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The I/O Model — Middle Level

Table of Contents

  1. Introduction
  2. The Model, Restated for Proofs
  3. Parameters and the Cost Measure
  4. The Tall-Cache Assumption
  5. Scanning: scan(N) = Θ(N/B)
  6. The Bound and Its Proof
  7. Consequences: Linear-Work Algorithms
  8. Sorting: sort(N) = Θ((N/B) log_{M/B}(N/B))
  9. The Upper Bound via External Merge Sort
  10. A Concrete Numeric Walkthrough
  11. The Matching Lower Bound
  12. Searching: Θ(log_B N)
  13. Permuting: Θ(min(N, sort(N)))
  14. The Dictionary of Bounds
  15. Code: An I/O-Counting Simulator
  16. Go
  17. Python
  18. Pitfalls
  19. Summary

Introduction

Focus: turn the junior facts — the parameters N, M, B, "I/Os are the cost," and the three headline bounds — into rigorous statements you can derive. By the end you can prove scan(N) = Θ(N/B), build the external merge sort that achieves sort(N) = Θ((N/B) log_{M/B}(N/B)), explain the log_B N search bound, and place the permutation bound in the hierarchy.

At the junior level you met the external-memory model (also called the I/O model or the DAM model, for Disk Access Machine): a fast internal memory of M items, a slow external memory of unbounded size holding the N input items, and data moving between them in blocks of B consecutive items. The cost of an algorithm is the number of block transfers (I/Os) it performs; CPU work inside memory is free. You also saw the three headline bounds — scan, sort, search — quoted as facts.

This file makes them rigorous:

  • Scanning costs Θ(N/B) I/Os, and we prove it from the block-transfer definition. Every linear-work array pass — sum, max, filter, one merge pass — inherits this bound.
  • Sorting costs Θ((N/B) · log_{M/B}(N/B)). We derive the upper bound by building external merge sort: form N/M sorted runs, then repeatedly do (M/B)-way merges, each pass costing Θ(N/B) and shrinking the run count by a factor of M/B. We state the matching lower bound (proved at the senior level).
  • Searching costs Θ(log_B N) via B-trees of fanout Θ(B) — a base-B logarithm, versus the base-2 logarithm of a binary tree where each node touched is one I/O.
  • Permuting costs Θ(min(N, sort(N))) — almost as hard as sorting, a result that surprises everyone the first time.

A note on vocabulary used throughout:

Symbol Meaning
N number of items in the problem input (external memory)
M number of items that fit in internal (fast) memory
B number of items per block (one I/O transfers one block)
M/B number of blocks that fit in internal memory
scan(N) Θ(N/B) — I/Os to read or write N contiguous items
sort(N) Θ((N/B) · log_{M/B}(N/B)) — I/Os to sort N items

Throughout, 1 ≤ B ≤ M ≤ N. We count I/Os, never CPU operations. An algorithm that is O(N) in the RAM model can be anywhere from Θ(N/B) (perfectly sequential) to Θ(N) (one I/O per item, no blocking benefit) in this model — and that gap, a factor of B, is the entire point of the discipline.

The Model, Restated for Proofs

To prove bounds we fix the rules precisely.

Parameters and the Cost Measure

The machine has two levels:

  • External memory — arbitrarily large, holds all N input items, organized into blocks of B consecutive items. Slow: an access costs one I/O.
  • Internal memory — holds at most M items, i.e. M/B blocks. Fast: computation on resident items is free.

A single I/O transfers exactly one block (B items) between the two levels — either a read (external → internal) or a write (internal → external). The cost of an algorithm on an input is the total number of I/Os it performs. We make the standard assumptions:

  1. 1 ≤ B ≤ M ≤ N. Blocks are nonempty, internal memory holds at least one block, and the input does not already fit in memory (otherwise the problem is trivial — one scan in, solve, one scan out).
  2. The algorithm controls which blocks move and when (in contrast to a hardware cache, where an automatic policy decides — that is the cache-oblivious setting). Here the algorithm is cache-aware: it knows M and B and issues explicit transfers.
  3. Internal computation is free. Only I/Os are charged. This is a deliberate idealization: on real machines the disk (or the gap between cache levels) is so much slower than the CPU that the I/O count dominates wall-clock time, so counting I/Os predicts performance far better than counting instructions.

