Van Eck Sequence

June 14, 2019

We maintain an a-list prev that stores, for each v in the list, the index i at which it appears. There will be one entry in the a-list for each member of the sequence; repeated sequence entries will appear multiple times, and assoc will return the most recent, which stores the associated index:

(define-generator (van-eck)
  (let loop ((v 0) (i 0) (prev (list)))
    (yield v)
    (let ((t v))
      (let ((v (- i (cond ((assoc v prev) => cdr) (else i)))))
        (loop v (+ i 1) (cons (cons t i) prev))))))

Here is the function in action:

> (take-gen 25 (van-eck))
(0 0 1 0 2 0 2 2 1 6 0 5 0 2 6 5 4 0 5 3 0 3 2 9 0)

You can run the program at https://ideone.com/tKuo7T.

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Posted by programmingpraxis
Filed in Exercises
4 Comments »

4 Responses to “Van Eck Sequence”

  1. Zack said
    June 14, 2019 at 11:39 AM

    Fun little sequence, though not much of a programming challenge. Here is my take on it using Julia 1.0:

    function main(n::Int64 = 100)
    x = zeros(Int64, n)

    for i = 3:n
    z = x[i-1]

    for j = (i-2):-1:1

        if x[j] == z
            x[i] = i - j - 1
            break
        end
    end
end

return x

    end

    Interestingly, the sequence yields larger gaps towards the higher numbers, for the samples tested. This makes the sequence “complete” for the early part of it, a pattern that is likely to continue for larger samples. Still not convinced about its usefulness, but a fun little exercise nevertheless. Have a nice weekend!

  2. Paul said
    June 14, 2019 at 1:20 PM

    In Python.

    def vaneck():
    L, x, n, prev = {}, 0, 0, -1
    while True:
        yield x
        L[prev], prev, n = n, x, n+1
        x = max(0, n - L.get(prev, n))
  3. chaw said
    June 14, 2019 at 1:55 PM

    Here is a R7RS Scheme + (srfi 69) solution that runs in linear time
    (or constant time per element if implemented as a generator), with the
    usual hash-table caveat.

    (import (scheme base)
        (scheme write)
        (only (srfi 69)
              make-hash-table hash-table-ref/default hash-table-set!
              hash-table-size))

(define (display-van-eck-sequence num-elements unique-count-only?)
  ;; ht[x] = max 0-based index i s.t. seq(i) = x, where seq is
  ;; sequence of elements already outputted, except the very last
  ;; (previous) one. ht[-1] = -1 is a special base case.
  (let ((ht (make-hash-table =)))
    ;; loop on current index and previous item:
    (let loop ((idx 0)
               (prev-item -1))
      (when (< idx num-elements)
        (let ((item (- idx
                       (hash-table-ref/default ht
                                               prev-item
                                               idx))))
          (unless unique-count-only? (display item))
          (hash-table-set! ht prev-item idx)
          (loop (+ idx 1)
                item))))
    (newline)
    (display (hash-table-size ht))
    (newline)))

(display-van-eck-sequence 20 #f)
(display-van-eck-sequence 1000 #t)
(display-van-eck-sequence #e1e6 #t)


    0 0 1 0 2 0 2 2 1 6 0 5 0 2 6 5 4 0 5 3
7

198

136529

  4. Daniel said
    June 30, 2019 at 10:31 PM

    Here’s a solution in C++.

    #include <cstdlib>
#include <iostream>
#include <unordered_map>

using namespace std;

int main(int argc, char* argv[]) {
  if (argc != 2) return EXIT_FAILURE;
  int n = stoi(argv[1]);
  int current = 0;
  int next = 0;
  unordered_map<int, int> map;
  for (int i = 0; i < n; ++i) {
    current = next;
    if (map.count(current)) {
      next = i - map[current];
    } else {
      next = 0;
    }
    cout << current << endl;
    map[current] = i;
  }
  return EXIT_SUCCESS;
}

    Example Usage:

    $ ./a.out 18
0
0
1
0
2
0
2
2
1
6
0
5
0
2
6
5
4
0

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