The specification generally relates to computing: computing hardware, random number generation in computation, random instructions, machine-implemented methods, machine-implemented systems, and self-modification of programs and hardware.
Consider two fundamental questions in computer science, which substantially influence the design of current digital computers and play a fundamental role in hardware and machine-implemented software of the prior art:
1. What can a computing machine compute?
2. How many computational steps does a computational machine require to solve an instance of the 3-SAT problem? The 3-SAT problem is the basis for the famous P NP problem [10].
In the prior art, the two questions are typically conceived and implemented with hardware and software that compute according to the Turing machine (TM) [24] (i.e., standard digital computer [17]) model, which is the standard model of computation [5, 6, 10, 15, 23] in the prior art.
In this invention(s), our embodiments advance far beyond the prior art, by applying new machine-implemented methods and new hardware to advance computation and in particular, machine learning and the computation of program correctness for standard digital computer programs. The machine embodiments bifurcate the first question into two questions. What is computation? What can computation compute? Our new computing machine adds two special types of instructions to the standard digital computer [17] instructions 110, as shown in
One type of special machine instruction is meta instruction 120
The other special instruction is a random instruction 140 in
Some of the ex-machine programs provided here compute beyond the Turing barrier (i.e., beyond the computing capabilities of the digital computer). Computing beyond this barrier has advantages over the prior art, particularly for embodiments of machine learning applications, cryptographic computation, and for verifying program correctness in standard digital computer programs. Furthermore, these embodiments provide machine programs for computing languages that a register machine, standard digital computer or Turing machine is not able to compute. In the disclosure of these inventions, a countable set of ex-machines are explicitly specified, using standard instructions, meta instructions and random instructions. Using the mathematics of probability, measure theory and Cantor's heirarchy of infinities, we prove that every one of these ex-machines can evolve to compute a Turing incomputable language with probability measure 1, whenever random measurements 130 (trials) behave like unbiased Bernoulli trials. (A Turing machine [24] or digital computer [17] cannot compute a Turing incomputable language.) For this reason, when we discuss to computational methods in the prior art, we will refer to a Turing machine program or digital computer program as an algorithm. Since the mechanical rules of an algorithm are fixed throughout the computation and the complexity of an algorithm stays constant during all executions of an algorithm, we will never refer to an ex-machine computation as an algorithm. Overall, the mathematics presented herein shows that the ex-machine inventions advance far beyond the prior art.
In the following figures, although they may depict various examples of the invention, the invention is not limited to the examples depicted in the figures.
f is described further in section 9, titled The ϕ Correspondence.
g is described further in section 9, titled The ϕ Correspondence.
Sigmoid functions are used in machine learning embodiments (
Our invention describes a quantum random, self-modifiable computer that adds two special types of instructions to standard digital computer instructions [5, 10, 14, 15, 17, 24]. Before the quantum random and meta instructions are defined, we present some preliminary notation, and specification for standard instructions.
denotes the integers. and + are the non-negative and positive integers, respectively. The finite set Q={0, 1, 2, . . . , n −1} ⊂ represents the machine states 114 in
Let ={a1, . . . , an}, where each ai represents a distinct memory value 118 stored in memory 116, represented as T. The set A={0, 1, #} ∪ consists of alphabet symbols (memory values), where # is the blank symbol and {0, 1, #} ∩ is the empty set. In some ex-machines, A={0, 1, #, Y, N, a}, where a1=Y, a2=N, a3=a. In some ex-machines, A={0, 1, #}, where is the empty set. The memory values are read from and written to memory (T). The ex-machine memory 116 is represented by function T:→A with an additional condition: before ex-machine execution starts, there exists an N>0 so that T(k)=# when |k|>N. In other words, this mathematical condition means all memory addresses contain blank or non-existent memory values, except for a finite number of memory addresses. When this condition holds for memory 116, we say that memory T is finitely bounded.
The standard ex-machine instructions S satisfy S ⊂ Q×A×Q×A×{−1, 0, 1} and a uniqueness condition: If (q1, α1, r1, a1, y1) ∈ S and (q2, α2, r2, a2, y2) ∈ S and (q1, α1, r1, a1, y1)≠(q2, α2, r2, a2, y2), then q1≠q2 or α1≠α2. A standard instruction I=(q, a, r, α, y) is similar to a Turing machine tuple [7, 33]. When the ex-machine is in state q and the memory head is scanning alphabet symbol a=T(k) at memory address k, instruction I is executed as follows:
The ex-machine state moves from state q to state r.
The ex-machine replaces alphabet symbol a with alphabet symbol α so that T(k)=α. The rest of the memory remains unchanged.
If y=−1, the ex-machine moves its memory head, pointing one memory cell to the left (lower) in memory and is subsequently scanning the memory value T(k −1) at memory address k−1.
If y=+1, the ex-machine moves its memory head, point one memory cell to the right (higher) in memory and is subsequently scanning the memory value T(k+1) at memory address k+1.
If y=0, the ex-machine does not moves its memory head and is subsequently scanning the memory value T(k)=α at memory address k.
In other embodiments, standard instructions 110 in
In some embodiments, random instruction 140 may measure a random bit, called random_bit and then non-deterministically execute according to following code:
In other embodiments, the standard instructions 110 may have a programming language syntax such as assembly language, C++, Fortran, JAVA, JAVA virtual machine instructions, Go, Haskell, RISC machine instructions, Ruby, LISP and execute on hardware 204, shown in
A digital computer program [17] or a Turing machine [24] has a fixed set of machine states Q, a finite alphabet A, a finitely bounded memory, and a finite set of standard ex-machine instructions that are executed according to specification 5.1. In other words, an ex-machine that uses only standard instructions is computationally equivalent to a digital computer [17]. An ex-machine with only standard instructions is called a standard machine or digital computer. A standard machine has no unpredictability because it contains no random instructions. A standard machine does not modify its instructions as it is computing.
Repeated independent trials are called random Bernoulli trials [9] if there are only two possible outcomes for each trial (i.e., random measurement 130 in
Consider the bit sequence (x1x2 . . . ) in the infinite product space . A single outcome xi of a bit sequence (x1x2 . . . ) generated by randomness is unbiased. The probability of measuring a 0 or a 1 are equal:
P(xi=1)=P(xi=0)=½.
History has no effect on the next random measurement. Each outcome xi is independent of the history. No correlation exists between previous or future outcomes. This is expressed in terms of the conditional probabilities: P(xi=1|x1=b1, . . . , xi-1=bi-1)=½ and P(xi=0|x1=b1, . . . , xi-1=bi-1)=½for each bi ∈ {0, 1}.
In some embodiments randomness generator 160 used by random instructions 140, 150 has measurement properties 1 and 2. In some embodiments, randomness generator 542 in
The random instructions are subsets of Q×A×Q×{−1, 0, 1}={(q, a, r, y): q, r are in Q and a in A and y in {−1, 0, 1} } that satisfy a uniqueness condition defined below.
In some embodiments, the random instructions satisfy ⊂ Q×A×Q×{−1, 0, 1} and the following uniqueness condition: If (q1, α1,r1, y1) ∈ and (q2, α2, r2, y2) ∈ and (q1, α1, r1, y1)≠(q2, α2, r2, y2), then q1≠q2 or α1≠α2. When the machine head is reading memory value a from memory T and the machine is in machine state q, the random instruction (q, a, r, y) executes as follows:
1. The machine reads random measurements 130 that returns a random bit b ∈ {0, 1}.
2. In memory, memory value a is replaced with random bit b.
(This is why A contains both symbols 0 and 1.)
3. The machine state changes to machine state r.
4. The machine moves its memory head left if y=−1, right if y=+1, or the memory head does not move if y=0.
In some embodiments, the source of non-determinism in random measurements 130 is a quantum event 170. In some embodiments, quantum event 170 is generated outside of the hardware of the ex-machine: for example, the arrival of a photon from the environment outside of the ex-machine, as shown in quantum event 547. These photons may arrive from our sun, or an external light build or by an LED 610, LED 620, or LED 630, mounted on a circuit board, outside the processor that executes the ex-machine. In an embodiment, LED 610 or LED 620 or more than one LED is mounted on a circuit board; phototransistor 544 or may be photodiode 554 may be mounted so that the photosensitive part of the semiconductor is facing LED 610 or LED 620. In another embodiment, quantum event 170 is generated by semiconductor LED 630, inside of the hardware that implements processor system 252 of the ex-machine; in an embodiment, semiconductor LED 630 is integrated into a semiconductor chip that implements processor system 252.
In some embodiments, quantum measurements satisfy random measurement property 1 and random measurement property 2. In some embodiments, random measurements 130 is implemented with randomness generator 542 in
In some embodiments, there are more than two outcomes of taking the physical measurement by random measurements 130. For example, 8 different outcomes can be generated by repeating 3 measurements with spin-1 source 560 (e.g., electrons), followed by a Sz splitter 570 and then a Sx splitter 580 as shown in
In an embodiment, after execution of a random instruction, at step 4, the machine may random access to a different part of memory. For example, before a random instruction is being executed, it may be reading memory address 0x4B6368895EDC0543418E8790738293B9 and after at step 4, it moves to reading memory address 0x29F2B761285674FB41DE074063529462.
