$ for each !
So the inverter computes $A_i(r){=}a_{Pr+v_i}$, for a random $P$
and all $i,r$. It uses Fast Fourier on $A_i$ to compute
$h_i(z)=\sum_rb_r(z)A_i(r)$. The sign of $h_i(z)$ is
the $i$-th bit of $z$-th member of output list. Q.E.D.
A very useful type of one-way permutations, called "trap-doors", has many
applications, especially in Cryptography. Here is a popular example:
Rabin's One-way Function. Pick random prime numbers $p,q$,
$\|p\|{=}\|q\|$ with two last bits ${=}1$, i.e. with odd $(p{-}1)(q{-}1)/4$.
Then $n=pq$ is called a Blum number. Its length should make factoring
infeasible. All residues below are $\bmod\,n$, i.e. in $Z_n$. Let $Q_n$
be the group of quadratic residues, i.e. squares in $\Z^*_n$.
Lemma. Let $n=pq$ be a Blum number, $F: x\mapsto x^2\pmod n$.
Then (1) $F$ is a permutation on $Q_n$ and (2) The ability to invert $F$
on random $x$ is equivalent to that of factoring $n$.
Proof. (1) $v\edf(p{-}1)(q{-}1)/4$ is odd, so $u{=}(v{+}1)/2$
is an integer. Let $x{=}F(z)$. Both $p{-}1$ and $q{-}1$ divide $2v$.
So, by Fermat's little theorem, both $p$, $q$ (and, thus $n$) divide
$x^v{-}1\equiv z^{2v}{-}1$. Then $F(x)^u\equiv x^{2u}=xx^v\equiv x$.
(2) The above $y^u$ inverts $F$. Conversely, let $F(A(y))=y$ for a
fraction $\e$ of $y\in Q_n$. Each $y{\in}Q_n$ has $x,x'{\ne}\pm x$ with
$F(x){=}F(x'){=}y$, both with equal chance to be chosen at random.
If $F(x)$ generates $y$ while $A(y){=}x'$ the Square Root Test
(Sec.5.1) has both $x,x'$ for factoring $n$ Q.E.D.
Picking random primes is easy: their density is $1/O(\|p\|)$.
Indeed, all primes in $[n,2n]$ divide $\binom{2n}n$ but none in
$[\frac23n,n]$ and no prime power $p^i{>}2n$ do. So,
$\log_n\binom{2n}n=2n/\log n-O(1)$ is an upper bound on the number of
primes in $[n,2n]$ and a lower bound on that in $[1,2n]$
(and in $[3n,6n]$ (i.e. in $[1,6n]\setminus[1,2n]$.
And fast VLSI exist to multiply long numbers and check primality.
Public Key Encryption. A perfect way to encrypt a message $m$
is to add it $\bmod\,2$ bit by bit to a random string $S$ of same length
$k$. The resulting encryption $m \oplus S$ has the same uniform
probability distribution, no matter what $m$ is. So it is useless to the
adversary for learning something about $m$, if $S$ is hidden. A
disadvantage is that the parties must secretly share $S$
as large as all messages to be exchanged, combined.
Public Key Cryptosystems use two keys. One, needed to encrypt the
messages, may be completely disclosed to the public. The secret
decryption key need not be sent to the encrypting party.
The same keys may be used repeatedly for many messages.
Such cryptosystem can be obtained [45] by
replacing the above random $S$ by pseudorandom $S_i= (x_i\cdot t)$;
$x_{i+1} =(x_i^2\ \bmod n)$. Here a Blum number $n=pq$ is chosen by the
Decryptor and is public, but $p,q$ are kept secret. The Encryptor chooses
$t\in\Z_2^{\|n\|},x_0\in\Z_n$ at random and sends $t,x_k, m{\oplus} S$.
Assuming factoring is intractable for the adversary, $S$ should be
indistinguishable from random strings (even with known $t,x_k$). Then
this scheme is as secure as if $S$ were random. The Decryptor knows $p,q$
and can compute $u,v$ (see above) and $w{=}(u^{k-1}\bmod v)$. So, he can
find $x_1{=}(x_k^w\bmod n)$, then $S$ and $m$.
Another use of the intractability of factoring is digital signatures
[44], [46] . Strings
$x$ can be released as authorizations of $y=(x^2\bmod n)$. Verifying $x$,
is easy but the ability of forging it for generic $y$ is equivalent to
that of factoring $n$.
You saw, most of our burning questions are still open. Take them on!
Start with reading recent results (FOCS/STOC is a good source). See
where you can improve them. Start writing, first notes just for your
friends, then real papers. Here is a little writing advice:
A well written paper has clear components: skeleton, muscles, etc. The
skeleton is an acyclic digraph of basic definitions and statements, with
cross-references. The meat consists of proofs (muscles) each
separately verifiable by competent graduate students having to read
no other parts but statements and definitions cited. Intuitive comments,
examples and other comfort items are fat and skin: a lack or excess will
not make the paper pretty. Proper scholarly references constitute
clothing, no paper should ever appear in public without! Trains of
thought which led to the discovery are blood and guts: keep them hidden.
Metaphors for other vital parts, like open problems, I skip out of
modesty.
Writing Contributions. Section
1 was originally prepared by
Elena Temin, Yong Gao, and Imre Kifor (BU), others by Berkeley students:
2.3 by Mark Sullivan,
3.1 by Eric Herrmann and Elena Eliashberg,
3.2 by Wayne Fenton and Peter Van Roy,
3.3 by
Carl Ludewig, Sean Flynn, and Francois Dumas,
4.1 by
Jeff Makaiwi Brian Jones and Carl Ludewig,
4.2 by David Leech and Peter Van Roy,
4.3 by Johnny and Siu-Ling Chan,
5.2 by Deborah Kordon,
5.3 by Carl Ludewig,
5.4 by Sean Flynn, Francois Dumas,
Eric Herrmann, 5.5 by Brian Jones.
An NP-Complete Problem: Tiling
$\begin{array}{cc}\begin{array}{|lr|}\hline a& x\\m& r\\\hline
\end{array}&\begin{array}{|lr|}\hline x& c\\r& z\\\hline\end{array}\\
\begin{array}{|lr|}\hline m& r\\n& s\\\hline\end{array}&
\begin{array}{|lr|}\hline r& z\\s& z\\\hline\end{array}\end{array}$.
Probability in Computing
A Monte-Carlo Primality Tester
Randomized Algorithms and Random Inputs
Arithmetization:
One-Player Games with Randomized Transition
Randomness
Randomness and Complexity
Rarity (Deficiency of Randomness).
Pseudo-randomness
Cryptography
Go On!
References
References for section 1:
Section 2:
Section 3:
Section 4:
Section 5:
Section 6: