G'MIC exercises

Reptorian · January 27, 2020, 1:08am

EDIT: Solved it, but new problem. Random numbers, and the effect doesn’t show up.

f i0
repeat $! l[$>]
    tcr=3
    cmyk_mode=0
    if $cmyk_mode tcr+=1 fi
    if s==2||s>$tcr
    
        ss={s-1}
        {w},{h},1,1,i(#-1,x,y,z,$ss)?1
        
        eval ${-math_lib}"
            const ww=w#-1;
            const hh=h#-1;
            orientation=1;
            direction=1;
            continue=1;
            if(orientation,
                if(direction,
                    for(xx=0,xx<ww-1,xx++,
                        N=0;
                        v=0;
                        for(yy=hh-1,yy>-1,yy--,
                            if(yy==hh-1,
                                old_data=i(#-1,xx,yy);
                                for(tyy=hh-1,tyy>-1,tyy--,
                                N+=i(#-1,xx,tyy);
                                );
                                N/=hh;
                                if(N==1||N==0,break(););
                                if(!old_data,
                                    nyy=yy;
                                    do( v++;
                                        nyy--;
                                    ,(!i(#-1,xx,nyy))&&nyy>-1
                                    );
                                    i(#-1,xx,yy)=v;
                                );
                                
                            ,
                                new_data=i(#-1,xx,yy);
                                if(new_data<old_data,
                                    nyy=yy;
                                    do( v++;
                                        nyy--;
                                    ,(!i(#-1,xx,nyy))&&nyy>-1
                                    );
                                    i(#-1,xx,yy)=v;
                                );
                                i(#-1,xx,yy)=v;
                                old_data=new_data;
                            );
                        );
                    );
                ,
                    for(xx=0,xx<ww-1,xx++,
                        N=0;
                        v=0;
                        for(yy=0,yy<hh,yy++,
                            if(yy==0,
                                old_data=i(#-1,xx,yy);
                                for(tyy=0,tyy<hh,tyy++,
                                N+=i(#-1,xx,tyy);
                                );
                                N/=hh;
                                if(N==1||N==0,break(););
                                if(!old_data,
                                    nyy=yy;
                                    do( v++;
                                        nyy--;
                                    ,(!i(#-1,xx,nyy))&&nyy>-1
                                    );
                                    i(#-1,xx,yy)=v;
                                );
                            ,
                                new_data=i(#-1,xx,yy);
                                if(new_data<old_data,
                                    nyy=0;
                                    do( v++;
                                        nyy++;
                                    ,(!i(#-1,xx,nyy))&&nyy<hh
                                    );
                                );
                                i(#-1,xx,yy)=v;
                                old_data=new_data;
                            );
                        );
                    );
                );
            ,
                if(direction,
                    for(yy=0,yy<hh,yy++,
                        N=0;
                        v=0;
                        for(xx=ww-1,xx>-1,xx--,
                            if(xx==ww-1,
                                old_data=i(#-1,xx,yy);
                                for(txx=ww-1,txx>-1,txx--,
                                N+=i(#-1,txx,yy);
                                );
                                N/=ww;
                                if(N==1||N==0,break(););
                                if(!old_data,
                                    nxx=xx;
                                    do( v++;
                                        nxx--;
                                    ,(!i(#-1,nxx,yy))&&nxx>-1
                                    );
                                    i(#-1,xx,yy)=v;
                                );
                            ,
                                new_data=i(#-1,xx,yy);
                                if(new_data<old_data,
                                    nxx=0;
                                    do( v++;
                                        nxx--;
                                    ,(!i(#-1,nxx,yy))&&nxx>-1
                                    );
                                    i(#-1,xx,yy)=v;
                                );
                                i(#-1,xx,yy)=v;
                                old_data=new_data;
                            );
                        );
                    );    
                ,
                    for(yy=0,yy<hh,yy++,
                        N=0;
                        v=0;
                        for(xx=0,xx<ww,xx++,
                            if(xx==0,
                                old_data=i(#-1,xx,yy);
                                for(txx=0,txx<ww,txx++,
                                N+=i(#-1,txx,yy);
                                );
                                N/=ww;
                                if(N==1||N==0,break(););
                                if(!old_data,
                                    nxx=xx;
                                    do( v++;
                                        nxx++;
                                    ,(!i(#-1,nxx,yy))&&nxx<ww
                                    );
                                    i(#-1,xx,yy)=v;
                                );
                            ,
                                new_data=i(#-1,xx,yy);
                                if(new_data<old_data,
                                    nxx=0;
                                    do( v++;
                                        nxx--;
                                    ,(!i(#-1,nxx,yy))&&nxx>-1
                                    );
                                );
                                i(#-1,xx,yy)=v;
                                old_data=new_data;
                            );
                        );
                    );  
                );
            );
        "
    fi
    if $cmyk_mode tcr-=1 fi
endl done

