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computation can arise from simple mechanical rules."},{"text":"Executing many instructions at the same time.","correct":false,"explanation":"Parallel execution is a feature of some computer architectures, not the essential idea represented here. A Turing machine captures following exact rules to manipulate symbols."},{"text":"Using binary numbers to represent every value.","correct":false,"explanation":"Turing machines can use different finite alphabets, so they are not defined by binary representation. They capture following exact rules to manipulate symbols."},{"text":"Connecting logic gates to form electronic circuits.","correct":false,"explanation":"Logic gates describe physical digital hardware rather than this abstract model of computation. A Turing machine captures following exact rules to manipulate symbols."}]},{"id":"ccf78d68-3692-450e-adfd-81bd8fa0bbfc","cardType":"cloze","points":null,"front":"For AQA, a Turing machine uses [ ] that is [ ].","back":"For AQA, a Turing machine uses one tape that is infinite to the right.","mcqOptions":[{"text":"For AQA, a Turing machine uses one tape that is infinite to the right.","correct":true,"explanation":"The AQA model has a single tape with a leftmost square and an unlimited supply of squares extending in the rightward direction."},{"text":"For AQA, a Turing machine uses two tapes that is infinite to the right.","correct":false,"explanation":"Multi-tape Turing machines exist as theoretical variants, but AQA specifies one tape. That tape has a leftmost square and is infinite to the right."},{"text":"For AQA, a Turing machine uses one tape that is finite to the right.","correct":false,"explanation":"Treating the tape as finite confuses a Turing machine with bounded physical memory. The AQA model uses one tape that is infinite to the right."},{"text":"For AQA, a Turing machine uses two tapes that is finite to the right.","correct":false,"explanation":"This incorrectly gives the abstract machine both multiple tapes and bounded storage. For AQA, it has one tape extending infinitely to the right."}]},{"id":"aacc042d-8362-4b90-960c-5af7859c469b","cardType":"cloze","points":null,"front":"A Turing machine has a finite set of [ ] and a finite alphabet of [ ].","back":"A Turing machine has a finite set of states and a finite alphabet of symbols.","mcqOptions":[{"text":"A Turing machine has a finite set of states and a finite alphabet of symbols.","correct":true,"explanation":"The control unit can occupy one of finitely many states, while each tape square contains a symbol chosen from a separately defined finite alphabet."},{"text":"A Turing machine has a finite set of symbols and a finite alphabet of states.","correct":false,"explanation":"This reverses the two categories: states describe the machine’s control condition, while symbols belong to the tape alphabet. The correct order is states, then symbols."},{"text":"A Turing machine has a finite set of transitions and a finite alphabet of cells.","correct":false,"explanation":"Cells are tape positions, not members of an alphabet, and the defining control set is the set of states. The machine has finite states and finite symbols."},{"text":"A Turing machine has a finite set of instructions and a finite alphabet of numbers.","correct":false,"explanation":"The alphabet is not restricted to numbers, and the named finite control components are states. A Turing machine has finite states and a finite alphabet of symbols."}]},{"id":"7b9ed423-2c04-4929-9a0e-9641f271ff29","cardType":"basic","points":1,"front":"Which state does a Turing-machine computation begin in?","back":"The start state.","mcqOptions":[{"text":"The start state.","correct":true,"explanation":"A Turing machine’s initial configuration places its control unit in the designated start state before any transition rule is applied."},{"text":"The halting state.","correct":false,"explanation":"A halting state ends a computation rather than beginning it. Before any symbols are processed, the machine is placed in its designated start state."},{"text":"The accept state.","correct":false,"explanation":"An accept state may represent a successful final outcome, but it is not the initial control condition. Computation begins in the start state."},{"text":"The previous state.","correct":false,"explanation":"There is no previous state before computation starts. The machine begins in its specifically designated start state and then follows applicable transitions."