Range consolidation
You are encouraged to solve this task according to the task description, using any language you may know.
Define a range of numbers R, with bounds b0 and b1 covering all numbers between and including both bounds. That range can be shown as:
[b0, b1]
or equally as:
[b1, b0].
Given two ranges, the act of consolidation between them compares the two ranges:
- If one range covers all of the other then the result is that encompassing range.
- If the ranges touch or intersect then the result is one new single range covering the overlapping ranges.
- Otherwise the act of consolidation is to return the two non-touching ranges.
Given N ranges where N>2 then the result is the same as repeatedly replacing all combinations of two ranges by their consolidation until no further consolidation between range pairs is possible. If N<2 then range consolidation has no strict meaning and the input can be returned.
- Example 1:
- Given the two ranges [1, 2.5] and [3, 4.2] then there is no
- common region between the ranges and the result is the same as the input.
- Example 2:
- Given the two ranges [1, 2.5] and [1.8, 4.7] then there is
- an overlap [2.5, 1.8] between the ranges and the result is the single
- range [1, 4.7]. Note that order of bounds in a range is not, (yet), stated.
- Example 3:
- Given the two ranges [6.1, 7.2] and [7.2, 8.3] then they
- touch at 7.2 and the result is the single range [6.1, 8.3].
- Example 4:
- Given the three ranges [1, 2] and [4, 8] and [2, 5]
- then there is no intersection of the ranges [1, 2] and [4, 8]
- but the ranges [1, 2] and [2, 5] overlap and consolidate to
- produce the range [1, 5]. This range, in turn, overlaps the other range
- [4, 8], and so consolidates to the final output of the single range
- [1, 8]
- Task:
Let a normalized range display show the smaller bound to the left; and show the range with the smaller lower bound to the left of other ranges when showing multiple ranges.
Output the normalised result of applying consolidation to these five sets of ranges:
[1.1, 2.2]
[6.1, 7.2], [7.2, 8.3]
[4, 3], [2, 1]
[4, 3], [2, 1], [-1, -2], [3.9, 10]
[1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6]
Show output here.
See also
Ada
<lang Ada>with Ada.Text_IO; with Ada.Containers.Vectors;
procedure Range_Consolidation is
type Set_Type is record
Left, Right : Float;
end record;
package Set_Vectors is
new Ada.Containers.Vectors (Positive, Set_Type);
procedure Normalize (Set : in out Set_Vectors.Vector) is
function Less_Than (Left, Right : Set_Type) return Boolean is
begin Return Left.Left < Right.Left; end;
package Set_Sorting is
new Set_Vectors.Generic_Sorting (Less_Than);
begin
for Elem of Set loop
Elem := (Left => Float'Min (Elem.Left, Elem.Right),
Right => Float'Max (Elem.Left, Elem.Right));
end loop;
Set_Sorting.Sort (Set);
end Normalize;
procedure Consolidate (Set : in out Set_Vectors.Vector) is
use Set_Vectors;
First : Cursor := Set.First;
Last : Cursor := Next (First);
begin
while Last /= No_Element loop
if Element (First).Right < Element (Last).Left then -- non-overlap
First := Last;
Last := Next (Last);
elsif Element (First).Right >= Element (Last).Left then -- overlap
Replace_Element (Set, First, (Left => Element (First).Left,
Right => Float'Max (Element (First).Right,
Element (Last) .Right)));
Delete (Set, Last);
Last := Next (First);
end if;
end loop;
end Consolidate;
procedure Put (Set : in Set_Vectors.Vector) is
package Float_IO is
new Ada.Text_IO.Float_IO (Float);
begin
Float_IO.Default_Exp := 0;
Float_IO.Default_Aft := 1;
Float_IO.Default_Fore := 3;
for Elem of Set loop
Ada.Text_IO.Put ("(");
Float_IO.Put (Elem.Left);
Float_IO.Put (Elem.Right);
Ada.Text_IO.Put (") ");
end loop;
