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{{task}} The sequence is generated by following this pseudo-code:
A: The first term is zero.
Repeatedly apply:
If the last term is *new* to the sequence so far then:
B: The next term is zero.
Otherwise:
C: The next term is how far back this last term occured previousely.
;Example:
Using A:
:0
Using B:
:0 0
Using C:
:0 0 1
Using B:
:0 0 1 0
Using C: (zero last occured two steps back - before the one)
:0 0 1 0 2
Using B:
:0 0 1 0 2 0
Using C: (two last occured two steps back - before the zero)
:0 0 1 0 2 0 2 2
Using C: (two last occured one step back)
:0 0 1 0 2 0 2 2 1
Using C: (one last appeared six steps back)
:0 0 1 0 2 0 2 2 1 6
...
;Task:
Create a function/proceedure/method/subroutine/... to generate the Van Eck sequence of numbers.
Use it to display here, on this page:
:# The first ten terms of the sequence. :# Terms 991 - to - 1000 of the sequence.
;References:
- [https://www.youtube.com/watch?v=etMJxB-igrc Don't Know (the Van Eck Sequence) - Numberphile video].
- [[wp:Van_Eck%27s_sequence|Wikipedia Article: Van Eck's Sequence]].
- [[oeis:A181391| OEIS sequence: A181391]].
AppleScript
AppleScript is not the tool for the job, but here is a quick assembly from ready-made parts:
use AppleScript version "2.4"
use scripting additions
-- vanEck :: Int -> [Int]
on vanEck(n)
script go
on |λ|(xxs)
maybe(0, elemIndex(item 1 of xxs, rest of xxs)) & xxs
end |λ|
end script
reverse of applyN(n - 1, go, {0})
end vanEck
-- TEST ---------------------------------------------------
on run
{vanEck(10), ¬
items 991 thru 1000 of vanEck(1000)}
end run
-- GENERIC ------------------------------------------------
-- Just :: a -> Maybe a
on Just(x)
-- Constructor for an inhabited Maybe (option type) value.
-- Wrapper containing the result of a computation.
{type:"Maybe", Nothing:false, Just:x}
end Just
-- Nothing :: Maybe a
on Nothing()
-- Constructor for an empty Maybe (option type) value.
-- Empty wrapper returned where a computation is not possible.
{type:"Maybe", Nothing:true}
end Nothing
-- applyN :: Int -> (a -> a) -> a -> a
on applyN(n, f, x)
script go
on |λ|(a, g)
|λ|(a) of mReturn(g)
end |λ|
end script
foldl(go, x, replicate(n, f))
end applyN
-- elemIndex :: Eq a => a -> [a] -> Maybe Int
on elemIndex(x, xs)
set lng to length of xs
repeat with i from 1 to lng
if x = (item i of xs) then return Just(i)
end repeat
return Nothing()
end elemIndex
-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl
-- maybe :: a -> Maybe a -> a
on maybe(v, mb)
if Nothing of mb then
v
else
Just of mb
end if
end maybe
-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
-- 2nd class handler function lifted into 1st class script wrapper.
if script is class of f then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- Egyptian multiplication - progressively doubling a list, appending
-- stages of doubling to an accumulator where needed for binary
-- assembly of a target length
-- replicate :: Int -> a -> [a]
on replicate(n, a)
set out to {}
if 1 > n then return out
set dbl to {a}
repeat while (1 < n)
if 0 < (n mod 2) then set out to out & dbl
set n to (n div 2)
set dbl to (dbl & dbl)
end repeat
return out & dbl
end replicate
{{Out}}
{{0, 0, 1, 0, 2, 0, 2, 2, 1, 6}, {4, 7, 30, 25, 67, 225, 488, 0, 10, 136}}
AWK
# syntax: GAWK -f VAN_ECK_SEQUENCE.AWK
# converted from Go
BEGIN {
limit = 1000
for (i=0; i<limit; i++) {
arr[i] = 0
}
for (n=0; n<limit-1; n++) {
for (m=n-1; m>=0; m--) {
if (arr[m] == arr[n]) {
arr[n+1] = n - m
break
}
}
}
printf("terms 1-10:")
for (i=0; i<10; i++) { printf(" %d",arr[i]) }
printf("\n")
printf("terms 991-1000:")
for (i=990; i<1000; i++) { printf(" %d",arr[i]) }
printf("\n")
exit(0)
}
{{out}}