Machine instructions 1 lists a random walk machine that has only standard instructions and random instructions. Alphabet A={0, 1, #, E}. The states are Q={0, 1, 2, 3, 4, 5, 6, h}, where the halting state h=7. A valid initial memory contains only blank symbols; that is, # ##. The valid initial state is 0.
There are three random instructions: (0, #, 0, 0), (1, #, 1, 0) and (4, #, 4, 0). The random instruction (0, #, 0, 0) is executed first. If the random source measures a 1, the machine jumps to state 4 and the memory head moves to the right of memory address 0. If the random source measures a 0, the machine jumps to state 1 and the memory head moves to the left of memory address 0. Instructions containing alphabet value E provide error checking for an invalid initial memory or initial state; in this case, the machine halts with an error.
;; Comments follow two semicolons.
;; Continue random walk to the left of memory address 0
;; Go back to state 0. Numbers of random 0's=number of random 1's.
;; Go back to state 1. Numbers of random 0's>number of random 1's.
;; Continue random walk to the right of memory address 0
;; Go back to state 0. Numbers of random 0's=number of random 1's.
;; Go back to state 4. Numbers of random 1's>number of random 0's.
Below are 31 computational steps of the ex-machine's first execution. This random walk machine never halts when the initial memory is blank and the initial state is 0. The first random instruction executed is (0, #, 0, 0), as a part of random instructions 140 in
Below are the first 31 steps of the ex-machine's second execution. As a part of random instructions 140 in
The first and second executions of the random walk ex-machine verify our statement in the introduction: in contrast with a standard machine or a digital computer, the execution behavior of the same ex-machine may be distinct at two different instances, even though each instance of the ex-machine starts its execution with the same input, stored in memory, the same initial states and same initial instructions. Hence, the ex-machine is a discrete, non-autonomous dynamical system, which enhances its computational capabilities.
Meta instructions are a second type of special instructions, as shown in 120 of
Define =S ∪ ∪, as the set of standard 110, random 140, and meta instructions 120 in
1. If (q1, α1, r1, β1, y1) and (q2, α2, r2, β2, y2) are both in S, then q1≠q2 or α1≠α2.
2. If (q1, α1, r1, β1, y1) ∈ S and (q2, α2, r2, y) ∈ , then q1≠q2 or α1≠α2.
3. If (q1, α1, r1, y1) and (q2, α2, r2, y2) are both in , then q1≠q2 or α1≠α2.
4. If (q1, α1, r1, y1) ∈ and (q2, α2, r2, a2, y2, J2) ∈ , then q1≠q2 or α1≠α2.
5. If (q1, α1, r1, β1, y1) ∈ S and (q2, α2, r2, a2, y2, J2) ∈ , then q1≠q2 or α1≠α2.
6. If (q1, α1, r1, a1, y1, J1) and (q2, α2, r2, a2, y2, J2) are both in , then q1≠q2 or α1≠α2.
Before a valid machine execution starts, a hardware machine, shown as a system in
Specification 5.3 is an embodiment of self-modification system 250 in
A meta instruction (q, a, r, α, y, J) in is executed as follows.
The first five coordinates (q, a, r, α, y) are executed as a standard instruction according to specification 5.1 with one caveat. State q may be expressed as |Q|−c1 and state r may be expressed as |Q| or |Q|−c2, where 0<c1, c2≤|Q|. When (q, a, r, α, y) is executed, if q is expressed as |Q|−c1, the value of q is instantiated to the current value of |Q| minus c1. Similarly, if r is expressed as |Q| or |Q|−c2, the value of state r is instantiated to the current value of |Q| or |Q| minus c2, respectively.
Subsequently, instruction J modifies , where instruction J has one of the three forms: J=(q, a, r, α, y) or J=(q, a, r, y) or J=(q, a, r, α, y, J1). In the third form, J is also a meta instruction.
For each of the three forms, if ∪ {J} still satisfies the unique state, scanning symbol condition, then is updated to ∪ {J}.
Otherwise, there is an instruction I in whose first two coordinates q, a, are equal to instruction J's first two coordinates. In this case, instruction J replaces instruction I in . That is, is updated to ∪ {J}−{I}.
In the third form, in meta instructions 120 of
Consider the meta instruction (q, a1, |Q|−1, α1, y1, J), where J=|Q|−1, a2, |Q|, α2, y2). After the standard instruction (q, a1, |Q|−1, α1, y1) is executed, this meta instruction adds one new state |Q| to the machine states Q and also adds the instruction J, instantiated with the current value of |Q|.
In other embodiments, meta instructions 120 in
Below is an example of how meta instructions 120 in
The aforementioned LISP code produces the following output when executed:
Before self-modifying lambda_fn=(lambda (y x) (+y x))
args: (y x)
number of arguments: 2
arg1: y
Before self-modification, lambda body: ((+y x))
After self-modification, lambda function double=(lambda (x) (+x x)) (double 5)=10
In the final computational step, lambda function (lambda (x) (+x x)) is returned.
Let be an ex-machine. The instantiation of |Q|−1 and |Q| in a meta instruction I (shown in
A simple meta instruction has one of the forms (q, a, |Q|−c2, α, y), (q, a, |Q|, α, y) (|Q|−c1, a, r, α, y) (|Q|−c1, a, |Q|−c2, α, y), (|Q|−c1, a, |Q|, α, y), where 0<c1, c2≤|Q|. The expressions |Q|−c1, |Q|−c2 and |Q| are instantiated to a state based on the current value of |Q| when the instruction is executed. In the embodiments in this section, ex-machines only self-reflect with the symbols |Q|−1 and |Q|. In other embodiments, an ex-machine may self-reflect with |Q|+c, where c is positive integer.
Let A={0, 1, #} and Q={0}. ex-machine has 3 simple meta instructions.
With an initial blank memory and starting state of 0, the first four computational steps are shown below. In the first step, 's memory head is scanning a # and the ex-machine state is 0. Since |Q|=1, simple meta instruction (|Q|−1, #, |Q|−1, 1, 0) instantiates to (0, #, 0, 1, 0), and executes.
In the second step, the memory head is scanning a 1 and the state is 0. Since |Q|=1, instruction (|Q|−1, 1, |Q|, 0, 1) instantiates to (0, 1, 1, 0, 1), executes and updates Q={0, 1}. In the third step, the memory head is scanning a # and the state is 1. Since |Q|=2, instruction (|Q|−1, #, |Q|−1, 1, 0) instantiates to (1, #, 1, 1, 0) and executes. In the fourth step, the memory head is scanning a 1 and the state is 1. Since |Q|=2, instruction (|Q|−1, 1, |Q|, 0, 1) instantiates to (1, 1, 2, 0, 1), executes and updates Q={0, 1, 2}. During these four steps, two simple meta instructions create four new instructions and add new states 1 and 2.
A machine is said to have finite initial conditions if the following conditions are satisfied by the computing hardware shown in hardware embodiments of
1. The number of machine states |Q| is finite.
2. The number of alphabet symbols |A| (i.e., memory values) is finite.
3. The number of machine instructions || is finite.
4. The memory is finitely bounded.
It may be useful to think about the initial conditions of an ex-machine as analogous to the boundary value conditions of a differential equation. While trivial to verify, the purpose of remark 5.1 is to assure computations by an ex-machine execute on different types of hardware such as semiconductor chips, lasers using photons, biological implementations using proteins, DNA and RNA, quantum computers that use entanglement and quantum superposition.
If the machine starts its execution with finite initial conditions, then after the machine has executed l instructions for any positive integer l, the current number of states Q(l) is finite and the current set of instructions (l) is finite. Also, the memory T is still finitely bounded and the number of measurements obtained from the random or non-deterministic source is finite.
Proof. The remark follows immediately from specification 5.5 of finite initial conditions and machine instruction specifications 5.1, 5.2, and 5.3. In particular, the execution of one meta instruction adds at most one new instruction and one new state to Q.
Specification 5.6 describes new ex-machines that can evolve from computations of prior ex-machines that have halted. The notion of evolving is useful because the random instructions 140 and meta instructions 120 can self-modify an ex-machine's instructions as it executes. In contrast with the ex-machine, after a digital computer program stops executing, its instructions have not changed.
This difference motivates the next specification, which is illustrated by the following. Consider an initial ex-machine 0 that has 9 initial states and 15 initial instructions. 0 starts executing on a finitely bounded memory T0 and halts. When the ex-machine halts, it (now called 1) has 14 states and 24 instructions and the current memory is S1. We say that ex-machine 0 with memory T0 evolves to ex-machine 1 with memory S1.
