Stupid Columnsort Tricks Geeta Chaudhry Tom Cormen Dartmouth College Department of Computer Science

Columnsort  Sorts N numbers  Organized as r  s mesh  Divisibility restriction: s must divide r  Height restriction: r ≥ 2s 2  8 steps 1.Sort each column 2.Transpose 3.Sort each column 4.Untranspose 5.Sort each column 6.Shift down 1/2 column 7.Sort each column 8.Shift up 1/2 column

Proof of Correctness  Columnsort is oblivious  Use 0-1 Principle: If an oblivious algorithm sorts all input sets consisting solely of 0s and 1s, then it sorts all input sets with arbitrary values.  After step 3, the mesh consists of  Clean rows of 0s at the top  Clean rows of 1s at the bottom  ≤ s dirty rows between the clean rows

Proof of Correctness (continued)  After step 4, the mesh consists of  Clean columns of 0s on the left  Clean columns of 1s on the right  A dirty area of size ≤ s 2 between the clean columns  r ≥ 2s 2 ==> s 2 ≤ r/2 ==> the dirty area is at most 1/2 a column large

Proof of Correctness (continued)  If, entering step 5, the dirty area is at most 1/2 a column large, then steps 5–8 complete the sorting  If the dirty area fits in a single column, step 5 cleans it, and steps 6–8 leave the mesh clean  If the dirty area spans two columns, then it’s in the bottom half of one column and the top half of the next column.  Step 5 does not change this  Step 6 gets the dirty area into one column  Step 7 cleans it  Step 8 moves all values back to where they belong

Removing the Divisibility Restriction  Step 1: Sort each column  Each column has ≤ 1 0  1 transition  ≤ s 0  1 transitions  There may be a 1  0 transition going from one column to the next  ≤ s–1 1  0 transitions  Step 2: Transpose  Within rows, ≤ s 0  1 transitions, ≤ s–1 1  0 transitions

Divisibility Restriction (continued)  After step 2, let  X = dirty rows with one 0  1 transition, no 1  0  Y = dirty rows with one 1  0 transition, no 0  1  Z = all other dirty rows (≥ 1 0  1 and ≥ 1 1  0)  Number of dirty rows = |X| + |Y| + |Z|  All other rows are clean  Claim: max(|X|, |Y|) + |Z| ≤ s  Every row of X, Z contains ≥ 1 0  1 => |X| + |Z| ≤ s  Every row of Y, Z contains ≥ 1 1  0 => |Y| + |Z| ≤ s–1  max(|X|, |Y|) = |X| => max(|X|, |Y|) + |Z| ≤ s  max(|X|, |Y|) = |Y| => max(|X|, |Y|) + |Z| ≤ s–1  In either case, max(|X|, |Y|) + |Z| ≤ s

Divisibility Restriction (continued)  After step 3  Clean rows of 0s move to the top  Clean rows of 1s move to the bottom  Pair up the min(|X|, |Y|) pairs of rows with one row in X and other row in Y more 0s than 1s more 1s than 0s equal 0s and 1s from X: from Y:

Divisibility Restriction (continued)  In all cases, ≥ 1 clean row is formed  ≥ min(|X|, |Y|) new clean rows are created  Dirty rows remaining ≤ |X| + |Y| + |Z| – min(|X|, |Y|)  |X| + |Y| – min(|X|, |Y|) = max(|X|, |Y|) ==> Dirty rows remaining ≤ max(|X|, |Y|) + |Z| ≤ s  From here, it’s the same as the original proof  Dirty area size ≤ s 2 ≤ r/2 (half a column) after step 4  Steps 5–8 clean up the dirty area

Subblock Distribution  Divide up the mesh into s 1/2  s 1/2 subblocks  Each subblock contains s values  Add two steps between steps 3 and 4  Step 3.1: Perform any fixed permutation that moves all values in each subblock into all s columns  Step 3.2: Sort each column  The resulting algorithm is subblock columnsort  Works with relaxed height restriction of r ≥ 4s 3/2 (assuming the divisibility restriction)

Subblock Columnsort Correctness  After step 3, the line dividing 0s and 1s goes left to right and bottom to top (southwest to northeast)  Never turns back to the left  Never turns back toward the bottom  To show, suffices to show that after step 2, # of 0s in each column ≥ # of 0s in column to its right  How could a column have # of 0s < # of 0s in column to its right?  There would have to be a 1  0 transition in a row  But divisibility restriction => there are no 1  0 transitions in rows after step 2

Subblock Columnsort Correctness (continued)  After step 3.1, the number of 0s in any two columns differs by ≤ 2s 1/2  The dirty area is confined to an area s rows high and s columns wide  In subblocks, s 1/2  s 1/2  Dividing line passes through ≤ s 1/2 + 1 subblocks vertically and ≤ s 1/2 subblocks horizontally ==> ≤ 2s 1/2 subblocks total  Don’t double-count the 1 extra subblock  Step 3.1 distributes each subblock to all s columns  All clean subblocks distribute their 0s and 1s uniformly to each column  Number of 0s between any two columns differs by ≤ number of dirty subblocks = 2s 1/2

Subblock Columnsort Correctness (continued)  After step 4, the mesh consists of  Clean columns of 0s on the left  Clean columns of 1s on the right  A dirty area of size ≤ 2s 3/2 between the clean columns  After sorting in step 3.2, the dirty area is ≤ 2s 1/2 rows high and ≤ s columns wide  Size of dirty area is ≤ 2s 3/2

Subblock Columnsort Correctness (continued)  Finish up by observing that r ≥ 4s 3/2 is equivalent to 2s 3/2 ≤ r/2  Now the dirty area is at most 1/2 a column  Steps 5–8 clean up the dirty area  Can remove the divisibility restriction at the cost of tightening the height restriction to r ≥ 6s 3/2

Slabpose Columnsort  Another variation on columnsort  Loosens height restriction to r ≥ 4s 3/2  Has 10 steps  Partitions the mesh into vertical slabs

Conclusion  Removed and relaxed restrictions on columnsort  Divisibility restriction: s divides r  Height restriction: r ≥ 2s 2  Divisibility restriction is not necessary  Subblock columnsort  With divisibility restriction: r ≥ 4s 3/2  Without divisibility restriction: r ≥ 6s 3/2  Slabpose columnsort: r ≥ 4s 3/2