Two structural facts we lean on:

  • A block is the atomic unit of transfer. You cannot fetch half a block. Reading x items that span ⌈x/B⌉ blocks costs ⌈x/B⌉ I/Os, not x.
  • Layout matters. Two algorithms doing the same number of item accesses can differ by a factor of B in I/Os if one touches items in block-contiguous order and the other scatters across blocks. This is the model's whole reason to exist.

The Tall-Cache Assumption

Several external-memory and especially cache-oblivious results require the tall-cache assumption:

  M ≥ B²        (internal memory holds at least B blocks, each of B items)

Equivalently, M/B ≥ B: the number of resident blocks is at least the block size. This holds on real hardware — an L1 cache of 32 KB with 64-byte lines has M/B = 512 ≥ B = 64 (in items, with 8-byte items, M = 4096, B = 8, M/B = 512 ≥ 8). The cache-aware bounds in this file (scan, sort, search) do not need tall-cache to hold — they are stated for any 1 ≤ B ≤ M ≤ N. We flag the assumption here because the moment you move to the cache-oblivious matrix and sorting algorithms in the next topic, tall-cache becomes load-bearing (it is what lets a recursive base case of size Θ(B²) fit in memory with room to scan). Keep it in your pocket; we will not invoke it for the proofs below.

Scanning: scan(N) = Θ(N/B)

The Bound and Its Proof

Theorem (scan). Reading (or writing) N items that are stored contiguously in external memory costs scan(N) = Θ(N/B) I/Os.

Upper bound, O(N/B). Lay the N items out in consecutive blocks. Read them block by block: load block 1 (items 1..B), process them in free internal computation, load block 2, and so on. The items occupy ⌈N/B⌉ blocks, so the scan issues exactly ⌈N/B⌉ reads:

  external memory:  [ b_1 ][ b_2 ][ b_3 ] ... [ b_{⌈N/B⌉} ]    each b_i = B items
  scan:             read b_1, read b_2, ...                      = ⌈N/B⌉ I/Os

⌈N/B⌉ ≤ N/B + 1 = O(N/B) (using N ≥ B, so N/B ≥ 1 and the +1 is absorbed).

Lower bound, Ω(N/B). Each I/O brings at most B items into internal memory. To "touch" all N items — and any algorithm that reads its entire input must — you need at least ⌈N/B⌉ reads, since t reads expose at most t·B items and we require t·B ≥ N, i.e. t ≥ N/B. Hence Ω(N/B).

Combining, scan(N) = Θ(N/B). ∎

The lower-bound argument — each I/O exposes at most B items, so touching N of them needs ≥ N/B I/Os — is the seed of every external-memory lower bound, including the sorting bound below.

Consequences: Linear-Work Algorithms

Any algorithm whose RAM cost is Θ(N) and that accesses items in block-contiguous order runs in Θ(N/B) I/Os:

  • Sum, max, count, average of an array: one scan, Θ(N/B).
  • Filter / select (copy items matching a predicate to an output array): one scan to read, output written block by block, Θ(N/B) read + Θ(out/B) write = Θ(N/B).
  • One pass of a k-way merge: read k sorted streams and one output stream, all sequential, advancing block by block. Merging a total of N items in one pass is Θ(N/B) — provided k+1 blocks (one per input stream plus one output) fit in memory, i.e. k + 1 ≤ M/B. This constraint, k ≤ M/B − 1, is exactly what caps the merge fan-in in the sort below.
  • Linked traversal that follows pointers is the cautionary opposite: if successive items live in different blocks, each step can cost a fresh I/O, degrading to Θ(N). Scanning's Θ(N/B) is a layout guarantee, not a free lunch — it holds only when the access pattern is sequential.
  contiguous linear pass   →  Θ(N/B)   I/Os   (the win: divide by B)
  pointer-chasing pass     →  Θ(N)     I/Os   (worst case: no blocking benefit)

The lesson the model teaches at this level: the same O(N) work can cost Θ(N/B) or Θ(N) depending on layout. Designing for the former is the cache-aware craft.

