Arrays and Iteration
Array Sum (Linear Scan)
Walk an array once, accumulating each element into a running total. This is
the canonical single-pass linear scan and the simplest possible loop
invariant: after step i, total equals the sum of arr[0..i].
Algorithm
The canonical input from the lesson spec is
arr = [3, 1, 4, 1, 5, 9, 2, 6]. After eight passes the running total is
31.
linear scan
Visit each element exactly once in index order.
running total
`total` accumulates the sum as the loop advances.
Basic Implementation
basic.rb
Replay: real traced execution (multi-file project)
arr = [3, 1, 4, 1, 5, 9, 2, 6]
total = 0
i = 0
while i < arr.length
total = total + arr[i]
i = i + 1
end
puts total
arr ← [3, 1, 4, 1, 5, 9, 2, 6]
1arr = [3, 1, 4, 1, 5, 9, 2, 6]2total = 0values this step[3, 1, 4, 1, 5, 9, 2, 6]arrtotal ← 0
1arr = [3, 1, 4, 1, 5, 9, 2, 6]2total = 03i = 0values this step0total[3, 1, 4, 1, 5, 9, 2, 6]arrtotal ← 3
4while i < arr.length5 total = total + arr[i]6 i = i + 1values this step0 → 3total0i3arr[i]total ← 4
4while i < arr.length5 total = total + arr[i]6 i = i + 1values this step3 → 4total1i1arr[i]total ← 8
4while i < arr.length5 total = total + arr[i]6 i = i + 1values this step4 → 8total2i4arr[i]total ← 9
4while i < arr.length5 total = total + arr[i]6 i = i + 1values this step8 → 9total3i1arr[i]total ← 14
4while i < arr.length5 total = total + arr[i]6 i = i + 1values this step9 → 14total4i5arr[i]total ← 23
4while i < arr.length5 total = total + arr[i]6 i = i + 1values this step14 → 23total5i9arr[i]total ← 25
4while i < arr.length5 total = total + arr[i]6 i = i + 1values this step23 → 25total6i2arr[i]total ← 31
4while i < arr.length5 total = total + arr[i]6 i = i + 1values this step25 → 31total7i6arr[i]
Trace Output
trace.rb
Replay: real traced execution (multi-file project)
arr = [3, 1, 4, 1, 5, 9, 2, 6]
total = 0
i = 0
while i < arr.length
before = total
total = total + arr[i]
puts "step #{i}: arr(#{i})=#{arr[i]} total #{before} -> #{total}"
i = i + 1
end
puts "final total = #{total}"
total ← 3, stdout ← step 0: arr(0)=3 total 0 -> 3
5before = total6total = total + arr[i]7puts "step #{i}: arr(#{i})=#{arr[i]} total #{before} -> #{total}"values this step3totalstep 0: arr(0)=3 total 0 -> 3stdout0before3arr[i]total ← 4, stdout ← step 1: arr(1)=1 total 3 -> 4
5before = total6total = total + arr[i]7puts "step #{i}: arr(#{i})=#{arr[i]} total #{before} -> #{total}"values this step4totalstep 1: arr(1)=1 total 3 -> 4stdout3before1arr[i]total ← 8, stdout ← step 2: arr(2)=4 total 4 -> 8
5before = total6total = total + arr[i]7puts "step #{i}: arr(#{i})=#{arr[i]} total #{before} -> #{total}"values this step8totalstep 2: arr(2)=4 total 4 -> 8stdout4before4arr[i]total ← 9, stdout ← step 3: arr(3)=1 total 8 -> 9
5before = total6total = total + arr[i]7puts "step #{i}: arr(#{i})=#{arr[i]} total #{before} -> #{total}"values this step9totalstep 3: arr(3)=1 total 8 -> 9stdout8before1arr[i]total ← 14, stdout ← step 4: arr(4)=5 total 9 -> 14
5before = total6total = total + arr[i]7puts "step #{i}: arr(#{i})=#{arr[i]} total #{before} -> #{total}"values this step14totalstep 4: arr(4)=5 total 9 -> 14stdout9before5arr[i]total ← 23, stdout ← step 5: arr(5)=9 total 14 -> 23
5before = total6total = total + arr[i]7puts "step #{i}: arr(#{i})=#{arr[i]} total #{before} -> #{total}"values this step23totalstep 5: arr(5)=9 total 14 -> 23stdout14before9arr[i]total ← 25, stdout ← step 6: arr(6)=2 total 23 -> 25
5before = total6total = total + arr[i]7puts "step #{i}: arr(#{i})=#{arr[i]} total #{before} -> #{total}"values this step25totalstep 6: arr(6)=2 total 23 -> 25stdout23before2arr[i]total ← 31, stdout ← step 7: arr(7)=6 total 25 -> 31
5before = total6total = total + arr[i]7puts "step #{i}: arr(#{i})=#{arr[i]} total #{before} -> #{total}"values this step31totalstep 7: arr(7)=6 total 25 -> 31stdout25before6arr[i]stdout ← final total = 31
9end10puts "final total = #{total}"values this stepfinal total = 31stdout31total
Complexity
- Time: O(n)
- Space: O(1)
Implementation notes
- Ruby: use the explicit
while i < arr.lengthloop withtotal = 0and a manual index. The stdlibarr.sumis fine for production but hides the loop the lesson spec is teaching. arr = [3, 1, 4, 1, 5, 9, 2, 6]documents the fixed-content array; the manuali = i + 1step keeps the iteration without leaning oneach_with_indexorinject(:+)that hide the running update.- The replay shows
i,arr[i], andtotalbefore and after each addition, matching the lesson spec's state-transition table.