afre · January 27, 2020, 2:44am

Could you make it more into a command? We have no understanding of the inputs and what to expect. The 1st version complained about cmyk_mode being unspecified. Now I am getting a grey scale image.

Reptorian · January 27, 2020, 2:48am

The only two inputs are orientation and direction. Those are what defines the end result. I can’t make it into a command unless this works.

What I want to do is something like this.

Step 1: Get the boolean values of alpha
1=greater than 0
0=0
Alpha values = 0,1,0,1,1,0,1,0,1
Step 2: From alpha values, for each time the new number is less than old number, use the amount of zeros next to the numbers. Add it up, and replace values accordingly.
New values = 1,1,2,2,2,3,3,4,4

afre · January 27, 2020, 2:54am

Your example is incomplete. I don’t know what to do with it…

But orientation is a direction… please elaborate.

Reptorian · February 28, 2020, 12:36am

Ok, how do I fix this? Only the two channels are modified with the distortion effect.

{w},{h},1,2,"
    begin(
    ww=w-1;
    hh=h-1;
    sd=max(w,h)/min(w,h);
    sx=w>h?sd:1;
    sy=w>h?1:sd;
    torus_val_1=1;
    torus_val_2=.5;
    new_min=min(torus_val_1,torus_val_2);
    new_max=max(torus_val_1,torus_val_2);
    old_max=max(1,0);
    old_min=min(1,0);
    nm(v)=((old_max-old_min)/(new_max-new_min))*(v-new_max)+new_max;
    oldcut(v)=v>old_max?old_max:(v<old_min?old_min:v);
    bndcut(v)=v>old_max||v<old_min?0:1;
);
xx=(x/ww-.5)*2*sx;
yy=(y/hh-.5)*2*sy;
radial_gradient=sqrt(xx^2+yy^2);
[oldcut(nm(radial_gradient)),bndcut(nm(radial_gradient))];
"
blur_angular. 1% 
n. 0,1
f. "begin(
dist=min(w,h)/2;
sd=max(w,h)/min(w,h);
sx=w>h?sd:1;
sy=w>h?1:sd;
ww=w-1;
hh=h-1;
max_ang=40;
ang2rad(v)=pi*(v/180);
rot_x(a,b,c)=a*cos(ang2rad(c))-b*sin(ang2rad(c));
rot_y(a,b,c)=a*sin(ang2rad(c))+b*cos(ang2rad(c));
softmode(v)=(cos(v*(2*pi)-pi)+1)/2;
midmode(v)=abs(cos(v*pi+pi/2));
hardmode(v)=sqrt(1-(abs(v-.5)*2)^2);
distroymode(v)=cos(v*pi)*i1;
invdistroymode(v)=(cos(v*pi)*-1)*i1;
);
z_depth=softmode(i0);
xx=(x/ww-.5)*2*sx;
yy=(y/hh-.5)*2*sy;
XX=rot_x(xx,yy,z_depth*max_ang);
YY=rot_y(xx,yy,z_depth*max_ang);
[(xx-XX)*dist,(yy-YY)*dist]
"
f[^-1] "j(i(#-1,x,y,z,0),i(#-1,x,y,z,1),z,c,1,1)"

The last line is what I need to fix for now. It appears that only two channels are modified for some reason.

