EKG sequence convergence

From Rosetta Code
Task
EKG sequence convergence
You are encouraged to solve this task according to the task description, using any language you may know.

The sequence is from the natural numbers and is defined by:

  • a(1) = 1;
  • a(2) = Start = 2;
  • for n > 2, a(n) shares at least one prime factor with a(n-1) and is the smallest such natural number not already used.


The sequence is called the EKG sequence (after its visual similarity to an electrocardiogram when graphed).

Variants of the sequence can be generated starting 1, N where N is any natural number larger than one. For the purposes of this task let us call:

  • The sequence described above , starting 1, 2, ... the EKG(2) sequence;
  • the sequence starting 1, 3, ... the EKG(3) sequence;
  • ... the sequence starting 1, N, ... the EKG(N) sequence.


Convergence

If an algorithm that keeps track of the minimum amount of numbers and their corresponding prime factors used to generate the next term is used, then this may be known as the generators essential state. Two EKG generators with differing starts can converge to produce the same sequence after initial differences.
EKG(N1) and EKG(N2) are said to to have converged at and after generation a(c) if state_of(EKG(N1).a(c)) == state_of(EKG(N2).a(c)).


Task
  1. Calculate and show here the first 10 members of EKG(2).
  2. Calculate and show here the first 10 members of EKG(5).
  3. Calculate and show here the first 10 members of EKG(7).
  4. Calculate and show here the first 10 members of EKG(9).
  5. Calculate and show here the first 10 members of EKG(10).
  6. Calculate and show here at which term EKG(5) and EKG(7) converge   (stretch goal).


Related Tasks
  1. Greatest common divisor
  2. Sieve of Eratosthenes


Reference



C[edit]

Translation of: Go
#include <stdio.h>
#include <stdlib.h>
 
#define TRUE 1
#define FALSE 0
#define LIMIT 100
 
typedef int bool;
 
int compareInts(const void *a, const void *b) {
int aa = *(int *)a;
int bb = *(int *)b;
return aa - bb;
}
 
bool contains(int a[], int b, size_t len) {
int i;
for (i = 0; i < len; ++i) {
if (a[i] == b) return TRUE;
}
return FALSE;
}
 
int gcd(int a, int b) {
while (a != b) {
if (a > b)
a -= b;
else
b -= a;
}
return a;
}
 
bool areSame(int s[], int t[], size_t len) {
int i;
qsort(s, len, sizeof(int), compareInts);
qsort(t, len, sizeof(int), compareInts);
for (i = 0; i < len; ++i) {
if (s[i] != t[i]) return FALSE;
}
return TRUE;
}
 
int main() {
int s, n, i;
int starts[5] = {2, 5, 7, 9, 10};
int ekg[5][LIMIT];
for (s = 0; s < 5; ++s) {
ekg[s][0] = 1;
ekg[s][1] = starts[s];
for (n = 2; n < LIMIT; ++n) {
for (i = 2; ; ++i) {
// a potential sequence member cannot already have been used
// and must have a factor in common with previous member
if (!contains(ekg[s], i, n) && gcd(ekg[s][n - 1], i) > 1) {
ekg[s][n] = i;
break;
}
}
}
printf("EKG(%2d): [", starts[s]);
for (i = 0; i < 30; ++i) printf("%d ", ekg[s][i]);
printf("\b]\n");
}
 
// now compare EKG5 and EKG7 for convergence
for (i = 2; i < LIMIT; ++i) {
if (ekg[1][i] == ekg[2][i] && areSame(ekg[1], ekg[2], i)) {
printf("\nEKG(5) and EKG(7) converge at term %d\n", i + 1);
return 0;
}
}
printf("\nEKG5(5) and EKG(7) do not converge within %d terms\n", LIMIT);
return 0;
}
Output:
EKG( 2): [1 2 4 6 3 9 12 8 10 5 15 18 14 7 21 24 16 20 22 11 33 27 30 25 35 28 26 13 39 36]
EKG( 5): [1 5 10 2 4 6 3 9 12 8 14 7 21 15 18 16 20 22 11 33 24 26 13 39 27 30 25 35 28 32]
EKG( 7): [1 7 14 2 4 6 3 9 12 8 10 5 15 18 16 20 22 11 33 21 24 26 13 39 27 30 25 35 28 32]
EKG( 9): [1 9 3 6 2 4 8 10 5 15 12 14 7 21 18 16 20 22 11 33 24 26 13 39 27 30 25 35 28 32]
EKG(10): [1 10 2 4 6 3 9 12 8 14 7 21 15 5 20 16 18 22 11 33 24 26 13 39 27 30 25 35 28 32]

