Solution

Correct option: 4)

Goal: Reconstruct a high-level program from the given three-address code (TAC).

100: t1 = y + 2
101: initial = x / t1
102: limit = 10
103: if i > j goto 105
104: goto 111
105: if num > b goto 107
106: goto 111
107: num = limit - 1
108: b = y + 2
109: i = i - num
110: goto 103
111: j = num
112: k = j + b

1) Identify blocks and the loop

• Lines 100–102: straight-line initializations.
• 110: goto 103 makes a back edge to 103 ⇒ there is a loop whose header is 103.
• From 103:
  • If i > j ⇒ go to 105 (stay in the loop).
  • Else (via 104: goto 111) ⇒ exit loop.
• At 105:
  • If num > b ⇒ go to 107 (execute the body and loop back).
  • Else (via 106: goto 111) ⇒ exit loop.
• Hence the loop continues only if both conditions are true: (i > j) AND (num > b).

2) Loop body

• Executed when both tests succeed:
  107: num = limit - 1
  108: b = y + 2
  109: i = i - num
  110: goto 103 (repeat)
• Note: y is never changed, so b = y + 2 could reuse t1, but that is an optimization detail.

3) After the loop

• When either test fails, control goes to 111:
  111: j = num
  112: k = j + b

4) High-level code reconstructed from the TAC

t1     = y + 2;
initial= x / t1;          // or: initial = x / (y + 2);
limit  = 10;

while (i > j && num > b) {
    num = limit - 1;
    b   = y + 2;         // can reuse t1
    i   = i - num;
}

j = num;
k = j + b;

5) Match with the options

• Option 1: uses an if with || (OR) — wrong control flow (no loop, wrong condition).
• Option 2: has a loop but uses || (OR) — the TAC requires && (AND).
• Option 3: splits the tests into an if/else with an inner while — not equivalent to the TAC.
• Option 4: the only one that captures the correct while (i > j && num > b) structure and the correct loop body; therefore it is the correct high-level code among the choices.

Answer: Option 4.