end Put;
procedure Show (Set : in out Set_Vectors.Vector) is
use Ada.Text_IO;
begin
Put (Set);
Normalize (Set);
Consolidate (Set);
Set_Col (70);
Put (Set);
New_Line;
end Show;
use Set_Vectors; Set_0 : Set_Vectors.Vector := Empty_Vector; Set_1 : Set_Vectors.Vector := Empty_Vector & (1.1, 2.2); Set_2 : Set_Vectors.Vector := (6.1, 7.2) & (7.2, 8.3); Set_3 : Set_Vectors.Vector := (4.0, 3.0) & (2.0, 1.0); Set_4 : Set_Vectors.Vector := (4.0, 3.0) & (2.0, 1.0) & (-1.0, -2.0) & (3.9, 10.0); Set_5 : Set_Vectors.Vector := (1.0, 3.0) & (-6.0, -1.0) & (-4.0, -5.0) & (8.0, 2.0) & (-6.0, -6.0);
begin
Show (Set_0); Show (Set_1); Show (Set_2); Show (Set_3); Show (Set_4); Show (Set_5);
end Range_Consolidation;</lang>
- Output:
( 1.1 2.2) ( 1.1 2.2) ( 6.1 7.2) ( 7.2 8.3) ( 6.1 8.3) ( 4.0 3.0) ( 2.0 1.0) ( 1.0 2.0) ( 3.0 4.0) ( 4.0 3.0) ( 2.0 1.0) ( -1.0 -2.0) ( 3.9 10.0) ( -2.0 -1.0) ( 1.0 2.0) ( 3.0 10.0) ( 1.0 3.0) ( -6.0 -1.0) ( -4.0 -5.0) ( 8.0 2.0) ( -6.0 -6.0) ( -6.0 -1.0) ( 1.0 8.0)
C#
<lang csharp>using static System.Math; using System.Linq; using System;
public static class RangeConsolidation {
public static void Main() {
foreach (var list in new [] {
new[] { (1.1, 2.2) }.ToList(),
new[] { (6.1, 7.2), (7.2, 8.3) }.ToList(),
new[] { (4d, 3d), (2, 1) }.ToList(),
new[] { (4d, 3d), (2, 1), (-1, 2), (3.9, 10) }.ToList(),
new[] { (1d, 3d), (-6, -1), (-4, -5), (8, 2), (-6, -6) }.ToList()
})
{
for (int z = list.Count-1; z >= 1; z--) {
for (int y = z - 1; y >= 0; y--) {
if (Overlap(list[z], list[y])) {
list[y] = Consolidate(list[z], list[y]);
list.RemoveAt(z);
break;
}
}
}
Console.WriteLine(string.Join(", ", list.Select(Normalize).OrderBy(range => range.s)));
}
}
private static bool Overlap((double s, double e) left, (double s, double e) right) =>
Max(left.s, left.e) > Max(right.s, right.e)
? Max(right.s, right.e) >= Min(left.s, left.e)
: Max(left.s, left.e) >= Min(right.s, right.e);
private static (double s, double e) Consolidate((double s, double e) left, (double s, double e) right) =>
(Min(Min(left.s, left.e), Min(right.s, right.e)), Max(Max(left.s, left.e), Max(right.s, right.e)));
private static (double s, double e) Normalize((double s, double e) range) =>
(Min(range.s, range.e), Max(range.s, range.e));
}</lang>
- Output:
(1.1, 2.2) (6.1, 8.3) (1, 2), (3, 4) (-1, 2), (3, 10) (-6, -1), (1, 8)
Dyalect
<lang dyalect>func max(x, y) {
if x > y {
x
} else {
y
}
}
func min(x, y) {
if x < y {
x
} else {
y
}
}
func overlap(left, right) {
if max(left.s, left.e) > max(right.s, right.e) {
max(right.s, right.e) >= min(left.s, left.e)
} else {
max(left.s, left.e) >= min(right.s, right.e)
}
}
func consolidate(left, right) {
(s: min(min(left.s, left.e), min(right.s, right.e)), e: max(max(left.s, left.e), max(right.s, right.e)))
}
func normalize(range) {
(s: min(range.s, range.e), e: max(range.s, range.e))
}
for list in [
[ (s: 1.1, e: 2.2) ], [ (s: 6.1, e: 7.2), (s: 7.2, e: 8.3) ], [ (s: 4.0, e: 3.0), (s: 2, e: 1) ], [ (s: 4.0, e: 3.0), (s: 2, e: 1), (s: -1, e: 2), (s: 3.9, e: 10) ], [ (s: 1.0, e: 3.0), (s: -6, e: -1), (s: -4, e: -5), (s: 8, e: 2), (s: -6, e: -6) ]
] {
var z = list.len()-1
while z >= 1 {
for y in (z - 1)..0 {
if overlap(list[z], list[y]) {
list[y] = consolidate(list[z], list[y])
list.removeAt(z)
break
}
}
z -= 1
}
for i in list.indices() {
list[i] = normalize(list[i])
}
list.sort((x,y) => x.s - y.s)
print(list)
} </lang>
- Output:
[(s: 1.1, e: 2.2)] [(s: 6.1, e: 8.3)] [(s: 1, e: 2), (s: 3, e: 4)] [(s: -1, e: 2), (s: 3, e: 10)] [(s: -6, e: -1), (s: 1, e: 8)]
Go
<lang go>package main
import (
"fmt" "math" "sort"
)
type Range struct{ Lower, Upper float64 }
func (r Range) Norm() Range {
if r.Lower > r.Upper {
return Range{r.Upper, r.Lower}
}
return r
