Which of the following is the Optimized version for the given TAC in the passage

1. A100:t1 =y+2 101 : initial = x/t1 102: limit = 10 103: if i >j goto 105 104 : goto 111 105: if num > b goto }10 106: goto 111 107:b=y+2 108: i=i - limit-1 109: goto 103 110:k=i+b
2. B100:t1=y+2 101 : initial = x/t1 102: limit = 10 103: if i >j goto 105 104: if num < b goto }11 105: num = limit-1 106:b = t1 107: i=i -num 108: goto 103 109:k=i+b
3. C100: t1 = y + 2 101: limit = 10 102: num = 9 103: b = t1 104: if i > j goto 106 105: goto 110 106: if num > b goto 108 107: goto 110 108: i = i - num 109: goto 104 110: j = i - num 111: k = j + b
4. D100:t1=y+2 101 : initial = x/t1 102: limit = 10 103: if i >j goto 105 104 : goto 111 105: if num < b goto 111 106: num = limit-1 107:b=t1 108: i=i -num 109 : goto 103 110:k= i+b

Pick an option to check yourself, or open the answer below.

Show answer & explanation

Correct answer: C. 100: t1 = y + 2 101: limit = 10 102: num = 9 103: b = t1 104: if i > j goto 106 105: goto 110 106: if num > b goto 108 107: goto 110 108: i = i - num 109: goto 104 110: j = i - num 111: k = j + b

Explanation

Option C removes the unused 'initial' computation, folds limit − 1 into the constant 9, replaces y + 2 with the already computed t1, and moves these loop-invariant assignments out of the loop, keeping the loop test and updates intact.

Note: Hoisting num = 9 before the first test assumes the loop body runs at least once; this follows the official key.