Let T0, T1, T2 . . . Ti-1 each be a finitely bounded memory. Consider ex-machine 0 with finite initial conditions in the ex-machine hardware. 0 starts executing with memory T0 and evolves to ex-machine 1 with memory S1. Subsequently, 1 starts executing with memory T1 and evolves to 2 with memory S2. This means that when ex-machine 1 starts executing on memory T1, its instructions are preserved after the halt with memory S1. The ex-machine evolution continues until i-1 starts executing with memory Ti-1 and evolves to ex-machine i with memory Si. One says that ex-machine 0 with finitely bounded memories T0, T1, T2 . . . Ti-1 evolves to ex-machine i after i halts.
When an ex-machine 0 evolves to 1 and subsequently 1 evolves to 2 and so on up to ex-machine n, then ex-machine i is called an ancestor of ex-machine j whenever 0≤i<j≤n. Similarly, ex-machine j is called a descendant of ex-machine i whenever 0≤i<j≤n. The sequence of ex-machines 0→1→ . . . →n . . . is called an evolutionary path.
In some embodiments, this sequence of ex-machines may be stored on distinct computing hardware, as shown in machines 214, 216, 218 and 220 of
In some embodiments, the computing machines described in this invention use non-determinism as a computational tool to make the computation unpredictable. In some embodiments, quantum random measurements are used as a computational tool. Quantum randomness is a type of non-determinism, based on measuring quantum events 170 by randomness generator 160, as shown in
Some embodiments of physically non-deterministic processes are as follows. In some embodiments that utilize non-determinism, photons can (i.e., quantum event 170, quantum event 547, quantum event 557) strike a semitransparent mirror and then take two or more paths in space. In one embodiment, if the photon is reflected by the semitransparent mirror, then it takes on one bit value b with a binary outcome: if the photon passes through by the semitransparent mirror (i.e., quantum event 170), then the non-deterministic process produces another bit value 1b.
In another embodiment, the spin 560 of an electron may be measured to generate the next non-deterministic bit. In another embodiment, a protein, composed of amino acids, spanning a cell membrane or artificial membrane, that has two or more conformations (quantum event 170 in
In sections 7 and 12, the execution of the standard instructions, random instructions and meta instructions uses the property that for any m, all 2m binary strings are equally likely to occur when a quantum random number generator takes m binary measurements. One, however, has to be careful not to misinterpret quantum random properties 1 and 2. In an embodiments, where the probability of one binary outcome is p≠½, then an unbiased source of randomness (i.e., each binary string is equally likely) can be produced with a simple standard machine that operates on more than one measurement of the binary outcome to produce a single, unbiased outcome. The number of measurements, taken by random measurments 130, needed to generated an unbiased binary outcome, depends upon how far p is from frac12.
Consider the Champernowne sequence 01 00 01 10 11 000 001 010 011 100 101 110 111 0000 , . . . , which is sometimes cited as a sequence that is Borel normal, yet still Turing computable. In [9], Feller discusses the mathematics of random walks. The Champernowne sequence catastrophically fails the expected number of changes of sign for a random walk as n→∞. Since all 2m strings are equally likely, the expected value of changes of sign follows from the reflection principle and simple counting arguments, as shown in section 111.5 of [9].
Furthermore, most of the 2m binary strings (i.e., binary strings of length m) have high Kolmogorov complexity. This fact leads to the following mathematical intuition that enables new computational behaviors that a standard digital computer cannot perform. The execution of quantum random instructions working together with meta instructions enables the ex-machine to increase its program complexity [28] as it evolves. In some cases, the increase in program complexity can increase the ex-machine's computational power as the ex-machine evolves. Also, notice the distinction here between the program complexity of the ex-machine and Kolmogorov complexity. The definition of Kolmogorov complexity only applies to standard machines. Moreover, the program complexity (e.g., the Shannon complexity |Q∥A|) stays fixed for standard machines. In contrast, an ex-machine's program complexity can increase without bound, when the ex-machine executes quantum random and meta instructions that productively work together. (For example, see ex-machine 2, called (x).)
In terms of the ex-machine computation performed, how one of these binary strings is generated from some particular type of non-deterministic process is not the critical issue. Suppose the quantum random generator demonstrated measuring the spin of particles, certified by the strong Kochen-Specker theorem [4, 13] outputs the 100-bit string a0a1 . . . a99=10110000101011110011001100111000100011100101010110111100 00000010011001000011010101101111001101010000 to ex-machine 1.
Suppose a distinct quantum random generator using radioactive decay outputs the same 100-bit string a0a1 . . . a99 to a distinct ex-machine z. Suppose 1 and 2 have identical programs with the same initial tapes and same initial state. Even though radioactive decay was discovered over 100 years ago and its physical basis is still phenomenological, the execution behavior of 1 and 2 are indistinguishable for the first 100 executions of their quantum random instructions. In other words, ex-machines 1 and 2 exhibit execution behaviors that are independent of the quantum process that generates these two identical binary strings.
Before some of the deeper theory on quantum randomness is reviewed, we take a step back to view randomness from a broader theoretical perspective. While we generally agree with the philosophy of Eagle [7] that randomness is unpredictablity, embodiment 3 helps sharpen the differences between indeterminism and unpredictability.
Our gedankenexperiment demonstrates a deterministic system which exhibits an extreme amount of unpredictability. This embodiment shows that a physical system whose mathematics is deterministic can still be an extremely unpredictable process if the measurements of the process has limited resolution. Some mathematical work is needed to define the dynamical system and summarize its mathematical properties before we can present the gedankenexperiment.
Consider the quadratic map f: →, where f(x)=9/2x(1−x). Set I0=[0, ⅓] and I1=[⅓, ⅔]. Set B=(⅓, ⅓). Define the set Λ={x ∈ [0, 1]: fn (x) ∈ I0 ∪ I1 for all n ∈ }. 0 is a fixed point of f and f2(⅓)=f2(⅔)=0, so the boundary points of B lie in Λ. Furthermore, whenever x ∈ B, then
This means all orbits that exit Λ head off to −∞.
The inverse image f−1(B) is two open intervals B0 ⊂ I0 and B1 ⊂ I1 such that f(B0)=f(B1)=B. Topologically, B0 behaves like Cantor's open middle third of I0 and B1 behaves like Cantor's open middle third of I1. Repeating the inverse image indefinitely, define the set
Using dynamical systems notation, set Σ2=. Define the shift map σ: Σ2→Σ2, where σ(a0a1 . . . )=(a1a2 . . . ). For each x in Λ, x's trajectory in I0 and I1 corresponds to a unique point in Σ2: define h: Λ→Σ2 as h(x)=(a0a1 . . . ) such that for each n ∈ , set an=0 if fn(x) ∈ I0 and an=1 if fn(x) ∈ I1.
For any two points (a0a1 . . . ) and (b0b1 . . . ) in Σ2, define the metric
Via the standard topology on inducing the subspace topology on Λ, it is straightforward to verify that h is a homeomorphism from Λ to Σ2. Moreover, h o f=σ o h, so h is a topological conjugacy. The set H and the topological conjugacy h enable us to verify that Λ is a Cantor set. This means that Λ is uncountable, totally disconnected, compact and every point of Λ is a limit point of Λ.
We are ready to pose our mathematical gedankenexperiment. We make the following assumption about our mathematical observer. When our observer takes a physical measurement of a point x in Λ2, she measures a 0 if x lies in I0 and measures a 1 if x lies in I1. We assume that she cannot make her observation any more accurate based on our idealization that is analogous to the following: measurements at the quantum level have limited resolution due to the wavelike properties of matter [3] Similarly, at the second observation, our observer measures a 0 if f (x) lies in I0 and 1 if f (x) lies in Il. Our observer continues to make these observations until she has measured whether F−1 (x) is in I0or in I1. Before making her k+1st observation, can our observer make an effective prediction whether fk(x) lies in I0 or I1 that is correct for more than 50% of her predictions?
The answer is no when h(x) is a generic point (i.e., in the sense of Lebesgue measure) in Σ2. Set to the Martin-Löf random points in Σ2. Then has Lebesgue measure 1 (Feller [9]) in Σ2, so its complement Σ2− has Lebesgue measure 0. For any x such that h(x) lies in , then our observer cannot predict the orbit of x with a Turing machine. Hence, via the topological conjugacy h, we see that for a generic point x in Λ, x's orbit between I0 and I1 is Martin-Löf random—even though f is mathematically deterministic and f is a Turing computable function.
Overall, the dynamical system (f, Λ) is mathematically deterministic and each real number x in Λ has a definite value. However, due to the lack of resolution in the observer's measurements, the orbit of generic point x is unpredictable—in the sense of Martin-Löf random.
The standard theory of quantum randomness stems from the seminal EPR paper [8]. Einstein, Podolsky and Rosen (EPR) propose a necessary condition for a complete theory of quantum mechanics: Every element of physical reality must have a counterpart in the physical theory. Furthermore, they state that elements of physical reality must be found by the results of experiments and measurements. While mentioning that there might be other ways of recognizing a physical reality, EPR propose the following as a reasonable criterion for a complete theory of quantum mechanics:
If, without in any way disturbing a system, one can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of physical reality corresponding to this physical quantity.
