# Viterbi distance measure

Started by May 5, 2005
```Hi,

Can someone help in proving/disproving the following?

if (a^2 > b^2)
then |a| > |b|.

can it be extended to say that

if (a^2 + b^2) > (c^2 + d^2)
then ( |a| + |b| ) > ( |c| + |d| )

This is to check if an euclidean distance measure can be replaced by an
absolute distance measure?

Thanks,
Raju

```
``` Now we can think of the argument as (|a|^2+|b|^2) > (|c|^2+|d|^2) =>
(|a|+|b|)^2 > (|c|+|d|)^2 + 2*(|ab|-|cd|) for all values of a,b,c,d.
therefore we can deduce that always
(|a|+|b|)^2 > (|c|+|d|)^2
now applying your first result we get
(|a|+|b|) > (|c|+|d|).

```
```Neo wrote:
> Now we can think of the argument as (|a|^2+|b|^2) > (|c|^2+|d|^2) =>
> (|a|+|b|)^2 > (|c|+|d|)^2 + 2*(|ab|-|cd|) for all values of a,b,c,d.
> therefore we can deduce that always
> (|a|+|b|)^2 > (|c|+|d|)^2
> now applying your first result we get
> (|a|+|b|) > (|c|+|d|).

On the other hand, consider

a = 3
b = 0
c = 2
d = 2

--
chris

```
```"chris" <thorpecp@yahoo.co.uk> writes:

> Neo wrote:
> > Now we can think of the argument as (|a|^2+|b|^2) > (|c|^2+|d|^2) =>
> > (|a|+|b|)^2 > (|c|+|d|)^2 + 2*(|ab|-|cd|) for all values of a,b,c,d.
> > therefore we can deduce that always
> > (|a|+|b|)^2 > (|c|+|d|)^2
> > now applying your first result we get
> > (|a|+|b|) > (|c|+|d|).
>
> On the other hand, consider
>
> a = 3
> b = 0
> c = 2
> d = 2

What if you were one-handed? ...

Nice counter-example, Chris.
--
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124
```
```>>>>> "Neo" == Neo  <zingafriend@yahoo.com> writes:

Neo>  Now we can think of the argument as (|a|^2+|b|^2) > (|c|^2+|d|^2) =>
Neo> (|a|+|b|)^2 > (|c|+|d|)^2 + 2*(|ab|-|cd|) for all values of a,b,c,d.
Neo> therefore we can deduce that always
Neo> (|a|+|b|)^2 > (|c|+|d|)^2
Neo> now applying your first result we get
Neo> (|a|+|b|) > (|c|+|d|).

a = 10, b = 0, c = 8, d = 5.

a^2 + b^2 = 100, c^2 + d^2 = 89

a+b = 10, c+d = 13

Ray
```
```On 5 May 2005 03:21:07 -0700, rajusr@sasken.com <rajusr@sasken.com> wrote:
> Hi,
>
> Can someone help in proving/disproving the following?
>
> if (a^2 > b^2)
>  then |a| > |b|.
>
> can it be extended to say that
>
> if (a^2 + b^2) > (c^2 + d^2)
>  then ( |a| + |b| ) > ( |c| + |d| )
>
> This is to check if an euclidean distance measure can be replaced by an
> absolute distance measure?
>

Infinitely many numeric counterexamples exist.

If sqrt(a^2 + b^2) > sqrt(c^2 + d^2)?

Then a^2 + b^2 > c^2 + d^2 (taking the positive square root)

```
```On the topic of Viterbi, if someone has an electronic version of "The
Viterbi Algorithm" by G.D Forney, IEEE Proceedings, 1973, I would
really appreciate if they would email me a copy. The address is
correct...

```
```Oh my, what a shameful blunder, I realized it when I though about it
after going home. serves me right for being so hasty. aah it hurts :(

```
```"Neo" <zingafriend@yahoo.com> writes:

> Oh my, what a shameful blunder, I realized it when I though about it
> after going home. serves me right for being so hasty. aah it hurts :(

I feel your pain, brother (sister?).
--
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124
```