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Some supporting mathematical notes
Note 1. The Greeks based their measurement upon fractions of the form
m/n, where m and n are integer numbers. Greatest common divisors are
important in comparing fractions: if m=Md and n=Nd, then M/N is the
fraction m/n "in its lowest terms".
Note 2. The sequence of numbers F[0], F[1], F[2],..., F[n] where F[k+1] =
F[k]+F[k-1] and F[0] = F[1] = 1, is named after Fibonacci, an Italian
mathematician who introduced it in 1202. The first few terms of the
sequence are:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...
We can derive a formula for the k-th Fibonacci number:
(a) find the roots, ( ('alpha') and ( ('beta') of the quadratic
equation x2 = x + 1
(b) find values for A and B so that A + B = A( + B( = 1
(c) F[k] can then be expressed as A(k + B(k
You can prove this by checking the following claims:
(1) Step (a) computes two numbers x such that xk+1 = xk + xk-1
(2) If A and B are any numbers, the sequence defined by f[k] = A(k
+ B(k satisfies
f[k+1] = f[k] + f[k-1]
(3) Step (b) computes A and B so that f[0] = f[1] = 1, and so
defines the Fibonacci sequence.
We can use the results of (a), (b) and (c) to calculate the k-th
Fibonacci number using tkeden:
/* the roots of the quadratic in (a) are alpha and beta */
alpha is (1 + sqrt(5))/2;
beta is (1 - sqrt(5))/2;
/* sqrt = square root */
/* the solutions to the equations in (b) are A and B */
B is (alpha - 1) / (alpha - beta);
A is (1 - beta) / (alpha - beta);
/* the formula for F[k] in (c) defines the k-th Fibonacci number */
fibk is int(A*pow(alpha, k)+B*pow(beta, k));
/* pow(x,n) returns 'x raised to the power n', int() rounds to an integer
*/
We can then check claim (1) by direct computation:
/* check that sequence pow(alpha,k) satisfies the Fibonacci recurrence:
pa1 is pow(alpha,1); pa2 is pow(alpha,2);
pa3 is pow(alpha,3); pa4 is pow(alpha,4);
pa12 is pa1+pa2; pa23 is pa2+pa3; pa34 is pa3+pa4;
*/
We can also confirm that computing the GCD of F[k+1] and F[k] leads to a
large number of division steps:
fibk is int(A*pow(alpha, k)+B*pow(beta, k));
fibkplus1 is int(A*pow(alpha, k+1)+B*pow(beta, k+1));
A2 is fibkplus1; B2 is fibk;
/* substitute different values for k to test the results */
k
k
k
k
=
=
=
=
4;
20;
23;
25;
Note 3. There is a link between the JUGS model and GCD computation:
A mathematical fact ...
gcd(m,n) is the smallest positive integer of the form Mm+Nn where M and N
are integer values
Proof: Let x be the smallest positive integer expressible as Mm+Nn. If we
divide n by x, we have an integer quotient Q and integer remainder r,
where 0<=r<x:
r = n - Qx = n - Q(Mm+Nn) = -QMm + (1-QN)n
This means that r is less than x, and is also expressible in the form
Mm+Nn. Since x is the smallest positive integer expressible in this form,
r = 0, and x divides n.
By a similar argument, x divides m. So x is a common divisor of m and n.
But if any number y divides m and n it also divides x = Mm+Nn. So x is
the greatest common divisor of m and n.
How this links to the JUGS model... Suppose that we have jugs Jug A and
Jug B of capacities a and b respectively, where a ( b. There are two
basic routines for manipulating the jugs:
(1)
Fill the Jug B
repeat {
Pour from Jug B to Jug A
Empty Jug A if Jug B is not empty
}
(2)
repeat {
if Jug B is not full {
Fill the Jug A
Pour from the Jug A to the Jug B
}
}
If p mod q denotes the remainder on dividing p by q, these routines are
respectively equivalent to:
computing b mod a, then (b - (a - b mod a)) mod a = 2b mod a etc
and
computing a - b mod a, then (a - (b - (a - b mod a)) mod a = -2b
mod a etc
Since gcd(a,b) is of the form Aa+Bb, and is less than a, it is
expressible as Bb mod a for some B. Hence the quantity of liquid gcd(a,
b) can be obtained using jugs of size a and b.