They consider a quantum-mechanical description of a particle, having one degree of freedom. After some analysis, they conclude that a definite value of the coordinate, for a particle in the state given by
is not, predictable, but may be obtained only by a direct measurement. However, such a measurement disturbs the particle and changes its state. They remind us that in quantum mechanics, when the momentum of the particle is known, its coordinate has no physical reality. This phenomenon has a more general mathematical condition that if the operators corresponding to two physical quantities, say A and B, do not commute, then a precise knowledge of one of them precludes a precise knowledge of the other. Hence, EPR reach the following conclusion:
(I) The quantum-mechanical description of physical reality given by the wave function is not complete. OR
(II) When the operators corresponding to two physical quantities (e.g., position and momentum) do not commute (i.e. AB≠BA), the two quantities cannot have the same simultaneous reality.
EPR justifies this conclusion by reasoning that if both physical quantities had a simultaneous reality and consequently definite values, then these definite values would be part of the complete description. Moreover, if the wave function provides a complete description of physical reality, then the wave function would contain these definite values and the definite values would be predictable.
From their conclusion of I OR II, EPR assumes the negation of I—that the wave function does give a complete description of physical reality. They analyze two systems that interact over a finite interval of time. And show by a thought experiment of measuring each system via wave packet reduction that it is possible to assign two different wave functions to the same physical reality. Upon further analysis of two wave functions that are eigenfunctions of two non-commuting operators, they arrive at the conclusion that two physical quantities, with non-commuting operators can have simultaneous reality. From this contradiction or paradox (depending on one's perspective), they conclude that the quantum-mechanical description of reality is not complete.
In [2], Neil Bohr responds to the EPR paper. Via an analysis involving single slit experiments and double slit (two or more) experiments, Bohr explains how during position measurements that momentum is transferred between the object being observed and the measurement apparatus. Similarly, Bohr explains that during momentum measurements the object is displaced. Bohr also makes a similar observation about time and energy: “it is excluded in principle to control the energy which goes into the clocks without interfering essentially with their use as time indicators”. Because at the quantum level it is impossible to control the interaction between the object being observed and the measurement apparatus, Bohr argues for a “final renunciation of the classical ideal of causality” and a “radical revision of physical reality”.
From his experimental analysis, Bohr concludes that the meaning of EPR's expression without in any way disturbing the system creates an ambiguity in their argument. Bohr states: “There is essentially the question of an influence on the very conditions which define the possible types of predictions regarding the future behavior of the system. Since these conditions constitute an inherent element of the description of any phenomenon to which the term physical reality can be properly attached, we see that the argumentation of the mentioned authors does not justify their conclusion that quantum-mechanical description is essentially incomplete.” Overall, the EPR versus Bohr-Born-Heisenberg position set the stage for understanding whether hidden variables can exist in the theory of quantum mechanics.
Embodiments of quantum random instructions utilize the lack of hidden variables because this contributes to the unpredictability of quantum random measurements. In some embodiments of the quantum random measurements used by the quantum random instructions, the lack of hidden variables are only associated with the quantum system being measured and in other embodiments the lack of hidden variables are associated with the measurement apparatus. In some embodiments, the value indefiniteness of measurement observables implies unpredictability. In some embodiments, the unpredictability of quantum measurements 130 in FIG. depend upon Kochen-Specker type theorems [4, 13] In some embodiments, the unpredictability of random measurements 130 in
In an embodiment of quantum random measurements 130 shown in
The Sx=±1 outcomes can be assigned 0 and 1, respectively. Moreover, since
neither of the Sx=±1 outcomes can have a pre-determined definite value. As a consequence, bits 0 and 1 are generated independently (stochastic independence) with a 50/50 probability (unbiased). These are quantum random properties 1 and 2.
A class of ex-machines are defined as evolutions of the fundamental ex-machine (x), whose 15 initial instructions are listed in ex-machine specification 2. These ex-machines compute languages L that are subsets of {a}*={an: n ∈ }. The expression an represents a string of n consecutive a's. For example, a5=aaaaa and a0 is the empty string. Define the set of languages
Machine Specification 7.1 defines a unique language in for each function f: N→{0,1}.
Consider any function f:→{0,1}. This means f is a member of the set Function f induces the language Lf={an: f(n)=1}. In other words, for each non-negative integer n, string an is in the language Lf if and only if f(n,)=1.
Trivially, Lf is a language in . Moreover, these functions f generate all of .
In order to define the halting syntax for the language in that an ex-machine computes, choose alphabet set A={#, 0, 1, N, Y, a}.
Let be an ex-machine. The language L in that computes is defined as follows. A valid initial tape has the form # #an#. The valid initial tape # ## represents the empty string. After machine starts executing with initial tape # #an#, string an is in 's language if ex-machine halts with tape #an# Y#. String an is not in 's language if halts with tape #an# N#. The use of special alphabet symbols (i.e., special memory values) Y and N—to decide whether an is in the language or not in the language—follows [14].
For a particular string # #am#, some ex-machine could first halt with #am# N# and in a second computation with input # #am# could halt with #am# Y#. This oscillation of halting outputs could continue indefinitely and in some cases the oscillation can be aperiodic. In this case, 's language would not be well-defined according to machine specification 7.2. These types of ex-machines will not be specified in this invention.
There is a subtle difference between (x) and an ex-machine whose halting output never stabilizes. In contrast to the Turing machine or a digital computer program, two different instances of the ex-machine (x) can evolve to two different machines and compute distinct languages according to machine specification 7.2. However, after (x) has evolved to a new machine (a0a1 . . . am x) as a result of a prior execution with input tape # #am#, then for each i with 0≤i≤m, machine (a0a1 . . . am x) always halts with the same output when presented with input tape # #ai#. In other words, (a0a1 . . . am x)'s halting output stabilizes on all input strings ai where 0≤i≤m. Furthermore, it is the ability of (x) to exploit the non-autonomous behavior of its two quantum random instructions that enables an evolution of (x) to compute languages that are Turing incomputable (i.e., not computable by a standard digital computer).
We designed ex-machines that compute subsets of {a}* rather than subsets of {0, 1}* because the resulting specification of (x) is much simpler and more elegant. It is straightforward to list a standard machine that bijectively translates each an to a binary string in {0, 1}* as follows. The empty string in {a}* maps to the empty string in {0,1}*. Let 0 represent this translation map. Hence,
and so on. Similarly, an inverse translation standard machine computes the inverse of ψ. Hence
aaa, wild so on. The translation and inverse translation computations immediately transfer any results about the ex-machine computation of subsets of {a}* to corresponding subsets of {0, 1}* via ψ. In particular, the following remark is relevant for our discussion.
Every subset of {a}* is computable by some ex-machine if and only if every subset of {0,1}* is computable by some ex-machine.
Proof. The remark immediately follows from the fact that the translation map ψ and the inverse translation map ψ−1 are computable with a standard machine.
When the quantum randomness in two quantum random instructions satisfy property 1 (unbiased bernoulli trials) and random measurement property 2 (stochastic independence), for each n − , all 2n finite paths of length n—in the infinite, binary tree of
Moreover, there is a one-to-one correspondence between a function f: →{0, 1} and an infinite downward path in the infinite binary tree of
A={#, 0, 1, N, Y, a}. States Q={0, h, n, y, t, v, w, x, 8} where halting state h=1, and states n=2, y=3, t=4, v=5, w=6, x=7. The initial state is always 0. The letters are used to represent machine states instead of explicit numbers because these states have special purposes. (This is for the reader's benefit.) State n indicates NO that the string is not in the machine's language. State y indicates YES that the string is in the machine's language. State x is used to generate a new random bit; this random bit determines the string corresponding to the current value of |Q|−1. The fifteen instructions of (x) are shown below.
With initial state 0 and initial memory # #aaaa##, an execution of machine (x) is shown below.
During this execution, (x) replaces instruction (8, #, x, #, 0) with (8, #, y, #, 1). Meta instruction (w, a, |Q|, a, 1, (|Q|−1, a, |Q|, a, 1)) executes and replaces (8, a, x, a, 0) with new instruction (8, a, 9, a, 1). Also, simple meta instruction (|Q|−1, a, x, a, 0) temporarily added instructions (9, a, x, a, 0), (10, a, x, a, 0), and (11, a, x, a, 0).
Subsequently, these new instructions were replaced by (9, a, 10, a, 1), (10, a, 11, a, 1), and (11, a, 12, a, 1), respectively. Similarly, simple meta instruction (|Q|−1, #, x, #, 0) added instruction (12, #, x, #, 0) and this instruction was replaced by instruction (12, #, n, #, 1). Lastly, instructions (9, #, y, #, 1), (10, #,
n, #, 1), (11, #, y, #, 1), and (12, a, 13, a, 1) were added.
Furthermore, five new states 9, 10, 11, 12 and 13 are added to Q. After this computation halts, the machine states are Q={0, h, n, y, t, v, w, x, 8, 9, 10, 11, 12, 13} and the resulting ex-machine evolved to has 24 instructions. It is called (11010 x).