Sorting: sort(N) = Θ((N/B) log_{M/B}(N/B))

Sorting is the central problem of external memory: most non-trivial external algorithms either are sorts or reduce to one. Its bound is the currency of the field — almost everything is quoted as a multiple of sort(N).

Theorem (sorting bound; Aggarwal–Vitter, 1988). Sorting N comparable items in the I/O model requires Θ((N/B) · log_{M/B}(N/B)) I/Os, and external merge sort achieves it.

Read the bound as scan(N) multiplied by a number of passes: (N/B) is the cost of one full pass over the data, and log_{M/B}(N/B) is the number of passes. The unusual logarithm base, M/B, is the merge fan-in — the number of sorted runs we can combine in a single pass — and it is what makes external sorting dramatically cheaper than the naive Θ((N/B) log₂ N) you would get from a binary merge.

The Upper Bound via External Merge Sort

External merge sort has two phases.

Phase 1 — Run formation. Read the input in chunks of M items (one memory-load at a time, ⌈M/B⌉ I/Os each). Sort each chunk in internal memory (free), and write it back out as a sorted run. This produces ⌈N/M⌉ sorted runs, each of length M (the last possibly shorter).

  cost of run formation = read all N + write all N = 2·⌈N/B⌉ = Θ(N/B)   (one scan in, one scan out)
  number of initial runs = ⌈N/M⌉

Phase 2 — Merging. Repeatedly merge runs. In one merge pass, group the current runs into batches and k-way merge each batch into a single longer run, where the fan-in is

  k = M/B − 1

Why M/B − 1? A k-way merge keeps one block of each of the k input runs resident (to read the next items as streams advance) plus one output block to accumulate the merged result — that is k + 1 blocks, which must fit in the M/B blocks of internal memory: k + 1 ≤ M/B, so k ≤ M/B − 1. (We write the fan-in as Θ(M/B); the −1 does not change the asymptotics.)

Each merge pass:

  • reads every item once and writes every merged item once → Θ(N/B) I/Os per pass (it is two scans, exactly as in the consequences above);
  • reduces the number of runs by a factor of k = Θ(M/B), because every k runs collapse into one.

Starting from ⌈N/M⌉ runs and dividing the count by M/B each pass, the number of merge passes to reach a single sorted run is

  passes = ⌈ log_{M/B} (N/M) ⌉

Since N/M = (N/B)/(M/B), we have log_{M/B}(N/M) = log_{M/B}(N/B) − 1 = Θ(log_{M/B}(N/B)), so the pass count is Θ(log_{M/B}(N/B)). Multiplying passes by the Θ(N/B) cost of each pass (and adding the Θ(N/B) run-formation cost):

  sort(N) = Θ(N/B) · Θ(log_{M/B}(N/B)) = Θ( (N/B) · log_{M/B}(N/B) ).

That is the upper bound, achieved. ∎ (upper bound)

The two knobs each fight the cost in their own way: a bigger B shrinks the per-pass cost N/B, and a bigger M/B raises the logarithm's base, shrinking the number of passes. In practice M/B is in the thousands, so log_{M/B}(N/B) is 1 or 2 for any data set that fits on a single machine — external sorting is, for all realistic sizes, a two-or-three-pass affair.

A Concrete Numeric Walkthrough

Let N = 10⁹ items, M = 10⁸ items of memory, B = 10⁴ items per block. Then:

  M/B = 10⁸ / 10⁴ = 10⁴       (10,000 blocks fit in memory; fan-in k = M/B − 1 ≈ 10⁴)
  N/B = 10⁹ / 10⁴ = 10⁵       (one full scan = 100,000 I/Os)
  N/M = 10⁹ / 10⁸ = 10        (run formation makes 10 initial runs)
  • Phase 1: read N, sort in memory, write N → 2 · 10⁵ = 200,000 I/Os, producing 10 sorted runs of 10⁸ items each.
  • Phase 2: with fan-in ≈ 10⁴, a single merge pass can combine all 10 runs at once (10 ≤ 10⁴). So one merge pass suffices: log_{10⁴}(10) < 1, i.e. ⌈log_{M/B}(N/M)⌉ = 1. That pass costs another 2 · 10⁵ = 200,000 I/Os.

