How to resolve the algorithm Successive prime differences step by step in the jq programming language
How to resolve the algorithm Successive prime differences step by step in the jq programming language
Table of Contents
Problem Statement
The series of increasing prime numbers begins: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ... The task applies a filter to the series returning groups of successive primes, (s'primes), that differ from the next by a given value or values. Example 1: Specifying that the difference between s'primes be 2 leads to the groups: (Known as Twin primes or Prime pairs) Example 2: Specifying more than one difference between s'primes leads to groups of size one greater than the number of differences. Differences of 2, 4 leads to the groups: In the first group 7 is two more than 5 and 11 is four more than 7; as well as 5, 7, and 11 being successive primes. Differences are checked in the order of the values given, (differences of 4, 2 would give different groups entirely). Note: Generation of a list of primes is a secondary aspect of the task. Use of a built in function, well known library, or importing/use of prime generators from other Rosetta Code tasks is encouraged.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Successive prime differences step by step in the jq programming language
Source code in the jq programming language
# Emit a stream of consecutive primes.
# The stream is unbounded if . is null or infinite,
# otherwise it continues up to but excluding `.`.
def primes:
(if . == null then infinite else . end) as $n
| 2, (range(3; $n; 2) | select(is_prime));
# s is a stream
# $deltas is an array
# Output: a stream of arrays, each corresponding to a selection of consecutive
# items from s satisfying the differences requirement.
def filter_differences(s; $deltas):
def diffs_equal: # i.e. equal to $deltas
. as $in
| all( range(1;length);
($in[.] - $in[.-1]) == $deltas[. - 1]);
($deltas|length + 1) as $n
| foreach s as $x ( {};
.emit = null
| .tuple += [$x]
| .tuple |= .[-$n:]
| if (.tuple|length) == $n
then if (.tuple|diffs_equal) then .emit = .tuple
else .
end
else .
end;
select(.emit).emit );
def report_first_last_count(s):
null | {first,last,count}
| reduce s as $x (.;
if .first == null then .first = $x else . end
| .count = .count + 1
| .last = $x ) ;
[pow(10;6) | primes] as $p1e6
| ([2], [1], [2,2], [2,4], [4,2], [6,4,2]) as $d
| ("\nFor deltas = \($d):", report_first_last_count(filter_differences($p1e6[]; $d ) ) )
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