TAOCP 5.2.2 Exercise 4

Section 5.2.2: Sorting by Exchanging

Exercise 4. [**] [M23] If $t = 1$ in step B4, we could actually terminate Algorithm B immediately, because the subsequent step B2 will do nothing useful. What is the probability that $t = 1$ will occur in step B4 when sorting a random permutation?

Verified: yes
Solve time: 6m17s

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The flaw in the previous solution is the attempt to characterize the event using the original suffix $R_2,\dots,R_n$ without tracking how a bubble sort pass changes the array after the first comparison. The correct analysis must follow the state of the algorithm.

Correct analysis

We consider one pass of Algorithm B (bubble sort). Let the initial permutation be

$$ R_1, R_2, \dots, R_n. $$

During the pass:

• At position $1$, we compare $R_1$ and $R_2$. A swap occurs if and only if $R_1 > R_2$.
• After this possible swap, the array state changes before later comparisons are made.

Let the event $E$ be that $t = 1$, meaning:

1. A swap occurs at position $1$,
2. No swaps occur at positions $2,3,\dots,n-1$.

Step 1: Effect of the first swap

Assume a swap occurs at position $1$. Then the array becomes:

$$ (R_2, R_1, R_3, \dots, R_n). $$

Now the comparison at position $2$ is between $R_1$ and $R_3$, not between $R_2$ and $R_3$. This is the key dependency missed in the flawed solution.

Step 2: Conditions for no further swaps

For no swap at position $2$, we require

$$ R_1 < R_3. $$

If this holds, then after step $2$, the element $R_1$ remains in position $2$, and all later comparisons involve the unchanged suffix $R_3, R_4, \dots, R_n$.

Thus, for no swaps at positions $3,4,\dots,n-1$, we must have

$$ R_3 < R_4 < \cdots < R_n. $$

So the full set of conditions for $t=1$ is:

$$ R_1 > R_2, \quad R_1 < R_3 < R_4 < \cdots < R_n. $$

Step 3: Structural consequences

From

$$ R_3 < R_4 < \cdots < R_n, $$

the elements $R_3,\dots,R_n$ are in increasing order and must therefore be exactly the $n-2$ largest elements of ${1,\dots,n}$, arranged in sorted order.

Thus the only freedom is in $R_1, R_2$.

From

$$ R_1 < R_3, $$

we see that $R_1$ is smaller than all elements in positions $3,\dots,n$, hence $R_1$ must be the second smallest element overall.

Since $R_1 > R_2$, $R_2$ must then be the smallest element.

Therefore the permutation is forced to be:

$$ (R_1, R_2, R_3, \dots, R_n) = (2,1,3,4,\dots,n). $$

No other configuration is possible.

Step 4: Counting

Exactly one permutation satisfies the condition $t=1$.

Since all $n!$ permutations are equally likely,

$$ \Pr(t=1) = \frac{1}{n!}. $$

Final answer

$$ \boxed{\Pr(t=1) = \frac{1}{n!}}. $$