Random-access stored-program machine
In theoretical computer science the random-access stored-program machine model is an abstract machine used for the purposes of algorithm development and algorithm complexity theory.
The RASP is a random-access machine model that, unlike the RAM, has its program in its "registers" together with its input. The registers are unbounded ; whether the number of registers is finite is model-specific. Thus the RASP is to the RAM as the Universal Turing machine is to the Turing machine. The RASP is an example of the von Neumann architecture whereas the RAM is an example of the Harvard architecture.
The RASP is closest of all the abstract models to the common notion of computer. But unlike actual computers the RASP model usually has a very simple instruction set, greatly reduced from those of CISC and even RISC processors to the simplest arithmetic, register-to-register "moves", and "test/jump" instructions. Some models have a few extra registers such as an accumulator.
Together with the register machine, the RAM, and the pointer machine the RASP makes up the four common sequential machine models, called this to distinguish them from the "parallel" models .
Informal definition: random-access stored-program model (RASP)
Nutshell description of a RASP:The reader will remember that the UTM is a Turing machine with a "universal" finite-state table of instructions that can interpret any well-formed "program" written on the tape as a string of Turing 5-tuples, hence its universality. While the classical UTM model expects to find Turing 5-tuples on its tape, any program-set imaginable can be put there given that the Turing machine expects to find them—given that its finite-state table can interpret them and convert them to the desired action. Along with the program, printed on the tape will be the input data/parameters/numbers, and eventually the output data/numbers. The "user" must position the Turing machine's head over the first instruction, and the input must be placed in a specified place and format appropriate to both the program-on-tape and the finite-state machine's instruction-table.
The RASP mimics this construction: it places the "program" and "data" in the holes. But unlike the UTM the RASP proceeds to "fetch" its instructions in a sequential manner, unless the conditional test sends it elsewhere.
A point of confusion: two sets of instructions: Unlike the UTM, the RASP model has two sets of instructions – the state machine table of instructions and the "program" in the holes. The two sets do not have to be drawn from the same set.
An example of a RAM working as a RASP
The following example of a program will move the contents of register #18 to register #19, erasing contents of #18 in the process.5: 03 18 15 JZ 18,15 ; if is zero, jump to 15 to end the program
02 18 DEC 18 ; Decrement
01 19 INC 19 ; Increment
03 15 05 JZ 15, 5 ; If is zero, jump to 5 to repeat the loop
15: 00 H ; Halt
18: n ; Source value to copy
19: ; Destination for copy
The program-instructions available in this RASP machine will be a simple set to keep the example short:
| Instruction | Mnemonic | Action on register "r" | Action on finite state machine's Instruction Register, IR |
| INCrement | INC | +1 → r | +1 → IR |
| DECrement | DEC | -1 → r | +1 → IR |
| Jump if Zero | JZ | none | IF = 0 THEN z → IR ELSE +1 → IR |
| Halt | H | none | → IR |
To ease the example we will equip the state machine of the RAM-as-RASP with the primitive instructions drawn from the same set, but augmented with two indirect copy instructions:
As the RASP machine's state machine interprets the program in the registers, what exactly will the state machine be doing? The column containing the exclamation mark ! will list in time sequence the state machine's actions as it "interprets" converts to action the program:
| PC | IR | ||||||||||||||||||
| hole # → | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 |
| program, parameters → | 5 | JZ | 18 | 15 | DEC | 18 | INC | 19 | JZ | 15 | 5 | H | n | ||||||
| encoded program → | 5 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | ||||||
| state machine instructions ↓ | |||||||||||||||||||
| ! |
Tradition divides the state-machine's actions into two major "phases" called Fetch and Execute. We will observe below that there are "sub-phases" within these two major phases. There is no agreed-to convention; every model will require its own precise description.
Fetch phase
The state machine has access to all the registers, both directly and indirectly. So it adopts #1 as "the program counter" PC. The role of the program counter will be to "keep the place" in the program's listing; the state machine has its own state register for its private use.Upon start, the state machine expects to find a number in the PC—the first "Program-Instruction" in the program.
The point of the above detour is to show that life is much easier when the state machine has access to two kinds of indirect copy:
- copy indirect from i and direct to j: CPY <
- copy direct from i and indirect to j: CPY
| PC | PIR | ||||||||||||||||||||
| hole # → | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | ||
| program, parameters → | 5 | JZ | 18 | 15 | DEC | 18 | INC | 19 | JZ | 15 | 5 | H | n | ||||||||
| encoded program → | 5 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | ||||||||
| step | state machine instructions ↓ | ||||||||||||||||||||
| 1 | fetch_instr: | CPY <<1>>,<2> | 5 i | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n |
Parse phase
Now that the number of the program-instruction is in register #2 -- the "Program-Instruction Register" PIR—the state machine proceeds to decrement the number until the IR is empty:If the IR were empty before decrement then the program-instruction would be 0 = HALT, and the machine would jump to its "HALT" routine. After the first decrement, if the hole were empty the instruction would be INC, and the machine would jump to instruction "inc_routine". After the second decrement, the empty IR would represent DEC, and the machine would jump to the "dec_routine". After the third decrement, the IR is indeed empty, and this causes a jump to the "JZ_routine" routine. If an unexpected number were still in the IR, then the machine would have detected an error and could HALT.