}
func (r Range) String() string {
return fmt.Sprintf("[%g, %g]", r.Lower, r.Upper)
}
func (r1 Range) Union(r2 Range) []Range {
if r1.Upper < r2.Lower {
return []Range{r1, r2}
}
r := Range{r1.Lower, math.Max(r1.Upper, r2.Upper)}
return []Range{r}
}
func consolidate(rs []Range) []Range {
for i := range rs {
rs[i] = rs[i].Norm()
}
le := len(rs)
if le < 2 {
return rs
}
sort.Slice(rs, func(i, j int) bool {
return rs[i].Lower < rs[j].Lower
})
if le == 2 {
return rs[0].Union(rs[1])
}
for i := 0; i < le-1; i++ {
for j := i + 1; j < le; j++ {
ru := rs[i].Union(rs[j])
if len(ru) == 1 {
rs[i] = ru[0]
copy(rs[j:], rs[j+1:])
rs = rs[:le-1]
le--
i--
break
}
}
}
return rs
}
func main() {
rss := [][]Range{
Template:1.1, 2.2,
{{6.1, 7.2}, {7.2, 8.3}},
{{4, 3}, {2, 1}},
{{4, 3}, {2, 1}, {-1, -2}, {3.9, 10}},
{{1, 3}, {-6, -1}, {-4, -5}, {8, 2}, {-6, -6}},
}
for _, rs := range rss {
s := fmt.Sprintf("%v", rs)
fmt.Printf("%40s => ", s[1:len(s)-1])
rs2 := consolidate(rs)
s = fmt.Sprintf("%v", rs2)
fmt.Println(s[1 : len(s)-1])
}
}</lang>
- Output:
[1.1, 2.2] => [1.1, 2.2]
[6.1, 7.2] [7.2, 8.3] => [6.1, 8.3]
[4, 3] [2, 1] => [1, 2] [3, 4]
[4, 3] [2, 1] [-1, -2] [3.9, 10] => [-2, -1] [1, 2] [3, 10]
[1, 3] [-6, -1] [-4, -5] [8, 2] [-6, -6] => [-6, -1] [1, 8]
Haskell
<lang haskell>import Data.List (intercalate, maximumBy, sort) import Data.Ord (comparing)
consolidated :: [(Float, Float)] -> [(Float, Float)] consolidated xs =
let go xy [] = [xy]
go xy@(x, y) abetc@((a, b):etc)
| y >= b = xy : etc
| y >= a = (x, b) : etc
| otherwise = xy : abetc
ab (a, b)
| a <= b = (a, b)
| otherwise = (b, a)
in foldr go [] (sort . fmap ab $ xs)
-- TEST ---------------------------------------------------
tests :: (Float, Float)
tests =
[ [] , [(1.1, 2.2)] , [(6.1, 7.2), (7.2, 8.3)] , [(4, 3), (2, 1)] , [(4, 3), (2, 1), (-1, -2), (3.9, 10)] , [(1, 3), (-6, -1), (-4, -5), (8, 2), (-6, -6)] ]
main :: IO () main =
putStrLn $ tabulated "Range consolidations:" showPairs showPairs consolidated tests
-- DISPLAY FORMATTING -------------------------------------
tabulated :: String -> (a -> String) -> (b -> String) -> (a -> b) -> [a] -> String tabulated s xShow fxShow f xs =
let w = length $ maximumBy (comparing length) (xShow <$> xs)
rjust n c s = drop (length s) (replicate n c ++ s)
in unlines $
s : fmap (((++) . rjust w ' ' . xShow) <*> ((" -> " ++) . fxShow . f)) xs
showPairs :: [(Float, Float)] -> String showPairs xs
| null xs = "[]" | otherwise = '[' : intercalate ", " (showPair <$> xs) ++ "]"
showPair :: (Float, Float) -> String showPair (a, b) = '(' : showNum a ++ ", " ++ showNum b ++ ")"
showNum :: Float -> String showNum n
| 0 == (n - fromIntegral (round n)) = show (round n) | otherwise = show n</lang>
- Output:
Range consolidations:
[] -> []
[(1.1, 2.2)] -> [(1.1, 2.2)]
[(6.1, 7.2), (7.2, 8.3)] -> [(6.1, 8.3)]
[(4, 3), (2, 1)] -> [(1, 2), (3, 4)]
[(4, 3), (2, 1), (-1, -2), (3.9, 10)] -> [(-2, -1), (1, 2), (3, 10)]
[(1, 3), (-6, -1), (-4, -5), (8, 2), (-6, -6)] -> [(-6, -1), (1, 8)]
J
Solution: <lang j>ensure2D=: ,:^:(1 = #@$) NB. if list make 1 row table normalise=: ([: /:~ /:~"1)@ensure2D NB. normalises list of ranges merge=: ,:`(<.&{. , >.&{:)@.(>:/&{: |.) NB. merge ranges x and y consolidate=: (}.@] ,~ (merge {.)) ensure2D</lang> Required Examples: <lang j> tests=: <@".;._2 noun define 1.1 2.2 6.1 7.2 ,: 7.2 8.3 4 3 ,: 2 1 4 3 , 2 1 , _1 _2 ,: 3.9 10 1 3 , _6 _1 , _4 _5 , 8 2 ,: _6 _6 )
consolidate/@normalise&.> tests
+-------+-------+---+-----+-----+ |1.1 2.2|6.1 8.3|1 2|_2 _1|_6 _1| | | |3 4| 1 2| 1 8| | | | | 3 10| | +-------+-------+---+-----+-----+</lang>
JavaScript
<lang javascript>(() => {
'use strict';
const main = () => {
// consolidated :: [(Float, Float)] -> [(Float, Float)]
const consolidated = xs =>
foldl((abetc, xy) =>
0 < abetc.length ? (() => {
const
etc = abetc.slice(1),
[a, b] = abetc[0],
[x, y] = xy;
return y >= b ? (
cons(xy, etc)
) : y >= a ? (
cons([x, b], etc)
) : cons(xy, abetc);
})() : [xy],
[],
sortBy(flip(comparing(fst)),
map(([a, b]) => a < b ? (
[a, b]
) : [b, a],
xs
)
)
);
// TEST -------------------------------------------
console.log(
tabulated(
'Range consolidations:',
JSON.stringify,
JSON.stringify,
consolidated,
[
[
[1.1, 2.2]
],
[
[6.1, 7.2],
[7.2, 8.3]
],
[
[4, 3],
[2, 1]
],
[
[4, 3],
[2, 1],
[-1, -2],
[3.9, 10]
],
[
[1, 3],
[-6, -1],
[-4, -5],
[8, 2],
[-6, -6]
]
]
)
);
};
// GENERIC FUNCTIONS ----------------------------
// comparing :: (a -> b) -> (a -> a -> Ordering)
const comparing = f =>
(x, y) => {
const
a = f(x),
b = f(y);
return a < b ? -1 : (a > b ? 1 : 0);
};
// compose (<<<) :: (b -> c) -> (a -> b) -> a -> c const compose = (f, g) => x => f(g(x));
// cons :: a -> [a] -> [a] const cons = (x, xs) => [x].concat(xs);
// flip :: (a -> b -> c) -> b -> a -> c
const flip = f =>
1 < f.length ? (
(a, b) => f(b, a)
) : (x => y => f(y)(x));
// foldl :: (a -> b -> a) -> a -> [b] -> a const foldl = (f, a, xs) => xs.reduce(f, a);
// fst :: (a, b) -> a const fst = tpl => tpl[0];
// justifyRight :: Int -> Char -> String -> String
const justifyRight = (n, cFiller, s) =>
n > s.length ? (
s.padStart(n, cFiller)
) : s;
// Returns Infinity over objects without finite length. // This enables zip and zipWith to choose the shorter // argument when one is non-finite, like cycle, repeat etc
// length :: [a] -> Int
const length = xs =>
(Array.isArray(xs) || 'string' === typeof xs) ? (
xs.length
) : Infinity;
// map :: (a -> b) -> [a] -> [b]
const map = (f, xs) =>
(Array.isArray(xs) ? (
xs
) : xs.split()).map(f);
// maximumBy :: (a -> a -> Ordering) -> [a] -> a
const maximumBy = (f, xs) =>
0 < xs.length ? (
xs.slice(1)
.reduce((a, x) => 0 < f(x, a) ? x : a, xs[0])
) : undefined;
// sortBy :: (a -> a -> Ordering) -> [a] -> [a]
const sortBy = (f, xs) =>
xs.slice()
.sort(f);
// tabulated :: String -> (a -> String) ->
// (b -> String) ->
// (a -> b) -> [a] -> String
const tabulated = (s, xShow, fxShow, f, xs) => {
// Heading -> x display function ->
// fx display function ->
// f -> values -> tabular string
const
ys = map(xShow, xs),
w = maximumBy(comparing(x => x.length), ys).length,
rows = zipWith(
(a, b) => justifyRight(w, ' ', a) + ' -> ' + b,
ys,
map(compose(fxShow, f), xs)
);
return s + '\n' + unlines(rows);
};
// take :: Int -> [a] -> [a]
// take :: Int -> String -> String
const take = (n, xs) =>
'GeneratorFunction' !== xs.constructor.constructor.name ? (
xs.slice(0, n)
) : [].concat.apply([], Array.from({
length: n
}, () => {
const x = xs.next();
return x.done ? [] : [x.value];
}));
// unlines :: [String] -> String
const unlines = xs => xs.join('\n');
// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
const zipWith = (f, xs, ys) => {
const
lng = Math.min(length(xs), length(ys)),
as = take(lng, xs),
bs = take(lng, ys);
return Array.from({
length: lng
}, (_, i) => f(as[i], bs[i], i));
};
// MAIN --- return main();
})();</lang>
- Output:
Range consolidations:
[[1.1,2.2]] -> [[1.1,2.2]]
[[6.1,7.2],[7.2,8.3]] -> [[6.1,8.3]]
[[4,3],[2,1]] -> [[1,2],[3,4]]
[[4,3],[2,1],[-1,-2],[3.9,10]] -> [[-2,-1],[1,2],[3,10]]
[[1,3],[-6,-1],[-4,-5],[8,2],[-6,-6]] -> [[-6,-1],[1,8]]
Julia
In Julia, a Range is a type of iterator, generally one over a specified interval. The task as specified is orthogonal to the iteration purpose of a Julia Range, since the task is about merging sets of numbers, not iterations. Therefore, a translation of the Python code is done, rather than using a native Julia Range.