EDIT: I used f[^-1] "i(x+i(#-1,x,y,z,0),y+i(#-1,x,y,z,0),z,c,2,2)" instead. It works now. It does however go out of bound, and I don’t know how to fix that like at all.

The whole point of this filter is to rotate the image, and use the torus map to define how much it gets rotated.

Another EDIT: If I take out mode in zdepth, the result and use 1 instead. The result is as expected. This is so odd because all the modes are always in 0-1 range. And yet if I keep mode. It doesn’t work as expected. Looks like my only real legit workaround is to create another layer as I don’t know what the heck is going on here.

EDIT: I fixed it, I just forgotten to change a number. SOLVED!

afre · March 7, 2020, 5:33pm

New exercise.

I have encountered more than one paper that requires calculations of distance within a neighbourhood. Does anyone (@Reptorian, @Iain, @David_Tschumperle?) have any recommendations on how to drive such algorithms in G’MIC? Well, starting with this example?

I am reading a paper that calculates the pixel weight as follows. Ln is the sum of distances between pixels along the path in the neighbourhood. In this case, red dot to yellow dot. It takes the min to the two routes of a corner. (I suppose all four corners would receive the same treatment for the kernel to be symmetrical… so min for 8 routes?) In Fig. 4, the min would be 0; i.e., the edge is too small for the kernel to detect.

Reptorian · March 7, 2020, 5:39pm

I’m not sure what is it your asking, but to find the distance between two pixels, you could use the hypotenuse route which is very easy. sqrt(x_length^2+y_length^2)

Now, if you’re asking to find the distance taking into account of pixels as blocks, that’s a bit more complicated as it’s not straight-forward as the hypotenuse route.

I would like you to clarify on this.

afre · March 7, 2020, 5:51pm

In Fig. 4, there are two routes, Lxypq1 is right and then down. If the range is [0,1], and the fig. only has white and black pixels, Lxypq1=2 because it encounters 2 edges. This can be calculated from the differences of every step added together: abs(0-1)+abs(1-2)+abs(2-3)+abs(3-4)+abs(4-5)+abs(5-6), the numbers being the ith pixel along the path, at least that is how I think of it. I am just having trouble seeing it in (G’MIC) code since I am not a programmer.

Reptorian · March 7, 2020, 6:01pm

Hold on, I think I actually start to understand what you want. You want to calculate the number of step it takes to get from one distance to another per axis, and blocks means 0?

afre · March 7, 2020, 6:07pm

It is simpler than you think. If I move from red dot one pixel, is there a change? No. Next pixel there is a change (or distance) of -1 but |-1|=1, etc. In total, there are two changes of 1, so Lxypq1=2. In the graph, there are two steps.

What I have so far, with my super poor coding skills, is

foo_d:
  (1,1,0,0;1,1,0,0;1,1,1,1;1,1,1,1) (0)
  f. "
    for(
      k=0,k<w#0,
      abs(i(#0,k++,0)-i(#0,k+++1,0))+
      abs(i(#0,w-1,k++)-i(#0,w-1,k+++1))
    )
  "

but I get L=1, which is incorrect.

Edit I am not looping properly. It looks like the calculations are inheriting the last values of the row and column.

Reptorian · March 7, 2020, 6:20pm

I was thinking more of this.

f "
begin(
    x_length=4;
    y_length=4;
    line_a_x(v)=(
        val=I(x,y,z,0,2);
        changes=0;
        for(n=1,n<v-1,n++,
            if(I(x+n,y,z,0,2)!=val,
            changes++;
            );
        );
        changes;
    );
);
line_a_x(x_length);
"

All you need to do is to add more macros, and then you have what you want.

