Run the machine.
Flip bits into two's complement, single-step a processor through fetch–decode–execute, and race linear against binary search while the comparison counters tell the truth.
Eight switches, two meanings
FIG. 01 · 4.5.4 · UNSIGNED & TWO'S COMPLEMENT0
The eight bits never change. Only the agreement about what the leftmost one is worth does.
Flip a switch, or negate to watch “invert every bit, then add one” happen for real.
What the exam wants. 4.5.4.3 asks you to convert between two's complement and denary both ways, and 4.5.4.9 asks when a result overflows. Every mark scheme accepts the same working you can see here: the leftmost column is worth −128, everything else is positive, and negating means invert-then-add-one. Shifting left is ×2 — until the sign flips, which is exactly the overflow they want you to spot.
One cycle at a time
FIG. 02 · 4.7.3.2 · FETCH–DECODE–EXECUTEReset. The PC points at address 0 — the first instruction.
The program is four instructions: load 12, add 30, store the answer, halt. Nothing is hidden — every register write is one step.
What the exam wants. 4.7.3.2 is nearly always “describe the fetch–execute cycle, referring to the registers” — and the marks are for the order: MAR ← PC, PC incremented, MBR ← memory, CIR ← MBR, then decode, then execute. Notice the PC increments before the instruction is even understood. Notice too that 0x15 in memory is just a number: it is an instruction only because the processor fetched it. That is the stored program concept (4.7.2.1) in one screen.
O(n) against O(log n)
FIG. 03 · 4.3.4 · 4.4.4 · SEARCH COMPLEXITYLinear search waiting
Binary search waiting
Pick a target and step. Linear checks the boxes in order; binary throws half of them away each time.
The list is sorted — that is the whole price of admission for binary search.
What the exam wants. 4.4.4.1–4.4.4.3 ask you to compare algorithms by order of complexity, not by seconds on your laptop. Search a 16-item list and binary wins by about three comparisons; search a million-item list and it wins by 999,980. Two marks usually go for the reason binary needs a sorted list, and one for saying that halving the search space is what makes it logarithmic.