New instructions (8, #, y, #, 1) , (9, #, y, #, 1), and (11, #, y, #, 1) help (11010 x) compute that the empty string, a and aaa are in its language, respectively. Similarly, the new instructions (10, #, n, #, 1) and (12, #, n, #, 1) help (11010 x) compute that aa and aaaa are not in its language, respectively.
The zeroeth, first, and third 1 in (11010 x)'s name indicate that the empty string, a and aaa are in (11010 x)'s language. The second and fourth 0 indicate strings aa and aaaa are not in its language. The symbol x indicates that all strings an with n≥5 have not yet been determined whether they are in (11010 x)'s language or not in its language.
Starting at state 0, ex-machine (11010 x) computes that the empty string is in ω(11010 x)'s language.
Starting at state 0, ex-machine (11010 x) computes that string a is in (11010 x)'s language.
Starting at state 0 (11010 x) computes that string aa is not in (11010 x)'s language.
Starting at state 0, (11010 x) computes that aaa is in (11010 x)'s language.
Starting at state 0, (11010 x) computes that aaaa is not in (11010 x)'s language.
Note that for each of these executions, no new states were added and no instructions were added or replaced. Thus, for all subsequent executions, ex-machine (11010 x) computes that the empty string, a and aaa are in its language. Similarly, strings aa and aaaa are not in (11010 x)'s language for all subsequent executions of (11010 x).
Starting at state 0, we examine an execution of ex-machine (11010 x) on input memory # #aaaaaaa##.
Overall, during this execution ex-machine (11010 x) evolved to ex-machine (11010 011 x). Three quantum random instructions were executed. The first quantum random instruction (x, a, t, 0) measured a 0 so it is shown above as (x, a, t, 0_qr, 0). The result of this 0 bit measurement adds the instruction (13, #, n, #, 1), so that in all subsequent executions of ex-machine (11010 011 x), string a5 is not in (11010 011 x)'s language. Similarly, the second quantum random instruction (x, a, t, 0) measured a 1 so it is shown above as (x, a, t, 1_qr, 0). The result of this 1 bit measurement adds the instruction (14, #, y, #, 1), so that in all subsequent executions, string a6 is in (11010 011 x)'s language. Finally, the third quantum random instruction (x, #, x, 0) measured a 1 so it is shown above as (x, #, x, 1_qr, 0). The result of this 1 bit measurement adds the instruction (15, #, y, #, 1), so that in all subsequent executions, string a7 is in (11010 011 x)'s language.
Lastly, starting at state 0, we examine a distinct execution of ex-machine (11010 x) on input memory # #aaaaaaa##. A distinct execution of (11010 x) evolves to ex-machine (11010 000 x).
Based on our previous examination of ex-machine (x) evolving to (11010 x) and then subsequently (11010 x) evolving to (11010 011 x), ex-machine 3 specifies (a0a1 . . . am x) in terms of initial machine states and initial machine instructions.
Let m ∈ Set Q={0, h, n, y, t, v, w, x, 8, 9, 10, . . . m+8, m+9 }. For 0≤i≤m, each ai is 0 or 1. ex-machine (a0a1 . . . am x)'s instructions are shown below. Symbol b832 y if a0=1. Otherwise, symbol b8=n if a0=0. Similarly, symbol b9=y if a1=1. Otherwise, symbol b9=n if a1=0. And so on until reaching the second to the last instruction (m+8, #, bm+8, #, 1), symbol bm+8=y if am=1. Otherwise, symbol bm+8=n if am=0.
Whenever i satisfies 0≤i≤m, string ai is in (a0a1 . . . am x)'s language if ai=1; string ai is not in (a0a1 . . . am x)'s language if ai=0. Whenever n>m, it has not yet been determined whether string an is in (a0a1 . . . am x)'s language or not in its language.
Proof. When 0≤i≤m, the first consequence follows immediately from the definition of ai being in (a0a1 . . . am x)'s language and from ex-machine 3. In instruction (i+8, #, bi+8, #, 1) the state value of bi+8 is y if ai=1 and bi+8 is n if ai=0.
For the indeterminacy of strings an when n>m, ex-machine (a0 . . . am x) executes its last instruction (m+8, a, m+9, a, 1) when it is scanning the mth a in an. Subsequently, for each a on the memory to the right (higher memory addresses) of #am, ex-machine (a0 . . . am x) executes the quantum random instruction (x, a, t, 0).
If the execution of (x, a, t, 0) measures a 0, the two meta instructions (t, 0, w, a, 0, (|Q|−1, #, n, #, 1)) and (w, a, |Q|, a, 1, (|Q|−1, a, |Q|, a, 1)) are executed. If the next alphabet symbol to the right is an a, then a new standard instruction is executed that is instantiated from the simple meta instruction (|Q|−1, a, x, a, 0) . If the memory address was pointing to the last a in an, then a new standard instruction is executed that is instantiated from the simple meta instruction (|Q|−1, #, x, 0) .
If the execution of (x, a, t, 0) measures a 1, the two meta instructions (t, 1, w, a, 0, (|Q|−1, #, y, #, 1)) and (w, a, |Q|, a, 1, (|Q|−1, a, |Q|, a, 1)) are executed. If the next alphabet symbol to the right is an a, then a new standard instruction is executed that is instantiated from the simple meta instruction (|Q|−1, a, x, a, 0). If the memory address was pointing to the last a in an, then a new standard instruction is executed that is instantiated from the simple meta instruction (|Q|−1, #, x, #, 0).
In this way, for each a on the memory to the right (higher memory addresses) of #am, the execution of the quantum random instruction (x, a, t, 0) determines whether each string am+k, satisfying 1≤k≤n−m, is in or not in (a0a1 . . . an x)'s language.
After the execution of (|Q|−1, #, x, #, 0) , the memory address is pointing to a blank symbol, so the quantum random instruction (x, #, x, 0) is executed. If a 0 is measured by the quantum random source, the meta instructions (x, 0, v, #, 0, (|Q|−1, #, n, #, 1)) and (v, #, n, #, (|Q|−1, a, |Q|, a, 1)) are executed. Then the last instruction executed is (a, #, h, N, 0) which indicates that an is not in (a0a1 . . . an x)'s language.
If the execution of (x, #, x, 0) measures a 1, the meta instructions (x, 1, w, #, 0, (|Q|−1, #, 1)) and (w, #, y, #, 1, (|Q|−1, a, |Q|, a, 1)) are executed. Then the last instruction executed is (y, #, h, Y, 0) which indicates that an is in (a0a1 . . . an x)'s language.
During the execution of the instructions, for each a on the memory to the right (higher memory addresses) of #am, ex-machine (a0a1 . . . am x) evolves to (a0a1 . . . an x) according to the specification in ex-machine 3, where one substitutes n for m.
When the binary string a0a1 . . . am is presented as input, the ex-machine instructions for (a0a1 . . . am x), specified in ex-machine 3, are constructible (i.e., can be printed) with a standard machine.
In contrast with lemma 7.1, (a0a1 . . . am x)'s instructions are not executable with a standard machine when the input memory # #ai# satisfies i>m because meta and quantum random instructions are required. Thus, remark 7.3 distinguishes the construction of (a0a1 . . . am x)'s instructions from the execution of (a0a1 . . . am x)'s instructions.
Proof. When given a finite list (a0 a1 . . . am), where each ai is 0 or 1, the code listing below constructs (a0a1 . . . am x)'s instructions. Starting with comment ; ; Qx_builder.lsp, the code listing is expressed in a dialect of LISP, called newLISP. (See www.newlisp.org). LISP was designed, based on the lambda calculus, developed by Alonzo Church. The appendix of [33] outlines a proof that the lambda calculus is computationally equivalent to digital computer instructions (i.e., standard machine instructions). The 3 instructions below print the ex-machine instructions for (11010 x), listed in machine instructions 1.
(set ′a0_a1_dots_am (list 1 1 0 1 0))
(set ′Qx_machine (build_Qx_machine a0_a1_dots_am))
(print_xmachine Qx_machine)
Define as the union of (x) and all ex-machines (a0 . . . am x) for each m ∈ and for each a0 . . . am in {0,1}m+1. In other words,
Each language Lf in can be computed by the evolving sequence of ex-machines (x), (f(0) x), (f(0)f(1) x) , . . . , (f(0)f(1) . . . f(n) x), . . . .
Proof. The theorem follows from ex-machine 2, ex-machine 3 and lemma 7.1. 0
Given function f: →{0, 1}, for any arbitrarily large n, the evolving sequence of ex-machines (f(0)f(1) . . . f(n) x), f(0)f(1) . . . f(n)f(n+1) x), . . . computes language Lf.
Moreover, for each n, all ex-machines (x), (f(0)x), (f(0)f(1) x), . . . , f(0)f(1) . . . f(n) x) combined have used only a finite amount of memory, finite number of states, finite number of instructions, finite number of executions of instructions and only a finite amount of quantum random information measured by the quantum random instructions.