}]},{"id":"a28119c0-0a16-4cfd-b372-92d27e537487","cardType":"basic","points":2,"front":"What happens when a Turing machine enters a halting state?","back":"It stops because there are no outgoing transitions.","mcqOptions":[{"text":"It stops because there are no outgoing transitions.","correct":true,"explanation":"A halting state has no transition rule leading onwards, so the machine cannot perform another read-write-move operation and the computation ends."},{"text":"It moves right because every halting transition advances the head.","correct":false,"explanation":"A halting state has no outgoing transition, so no further head movement occurs. The machine stops as soon as that state is entered."},{"text":"It returns to the start state before reading the tape again.","correct":false,"explanation":"Halting does not restart or loop the computation. With no outgoing transitions available, the machine stops rather than returning to its start state."},{"text":"It erases the tape because the computation has finished.","correct":false,"explanation":"Finishing a computation does not automatically alter every tape square. Entering a halting state simply stops the machine because no outgoing transitions exist."}]},{"id":"613301df-c367-4127-b975-4c1ed8bffb1e","cardType":"cloze","points":null,"front":"One Turing-machine step depends only on the current [ ] and the [ ] being read.","back":"One Turing-machine step depends only on the current state and the symbol being read.","mcqOptions":[{"text":"One Turing-machine step depends only on the current state and the symbol being read.","correct":true,"explanation":"The transition function uses the machine’s current control state and the symbol under the head to determine what to write, how to move, and the next state."},{"text":"One Turing-machine step depends only on the current tape position and the symbol being read.","correct":false,"explanation":"The absolute tape position does not determine a transition; identical state-symbol pairs produce the same action anywhere. Each step depends on the current state and read symbol."},{"text":"One Turing-machine step depends only on the current state and the next symbol being read.","correct":false,"explanation":"The machine does not look ahead to a neighbouring square before choosing a transition. It uses its current state and the symbol presently under the head."},{"text":"One Turing-machine step depends only on the current tape position and the next symbol being read.","correct":false,"explanation":"This combines two look-ahead misconceptions: neither absolute position nor a future symbol selects the action. The current state and currently read symbol determine the step."}]},{"id":"5bb88486-e55e-4125-b11a-214902cd21a9","cardType":"basic","points":1,"front":"Why does the blank symbol matter on a Turing-machine tape?","back":"Most of the infinite tape is initially blank.","mcqOptions":[{"text":"Most of the infinite tape is initially blank.","correct":true,"explanation":"Only a finite region normally contains the input, so the blank symbol identifies all the otherwise unused squares across the machine’s infinite tape."},{"text":"The head must halt whenever it reads a blank.","correct":false,"explanation":"A blank is an ordinary alphabet symbol and may have an outgoing transition defined for it. Its significance is that most of the infinite tape initially contains blanks."},{"text":"Input symbols must always be separated by blank squares.","correct":false,"explanation":"Blank separators may be used by a particular encoding, but they are not universally required. The blank symbol primarily fills most of the tape initially."},{"text":"Blank squares cannot be changed by the machine.","correct":false,"explanation":"The read-write head can overwrite a blank just like any other tape symbol. Blanks matter because they initially occupy most of the infinite tape."}]},{"id":"9bb3f09f-014d-4676-afba-afbe230381dd","cardType":"basic","points":1,"front":"How far can the read-write head move in one transition?","back":"One square left or right.","mcqOptions":[{"text":"One square left or right.","correct":true,"explanation":"Each transition permits a single local movement of the read-write head, shifting it exactly one tape square either left or right."},{"text":"Two squares left or right.","correct":false,"explanation":"This incorrectly combines two movements into one transition. A single transition moves the read-write head exactly one square to the left or right."