// Last changed 1/27/2003 at 10:20am
// Assume that the program was invoked with 10 slots in the sequence
// array (indexed from 0 to 9).

insert 2 DNA AGG

// Now we have a sequence inserted

  insert 3   DNA  CUP

// This is an error, since P is not a legal letter in the DNA alphabet

print

// Print out all the sequences in the sequence array.
//
// 2 DNA AGG

print 2

// Print out only the sequence in position 2 of the sequence array
//
// 2 DNA AGG

remove 2

// Remove the sequence in slot 2

print

// There is nothing left... should print a message to this effect

insert 1  RNA ACCGA

// Now we have one sequence, at position 1

    copy   1 0

// Now we have a copy in position 0.

clip  1 2 4

// We now replace ACCGA with CGA

clip 1 1

// We now replace CGA with GA

print

// 0 RNA ACCGA
// 1 RNA GA

insert 3 RNA AAAC

// Now we have 3 sequences, at positions 0, 1, and 3

print

// 0 RNA ACCGA
// 1 RNA GA
// 3 RNA AAAC

swap 1 2 3 1

// Position 2 in the sequence GA is the null, and position 1 in
// the sequence AAAC is AAC. So the result is shown by the next print
// statement.

print

// 0 RNA ACCGA
// 1 RNA GAAAC
// 3 RNA A

overlap 0 1

// The best overlap is GA at the end of 0 and at the beginning of 1.

insert 3 DNA CAAGT

// This replaces the current sequence at position 3

print

// 0 RNA ACCGA
// 1 RNA GAAAC
// 3 DNA CAAGT

// Be careful, the blank line that comes next has some spaces in it.

transcribe 3

// We see the result of transcribing 3 in the next print statement

print 3

// 3 RNA ACUUG

// Be sure that you deal with the fact that the input file has a
// line at the end that contains only a space.