Diminishing Gap Sort

August 19, 2016

Today’s exercise comes from my friend Dan:

I don’t have time to think about this now (very busy these days), so I’m just throwing the idea at you. I’m sure there’s some sort of mathematical theorem that relates, but I’m sure you’ll probably know off the top of your head.

I was fascinated not too long ago by the consulting firm we’ve hired to re-vamp our website. I think they were given a tip-off that the color selection for the website theme could be a sticky wicket. So their plan was to pull the colors dynamically from the main picture selected on the home page (and/or departmental pages), which could be changed easily or automatically. They were going to develop something that would pick out the most prominent 6 to 10 colors — even if there wasn’t much contrast (and even black & white photos would supposedly work) — which would be then used for the menus and general color scheme.

Anyway, all this got me thinking about sorting a list of numbers (if these were color codes, there could be lots) such that the greatest gaps would float to the top. I’m sure you get the idea, but I’ll give an example anyway — which I figured out in Excel (below):

Given a number set of 5, 12, 14, 15, 80, 121, 134, 144, 256 the descending-by-highest-difference sort would yield: 256, 80, 121, 134, 144, 12, 14, 15, 5. So in this case, the top 3 biggest gaps would be between 80, 121, and 256.
     Num  Diff              Num  Diff
    ----  ----             ----  ----
     256   112              256   112
     144    10               80    65
     134    13              121    41
     121    41              134    13
      80    65              144    10
      15     1               12     7
      14     2                5     5
      12     7               14     2
       5     5               15     1
I have no idea if this makes sense for the colors thing — it just seemed like the type of problem you frequently do on Programming Praxis.

Hope all is well with you and your family!

-Dan-

Your task is to write a program that sorts a set of points in order by the diminishing gaps between them. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

Posted by programmingpraxis

Filed in Exercises

8 Comments »

8 Responses to “Diminishing Gap Sort”

-Dan- said

August 19, 2016 at 8:32 PM

Ok, Phil. We had a major VMWare outage this afternoon that impacted a bunch of stuff, so I was C-drive-bound for a while (plus your email shamed me into writing this). When I was finally able to connect to our databases, we still had no shared drives and the application servers were mostly out to lunch. So I decided to see if I could implement the color-sort idea, with red/green/blue values – 3 values instead of just one. I sorted by red/green/blue, but I needed to get the absolute values of the aggregated gaps, since the sort yielded green and blue values going up and down. (I don’t know if that sort makes sense color-wise, but it seems logical to me.) I also took a minute to research the “Gnome sort” – or “Stupid sort”, which sounded like something made for me! So that’s what I implemented here (I can’t remember the last time I actually coded a sort algorithm!). Oracle PL/SQL code and output pasted below, which I hope is formatted so it’s readable. Like when we worked together, I fully expect you to find a bug or recommend an improvement! Take care, Phil.

declare
    type val_rec is record (
        red     number(3),
        green   number(3),
        blue    number(3),
        gap     number(6));
    type tblnum_type is table of val_rec index by pls_integer;
    tbl    tblnum_type;
    rectmp val_rec;
    pos    pls_integer;
    diff   pls_integer;

    procedure prttbl(hdg in varchar2) is
    begin
        dbms_output.put_line(chr(10)||'*** '||hdg);
        dbms_output.put_line('red green  blue   gap');
        dbms_output.put_line('--- -----  ----   ---');
        for i in 1..tbl.count loop
            dbms_output.put_line(
                to_char(tbl(i).red,'fm000')||
                '   '||to_char(tbl(i).green,'fm000')||
                '   '||to_char(tbl(i).blue,'fm000')||
                '   '||to_char(tbl(i).gap,'fm000'));
        end loop;
    end prttbl;

begin
    dbms_output.enable(1000000);

    -- randomly generate colors
    for i in 1..10 loop
        tbl(tbl.count+1).red := trunc(dbms_random.value(0, 256));
        tbl(tbl.count).green := trunc(dbms_random.value(0, 256));
        tbl(tbl.count).blue  := trunc(dbms_random.value(0, 256));
        tbl(tbl.count).gap   := 0;
    end loop;
    -- seed bottom value (assume black always included)
    tbl(tbl.count+1).red := 0;
    tbl(tbl.count).green := 0;
    tbl(tbl.count).blue  := 0;
    tbl(tbl.count).gap   := 0;
    prttbl('initial state');

    -- sort by red, green, blue
    pos := 1;
    loop
        exit when pos >= tbl.count;
        if pos = 1 or (to_char(tbl(pos-1).red,'fm000')||to_char(tbl(pos-1).green,'fm000')||to_char(tbl(pos-1).blue,'fm000') >=
                       to_char(tbl(pos).red,'fm000')||to_char(tbl(pos).green,'fm000')||to_char(tbl(pos).blue,'fm000')) then
            pos := pos+1;
        else
            rectmp := tbl(pos);
            tbl(pos) := tbl(pos-1);
            tbl(pos-1) := rectmp;
            pos := pos-1;
        end if;
    end loop;

