It won't work.
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1. convert the binary to many base 10 data blocks
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No need; a normal number, such as pi is suspected to be, has an equal distribution of digits in
all bases. You could do it in hexadecimal, base 256, or binary.
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if the size of a1+a2 >= the data block itself, abort the rest steps and start the next data block matching,
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Here's where it will fall down: in order to differentiate between raw data blocks and compressed data blocks, you will need to add additional bytes (or at the very least, bits). These will outweigh the miniscule savings you'll make even if you do happen to find matching blocks early enough in
pi (which in almost all cases you won't).