},{"text":"One square left only.","correct":false,"explanation":"The distance is correct, but the direction is unnecessarily restricted. Depending on the transition rule, the head can move one square left or right."},{"text":"One square right only.","correct":false,"explanation":"The head is not limited to rightward travel after each operation. A transition can move it exactly one square in either the left or right direction."}]},{"id":"1ffce4df-d99d-4e55-80ef-4ced04d83979","cardType":"basic","points":1,"front":"What happens if no transition rule applies in a non-halting state?","back":"The machine stops because no valid transition applies.","mcqOptions":[{"text":"The machine stops because no valid transition applies.","correct":true,"explanation":"A Turing machine can continue only when a transition is defined for its current state and scanned symbol. With no applicable rule, it stops even though the state is not designated as halting."},{"text":"The machine continues using the previous valid transition.","correct":false,"explanation":"Transition rules are not reused automatically. If no rule matches the current state and scanned symbol, the machine stops because it has no defined action to perform."},{"text":"The machine halts by moving to its designated halting state.","correct":false,"explanation":"Stopping due to an undefined transition does not cause a move into a halting state. The machine simply stops because no valid transition specifies its next action."},{"text":"The machine waits until a valid transition becomes available.","correct":false,"explanation":"A Turing machine does not wait for its transition function to change. Its rules are fixed, so it stops immediately when no transition applies."}]},{"id":"0dfbf068-595f-4dda-b037-b35984ab20f5","cardType":"basic","points":1,"front":"In a transition, does the machine write before or after moving?","back":"It writes before moving.","mcqOptions":[{"text":"It writes before moving.","correct":true,"explanation":"A transition first replaces the symbol in the current tape cell, then moves the head one position in the specified direction and enters the next state."},{"text":"It moves before writing.","correct":false,"explanation":"Reversing the order would write to the adjacent cell rather than the scanned cell. The machine writes in its current position before moving the tape head."},{"text":"It writes while moving.","correct":false,"explanation":"Writing and movement are distinct ordered actions, not simultaneous operations. The scanned cell is updated first, and the tape head then moves left or right."},{"text":"It moves without writing.","correct":false,"explanation":"Every transition specifies a symbol to write, although it may be identical to the symbol already present. Writing occurs before the head moves."}]},{"id":"dd8c2a3f-7d9c-4c03-b202-902fef5a3db2","cardType":"cloze","points":null,"front":"A transition function maps (current state, current symbol) to the [ ], the [ ], and the movement direction.","back":"A transition function maps (current state, current symbol) to the next state, the symbol to write, and the movement direction.","mcqOptions":[{"text":"A transition function maps (current state, current symbol) to the next state, the symbol to write, and the movement direction.","correct":true,"explanation":"A transition determines the machine’s complete next action: which state to enter, which symbol replaces the scanned symbol, and which direction the head moves."},{"text":"A transition function maps (current state, current symbol) to the previous state, the symbol to write, and the movement direction.","correct":false,"explanation":"A transition specifies the next state rather than recording the previous state. Its outputs are the next state, the symbol to write, and the movement direction."},{"text":"A transition function maps (current state, current symbol) to the next state, the symbol just read, and the movement direction.","correct":false,"explanation":"The symbol just read is already an input to the function, not an output. The outputs are the next state, the symbol to write, and the movement direction."},{"text":"A transition function maps (current state, current symbol) to the halting state, the movement distance, and the movement direction.","correct":false,"explanation":"Transitions need not enter a halting state, and standard movement is already determined by its direction. The function outputs the next state and symbol to write."}]},{"id":"0bd16528-93eb-467d-a99e-cc51a65fd313","cardType":"basic","points":3,"front":"In $\\delta(q_1,0)=(q_2,1,\\text{L})$, what does the right-hand side mean?","back":"