<lang julia>normalize(s) = sort([sort(bounds) for bounds in s])
function consolidate(ranges)
norm = normalize(ranges)
for (i, r1) in enumerate(norm)
if !isempty(r1)
for r2 in norm[i+1:end]
if !isempty(r2) && r1[end] >= r2[1] # intersect?
r1 .= [r1[1], max(r1[end], r2[end])]
empty!(r2)
end
end
end
end
[r for r in norm if !isempty(r)]
end
function testranges()
for s in [[[1.1, 2.2]], [[6.1, 7.2], [7.2, 8.3]], [[4, 3], [2, 1]],
[[4, 3], [2, 1], [-1, -2], [3.9, 10]],
[[1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6]]]
println("$s => $(consolidate(s))")
end
end
testranges()
</lang>
- Output:
Array{Float64,1}[[1.1, 2.2]] => Array{Float64,1}[[1.1, 2.2]]
Array{Float64,1}[[6.1, 7.2], [7.2, 8.3]] => Array{Float64,1}[[6.1, 8.3]]
Array{Float64,1}[[4.0, 3.0], [2.0, 1.0]] => Array{Float64,1}[[1.0, 2.0], [3.0, 4.0]]
Array{Float64,1}[[4.0, 3.0], [2.0, 1.0], [-1.0, -2.0], [3.9, 10.0]] => Array{Float64,1}[[-2.0, -1.0], [1.0, 2.0], [3.0, 10.0]]
Array{Float64,1}[[1.0, 3.0], [-6.0, -1.0], [-4.0, -5.0], [8.0, 2.0], [-6.0, -6.0]] => Array{Float64,1}[[-6.0, -1.0], [1.0, 8.0]]
Perl
Note: the output is shown in the standard Perl notation for Ranges.
<lang perl>use strict; use warnings;
use List::Util qw(min max);
sub consolidate {
our @arr; local *arr = shift;
my @sorted = sort { @$a[0] <=> @$b[0] } map { [sort { $a <=> $b } @$_] } @arr;
my @merge = shift @sorted;
for my $i (@sorted) {
if ($merge[-1][1] >= @$i[0]) {
$merge[-1][0] = min($merge[-1][0], @$i[0]);
$merge[-1][1] = max($merge[-1][1], @$i[1]);
} else {
push @merge, $i;
}
}
return @merge;
}
for my $intervals (
[[1.1, 2.2],],
[[6.1, 7.2], [7.2, 8.3]],
[[4, 3], [2, 1]],
[[4, 3], [2, 1], [-1, -2], [3.9, 10]],
[[1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6]]) {
my($in,$out);
$in = join ', ', map { '[' . join(', ', @$_) . ']' } @$intervals;
$out .= join('..', @$_). ' ' for consolidate($intervals);
printf "%44s => %s\n", $in, $out;
}</lang>
- Output:
[1.1, 2.2] => 1.1..2.2
[6.1, 7.2], [7.2, 8.3] => 6.1..8.3
[4, 3], [2, 1] => 1..2 3..4
[4, 3], [2, 1], [-1, -2], [3.9, 10] => -2..-1 1..2 3..10
[1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6] => -6..-1 1..8
Perl 6
In Perl 6, a Range is a first class object with its own specialized notation. Perl 6 Ranges allow for exclusion of the boundary numbers. This example doesn't since it isn't a requirement in this task. Much of the logic is lifted from the Set_of_real_numbers task with simplified logic for the much simpler requirements.
Note: the output is in standard Perl 6 notation for Ranges.
