Thanks for reply.
I ran some artificial data for calculating WUW matrix, in matlab.
In the HTS codes, the 'wuw' matrix is set length*width, and width=2*max_width+1.
By default, width=3, so 'wuw' is a TX3 matrix.
By observing the data in matlab, I found that the number of non-zero elements in each row of WUW is 5, but the matrix is symmetrical, so is it the reason why 'wuw' was set to restore only the upper or lower block of the complete matrix WUW?
And was the Cholesky decomposition and following substitutions specially implemented for this data structure?
Thank you very much.
Best,
At 2013-02-15 01:25:47,"Heiga ZEN (Byung Ha CHUN)" <heigazen@xxxxxxxxxx> wrote:
>They are the same. However, to avoid confusions, I recommend you to use
>
>5 1.0 1.0 [-2.0] 1.0 0.0
>
>for this case.
>
>I personally think that it was my mistake to introduce the following code.
>
> if (fsize % 2 == 0)
> win->r_width[i]--;
>
>Regards,
>
>Heiga
>
>
>2013/2/14 那兴宇 <nxy-yzqs@xxxxxxx>:
>> Hi all,
>>
>> In current MLPG implementations, either HMGenS or hts_engine_API, the default dynamic window is symmetrical.
>> But I wonder what would happen if I set unsymmetrical window.
>> So I checked the HGen(in HTS-2.2) and HTS_model(in hts_engine_API-1.06), found similar codes below;
>> + /* set pointer */
>> + length = fsize / 2;
>> + win->coefficient[i] += length;
>> + win->l_width[i] = -length;
>> + win->r_width[i] = length;
>> + if (fsize % 2 == 0)
>> + win->r_width[i]--;
>>
>> I suppose it means I can set unsymmetrical window like
>> + 4 1.0 1.0 [-2.0] 1.0
>> with '-2.0' be the coeff. for current frame, right?
>> But what if I set
>> + 5 1.0 1.0 [-2.0] 1.0 0.0
>> as unsymmetrical window in a symmetrical form? Is it the same?
>> Please drop in any opinion.
>> Thank you.
>>
>> Best,
>> --
>> Xingyu Na (那兴宇)
>> http://www.idiap.ch/~xna/
>> Beijing Institute of Technology [naxy(at)bit.edu.cn]
>> Idiap Research Institute [xingyu.na(at)idiap.ch]
>>
>
>
>
>--
>Heiga ZEN (in Japanese)
>Byung Ha CHUN (in Korean)
><heigazen@xxxxxxxxxx>
>