terms 1-10: 0 0 1 0 2 0 2 2 1 6
terms 991-1000: 4 7 30 25 67 225 488 0 10 136
C
#include <iostream>
#include <stdio.h>
int main(int argc, const char *argv[]) {
const int max = 1000;
int *a = malloc(max * sizeof(int));
for (int n = 0; n < max - 1; n ++) {
for (int m = n - 1; m >= 0; m --) {
if (a[m] == a[n]) {
a[n+1] = n - m;
break;
}
}
}
printf("The first ten terms of the Van Eck sequence are:\n");
for (int i = 0; i < 10; i ++) printf("%d ", a[i]);
printf("\n\nTerms 991 to 1000 of the sequence are:\n");
for (int i = 990; i < 1000; i ++) printf("%d ", a[i]);
putchar('\n');
return 0;
}
{{Out}}
The first ten terms of the Van Eck sequence are:
0 0 1 0 2 0 2 2 1 6
Terms 991 to 1000 of the sequence are:
4 7 30 25 67 225 488 0 10 136
Clojure
(defn van-eck
([] (van-eck 0 0 {}))
([val n seen]
(lazy-seq
(cons val
(let [next (- n (get seen val n))]
(van-eck next
(inc n)
(assoc seen val n)))))))
(println "First 10 terms:" (take 10 (van-eck)))
(println "Terms 991 to 1000 terms:" (take 10 (drop 990 (van-eck))))
{{out}}
First 10 terms: (0 0 1 0 2 0 2 2 1 6)
Terms 991 to 1000 terms: (4 7 30 25 67 225 488 0 10 136)
Common Lisp
;;Tested using CLISP
(defun VanEck (x) (reverse (VanEckh x 0 0 '(0))))
(defun VanEckh (final index curr lst)
(if (eq index final)
lst
(VanEckh final (+ index 1) (howfar curr lst) (cons curr lst))))
(defun howfar (x lst) (howfarh x lst 0))
(defun howfarh (x lst runningtotal)
(cond
((null lst) 0)
((eq x (car lst)) (+ runningtotal 1))
(t (howfarh x (cdr lst) (+ runningtotal 1)))))
(format t "The first 10 elements are ~a~%" (VanEck 9))
(format t "The 990-1000th elements are ~a~%" (nthcdr 990 (VanEck 999)))
{{out}}
The first 10 elements are (0 0 1 0 2 0 2 2 1 6)
The 990-1000th elements are (4 7 30 25 67 225 488 0 10 136)
Dyalect
{{trans|Go}}
const max = 1000
var a = Array.empty(max, 0)
for n in 0..(max-2) {
var m = n - 1
while m >= 0 {
if a[m] == a[n] {
a[n+1] = n - m
break
}
m -= 1
}
}
print("The first ten terms of the Van Eck sequence are: \(a[0..10])")
print("Terms 991 to 1000 of the sequence are: \(a[991..1000])")
{{out}}
The first ten terms of the Van Eck sequence are: {0, 0, 1, 0, 2, 0, 2, 2, 1, 6}
Terms 991 to 1000 of the sequence are: {7, 30, 25, 67, 225, 488, 0, 10, 136}
=={{header|F_Sharp|F#}}== {{incomplete|F_Sharp|F#|Too much output given - see talk page}}
The function
// Generate Van Eck's Sequence. Nigel Galloway: June 19th., 2019
let ecK()=let n=System.Collections.Generic.Dictionary<int,int>()
Seq.unfold(fun (g,e)->Some(g,((if n.ContainsKey g then let i=n.[g] in n.[g]<-e;e-i else n.[g]<-e;0),e+1)))(0,0)
The Task
;First 50
ecK() |> Seq.take 50 |> Seq.iter(printf "%d "); printfn "";;
{{out}}
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 4 9 3 6 14 0 6 3 5 15 0 5 3 5 2 17 0 6 11 0 3 8 0 3 3
;50 from 991
ecK() |> Seq.skip 990 |> Seq.take 50|> Seq.iter(printf "%d "); printfn "";;
{{out}}
4 7 30 25 67 225 488 0 10 136 61 0 4 12 72 0 4 4 1 24 41 385 0 7 22 25 22 2 84 68 282 464 0 10 25 9 151 697 0 6 41 20 257 539 0 6 6 1 29 465
;I thought the longest sequence of non zeroes in the first 100 million items might be interesting It occurs between 32381749 and 32381774:
9 47 47 1 10 33 27 548 548 1 6 33 6 2 154 15657 695734 270964 235721 238076 4896139 655158 7901804 146089 977945 21475977
Factor
USING: assocs fry kernel make math namespaces prettyprint
sequences ;
: van-eck ( n -- seq )
[
0 , 1 - H{ } clone '[
building get [ length 1 - ] [ last ] bi _ 3dup
2dup key? [ at - ] [ 3drop 0 ] if , set-at
] times
] { } make ;
1000 van-eck 10 [ head ] [ tail* ] 2bi [ . ] bi@
{{out}}
{ 0 0 1 0 2 0 2 2 1 6 }
{ 4 7 30 25 67 225 488 0 10 136 }
=={{header|Fōrmulæ}}==
In [https://wiki.formulae.org/Van_Eck_sequence this] page you can see the solution of this task.