Go to $q_2$
Write $1$
Move left

","mcqOptions":[{"text":"

Go to $q_2$
Write $1$
Move left

","correct":true,"explanation":"The tuple on the right gives the next state, replacement symbol, and movement direction in that order. Therefore the machine enters $q_2$, writes $1$, and moves left."},{"text":"

Go to $q_1$
Write $0$
Move left

","correct":false,"explanation":"This incorrectly treats the input state and scanned symbol as the resulting actions. The right-hand tuple means enter $q_2$, write $1$, and move left."},{"text":"

Go to $q_2$
Write $0$
Move right

","correct":false,"explanation":"The input symbol $0$ is read, not necessarily retained, and L means left rather than right. The machine enters $q_2$, writes $1$, and moves left."},{"text":"

Go to $q_1$
Write $1$
Move right

","correct":false,"explanation":"This retains the current state and reverses the direction. The tuple explicitly gives $q_2$ as the next state, $1$ as the written symbol, and L as left."}]},{"id":"091ee8ab-d6e0-4175-adc2-549b337e7312","cardType":"cloze","points":null,"front":"A state-diagram label is read symbol / write symbol / movement; B / 1 / R reads [ ] and writes [ ].","back":"A state-diagram label is read symbol / write symbol / movement; B / 1 / R reads B and writes 1.","mcqOptions":[{"text":"A state-diagram label is read symbol / write symbol / movement; B / 1 / R reads B and writes 1.","correct":true,"explanation":"The three label components are interpreted in order. B is the scanned symbol, 1 is written into that cell, and R instructs the head to move right."},{"text":"A state-diagram label is read symbol / write symbol / movement; B / 1 / R reads 1 and writes B.","correct":false,"explanation":"This reverses the read and write fields. In B / 1 / R, the machine reads B, replaces it with 1, and then moves right."},{"text":"A state-diagram label is read symbol / write symbol / movement; B / 1 / R reads B and writes R.","correct":false,"explanation":"R is the movement direction, not a tape symbol to be written. The first field B is read, while the second field 1 is written."},{"text":"A state-diagram label is read symbol / write symbol / movement; B / 1 / R reads 1 and writes R.","correct":false,"explanation":"This shifts both fields out of position: 1 is the written symbol and R is the direction. The machine reads B and writes 1."}]},{"id":"618da58c-5559-420e-980c-bd6f7d6ab1e0","cardType":"basic","points":1,"front":"For $\\delta(q_0,B)=(q_H,1,\\text{R})$, where does the diagram arrow go?","back":"From $q_0$ to $q_H$.","mcqOptions":[{"text":"From $q_0$ to $q_H$.","correct":true,"explanation":"The state before the comma on the left is the source state, while the first item on the right is the destination. The arrow therefore runs from $q_0$ to $q_H$."},{"text":"From $q_H$ to $q_0$.","correct":false,"explanation":"This reverses the transition arrow. The function is evaluated in state $q_0$ and gives $q_H$ as the next state, so the arrow goes from $q_0$ to $q_H$."},{"text":"From $q_0$ to $q_0$.","correct":false,"explanation":"A self-loop would require $q_0$ to appear as the next state on the right. Here the next state is $q_H$, so the arrow ends there."},{"text":"From $q_H$ to $q_H$.","correct":false,"explanation":"The left-hand state identifies where the transition begins, and that state is $q_0$, not $q_H$. The arrow runs from $q_0$ to $q_H$."}]},{"id":"63beaac7-6411-43d3-9c0b-b685dec31681","cardType":"basic","points":3,"front":"During hand-tracing, what must a configuration record?","back":"

Current state
Tape contents
Head position

","mcqOptions":[{"text":"

Current state
Tape contents
Head position

","correct":true,"explanation":"These three items completely describe a Turing machine at one moment. Together they determine which symbol is scanned and which transition, if any, applies next."},{"text":"

Previous state
Tape contents
Step number

","correct":false,"explanation":"The previous state and step number do not determine the next transition, and the head location is missing. A configuration records the current state, tape contents, and head position."},{"text":"

Current state
Transition table
Final output

","correct":false,"explanation":"The transition table defines the machine rather than its current configuration, while the final output may not yet exist. Record the current state, tape contents, and head position."},{"text":"

