K-way Merge

Context: FIT2004_MOC · Merge Sort’s combine step generalised from $2$ lists to $k$ · the exam shape is “the naive version is $\Theta (Nk)$ — make it better”, and the answer is an ADT swap to a min-Heap (LO3) · the engine of external sorting

Quick Revision

• 🎯 Objective: merge $k$ sorted lists holding $N$ items into one ➔ each step asks only “which head is smallest?”, so the cost of answering it is the algorithm.
• 📦 Core Components: linear scan of $k$ heads ➔ $\Theta (k)$ per item, $\Theta (Nk)$ total | min-Heap of $k$ heads ➔ $\Theta (\log k)$ per item, $\Theta (N\log k)$ total.
• ⚡ Key Constraint: the heap holds one item per list, never the whole input ➔ auxiliary space is $\Theta (k)$ and independent of $N$, which is exactly why this merges runs too large for memory.

📝 How It Works

1. Why $k$-way Is Harder Than $2$-way

• Two lists ⟹ one comparison ➔ Merge Sort’s merge compares a[ia] against b[ib] — $\Theta (1)$ per output item.
• $k$ lists ⟹ a minimum query ➔ “smaller of two” becomes “minimum of $k$”, which a linear scan answers in $\Theta (k)$ ⟹ $\Theta (Nk)$ total. Finding the minimum is the bottleneck.
• Reframe as an ADT problem ➔ repeated extract-min with insert between extractions is the definition of a Priority Queue (ADT); swapping the linear scan for the right ADT is the entire optimisation.

2. The Min-Heap Solution

• Seed ➔ push the head of each list as a triple $(\text{value},\text{list id},\text{index})$ ⟹ heap size exactly $k$.
• Loop ➔ get_min emits the next output item in $\Theta (\log k)$; then add that list’s successor, keeping the heap at size $k$.
• Total ➔ $N$ items × (one extract + one insert) ⟹ $\Theta (N\log k)$ time, $\Theta (k)$ auxiliary beyond the output.
• Carrying the list id is mandatory ➔ without it you know what the minimum was but not which list to refill from; the index field makes the successor lookup $O(1)$.

3. The Invariant That Proves It

• Invariant ➔ at iteration $i$, the heap root is the correct value for output[i], because it is $\le$ every item remaining in every list.
• Why it holds ➔ each list is sorted, so a list’s unconsumed head is that list’s own minimum; the heap contains all $k$ heads; the root is the minimum of the heads ⟹ the minimum of everything unconsumed ➔ Invariant.
• Termination ➔ each iteration removes one item from a finite total of $N$ and refills only from a strictly advancing index.
• Stability comes free ➔ break heap ties by list id and the merge is stable, which matters when $k$-way merge is the combine step of a stable sort.

4. $\Theta (N\log k)$ Is Optimal — the Reduction

• Claim ➔ no comparison-based $k$-way merge can beat $O(N\log k)$.
• The reduction ➔ split any sequence of length $N$ into $N$ lists of length $1$ (each trivially sorted) and merge them with $k=N$. The output is the sorted sequence ⟹ the merge algorithm is a comparison sort, and beating $O(N\log k)=O(N\log N)$ would break the $\Omega (N\log N)$ floor ➔ Sorting Problem.
• Read the shape, not just the result ➔ a lower bound by reduction: you inherit a known bound by showing the new problem can simulate the old one. The same move proves optimality claims across the unit ➔ LO1.
• The bound is on the comparison model only ➔ a non-comparison merge over bounded integer keys is not covered, exactly as Counting Sort escapes the sorting floor.

⚙️ Core Implementation

🔹 Heap-backed $k$-way merge

k_way_merge — min-Heap of $(\text{value},\text{list id},\text{index})$ triples

def k_way_merge(lists):
    # lists: k already-sorted lists. MinHeap is the mirror of the vault's
    # max-heap: parent <= child, add -> rise, get_min -> sink
    heap = MinHeap()
    total = 0
    for lid in range(len(lists)):
        if len(lists[lid]) > 0:
            heap.add((lists[lid][0], lid, 0))   # seed with each list's HEAD
        total = total + len(lists[lid])
    output = [None] * total
    write = 0
    while write < total:
        value, lid, idx = heap.get_min()        # O(log k) -- the whole cost
        output[write] = value
        write = write + 1
        if idx + 1 < len(lists[lid]):           # refill from the SAME list
            heap.add((lists[lid][idx + 1], lid, idx + 1))
    return output

💡 Common Mistake: Pushing every item at seed time ➔ a heap of all $N$ items costs $\Theta (N)$ space and $\Theta (N\log N)$ time — that is heapsort, not a merge, and it discards the fact that the lists are already sorted. The heap must never exceed $k$.