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text ([http://wiki.formulae.org/Editing_F%C5%8Drmul%C3%A6_expressions more info]). Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for transportation effects more than visualization and edition.
The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code.
FreeBASIC
Const limite = 1000
Dim As Integer a(limite), n, m, i
For n = 0 To limite-1
For m = n-1 To 0 Step -1
If a(m) = a(n) Then a(n+1) = n-m: Exit For
Next m
Next n
Print "Secuencia de Van Eck:" &Chr(10)
Print "Primeros 10 terminos: ";
For i = 0 To 9
Print a(i) &" ";
Next i
Print Chr(10) & "Terminos 991 al 1000: ";
For i = 990 To 999
Print a(i) &" ";
Next i
End
{{out}}
Secuencia de Van Eck:
Primeros 10 terminos: 0 0 1 0 2 0 2 2 1 6
Terminos 991 al 1000: 4 7 30 25 67 225 488 0 10 136
Go
package main
import "fmt"
func main() {
const max = 1000
a := make([]int, max) // all zero by default
for n := 0; n < max-1; n++ {
for m := n - 1; m >= 0; m-- {
if a[m] == a[n] {
a[n+1] = n - m
break
}
}
}
fmt.Println("The first ten terms of the Van Eck sequence are:")
fmt.Println(a[:10])
fmt.Println("\nTerms 991 to 1000 of the sequence are:")
fmt.Println(a[990:])
}
{{out}}
The first ten terms of the Van Eck sequence are:
[0 0 1 0 2 0 2 2 1 6]
Terms 991 to 1000 of the sequence are:
[4 7 30 25 67 225 488 0 10 136]
Alternatively, using a map to store the latest index of terms previously seen (output as before):
package main
import "fmt"
func main() {
const max = 1000
a := make([]int, max) // all zero by default
seen := make(map[int]int)
for n := 0; n < max-1; n++ {
if m, ok := seen[a[n]]; ok {
a[n+1] = n - m
}
seen[a[n]] = n
}
fmt.Println("The first ten terms of the Van Eck sequence are:")
fmt.Println(a[:10])
fmt.Println("\nTerms 991 to 1000 of the sequence are:")
fmt.Println(a[990:])
}
Haskell
import Data.List (elemIndex)
import Data.Maybe (maybe)
vanEck :: Int -> [Int]
vanEck n = reverse $ iterate go [] !! n
where
go [] = [0]
go xxs@(x:xs) = maybe 0 succ (elemIndex x xs) : xxs
main :: IO ()
main = do
print $ vanEck 10
print $ drop 990 (vanEck 1000)
{{Out}}
[0,0,1,0,2,0,2,2,1,6]
[4,7,30,25,67,225,488,0,10,136]
And if we wanted to look a little further than the 1000th term, we could accumulate a Map of most recently seen positions to improve performance:
import qualified Data.Map.Strict as M hiding (drop)
import Data.List (mapAccumL)
import Data.Maybe (maybe)
vanEck :: [Int]
vanEck = 0 : snd (mapAccumL go (0, M.empty) [1 ..])