<lang perl6># Union sub infix:<∪> (Range $a, Range $b) { Range.new($a.min,max($a.max,$b.max)) }
- Intersection
sub infix:<∩> (Range $a, Range $b) { so $a.max >= $b.min }
multi consolidate() { () }
multi consolidate($this is copy, **@those) {
gather {
for consolidate |@those -> $that {
if $this ∩ $that { $this ∪= $that }
else { take $that }
}
take $this;
}
}
for [[1.1, 2.2],],
[[6.1, 7.2], [7.2, 8.3]], [[4, 3], [2, 1]], [[4, 3], [2, 1], [-1, -2], [3.9, 10]], [[1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6]]
-> @intervals {
printf "%46s => ", @intervals.perl;
say reverse consolidate |@intervals.grep(*.elems)».sort.sort({ [.[0], .[*-1]] }).map: { Range.new(.[0], .[*-1]) }
}</lang>
- Output:
[[1.1, 2.2],] => (1.1..2.2)
[[6.1, 7.2], [7.2, 8.3]] => (6.1..8.3)
[[4, 3], [2, 1]] => (1..2 3..4)
[[4, 3], [2, 1], [-1, -2], [3.9, 10]] => (-2..-1 1..2 3..10)
[[1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6]] => (-6..-1 1..8)
Phix
<lang Phix>function consolidate(sequence sets)
for i=length(sets) to 1 by -1 do
sets[i] = sort(sets[i])
atom {is,ie} = sets[i]
for j=length(sets) to i+1 by -1 do
atom {js,je} = sets[j]
bool overlap = iff(is<=js?js<=ie:is<=je)
if overlap then
sets[i] = {min(is,js),max(ie,je)}
sets[j..j] = {}
end if
end for
end for
return sort(sets)
end function
procedure test(sequence set)
printf(1,"%40v => %v\n",{set,consolidate(set)})
end procedure
test(Template:1.1,2.2) test({{6.1,7.2},{7.2,8.3}}) test({{4,3},{2,1}}) test({{4,3},{2,1},{-1,-2},{3.9,10}}) test({{1,3},{-6,-1},{-4,-5},{8,2},{-6,-6}})</lang>
- Output:
{{1.1,2.2}} => {{1.1,2.2}}
{{6.1,7.2},{7.2,8.3}} => {{6.1,8.3}}
{{4,3},{2,1}} => {{1,2},{3,4}}
{{4,3},{2,1},{-1,-2},{3.9,10}} => {{-2,-1},{1,2},{3,10}}
{{1,3},{-6,-1},{-4,-5},{8,2},{-6,-6}} => {{-6,-1},{1,8}}
Python
Procedural
<lang python>def normalize(s):
return sorted(sorted(bounds) for bounds in s if bounds)
def consolidate(ranges):
norm = normalize(ranges)
for i, r1 in enumerate(norm):
if r1:
for r2 in norm[i+1:]:
if r2 and r1[-1] >= r2[0]: # intersect?
r1[:] = [r1[0], max(r1[-1], r2[-1])]
r2.clear()
return [rnge for rnge in norm if rnge]
if __name__ == '__main__':
for s in [
1.1, 2.2,
[[6.1, 7.2], [7.2, 8.3]],
[[4, 3], [2, 1]],
[[4, 3], [2, 1], [-1, -2], [3.9, 10]],
[[1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6]],
]:
print(f"{str(s)[1:-1]} => {str(consolidate(s))[1:-1]}")
</lang>
- Output:
[1.1, 2.2] => [1.1, 2.2] [6.1, 7.2], [7.2, 8.3] => [6.1, 8.3] [4, 3], [2, 1] => [1, 2], [3, 4] [4, 3], [2, 1], [-1, -2], [3.9, 10] => [-2, -1], [1, 2], [3, 10] [1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6] => [-6, -1], [1, 8]
Functional
Defining consolidation as a fold over a list of tuples:
<lang python>Range consolidation
from functools import reduce
- consolidated :: [(Float, Float)] -> [(Float, Float)]
def consolidated(xs):
A consolidated list of
[(Float, Float)] ranges.
def go(abetc, xy):
A copy of the accumulator abetc,
with its head range ab either:
1. replaced by or
2. merged with
the next range xy, or
with xy simply prepended.
if abetc:
a, b = abetc[0]
etc = abetc[1:]
x, y = xy
return [xy] + etc if y >= b else ( # ab replaced.
[(x, b)] + etc if y >= a else ( # xy + ab merged.
[xy] + abetc # xy simply prepended.
)
)
else:
return [xy]
def tupleSort(ab):
a, b = ab
return ab if a <= b else (b, a)
return reduce(
go,
sorted(map(tupleSort, xs), reverse=True),
[]
)
- TEST ----------------------------------------------------
- main :: IO ()
def main():
Tests
print(
tabulated('Consolidation of numeric ranges:')(str)(str)(
consolidated
)([
[(1.1, 2.2)],
[(6.1, 7.2), (7.2, 8.3)],
[(4, 3), (2, 1)],
[(4, 3), (2, 1), (-1, -2), (3.9, 10)],
[(1, 3), (-6, -1), (-4, -5), (8, 2), (-6, -6)]
])
)
- GENERIC FUNCTIONS FOR DISPLAY ---------------------------
- compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
def compose(g):
Right to left function composition. return lambda f: lambda x: g(f(x))
- tabulated :: String -> (a -> String) ->
- (b -> String) ->
- (a -> b) -> [a] -> String
def tabulated(s):
Heading -> x display function -> fx display function ->
f -> value list -> tabular string.