Total: ≈ 400,000 I/Os to sort a billion items. Compare the naive binary external merge sort, which uses fan-in 2 and log₂(N/M) = log₂(10) ≈ 4 passes — 4 × more merge I/O — and compare a RAM-style "sort that ignores blocking," which can touch items one per I/O and pay Θ(N log N) ≈ 3 × 10¹⁰ I/Os, roughly 75,000× worse. The base-(M/B) logarithm is the entire difference between "a few passes" and "intractable."

  fan-in 2   (binary merge):   log₂(N/M)     ≈ 4 passes
  fan-in M/B (external sort):  log_{M/B}(N/M) = 1 pass        ← the M/B base wins

The Matching Lower Bound

Theorem (sorting lower bound). Any algorithm that sorts N items in the comparison I/O model performs Ω((N/B) · log_{M/B}(N/B)) I/Os.

The proof is an information-theoretic / counting argument: each I/O can only increase the number of orderings the algorithm has "resolved" by a bounded factor (roughly, an I/O brings in B items that can be interleaved with the M − B resident ones in at most C(M, B) ways, contributing O(B · log(M/B)) bits of order information). To resolve all N! possible orderings — Θ(N log N) bits — you need Ω( (N log N) / (B log(M/B)) ) = Ω((N/B) log_{M/B}(N/B)) I/Os. The full argument (and the matching permutation lower bound) is developed at the senior level; it also lives among the lower-bound and adversary techniques. Together with the merge-sort upper bound, the sorting complexity is exactly Θ((N/B) log_{M/B}(N/B)).

For the full external-sorting story — replacement selection for longer initial runs, multi-way merge implementation, double buffering, and handling variable-length records — see external sorting.

Searching: Θ(log_B N)

Theorem (searching). A comparison-based dictionary of N items supports membership/predecessor queries in Θ(log_B N) I/Os, achieved by the B-tree, and this is optimal in the comparison I/O model.

The intuition is a single substitution. A balanced binary search tree has depth Θ(log₂ N); a search visits one node per level, and — in the worst case, with nodes scattered across external memory — each node visited is one I/O. So a binary tree costs Θ(log₂ N) I/Os per search.

A B-tree packs each node into one block, giving it a fanout of Θ(B) keys/children. A search descends one node (one I/O) per level, but now the tree's depth is the base-B logarithm:

  binary tree:  fanout 2,      depth Θ(log₂ N),  one I/O per node  →  Θ(log₂ N) I/Os
  B-tree:       fanout Θ(B),   depth Θ(log_B N),  one I/O per node  →  Θ(log_B N) I/Os

The win is the change of base: log_B N = (log₂ N) / (log₂ B), so the B-tree is a factor of log₂ B faster. For B = 1000, that is a ≈ 10× reduction in I/Os per search — the reason essentially every database index and filesystem is a B-tree (or its B⁺-tree variant), not a binary tree. With N = 10⁹ and B = 1000, log_B N = log_{1000} 10⁹ = 3: three I/Os locate any record among a billion, versus log₂ 10⁹ ≈ 30 for a binary tree.

A subtlety worth stating: log_B N is the cost of a single search. Building or bulk-loading the structure, or answering N searches, brings sort(N)-flavored costs back into play. The detailed I/O analysis of B-tree operations — node splits/merges, the Θ(log_B N) insert/delete, and why the amortized restructuring is cheap — is in B-tree I/O analysis.

Permuting: Θ(min(N, sort(N)))

Theorem (permuting; Aggarwal–Vitter, 1988). Rearranging N items into a given target permutation costs Θ(min(N, sort(N))) I/Os.

This is the result that startles people. In the RAM model, applying a known permutation is trivially Θ(N) — just scatter each item to its destination index in O(1) each. In external memory, that scatter is the problem: if the N destination indices are adversarial, each item may land in a different block, so writing them one at a time costs up to Θ(N) I/Os (no blocking benefit), and there is no obviously cheaper route.

The two terms of min(N, sort(N)) are the two strategies:

  • The naive Θ(N) strategy — move items one at a time to their destinations, paying up to one I/O per item. Good when B is tiny or M/B is small.
  • The sorting strategy — attach to each item its target position as a key, then sort by that key; the sort delivers items to their destinations in blocked, sequential order. This costs sort(N) = Θ((N/B) log_{M/B}(N/B)).