where
go (x, dct) i =
let v = maybe 0 (i -) (M.lookup x dct)
in ((v, M.insert x i dct), v)
main :: IO ()
main =
mapM_ print $
(drop . subtract 10 <*> flip take vanEck) <$>
[10, 1000, 10000, 100000, 1000000]
{{Out}}
[0,0,1,0,2,0,2,2,1,6]
[4,7,30,25,67,225,488,0,10,136]
[7,43,190,396,2576,3142,0,7,7,1]
[92,893,1125,47187,0,7,34,113,140,2984]
[8,86,172,8878,172447,0,6,30,874,34143]
J
===The tacit verb (function)===
VanEck=. (, (<:@:# - }: i: {:))^:(]`0:)
The output
VanEck 9
0 0 1 0 2 0 2 2 1 6
990 }. VanEck 999
4 7 30 25 67 225 488 0 10 136
===A structured derivation of the verb (function)===
next =. <:@:# - }: i: {: NB. Next term of the sequence
VanEck=. (, next)^:(]`0:) f. NB. Appending terms and fixing the verb
JavaScript
Either declaratively, without premature optimization: {{Trans|Python}}
(() => {
'use strict';
// vanEck :: Int -> [Int]
const vanEck = n =>
reverse(
churchNumeral(n)(
xs => 0 < xs.length ? cons(
maybe(
0, succ,
elemIndex(xs[0], xs.slice(1))
),
xs
) : [0]
)([])
);
// TEST -----------------------------------------------
const main = () => {
console.log('VanEck series:\n')
showLog('First 10 terms', vanEck(10))
showLog('Terms 991-1000', vanEck(1000).slice(990))
};
// GENERIC FUNCTIONS ----------------------------------
// Just :: a -> Maybe a
const Just = x => ({
type: 'Maybe',
Nothing: false,
Just: x
});
// Nothing :: Maybe a
const Nothing = () => ({
type: 'Maybe',
Nothing: true,
});
// churchNumeral :: Int -> (a -> a) -> a -> a
const churchNumeral = n => f => x =>
Array.from({
length: n
}, () => f)
.reduce((a, g) => g(a), x)
// cons :: a -> [a] -> [a]
const cons = (x, xs) => [x].concat(xs)
// elemIndex :: Eq a => a -> [a] -> Maybe Int
const elemIndex = (x, xs) => {
const i = xs.indexOf(x);
return -1 === i ? (
Nothing()
) : Just(i);
};
// maybe :: b -> (a -> b) -> Maybe a -> b
const maybe = (v, f, m) =>
m.Nothing ? v : f(m.Just);
// reverse :: [a] -> [a]
const reverse = xs =>
'string' !== typeof xs ? (
xs.slice(0).reverse()
) : xs.split('').reverse().join('');
// showLog :: a -> IO ()
const showLog = (...args) =>
console.log(
args
.map(JSON.stringify)
.join(' -> ')
);
// succ :: Int -> Int
const succ = x => 1 + x;
// MAIN ---
return main();
})();
{{Out}}
VanEck series:
"First 10 terms" -> [0,0,1,0,2,0,2,2,1,6]
"Terms 991-1000" -> [4,7,30,25,67,225,488,0,10,136]
or as a map-accumulation, building a look-up table: {{Trans|Python}}
(() => {
'use strict';
// vanEck :: Int -> [Int]
const vanEck = n =>
// First n terms of the vanEck series.
[0].concat(mapAccumL(
([x, seen], i) => {
const
prev = seen[x],
v = 0 !== prev ? (
i - prev
) : 0;
return [
[v, (seen[x] = i, seen)], v
];
}, [0, replicate(n - 1, 0)],
enumFromTo(1, n - 1)
)[1]);
// TEST -----------------------------------------------
const main = () =>
console.log(fTable(
'Terms of the VanEck series:\n',
n => str(n - 10) + '-' + str(n),
xs => JSON.stringify(xs.slice(-10)),
vanEck,
[10, 1000, 10000]
))
// GENERIC FUNCTIONS ----------------------------------
// enumFromTo :: Int -> Int -> [Int]
const enumFromTo = (m, n) =>
Array.from({
length: 1 + n - m
}, (_, i) => m + i);
// fTable :: String -> (a -> String) -> (b -> String) ->
// (a -> b) -> [a] -> String
const fTable = (s, xShow, fxShow, f, xs) => {
// Heading -> x display function ->
// fx display function ->
// f -> values -> tabular string
const
ys = xs.map(xShow),
w = Math.max(...ys.map(x => x.length));
return s + '\n' + zipWith(
(a, b) => a.padStart(w, ' ') + ' -> ' + b,
ys,
xs.map(x => fxShow(f(x)))
).join('\n');
};
// Map-accumulation is a combination of map and a catamorphism;
// it applies a function to each element of a list, passing an accumulating
// parameter from left to right, and returning a final value of this
// accumulator together with the new list.