Proof. For each n, the finite use of computational resources follows immediately from remark 2, specification 7 and the specification of ex-machine 3.
A set X is called countable if there exists a bijection between X and . Since the set of all Turing machines is countable and each Turing machine only recognizes a single language most (in the sense of Cantor's heirarchy of infinities) languages Lf are not computable with a Turing machine. More precisely, the cardinality of the set of languages Lf computable with a Turing machine is , while the cardinality of the set of all languages is .
For each non-negative integer n, define the language tree (a0a1 . . . an)={Lf: f ∈ and f(i)=ai for i satisfying 0≤i≤n}. Define the corresponding subset of as S(a0a1 . . . an)={f ∈ : f(i)=ai for i satisfying 0≤i≤n}. Let Ψ denote this 1-to-1 correspondence, where
Since random measurement property 1 and random measurement property 2 are both satisfied, each finite path f(0)f(1) . . . f(n) is equally likely and there are 2n+1 of these paths. Thus, each path of length n+1 has probability 2−(n+1). These uniform probabilities on finite strings of the same length can be extended to the Lebesgue measure μ on probability space . Hence, each subset S(a0a1 . . . an) has measure 2−(n+1). That is, μ(S(a0a1 . . . an))=2−(n+1) and μ()=1. Via the Ψ correspondence between each language tree (a0a1 . . . an) and subset S(a0a1 . . . an), uniform probability measure μ induces a uniform probability measure on , where v (a0a1 . . . an))=2−(n+1) and v()=1.
For functions f: {0, 1}, the probability that language Lf is Turing incomputable has measure 1 in (v,).
Proof. The Turing machines are countable and therefore the number of functions f: →{0, 1} that are Turing computable is countable. Hence, via the Ψ correspondence, the Turing computable languages Lf have measure 0 in .
Moreover, the Martin-Löf random sequences f: →{0,1} have Lebesgue measure 1 in and are a proper subset of the Turing incomputable sequences.
(x) is not a Turing machine. Each ex-machine (a0a1 . . . am x) in is not a Turing machine.
Proof (x) can evolve to compute Turing incomputable languages on a set of probability measure 1 with respect to (v, ). Also, (a0a1 . . . am x) can evolve to compute Turing incomputable languages on a set of measure 2−(m+1) with respect to (v, ). In contrast, each Turing machine only recognizes a single language, which has measure 0. In fact, the measure of all Turing computable languages is 0 in .
These corollaries are remarkable because the language that an ex-machine can compute reflects it computing capabilities. This means practical applications that were not possible with digital computer programs are feasible with an ex-machine.
In [24], Alan Turing posed the halting problem for Turing machines. Does there exist a Turing machine that can determine for any given Turing machine and finite initial tape T whether 's execution on tape T eventually halts? In the same paper [24], Turing proved that no single Turing machine could solve his halting problem.
Next, we explain what Turing's seminal result relies upon in terms of abstract computational resources. Turing's result means that there does not exist a single Turing machine —regardless of the size of 's finite state set Q and finite alphabet set A—so that when this special machine is presented with any Turing machine with a finite tape T and initial state q0, then can execute a finite number of computational steps, halt and correctly determine whether halts or does not halt with a tape T and initial state q0. In terms of definability, the statement of Turing's halting problem ranges over all possible Turing machines and all possible finite tapes. This means: for each tape T and machine , there are finite initial conditions imposed on tape T and machine . However, as tape T and machine range over all possibilities, the computational resources required for tape T and machine are unbounded. Thus, the computational resources required by are unbounded as its input ranges over all finite tapes T and machines .
The previous paragraph provides some observations about Turing's halting problem because any philosophical objection to (x)'s unbounded computational resources during an evolution should also present a similar philosophical objection to Turing's assumptions in his statement and proof of his halting problem. Notice that corollary 7.4 supports our claim.
Since (x) and every other x-machine (a0a1 . . . am x) in is not a Turing machine, there is a natural extension of Turing's halting problem. Does there exist an x-machine (x) such that for any given Turing machine ¿ and finite initial tape T, then (x) can sometimes compute whether ¿'s execution on tape T will eventually halt? Before we call this the x-machine halting problem, the phrase can sometimes compute whether must be defined so that this problem is well-posed. A reasonable definition requires some work.
From the universal Turing machine/enumeration theorem [21], there exists a Turing computable enumeration ε: →{all Turing machines }×{Each of 's states as an initial state} of every Turing machine. Similar to x-machines, for each machine , the set {Each of 's states as an initial state} can be realized as a finite subset {0, . . . , n−1} of . Since E(n) is an ordered pair, the phrase “Turing machine E(n)” refers to the first coordinate of E(n). Similarly, the “initial state E(n)” refers to the second coordinate of E(n).
Recall that the Turing machine halting problem is equivalent to the blank-tape halting problem. (See pages 150-151 of [15]). For our discussion, the blank-tape halting problem translates to: for each Turing machine E(n), does Turing machine E(n) halt when E(n) begins its execution with a blank initial tape and initial state E(n)?
Lemma 7.1 implies that the same initial x-machine can evolve to two different x-machines; furthermore, these two x-machines will never compute the same language no matter what descendants they evolve to. For example, (0 x) and (1 x) can never compute the same language in . Hence, sometimes means that for each n, there exists an evolution of (x) to (a0x), and then to (a0a1x) and so on up to (a0a1 . . . an x) , where for each i with 0≤i≤n, then (a0a1 . . . an x) correctly computes whether Turing machine E(n)−executing on an initial blank tape with initial state E(n)−halts or does not halt.
In the prior sentence, the word computes means that (a0a1 . . . ai x) halts after a finite number of instructions have been executed and the halting output written by (a0a1 . . . ai x) on its tape indicates whether machine E(n) halts or does not halt. For example, if the input tape is # #ai#, then enumeration machine E writes the representation of E(i) on the tape, and then (a0a1 . . . am x) with m≥i halts with # Y# written to the right of the representation for machine E(i). Alternatively (a0a1 . . . am x) with m≥i halts with # N# written to the right of the representation for machine E(i). The word correctly means that x-machine (a0ai . . . am x) halts with # Y# written on the tape if machine E(i) halts and x-machine (a0a1 . . . x) halts with # N# written on the tape if machine E(i) does not halt.
Next, our goal is to transform the x-machine halting problem to a form so that the results from the previous section can be applied. Choose the alphabet as ={#, 0, 1, a, A, B, M, N, S, X, Y}. As before, for each Turing machine, it is helpful to identify the set of machine states Q as a finite subset of . Let E be the Turing machine that computes a Turing computable enumeration as Ea: →{}* ×, where the tape # #an# represents natural number n. Each Ea(n) is an ordered pair where the first coordinate is the Turing machine and the second coordinate is an initial state chosen from one of Ea(n)'s states. (Chapter 7 of [15] provides explicit details of encoding quintuples with a particular universal Turing machine. Alphabet was selected so that it is compatible with this encoding. A careful study of chapter 7 provides a clear path of how E's instructions can be specified to implement Ea.)
For each n ∈ , with blank initial tape and initial state Ea(n), then Turing machine Ea(n) either halts or does not halt.
Proof. The execution behavior of Turing machine computation is unambiguous. For each n, there are only two possibilities.
For our particular instance of Ea, define the halting function hE
The x-machine (x) has an evolutionary path that computes halting language
This evolutionary path is (hE
Proof. Theorem 8.1 follows from the previous discussion, including the definition of halting function hE
and theorem 7.2.
Theorem 8.1 provides an affirmative answer to the x-machine halting problem, but in practice, a particular execution of (x) will not, from a probabilistic perspective, evolve to compute Lh(Ea). For example, the probability is 2−128 that a particular execution of (x) will evolve to (a0a1 . . . a127 x) so that (a0a1 . . . a99 x) correctly computes whether each string λ, a, a2 . . . a127 is a member of Lh(E
Furthermore, theorem 8.1 provides no general method for infallibly testing (proving) that an evolution of (x) to some new machine (a0a1 . . . am x) satisfies ai=hE
The existence of the path shows that it is logically possible for this evolution to happen, since (x) can in principle evolve to compute any language Lf in . In other words, every infinite, downward path in the infinite binary tree of
In a sense, theorem 8.1 has an analogous result in pure mathematics. The Brouwer fixed point theorem guarantees that a continuous map from an n-simplex to an n-simplex has at least one fixed point and demonstrates the power of algebraic topology. However, the early proofs were indirect and provided no contructive methods for finding the fixed point(s). The parallel here is that theorem 8.1 guarantees that an evolutionary path exists, but the proof provides no general method for infallibly testing that for an evolution up to stage m, then (a0al . . . am x) satisfies ai=he
In the previous section, entitled “The Ex-Machine Halting Problem”, the ex-machine is not limited by the halting problem due to its meta instructions and random instructions that use self-modification and randomness. This capability of the ex-machine has some extremely practical applications. In the prior art, program correctness of standard program is unsolvable by a standard machine or digital computer, as a consequence of Turing's halting problem. Program correctness is extremely important not only due to the problem of malware [28] infecting mission critical software. However, program correctness is also critical to any hardware/software system —regardless of whether it is infected with malware—that operates air traffic control, the electric grid, the Internet, GPS [16], an autonomous driving network, and so on.
