Radix Sort

Context: FIT2004_MOC · the fix for Counting Sort’s $M$ blow-up — stop treating the key as one huge number, treat it as $K$ narrow columns · still non-comparison, so the $\Omega (N\log N)$ floor in Sorting Problem does not bind

Quick Revision

• 🎯 Objective: run a stable Counting Sort once per digit column, least-significant first ➔ $\Theta (KN+KM)$, beating Merge Sort’s $\Theta (kN\log N)$ for fixed-width keys.
• 📦 Core Components: $N$ items $\times$ $K$ columns | base $M$ (digits $10$ · bits $2$ · letters $26$) | stable subsort per column.
• ⚡ Key Constraint: the subsort must be stable — an unstable column pass destroys all order won by previous passes, and the output is wrong, not merely unsorted.

📝 How It Works

1. The Reframing

• Key as a grid ➔ items are rows, digits are $K$ columns; rightmost = least significant.
• Why $M$ collapses ➔ each column holds only base-$M$ symbols, so every subsort sees $M=10$ (decimal), $2$ (binary), $26$ (alphabet) — never the $981$ or $2^{32}$ that ruined plain Counting Sort.
• $K$ is a length, not a count ➔ $K=\lceil {\log }_M(\text{largest key})\rceil$ ⟹ raising the base shrinks $K$ but grows $M$; that trade is the exam’s design question.
• Lecture notation map — memorise both ➔ this note’s $K$ (columns) is the lecture’s $c$; this note’s $M$ (base) is the lecture’s $b$; the lecture reserves $M$ for the largest item. In lecture symbols: $c=\lfloor {\log }_{b}M\rfloor +1$, time $O(cN+cb)$, auxiliary $O(N+b)$.
• Digit extraction is arithmetic, not string slicing ➔ column $j$ (from $0$ at the least significant end) of key $x$ is $\lfloor x/b^j\rfloor \mathrm{mod}b$ ⟹ $O(1)$ per digit.

2. Least-Significant-Digit Order + Stability

• Pass order ➔ sort on column $K$ (rightmost), then $K-1$, …, then column $1$ ⟹ after pass $j$ the list is correctly sorted on the last $j$ digits.
• Stability carries the earlier work ➔ when column $j$ ties, a stable subsort preserves the previous pass’s order, which is the ordering on the lower-significance digits ⟹ the invariant extends.
• Unstable ⟹ catastrophe ➔ with $200$ before $291$ from pass 2, an unstable hundreds pass may emit $291$ first; both share hundreds digit $2$, so nothing later repairs it.

3. Choosing the Base — Where Linearity Comes From

• The base is a free design parameter ➔ nothing forces $b=10$; that choice alone decides whether the sort is linear.
• Raise $b$ up to $N$ ➔ per-pass cost is $\Theta (N+b)$, so any $b\le N$ leaves it at $\Theta (N)$ ⟹ total $\Theta (cN)$; $b=N$ buys the smallest $c$ at no asymptotic cost.
• Never let $b>N$ ➔ the size-$b$ count array then dominates the $N$ items and the sort is linear in the base, not in $N$.
• The linearity theorem ➔ with $b=N$, $c=\lfloor {\log }_{N}M\rfloor +1$; so if the largest item satisfies $M=O(N^d)$ for constant $d$, then $c=\Theta (1)$ and radix sort is $\Theta (N)$:

$$b=N,M=O(N^d)⟹c={\log }_{N}M=\Theta (d)⟹T=\Theta (cN+cb)=\Theta (N)$$

• Where it stops being linear ➔ $M$ super-polynomial in $N$ ($N^N$, $N!$) makes $c$ grow with $N$ — the distinct-keys argument reached from the base side.
• $b=N$ is the asymptotic rule, NOT the constant-factor optimum [D] ➔ for $N$ integers of $W$ bits, $c=W/{\log }_{2}b$ and the operation count $f(b)=\frac{W}{{\log }_{2}b}(N+b)$ is minimised well below $N$: at $W=64,N=10^6$ the minimum sits near $b\approx 95,536$, so the practical choice is $b=2^{16}$ — the nearest power of two whose exponent divides the word width, giving exactly $c=4$ passes.
• Powers of two win twice ➔ base $2^t$ makes digit extraction a shift and mask instead of a division, and $t\mid W$ avoids a ragged final column; timings degrade sharply for $b=2^{20}$ and $2^{24}$ because $20$ and $24$ do not divide $64$.

4. Ragged Keys

• Unequal lengths break the column grid ➔ banhammer and kappa have no common $K$.
• Pad to a common width ➔ insert a filler that sorts below every real symbol; numbers pad with leading zeros (right-aligned), strings with spaces.
• Alignment decides the order semantics ➔ right-aligned padding gives numeric order, left-aligned gives lexicographic — choose deliberately, they differ.