// mapAccumL :: (acc -> x -> (acc, y)) -> acc -> [x] -> (acc, [y])
const mapAccumL = (f, acc, xs) =>
xs.reduce((a, x, i) => {
const pair = f(a[0], x, i);
return [pair[0], a[1].concat(pair[1])];
}, [acc, []]);
// replicate :: Int -> a -> [a]
const replicate = (n, x) =>
Array.from({
length: n
}, () => x);
// str :: a -> String
const str = x => x.toString();
// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
const zipWith = (f, xs, ys) => {
const
lng = Math.min(xs.length, ys.length),
as = xs.slice(0, lng),
bs = ys.slice(0, lng);
return Array.from({
length: lng
}, (_, i) => f(as[i], bs[i], i));
};
// MAIN ---
return main();
})();
{{Out}}
Terms of the VanEck series:
0-10 -> [0,0,1,0,2,0,2,2,1,6]
990-1000 -> [4,7,30,25,67,225,488,0,10,136]
9990-10000 -> [7,43,190,396,2576,3142,0,7,7,1]
Julia
function vanecksequence(N, startval=0)
ret = zeros(Int, N)
ret[1] = startval
for i in 1:N-1
lastseen = findlast(x -> x == ret[i], ret[1:i-1])
if lastseen != nothing
ret[i + 1] = i - lastseen
end
end
ret
end
println(vanecksequence(10))
println(vanecksequence(1000)[991:1000])
{{out}}
[0, 0, 1, 0, 2, 0, 2, 2, 1, 6]
[4, 7, 30, 25, 67, 225, 488, 0, 10, 136]
Alternate version, with a Dict for memoization (output is the same):
function vanecksequence(N, startval=0)
ret = zeros(Int, N)
ret[1] = startval
lastseen = Dict{Int, Int}()
for i in 1:N-1
if haskey(lastseen, ret[i])
ret[i + 1] = i - lastseen[ret[i]]
end
lastseen[ret[i]] = i
end
ret
end
Lua
-- Return a table of the first n values of the Van Eck sequence
function vanEck (n)
local seq, foundAt = {0}
while #seq < n do
foundAt = nil
for pos = #seq - 1, 1, -1 do
if seq[pos] == seq[#seq] then
foundAt = pos
break
end
end
if foundAt then
table.insert(seq, #seq - foundAt)
else
table.insert(seq, 0)
end
end
return seq
end
-- Show the set of values in table t from key numbers lo to hi
function showValues (t, lo, hi)
for i = lo, hi do
io.write(t[i] .. " ")
end
print()
end
-- Main procedure
local sequence = vanEck(1000)
showValues(sequence, 1, 10)
showValues(sequence, 991, 1000)
{{out}}
0 0 1 0 2 0 2 2 1 6
4 7 30 25 67 225 488 0 10 136
Pascal
I memorize the last position of each number that occured and use a circular buffer to remember last values. Running once through the list of last positions maybe faster [https://tio.run/##jVTbThsxEH3frxipDyR0k2xAFWjToAIFNVKBqoEKKULIOLPEZfFubW9CqPrrTT129sKlVf2wlzlzOz5jZzffkZtOzjRnaSfJ@WqVq@xWsXv4xuQRvxsEP4NN2I8BzmcIiVDagEF1D0LDI6qsG4BdXzFHZnCaLoHlebqMnZXWKAFjA1PWiNuUuNgEkzlE448CJbcfGSRMkU3GwUEM1aLCEh9eq0vrzEaohdAYB4f/ipplC1fghvE7W8Va6qYyzguFU8gVzkVWaEyX3V@BfWtbRi91YUSqBwHPpDbWcrJ/eXpxAkPY3tre7e/svBv0ev0oijarx8B7jY@OTkcfL61nqw96lsJOu9MfBHOmLP4l0weYZAohBqYUW0KWwMVImu0tCv9suxsjSijhSdTt1jmvaudeD7hQvEiJXZEkSNkpdDR9CD@xObo0MVwInzuwEnOcWsZwVrgarUNp4lNmxBxtynbZ4SgcX9scNrTCBsEB3gpJ4DHYKNhr8iTxnDCExMMGRIzsLKxDqqZeRJQI@a@LD0suT2wt99U5dO26bjz0HqI6qZDc@4V1J879OFMgKE2f5pBqTzNrLrkBLJQw2NqAjbDUYeISXbnwdR9VG2/7bdg//fiML8opvVyq1DJy/43NP0dtWifsgd4kgNen2n0RMm5kcR/aQRFeiVpDgNzPT5h7dfMSKTkcizTlM6ZaJYFwLB7xLKn@2@GbyLGpR2QI0Xoz67b2ynGvdvXocnRObr2eRpOivDUzYHIKPEXXeGVtVSMe@kqvIXUl61LzomY@VF6T6Mpx9kPjsEoXD1WD4Rh4benL72EFKHdTORr22GjLHOYsLdBZ8obOFOFj/19wyokP7F5IunsWPrM91cY2nRWGzjdnfOaLkapUZM134otd@TzPjOQnPGK1cZHD5pw/oeksmGp8AYkOhQ6qoyGcKIW0t5vdsb2G6M/HQnTcZq4PojuT1@5E@pkuZ85NdD9qAwyqq4WuQ6/@GqW/BlpDfs4a4NbuX6eGwmzt7mr1mycpu9Wrztn2Hw Try it online!] takes only 1.4 s for 32,381,775
program VanEck;
{
* A: The first term is zero.