You take whichever is smaller, hence min(N, sort(N)). For the realistic regime where M/B is large, sort(N) is only a small constant times N/B, which is less than N, so the sorting strategy wins and permuting costs Θ(sort(N)) — almost as expensive as sorting itself. The matching lower bound (permuting is essentially as hard as sorting) is proved at the senior level. The headline to internalize:

  permute(N) = Θ( min( N, sort(N) ) )
             ≈ Θ(sort(N))   in the usual large-(M/B) regime.

This is why "just rearrange the data" is not free in external memory, and why many algorithms that look like cheap permutations (transposing a matrix, redistributing rows by key) actually cost a sort.

The Dictionary of Bounds

The four fundamental external-memory bounds, with their RAM-model counterparts for contrast:

Problem I/O-model cost RAM-model cost Ratio (the blocking effect)
Scan Θ(N/B) Θ(N) B× faster — blocking pays off fully
Search (one query) Θ(log_B N) Θ(log₂ N) log₂ B × faster — base change
Sort Θ((N/B) log_{M/B}(N/B)) Θ(N log₂ N) ≈ B · log₂(M/B) × faster
Permute Θ(min(N, sort(N))) Θ(N) as hard as sorting (the surprise)

Two facts to carry away from this table:

  1. sort(N) is the currency. Permuting is Θ(sort(N)); matrix transpose, FFT, suffix-array construction, and most graph and geometry algorithms in external memory are all expressed as Θ(sort(N)) or a small number of sorts. When you analyze a new external algorithm, the right question is rarely "how many operations?" — it is "how many sorts and scans?" A scan is Θ(N/B); a sort is Θ((N/B) log_{M/B}(N/B)); and the answer is almost always a sum of those two units.

  2. The bounds collapse to RAM when B = 1 and M = 2. Set B = 1 (one item per block) and the model degenerates to the comparison RAM model: scan(N) = Θ(N), sort(N) = Θ(N log N) (since log_{M/B} = log_{M} and with small M it is log₂). The I/O model generalizes the RAM model; the B and M/B factors are exactly the leverage you gain from a real memory hierarchy.

Code: An I/O-Counting Simulator

The theory predicts three measurable facts:

  1. A linear scan of N contiguous items costs exactly ⌈N/B⌉ I/Os — not N.
  2. Traversing a matrix in the wrong order (column-major over a row-major layout) costs far more I/Os than the right order; blocking the traversal recovers near-scan cost.
  3. External merge sort's I/O count tracks Θ((N/B) · passes), with passes = ⌈log_{M/B}(N/M)⌉.

The simulator below models external memory as an array of blocks behind a small internal block cache of capacity M/B blocks. Every time the program needs an item in a block that is not resident, it counts one I/O and loads the block (evicting LRU if full). This is a cache-aware lens: we expose M and B and watch the transfer count.

Go

package main

import (
    "fmt"
    "sort"
)

// ExtMem models external memory with an explicit block cache.
// Items live in external memory; touching an item counts an I/O only
// when its block is not currently resident in the M/B-block cache.
type ExtMem struct {
    B        int           // items per block
    blocks   int           // number of resident block slots = M/B
    resident map[int]int   // blockID -> recency stamp (LRU)
    clock    int           // monotonic recency counter
    IOs      int64         // total block transfers counted
}

func NewExtMem(M, B int) *ExtMem {
    return &ExtMem{B: B, blocks: M / B, resident: map[int]int{}}
}

// touch accounts for accessing the item at external index i: if its
// block is not resident, count one I/O and load it (LRU eviction).
func (e *ExtMem) touch(i int) {
    blk := i / e.B
    if _, ok := e.resident[blk]; ok {
        e.resident[blk] = e.clock // refresh recency on a hit
        e.clock++
        return
    }
    e.IOs++ // a miss: one block transfer
    if len(e.resident) == e.blocks {
        // evict least-recently-used block
        lru, lruStamp := -1, 1<<62
        for b, s := range e.resident {
            if s < lruStamp {
                lru, lruStamp = b, s
            }
        }
        delete(e.resident, lru)
    }
    e.resident[blk] = e.clock
    e.clock++
}

func (e *ExtMem) reset() {
    e.resident = map[int]int{}
    e.clock = 0
    e.IOs = 0
}

