08 November 2020  Challenge 085 
Interwoven numbers
The task
You are given an array of real numbers greater than zero.
Write a script to find if there exists a triplet (a,b,c)
such that
S) 1 < a + b + c < 2
Print 1 if you succeed otherwise 0.
Let’s call the inequality chain S) the sum conditions
.
Motivation
The first attempt to solve this task was a simple combinatorical
approach as given in triplet_sum_combine
.
To reduce the search space, some sanity checks were added that would
possibly exclude some of the elements.
The idea behind it: If a number plus the smallest sum of two other
numbers exceeds 2, this number cannot be part of a valid triplet.
And the other way round: If a number plus the largest sum of two other
numbers is below one, it cannot be part of a valid triplet neither.
This worked well from the beginning, but one piece was missing: There was no test case with a proper list of reduced numbers that failed the combinatorical part. Finding such numbers appeared difficult at first and turned out to be impossible, leading to an unexpected solution to the task.
Trying to find such a nonexistent combination of numbers satisfying the
filter conditions but violating the sum conditions revealed the complex
dependencies between these numbers.
Driving one number towards a certain limit never led to a violation of
a sum condition but instead to the existence of a new valid triplet or
the removal of a number from the reduced list.
This experience led to the title Interwoven numbers
.
List reduction
Let sps
be the smallest partial sum of two of the given numbers
and lps
be the largest partial sum.
These are easily found after the numbers have been sorted.
The list of numbers is then reduced by (repeatedly) applying the filter
m > 1  lps && m < 2  sps
on the members m
of the list.
The filter condition consists of two parts:
m < 2  sps
m > 1  lps
For the second expression to take effect, lps
has to be
smaller than one, requiring all numbers and sps
to be smaller
than one, too.
But in this case all numbers comply to the first part of the filter
condition, making it a noop, which in turn leaves lps
unchanged as
there is no modification at the larger values of the list.
In summary, a list modification at the lower values can occur only once.
As a consequence, the value of sps
in the first expression may change
only once and thus the whole filter will stay unmodified after it has
been applied twice, making two the maximum number of required filter
applications.
Note: The maximum loop count of two holds provided that the list has a length of three or more. (Otherwise lower and larger values coincide, invalidating the reasoning.) Shorter lists can be ignored as these reveal the nonexistence of a solution anyway.
Characteristics of a reduced list
A list shall be called reduced list
if it is stable under the list
reduction filter.
When the filter has been applied twice to a list, it is a reduced list.
If a reduced list consists of less than three elements, then there is obviously no triplet conforming to the sum conditions.
It is easy to see that a reduced list consisting of three elements represents a valid triplet.
So we need to analyse a reduced list consisting of four or more elements.
Let a
, b
, c
and d
be the smallest and the largest two numbers
of the reduced list.
Then we have:
sps = a + b
lps = c + d
Considering the filter conditions of a reduced list and the order of the list elements leads to the following chain of inequalities:
C) 1  c  d < a <= b <= c <= d < 2  a  b
There are three fundamental triplets built from these four elements that shall be identified by the missing part as follows:
Xb: (a, c, d) s_xb = a + c + d
Xc: (a, b, d) s_xc = a + b + d
Xd: (a, b, c) s_xd = a + b + c
We now distinguish four cases depending on the values of d
and
s_xb
:

Case 1:
d > 1
From C) and the condition of case 1 follows:
a + b + d < 2 a + b + d > d > 1
i.e.
1 < s_xc < 2
Thus
Xc
is a valid triplet. 
Case 2:
d < 2/3
From C) and the condition of case 2 follows:
a + c + d > 1 a + c + d <= 3 * d < 3 * 2/3 = 2
i.e.
1 < s_xb < 2
Hence
Xb
is a valid triplet. 
Case 3:
2/3 <= d <= 1
ands_xb < 2
Here we have from C) and the condition above:
a + c + d > 1 a + c + d = s_xb < 2
This identifies
Xb
as a valid triplet. 
Case 4:
2/3 <= d <= 1
ands_xb >= 2
This is the most interesting and complex case. Note that there are numbers
a
,b
andc
satisfying the conditions_xb >= 2
because of2/3 <= d
.From C) and the conditions of case 4 we conclude:
2 <= s_xb = a + c + d < a + c + 1 1 < a + c
which in turn leads to:
2 > a + b + d >= a + b + c = a + c + b > 1 + b > 1 2 > a + b + c = s_xd > 1
This reveals
Xd
as a valid triplet.
In summary, solely from the existence of a reduced list consisting of three or more members we may conclude the existence of a valid triplet. Furthermore, there is a fixed set of three triplets that contains at least one valid triplet. This provides a source for a solution to the given task.
There are some test cases at the end of the script comparing the results from the reduced set implementation with a combinatorical approach for a number of random sets.
Conclusion
Utilizing these findings, the task can be solved by grep’ing twice over the list of numbers and then simply comparing the result’s length against three.
If you have a question about this post or if you like to comment on it, feel free to open an issue in my github repository.