How to Compare two multiplications without multiplying?

Question

How to check if two multiplications are equal to each other or greater or lesser without actually multiplying them?
For example, compare (254)(847) and (383)(536)

EDIT:
While trying to find a rule i got one
(5)(11) < (6)(10)
or
(x)(y) < (x+1)(y-1) when y > x > 0
and another rule is that if adding and subtracting 1 equates them the difference is one
(3)(5) + 1 = (4)(4)
(x)(y) + 1 = (x+1)(y-1) when y + 2 = x , y > x >= 0

Answer 1

Generally such comparisons can be done efficiently via continued fractions, e.g.

$$\rm\displaystyle a = \frac{847}{383}\ =\ 2 + \cfrac{1}{4 + \cfrac{1}{1 + \cdots}}\ \ \Rightarrow\ \ 2+\frac{1}{4+1} < a < 2+\frac{1}4$$

$$\rm\displaystyle b = \frac{536}{254}\ =\ 2 + \cfrac{1}{9 + \cdots}\ \ \Rightarrow\ \ 2 < b < 2 + \frac{1}9 < 2+\frac{1}5 < a$$

The comparison of the continued fraction coefficients can be done in parallel with the computation of the continued fraction. Namely, compare the integer parts. If they are unequal then that will determine the inequality. Otherwise, recurse on the (inverses of) the fractional parts (and note that inversion reverses the inequality). For the above example this yields:

$$\rm\displaystyle\ \frac{847}{383} > \frac{536}{254}\ \iff\ \frac{81}{383}>\frac{28}{254}\ \iff\ \frac{254}{28}>\frac{383}{81}\ \Leftarrow\ \ 9 > 4$$

In words: to test if $\rm\:847/383 > 536/254\: $ we first compare their integer parts (floor). They both have integer part $\:2\:$ so we subtract $\:2\:$ from both and reduce to comparing their fractional parts $\rm\ 81/383,\ \ 28/254\:.$ To do this we invert them and recurse. But since inversion reverses inequalities $\rm\ x < y\ \iff\ 1/y < 1/x\ $ (for $\rm\:xy>0\:$), the equivalent inequality to test is if $\rm\ 254/28 > 383/81\:.\ $ Comparing their integer parts $\rm\:m,\:n\:$ we see since $\rm\ m > 5 > n\:$ so the inequality is true. This leads to the following simple algorithm that essentially compares any two real numbers via the lex order of their continued fraction coefficients (note: it will not terminate if given two equal reals with infinite continued fraction).

$\rm compare\_reals\:(A,\: B)\ := \qquad\qquad\qquad\quad\ \ \color{blue}{\ // \ computes\ \ sgn(A - B) }$

$\rm\quad\ let\ [\ n_1\leftarrow \lfloor A\rfloor\:;\ \ \ n_2\leftarrow \lfloor B\rfloor\ ] \ \qquad\qquad\quad \color{blue}{\ //\ compare\ integer\ parts}$

$\rm\quad\quad if\ \ n_1 \ne n_2\ \ then\ \ return\ \ sgn(n_1 - n_2)\:;$

$\rm\quad\quad let\ [\ a \leftarrow A - n_1\:;\ \ \ b \leftarrow B - n_2\ ] \quad\quad\quad \color{blue}{//\ compute\ fractional\ parts\ \ a,\: b }$

$\rm\quad\quad\quad if\ \ a\:b=0\ \ then\ \ return\ \ sgn(a-b)\:;$

$\rm\quad\quad\quad compare\_reals(b^{-1}\:, a^{-1})\:;\qquad\qquad \color{blue}{\ //\ \text{recurse on inverses of fractional parts}}$

Equivalently one can employ Farey fractions and mediants. Generally such approaches will be quite efficient due to the best approximations properties of the algorithm. For a nice example see my post here using continued fractions to solve the old chestnut of comparing $\ 7^\sqrt 8\ $ to $\ 8^\sqrt 7\ $ and see also this thread where some folks struggled to prove this by calculus.

