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 witha(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, ...
theEKG(2)
sequence; - the sequence starting
1, 3, ...
theEKG(3)
sequence; - ... the sequence starting
1, N, ...
theEKG(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
- Calculate and show here the first 30 members of
EKG(2)
. - Calculate and show here the first 30 members of
EKG(5)
. - Calculate and show here the first 30 members of
EKG(7)
. - Calculate and show here the first 30 members of
EKG(9)
. - Calculate and show here the first 30 members of
EKG(10)
. - Stretch goal: Calculate and show here at which term EKG(5) and EKG(7) converge.
- Reference
- The EKG Sequence and the Tree of Numbers. (Video).
C
<lang c>#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[3] = {2, 5, 7}; int ekg[3][LIMIT]; for (s = 0; s < 3; ++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(%d): [", starts[s]); for (i = 0; i < 10; ++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;
}</lang>
- 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(5) and EKG(7) converge at term 21
Go
<lang go>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 := [3]int{2, 5, 7} var ekg [3][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(%d): %v\n", start, ekg[s][:10]) }
// 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")
}</lang>
- 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(5) and EKG(7) converge at term 21
Kotlin
<lang scala>// 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) val ekg = Array(3) { 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(%d): %s\n", start, ekg[s].slice(0 until 10)) }
// 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 $LIMIT terms")
}</lang>
- Output:
EKG(2): [1, 2, 4, 6, 3, 9, 12, 8, 10, 5] EKG(5): [1, 5, 10, 2, 4, 6, 3, 9, 12, 8] EKG(7): [1, 7, 14, 2, 4, 6, 3, 9, 12, 8] EKG(5) and EKG(7) converge at term 21
Perl 6
<lang perl6>sub infix:<shares-divisors-with> { ($^a gcd $^b) > 1 }
sub next-EKG ( *@s ) {
return first { @s ∌ $_ and @s.tail shares-divisors-with $_ }, 2..*;
}
sub EKG ( Int $start ) { 1, $start, &next-EKG … * }
say "EKG($_): ", .&EKG.head(10) for 2, 5, 7;</lang>
- 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)
Python
Python: Using prime factor generation
<lang python>from itertools import count, islice, takewhile from functools import lru_cache
primes = [2, 3, 5, 7, 11, 13, 17] # Will be extended
@lru_cache(maxsize=2000) def pfactor(n):
# From; http://rosettacode.org/wiki/Count_in_factors if n == 1: return [1] n2 = n // 2 + 1 for p in primes: if p <= n2: d, m = divmod(n, p) if m == 0: if d > 1: return [p] + pfactor(d) else: return [p] else: if n > primes[-1]: primes.append(n) return [n]
def next_pfactor_gen():
"Generate (n, set(pfactor(n))) pairs for n == 2, ..." for n in count(2): yield n, set(pfactor(n))
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). """ def min_share_pf_so_far(pf, so_far): "searches the unused smaller number prime factors" nonlocal found for index, (num, pfs) in enumerate(so_far): if pf & pfs: found = so_far.pop(index) break else: found = () return found
yield 1, [] last, last_pf = start, set(pfactor(start)) pf_gen = next_pfactor_gen() # minimum list of prime factors needed so far so_far = list(islice(pf_gen, start)) found = () while True: while not min_share_pf_so_far(last_pf, so_far): so_far.append(next(pf_gen)) yield found[0], [n for n, _ in so_far] last, last_pf = found
- %%
def find_convergence(ekgs=(5,7), limit=1_000):
"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 differ = list(takewhile(lambda state: not all(state[0] == s for s in state[1:]), islice(zip(*ekg), limit-1))) ldiff = len(differ) return ldiff + 2 if ldiff < limit-1 else 0
- %%
if __name__ == '__main__':
for start in 2, 5, 7: 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))}!")</lang>
- 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(5) and EKG(7) converge at term 21!
Python: Using math.gcd
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. <lang python>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: 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))}!")</lang>
- 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(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.
<lang python># 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)]))</lang>
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])]
zkl
Using gcd hint from Go. <lang zkl>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)
}</lang> <lang zkl>foreach n in (T(2,5,7,10)){ println("EKG(%d): %s".fmt(n,ekgW(n).walk(10))) }</lang>
- Output:
EKG(2): L(1,2,4,6,3,9,12,8,10,5) EKG(5): L(1,5,10,2,4,6,3,9,12,8) EKG(7): L(1,7,14,2,4,6,3,9,12,8) EKG(10): L(1,10,2,4,6,3,9,12,8,14)
<lang zkl>ekg5,ekg7, ekg5W,ekg7W := List(),List(), ekgW(5),ekgW(7); ekg5W.next(); ekg7W.next(); // pop initial 1 foreach e5,e7 in (ekg5W.zip(ekg7W)){
ekg5.merge(e5); ekg7.merge(e7); // keep terms sorted if(e5==e7 and ekg5==ekg7){ // a(n) are ==, both sequences have same terms println("EKG(5) and EKG(7) converge at term ",ekg7.len() + 1); break; } // should put limiter here
}</lang>
- Output:
EKG(5) and EKG(7) converge at term 21