Never mind. Thought of a way to minimize loop to find objects centers

afre · May 20, 2020, 7:09pm

Could someone explain how pattern works? E.g., polygon uses pattern.

polygon (+):
    N>=1,x1[%],y1[%],...,xN[%],yN[%],_opacity,_pattern,_color1,...

David_Tschumperle · May 20, 2020, 8:42pm

Pattern is an hexadecimal number with 8 digits, like 0xCCCCFFFF.
It has then 64 bits (0 or 1) that defines the pattern uses to draw the outline of the polygon.
Commonly used patterns are:

0xFFFFFFFF : plain line. Use it to draw an outlined polygon.
0xF0F0F0F0 : dashed line with 8px plain, 8px empty:
0xCCCCCCCC : dashed line with 2px plain, 2px empty:
0xAAAAAAAA: dashed line with 1px plain, 1px empty:

It’s quite flexible, as you can choose exactly what pattern you want.
For instance:

0xF99FF99F gives something like this:

afre · May 20, 2020, 9:17pm

Sorry, I have no idea how to decipher the hexadecimal value and how it contributes to the pattern. Could you please elaborate?

(Looks like more than 8 digits 0xccccccccc produces a solid line.)

David_Tschumperle · May 21, 2020, 3:59pm

You just write the sequence of 1 and 0 that defines your pattern (64 digits), and then use a binary to hexadecimal converter, as: https://www.rapidtables.com/convert/number/binary-to-hex.html![bin2hex|638x318](upload://iuD3o7yAr1VzYRVuG7HKvE4cZPH.png)

afre · July 6, 2020, 5:35am

To work on a neighbourhood of pixels (N), I would do

foo_k : check ${1=1}>=1
  f "begin(const boundary=1; const S=$1*2+1);
    ref(crop(x-$1,y-$1,0,c,S,S,1,1),N)"

How about a non-rectangular neighbourhood; e.g. an x shape?

xmarksthespot

David_Tschumperle · July 6, 2020, 7:11am

In that case, you have to define your own pixel extraction macro, e.g.

xcross() = (
  ref(vector5(),res);
  res[0] = j(-1,-1);
  res[1] = j(1,-1);
  res[2] = i;
  res[3] = j(-1,1);
  res[4] = j(1,1);
  res;
);

If needed, it is possible to write some commands to auto-generate macros that do something similar according to a certain shape. I’ve done something like this for command percentile, with:

  # Generate code for masking.
  pass$1 0 !=. 0 N={is} if !$N rm. return fi 128,$N
  eval.. ">
    begin(
      p = 0;
      const w2 = int(w/2);
      const h2 = int(h/2);
    );
    i?(
      out = string('N[',p,']=j(',x - w2,',',y - h2,');');
      copy(i(#-1,0,p++),out,size(out));
    )"
  discard. 0 code={t} rm[-2,-1]

Reptorian · July 11, 2020, 6:30pm

How exactly should I format this? To expand, let’s say there’s 6 variables next to formulamode with formulamode set to 1, and 4 variables next to formulamode with formulamode set to 0. How should I format it so that cli users can understand there are 4 variables next to formulamode if it is set to 4, and 6 if it is set to 1?

#@cli rep_popcorn_fractal  : _pts_per_pixels>0,0<_density<=1,_H,_K,_zoom>0,_rotation_angle,_formulamode

afre · July 11, 2020, 8:02pm

If you would like to go by convention, you have two choices.

