How to resolve the algorithm Successive prime differences step by step in the XPL0 programming language

Published on 12 May 2024 09:40 PM
#Xpl0

How to resolve the algorithm Successive prime differences step by step in the XPL0 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 XPL0 programming language

Source code in the xpl0 programming language

func IsPrime(N);        \Return 'true' if N is prime
int  N, D;
[if N < 2 then return false;
if (N&1) = 0 then return N = 2;
if rem(N/3) = 0 then return N = 3;
D:= 5;
while D*D <= N do
        [if rem(N/D) = 0 then return false;
        D:= D+2;
        if rem(N/D) = 0 then return false;
        D:= D+4;
        ];
return true;
];

char Prime(1_000_000), S;
int  Diff, Diffs, Count, N, I, First, Last;
int  Str, Len;
[for N:= 0 to 1_000_000-1 do
    Prime(N):= if IsPrime(N) then ^1 else ^0;
Diffs:= [[2, 0, 0], [1, 0, 0], [2, 2+2, 0], [2, 4+2, 0], [4, 2+4, 0], [6, 4+6, 2+4+6]];
Str:= [ "101 ", "11 ", "10101 ", "1010001 ", "1000101 ", "1000001000101 "];
Len:= [      3,     2,        5,          7,          7,               13];
for Diff:= 0 to 6-1 do
    [Count:= 0;
    for N:= 0 to 1_000_000-1 -13 do
        [S:= Str(Diff);
        for I:= 0 to Len(Diff)-1 do
            if S(I) # Prime(N+I) then I:= 100;
        if I < 100 then         \have match
            [Count:= Count+1;
            if Count = 1 then First:= N;
            Last:= N;
            ];
        ];
    Text(0, "First: ");  IntOut(0, First);
    for I:= 0 to 2 do
        if Diffs(Diff,I) # 0 then
            [ChOut(0, ^ );  IntOut(0, First+Diffs(Diff,I))];
    Text(0, "  Last: ");  IntOut(0, Last);
    for I:= 0 to 2 do
        if Diffs(Diff,I) # 0 then
            [ChOut(0, ^ );  IntOut(0, Last+Diffs(Diff,I))];
    Text(0, " Groups: ");  IntOut(0, Count);  CrLf(0);
    ];
]

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