### TASK #1

#### THE WEEKLY CHALLENGE – PERL & RAKU #50

#### Merge Intervals

Write a script to merge the given intervals where ever possible.

[2,7], [3,9], [10,12], [15,19], [18,22]

The script should merge **[2, 7]** and **[3, 9]** together to return **[2, 9]**.

Similarly it should also merge **[15, 19]** and **[18, 22]** together to return **[15, 22]**.

The final result should be something like below:

[2, 9], [10, 12], [15, 22]

#### Analysis

Intervals can be merged if the start or end of one interval is contined within the span of another. If the other boundary of the second interval is outside the range of the first, a new interval is created encompassing the combined span of the two.^{1}

This process of integration can be chained, with the new interval becoming increasingly larger, as long as there are intervals that overlap the aggregate. Amoeba-like, the intervals expand as we consolidate the intersecting areas, resulting in a remaining field of discontinuous elements.

We will choose to to interpret the intervals expressed as an absolute value, a scalar without a vector. The interval [6,4] will be considered equivalent to the interval [4,6] as they encompass the same span; the quantity of difference is what is being measured, rather than a specific sequential progression. With this stipulation the intervals can be always be coerced into ascending order; without it, it’s difficult to understand how the idea of merging the intervals can meaningfully done without damaging the data. That task would be more akin to summing vectors, or perhaps finding the area under a graph. In any case it’s outside the purview of this challenge.

^{1}If the outer boundry is also contained within the range of the first, the second interval is swallowed whole into the larger and disappears.

#### Method

In preparation for merging, the interval pairs are sorted ascending, first internally and then by their lower bound. Stepwise, the lowest-bounded (leftmost) interval is shifted off the list. If the next interval start is contained within the existing, it is shifted off the list and the upper bound of the current interval is increased or retained acordingly. This process is continued until the next interval lower bound is outside the range of the current. The current, expanded interval is then output, and the process is begun again with the next, until we run out of data.

##### PERL 5 SOLUTION

```
use warnings;
use strict;
use feature ":5.26";
## ## ## ## ## MAIN:
## sort and order the data before commencing
my @intervals = ([2,7], [3,9], [19,15], [18,22], [10,12]);
my @remapped = map { $_->[0] <= $_->[1] ? $_
: [reverse $_->@*] } @intervals;
my @sorted = sort { $a->[0] <=> $b->[0] } @remapped;
my @output;
while ( my $current = shift @sorted ){
## iterate through the intervals until a lower is greater than the current
## upper bound
while (scalar @sorted && ($sorted[0]->[0] <= $current->[1])) {
my $next = shift @sorted;
$current->[1] = $next->[1] if $next->[1] > $current->[1];
}
## once out of there we add to the output list, loop and and start again
## with the next discontinuous interval
push @output, $current;
}
## output
say join ', ', map { "[" . (join ", ", $_->@*) . "]" } @output;
```

##### raku solution

```
sub MAIN () {
my @intervals = [ [2,7], [3,9], [19,15], [18,22], [10,12] ];
## present the raw input intervals
say @intervals;
@intervals .= map( {$_[1] > $_[0] ?? $_ !! [$_.reverse]} );
@intervals .= sort( { $^a[0] <=> $^b[0] } );
my @output;
## take a peek after transformation but before amalgamation
say @intervals;
while @intervals.elems {
## shift the first element and establish the bounds
my $current = @intervals.shift;
## iterate through the intervals until a lower is greater than the
## current upper bound
while @intervals.elems && (@intervals[0][0] <= $current[1]) {
my $next = @intervals.shift;
$current[1] = $next[1] if $next[1] > $current[1];
}
## once out of there we add to the output list and start again with the
## next discontinuous interval
@output.push( $current );
}
## output the combined intervals
say (map {"[" ~ .join(', ') ~ "]"}, @output).join(' ');
}
```

### TASK #2

**Challenge by Ryan Thompson **

#### Noble Integer

You are given a list, `@L`

, of three or more random integers between 1 and 50. A Noble Integer is an integer *N* in `@L`

, such that there are exactly *N* integers greater than *N* in `@L`

. Output any Noble Integer found in `@L`

, or an empty list if none were found.

An interesting question is whether or not there can be multiple Noble Integers in a list.

For example,

Suppose we have list of **4** integers **[2, 6, 1, 3]**.

Here we have **2** in the above list, known as **Noble Integer**, since there are exactly **2** integers in the list i.e.**3** and **6**, which are greater than **2**.

Therefore the script would print **2**.

#### Analysis

First off, just to get it out of the way, we can address the open question: whether or not there can be multiple Noble Integers in a list. To do this we first need to decide the unspecified criteria of whether or not to allow multiple instances of any integer in the random list, and if we do, whether we count these instances as different candidates. If we accept both of these then trivially duplicating any element that satifies the criteria will produce another Noble Integer. Aside from this somewhat pathological case, it turns out that duplication does not otherwise affect the outcome. Elements of the list are only evaluated as to whether they are greater than a given item or not; their precise value is irrelevant beyond this scope. Whether or not they are the same as another element makes no difference.