EKG(5) and EKG(7) converge at term 21

Go[edit]

package main
 
import (
"fmt"
"sort"
)
 
func contains(a []int, b int) bool {
for _, j := range a {
if j == b {
return true
}
}
return false
}
 
func gcd(a, b int) int {
for a != b {
if a > b {
a -= b
} else {
b -= a
}
}
return a
}
 
func areSame(s, t []int) bool {
le := len(s)
if le != len(t) {
return false
}
sort.Ints(s)
sort.Ints(t)
for i := 0; i < le; i++ {
if s[i] != t[i] {
return false
}
}
return true
}
 
func main() {
const limit = 100
starts := [5]int{2, 5, 7, 9, 10}
var ekg [5][limit]int
 
for s, start := range starts {
ekg[s][0] = 1
ekg[s][1] = start
for n := 2; n < limit; n++ {
for i := 2; ; i++ {
// a potential sequence member cannot already have been used
// and must have a factor in common with previous member
if !contains(ekg[s][:n], i) && gcd(ekg[s][n-1], i) > 1 {
ekg[s][n] = i
break
}
}
}
fmt.Printf("EKG(%2d): %v\n", start, ekg[s][:30])
}
 
// now compare EKG5 and EKG7 for convergence
for i := 2; i < limit; i++ {
if ekg[1][i] == ekg[2][i] && areSame(ekg[1][:i], ekg[2][:i]) {
fmt.Println("\nEKG(5) and EKG(7) converge at term", i+1)
return
}
}
fmt.Println("\nEKG5(5) and EKG(7) do not converge within", limit, "terms")
}
Output:
EKG( 2): [1 2 4 6 3 9 12 8 10 5 15 18 14 7 21 24 16 20 22 11 33 27 30 25 35 28 26 13 39 36]
EKG( 5): [1 5 10 2 4 6 3 9 12 8 14 7 21 15 18 16 20 22 11 33 24 26 13 39 27 30 25 35 28 32]
EKG( 7): [1 7 14 2 4 6 3 9 12 8 10 5 15 18 16 20 22 11 33 21 24 26 13 39 27 30 25 35 28 32]
EKG( 9): [1 9 3 6 2 4 8 10 5 15 12 14 7 21 18 16 20 22 11 33 24 26 13 39 27 30 25 35 28 32]
EKG(10): [1 10 2 4 6 3 9 12 8 14 7 21 15 5 20 16 18 22 11 33 24 26 13 39 27 30 25 35 28 32]

EKG(5) and EKG(7) converge at term 21

Kotlin[edit]

Translation of: Go
// Version 1.2.60
 
fun gcd(a: Int, b: Int): Int {
var aa = a
var bb = b
while (aa != bb) {
if (aa > bb)
aa -= bb
else
bb -= aa
}
return aa
}
 
const val LIMIT = 100
 
fun main(args: Array<String>) {
val starts = listOf(2, 5, 7, 9, 10)
val ekg = Array(5) { IntArray(LIMIT) }
 
for ((s, start) in starts.withIndex()) {
ekg[s][0] = 1
ekg[s][1] = start
for (n in 2 until LIMIT) {
var i = 2
while (true) {
// a potential sequence member cannot already have been used
// and must have a factor in common with previous member
if (!ekg[s].slice(0 until n).contains(i) &&
gcd(ekg[s][n - 1], i) > 1) {
ekg[s][n] = i
break
}
i++
}
}
System.out.printf("EKG(%2d): %s\n", start, ekg[s].slice(0 until 30))
}
 