def go(xShow, fxShow, f, xs):
w = max(map(compose(len)(xShow), xs))
return s + '\n' + '\n'.join([
xShow(x).rjust(w, ' ') + ' -> ' + fxShow(f(x)) for x in xs
])
return lambda xShow: lambda fxShow: (
lambda f: lambda xs: go(
xShow, fxShow, f, xs
)
)
- MAIN ---
if __name__ == '__main__':
main()</lang>
- Output:
Consolidation of numeric ranges:
[(1.1, 2.2)] -> [(1.1, 2.2)]
[(6.1, 7.2), (7.2, 8.3)] -> [(6.1, 8.3)]
[(4, 3), (2, 1)] -> [(1, 2), (3, 4)]
[(4, 3), (2, 1), (-1, -2), (3.9, 10)] -> [(-2, -1), (1, 2), (3, 10)]
[(1, 3), (-6, -1), (-4, -5), (8, 2), (-6, -6)] -> [(-6, -1), (1, 8)]
Racket
<lang racket>#lang racket
- Racket's max and min allow inexact numbers to contaminate exact numbers
- Use argmax and argmin instead, as they don't have this problem
(define (max . xs) (argmax identity xs)) (define (min . xs) (argmin identity xs))
- a bag is a list of disjoint intervals
(define ((irrelevant? x y) item) (or (< (second item) x) (> (first item) y)))
(define (insert bag x y)
(define-values (irrelevant relevant) (partition (irrelevant? x y) bag))
(cons (list (apply min x (map first relevant))
(apply max y (map second relevant))) irrelevant))
(define (solve xs)
(sort (for/fold ([bag '()]) ([x (in-list xs)])
(insert bag (apply min x) (apply max x))) < #:key first))
(define inputs '(([1.1 2.2])
([6.1 7.2] [7.2 8.3])
([4 3] [2 1])
([4 3] [2 1] [-1 -2] [3.9 10])
([1 3] [-6 -1] [-4 -5] [8 2] [-6 -6])))
(for ([xs (in-list inputs)]) (printf "~a => ~a\n" xs (solve xs)))</lang>
- Output:
((1.1 2.2)) => ((1.1 2.2)) ((6.1 7.2) (7.2 8.3)) => ((6.1 8.3)) ((4 3) (2 1)) => ((1 2) (3 4)) ((4 3) (2 1) (-1 -2) (3.9 10)) => ((-2 -1) (1 2) (3 10)) ((1 3) (-6 -1) (-4 -5) (8 2) (-6 -6)) => ((-6 -1) (1 8))
REXX
Most of the REXX code was testing (and rebuilding) the syntax (insuring blanks after commas), and handling of a null set.
The actual logic for the range consolidation is marked with the comments: /*■■■■►*/ <lang rexx>/*REXX program performs range consolidation (they can be [equal] ascending/descending). */
- .= /*define the default for range sets. */
parse arg #.1 /*obtain optional arguments from the CL*/ if #.1= then do /*Not specified? Then use the defaults*/
#.1= '[1.1, 2.2]'
#.2= '[6.1, 7.2], [7.2, 8.3]'
#.3= '[4, 3], [2, 1]'
#.4= '[4, 3], [2, 1], [-1, -2], [3.9, 10]'
#.5= '[1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6]'
#.6= '[]'
end
do j=1 while #.j\==; $= #.j /*process each of the range sets. */
say copies('═', 75) /*display a fence between range sets. */
say ' original ranges:' $ /*display the original range set. */
$= order($) /*order low and high ranges; normalize.*/
call xSort words($) /*sort the ranges using a simple sort. */
$= merge($) /*consolidate the ranges. */
say ' consolidated ranges:' $ /*display the consolidated range set. */
end /*j*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ merge: procedure expose @.; parse arg y
if words(y)<2 then signal build /*Null or only 1 range? Skip merging. */
do j=1 to @.0-1; if @.j== then iterate /*skip deleted ranges.*/
do k=j+1 to @.0; if @.k== then iterate /* " " " */
parse var @.j a b; parse var @.k aa bb /*extract low and high*/
/*■■■■►*/ if a<=aa & b>=bb then do; @.k=; iterate; end /*within a range*/ /*■■■■►*/ if a<=aa & b>=aa then do; @.j= a bb; @.k=; iterate; end /*abutted ranges*/
end /*k*/
end /*j*/
build: z=
do r=1 for @.0; z= z translate(@.r, ',', " "); end /*r*/ /*add comma*/
f=; do s=1 for words(z); f= f '['word(z, s)"], "; end /*s*/ /*add [ ], */
if f== then return '[]' /*null set.*/
return space( changestr(',', strip( space(f), 'T', ","), ", ") ) /*add blank*/
/*──────────────────────────────────────────────────────────────────────────────────────*/ order: procedure expose @.; parse arg y,,z; @.= /*obtain arguments from the invocation.*/
y= space(y, 0) /*elide superfluous blanks in the sets.*/