This section shows how to convert any standard program to a geometric problem; then novel self-modification procedures and/or traditional machine learning methods, implemented with the ex-machine's self-modification capability, can be used to solve the geometric problem. For example, the detection of an infinite loop in a standard program is a special case of the halting problem, and the detection of infinite loops is a subset of program correctness.
With this in mind, the ϕ correspondence can translate any standard digital computer program—no matter how big —to a geometric problem in the complex plane . Consider standard machine M. A transformation ϕ is defined such that each standard instruction in M is mapped to an affine function in the complex plane . Let machine states Q={q1, . . . , qm}. Let alphabet A={a1, . . . , an}, where a1 is the blank symbol #. Halt state h is not in Q. Function η: Q×A→Q ∪ {h}×A×{−1,+1} represents the standard machine instructions, where n(q, a)=(r, b, x) corresponds to standard instruction (q, a, r, b, x). Set B=|A|+|Q|+1. Define symbol value function v: A ∪ {h} ∪ Q→ as v(a1)=0, v(a2)=1, . . . , v(ai)=i−1, v(h)=|A|, v(q1)=|A|+1, . . . , v(qi)=|A|+i.
Tk is the memory value in the kth memory address. The machine configuration (q, k, T) lies in Q××AZ. Machine configuration (q, k,T) is mapped to the complex number
The real and imaginary parts of ϕ(q , k, T) are rational because the finitely bounded memory condition (i.e., finite memory) for standard machines implies that only a finite number of memory cells contain non-blank values.
If two machine configurations (q, l, T) and (r, k, S) are not equivalent, then ϕ(q, l, T) ≠ ϕ(r, k, S). That is, ϕ is one-to-one on the equivalence classes of machine configurations.
Proof. By translating our memory representation T via R(j)=T(j+l−k), W.L.O.G. we may assume that l=k. Thus, we are comparing (q, k,T) and (r, k, S). Suppose
Then the real parts must be equal. This implies for all j≥−1 that Tk+j+1=Sk+j+1. Hence, Tj=Sj for all j≥k.
For the imaginary parts, |Bv(q)−Bv(r)|≥B whenever q≠r. Also,
Since B>|A|, this implies q=r and Tk−j−1=Sk−j−1 for all j≥0. This means Tj=Sj for all j≤k−1. Thus, (q, k, T)=(r, k, S).
This next part defines a one-to-one mapping ϕ from a standard digital computer program η to a finite set of affine functions, whose domain is a bounded subset of . From
Before executing instruction (q, Tk, r, b, +1), the machine is in configuration 710 as shown in
From equation 1,
Equation 3 defines the real part of the affine map for instruction (q, Tk, r, b, +1).
f
1(x)=|A|x+(|A|−1)v(Tk+1)−|A|2v(Tk) (3)
From equation 2,
Equation 4 defines the imaginary part of right affine map corresponding to instruction 720, where (q, Tk, r, b, +1) is shown in
We compute the formula for ϕ mapping instruction (q, Tk, r, b, −1) to left affine function g(x+yi)=g1(x)+g2(y) i, where
First,
Before (q, Tk, r, b, −1) executes, equation 1 holds. Also,
For the imaginary part,
Before instruction (q, Tk, r, b, −1) executes, equation 2 holds. Multiplying by |A| yields
Hence, n=Bv(r)−|A|Bv(q)−|A|v(Tk-1).
g
2(y)=|A|y+Bv(r)−|A|Bv(q)−|A|v(Tk-1) (6)
Define halting attractor h={x+yi∈: 0≤x≤|A|2} and B|A|≤y≤(B+1)|A|}. The set of all points in that correspond to halting configurations (h, k, T) are called halting points. Remarks 9.2 and 9.3 establish that the halting points are a subset of h. Define the halting map : h→h such that h(x+yi)=x+yi on h. This means every point in the halting attractor is a fixed point.
Every halting point xh+yhi in h satisfies B|A|≤y≤(B+1)|A|.
PROOF. Let xh+yhi be a halting point. Then
Thus, yh≥B|A| since a tape of all blanks yields 0 for the infinite sum. If A={#}, then the halting behavior is trivial, so we assume |A|≥2 for this next part. If v(Tk−j−1)=|A|−1 for all j≥0, then the sum of this geometric series is
Every halting point xh+yhi in h satisfies 0≤xh<|A|2.
PROOF. |x|≥0 because every term is ≥0. Analyzing equation 1
Each affine function's domain is a subset of a unique unit square {x+yi ∈: m≤x≤m+1 and n≤y≤n+1}, where m and n are integers. Hence, ϕ transforms Turing's halting problem to the following dynamical systems problem, which is a type of geometric problem. If machine configuration (q, k, T) halts after n computational steps, then the orbit of ϕ (q, k, T) exits one of the unit squares on the nth iteration and enters halting attractor h. In other words, we say that the execution of computer program η with initial configuration (q, k, T) enters the halting attractor h after a finite number of execution steps, where it is understood that computer program η has been mapped to a finite set of affine maps, and initial configuration (q, k, T) has been mapped to an initial point ϕ (q, k, T) in .
If machine configuration (r, j, S) is immortal (i.e., never halts), then the orbit of ϕ(r, j, S) remains in these finite number of unit squares forever and never enters the halting attractor h. In other words, the execution of computer program η with initial configuration (q, k, T) never enters halting attractor h.
For a particular standard machine, set X equal to the union of all the closed unit squares induced by ϕ along with halting attractor h. Closed means the unit squares contain their boundary. Because there are a finite number of unit squares and h is a compact set, there is a set D that is compact (i.e., closed and bounded) and path connected such that D contains the closed unit squares and halting attactor h. Define F: D→ as a function extension of the finite number of right and left affine functions f and g, specified by equations 3, 4, 5, 6 and also an extension of halting function .
One of the main inventions, provided in this section, is that the correctness of a computer program with an initial configuration can be computed based on whether the program's execution enters the program's halting attractor h in the complex plane . In other words, a new computational approach, implemented in hardware and software, to a deep, theoretical mathematical problem has many practical software and hardware applications and inventions that currently are believed to be unsolvable by the prior art.
In an embodiment, ex-machine receives as input standard instructions 110 (
Ex-machine executes machine instructions 100 with processor system 252 and computes the left affine maps and right affine maps of M according to equations 3, 4, 5, and 6. Then ex-machine stores the left and right affine maps of M (i.e., the computational results of equations 3, 4, 5, and 6) in memory system 246.
In an embodiment, the ex-machine receives standard instructions 110 (
In the prior C listing, the following comprise the initial configuration of this C program: the initial value of constant NUM_TESTS is set to 3; the initial value of the array variable a [NUM_TESTS] is set to 59, 117, and 197 in memory system 246; the initial value of the variable b [NUM_TESTS] is set to 237, 88, 84 in memory system 246; the initial value of the variable correct_result [NUM_TESTS] is set to 243, 208, 17 in memory system 246; the initial value of the variable test_result is set to 0 in memory system 246.
In an embodiment, the ex-machine receives as input standard instructions 110 (
In another embodiment, the ex-machine receives as input standard instructions 110 (
The purpose of the following program is to prove that the Goldbach conjecture is true. The Goldbach conjecture states that every even number greater than or equal to 4 is the sum of two primes. For example, 4=2+2, 6=3+3, 8=3+5, 10=3+7, 12=5+7, 14=3+11 and so on. The following code, named Goldbach Machine, is a correct program if its execution never reaches the halting attractor; that is, if its execution stays in the while (g==true) loop forever and never reaches the halt instruction. The code in the Goldbach machine can be transformed to a finite set of affine maps in the complex plane as described previously.
We describe another embodiment of a program that can be checked for correctness. In hardware design, there are a finite number of timing constraints that determine if race conditions exist for a hardware circuit. Race conditions are usually considered to be a program error or hardware design error. In some cases, a computer program can be specified that checks each of these timing constraints; if one of the timing constraints is violated then the computer program can enter a trivial infinite loop such as
If none of the timing constraints are violated then the program enters a halting state, indicating that the computer program testing the timing constraints is correct. In this case, the entry of the computer program into the halting attractor indicates that the computer program is correct, and thus that there are no race conditions in the circuit design. In this case, in contrast to the Goldbach machine, the meaning of the computer program's execution entering the halting attractor is inverted from the program is incorrect to the program is correct.
Machine embodiment 4 and machine embodiment 5 teach us the following: for some computer programs, entering the halting attractor means that the program has an error; for other computer programs, entering the halting attractor means the program is correct. The meaning of a computer program entering the halting attractor depends upon the particular computer program and must be understood on a case by case basis.