Repeatedly apply:
If the last term is *new* to the sequence so far then:
B: The next term is zero.
Otherwise:
C: The next term is how far back this last term occured previousely.}
uses
sysutils;
const
MAXNUM = 32381775;//1000*1000*1000;
MAXSEENIDX = (1 shl 7)-1;
var
PosBefore : array of UInt32;
LastSeen : array[0..MAXSEENIDX]of UInt32;// circular buffer
SeenIdx,HaveSeen : Uint32;
procedure OutSeen(Cnt:NativeInt);
var
I,S_Idx : NativeInt;
Begin
IF Cnt > MAXSEENIDX then
Cnt := MAXSEENIDX;
If Cnt > HaveSeen then
Cnt := HaveSeen;
S_Idx := SeenIdx;
S_Idx := (S_Idx-Cnt);
IF S_Idx < 0 then
inc(S_Idx,MAXSEENIDX);
For i := 1 to Cnt do
Begin
write(' ',LastSeen[S_Idx]);
S_Idx:= (S_Idx+1) AND MAXSEENIDX;
end;
writeln;
end;
procedure Test(MaxTestCnt:Uint32);
var
i,actnum,Posi,S_Idx: Uint32;
pPosBef,pSeen :pUint32;
Begin
Fillchar(LastSeen,SizeOf(LastSeen),#0);
HaveSeen := 0;
IF MaxTestCnt> MAXNUM then
EXIT;
//setlength and clear
setlength(PosBefore,0);
setlength(PosBefore,MaxTestCnt);
pPosBef := @PosBefore[0];
pSeen := @LastSeen[0];
S_Idx := 0;
i := 1;
actnum := 0;
repeat
// save value
pSeen[S_Idx] := actnum;
S_Idx:= (S_Idx+1) AND MAXSEENIDX;
//examine new value often out of cache
Posi := pPosBef[actnum];
pPosBef[actnum] := i;
// if Posi=0 ? actnum = 0:actnum = i-Posi
IF Posi = 0 then
actnum := 0
else
actnum := i-Posi;
inc(i);
until i > MaxTestCnt;
HaveSeen := i-1;
SeenIdx := S_Idx;
end;
Begin
Test(10) ; OutSeen(10000);
Test(1000); OutSeen(10);
Test(MAXNUM); OutSeen(28);
setlength(PosBefore,0);
end.
{{out}}
0 0 1 0 2 0 2 2 1 6
4 7 30 25 67 225 488 0 10 136
0 9 47 47 1 10 33 27 548 548 1 6 33 6 2 154 15657 695734 270964 235721 238076 4896139 655158 7901804 146089 977945 21475977 0
Perl
{{trans|Perl 6}}
use strict;
use warnings;
use feature 'say';
sub van_eck {
my($init,$max) = @_;
my(%v,$k);
my @V = my $i = $init;
for (1..$max) {
$k++;
my $t = $v{$i} ? $k - $v{$i} : 0;
$v{$i} = $k;
push @V, $i = $t;
}
@V;
}
for (
['A181391', 0],
['A171911', 1],
['A171912', 2],
['A171913', 3],
['A171914', 4],
['A171915', 5],
['A171916', 6],
['A171917', 7],
['A171918', 8],
) {
my($seq, $start) = @$_;
my @seq = van_eck($start,1000);
say <<~"END";
Van Eck sequence OEIS:$seq; with the first term: $start
First 10 terms: @{[@seq[0 .. 9]]}
Terms 991 through 1000: @{[@seq[990..999]]}
END
}
{{out}}
Van Eck sequence OEIS:A181391; with the first term: 0
First 10 terms: 0 0 1 0 2 0 2 2 1 6
Terms 991 through 1000: 4 7 30 25 67 225 488 0 10 136
Van Eck sequence OEIS:A171911; with the first term: 1
First 10 terms: 1 0 0 1 3 0 3 2 0 3
Terms 991 through 1000: 0 6 53 114 302 0 5 9 22 71
Van Eck sequence OEIS:A171912; with the first term: 2
First 10 terms: 2 0 0 1 0 2 5 0 3 0
Terms 991 through 1000: 8 92 186 0 5 19 41 413 0 5
Van Eck sequence OEIS:A171913; with the first term: 3
First 10 terms: 3 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 5 5 1 17 192 0 6 34 38 179
Van Eck sequence OEIS:A171914; with the first term: 4
First 10 terms: 4 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 33 410 0 6 149 0 3 267 0 3
Van Eck sequence OEIS:A171915; with the first term: 5
First 10 terms: 5 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 60 459 0 7 13 243 0 4 10 211
Van Eck sequence OEIS:A171916; with the first term: 6
First 10 terms: 6 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 6 19 11 59 292 0 6 6 1 12
Van Eck sequence OEIS:A171917; with the first term: 7
First 10 terms: 7 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 11 7 2 7 2 2 1 34 24 238
Van Eck sequence OEIS:A171918; with the first term: 8
First 10 terms: 8 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 16 183 0 6 21 10 249 0 5 48
```
## Perl 6
There is not '''a''' Van Eck sequence, rather a series of related sequences that differ in their starting value. This task is nominally for the sequence starting with the value 0. This Perl 6 implementation will handle any integer starting value.