5. Optimising for Strings — the $\Theta (n)$ Algorithm (applied Problem 4 — [D, HD])

• Applied-sheet notation ➔ $k$ = number of strings · $\ell$ = length of the longest · $n={\sum }_i\text{len}(S_i)$ = total characters. The optimal bound is $\Theta (n)$ in that $n$.
• Naive padding is $\Theta (\ell k)$ ➔ every string is scanned $\ell$ times regardless of its own length ⟹ $10^6$ strings of length $10$ plus one of length $10^6$ costs $10^{12}$ operations against $n\approx 2\times 10^7$ actual characters.
• The fix — never look at a string before it has a character to contribute ➔ a string of length $i$ only matters in the final $i$ iterations of an LSD sweep.
• Step 1 — counting-sort by length, ASCENDING ➔ key $=\text{len}(S)$, range $\ell$ ⟹ $\Theta (\ell +k)$. Ascending is load-bearing: shorter strings must precede longer ones that share a prefix (cat before cats) — exactly alphabetical order.
• Step 2 — sweep columns $i=\ell$ down to $1$ with a live pointer ➔ maintain $j$ so that $S[j\dots k]$ are the strings of length $\ge i$; on each decrement of $i$, decrement $j$ while $\text{len}(S[j-1])\ge i$ ⟹ strings enter the window exactly when they acquire a character at position $i$, and never leave.
• Step 3 — run the stable counting subsort on $S[j\dots k]$ only ➔ base $26$, so no pass touches a non-existent character.
• Why it totals $\Theta (n)$ ➔ string $S_i$ participates in exactly $\text{len}(S_i)$ passes ⟹ work $={\sum }_i\text{len}(S_i)=n$, plus $\Theta (\ell +k)$ ⟹ optimal, since every character must be read once.
• This scheme is left-aligned/lexicographic ➔ position $1$ is the first character; right-aligned padding is the numeric reading and the wrong model for words ➔ §4.

⚙️ Core Implementation

🔹 LSD radix sort over base $M$

radix_sort — stable counting subsort per column, raw index manipulation

def radix_sort(my_list, base=10):
    if len(my_list) == 0:
        return my_list
    maximum = my_list[0]
    for i in range(1, len(my_list)):
        if my_list[i] > maximum:
            maximum = my_list[i]
    exp = 1                                   # column selector: 1, base, base^2, ...
    while maximum // exp > 0:                 # K iterations
        my_list = counting_pass(my_list, exp, base)
        exp = exp * base
    return my_list
 
def counting_pass(my_list, exp, base):
    count = [0] * base
    for i in range(len(my_list)):
        digit = (my_list[i] // exp) % base
        count[digit] += 1
    position = [0] * base                     # prefix sum -> block start of each digit
    for d in range(1, base):
        position[d] = position[d - 1] + count[d - 1]
    output = [0] * len(my_list)
    for i in range(len(my_list)):             # INPUT ORDER => stable
        digit = (my_list[i] // exp) % base
        output[position[digit]] = my_list[i]
        position[digit] += 1
    return output

💡 Common Mistake: Iterating the placement loop backwards or reusing count as position ➔ silently makes the pass unstable; the first column still looks right, so the bug only surfaces on multi-digit input.

🔹 $\Theta (n)$ radix sort for variable-length strings

radix_sort_strings — length sort, then a live-window sweep (applied Problem 4)

def radix_sort_strings(S):
    # S: non-empty lowercase strings. n = SUM of all lengths.
    k = len(S)
    longest = 0
    for i in range(k):
        if len(S[i]) > longest:
            longest = len(S[i])
    S = counting_sort_by_length(S, longest)   # ASCENDING; O(longest + k)
    j = k                                     # invariant: S[j..k-1] have length >= i
    for i in range(longest, 0, -1):           # LSD sweep over character positions
        while j > 0 and len(S[j - 1]) >= i:
            j = j - 1                         # admit strings that now HAVE a char i
        count = [0] * 26
        for p in range(j, k):
            count[ord(S[p][i - 1]) - 97] += 1
        position = [0] * 26                   # prefix sum -> block starts
        for c in range(1, 26):
            position[c] = position[c - 1] + count[c - 1]
        temp = [None] * (k - j)
        for p in range(j, k):                 # INPUT ORDER => stable
            d = ord(S[p][i - 1]) - 97
            temp[position[d]] = S[p]
            position[d] += 1
        for p in range(j, k):                 # write the window back
            S[p] = temp[p - j]
    return S

💡 Common Mistake: Sorting the lengths descending ➔ then cats precedes cat, which is not alphabetical. Ascending is what makes a shorter string win when it is a prefix of a longer one — the same rule that makes the pad symbol sort below every real character ➔ §4.

⚖️ Core Decision Matrix

Algorithm	Time	Auxiliary space	Stable	Selection rule
Counting Sort	$\Theta (N+M)$	$\Theta (M+N)$	engineered	keys are integers with $M\ll N$
Radix Sort	$\Theta (KN+KM)\approx \Theta (KN)$	$\Theta (M+N)\approx \Theta (N)$	required	keys decompose into $K$ narrow columns, $K$ small/fixed
Merge Sort	$\Theta (N\log N)\cdot O(k)$	$\Theta (N)$	yes	keys not integer-decomposable, or $K$ grows with $N$
Quick Sort	$\Theta (N\log N)$ avg, $\Theta (N^2)$ worst	$O(\log N)$	no	in-place matters more than the guarantee

When It Flips: radix beats Merge Sort while $K<{\log }_{2}N$. Fixed-width keys ($32$-bit ints, $K=10$ decimal digits) ⟹ $K$ is a constant ⟹ $\Theta (N)$. Distinct keys force $K\ge {\log }_{M}N$, degenerating $\Theta (KN)$ back to $\Theta (N\log N)$ — so the advantage shrinks exactly when keys become long; raising base $M$ buys $K$ down at $\Theta (KM)$ time and $\Theta (M)$ space, and stops paying once $M$ approaches $N$.