| PC | IR | ||||||||||||||||||||
| hole # → | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | ||
| program, parameters → | 5 | JZ | 18 | 15 | DEC | 18 | INC | 19 | JZ | 15 | 5 | H | n | ||||||||
| encoded program → | 5 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | ||||||||
| state machine instructions ↓ | |||||||||||||||||||||
| CPY <<1>>,<2> | 5 i | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | |||||||||
| JZ 2,halt | 5 | 3 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 19 | 5 | 0 | n | |||||||
| 3 | DEC 2 | 5 | 2 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | ||||||
| 4 | JZ 2,inc_routine: | 5 | 2 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | ||||||
| 5 | DEC 2 | 5 | 1 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | ||||||
| 6 | JZ 2,dec_routine | 5 | 1 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | ||||||
| 7 | DEC 2 | 5 | 0 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | ||||||
| 8 | JZ 2, JZ_routine | 5 | 0 ! | ||||||||||||||||||
| halt: | HALT | 5 | 3 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | ||||||
| inc_routine: | etc. | 5 | 3 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | ||||||
| dec_routine: | etc. | 5 | 3 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | ||||||
| 9 | JZ_routine: | etc. | 5 | 3 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n |
Execute phase, JZ_routine
Now the state machine knows what program-instruction to execute; indeed it has jumped to the "JZ_routine" sequence of instructions. The JZ instruction has 2 operands the number of the register to test, and the address to go to if the test is successful.Operand fetch which register to test for empty?: Analogous to the fetch phase, the finite state machine moves the contents of the register pointed to by the PC, i.e. hole #6, into the Program-Instruction Register PIR #2. It then uses the contents of register #2 to point to the register to be tested for zero, i.e. register #18. Hole #18 contains a number "n". To do the test, now the state machine uses the contents of the PIR to indirectly copy the contents of register #18 into a spare register, #3. So there are two eventualities, register #18 is empty, register #18 is not empty.
: If register #3 is empty then the state machine jumps to Second operand fetch fetch the jump-to address.
: If register #3 is not empty then the state machine can skip Second operand fetch. It simply increments twice the PC and then unconditionally jumps back to the instruction-fetch phase, where it fetches program-instruction #8.
Operand fetch jump-to address. If register #3 is empty, the state machine proceeds to use the PC to indirectly copy the contents of the register it points to into itself. Now the PC holds the jump-to address 15. Then the state machine unconditionally goes back to the instruction fetch phase, where it fetches program-instruction #15.
| PC | IR | ||||||||||||||||||||
| hole # → | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | ||
| program, parameters → | 5 | JZ | 18 | 15 | DEC | 18 | INC | 19 | JZ | 15 | 5 | H | n | ||||||||
| encoded program → | 5 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | ||||||||
| step | state machine instructions ↓ | ||||||||||||||||||||
| 9 | JZ_routine | INC 1 | 3 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | ||||||
| 10 | CPY <<1>>,<2> | 6 i | 3 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | ||||||||
| 11 | test hole: | CPY <<2>>,<3> | 6 | 18 i | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | ||||||
| 12 | test hole: | JZ 3, jump | 6 | 18 i | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | |||||
| n | n | ||||||||||||||||||||
| 13 | no_jump: | INC 1 | 18 | n | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | |||||
| 14 | INC 1 | 18 | n | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | ||||||
| 15 | J fetch_instr | 8 | 18 | n | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | |||||
| 1 | fetch_instr: | CPY <<1>>,<2> | 8 i | n | 3 | 18 | 15 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | ||||||
| 2 | parse: | etc. | |||||||||||||||||||
| 13 | jump: | INC 1 | 18 | n | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | |||||
| 14 | CPY <<1>>,<1> | 18 | n | 3 | 18 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | |||||||
| 15 | J fetch_instr | 15 | 18 | n | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | |||||
| 1 | fetch_instr: | CPY <<1>>,<2> | 15 i | n | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | n | ||||||
| 2 | parse: | etc. |
Execute phase INC, DEC
The following completes the RAM's state-machine interpretation of program-instructions, INC h, DEC h and thus completes the demonstration of how a RAM can "impersonate" a RASP:Without indirect state-machine instructions INCi and DECi, to execute the INC and DEC program-instructions the state machine must use indirect copy to get the contents of the pointed-to register into spare register #3, DEC or INC it, and then use indirect copy to send it back to the pointed-to register.