Specifically handles:
* [[oeis:A181391| OEIS:A181391 - Van Eck sequence starting with 0]]
* [[oeis:A171911| OEIS:A171911 - Van Eck sequence starting with 1]]
* [[oeis:A171912| OEIS:A171912 - Van Eck sequence starting with 2]]
* [[oeis:A171913| OEIS:A171913 - Van Eck sequence starting with 3]]
* [[oeis:A171914| OEIS:A171914 - Van Eck sequence starting with 4]]
* [[oeis:A171915| OEIS:A171915 - Van Eck sequence starting with 5]]
* [[oeis:A171916| OEIS:A171916 - Van Eck sequence starting with 6]]
* [[oeis:A171917| OEIS:A171917 - Van Eck sequence starting with 7]]
* [[oeis:A171918| OEIS:A171918 - Van Eck sequence starting with 8]]
among others.
Implemented as lazy, extendable lists.
```perl6
sub n-van-ecks ($init) {
$init, -> $i, {
state %v;
state $k;
$k++;
my $t = %v{$i}.defined ?? $k - %v{$i} !! 0;
%v{$i} = $k;
$t
} ... *
}
for <
A181391 0
A171911 1
A171912 2
A171913 3
A171914 4
A171915 5
A171916 6
A171917 7
A171918 8
> -> $seq, $start {
my @seq = n-van-ecks($start);
# The task
put qq:to/END/
Van Eck sequence OEIS:$seq; with the first term: $start
First 10 terms: {@seq[^10]}
Terms 991 through 1000: {@seq[990..999]}
END
}
```
{{out}}
```txt
Van Eck sequence OEIS:A181391; with the first term: 0
First 10 terms: 0 0 1 0 2 0 2 2 1 6
Terms 991 through 1000: 4 7 30 25 67 225 488 0 10 136
Van Eck sequence OEIS:A171911; with the first term: 1
First 10 terms: 1 0 0 1 3 0 3 2 0 3
Terms 991 through 1000: 0 6 53 114 302 0 5 9 22 71
Van Eck sequence OEIS:A171912; with the first term: 2
First 10 terms: 2 0 0 1 0 2 5 0 3 0
Terms 991 through 1000: 8 92 186 0 5 19 41 413 0 5
Van Eck sequence OEIS:A171913; with the first term: 3
First 10 terms: 3 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 5 5 1 17 192 0 6 34 38 179
Van Eck sequence OEIS:A171914; with the first term: 4
First 10 terms: 4 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 33 410 0 6 149 0 3 267 0 3
Van Eck sequence OEIS:A171915; with the first term: 5
First 10 terms: 5 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 60 459 0 7 13 243 0 4 10 211
Van Eck sequence OEIS:A171916; with the first term: 6
First 10 terms: 6 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 6 19 11 59 292 0 6 6 1 12
Van Eck sequence OEIS:A171917; with the first term: 7
First 10 terms: 7 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 11 7 2 7 2 2 1 34 24 238
Van Eck sequence OEIS:A171918; with the first term: 8
First 10 terms: 8 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 16 183 0 6 21 10 249 0 5 48
```
## Phix
Just like the pascal entry, instead of searching/dictionaries use a fast direct/parallel lookup table,
and likewise this can easily create a 32-million-long table in under 2s.
While dictionaries are pretty fast, there is a huge overhead adding/updating millions of entries compared to a flat list of int.
