Consider a linear arrangement of seven bulbs, each of which can be in the ON or OFF states. The initial configuration of the bulbs is shown in the figure. In every Step, the states of the bulbs are changed based on the following rules:
Any OFF bulb with exactly one ON neighbor at the end of the previous Step is turned ON.
Any ON bulb with both neighbors ON at the end of the previous Step is turned OFF.
The state of any bulb not meeting the conditions above is left unchanged.
The state of bulbs at the end of Step 1 and Step 2 are also shown in the figure. The number of bulbs which are ON at the end of Step 8 is ______
Show Hint
Track the bulb states with 0 for OFF and 1 for ON; the pattern locks into a repeating state after a few steps.
Step 1: Set up the states and the update rules. Label the seven bulbs \(1\) to \(7\) from left to right. Write OFF as \(0\) and ON as \(1\). The Initial state is \((0,0,0,1,0,0,0)\), since only bulb \(4\) is ON. Each Step is a synchronous update using the previous Step's full state: an OFF bulb with exactly one ON neighbour turns ON; an ON bulb with both neighbours ON turns OFF; every other bulb stays the same. The end bulbs (\(1\) and \(7\)) have only one neighbour, so they can never satisfy 'both neighbours ON', and once an end bulb turns ON it can never be switched OFF again.
Step 2: Verify the rule against the given Step 1 and Step 2. Applying the rule to \((0,0,0,1,0,0,0)\): bulb \(3\) picks up one ON neighbour (bulb \(4\)) and turns ON; bulb \(5\) picks up one ON neighbour (bulb \(4\)) and turns ON; bulb \(4\) has both neighbours OFF, so it is unchanged and stays ON; all others have zero ON neighbours and stay OFF. This gives Step 1 \(= (0,0,1,1,1,0,0)\), matching the figure. Applying the rule again to \((0,0,1,1,1,0,0)\): bulb \(2\) gets one ON neighbour (bulb \(3\)) and turns ON; bulb \(4\) has both neighbours (bulb \(3\) and bulb \(5\)) ON, so it turns OFF; bulb \(6\) gets one ON neighbour (bulb \(5\)) and turns ON; bulbs \(3\) and \(5\) each keep only one ON neighbour and stay ON. This gives Step 2 \(= (0,1,1,0,1,1,0)\), which also matches the figure, confirming the rule is being applied correctly.
Step 3: Continue the update to Step 3 and Step 4. From Step 2 \((0,1,1,0,1,1,0)\): bulb \(1\) picks up one ON neighbour (bulb \(2\)) and turns ON; bulb \(4\) has two ON neighbours (bulb \(3\), bulb \(5\)), which is not exactly one, so it stays OFF; bulb \(7\) picks up one ON neighbour (bulb \(6\)) and turns ON; bulbs \(2, 3, 5, 6\) each keep only one ON neighbour and stay ON. Step 3 \(= (1,1,1,0,1,1,1)\). From Step 3 \((1,1,1,0,1,1,1)\): bulb \(2\) now sits between two ON bulbs (bulb \(1\) and bulb \(3\)), so it turns OFF; bulb \(6\) also sits between two ON bulbs (bulb \(5\) and bulb \(7\)), so it turns OFF; bulbs \(1\) and \(7\) are end bulbs and stay ON; bulb \(4\) still has two ON neighbours and stays OFF; bulbs \(3\) and \(5\) each have only one ON neighbour in the Step 3 state, so they stay ON. Step 4 \(= (1,0,1,0,1,0,1)\).
Step 4: Check that Step 4 is a fixed (stable) state. Apply the rule once more to \((1,0,1,0,1,0,1)\). Bulbs \(2, 4, 6\) are OFF, and each has both neighbours ON (for example bulb \(2\)'s neighbours are bulb \(1=1\) and bulb \(3=1\)), so 'exactly one ON neighbour' fails and they stay OFF. Bulbs \(1, 3, 5, 7\) are ON; bulbs \(1\) and \(7\) are end bulbs and can never turn OFF; bulbs \(3\) and \(5\) each have both neighbours OFF, not both ON, so they stay ON. The result is again \((1,0,1,0,1,0,1)\), identical to Step 4. Once a state maps to itself, every later Step repeats it forever.
Final Answer: Since Step 4 = Step 5 = ... = Step 8, the state at Step 8 is \((1,0,1,0,1,0,1)\), which has \(4\) bulbs ON (bulbs \(1, 3, 5, 7\)). \[ \boxed{4} \]
