A near Young tableau is a filling of a Young diagram by distinct entries from a totally ordered alphabet, increasing along rows and down columns; the entries need not be . For row insertion of , scan the first row for its leftmost entry larger than . If one exists, replace it by and insert the displaced entry into the next row by the same rule. Otherwise append to the row and stop. Continue until a new outer-corner box is appended.
The Robinson–Schensted correspondence inserts the letters of a permutation successively to form . Whenever the th insertion creates a new cell, put in that cell of a second tableau . Row insertion preserves increasing rows and columns, so is standard; the sequence of growing diagrams makes standard, with the same shape. To reverse the construction, remove the box carrying the largest label in , and reverse the bumping in : in each row above, exchange the carried entry with the rightmost smaller entry, continuing up to the first row. The final expelled entry is the last letter of the permutation. Repeating recovers the whole permutation, establishing the bijection with pairs of standard Young tableaux of a common shape.
For the bumping inequality, each displaced entry is the first entry strictly larger than the incoming one. Thus every step replaces a larger entry with a smaller entry and carries that larger value downwards. Consequently the bumped values strictly increase:
Suppose an entry is bumped from column of row . If row already contains a cell in column , its entry is greater than by the strict column condition in the original near Young tableau. When is inserted into that row, the first entry greater than therefore occurs in a column at most .
If the next row is shorter than , either a bump occurs earlier or is appended in column . This also handles the final appended box. Each successive bumping position is therefore weakly to the left of the previous one. The bumping columns satisfyThis is the row-insertion bumping-path monotonicity needed to preserve the Young-diagram shape and column order.
Every old box either remains untouched or has its entry replaced by a strictly smaller incoming value. Thus on all boxes of the original diagram,The appended box has no old entry to compare; equivalently, extend the old filling by outside its diagram if the inequality is to be written for every position.
We now prove the unheaded longest increasing subsequence assertion. After any prefix of the permutation has been inserted, write the first-row entries as . The useful invariant is: is the smallest possible final value of an increasing subsequence of length in that prefix, and is the greatest possible subsequence length. It holds for the empty prefix.
Suppose the next letter enters the first row in column . Then , if , while if column already exists. A subsequence of length ending at can be extended by , giving a subsequence ending at of length ; for , the singleton suffices. A longer subsequence ending at would have an earlier length- prefix ending below , contradicting the minimality of . If , the previous prefix has no subsequence longer than , giving the same upper bound.
Replacing by , or appending it at , preserves the invariant. It lowers the smallest final value at length . At smaller lengths, existing minima are already below , so they cannot improve; no subsequence of greater length can end at . The lower rows do not change this first-row update. Induction proves that the longest increasing subsequence ending in has length exactly the column in which first enters.
Articles by others on the same topic
There are currently no matching articles.