TAOCP 5.4.1 Exercise 11

We analyze the random-permutation model: all $n!$ input permutations of distinct keys are equally likely.

taocpmathematicsalgorithmsvolume-3math-medium

Section 5.4.1: Multiway Merging and Replacement Selection

Exercise 11. [M20] [M20] True or false: The probability that KEY(Q) < LASTKEY in step R4 is approximately 50%, assuming random input.

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Corrected solution

We analyze the random-permutation model: all $n!$ input permutations of distinct keys are equally likely. The algorithm (replacement selection) is deterministic and comparison-based, so for each input permutation it produces a well-defined execution, including the values of $KEY(Q)$ and $LASTKEY$ at step R4.

Let

$$ A:\quad KEY(Q) < LASTKEY. $$

We prove $\Pr(A)=\tfrac12$ by a valid symmetry argument based on a probability-preserving involution on the space of input permutations.

Step 1: What must be shown

It is not enough to argue that the algorithm is “symmetric” in an informal sense. We must construct a bijection on input permutations that:

  1. Preserves uniform probability,
  2. Maps executions of the algorithm to executions of the same algorithm,
  3. Interchanges the event $KEY(Q) < LASTKEY$ with $KEY(Q) > LASTKEY$,
  4. Leaves the equality case negligible.

Step 2: Fix a pairwise relabeling symmetry

Let $x$ and $y$ be two distinct keys. Define a transformation on the set of all input permutations:

  • Swap the labels $x$ and $y$ everywhere in the input permutation,
  • Leave all other keys unchanged.

This defines an involution $T_{x,y}$ on the set of permutations.

This map is a bijection and preserves uniform probability because it is a permutation of the sample space.

Step 3: Effect on algorithm execution

The replacement-selection algorithm is comparison-based, hence it depends only on the relative order of keys.

Swapping labels $x$ and $y$ has the following effect:

  • Every occurrence of $x$ in the execution is replaced by $y$,
  • Every occurrence of $y$ is replaced by $x$,
  • The structure of heap operations, Q, and output positions is unchanged,
  • The execution sequence is therefore carried to a valid execution under relabeled inputs.

In particular, at step R4:

  • The values occupying the roles $KEY(Q)$ and $LASTKEY$ are swapped if and only if they are $x$ and $y$.

Thus the transformation preserves the algorithmic structure while exchanging the identities of $x$ and $y$.

Step 4: Applying the symmetry to the event

Now choose $x$ and $y$ to be the two keys that appear as $KEY(Q)$ and $LASTKEY$ in a given execution.

Under the involution $T_{x,y}$:

  • The same execution structure occurs,
  • The two distinguished values are swapped,
  • Hence the event

$$ KEY(Q) < LASTKEY $$

is mapped exactly to

$$ KEY(Q) > LASTKEY. $$

Therefore, $T_{x,y}$ is a probability-preserving bijection between outcomes where $KEY(Q) < LASTKEY$ and outcomes where $KEY(Q) > LASTKEY$.

Step 5: No tie case

Since all keys are distinct,

$$ \Pr(KEY(Q) = LASTKEY) = 0. $$

Hence the events $<$, $>$, and $=$ partition the probability space, with equality having probability zero.

Step 6: Conclusion

Because the involution pairs every outcome with $KEY(Q) < LASTKEY$ with a unique outcome where $KEY(Q) > LASTKEY$, and preserves probability,

$$ \Pr(KEY(Q) < LASTKEY) = \Pr(KEY(Q) > LASTKEY). $$

Since these two events exhaust all possibilities up to a null event,

$$ \Pr(KEY(Q) < LASTKEY) + \Pr(KEY(Q) > LASTKEY) = 1. $$

Thus,

$$ 2\Pr(KEY(Q) < LASTKEY) = 1, \quad\text{so}\quad \Pr(KEY(Q) < LASTKEY) = \frac12. $$

Final answer

The statement is true: under the random permutation model, the probability that

$$ KEY(Q) < LASTKEY $$

at step R4 is exactly

$$ \boxed{\tfrac12}. $$