Re: $20 Prize for Custom Show-Me

[Jim Osborn <jimo@xxxxxxxxxx>]

Richard Josslin <oldfogey@xxxxxxxxxx> asked:
>I offer $20 for the best submission (in my judgment) for Easy Language
>program code (for use in TS 3.5) for the following Show-Me study:
>
>1.  First, identifies (by a dot on the low of the bar) "Low Pivots": any bar
>whose Low is lower than the Lows of the two immediately adjacent bars to its
>left and the two immediately adjacent bars to its right, with the proviso
>that two immediately adjacent bars having identical Lows shall be considered
>to be one bar.
>
>2.  Second, once a Low Pivot has been identified, identifies (by a dot on
>the high of the bar) its "Second Higher High": the nearest bar to the left
>of the Low Pivot whose High is higher than the "First Higher High": the
>nearest bar to the left of the Low Pivot whose High is higher than the High
>of the Low Pivot.
>
>3.  Ditto for High Pivot and its Second Lower Low, using reciprocal rules.


The code below should do the trick:

{* Search for pivot, defined as a low with two higher lows on either side
 * then mark the Nth previous high that's higher than the high of the pivot bar.
 * The pivot may consist of two equal lows.  
 * Reciprocal specs apply to high pivot, and lower lows.
 * From an idea by Richard Josslin
 * jimo, Sun Feb 22 20:05:31 1998
 *}
input: Nth(2);                  { How many highs/lows to search for }
var: x(0), n(0),                { general index }
     break(0),                  { loop breaker }
     px(0),                     { index of leftmost pivot bar }
     ppx(0),                    { index of pivot bar with extreme High/Low }
     lastextreme(0);            { running extreme High/Low }

{*
 * Low pivot:
 *}
if L > L[1] and L[1] > L[2]     { Find pivot pattern }
 and ((L[2] < L[3] and L[3] < L[4])
   or (L[2] = L[3] and L[3] < L[4] and L[4] < L[5]))
then begin
    if L[2] = L[3] then begin   { Account for equal-low pivot bars }
        px = 3;
        if H[2] > H[3] then
            ppx = 2
        else 
            ppx = 3;
    end else begin
        px = 2;
        ppx = 2;
    end;
    lastextreme = H[ppx];       { Remember pivot High }

    x = px;                     { Search for Nth Highest High }
    break = 0;
    n = 1;
    while break = 0 and x < MaxBarsBack begin
        x = x + 1;
        if H[x] > lastextreme then begin
            lastextreme = H[x];
            n = n + 1;
            if n > Nth then
                break = 1;
        end;
    end;
    plot1[ppx](L[ppx], "Lopivot");
    if x < MaxBarsBack then
        plot2[x](H[x], "NthHigh");
end;

{*
 * High pivot:
 *}
if H < H[1] and H[1] < H[2]     { Find pivot pattern }
 and ((H[2] > H[3] and H[3] > H[4])
   or (H[2] = H[3] and H[3] > H[4] and H[4] > H[5]))
then begin
    if H[2] = H[3] then begin   { Account for equal-low pivot bars }
        px = 3;
        if L[2] < L[3] then
            ppx = 2
        else 
            ppx = 3;
    end else begin
        px = 2;
        ppx = 2;
    end;
    lastextreme = L[ppx];       { Remember pivot Low }

    x = px;                     { Search for Nth Lowest Low }
    break = 0;
    n = 1;
    while break = 0 and x < MaxBarsBack begin
        x = x + 1;
        if L[x] < lastextreme then begin
            lastextreme = L[x];
            n = n + 1;
            if n > Nth then
                break = 1;
        end;
    end;
    plot3[ppx](H[ppx], "Hipivot");
    if x < MaxBarsBack then
        plot4[x](L[x], "NthLow");
end;