TAOCP 5.4.7 Exercise 5

Exercise 5. [15] [15] Draw the 4-lifo tree corresponding to Mauchly’s read-backwards radix sort for 10 keys.

[20] A certain file contains two-digit keys 00, 01, ..., 99. After performing Mauchly’s radix sort on the least significant digits, we can repeat the same scheme on the most significant digits, interchanging the roles of tapes T2 and T4. In what order will the keys finally appear on T2?

Verified: no
Solve time: 3m47s

The original submission contains no construction or argument, so the solution must be rebuilt from the definitions of Mauchly’s read-backwards radix sort and the 4-LIFO representation used in Section 5.4.7.

5. The 4-LIFO tree for Mauchly’s read-backwards radix sort on 10 keys

In Mauchly’s read-backwards radix sort, the “tree” is not a binary search tree but a 4-LIFO structure that records how items are repeatedly distributed among four tapes (or stacks) $T_1, T_2, T_3, T_4$ in cyclic LIFO fashion while processing digits from least significant to most significant.

The construction for 10 keys depends only on the cyclic dealing pattern in the initial (least significant digit) distribution phase. With 4 tapes, keys are dealt in order, cycling through tapes:

$$ T_1, T_2, T_3, T_4, T_1, T_2, \dots $$

Assuming the keys are $1,2,\dots,10$, the distribution is:

$T_1:; 1,5,9$
$T_2:; 2,6,10$
$T_3:; 3,7$
$T_4:; 4,8$

Because each tape operates in LIFO order, each tape corresponds to a stack, and the 4-LIFO tree is the forest of four downward paths (each path representing a stack growing upward from its first insertion).

Thus the 4-LIFO tree can be described as four rooted chains:

$T_1:; 1 \rightarrow 5 \rightarrow 9$
$T_2:; 2 \rightarrow 6 \rightarrow 10$
$T_3:; 3 \rightarrow 7$
$T_4:; 4 \rightarrow 8$

Interpreting these as a single conceptual tree with a common origin, the root branches into four LIFO chains, each chain ordered by reverse insertion time (top of stack last).

This is the required 4-LIFO structure: a 4-way branching root with four descending LIFO paths determined by cyclic distribution.

6. Final order of two-digit keys after LSD then MSD passes with tape interchange

We now consider keys $00,01,\dots,99$. The algorithm performs:

A radix pass on the least significant digit (LSD).
A second radix pass on the most significant digit (MSD), with the roles of tapes $T_2$ and $T_4$ interchanged.

First pass (LSD)

After the first pass, keys are stably grouped by their last digit. This produces 10 ordered buckets:

$$ (00,10,20,\dots,90),\ (01,11,21,\dots,91),\ \dots,\ (09,19,\dots,99) $$

Concatenated in increasing order of the last digit.

This ordering is stable with respect to the LSD.

Second pass (MSD with $T_2 \leftrightarrow T_4$)

The second pass groups by the most significant digit, but stability is preserved because:

Each MSD group (00–09, 10–19, etc.) is already internally ordered by LSD.
Swapping $T_2$ and $T_4$ only changes which physical tape acts as output/input during merging, not the stability of ordering within digit groups.

Thus the MSD pass performs a stable radix refinement of already stably ordered buckets, yielding full lexicographic ordering by:

$$ (\text{MSD}, \text{LSD}) $$

Final result

The keys therefore appear on $T_2$ in increasing numerical order:

$$ 00, 01, 02, \dots, 99 $$

Conclusion

The 4-LIFO tree for 10 keys consists of four LIFO chains induced by cyclic distribution among four tapes.
The two-pass radix procedure remains stable under the tape interchange and produces the standard lexicographic order of all two-digit keys.

$$ \boxed{00, 01, 02, \dots, 99} $$