A counter machine is an abstract machine used in a formal logic and theoretical computer science to model computation. It is the most primitive of the four types of register machines. A counter machine comprises a set of one or more unbounded registers, each of which can hold a single non-negative integer, and a list of (usually sequential) arithmetic and control instructions for the machine to follow. The counter machine is typically used in the process of designing parallel algorithms in relation to the mutual exclusion principle. When used in this manner, the counter machine is used to model the discrete time-steps of a computational system in relation to memory accesses. By modeling computations in relation to the memory accesses for each respective computational step, parallel algorithms may be designed in such a matter to avoid interlocking, the simultaneous writing operation by two (or more) threads to the same memory address.
For a given counter machine model the instruction set is tiny—from just one to six or seven instructions. Most models contain a few arithmetic operations and at least one conditional operation (if condition is true, then jump). Three base models, each using three instructions, are drawn from the following collection. (The abbreviations are arbitrary.)
In addition, a machine usually has a HALT instruction, which stops the machine (normally after the result has been computed).
Using the instructions mentioned above, various authors have discussed certain counter machines:
The three counter machine base models have the same computational power since the instructions of one model can be derived from those of another. All are equivalent to the computational power of Turing machines. Due to their unary processing style, counter machines are typically exponentially slower than comparable Turing machines.
Main article: Counter-machine model |
The counter machine models go by a number of different names that may help to distinguish them by their peculiarities. In the following the instruction "JZDEC ( r )" is a compound instruction that tests to see if a register r is empty; if so then jump to instruction I_{z}, else if not then DECrement the contents of r:
A counter machine consists of:
This example shows how to create three more useful instructions: clear, unconditional jump, and copy.
Afterward r_{s} will contain its original count (unlike MOVE which empties the source register, i.e., clears it to zero).
The basic set (1) is used as defined here:
Instruction | Effect on register "j" | Effect on Instruction Counter Register ICR | Summary |
---|---|---|---|
INC ( j ) | [j] +1 → j | [IC] +1 → IC | Increment contents of register j; next instruction |
DEC ( j ) | [j] -1 → j | [IC] +1 → IC | Decrement contents of register j; next instruction |
JZ ( j, z) | IF [j] = 0 THEN I_{z} → IC ELSE [IC] +1 → IC | If contents of register j=0 then instruction z else next instruction | |
HALT |
Initially, register #2 contains "2". Registers #0, #1 and #3 are empty (contain "0"). Register #0 remains unchanged throughout calculations because it is used for the unconditional jump. Register #1 is a scratch pad. The program begins with instruction 1.
The program HALTs with the contents of register #2 at its original count and the contents of register #3 equal to the original contents of register #2, i.e.,
The program COPY ( #2, #3) has two parts. In the first part the program moves the contents of source register #2 to both scratch-pad #1 and to destination register #3; thus #1 and #3 will be copies of one another and of the original count in #2, but #2 is cleared in the process of decrementing it to zero. Unconditional jumps J (z) are done by tests of register #0, which always contains the number 0:
In the second the part the program moves (returns, restores) the contents of scratch-pad #1 back to #2, clearing the scratch-pad #1 in the process:
The program, highlighted in yellow, is shown written left-to-right in the upper right.