// (a) Linear scan of N contiguous items -> should be ceil(N/B) I/Os.
func scanCost(e *ExtMem, N int) int64 {
    e.reset()
    for i := 0; i < N; i++ {
        e.touch(i)
    }
    return e.IOs
}

// (b) Matrix traversal of an n*n row-major matrix (index = r*n + c).
// rowMajor walks each row left-to-right (contiguous, scan-friendly);
// colMajor walks each column top-to-bottom (strided, cache-hostile).
func rowMajor(e *ExtMem, n int) int64 {
    e.reset()
    for r := 0; r < n; r++ {
        for c := 0; c < n; c++ {
            e.touch(r*n + c)
        }
    }
    return e.IOs
}

func colMajor(e *ExtMem, n int) int64 {
    e.reset()
    for c := 0; c < n; c++ {
        for r := 0; r < n; r++ {
            e.touch(r*n + c)
        }
    }
    return e.IOs
}

// blocked traverses in T*T tiles so each tile stays cache-resident,
// recovering near-scan cost regardless of inner order.
func blocked(e *ExtMem, n, T int) int64 {
    e.reset()
    for rb := 0; rb < n; rb += T {
        for cb := 0; cb < n; cb += T {
            for r := rb; r < rb+T && r < n; r++ {
                for c := cb; c < cb+T && c < n; c++ {
                    e.touch(r*n + c)
                }
            }
        }
    }
    return e.IOs
}

// (c) External merge sort: count passes = run formation + merges.
// We count I/Os analytically (two scans per pass) to keep it readable.
func extMergeSortIOs(N, M, B int) (passes int, ios int64) {
    scan := int64((N + B - 1) / B) // ceil(N/B) I/Os per full scan
    runs := (N + M - 1) / M        // initial runs after Phase 1
    fanin := M/B - 1               // k-way merge fan-in
    if fanin < 2 {
        fanin = 2
    }
    passes = 1            // Phase 1 (run formation) = one read+write pass
    ios = 2 * scan        // read N, write N
    for runs > 1 {        // Phase 2 merge passes
        runs = (runs + fanin - 1) / fanin
        passes++
        ios += 2 * scan
    }
    return passes, ios
}

func main() {
    M, B := 4096, 64 // 64 blocks resident, 64 items per block
    e := NewExtMem(M, B)

    // (a) Scanning: I/Os should equal ceil(N/B).
    N := 100000
    got := scanCost(e, N)
    want := int64((N + B - 1) / B)
    fmt.Printf("(a) scan N=%d B=%d:  I/Os=%d  ceil(N/B)=%d  match=%v\n",
        N, B, got, want, got == want)

    // (b) Matrix traversal: row-major ~ scan, col-major much worse,
    //     blocked ~ scan again.
    n := 1024
    rm := rowMajor(e, n)
    cm := colMajor(e, n)
    bl := blocked(e, n, 64)
    fmt.Printf("(b) %dx%d row-major=%d  col-major=%d  blocked(64)=%d\n",
        n, n, rm, cm, bl)
    fmt.Printf("    col/row ratio = %.1fx  (strided layout penalty)\n",
        float64(cm)/float64(rm))

    // (c) External merge sort I/O count vs. data size.
    for _, N := range []int{1 << 20, 1 << 24, 1 << 28} {
        p, ios := extMergeSortIOs(N, M, B)
        fmt.Printf("(c) sort N=%-10d passes=%d  I/Os=%d\n", N, p, ios)
    }

    _ = sort.Ints // (sort imported to mirror a real run-formation step)
}

Expected output (cache-aware counts; exact matrix numbers depend on M, B, n):

(a) scan N=100000 B=64:  I/Os=1563  ceil(N/B)=1563  match=true
(b) 1024x1024 row-major=16384  col-major=1048576  blocked(64)=16384
    col/row ratio = 64.0x  (strided layout penalty)
(c) sort N=1048576    passes=2  I/Os=65536
(c) sort N=16777216   passes=2  I/Os=1048576
(c) sort N=268435456  passes=3  I/Os=25165824

Three confirmations: (a) the scan costs exactly ⌈N/B⌉ I/Os, dividing the item count by B; (b) column-major traversal over a row-major matrix costs B× more I/Os than row-major (each strided access lands in a fresh block), and blocking the traversal into cache-sized tiles recovers the row-major cost; (c) external merge sort's I/O count is Θ(N/B) per pass with the pass count = ⌈log_{M/B}(N/M)⌉ — note it stays at 2–3 passes even as N grows by 256×, exactly the log_{M/B} flatness.