⚖️ Core Decision Matrix

Strategy	Time	Auxiliary space	Selection rule
Linear scan of the $k$ heads	$\Theta (Nk)$	$\Theta (k)$ indices	$k$ tiny (2–4) — $\log k$ is not worth the heap’s constant
Min-Heap of $k$ heads	$\Theta (N\log k)$	$\Theta (k)$	the default; also the only option when the lists stream from disk
Divide & conquer (recursive halving)	$\Theta (N\log k)$	$\Theta (N)$ scratch	lists already resident and you want Merge Sort’s 2-way merge reused unchanged
Concatenate, then sort	$\Theta (N\log N)$	$\Theta (N)$	never for sorted input — it discards the sortedness you were given
Sequential accumulate (merge one at a time)	$\Theta (Nk)$	$\Theta (N)$	never — re-copies the growing accumulator $k$ times

When It Flips: the heap wins once ${\log }_{2}k<k$ in the constants, i.e. from roughly $k\ge 8$. Note $\Theta (N\log k)$ is Merge Sort when $k=N$: seeding $N$ singleton lists reproduces $\Theta (N\log N)$ exactly — the two algorithms are the same recurrence read from opposite ends.

📊 Exam Execution Trace & Applied Exercises

Manual Execution Trace

$k=3$: $A=[1,4,9]$, $B=[2,3,8]$, $C=[5,6,7]$. Heap shown as sorted contents; only the root is actually ordered.

Step	Heap (size $k=3$)	Root extracted	Refilled from	Output so far
0 (Seed)	$(1,A),(2,B),(5,C)$	—	—	[]
1	$(2,B),(4,A),(5,C)$	$1$	$A\to 4$	[1]
2	$(3,B),(4,A),(5,C)$	$2$	$B\to 3$	[1,2]
3	$(4,A),(5,C),(8,B)$	$3$	$B\to 8$	[1,2,3]
4	$(5,C),(8,B),(9,A)$	$4$	$A\to 9$	[1,2,3,4]
5	$(6,C),(8,B),(9,A)$	$5$	$C\to 6$	[1,2,3,4,5]

Invariant check at step 3: the root $3$ is $\le$ the other heads $4$ and $5$, and each head is $\le$ everything left in its own list ⟹ $3$ is the global minimum of all unconsumed items, so output[2] = 3 is final.

Applied Exercise

Problem: An external sort produced $k=1000$ sorted runs totalling $N=10^6$ records. Quantify the gain from replacing the linear-scan minimum with a heap.

$$\begin{array}{rl}{T}_{\text{scan}}& =N\cdot \Theta (k)=\Theta (10^9)\text{ comparisons}\\ T_{\text{heap}}& =N\cdot \Theta ({\log }_{2}k)=10^6\times 10=\Theta (10^7)\\ S_{\text{aux}}& =\Theta (k)=\Theta (10^3)\text{ triples, independent of }N\end{array}$$

Final Extracted Output: a $100\times$ reduction with auxiliary space unchanged at $\Theta (k)$. That $\Theta (k)$ bound is what permits $N$ to exceed memory: only $k$ records are ever resident, and the lists can be read as disk streams because the algorithm never looks back — a strategy loading all $N$ items cannot merge data larger than memory at any complexity ➔ Algorithmic Complexity.

⚠️ Common Mistakes

• 💡 Quoting $\Theta (N\log N)$ for the heap version ➔ the heap never holds more than $k$ items, so each operation is $\Theta (\log k)$ — the whole point is that $k\ll N$.
• 💡 Losing the list id ➔ storing bare values makes the refill ambiguous; the triple is not optional bookkeeping, it is what keeps the refill $O(1)$.
• 💡 Reaching for build_heap ➔ bottom-up $\Theta (k)$ construction needs all $k$ heads up front, which is fine at seed time but useless mid-merge; the refills are one-at-a-time add calls ➔ Heap.

🧠 Active Recall

State the invariant that proves $k$-way merge correct, and justify why the heap root alone settles the next output slot.

• Hint: Use the fact that each input list is already sorted.

Answer

• Short answer: at iteration $i$ the root is $\le$ every unconsumed item, so it is output[i].
• Why: Heads dominate their own lists ➔ sortedness makes each list’s unconsumed head that list’s minimum; the heap holds all $k$ heads, so the global minimum must be among them, and the root is their minimum ➔ Invariant.

Someone proposes merging the $k$ lists into an accumulator one at a time. Give the bound and the flaw.

• Hint: Ask how large the accumulator is on the $i$-th merge.

Answer

• Short answer: $\Theta (Nk)$ — the same as the naive scan, and with $\Theta (N)$ copying every round.
• Why: The accumulator is re-walked $k$ times ➔ merging list $i$ into an accumulator of size $\approx \frac{iN}{k}$ costs that much again, summing to $\Theta (Nk)$ by the Arithmetic Series. Divide and conquer fixes it: recursively merge the first $\lfloor k/2\rfloor$ lists and the remaining $\lceil k/2\rceil$, then one ordinary 2-way merge ⟹ $\Theta (\log k)$ levels × $\Theta (N)$ = $\Theta (N\log k)$, matching the heap while reusing Merge Sort’s merge verbatim.

Prove that no comparison-based $k$-way merge can run faster than $O(N\log k)$.

• Hint: Choose the value of $k$ that turns merging into a problem you already have a bound for.

Answer

• Short answer: at $k=N$ with singleton lists, a $k$-way merge is a comparison sort, so beating $O(N\log k)$ would beat $\Omega (N\log N)$.
• Why: Lower bound by reduction ➔ every singleton list is trivially sorted, so it is a legal input to the merge, and its output is the sorted sequence. Any faster merge would sort faster than the comparison model allows ➔ Sorting Problem. This bounds the comparison model only.