Proto-Euclidean algorithm - MathOverflow

Proto-Euclidean algorithm

2012-05-03T11:05:15Z

Consider the Euclidean algorithm (EA) as a way to measure the relative length $b/a$ of a shorter stick $b$ compared to a longer one $a$ by recursively determining

$$q_i = \left\lfloor \frac{r_i}{r_{i+1}} \right\rfloor\qquad (*)$$

$$r_{i+2} = r_i\bmod r_{i+1} $$

with $r_0 = a$, $r_1 = b$. The relative length $b/a$ is then given by the (finite or infinite) continued fraction

$$\cfrac{1}{q_0 + \cfrac{1}{q_1 + \cfrac{1}{q_2 + \cfrac{1}{\ddots }}}} =:\ [ q_0, q_1, q_2, \ldots ]^{-1}$$

A rather similar and somehow simpler algorithm is the following which I call proto-Euclidean algorithm (PEA):

$$q_i = \left\lfloor \frac{r_0}{r_{i+1}} \right\rfloor $$

$$r_{i+2} = r_0\bmod r_{i+1} $$

The relative length $b/a$ is then given by the (finite or infinite) continued product

$$\frac{1}{q_0}(1- \frac{1}{q_1}(1- \frac{1}{q_2}(1-\cdots))) =:\ \langle q_0, q_1, q_2, \ldots \rangle$$

[Update: The one and crucial difference between the two algorithms is the numerator in $(*)$ which represents the reference length against which the current "remainder" is measured: in EA it is adjusted in every step to the last "remainder", in PEA it is held fixed to $r_0$.]

For comparison’s sake, with $a=1071$, $b=462$ , the Euclidean algorithm yields

$$[2, 3, 7]^{-1} = \cfrac{1}{2 + \cfrac{1}{3 + \cfrac{1}{7}}} = \frac{22}{51} $$

while the proto-Euclidean algorithm yields

$$\langle2,7,25,51\rangle = \frac{1}{2}(1- \frac{1}{7}(1- \frac{1}{25}(1-\frac{1}{51}))) = \frac{22}{51} $$.

Under which name is the proto-Euclidean algorithm known? Where is it investigated and compared to the Euclidean algorithm? Or is it just folklore?

I am especially interested in the following questions:

How fast does PEA converge compared to EA?

(Just a side note: the first approximations in the sample above are equal: $[2, 3]^{-1} = \frac{3}{7} = \langle2,7\rangle $).

One advantage of EA over PEA seems to be that it takes fewer steps, and smaller numbers are involved in the course of calculation, since the numerator in $(*)$ decreases.

Is PEA significantly less efficient than EA?

Answer by Aeryk for Proto-Euclidean algorithm

2012-05-03T13:26:29Z

Your proto-Euclidean algorithm is basically equivalent to the Greedy algorithm for finding the alternating Egyptian fraction representation of a rational. For instance, in your example, if we expand the nested parentheses we get: $$\frac{22}{51} = \frac{1}{2}(1-\frac{1}{7}(1-\frac{1}{25}(1-\frac{1}{51}))) = \frac{1}{2}-\frac{1}{14}+\frac{1}{350}-\frac{1}{17850}$$

UPDATE: This is almost the Fibonacci-Sylvester Algorithm for finding Egyptian Fractions. The difference being that alternating signs between the fractions that the proto-Euclidean algorithm creates. I'm not sure how that affects the rate of convergence and such. You could probably eliminate the sign changes by choosing the signs on the $r_i$ and/or adding/subtracting 1 from each of them. (This is what I had to do for a similar project but can't remember which turned out to give the right answer.) Some heuristics on the F-S method can be found here.

UPDATE #2: Here's a more detailed explanation of the similarity. The Greedy/Fibonacci-Sylvester algorithm can be rephrased to look like a Euclidean-ish Algorithm. Here is the example above: $$ 51 = 3 \cdot 22 - 15$$ $$51 \cdot 3 = 11 \cdot 15 - 12$$ $$51 \cdot 3 \cdot 11 = 141\cdot 12-9$$ $$51\cdot 3 \cdot 11 \cdot 141 = 26367 \cdot 9 - 0$$ so the Greedy/F-S algorithm gives $$\frac{22}{51} = \frac{1}{3}+\frac{1}{11}+\frac{1}{141}+\frac{1}{26367}$$ So the Greedy/F-S algorithm for $a/b$ at the $n$th step is doing a modified division algorithm with $bq_1q_2q_3\cdots q_{n-1}$ as the dividend and $r_{n-1}$ as the divisor (where $q_i$ is the $i$th quotient and $r_i$ is the $i$th remainder) and the Egyptian fraction is given by $\sum 1/q_i$. I say "modified division algorithm" because instead of the usual $b=aq+r$, the $+$ is replaced by a $-$. In your PEA (I think), you just kept the plus.

This is why I conjecture that the heuristics and such are the same. It seems like for every long $a$, $b$ pair in the Greedy/F-S algorithm, there should be an analogous long $a$, $b$ pair for the PEA. I don't have anything at this time other than a gut feeling to back me up. Maybe I'll try to construct an example...

Answer by Aaron Meyerowitz for Proto-Euclidean algorithm

2012-05-03T18:03:31Z

Sometimes your method is much faster. For the golden ratio $\tau=\frac{1+\sqrt5}{2},$ the Euclidean algorithm gives all quotients $1$ so $[1,1,1,1,1,1,\cdots]$. Your method gives $<1, 2, 4, 17, 19, 5777, 5779, 192900153617, 192900153619, \cdots>$ where the terms after the first appear to come in pairs $\lceil \tau^{2\cdot3^j} \rceil-1,\lceil \tau^{2\cdot3^j} \rceil+1$.

So taking $b,a$ to be successive Fibonacci numbers can sometimes give a large advantage to your method. Actually a ratio of $\tau+1$ is slightly more dramatic. By my calculations $b,a=F_{53},F_{51}=86267571272, 32951280099$ gives $6$ terms $<2,4,17,19,5777,5779>$ vs $51$ terms $[2,1,1,\cdots,1,2]$.

At the other extreme, the Euclidean algorithm gives $[n-1,1,L-1]$ for $\frac{nL-1}{L}.$ It would appear that taking $L=\frac{\mathop{lcm}(1,2,\cdots,n)}{n}$ requires $n-2$ terms for your method. Hence with $n=12$ and $L=2310$ one has for $\frac{27719}{2310}$ the expansions

$[11,1,2309]$ vs $<11, 12, 2519, 2771, 3079, 3464, 3959, 4619, 5543, 6929>.$

Answer by Hans Stricker for Proto-Euclidean algorithm

2012-05-03T23:19:55Z

Let me point out that PEA is sometimes considerably "better" and more "to the point" than Fibonacci-Sylvester (see here):

By FS:

$$\frac{5}{91} = \frac{1}{19} + \frac{1}{433} + \frac{1}{249553} + \frac{1}{93414800161} + \frac{1}{17452649778145716451681}$$

$$\frac{5}{121} = \frac{1}{25} + \frac{1}{757} + \frac{1}{763309} + \frac{1}{873960180913} + \frac{1}{1527612795642093418846225}$$

By PEA:

$$\frac{5}{91} = \frac{1}{18} - \frac{1}{1638}$$

$$\frac{5}{121} = \frac{1}{24} - \frac{1}{2904}$$

The claim that PEA is "basically equivalent to the Greedy algorithm" which in turn is "almost the Fibonacci-Sylvester Algorithm" needs further explanation.