do k=1 while y\== & y\=='[]' /*process ranges while range not blank.*/
if left(y,1)==',' then y= substr(y,2) /*elide commas between sets of ranges. */
parse var y '[' L "," H ']' y /*extract the "low" and "high" values.*/
if H<L then parse value L H with H L /*order " " " " " */
L= L / 1; H= H / 1 /*normalize the L and the H values.*/
@.k= L H; z= z L','H /*re─build the set w/o and with commas.*/
end /*k*/ /* [↓] at this point, K is one to big.*/
@.0= k - 1 /*keep track of the number of ranges. */
return strip(z) /*elide the extra leading blank in set.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/ xSort: procedure expose @.; parse arg n /*a simple sort for small set of ranges*/
do j=1 to n-1; _= @.j
do k=j+1 to n; if word(@.k,1)>=word(_,1) then iterate; @.j=@.k; @.k=_; _=@.j
end /*k*/</lang>
- output when using the default inputs:
═══════════════════════════════════════════════════════════════════════════
original ranges: [1.1, 2.2]
consolidated ranges: [1.1, 2.2]
═══════════════════════════════════════════════════════════════════════════
original ranges: [6.1, 7.2], [7.2, 8.3]
consolidated ranges: [6.1, 8.3]
═══════════════════════════════════════════════════════════════════════════
original ranges: [4, 3], [2, 1]
consolidated ranges: [1, 2], [3, 4]
═══════════════════════════════════════════════════════════════════════════
original ranges: [4, 3], [2, 1], [-1, -2], [3.9, 10]
consolidated ranges: [-2, -1], [1, 2], [3, 10]
═══════════════════════════════════════════════════════════════════════════
original ranges: [1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6]
consolidated ranges: [-6, -1], [1, 8]
═══════════════════════════════════════════════════════════════════════════
original ranges: []
consolidated ranges: []
Yabasic
<lang Yabasic>sub sort(tabla())
local items, i, t1, t2, s
items = arraysize(tabla(), 1)
repeat
s = true
for i = 1 to items-1
if tabla(i, 1) > tabla(i+1, 1) then
t1 = tabla(i, 1) : t2 = tabla(i, 2)
tabla(i, 1) = tabla(i + 1, 1) : tabla(i, 2) = tabla(i + 1, 2)
tabla(i + 1, 1) = t1 : tabla(i + 1, 2) = t2
s = false
end if
next
until(s)
end sub
sub normalize(tabla())
local items, i, t
items = arraysize(tabla(), 1)
for i = 1 to items
if tabla(i, 1) > tabla(i, 2) then
t = tabla(i, 1)
tabla(i, 1) = tabla(i, 2)
tabla(i, 2) = t
end if
next
sort(tabla())
end sub
sub consolidate(tabla())
local items, i
normalize(tabla())
items = arraysize(tabla(), 1)
for i = 1 to items - 1
if tabla(i + 1, 1) <= tabla(i, 2) then
tabla(i + 1, 1) = tabla(i, 1)
if tabla(i + 1, 2) <= tabla(i, 2) then
tabla(i + 1, 2) = tabla(i, 2)
end if
tabla(i, 1) = void : tabla(i, 2) = void
end if
next
end sub
// data 1, 1.1, 2.2 // data 2, 6.1, 7.2, 7.2, 8.3 // data 2, 4, 3, 2, 1 // data 4, 4, 3, 2, 1, -1, -2, 3.9, 10
data 5, 1,3, -6,-1, -4,-5, 8,2, -6,-6
void = 10^30 read items
dim tabla(items, 2)
for i = 1 to items
read tabla(i, 1), tabla(i, 2)
next
consolidate(tabla())
for i = 1 to items
if tabla(i, 1) <> void print tabla(i, 1), "..", tabla(i, 2);
next</lang>
zkl
<lang zkl>fcn consolidate(rs){
(s:=List()).append(
normalize(rs).reduce('wrap(ab,cd){
if(ab[1]>=cd[0]) L(ab[0],ab[1].max(cd[1])); // consolidate else{ s.append(ab); cd } // no overlap
}) )
} fcn normalize(s){ s.apply("sort").sort(fcn(a,b){ a[0]<b[0] }) }</lang> <lang zkl>foreach rs in (L(
L(L(1.1, 2.2)), L(L(6.1, 7.2), L(7.2, 8.3)), L(L(4, 3), L(2, 1)),
L(L(4.0, 3.0), L(2.0, 1.0), L(-1.0, -2.0), L(3.9, 10.0)),
L(L(1, 3), L(-6, -1), L(-4, -5), L(8, 2), L(-6, -6)),
)){ println(ppp(rs),"--> ",ppp(consolidate(rs))) }
fcn ppp(ll){ ll.pump(String,fcn(list){ list.concat(", ", "[", "] ") }) }</lang>
- Output:
[1.1, 2.2] --> [1.1, 2.2] [6.1, 7.2] [7.2, 8.3] --> [6.1, 8.3] [4, 3] [2, 1] --> [1, 2] [3, 4] [4, 3] [2, 1] [-1, -2] [3.9, 10] --> [-2, -1] [1, 2] [3, 10] [1, 3] [-6, -1] [-4, -5] [8, 2] [-6, -6] --> [-6, -1] [1, 8]