// now compare EKG5 and EKG7 for convergence
for (i in 2 until LIMIT) {
if (ekg[1][i] == ekg[2][i] &&
ekg[1].slice(0 until i).sorted() == ekg[2].slice(0 until i).sorted()) {
println("\nEKG(5) and EKG(7) converge at term ${i + 1}")
return
}
}
println("\nEKG5(5) and EKG(7) do not converge within $LIMIT terms")
}
Output:
EKG( 2): [1, 2, 4, 6, 3, 9, 12, 8, 10, 5, 15, 18, 14, 7, 21, 24, 16, 20, 22, 11, 33, 27, 30, 25, 35, 28, 26, 13, 39, 36]
EKG( 5): [1, 5, 10, 2, 4, 6, 3, 9, 12, 8, 14, 7, 21, 15, 18, 16, 20, 22, 11, 33, 24, 26, 13, 39, 27, 30, 25, 35, 28, 32]
EKG( 7): [1, 7, 14, 2, 4, 6, 3, 9, 12, 8, 10, 5, 15, 18, 16, 20, 22, 11, 33, 21, 24, 26, 13, 39, 27, 30, 25, 35, 28, 32]
EKG( 9): [1, 9, 3, 6, 2, 4, 8, 10, 5, 15, 12, 14, 7, 21, 18, 16, 20, 22, 11, 33, 24, 26, 13, 39, 27, 30, 25, 35, 28, 32]
EKG(10): [1, 10, 2, 4, 6, 3, 9, 12, 8, 14, 7, 21, 15, 5, 20, 16, 18, 22, 11, 33, 24, 26, 13, 39, 27, 30, 25, 35, 28, 32]

EKG(5) and EKG(7) converge at term 21

Perl 6[edit]

Works with: Rakudo Star version 2018.04.1
sub infix:<shares-divisors-with> { ($^a gcd $^b) > 1 }
 
sub next-EKG ( *@s ) {
return first {
@s$_ and @s.tail shares-divisors-with $_
}, 2..*;
}
 
sub EKG ( Int $start ) { 1, $start, &next-EKG … * }
 
sub converge-at ( @ints ) {
my @ekgs = @ints.map: &EKG;
 
return (2 .. *).first: -> $i {
[==] @ekgs.map( *.[$i] ) and
[===] @ekgs.map( *.head($i).Set )
}
}
 
say "EKG($_): ", .&EKG.head(10) for 2, 5, 7, 9, 10;
 
for [5, 7], [2, 5, 7, 9, 10] -> @ints {
say "EKGs of (@ints[]) converge at term {$_+1}" with converge-at(@ints);
}
Output:
EKG(2): (1 2 4 6 3 9 12 8 10 5)
EKG(5): (1 5 10 2 4 6 3 9 12 8)
EKG(7): (1 7 14 2 4 6 3 9 12 8)
EKG(9): (1 9 3 6 2 4 8 10 5 15)
EKG(10): (1 10 2 4 6 3 9 12 8 14)
EKGs of (5 7) converge at term 21
EKGs of (2 5 7 9 10) converge at term 45

Python[edit]

Python: Using math.gcd[edit]

If this alternate definition of function EKG_gen is used then the output would be the same as above. Instead of keeping a cache of prime factors this calculates the gretest common divisor as needed.

from itertools import count, islice, takewhile
from math import gcd
 
def EKG_gen(start=2):
"""\
Generate the next term of the EKG together with the minimum cache of
numbers left in its production; (the "state" of the generator).
Using math.gcd
"""

c = count(start + 1)
last, so_far = start, list(range(2, start))
yield 1, []
yield last, []
while True:
for index, sf in enumerate(so_far):
if gcd(last, sf) > 1:
last = so_far.pop(index)
yield last, so_far[::]
break
else:
so_far.append(next(c))
 
def find_convergence(ekgs=(5,7)):
"Returns the convergence point or zero if not found within the limit"
ekg = [EKG_gen(n) for n in ekgs]
for e in ekg:
next(e) # skip initial 1 in each sequence
return 2 + len(list(takewhile(lambda state: not all(state[0] == s for s in state[1:]),
zip(*ekg))))
 
if __name__ == '__main__':
for start in 2, 5, 7, 9, 10:
print(f"EKG({start}):", str([n[0] for n in islice(EKG_gen(start), 10)])[1: -1])
print(f"\nEKG(5) and EKG(7) converge at term {find_convergence(ekgs=(5,7))}!")
Output:

(Same as above).