| PC | IR | ||||||||||||||||||||
| hole # → | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | ||
| program, parameters → | 5 | JZ | 18 | 15 | DEC | 18 | INC | 19 | JZ | 15 | 5 | H | n | ||||||||
| encoded program → | 5 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | ||||||||
| state machine instructions ↓ | |||||||||||||||||||||
| 15 | J fetch_instr | 8 | 18 | n | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | |||||
| 16 | fetch_instr: | CPY <<1>>,<2> | 8 i | n | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | |||||
| 17 | parse: | JZ 2,halt | 8 | 2 | n | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | ||||
| 18 | DEC 2 | 8 | n | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | ||||||
| 19 | JZ 2, inc_routine: | 8 | 1 | n | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | |||||
| 20 | DEC 2 | 8 | n | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | ||||||
| 21 | JZ 2, dec_routine: | 8 | 0 ! | n | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | |||||
| 22 | dec_routine: | INC 1 | 9 | 0 | n | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | ||||
| 23 | CPY <<1>>,<2> | 9 i | 18 | n | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | |||||
| 24 | CPY <<2>>,<3> | 9 | 18 i | n | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | |||||
| 25 | JZ 3,*+2 | 9 | 18 | n | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | |||||
| 26 | DEC 3 | 9 | 18 | n-1 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n | |||||
| 27 | CPY <3>,<<2>> | 9 | 18 i | n-1 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n-1 | |||||
| 28 | INC 1 | 10 | 18 | n-1 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n-1 | |||||
| 29 | J fetch_instr | 10 | 18 | n-1 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n-1 | |||||
| 30 | fetch_instr: | CPY <<1>>,<2> | 10 i | 1 | n-1 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n-1 | ||||
| 31 | parse: | JZ 2,halt | 10 | 1 | n-1 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n-1 | ||||
| 32 | DEC 2 | 10 | 0 | n-1 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n-1 | |||||
| 33 | JZ 2,inc_routine: | 10 | 0 ! | n-1 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n-1 | |||||
| 34 | inc_routine: | INC 1 | 11 | 0 | n-1 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n-1 | ||||
| 35 | CPY <<1>>,<2> | 11 i | 19 | n-1 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n-1 | |||||
| 36 | CPY <<2>>,<3> | 11 | 19 i | 0 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n-1 | 0 | ||||
| 37 | INC 3 | 11 | 19 | 1 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n-1 | 0 | ||||
| 38 | CPY <3>,<<2>> | 11 | 19 i | 1 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n-1 | 1 | ||||
| 39 | INC 1 | 12 | 19 | 1 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n-1 | 0 | ||||
| 40 | J fetch_instr | 12 | 19 | 1 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n-1 | 0 | ||||
| 41 | fetch_instr: | etc. | 12 | 19 | 1 | 3 | 18 | 15 | 2 | 18 | 1 | 19 | 3 | 15 | 5 | 0 | n-1 | 0 |
Alternate instructions: Although the demonstration resulted in a primitive RASP of only four instructions, the reader might imagine how an additional instruction such as "ADD
Self-Modifying RASP programs
When a RAM is acting as a RASP, something new has been gained: unlike the RAM, the RASP has the capacity for self-modification of its program-instructions. Cook-Reckhow comment on this in their description of their RASP model, as does HartmanisAn early description of this notion can be found in Goldstine-von Neumann :
Such an ability makes the following possible:
- subroutines -- the calling routine stores the return address "return_address" in the subroutine's last command, i.e. "JMP return_address"
- so-called JUMP-tables
- self-modifying code
RASP program-instruction set of Cook and Reckhow (1973)
Their purpose was to compare execution-times of the various models: RAM, RASP and multi-tape Turing machine for use in the theory of complexity analysis.
The salient feature of their RASP model is no provision for indirect program-instructions. This they achieve by requiring the program to modify itself: if necessary an instruction can modify the "parameter" of a particular instruction. They have designed their model so each "instruction" uses two consecutive registers, one for the "operation code" and the parameter "either an address or an integer constant".
Their RASP's registers are unbounded in capacity and unbounded in number; likewise their accumulator AC and instruction counter IC are unbounded. The instruction set is the following:
| operation | mnemonic | operation code | description |
| load constant | LOD, k | 1 | put constant k into accumulator |
| add | ADD, j | 2 | add contents of register j to accumulator |
| subtract | SUB, j | 3 | subtract contents of register j from accumulator |
| store | STO, j | 4 | copy contents of accumulator into register j |
| branch on positive accumulator | BPA, xxx | 5 | IF contents of accumulator > 0 THEN jump to xxx ELSE next instruction |
| read | RD, j | 6 | next input into register j |
| PRI, j | 7 | output contents of register j | |
| halt | HLT | any other - or + integer | stop |