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

# Viterbi distance measure

Started by ●May 5, 2005

Reply by ●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

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 ●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 = 2What 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 ●May 5, 20052005-05-05

>>>>> "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

Reply by ●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)

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 ●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 6, 20052005-05-06

"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