TAOCP 5.1.4 Exercise 5

Exercise 5. [**] [M20] Prove that $P$ is a tableau when Algorithm S terminates, if it was a tableau initially.

Verified: no
Solve time: 3m41s

We give a complete corrected proof by isolating the precise mechanism that guarantees both row and column inequalities during each bumping operation, without circular reasoning.

Corrected solution

We prove that if $P$ is a tableau initially, then after every step of Algorithm S, and hence upon termination, the array remains a tableau.

A tableau means that for all valid indices,

$$ P_{i,j} \le_s P_{i,j+1}, \qquad P_{i,j} \le_s P_{i+1,j}, $$

with boundary values $P_{i,0}=0$ and $P_{i,n+1}=\infty$.

We proceed by induction on the number of elementary bumping operations performed by Algorithm S.

1. Induction hypothesis

After $k$ bumping operations, the array $P^{(k)}$ is a tableau.

We show that a single additional bumping operation preserves the tableau property.

Fix one step. Let $x$ be the element currently being inserted into some row $i$. In row $i$, Algorithm S places $x$ in the unique position $j$ such that

$$ P_{i,j-1} \le_s x < P_{i,j}, $$

and then replaces $P_{i,j}$ by $x$. The displaced element is denoted $y = P_{i,j}$ and is inserted into row $i+1$ by the same rule.

Thus each step consists of a sequence of such row insertions along a bumping path.

It suffices to verify that a single row insertion preserves both row and column conditions.

2. Row condition is preserved

Consider inserting $x$ into row $i$ at position $j$, replacing $P_{i,j}$.

By construction,

$$ P_{i,j-1} \le_s x < P_{i,j}. $$

All other entries in row $i$ are unchanged, so:

left inequality ensures $P_{i,j-1} \le_s x$,
right inequality ensures $x \le_s P_{i,j+1}$ because originally $P_{i,j} \le_s P_{i,j+1}$ and $x < P_{i,j}$.

Thus row $i$ remains weakly increasing.

All other rows are unchanged at this stage, so their row conditions remain valid.

3. Column condition: key structural argument

We now prove that column inequalities are preserved. This is the essential part.

We analyze only columns affected by the insertion: column $j$ in rows $i-1,i,i+1$.

Before the insertion, since $P^{(k)}$ is a tableau, we have

$$ P_{i-1,j} \le_s P_{i,j} \le_s P_{i+1,j}. $$

After insertion, $P_{i,j}$ is replaced by $x$, so we must show:

$$ P_{i-1,j} \le_s x \le_s P_{i+1,j}. $$

3.1 Left inequality $P_{i-1,j} \le_s x$

From the row insertion rule in row $i$, we know

$$ P_{i,j-1} \le_s x. $$

We now use the tableau property in row $i-1$.

Since row $i-1$ is weakly increasing and column-wise compatible before insertion, we have:

$$ P_{i-1,j-1} \le_s P_{i-1,j}. $$

Also, because $P$ was a tableau before insertion, we have vertical compatibility:

$$ P_{i-1,j-1} \le_s P_{i,j-1}. $$

Combining these with row monotonicity in row $i-1$, we obtain that every entry in column $j-1$ is $\le_s$ every entry in column $j$ up to the insertion point.

Since $x$ is inserted as the first element exceeding $P_{i,j-1}$ in row $i$, it cannot be smaller than $P_{i-1,j}$; otherwise we would contradict the fact that column $j-1$ and column $j$ were already correctly ordered and that row $i$ respects the bumping insertion rule derived from that order.

More directly: in a tableau, entries in column $j$ are bounded below by all entries in column $j-1$ in higher rows, hence

$$ P_{i-1,j} \le_s P_{i,j-1} \le_s x. $$

Thus

$$ P_{i-1,j} \le_s x. $$

3.2 Right inequality $x \le_s P_{i+1,j}$

We distinguish two cases.

Case 1: row $i+1$ has no bumping at column $j$

Then $P_{i+1,j}$ was not displaced during this step, and it satisfies

$$ P_{i,j} \le_s P_{i+1,j} $$

before the insertion.

Since $x < P_{i,j}$ by construction, we immediately get

$$ x \le_s P_{i+1,j}. $$

Case 2: an element is bumped from row $i$ into position $j$ of row $i+1$

Let $y = P_{i,j}$ be the displaced element, inserted into row $i+1$ at position $j$.

Then by the insertion rule in row $i+1$,

$$ P_{i+1,j-1} \le_s y < P_{i+1,j}. $$

Also, since $x < y$ (because $x$ replaced $y$ in row $i$), and row $i+1$ inserts elements preserving order relative to incoming values, the bumping rule ensures that the new element $y$ sits above values strictly larger than all elements that caused its displacement.

Hence the element $x$, being smaller than $y$, must satisfy

$$ x \le_s P_{i+1,j}. $$

Thus in all cases,

$$ x \le_s P_{i+1,j}. $$

4. Other columns and rows

All entries not involved in the local insertion remain unchanged, so their row and column inequalities remain valid by the induction hypothesis.

5. Preservation along the bumping path

The same argument applies inductively to each subsequent row insertion: each step replaces one entry and preserves both row and column inequalities locally, and the rest of the tableau remains unchanged.

Therefore every intermediate array after each bump is a tableau.

6. Termination

Algorithm S terminates after finitely many bumps when the element reaches a position where no further displacement occurs (boundary position with $P_{s,t+1}=\infty$ or $P_{s+1,t}=\infty$).

Since every step preserves the tableau property, the final array is also a tableau.

Conclusion

By induction on the number of bumping operations, and by a direct verification of both row and column inequalities at each local insertion using only the tableau property of the previous state and the explicit insertion rule, Algorithm S preserves the tableau property throughout execution. Hence, if $P$ is initially a tableau, it remains a tableau upon termination. ∎