1 Easy: make 2 commands; one per mode.
2 Medium: allow 2 sets of parameters (see resize).

#@cli resize : [image],_interpolation,_boundary_conditions,_ax,_ay,_az,_ac : \
# {[image_w] | width>0[%]},_{[image_h] | height>0[%]},_{[image_d] | depth>0[%]},\
# _{[image_s] | spectrum>0[%]},_interpolation,_boundary_conditions,_ax,_ay,_az,_ac : (+)

Reptorian · July 11, 2020, 8:51pm

@afre

This doesn’t look nice.

#@cli rep_pfrac : eq. to 'rep_popcorn_fractal' : (+)
#@cli rep_popcorn_fractal: _pts_per_pixels>0,_density>0,_H,_K,_zoom,_rotation_angle,_origin_x,_origin_y,_formulamode : \
# _pts_per_pixels>0,_density>0,_H,_K,_zoom,_rotation_angle,_origin_x,_origin_y,_formulamode=0, _formulafunc_a={ 0=sin | 1=cos | 2=tan | 3=atan},, _formulafunc_b={ 0=sin | 1=cos | 2=tan | 3=atan}, _formulafunc_c={ 0=sin | 1=cos | 2=tan | 3=atan},_formulafunc_d={ 0=sin | 1=cos | 2=tan | 3=atan} : \
# _pts_per_pixels>0,_density>0,_H,_K,_zoom,_rotation_angle,_origin_x,_origin_y,_formulamode=0, _formulafunc_a={ 0=sin | 1=cos | 2=tan | 3=atan},, _formulafunc_b={ 0=sin | 1=cos | 2=tan | 3=atan}, _formulafunc_c={ 0=sin | 1=cos | 2=tan | 3=atan},_formulafunc_d={ 0=sin | 1=cos | 2=tan | 3=atan},_formulafunc_e={ 0=sin | 1=cos | 2=tan | 3=atan},_formulafunc_d={ 0=sin | 1=cos | 2=tan | 3=atan}

Output:

    rep_pfrac: Shortcut for command 'rep_popcorn_fractal'.

    rep_popcorn_fractal:
        _pts_per_pixels>0,_density>0,_H,_K,_zoom,_rotation_angle,_origin_x,_origin_y,\
         _formulamode |
        _pts_per_pixels>0,_density>0,_H,_K,_zoom,_rotation_angle,_origin_x,_origin_y,\
         _formulamode=0, _formulafunc_a={ 0=sin | 1=cos | 2=tan | 3=atan},, \
         _formulafunc_b={ 0=sin | 1=cos | 2=tan | 3=atan}, _formulafunc_c={ 0=sin | \
         1=cos | 2=tan | 3=atan},_formulafunc_d={ 0=sin | 1=cos | 2=tan | 3=atan} |
        _pts_per_pixels>0,_density>0,_H,_K,_zoom,_rotation_angle,_origin_x,_origin_y,\
         _formulamode=0, _formulafunc_a={ 0=sin | 1=cos | 2=tan | 3=atan},, \
         _formulafunc_b={ 0=sin | 1=cos | 2=tan | 3=atan}, _formulafunc_c={ 0=sin | \
         1=cos | 2=tan | 3=atan},_formulafunc_d={ 0=sin | 1=cos | 2=tan | 3=atan},\
         _formulafunc_e={ 0=sin | 1=cos | 2=tan | 3=atan},_formulafunc_d={ 0=sin | 1=cos \
         | 2=tan | 3=atan}

In fact, I think this is a good solution with clarification added.

#@cli rep_pfrac : eq. to 'rep_popcorn_fractal' : (+)
rep_pfrac: rep_popcorn_fractal $*
#@cli rep_popcorn_fractal: _pts_per_pixels>0,_density>0,_H,_K,_zoom,_rotation_angle,_origin_x,_origin_y,_formulamoda,_formulafunc_1.._formulafunc_n

Output

   rep_popcorn_fractal:
        _pts_per_pixels>0,_density>0,_H,_K,_zoom,_rotation_angle,_origin_x,_origin_y,\
         _formulamoda,_formulafunc_1.._formulafunc_n