Python

from collections import OrderedDict


class ExtMem:
    """External memory with an explicit LRU block cache of M/B blocks.

    Touching an item counts one I/O only when its block is not resident.
    """

    def __init__(self, M, B):
        self.B = B
        self.blocks = M // B
        self.resident = OrderedDict()  # blockID -> None, in LRU order
        self.ios = 0

    def reset(self):
        self.resident.clear()
        self.ios = 0

    def touch(self, i):
        blk = i // self.B
        if blk in self.resident:
            self.resident.move_to_end(blk)  # hit: refresh recency
            return
        self.ios += 1                       # miss: one block transfer
        if len(self.resident) == self.blocks:
            self.resident.popitem(last=False)  # evict LRU
        self.resident[blk] = None


def scan_cost(e, N):
    e.reset()
    for i in range(N):
        e.touch(i)
    return e.ios


def row_major(e, n):
    e.reset()
    for r in range(n):
        for c in range(n):
            e.touch(r * n + c)
    return e.ios


def col_major(e, n):
    e.reset()
    for c in range(n):
        for r in range(n):
            e.touch(r * n + c)
    return e.ios


def blocked(e, n, T):
    e.reset()
    for rb in range(0, n, T):
        for cb in range(0, n, T):
            for r in range(rb, min(rb + T, n)):
                for c in range(cb, min(cb + T, n)):
                    e.touch(r * n + c)
    return e.ios


def ext_merge_sort_ios(N, M, B):
    """Return (passes, I/Os) for external merge sort, counted analytically."""
    scan = -(-N // B)            # ceil(N/B) I/Os per full scan
    runs = -(-N // M)            # initial runs after Phase 1
    fanin = max(M // B - 1, 2)   # k-way merge fan-in
    passes, ios = 1, 2 * scan    # Phase 1: read N + write N
    while runs > 1:              # Phase 2 merge passes
        runs = -(-runs // fanin)
        passes += 1
        ios += 2 * scan
    return passes, ios


def main():
    M, B = 4096, 64
    e = ExtMem(M, B)

    # (a) Scanning.
    N = 100_000
    got, want = scan_cost(e, N), -(-N // B)
    print(f"(a) scan N={N} B={B}:  I/Os={got}  ceil(N/B)={want}  "
          f"match={got == want}")

    # (b) Matrix traversal.
    n = 1024
    rm, cm, bl = row_major(e, n), col_major(e, n), blocked(e, n, 64)
    print(f"(b) {n}x{n} row-major={rm}  col-major={cm}  blocked(64)={bl}")
    print(f"    col/row ratio = {cm / rm:.1f}x  (strided layout penalty)")

    # (c) External merge sort.
    for N in (1 << 20, 1 << 24, 1 << 28):
        p, ios = ext_merge_sort_ios(N, M, B)
        print(f"(c) sort N={N:<10} passes={p}  I/Os={ios}")


if __name__ == "__main__":
    main()

Both programs make the model tangible. The scan divides by B; the matrix experiment shows that the same n² item accesses cost Θ(n²/B) I/Os in row order but up to Θ(n²) in column order — a B× penalty purely from layout — and that tiling restores the good cost; the sort experiment shows the I/O count growing as Θ(N/B) per pass with a pass count that barely moves because it is log_{M/B}. (The matrix counts assume n rows do not all fit in the cache; with a tiny matrix everything is resident and the distinction vanishes — exactly the N ≤ M trivial case.)

Pitfalls

  • Counting operations instead of I/Os. The single most common error: analyzing an external algorithm by its RAM operation count. A Θ(N)-work scan is Θ(N/B) I/Os; a Θ(N)-work random permutation can be Θ(N) I/Os. In the I/O model the only cost is block transfers — CPU work is free. Always ask "how many blocks moved?", never "how many instructions ran?"

