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PROFESSOR: So now we're ready
to put together the facts

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that the gcd is a
linear combination

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to prove two cool
results-- one fun,

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and the other
important and serious.

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Let's begin with the
Die Hard example.

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So we looked at the
Die Hard state machine,

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and we figured out
the behavior of it

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with jugs of size
3 and 5 gallons,

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and also with jugs of
size 3 and 6 gallons.

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Let's look at the
general case now.

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Suppose that I have
jugs of a gallons

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and b gallons, where a and
b are positive integers.

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Now, when we looked
at the state machine,

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we figured out that
under the Die Hard rules,

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the number of gallons in
each bucket at any stage

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is a linear combination
of the bucket sizes.

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So at any point after any
sequence of moves of Die Hard

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moves, in each bucket there will
be a linear combination of a

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and b.

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Now, the point is that linear
combinations of a and b

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are the same as
multiples of the gcd.

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The reason is that the gcd is a
divisor of a and b, of course.

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It's a common divisor,
and therefore it

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divides any linear
combination of a and b.

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So any linear combination of a
and b is a multiple of the gcd,

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and the gcd is itself
a linear combination.

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So linear combinations
of a and b

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are the same as
multiples of gcd.

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So that gives us a
pretty good understanding

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of what the amounts that we can
get in the various buckets are.

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We can only get multiples
of gcd's, but in fact, you

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can get any multiple of the
gcd of a and b into a bucket,

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providing it will
fit in the bucket.

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That's the same
as saying that you

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can get any linear combination
amount of a and b into a bucket

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if there is room for
it in the bucket.

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So let's see how to do that.

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So suppose I have a linear
combination of a and b,

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sa plus tb, that
will fit in bucket

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b, meaning it's greater
than or equal to 0,

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and it's less than b.

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So it's a number of gallons
that could fit into bucket b.

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How do I get that
amount into bucket b?

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And here's how.

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We can assume that
s is positive.

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We've already seen that we can
arrange that to be the case.

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And so what we're going to do is
repeat the following procedure

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s times.

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I'm going to fill up bucket
a and pour it into bucket b.

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Whenever b gets
filled up, I'll just

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dump it so that it's empty, and
I can keep filling up bucket a

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and pouring it into bucket b.

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And I repeat that s times.

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Now, when I do that,
the total number

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of times that I've filled
bucket a is s times.

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So the total amount
of water that I

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have taken from the faucet,
or from the fountain,

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is s times a.

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And I've poured it
into b and then dumped

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it, leaving only some amount
that's in b that's less than b.

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So the amount that's left
after pouring in sa gallons

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and dumping out
what won't fit, I'm

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left with some amount that's
non-negative and less than b

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in bucket b.

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OK.

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Now, the point is that the
number of emptyings of bucket b

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must be exactly t, which is why
the amount of water that's left

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in bucket b is sa minus tb.

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And the reason why
it has to be minus t

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is that if I've got sa there,
if I had more than t emptyings,

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I would have had
bucket b go negative.

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There just isn't
enough room for it.

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And if I had fewer
than t emptyings,

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then the bucket would have an
amount larger than b in it.

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So the only possible number
of emptyings of b is minus t.

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Remember, t is negative, so
minus t is a positive number.

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And that means that I've
put in sa and taken out tb,

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and I'm left with exactly the
linear combination sa minus tb.

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So in fact, there's
no need to count,

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because you don't need
to know what s and t are.

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Because knowing that you can
get any desired amount that's

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a multiple of the
gcd into bucket b,

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you just keep doing
this process until you

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get the amount that you want.

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So you fill bucket a.

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You pour it into b.

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When b fills, you empty it.

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You just keep track of how many
gallons there are in bucket b,

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and you keep going until you
get the amount that you want.

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And then you're done.