As it seems a stretch to consider different instances of the number 2 under evaluation for Nobility as different integers, we will not, and move on. The given term ‘random’ in the challenge put forward implies freedom from constraints, to be any value; we will not impose such a constraint here.

**def:** Given a list of integers (l, m, n, a, b, c…), the integer

n is a Noble Integer if and only if the quantity of elements in

the list greater than n is equal to n.

**so:** Given the two elements {m, n} : m ≠ n, these can be arbitrarily reordered such that n > m

m ∈ ℤ’ –> |{n, p, q, s…}| : (n,p,q,s,… > m) = m

- ∀ n > m ,

n ≯ n –> |{ p, q, s…}| : ( p,q,s,… > n) < m < n

∴ n ∉ ℤ’ - ∀ l < m , m > l –> |{m, n, p, q, s…}| : (m,n,p,q,s,… > n) > m > l

∴ l ∉ ℤ’

the contradiction is that n is contained within the set of elements > m, yet not contained within the set of elements greater than itself. Therefore the set of elements greater than n is always smaller than the set of elements greater than m. Yet n > m, and the number of elements greater than n must be larger than m for n to be Noble. So if m is Noble, no number n > m can also satisfy the criterium that the number of list elements greater than the number is equal to that number. Similarly, for any l < m, the number of list elements greater than l will contain m, and as such be greater than the list for m (which will not contain itself), and therefore be greater than l. So for any Noble Integer m, no number n greater than m can be Noble, and no number l less than m can be Noble.

Thus, for any set that contains a Noble Integer, in the words of the Highlander,

*“There can be only one.” *

All Hail our Monarch King Int, most Regal, Finite in Quantity yet Infinite in Wisdom, Singular and Omnipotent Ruler of his Domain!

It is not exactly clear *why* satisfying the given criteria allows the Noble Integer to reign alone, but all in all this is not uncommon throughout history, which is riddled with obtuse justifcations and backformation rationalizations by rulers. If looking for a reason, it may be best to consider the words: **“That’s why I’m up on the truck and you’re down there digging the ditch.” **Such is life.

#### method

Because it is known there can be only 1 or 0 Noble integers, we can create a new list using grep, comparing the topic to the length of a list of items greater than the topic, using grep again. The result list if populated will only have one element. This can be done in one line:

`my ($noble) = grep { my $ele = $_; scalar( grep { $ele < $_ } @list ) == $ele } @list;`

but with a separate `validate()`

subroutine the nested greps are easier to follow.

We will also decide “integers between 1 and 50” to mean 1..50 inclusive. It is not specified how long the list is, so we will pick a random length less than 100 elements, or twice the top bound.

##### PERL 5 SOLUTION

```
use warnings;
use strict;
use feature ":5.26";
## ## ## ## ## MAIN
## prepare a random list
my @list = make_list();
## there is only one Noble Integer if present, so the list of solutions will
## have at maximum one element
my ($noble) = grep { validate($_, @list) } @list;
## output
say scalar @list, " elements generated";
say join ', ', @list;
say $noble ? "the number $noble is the Noble Integer"
: "there is no Noble Integer for this list";
## ## ## ## ## SUBS
sub validate {
## given a scalar and a list, returns true if the number of list elements
## greater than the scalar is equal to the scalar
my ($candidate, @list) = @_;
return scalar( grep { $candidate < $_ } @list ) == $candidate ? 1 : 0;
}
sub make_list {
## makes a list of between three to 100 random integers between 1 and 50
## inclusive
my @list;
my $elems = int (rand(98)) + 3;
for (1..$elems) {
push @list, int rand(50) + 1;
}
return @list;
}
```

##### raku solution

```
sub MAIN () {
## create a list of 3..100 elements between 1..50 inclusive
my @list = (^(2..99).pick).map({ (1..50).pick }) ;
## determine whether there is a noble integer and note it
my @noble = @list.grep({ my $i = $_; @list.grep({ $i < $_ }).elems == $i ?? 1
!! 0 });
## say the list created, the noble integer if present, or alternately a
## victory cry for "Liberté, égalité, fraternité" because presumably the
## heads of the nobles are all over there in a basket. Sometimes life takes
## a turn.
@list.say;
say @noble[0] ?? "the integer @noble[0] is Noble"
!! "Vive la France! Vive la révolution!";
}
```

*The Perl Weekly Challenge, that idyllic glade wherein we stumble upon the holes for these sweet descents, is now known as *

**The Weekly Challenge – Perl and Raku**

*It is the creation of the lovely Mohammad Sajid Anwar and a veritable swarm of contributors from all over the world, who gather, as might be expected, weekly online to solve puzzles. Everyone is encouraged to visit, learn and contribute at *

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