A "run" of the program is shown below. Time runs down the page. The instructions are in yellow, the registers in blue. The program is flipped 90 degrees, with the instruction numbers (addresses) along the top, the instruction mnemonics under the addresses, and the instruction parameters under the mnemonics (one per cell):
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | ← Instruction number (address) | |||||||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
JZ | DEC | INC | INC | JZ | JZ | DEC | INC | JZ | H | ← Instruction | |||||||||||||||
2 | 2 | 3 | 1 | 0 | 1 | 1 | 2 | 0 | ← Register number | ||||||||||||||||
6 | 1 | 10 | 6 | ← Jump-to instruction number | |||||||||||||||||||||
step | IC | Inst # | reg # | J-addr | reg0 | reg1 | reg2 | reg3 | reg4 | IC | |||||||||||||||
start | 0 | 0 | 2 | 0 | 0 | 1 | move [#2] to #1 and #3: | ||||||||||||||||||
1 | 1 | JZ | 2 | 6 | 0 | 0 | 2 | 0 | 0 | 1→2 | JZ | Jump fails: register #2 contains 2 | |||||||||||||
2 | 2 | DEC | 2 | 0 | 0 | 0 | 2→1 | 0 | 0 | 2→3 | DEC | Decrement register #2 from 2 to 1 | |||||||||||||
3 | 3 | INC | 3 | 0 | 0 | 0 | 1 | 0→1 | 0 | 3→4 | INC | Increment register #3 from 0 to 1 | |||||||||||||
4 | 4 | INC | 1 | 0 | 0 | 0→1 | 1 | 1 | 0 | 4→5 | INC | Increment register #1 from 0 to 1 | |||||||||||||
5 | 5 | JZ | 0 | 1 | 0 | 1 | 1 | 1 | 0 | 5→1 | JZ | U-Jump: register #0 is empty | |||||||||||||
6 | 1 | JZ | 2 | 6 | 0 | 1 | 1 | 1 | 0 | 1→2 | JZ | Jump fails: register #2 contains 1 | |||||||||||||
7 | 2 | DEC | 2 | 0 | 0 | 1 | 1→0 | 1 | 0 | 2→3 | DEC | Decrement register #2 from 1 to 0 | |||||||||||||
8 | 3 | INC | 3 | 0 | 0 | 1 | 0 | 1→2 | 0 | 3→4 | INC | Increment register #3 from 1 to 2 | |||||||||||||
9 | 4 | INC | 1 | 0 | 0 | 1→2 | 0 | 2 | 0 | 4→5 | INC | Increment register #1 from 1 to 2 | |||||||||||||
10 | 5 | JZ | 0 | 1 | 0 | 2 | 0 | 2 | 0 | 5→1 | JZ | U-Jump: register #0 is empty | |||||||||||||
11 | 1 | JZ | 2 | 6 | 0 | 2 | 0 | 2 | 0 | 1→6 | JZ | Jump !: register #2 is empty | |||||||||||||
move [1] to 2: | |||||||||||||||||||||||||
12 | 6 | JZ | 1 | 10 | 0 | 2 | 0 | 2 | 0 | 6→7 | JZ | Jump fails: register #1 contains 2 | |||||||||||||
13 | 7 | DEC | 1 | 0 | 0 | 2→1 | 0 | 2 | 0 | 7→8 | DEC | Decrement register #1 from 2 to 1 | |||||||||||||
14 | 8 | INC | 2 | 0 | 0 | 1 | 0→1 | 2 | 0 | 8→9 | INC | Increment register #2 from 0 to 1 | |||||||||||||
15 | 9 | JZ | 0 | 6 | 0 | 1 | 1 | 2 | 0 | 9→6 | JZ | U-Jump: register #0 is empty | |||||||||||||
16 | 6 | JZ | 1 | 10 | 0 | 1 | 1 | 2 | 0 | 6→7 | JZ | Jump fails: register #1 contains 1 | |||||||||||||
17 | 7 | DEC | 1 | 0 | 0 | 1→0 | 1 | 2 | 0 | 7→8 | DEC | Decrement register #1 from 1 to 0 | |||||||||||||
18 | 8 | INC | 2 | 0 | 0 | 0 | 1→2 | 2 | 0 | 8→9 | INC | Increment register #2 from 1 to 2 | |||||||||||||
19 | 9 | JZ | 0 | 6 | 0 | 0 | 2 | 2 | 0 | 9→6 | JZ | U-Jump: register #0 is empty | |||||||||||||
20 | 6 | JZ | 1 | 10 | 0 | 0 | 2 | 2 | 0 | 6→10 | JZ | Jump !: register #1 is empty | |||||||||||||
21 | 10 | H | 0 | 0 | 0 | 0 | 2 | 2 | 0 | 10→10 | H | HALT |
The example above demonstrates how the first basic instructions { INC, DEC, JZ } can create three more instructions—unconditional jump J, CLR, CPY. In a sense CPY used both CLR and J plus the base set. If register #3 had had contents initially, the sum of contents of #2 and #3 would have ended up in #3. So to be fully accurate CPY program should have preceded its moves with CLR (1) and CLR (3).
