CPTS350 HW1 – Bonus solution

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There are three things that we have learned in this week that are really
important:
(1). Input size. (All complexity functions are measured on input sizes
hence you need figure out input sizes correctly.)

(2). Input doesn’t take memory when it is read-only. Memory (in the
context of space complexity of an algorithm) refers to “additional” memory
needed.

(3). I keep saying that C-programs are algorithms but actually this is
not completely right. I say this only for those students who hate Turing machines. Algorithms are Turing machines that halt on all inputs.

Therefore,
If you are given an algorithm, you have many many C programs to implement it; some are good some are bad. However, if I give you a C program,
you usually don’t have many drastically different “programs” to implement
the C program (at most you use different compilers, but the assembly code
you get is roughly the same.)

Many people also think to make an algorithm any times faster (on almost
all inputs! not just a single one, not necessarily all inputs) is a mystery
(Blum’s speedup theorem). Of course, if you can make it two times faster,
you can make it any times faster. Now, let us look at an idea that makes
an algorithm two times faster.

For a natural number x represented in digits (so the size of input is
log10 x), we learned in our grammar school how to do arithmetic operations
like addition, subtraction, multiplication, etc., on those digits. Being in
digits, you are talking about alphabet {0, 1, 2, …, 9}. Because of this, we
were taught to do those arithmetics digit-by-digit.

Now, suppose that you
do two digits a time. Hence, in order to multiply
1234 × 5678
you do this
(12)(34) × (56)(78)
and now your alphabet is {00, …, 99} (if we humans have 100 fingers to begin
with, then we would do math in this way thousands years ago) — you might
have already noticed that the input size now is log100 x where x is the natural
number mentioned at the beginning).

This will make our multiplication at
least two times faster. Can you figure it out and sketch how we would do
this for n digit by n digit multiplication and why it is at least two times
faster? This is why Blum’s theorem works.