Input string
Current symbol
Movement direction

","correct":false,"explanation":"The original input and latest direction do not capture later tape changes or the machine’s state. A full configuration needs the current state, complete tape contents, and head position."}]},{"id":"b6ee8985-9196-4655-9cad-adff8007716d","cardType":"basic","points":1,"front":"In the configuration 1 [q0] 1 B B ..., which symbol is being read?","back":"The second 1.","mcqOptions":[{"text":"The second 1.","correct":true,"explanation":"In this configuration notation, the bracketed state is placed immediately before the cell under the tape head. Since [q0] precedes the second 1, that 1 is being read."},{"text":"The first 1.","correct":false,"explanation":"The state marker does not follow the scanned symbol; it appears immediately before it. Therefore [q0] marks the second 1, not the first, as the symbol being read."},{"text":"The first B.","correct":false,"explanation":"The head is not inferred from the first blank cell. Its position is shown by [q0], which appears immediately before the second 1, so that 1 is scanned."},{"text":"The state symbol q0.","correct":false,"explanation":"q0 represents the machine’s current state and is not stored as a tape symbol. The symbol immediately after [q0], the second 1, is being read."}]},{"id":"baffcf1e-eb2b-437f-b1c0-6815e4569dc0","cardType":"basic","points":2,"front":"How does unary increment find where to add the new 1?","back":"It scans right until the first blank.","mcqOptions":[{"text":"It scans right until the first blank.","correct":true,"explanation":"The machine moves across the existing unary 1s until it reaches the first blank cell, which marks the end of the current unary number."},{"text":"It scans left until the first blank.","correct":false,"explanation":"Scanning left moves towards the beginning of the unary number rather than its end. The machine must scan right until the first blank, where the new 1 is added."},{"text":"It counts the 1s before moving.","correct":false,"explanation":"The machine does not need to store or calculate the number of existing 1s. It simply scans right until the first blank cell identifies the end."},{"text":"It stops at the first 1.","correct":false,"explanation":"Stopping at the first 1 would locate the beginning, not the end, of the unary number. The machine scans right until it encounters the first blank."}]},{"id":"4853c062-140d-4223-84ed-e4ebdfb2879d","cardType":"basic","points":3,"front":"In unary increment, what happens when the first blank is read?","back":"

Write 1
Move right
Enter $q_H$

","mcqOptions":[{"text":"

Write 1
Move right
Enter $q_H$

","correct":true,"explanation":"Writing 1 in the first blank appends one unary digit, increasing the represented value by one. The machine then moves right and enters the halting state $q_H$."},{"text":"

Write 0
Move right
Enter $q_H$

","correct":false,"explanation":"Writing 0 would not append another unary 1, so the represented number would not be incremented correctly. The machine must write 1, move right and enter $q_H$."},{"text":"

Write 1
Move left
Enter $q_H$

","correct":false,"explanation":"Writing 1 performs the increment, but the specified transition moves right after writing it. Moving left would leave the head on the original unary number rather than beyond it."},{"text":"