Answer 2

So, you want to compare (254)(847) and (383)(536) without actually computing the products, look at

$$\frac{847}{383} \; ? \; \frac{536}{254}$$

Multiplying both denominators by 2

$$\frac{847}{766} \; ? \; \frac{536}{508}$$

or

$$1+\frac{81}{766} \; ? \; 1+\frac{28}{508}$$

You are now left with comparing

$$\frac{81}{766} \; ? \; \frac{28}{508}$$

Applying the same trick as before, consider

$$\frac{81}{28} \; ? \; \frac{766}{508}$$

The left hand side is obviously bigger than $2$ while the right hand side is smaller, therefore, you can conclude that the original left hand side product was the bigger one, since none of the operations performed inverted the order of the inequalities.

Answer 3

Let $(a,b,c,d)$ be a quadruple of positive integers, and let us want to check if $ab=cd$, $ab>cd$ or $ab<cd$. The case when some of the numbers $a,b$ equals some of the numbers $c,d$ is trivial, so let us assume that $a\ne c$, $a\ne d$, $b\ne c$, $b\ne d$. If $a>c$ and $b>d$ then the situation is also trivial, and so is it in the case when $a<c$ and $b<d$. If $a>c$, but $b<d$, then $(a-c,b,c,d-b)$ is a quadruple of positive integers whose sum is less than $a+b+c+d$, and the equality $$(a-c)b-c(d-b)=ab-cd$$ holds. If $a<c$, but $b>d$, then $(a,b-d,c-a,d)$ is a quadruple of positive integers whose sum is less than $a+b+c+d$, and the equality $$a(b-d)-(c-a)d=ab-cd$$ holds. This obviously shows a solution of the problem by an algorithm using only comparing of positive integers and subtracting them in the case of positive difference. The algorithm can be somewhat improved by using that the following two cases are also trivial: the one of $a>d$, $b>c$, and the one of $a<d$, $b<c$ (another reduction similar to the above one can be performed if none of these two cases is present).

Example. Having in mind the initial question in this thread, let us apply the above-mentioned algorithm to the quadruple $(254,847,383,536)$. We get consecutively the quadruples $(254,311,129,536)$, $(125,311,129,225)$, $(125,86,4,225)$, $(121,86,4,139)$, $(117,86,4,53)$, and since $117>4$ and $86>53$, it follows that the product of $254$ and $847$ is greater than the product of $383$ and $536$.

Remark. One can get the result in the above example by means of several shorter sequences of quadruples if sometimes the other reduction possibility is used instead of the one which was described above. Here are some of these sequences:
$(254,311,129,536)$, $(125,311,129,225)$, $(125,86,4,225)$, $(125,82,4,100)$;
$(254,311,129,536)$, $(125,311,129,225)$, $(125,182,129,100)$;
$(254,311,129,536)$, $(254,182,129,282)$, $(125,182,129,100)$;
$(254,464,383,282)$, $(254,182,129,282)$, $(254,53,129,28)$.
The first of them would be actually produced by the transformations done in Arjang's answer from Dec 16'10 after the correction of an error made there (it must be 311 instead of 310).

Problem. Is it possible to design an algorithm using the same primitive operations which compares $x_1x_2...x_n$ and $y_1y_2...y_n$, whenever $n,x_1,x_2,...,x_n,y_1,y_2,...,y_n$ are positive integers?

Answer 4

In the worst case, where the comparison is close, you will always end up having to do as much work as multiplication. The performance gains from any non-worst-case improvements would have to be pretty good to make the programming effort worthwhile. Why do you need this?

Answer 5

Here is a simple example that does the job without any multiplication: $5 \times 12$ vs $6 \times 11$ , rewrite both sides as $5\times(11+1)$ and $(5+1)\times11$, then we get $5\times11+5\times1$ vs $5\times11 + 1 \times 11$ subtracting $5\times11$ from both sides we get : $1\times5 vs 1\times11$ which does not require any multiplication at all. The same technique can be used on (254)(847) and (383)(536). $ 254 \times 847 vs 383 \times 536 $
$ 254 \times ( 536 + 310 ) vs (254+129)\times536$
$ 254 \times 310 vs 129\times536$ continuing in similar fashion we end up with
$125\times 80 vs 4 \times 101$ although obvious but just for fun :
$ (101+24)\times(76+4) vs 4 \times 101$ where in the next line we get :
$ 4 \times 101 + 101\times 76 + 24\times76 + 24\times 4 vs 4\times 101 $

Answer 6

Use Russian peasant multiplication? Which just involves addition and left-shifting. Then you aren't actually doing any multiplication.