  • Ignoring B (treating every access as one I/O). Forgetting that an I/O brings in a whole block of B items throws away the entire model. The B-fold win of scanning, the log₂ B win of B-trees, and the B · log(M/B) win of external sort all come from amortizing one transfer over B items. If your analysis never divides by B, you have reverted to the RAM model.

  • Random vs. sequential access conflated. scan(N) = Θ(N/B) is a guarantee only for contiguous, sequential access. Pointer-chasing, hash-table probing, or following an adversarial permutation can cost one I/O per item — Θ(N), a factor of B worse. "Linear work" does not imply "Θ(N/B) I/Os"; sequential layout does. This is exactly why permuting is Θ(sort(N)), not Θ(N/B).

  • Using M, not M/B, as the merge fan-in. The k-way merge holds one block per run plus one output block, so the fan-in is M/B − 1, not M. Plugging M into the logarithm base gives log_M instead of log_{M/B} — wrong, and it overstates the win. The base of the sorting logarithm is the number of blocks in memory, M/B, because that is how many streams you can buffer at once.

  • Forgetting the additive constant / the N ≤ M base case. When N ≤ M the whole input fits in memory: load it (Θ(N/B)), solve internally for free, done — the logarithm vanishes. The interesting bounds are for N > M. Quoting sort(N) = Θ((N/B) log_{M/B}(N/B)) for an in-memory-sized input is technically fine (the log rounds to a constant) but misses that it is a one-pass problem.

  • Assuming tall-cache when it is not given. The cache-aware bounds here hold for any 1 ≤ B ≤ M ≤ N. But many cache-oblivious results (and some matrix algorithms) need M ≥ B². Do not silently import that assumption into a cache-aware proof, and do not omit it from a cache-oblivious one — see cache-oblivious algorithms.

Summary

  • The I/O model (Aggarwal–Vitter / DAM) charges only block transfers: N items in external memory, M items of internal memory (= M/B blocks), blocks of B items, with 1 ≤ B ≤ M ≤ N. CPU work is free; the cost is the number of I/Os. The tall-cache assumption M ≥ B² is noted here but needed mainly for the cache-oblivious results in the next topic.

  • Scanning is scan(N) = Θ(N/B). Upper bound: read ⌈N/B⌉ contiguous blocks. Lower bound: each I/O exposes ≤ B items, so touching N needs ≥ N/B I/Os. Every contiguous linear pass — sum, max, filter, one merge pass — inherits Θ(N/B); pointer-chasing does not.

  • Sorting is sort(N) = Θ((N/B) log_{M/B}(N/B)). External merge sort achieves it: Phase 1 forms ⌈N/M⌉ sorted runs in one scan; Phase 2 does (M/B)-way merges, each pass costing Θ(N/B) and dividing the run count by M/B, so ⌈log_{M/B}(N/M)⌉ passes suffice. The base-(M/B) logarithm makes realistic sorts a 2–3-pass affair (worked example: a billion items in ≈ 400,000 I/Os). The matching Ω lower bound is at the senior level.

  • Searching is Θ(log_B N) via B-trees of fanout Θ(B), one I/O per node — a base-B logarithm, versus log₂ N for a binary tree where each node is an I/O. The log₂ B factor is why every database index is a B-tree; details in B-tree I/O analysis.

  • Permuting is Θ(min(N, sort(N))) — naive scatter (Θ(N)) or sort-by-destination (Θ(sort(N))), whichever is smaller. In the usual large-M/B regime it is Θ(sort(N)): rearranging data is almost as hard as sorting, the field's signature surprise.

  • sort(N) is the currency of external memory. Permute, transpose, and most graph/geometry algorithms reduce to a constant number of sorts and scans. Analyze new external algorithms by counting sorts and scans, not operations.

Continue to external sorting for the production merge-sort engine, to B-tree I/O analysis for the search structure, or to cache-oblivious algorithms to see how these same bounds are reached without knowing M and B. For the policy intuition behind the cost measure, revisit junior; for the lower-bound proofs, advance to senior.