Write 1
Move right
Enter $q_0$

","correct":false,"explanation":"Returning to $q_0$ would restart the scanning process instead of completing the operation. After writing the new 1 and moving right, the machine must enter $q_H$."}]},{"id":"535d67ce-401e-48fc-b48d-209e48de263c","cardType":"cloze","points":null,"front":"A problem is computable if some Turing machine solves every valid input and eventually [ ] with the [ ].","back":"A problem is computable if some Turing machine solves every valid input and eventually halts with the correct result.","mcqOptions":[{"text":"A problem is computable if some Turing machine solves every valid input and eventually halts with the correct result.","correct":true,"explanation":"Computability requires an algorithmic Turing machine that terminates for every valid input and produces the correct answer, rather than merely working for selected inputs or running indefinitely."},{"text":"A problem is computable if some Turing machine solves every valid input and eventually restarts with the correct result.","correct":false,"explanation":"Restarting is not successful termination and could cause the computation to repeat indefinitely. A computable problem requires a Turing machine that eventually halts with the correct result."},{"text":"A problem is computable if some Turing machine solves every valid input and eventually halts with the approximate result.","correct":false,"explanation":"An approximate answer does not satisfy the definition unless approximation is itself the specified problem. The machine must halt for every valid input with the correct result."},{"text":"A problem is computable if some Turing machine solves every valid input and eventually times out with the partial result.","correct":false,"explanation":"Timing out with incomplete output does not solve the problem and is not equivalent to halting successfully. The required machine must halt with the correct result for every valid input."}]},{"id":"d7eda046-ab95-4e99-ad85-673ba751d14d","cardType":"basic","points":1,"front":"Which issue is the Turing-machine model mainly about, rather than efficiency?","back":"Whether a problem is computable in principle.","mcqOptions":[{"text":"Whether a problem is computable in principle.","correct":true,"explanation":"The Turing-machine model is primarily used to determine whether any algorithm can solve a problem at all, regardless of how much time or memory that computation requires."},{"text":"Whether a problem is computable efficiently.","correct":false,"explanation":"Efficiency concerns the time or space resources used by an algorithm, which belongs to complexity analysis. The central issue here is whether the problem is computable in principle."},{"text":"Whether a program uses the minimum memory.","correct":false,"explanation":"Minimising memory is a question of space efficiency rather than basic computability. Turing machines chiefly establish whether a problem can be solved algorithmically in principle."},{"text":"Whether an answer is easy for humans.","correct":false,"explanation":"Human difficulty is not the criterion used by the Turing-machine model. It considers whether a mechanical procedure can compute the correct result in principle."}]},{"id":"0807b891-11cf-45ab-8f63-572cf7b11b68","cardType":"cloze","points":null,"front":"A Universal Turing machine can simulate another Turing machine when given its [ ] and the [ ].","back":"A Universal Turing machine can simulate another Turing machine when given its encoded description and the input.","mcqOptions":[{"text":"A Universal Turing machine can simulate another Turing machine when given its encoded description and the input.","correct":true,"explanation":"The encoded description specifies the simulated machine's states and transitions, while the input supplies the data on which that simulated machine is to operate."},{"text":"A Universal Turing machine can simulate another Turing machine when given its current state and the input.","correct":false,"explanation":"A current state alone does not specify the simulated machine's complete transition rules. The Universal Turing machine needs the machine's encoded description together with its input."},{"text":"A Universal Turing machine can simulate another Turing machine when given its encoded description and the output.","correct":false,"explanation":"The output is produced by a computation and is not the data on which simulation begins. The Universal Turing machine requires the encoded description and the input."},{"text":"A Universal Turing machine can simulate another Turing machine when given its hardware specification and the expected result.","correct":false,"explanation":"A Turing machine is defined by its encoded states and transition rules, not a physical hardware specification. Simulation requires that encoded description and the input to process."}]},{"id":"2cb659a3-1e94-46e2-8437-2cecfad38b84","cardType":"basic","points":2,"front":"What modern-computing idea does the Universal Turing machine capture?","back":"A machine can treat a program as data and execute it.","mcqOptions":[{"text":"A machine can treat a program as data and execute it.","correct":true,"explanation":"A Universal Turing machine reads an encoded program as part of its data and then simulates that program, capturing the stored-program principle used by modern general-purpose computers."},{"text":"A machine can treat a program as hardware and rebuild itself.","correct":false,"explanation":"The Universal Turing machine does not physically rebuild its components for each program. It treats an encoded program as data and executes it through simulation."},{"text":"A machine can execute only a single fixed program.","correct":false,"explanation":"Executing only one fixed program describes a special-purpose machine, not a universal one. A Universal Turing machine can receive different encoded programs as data and execute them."},{"text":"A machine must keep programs separate from data at all times.","correct":false,"explanation":"This reverses the stored-program principle represented by universal computation. A Universal Turing machine can treat the encoded description of a program as data and execute it."}]}]}],"examCountdown":{"examDate":"2027-06-07","paperLabel":"Paper 1"},"heading":"A model of computation","title":"A model of computation Flashcards","topbar":{"backHref":"/a-level/computer-science/aqa/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"0fbd4e6f-7573-5fe4-98ff-c912c37b6c82","walled":"$L38"}],"$L39","$L3a"]