# importance of alternate storage in external sort

I'm trying to understand how external sorting works from the book "Data Structures and Algorithm Analysis" by Mark Allen Weiss.

The author illustrates an example where he tries to sort 81,94,11,96,12,35,17,99,28,58,41 with a memory which can hold upto M=3 numbers. He uses two input tapes Ta1,Ta2 and two output tapes Tb1,Tb2.

Ta1    81,94,11,96,12,35,17,99,28,58,41
Ta2
Tb1
Tb2


In the first step M numbers are read into memory then, sorted and written to Tb1 and Tb2 alternatively.

Ta1
Ta2
Tb1    11,81,94   17,28,99   15
Tb2    12,35,96   41,58,75


I understand how the algorithm works but, don't understand why it is important to write results alternatively. Couldn't this be achieved without alternating?

• I understand how the algorithm works please enlighten me how it arrives at 75 and 15. – greybeard Feb 21 '17 at 6:27