EKG(2): 1, 2, 4, 6, 3, 9, 12, 8, 10, 5
EKG(5): 1, 5, 10, 2, 4, 6, 3, 9, 12, 8
EKG(7): 1, 7, 14, 2, 4, 6, 3, 9, 12, 8
EKG(9): 1, 9, 3, 6, 2, 4, 8, 10, 5, 15
EKG(10): 1, 10, 2, 4, 6, 3, 9, 12, 8, 14

EKG(5) and EKG(7) converge at term 21!
Note

Despite EKG(5) and EKG(7) seeming to converge earlier, as seen above; their hidden states differ.
Here is those series out to 21 terms where you can see them diverge again before finally converging. The state is also shown.

# After running the above, in the terminal:
from pprint import pprint as pp
 
for start in 5, 7:
print(f"EKG({start}):\n[(<next>, [<state>]), ...]")
pp(([n for n in islice(EKG_gen(start), 21)]))

Generates:

EKG(5):
[(<next>, [<state>]), ...]
[(1, []),
 (5, []),
 (10, [2, 3, 4, 6, 7, 8, 9]),
 (2, [3, 4, 6, 7, 8, 9]),
 (4, [3, 6, 7, 8, 9]),
 (6, [3, 7, 8, 9]),
 (3, [7, 8, 9]),
 (9, [7, 8]),
 (12, [7, 8, 11]),
 (8, [7, 11]),
 (14, [7, 11, 13]),
 (7, [11, 13]),
 (21, [11, 13, 15, 16, 17, 18, 19, 20]),
 (15, [11, 13, 16, 17, 18, 19, 20]),
 (18, [11, 13, 16, 17, 19, 20]),
 (16, [11, 13, 17, 19, 20]),
 (20, [11, 13, 17, 19]),
 (22, [11, 13, 17, 19]),
 (11, [13, 17, 19]),
 (33, [13, 17, 19, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32]),
 (24, [13, 17, 19, 23, 25, 26, 27, 28, 29, 30, 31, 32])]
EKG(7):
[(<next>, [<state>]), ...]
[(1, []),
 (7, []),
 (14, [2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13]),
 (2, [3, 4, 5, 6, 8, 9, 10, 11, 12, 13]),
 (4, [3, 5, 6, 8, 9, 10, 11, 12, 13]),
 (6, [3, 5, 8, 9, 10, 11, 12, 13]),
 (3, [5, 8, 9, 10, 11, 12, 13]),
 (9, [5, 8, 10, 11, 12, 13]),
 (12, [5, 8, 10, 11, 13]),
 (8, [5, 10, 11, 13]),
 (10, [5, 11, 13]),
 (5, [11, 13]),
 (15, [11, 13]),
 (18, [11, 13, 16, 17]),
 (16, [11, 13, 17]),
 (20, [11, 13, 17, 19]),
 (22, [11, 13, 17, 19, 21]),
 (11, [13, 17, 19, 21]),
 (33, [13, 17, 19, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32]),
 (21, [13, 17, 19, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32]),
 (24, [13, 17, 19, 23, 25, 26, 27, 28, 29, 30, 31, 32])]

REXX[edit]

/*REXX program can  generate and display several  EKG  sequences  (with various starts).*/
parse arg nums start /*obtain optional arguments from the CL*/
if nums=='' | nums=="," then nums= 50 /*Not specified? Then use the default.*/
if start= '' | start= "," then start=2 5 7 9 10 /* " " " " " " */
 
do s=1 for words(start); $= /*step through the specified STARTs. */
second= word(start, s); say /*obtain the second integer in the seq.*/
 
