> 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
Reply by Neo●May 6, 20052005-05-06
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 :(
Reply by ●May 5, 20052005-05-05
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...
Reply by Charles Krug●May 5, 20052005-05-05
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 you want to save cycles how about this?
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)
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
Reply by ●May 5, 20052005-05-05
"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
Reply by chris●May 5, 20052005-05-05
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
Reply by Neo●May 5, 20052005-05-05
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|).
Reply by ●May 5, 20052005-05-05
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