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Task 1 ( 2 P ) We have more than 3 2 but less than 6 3 characters, so we only need 6 bits instead

Task 1(2P) We have more than 32 but less than 63 characters, so we only need 6 bits instead of 8 bits to
represent all the characters. The first method thus reduces the size. Each character of 8 bits is replaced by a
character of 6 bits and all bits come directly after each other. Depending on the number of characters, this
may mean that not exactly one byte comes out at the end. We then simply fill it up with 0-en. The only case in
which we can have an ambiguity is if the last 6-bit word extends into the next byte with 2 bits, leaving exactly 6
bits. Then the last 6 bits are potentially another 6-bit word. However, if we make sure that the 0-bit does
not occur 6 times as a correct character, then we can recognize the case. To do this, we simply have to
start with 000001 instead of 000000 to exclude 000000 as a valid 6-bit word. We use the bit
patterns in ascending order as above, so a is equivalent to 000001 and so on. For example, the text abc
is used as the bit pattern 011000010110001001100011 to 000001000010000011000000, with
the spaces inserted after 8 bits for readability only. In the example, 60's have been inserted at the end
If you implement the procedure using a vector, you will learn how to handle bit sequences. We
compress with the -a flag and decompress with the -b flag. When compressing, we simply add the extension
.kompein to the file name if the output file is not specified. When decompressing, we use out.txt if
the output file name is not specified. We abort with an error message if the input is not expected.
explain this task further and more deeply

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