TAOCP 5.2.2 Exercise 53

Radix exchange sort (Algorithm R in Section 5.

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Section 5.2.2: Sorting by Exchanging

Exercise 53. [HM42] Analyze the average number of bit inspections and exchanges done by radix exchange when the input data consists of infinite-precision binary numbers in $[0\ldots 1)$, each of whose bits is independently equal to 1 with probability $p$. (Only the case $p = \frac{1}{2}$ is discussed in the text; the methods we have used can be generalized to arbitrary $p$.) Consider in particular the case $p = 1/\phi = .61803\ldots$

Verified: no
Solve time: 21m11s

Setup

Radix exchange sort (Algorithm R in Section 5.2.2) partitions the file by examining the most significant bits.
We are given (N) independent infinite binary strings in ([0,1)); each bit is (1) with probability (p) and (0) with probability (q=1-p).

The algorithm partitions the current subarray by the current bit, exchanging records so that all (0)'s precede all (1)'s, then recurses on the two parts with the next bit.
The process stops when a part contains at most one record.

  • A bit inspection occurs each time a record’s current bit is examined.
  • An exchange is a swap of two records during partitioning.

The recursion builds a binary trie from the (N) strings.
The total bit inspections equal the external path length of the trie.
The total exchanges equal the sum over all internal nodes of the number of inversions in that node’s current bit sequence.

Solution

1. Average number of exchanges

In a partition of size (m) the bits are independent (\text{Bernoulli}(p)) variables.
The number of exchanges equals the number of inversions ((i<j) with bit (1) followed by bit (0)).
For a random sequence of (m) such bits the expected number of inversions is

[ \frac{m(m-1)}{2},pq. ]

Summing over all internal nodes of the trie gives the expected total exchanges

[ E_N = pq; \mathbb{E}\Bigl[\sum_{\text{internal nodes}} \binom{m}{2}\Bigr]. ]

Alternatively, consider any pair of records (i<j).
They remain in the same part until the first bit where they differ.
At that bit the left record has (1) and the right record has (0) with probability

[ \frac{pq}{pq+qp} = \frac12, ]

independent of (p) (for (0<p<1)).
Hence each pair contributes exactly (\frac12) to the expectation, yielding the simple exact formula

[ \boxed{E_N = \frac{N(N-1)}{4}.} ]

2. Average number of bit inspections

Let (r = p^2+q^2) be the probability that two independent bits are equal.
A given record requires a ((k+1))-st bit inspection iff its prefix of length (k) is shared by at least one other record.
The probability of this event is (1-(1-r^k)^{N-1}).
By linearity of expectation the total expected bit inspections is

[ C_N = N\sum_{k=0}^{\infty} \bigl[1-(1-r^k)^{N-1}\bigr]. \tag{1} ]

Using the binomial theorem this can be rewritten as

[ C_N = N\sum_{j=1}^{N-1} \binom{N-1}{j}(-1)^{j+1}\frac{1}{1-r^j}. \tag{2} ]

3. Asymptotic analysis of (C_N)

Set (f(x)=1-(1-x)^{N-1}).
Equation (1) is (C_N = N\sum_{k\ge 0} f(r^k)).
For large (N) the sum is concentrated where (r^k \approx 1/N).
Using the Mellin transform (or Euler-Maclaurin with a careful treatment of the boundary layer) one obtains the asymptotic expansion

[ \sum_{k=0}^{\infty} f(r^k) = \frac{\ln N}{\ln(1/r)} + \frac{\gamma}{\ln(1/r)} + \frac12

  • \delta\bigl(\log_{1/r} N\bigr) + o(1), ]

where (\gamma) is Euler’s constant and (\delta) is a continuous periodic function of period (1) with mean zero, given by the Fourier series

[ \delta(x) = \frac{1}{\ln(1/r)} \sum_{m\ne 0} \Gamma!\left(\frac{2\pi i m}{\ln(1/r)}\right) e^{2\pi i m x}. ]

The amplitude of (\delta) is extremely small (e.g., (<10^{-6}) for (p=\frac12)).
Multiplying by (N) gives the final asymptotic formula

[ \boxed{C_N = N,\frac{\ln N}{\ln(1/r)}

  • N!\left(\frac{\gamma}{\ln(1/r)}+\frac12\right)
  • N,\delta\bigl(\log_{1/r} N\bigr) + o(N).} \tag{3} ]

4. Special case (p = 1/\phi)

Here (\phi = (1+\sqrt5)/2 \approx 1.61803).
We have

[ p = \frac1\phi = \phi-1 \approx 0.618034,\qquad q = 1-p = \frac1{\phi^2} \approx 0.381966. ]

Then

[ r = p^2+q^2 = \frac1{\phi^2}+\frac1{\phi^4} = \frac{\phi^2+1}{\phi^4}. ]

Using (\phi^2 = \phi+1) and (\phi^4 = 3\phi+2), this simplifies to

[ r = \frac{\phi+2}{3\phi+2} \approx 0.527864. ]

Consequently

[ \frac{1}{r} \approx 1.89443,\qquad \ln(1/r) \approx 0.6390. ]

The leading asymptotic behaviour of the bit inspections is

[ C_N \sim N,\frac{\ln N}{0.6390}. ]

The constant term (excluding the periodic part) is

[ \frac{\gamma}{\ln(1/r)}+\frac12 \approx \frac{0.5772}{0.6390}+0.5 \approx 1.403. ]

The average number of exchanges remains (N(N-1)/4).

Verification

  • For (N=2): (E_2 = 1/2), which equals the probability that the first differing bit has left (=1), right (=0).
    (C_2 = 2\sum_{k\ge0} r^k = 2/(1-r) = 1/(pq)), matching the exact formula (1).
  • For (p=\frac12): (r=\frac12), (\ln(1/r)=\ln2). Formula (3) gives
    (C_N = N\log_2 N + N(\gamma/\ln2 + 1/2) + N\delta(\log_2 N) + o(N)).
    Numerically (\gamma/\ln2+1/2 \approx 1.3327). For (N=4) the exact value is (C_4 = 12.5714);
    (N\log_2 N = 8); the difference per record is (1.143), converging to (1.3327).
  • The exchange formula (E_N = N(N-1)/4) is exact for all (p) and (N); it was independently verified by the pair‑wise argument.

This completes the proof. ∎