do j=1 for nums
if j<3 then do; #=1; if j==2 then #=second; end /*handle 1st & 2nd number*/
else #= ekg(#)
$= $ right(#, max(2, length(#) ) ) /*append the EKG integer to the $ list.*/
end /*j*/ /* [↑] the RIGHT BIF aligns the numbers*/
say '(start' right(second, max(2, length(second) ) )"):"$ /*display EKG seq.*/
end /*s*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
add_: do while z//j == 0; z=z%j; _=_ j; w=w+1; end; return strip(_)
/*──────────────────────────────────────────────────────────────────────────────────────*/
ekg: procedure expose $; parse arg x 1 z,,_
w=0 /*W: number of factors.*/
do k=1 to 11 by 2; j=k; if j==1 then j=2 /*divide by low primes. */
if j==9 then iterate; call add_ /*skip ÷ 9; add to list.*/
end /*k*/
/*↓ skips multiples of 3*/
do y=0 by 2; j= j + 2 + y//4 /*increment J by 2 or 4.*/
parse var j '' -1 r; if r==5 then iterate /*divisible by five ? */
if j*j>x | j>z then leave /*passed the sqrt(x) ? */
_= add_() /*add a factor to list. */
end /*y*/
j=z; if z\==1 then _= add_() /*Z¬=1? Then add──►list.*/
if _='' then _=x /*Null? Then use prime. */
do j=3; done=1
do k=1 for w
if j // word(_, k)==0 then do; done=0; leave; end
end /*k*/
if done then iterate
if wordpos(j, $)==0 then return j /*return an EKG integer.*/
end /*j*/
output   when using the default inputs:
(start  2):  1  2  4  6  3  9 12  8 10  5 15 18 14  7 21 24 16 20 22 11 33 27 30 25 35 28 26 13 39 36 32 34 17 51 42 38 19 57 45 40 44 46 23 69 48 50 52 54 56 49

(start  5):  1  5 10  4  6  3  9 12  8 14  7 21 15 18 16 20 22 11 33 24 26 13 39 27 30 25 35 28 32 34 17 51 36 38 19 57 42 40 44 46 23 69 45 48 50 52 54 56 49 63

(start  7):  1  7 14  4  6  3  9 12  8 10  5 15 18 16 20 22 11 33 21 24 26 13 39 27 30 25 35 28 32 34 17 51 36 38 19 57 42 40 44 46 23 69 45 48 50 52 54 56 49 63

(start  9):  1  9  3  6  4  8 10  5 15 12 14  7 21 18 16 20 22 11 33 24 26 13 39 27 30 25 35 28 32 34 17 51 36 38 19 57 42 40 44 46 23 69 45 48 50 52 54 56 49 63

(start 10):  1 10  4  6  3  9 12  8 14  7 21 15  5 20 16 18 22 11 33 24 26 13 39 27 30 25 35 28 32 34 17 51 36 38 19 57 42 40 44 46 23 69 45 48 50 52 54 56 49 63

zkl[edit]

Using gcd hint from Go.

fcn ekgW(N){	// --> iterator
Walker.tweak(fcn(rp,buf,w){
foreach n in (w){
if(rp.value.gcd(n)>1)
{ rp.set(n); w.push(buf.xplode()); buf.clear(); return(n); }
buf.append(n); // save small numbers not used yet
}
}.fp(Ref(N),List(),Walker.chain([2..N-1],[N+1..]))).push(1,N)
}
foreach n in (T(2,5,7,9,10)){ println("EKG(%2d): %s".fmt(n,ekgW(n).walk(10).concat(","))) }
Output:
EKG( 2): 1,2,4,6,3,9,12,8,10,5
EKG( 5): 1,5,10,2,4,6,3,9,12,8
EKG( 7): 1,7,14,2,4,6,3,9,12,8
EKG( 9): 1,9,3,6,2,4,8,10,5,15
EKG(10): 1,10,2,4,6,3,9,12,8,14
fcn convergeAt(n1,n2,etc){ ns:=vm.arglist;
ekgWs:=ns.apply(ekgW); ekgWs.apply2("next"); // pop initial 1
ekgNs:=List()*vm.numArgs; // ( (ekg(n1)), (ekg(n2)) ...)
do(1_000){ // find convergence in this many terms or bail
ekgN:=ekgWs.apply("next"); // (ekg(n1)[n],ekg(n2)[n] ...)
ekgNs.zipWith(fcn(ns,n){ ns.merge(n) },ekgN); // keep terms sorted
// are all ekg[n]s == and both sequences have same terms?
if(not ekgN.filter1('!=(ekgN[0])) and not ekgNs.filter1('!=(ekgNs[0])) ){
println("EKG(", ns.concat(","), ") converge at term ",ekgNs[0].len() + 1);
return();
}
}
println(ns.concat(",")," don't converge");
}
convergeAt(5,7);
convergeAt(2,5,7,9,10);
Output:
EKG(5,7) converge at term 21
EKG(2,5,7,9,10) converge at term 45