However, we do see that ADD would have been possible, easily. And in fact the following is summary of how the primitive recursive functions such as ADD, MULtiply and EXPonent can come about (see Boolos-Burgess-Jeffrey (2002) p. 45-51).
In general, we can build any partial- or total- primitive recursive function that we wish, by using the same methods. Indeed, Minsky (1967), Shepherdson-Sturgis (1963) and Boolos-Burgess-Jeffrey (2002) give demonstrations of how to form the five primitive recursive function "operators" (1-5 below) from the base set (1).
But what about full Turing equivalence? We need to add the sixth operator—the μ operator—to obtain the full equivalence, capable of creating the total- and partial- recursive functions:
The authors show that this is done easily within any of the available base sets (1, 2, or 3) (an example can be found at μ operator). This means that any mu recursive function can be implemented as a counter machine,^{[1]} despite the finite instruction set and program size of the counter machine design. However, the required construction may be counter-intuitive, even for functions that are relatively easy to define in more complex register machines such as the random-access machine. This is because the μ operator can iterate an unbounded number of times, but any given counter machine cannot address an unbounded number of distinct registers due to the finite size of its instruction list.
For instance, the above hierarchy of primitive recursive operators can be further extended past exponentiation into higher-ordered arrow operations in Knuth's up-arrow notation. For any fixed , the function is primitive recursive, and can be implemented as a counter machine in a straightforward way. But the function is not primitive recursive. One may be tempted to implement the up-arrow operator using a construction similar to the successor, addition, multiplication, and exponentiation instructions above, by implementing a call stack so that the function can be applied recursively on smaller values of . This idea is similar to how one may implement the function in practice in many programming languages. However, the counter machine cannot use an unbounded number of registers in its computation, which would be necessary to implement a call stack that can grow arbitrarily large. The up-arrow operation can still be implemented as a counter machine since it is mu recursive, however the function would be implemented by encoding an unbounded amount of information inside a finite number of registers, such as by using Gödel numbering.
(1) Unbounded capacities of registers versus bounded capacities of state-machine instructions: How will the machine create constants larger than the capacity of its finite state machine?
(2) Unbounded numbers of registers versus bounded numbers of state-machine instructions: How will the machine access registers with address-numbers beyond the reach/capability of its finite state machine?
(3) The fully reduced models are cumbersome:
Shepherdson and Sturgis (1963) are unapologetic about their 6-instruction set. They have made their choice based on "ease of programming... rather than economy" (p. 219 footnote 1).
Shepherdson and Sturgis' instructions ( [r] indicates "contents of register r"):
Minsky (1967) expanded his 2-instruction set { INC (z), JZDEC (r, I_{z}) } to { CLR (r), INC (r), JZDEC (r, I_{z}), J (I_{z}) } before his proof that a "Universal Program Machine" can be built with only two registers (p. 255ff).
For every Turing machine, there is a 2CM that simulates it, given that the 2CM's input and output are properly encoded. This is proved in Minsky's book (Computation, 1967, p. 255-258), and an alternative proof is sketched below in three steps. First, a Turing machine can be simulated by a finite-state machine (FSM) equipped with two stacks. Then, two stacks can be simulated by four counters. Finally, four counters can be simulated by two counters. The two counters machine use the set of instruction { INC ( r, z ), JZDEC ( r, z_{true}, z_{false}) }.
A Turing machine consists of an FSM and an infinite tape, initially filled with zeros, upon which the machine can write ones and zeros. At any time, the read/write head of the machine points to one cell on the tape. This tape can be conceptually cut in half at that point. Each half of the tape can be treated as a stack, where the top is the cell nearest the read/write head, and the bottom is some distance away from the head, with all zeros on the tape beyond the bottom. Accordingly, a Turing machine can be simulated by an FSM plus two stacks. Moving the head left or right is equivalent to popping a bit from one stack and pushing it onto the other. Writing is equivalent to changing the bit before pushing it.
