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A least-significant-digit radix sort processes numerical digits from smallest to largest place value. Fix integers $R\ge 2$ and $d\ge 1$, and take $n$ complete sorting records whose sort keys satisfy $0\le k<R^d$. Use exactly $d$ positions in positional notation with radix $R$, including leading zeros. At pass $j=0,\dots ,d-1$, apply a stable digit sort to the current array usingThe floor function selects the integer quotient, and the remainder selects the current numerical digit. Move the complete sorting record, including its associated data. There are exactly $d$ passes, so this procedure terminates even for an empty array. The radix sort suffix invariant supplies its algorithm correctness argument.

For an original input, let the letters identify the attached data:Here $R=10$, $d=3$, and leading zeros only align place values; $005$ represents the integer five. The exact pass ledger is:After the units pass, the numerical digits are $0,2,4,5,5,7,7,7$. The ties retain $C$ before $F$ and $B$ before $E$ before $G$. This array is not yet ordered by its whole integer sort keys: $314$ still precedes $305$. After the tens pass, the two-position suffixes are $05,05,14,17,27,27,40,42$. In particular, $217_E$ now precedes $027_B$: their tens numerical digits differ, so stable sorting does not require retaining their previous order. The hundreds pass produces the final sorted array, with all attached letters preserved.

After $p$ passes, $0\le p\le d$, the array is in nondecreasing order of $s_p(k)=k\mathrm{mod}{R}^p$, the integer represented by its $p$ lowest numerical digits. Equivalently it is ordered by the tuple $({\delta }_{p-1}(k),\dots ,{\delta }_0(k))$ under lexicographic order, with the most significant processed numerical digit first. Among equal suffixes it retains the original order of the complete sorting records.

The loop invariant has a simple mathematical induction proof. Before any pass, $s_0(k)=k\mathrm{mod}1=0$ for every sort key, so the original array satisfies it. Suppose it holds after $p$ passes. The next stable digit sort puts smaller ${\delta }_p$ groups first and preserves the existing $s_p$ order within each tie. Sincethe new array is ordered by $s_{p+1}$. Equal new suffixes have equal current numerical digits and equal old suffixes, so their original order is also preserved. Every pass is a permutation of complete sorting records, establishing sorting permutation preservation. At termination $p=d$, the assumption $k<R^d$ gives $s_d(k)=k$: the invariant becomes whole-key sortedness, together with stable sorting.

A stable digit sort applies stable sorting to complete sorting records using one specified numerical digit of their encoded sort keys. Counting sort can supply it; stable record placement in counting sort moves records without discarding their associated data. Stability concerns equal current numerical digits, even when the full sort keys differ. This is the information-preservation step in least-significant-digit radix sort.^[ref]

An original two-record counterexample exposes the need for the tie rule. Start with $(42_A,41_B)$ at width three. A units pass gives $(041_B,042_A)$. Both tens numerical digits are four. A deliberately unstable tens pass may reverse their tie and return $(042_A,041_B)$: it is sorted by the tens numerical digit, yet its two-position suffixes $42,41$ are decreasing. The final stable hundreds pass sees two zeros and keeps this wrong order. Sorting permutation preservation still holds, but sortedness fails. Stability in the tens pass would have retained the units information.

With counting sort as the stable digit sort, each pass initializes $R$ counters, counts $n$ current numerical digits, constructs their prefix sums, and places $n$ complete sorting records. Thus each pass costs $O(n+R)$, and $d$ passes cost $O(d(n+R))$ in Big O notation. Reuse the counters and two arrays of record references between passes: auxiliary space complexity is $O(n+R)$, rather than $O(d(n+R))$. Assume constant-cost numerical digit extraction, counter arithmetic, array indexing and record-reference movement; the bound counts these operations, not the cost of copying arbitrarily large payloads or performing arbitrary-precision integer arithmetic.

The benefit is avoiding a counter for every possible whole sort key: width $d$ has $R^d$ possibilities, while each pass needs only $R$ counters. A larger radix reduces the number of passes for a fixed key range but enlarges the counter array. The comparison sorting lower bound does not apply to these direct array indices; they reveal more than a two-way comparison. Nor does fixing $R$ make $d$ disappear: growing key width still increases work.