    -- calculate gaps
    for i in 1..tbl.count-1 loop
        tbl(i).gap := (abs(tbl(i).red-tbl(i+1).red) + abs(tbl(i).green-tbl(i+1).green) + abs(tbl(i).blue-tbl(i+1).blue));
    end loop;
    prttbl('sorted by red, green, blue with gaps calculated');

    -- diminishing gap sort
    pos := 1;
    loop
        exit when pos >= tbl.count;
        if pos = 1 or tbl(pos-1).gap >= tbl(pos).gap then
            pos := pos+1;
        else
            rectmp := tbl(pos);
            tbl(pos) := tbl(pos-1);
            tbl(pos-1) := rectmp;
            pos := pos-1;
        end if;
    end loop;
    prttbl('sorted by gap');
end;

Well, that turned out to be a bit more fun than I expected! Your straight-SQL solution is (as usual) both elegant and straight-forward. Here’s my output (I always like to see the starting point, interim state, and final results!):

*** initial state
red green  blue   gap
--- -----  ----   ---
153   251   231   000
049   117   172   000
041   202   192   000
226   189   051   000
122   044   103   000
217   239   017   000
052   149   102   000
011   072   196   000
140   151   070   000
105   243   027   000
000   000   000   000

*** sorted by red, green, blue with gaps calculated
red green  blue   gap
--- -----  ----   ---
226   189   051   093
217   239   017   290
153   251   231   274
140   151   070   158
122   044   103   292
105   243   027   222
052   149   102   105
049   117   172   113
041   202   192   164
011   072   196   279
000   000   000   000

*** sorted by gap
red green  blue   gap
--- -----  ----   ---
122   044   103   292
217   239   017   290
011   072   196   279
153   251   231   274
105   243   027   222
041   202   192   164
140   151   070   158
049   117   172   113
052   149   102   105
226   189   051   093
000   000   000   000

Rainer said

August 20, 2016 at 6:09 AM

with data (n) as (values                            
(256),(144),(134),(121),(80),(15),(14),(12),(5)),   
     neighbour (n,next) as                          
(select d.n, coalesce((select max(x.n)              
                       from   data x                
                       where  x.n < d.n), 0) as nxt 
 from   data d)                                     
select n,n - next as diff                           
from   neighbour                                    
order  by diff desc                                 

==>

  N            DIFF
256             112
 80              65
121              41
134              13
144              10
 12               7
  5               5
 14               2
 15               1

Milbrae said

August 21, 2016 at 2:21 PM

In D…probably not suitable for large arrays due to Bubble Sort

import std.algorithm;
import std.stdio;

void main()
{
    uint i, len;
    
    /* create arrays, sort, and get diffs */
    uint[] num = [5, 12, 14, 15, 80, 121, 134, 144, 256];
    num.sort();
    len = cast(uint)num.length;

    uint[] diff = new uint[](num.length);
    diff[0] = num[0];
    for (i = 1; i < len; i++) {
        diff[i] = num[i] - num[i-1];
    }

    writefln("Num:  %s", num);
    writefln("Diff: %s", diff);
    writeln;

    /* slow bubble sort of both arrays in parallel */
    bool sorted;
    do {
        sorted = true;
        for (i = 0; i < len-1; i++) {
            if (diff[i] < diff[i+1]) {
                swap(diff[i], diff[i+1]);
                swap(num[i], num[i+1]);
                sorted = false;
            }
        }
    } while (!sorted);

    writefln("Num:  %s", num);
    writefln("Diff: %s", diff);
}

Output:
Num: [5, 12, 14, 15, 80, 121, 134, 144, 256]
Diff: [5, 7, 2, 1, 65, 41, 13, 10, 112]

Num: [256, 80, 121, 134, 144, 12, 5, 14, 15]
Diff: [112, 65, 41, 13, 10, 7, 5, 2, 1]

Josh said
August 21, 2016 at 10:13 PM
Written in Common Lisp
This is my first public answer to a question. Fingers crossed I did good :D
Josh said
August 21, 2016 at 10:15 PM
http://paste.lisp.org/display/323870
It didn’t post the link…sorry.

Paul said

August 22, 2016 at 8:01 AM

In Python.

from util.itert import pairwise  # from itertools doc
from operator import itemgetter

def dgs(seq):
    gaps = [(a, a - b) for a, b in pairwise(sorted(seq, reverse=True) + [0])]
    return sorted(gaps, key=itemgetter(1), reverse=True)

r. clayton said
August 22, 2016 at 4:47 PM
A solution in Racket.
Daniel said
September 1, 2016 at 12:29 AM
> “They were going to develop something that would pick out the most prominent 6 to 10 colors…”

One approach to that problem is to use a clustering algorithm on points of RGB values. For example:
http://charlesleifer.com/blog/using-python-and-k-means-to-find-the-dominant-colors-in-images/

Back to the problem…

Here’s a solution in matlab.
```
l = [5, 12, 14, 15, 80, 121, 134, 144, 256];
sorted = sort([l, 0]);
[~,I] = sort(diff(sorted), 'descend');
l(I)
```
Output:
```
>> diminishing_gap_sort

ans =

   256    80   121   134   144    12     5    14    15
```

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