A stack containing zeros and ones can be simulated by two counters when the bits on the stack are thought of as representing a binary number (the topmost bit on the stack being the least significant bit). Pushing a zero onto the stack is equivalent to doubling the number. Pushing a one is equivalent to doubling and adding 1. Popping is equivalent to dividing by 2, where the remainder is the bit that was popped. Two counters can simulate this stack, in which one of the counters holds a number whose binary representation represents the bits on the stack, and the other counter is used as a scratchpad. To double the number in the first counter, the FSM can initialize the second counter to zero, then repeatedly decrement the first counter once and increment the second counter twice. This continues until the first counter reaches zero. At that point, the second counter will hold the doubled number. Halving is performed by decrementing one counter twice and increment the other once, and repeating until the first counter reaches zero. The remainder can be determined by whether it reached zero after an even or an odd number of steps, where the parity of the number of steps is encoded in the state of the FSM.
As before, one of the counters is used as scratchpad. The other holds an integer whose prime factorization is 2^{a}3^{b}5^{c}7^{d}. The exponents a, b, c, and d can be thought of as four virtual counters that are being packed (via Gödel numbering) into a single real counter. If the real counter is set to zero then incremented once, that is equivalent to setting all the virtual counters to zero. If the real counter is doubled, that is equivalent to incrementing a, and if it's halved, that's equivalent to decrementing a. By a similar procedure, it can be multiplied or divided by 3, which is equivalent to incrementing or decrementing b. Similarly, c and d can be incremented or decremented. To check if a virtual counter such as c is equal to zero, just divide the real counter by 5, see what the remainder is, then multiply by 5 and add back the remainder. That leaves the real counter unchanged. The remainder will have been nonzero if and only if c was zero.
As a result, an FSM with two counters can simulate four counters, which are in turn simulating two stacks, which are simulating a Turing machine. Therefore, an FSM plus two counters is at least as powerful as a Turing machine. A Turing machine can easily simulate an FSM with two counters, therefore the two machines have equivalent power.
This result, together with a list of other functions of N that are not calculable by a two-counter machine — when initialised with N in one counter and 0 in the other — such as N^{2}, sqrt(N), log_{2}(N), etc., appears in a paper by Schroeppel (1972). The result is not surprising, because the two-counter machine model was proved (by Minsky) to be universal only when the argument N is appropriately encoded (by Gödelization) to simulate a Turing machine whose initial tape contains N encoded in unary; moreover, the output of the two-counter machine will be similarly encoded. This phenomenon is typical of very small bases of computation whose universality is proved only by simulation (e.g., many Turing tarpits, the smallest-known universal Turing machines, etc.).
The proof is preceded by some interesting theorems:
With regard to the second theorem that "A 3CM can compute any partial recursive function" the author challenges the reader with a "Hard Problem: Multiply two numbers using only three counters" (p. 2). The main proof involves the notion that two-counter machines cannot compute arithmetic sequences with non-linear growth rates (p. 15) i.e. "the function 2^{X} grows more rapidly than any arithmetic progression." (p. 11).
The Friden EC-130 calculator had no adder logic as such. Its logic was highly serial, doing arithmetic by counting. Internally, decimal digits were radix-1 — for instance, a six was represented by six consecutive pulses within the time slot allocated for that digit. Each time slot carried one digit, least significant first. Carries set a flip-flop which caused one count to be added to the digit in the next time slot.
Adding was done by an up-counter, while subtracting was done by a down-counter, with a similar scheme for dealing with borrows.
The time slot scheme defined six registers of 13 decimal digits, each with a sign bit. Multiplication and division were done basically by repeated addition and subtraction. The square root version, the EC-132, effectively subtracted consecutive odd integers, each decrement requiring two consecutive subtractions. After the first, the minuend was incremented by one before the second subtraction.