In this section, we describe the inventions of using self-modification and randomness to advance machine learning procedures. In the prior art, machine learning procedures are implemented with algorithms that are computationally equivalent to a Turing machine algorithm [5, 14, 15, 24]. As was shown in sections 7 and 8, machine computation that adds randomness and self-modification to the standard digital computer instructions has more computing capability than a standard digital computer. This capability enables more advanced machine learning procedures where in some embodiments meta instructions 120 and random instructions 140 improve the machine learning procedure.
In the prior art, some machine learning algorithms use multi-layer methods of building functions, based on the use of weighted sums of sigmoidal functions (
is our prototypical sigmoidal function 810, as shown in
When the nonlinear functions are sigmoidal functions 810, we will call this a neural network, as shown in
In our embodiments, our almost step functions will generally be piecewise linear approximations of the form ga,b,c(x)=exp (−b(x−a)c), where b and c are parameters that adjust the width of where the step function maps that interval to 1 and a translates the center of the step function. In
such that
on a set of Lebesgue measure greater than 1−∈.
In some embodiments, our machine learning procedure, comprised of standard, random and meta instructions, starts with a gradient descent methods that can be used on networks of almost step and sigmoidal functions. The meta instructions can be used to modify our initial gradient descent procedure and also modify how the gradient descent is understanding the differential geometry of our network of nonlinear functions. In some embodiments, the self-modification of the initial gradient descent will depend upon differential forms, curvature tensors, and the curvature of saddle points so that standard instructions 940, meta instructions 930 and random instructions 920 can reason about the performance of machine learning procedure 950 that is evolving, based on its training data set. Meta instructions 930 and random instructions 920 help self-modify machine learning procedure, based on the reasoning that depends upon the geometric concepts of differential forms, curvature tensors, and the curvature of saddle points. How the machine learning procedure evolves will sometimes depend upon the idiosyncracies of a particular data set where the training is performed upon.
In some embodiments, we use almost step functions 820, shown in our
In some embodiments, two subscripts indicate the layer and reference the particular nonlinear function 850 at that location. In an embodiment, there are n different nonlinear functions that we use at each node of the network and they are named ={f1(x), f2(x), . . . fn(x) }. In
In
arctan (x) and
The use of multi-layer algorithmic machine learning is based on weighted sigmoidal functions 810, or some other non-linear function that looks like a hockey stick constructed from two linear pieces.
In a machine learning procedure 950 embodiment that uses standard instructions 940, random instructions 920 and meta instructions 930, as shown in
wi,j,l, as shown in
A brief, intuitive summary of a non-deterministic execution machine is provided first before the formal description in machine specification 12.1. Our non-deterministic execution machine should not be confused with the nondeterministic Turing machine that is described on pages 30 and 31 of [12].
Part of the motivation for our approach is that procedure 5 is easier to analyze and implement. Some embodiments of procedure 5 can be implemented using randomness generator 548 in
In a standard digital computer, the computer program can be specified as a function; from a current machine configuration (q, k, T), where q is the machine state, k is the memory address being scanned and T is the memory, there is exactly one instruction to execute next or the Turing machine halts on this machine configuration.
Instead of a function, in embodiments of our invention(s) of a non-deterministic execution machine, the program is a relation. From a particular machine configuration, there is ambiguity on the next instruction I in to execute. This is computational advantage when the adversary does not know the purpose of the program . Measuring randomness with random measurements 130 in
Part of our non-deterministic execution machine specification is that each possible instruction has a likelihood of being executed. Some quantum randomness is measured with 130 in
Before machine specifications are provided, some symbol definitions are reviewed. Symbol 30 is the positive integers. Recall that represents the rational numbers. [0, 1] is the closed interval of real numbers x such that 0≤x≤1. The expression x ∉ X means element x is NOT a member of X. For example,
The symbol ∩ means intersection. Note
A non-deterministic machine is defined as follows:
Q is a finite set of states.
is a set of final states and ⊂
When machine execution begins, the machine is in an initial state q0 and q0 ∈ Q.
is a finite set of alphabet symbols that are read from and written to memory.
# is a special alphabet symbol called the blank symbol that lies in .
The machine's memory T is represented as a function T: →Ð. The alphabet symbol stored in the kth memory cell is T(k).
is a next-instruction relation, which is a subset of Q××Q ××{−1, 0, +1}. acts as a non-deterministic program where the non-determinism arises from how an instruction is selected from before it is executed. Each instruction I that lies in the set specifies how machine executes a computational step. When machine is in state q and scanning alphabet symbol α=T(k) at memory cell (address) k, an instruction (such that the first two coordinates of I are q and α, respectively) is non-deterministically selected. If instruction I=(q, α, r, β, y) is selected, then instruction I is executed as follows:
Machine changes from machine state q to machine state r.
Machine replaces alphabet symbol α with alphabet symbol β so that T(k)=β. The rest of the memory remains unchanged.
If y=1, then machine starts scanning one memory cell downward in memory and is subsequently scanning the alphabet symbol T(k−1), stored in memory address k−1.
Ify=+1, then machine starts scanning one memory cell upward in memory and is subsequently scanning the alphabet symbol T(k+1), stored in memory address k+1.
If y=0, then machine continues to scan the same memory cell and subsequently is scanning alphabet symbol T(k)=β, stored in memory address k.
To assure is a physically realizable machine, before machine starts executing, there exist M>0 such that the memory contains only blank symbols (i.e., T(k)=#) in all memory addresses, satisfying |k|>M.
={p1p2, . . . , } is a finite set of rational numbers, contained in ∩ [0,1], such that = where is the number of instructions in .
v: → is a bijective function. Let I be an instruction in . The purpose of v(I) is to help derive the likelihood of non-deterministically selecting instruction I as the next instruction to execute.
Non-deterministic machine 12.1 can be used to execute two different instances of a procedure with a different sequence of machine instructions 100, as shown in
A machine configuration i is a triplet (q, k, Ti) where q is a machine state, k is an integer which is the current memory address being scanned and Ti represents the memory. Consider machine configuration i+1=(r, l, Ti+1). Per machine specification 12.1, then
is a valid compuuamolial step if instruction I=(q, α, r, ,βy) in ħ and memories Ti and Ti+1 satisfy the following four conditions:
1. v(I)>0.
2. Ti(k)=α.
3. Ti+1(k)=β and Ti+1(j)=Ti(j) for all j≠k.
4. l=k+y.
Symbol + is the positive integers. Recall that represents the rational numbers. [0, 1] is the closed interval of real numbers x such that 0≤x≤1. The expression x ∉ X means element x is NOT a member of X. For example,
The symbol ∩ means intersection. Note
The purpose of machine procedure 4 is to describe how a non-deterministic machine selects a machine instruction from the collection of machine instructions 6700 as shown in
In some embodiments, machine instructions (procedure) 5 can be implemented with standard instructions 110, meta instructions 120, and random instructions 140, as shown in
In some embodiments, machine instructions non-deterministically selected may be represented in a C programming language syntax. In some embodiments, some of the instructions non-deterministically selected to execute have the C syntax such as x=1; or z=x * y; . In some embodiments, one of the instructions selected may be a loop with a body of machine instructions such as:
or function such as
In another embodiment, the instructions non-deterministically selected can be performed at the hardware level, using the VHDL dataflow language.
In some embodiments, a random instruction may measure a random bit, called random_bit and then non-deterministically execute according to the following code:
In other embodiments, the machine instructions may have a programming language syntax such as JAVA, Go Haskell, C++, RISC machine instructions, JAVA virtual machine, Ruby, LISP, or a hardware language such as VHDL.
In machine instructions (procedure) 5, the ex-machine that implements the pseudocode executes a particular non-deterministic machine a finite number of times and if a final state is reached, the procedure increments the corresponding final state score. After the final state scores have been tallied such that they converge inside the reliability interval (0, ∈), then the ex-machine execution is completed and the ex-machine halts.
Machine specification 6 is useful for executing a computational procedure so that each instance of the procedure executes a different sequence of instructions and in a different order. This unpredictability of the execution of instructions makes it difficult for an adversary to comprehend the computation. Machine specification 6 can be executed with a finite sequence of standard, meta and random instructions as shown in
In the following analysis, the likelihood of procedure (machine instructions) 5 finding the maximum is estimated.
For each x, let vx=v(Ix). Consider the acceptable computational path
such that machine configuration N is in final state qf and p=vj1 * vj2 * . . . * vjN is a maximum. The expected number of times that inner loop in procedure 6 executes acceptable path
is 0 when n<N and at least rp when n≥N.
Proof. When n<N, it is impossible for the inner loop to execute the path
so the expected number of times is 0. When n=N, the independence of each random sample that selects instruction Iji and the fact that
is an acceptable path implies that the probability of executing this computational path is p=vj1 * vj2 * . . . *vjN. When n>N, the probability of executing this path is still p because the inner loop exits if the final state in N is reached.
This application claims priority benefit of U.S. Provisional Patent Application Ser. No. 62/682,979, entitled Quantum Random Self-Modifiable Computer, filed Jun. 10, 2018.
Number | Date | Country | |
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62682979 | Jun 2018 | US |