📊 Exam Execution Trace & Applied Exercises

Manual Execution Trace

[200, 151, 291, 981, 369, 421, 671], base $10$, $K=3$, LSD first:

Pass	Column	Digits read	List after the pass
0 (Init)	—	—	200 151 291 981 369 421 671
1	units	$0,1,1,1,9,1,1$	200 151 291 981 421 671 369
2	tens	$0,5,9,8,2,7,6$	200 421 151 369 671 981 291
3	hundreds	$2,4,1,3,6,9,2$	151 200 291 369 421 671 981

Stability check on pass 3: $200$ and $291$ both have hundreds digit $2$; pass 2 left $200$ before $291$, and the stable subsort keeps it ⟹ correct. An unstable pass could emit 151 291 200 … — wrong final answer.

Applied Exercise

Problem: Derive radix sort’s time and auxiliary space from Counting Sort’s $\Theta (N+M)$, then specialise to base-$10$ integers.

$$\begin{array}{rl}T& =\underset{\text{columns}}{\underbrace{K}}\times \underset{\text{one stable counting sort}}{\underbrace{\Theta (N+M)}}=\Theta (KN+KM)\\ S& =\underset{\text{input, }K\text{ symbols per item}}{\underbrace{\Theta (KN)}}+\underset{\text{one subsort, reused}}{\underbrace{\Theta (M+N)}}=\Theta (KN+M+N)\end{array}$$

Final Extracted Output: $\Theta (KN+KM)$ time, $\Theta (KN+M+N)$ total space, $\Theta (M+N)$ auxiliary — auxiliary carries no $K$, because the same count/position/output arrays are reused every pass rather than allocated per column. At $M=10$: $T=\Theta (KN)$, $S_{\text{aux}}=\Theta (N)$.

⚠️ Common Mistakes

• 💡 Multiplying $K$ into the auxiliary space ➔ the subsort arrays are reused across passes ⟹ auxiliary is $\Theta (M+N)$; only the input carries the $K$ factor.
• 💡 Reusing the two meanings of $M$ ➔ in Counting Sort $M$ is the maximum key; here $M$ is the base. Name which one you mean before quoting a bound.
• 💡 Calling radix sort $\Theta (N)$ with no conditions ➔ true only once you state that $K$ and $M$ are capped; distinct keys force $K=\Omega ({\log }_{M}N)$, recovering $\Theta (N\log N)$.
• 💡 Sizing the count array from max() ➔ that is Counting Sort’s rule; radix sizes it from the base and uses max only for the column count $c=\lfloor {\log }_{b}M\rfloor +1$. Lecturer-flagged as the recurring code-review error.

🧠 Active Recall

Why must radix sort's per-column subsort be stable, and why is LSD order the one that needs it?

• Hint: Ask what the previous passes have already established when a tie occurs.

Answer

• Short answer: ties on the current digit must fall back on the order from the lower-significance passes — stability is precisely that fallback.
• Why: Invariant ➔ after pass $j$ the list is sorted on the last $j$ digits; a stable pass $j+1$ preserves that order within each new digit group, so the invariant extends. An unstable pass discards it and no later pass can recover it.

Radix sort is $\Theta (KN)$ and merge sort $\Theta (N\log N)$ — is radix sort therefore always better?

• Hint: Ask how small $K$ can be when all $N$ keys are distinct.

Answer

• Short answer: no — $N$ distinct base-$M$ keys need $K\ge {\log }_{M}N$, so $\Theta (KN)$ degenerates to $\Theta (N\log N)$.
• Why: $K$ is only constant when the key width is capped ➔ fixed-width $32$-bit integers give $K=\Theta (1)$ and a genuine $\Theta (N)$; unbounded keys do not, and radix additionally requires integer-decomposable keys where Merge Sort accepts any orderable type.

You may choose the base freely. What choice makes radix sort provably $\Theta (N)$, and where does that stop working?

• Hint: $K$ and $M$ move in opposite directions; put a ceiling on the base.

Answer

• Short answer: choose $b=N$ — per-pass cost stays $\Theta (N+b)=\Theta (N)$ and $c=\lfloor {\log }_{N}M\rfloor +1$, so if the largest key is $O(N^d)$ for constant $d$ then $c=\Theta (1)$ and $T=\Theta (N)$.
• Why: Past $b=N$ the count array dominates ➔ the sort becomes linear in the base rather than in $N$; and for $M$ super-polynomial in $N$, $c$ grows with $N$ and linearity is lost from the other side.