```Phix
constant lim = 1000
sequence van_eck = repeat(0,lim),
pos_before = repeat(0,lim)
for n=1 to lim-1 do
integer vn = van_eck[n]+1,
prev = pos_before[vn]
if prev!=0 then
van_eck[n+1] = n - prev
end if
pos_before[vn] = n
end for
printf(1,"The first ten terms of the Van Eck sequence are:%v\n",{van_eck[1..10]})
printf(1,"Terms 991 to 1000 of the sequence are:%v\n",{van_eck[991..1000]})
```
{{out}}
```txt
The first ten terms of the Van Eck sequence are:{0,0,1,0,2,0,2,2,1,6}
Terms 991 to 1000 of the sequence are:{4,7,30,25,67,225,488,0,10,136}
```
## Python
### Python: Using a dict
```python
def van_eck():
n, seen, val = 0, {}, 0
while True:
yield val
last = {val: n}
val = n - seen.get(val, n)
seen.update(last)
n += 1
#%%
if __name__ == '__main__':
print("Van Eck: first 10 terms: ", list(islice(van_eck(), 10)))
print("Van Eck: terms 991 - 1000:", list(islice(van_eck(), 1000))[-10:])
```
{{out}}
```txt
Van Eck: first 10 terms: [0, 0, 1, 0, 2, 0, 2, 2, 1, 6]
Van Eck: terms 991 - 1000: [4, 7, 30, 25, 67, 225, 488, 0, 10, 136]
```
### Python: List based
The following alternative stores the sequence so far in a list seen rather than the first example that just stores ''last occurrences'' in a dict.
```python
def van_eck():
n = 0
seen = [0]
val = 0
while True:
yield val
if val in seen[1:]:
val = seen.index(val, 1)
else:
val = 0
seen.insert(0, val)
n += 1
```
{{out}}
As before.
### Python: Composition of pure functions
As an alternative to the use of generators, a declarative definition in terms of a Church numeral function:
{{Works with|Python|3.7}}
```python
'''Van Eck sequence'''
from functools import reduce
from itertools import repeat
# vanEck :: Int -> [Int]
def vanEck(n):
'''First n terms of the van Eck sequence.'''
return churchNumeral(n)(
lambda xs: cons(
maybe(0)(succ)(
elemIndex(xs[0])(xs[1:])
)
)(xs) if xs else [0]
)([])[::-1]
# TEST ----------------------------------------------------
def main():
'''Terms of the Van Eck sequence'''
print(
main.__doc__ + ':\n\n' +
'First 10: '.rjust(18, ' ') + repr(vanEck(10)) + '\n' +
'991 - 1000: '.rjust(18, ' ') + repr(vanEck(1000)[990:])
)
# GENERIC -------------------------------------------------
# Just :: a -> Maybe a
def Just(x):
'''Constructor for an inhabited Maybe (option type) value.
Wrapper containing the result of a computation.
'''
return {'type': 'Maybe', 'Nothing': False, 'Just': x}
# Nothing :: Maybe a
def Nothing():
'''Constructor for an empty Maybe (option type) value.
Empty wrapper returned where a computation is not possible.
'''
return {'type': 'Maybe', 'Nothing': True}
# churchNumeral :: Int -> (a -> a) -> a -> a
def churchNumeral(n):
'''n applications of a function
'''
return lambda f: lambda x: reduce(
lambda a, g: g(a), repeat(f, n), x
)
# cons :: a -> [a] -> [a]
def cons(x):
'''Construction of a list from a head and a tail.
'''
return lambda xs: [x] + xs
# elemIndex :: Eq a => a -> [a] -> Maybe Int
def elemIndex(x):
'''Just the index of the first element in xs
which is equal to x,
or Nothing if there is no such element.
'''
def go(xs):
try:
return Just(xs.index(x))
except ValueError:
return Nothing()
return lambda xs: go(xs)
# maybe :: b -> (a -> b) -> Maybe a -> b
def maybe(v):
'''Either the default value v, if m is Nothing,
or the application of f to x,
where m is Just(x).
'''
return lambda f: lambda m: v if None is m or m.get('Nothing') else (
f(m.get('Just'))
)
# succ :: Enum a => a -> a
def succ(x):
'''The successor of a value.
For numeric types, (1 +).
'''
return 1 + x if isinstance(x, int) else (
chr(1 + ord(x))
)
# MAIN ---
if __name__ == '__main__':
main()
```
{{Out}}
```txt
Terms of the Van Eck sequence:
First 10: [0, 0, 1, 0, 2, 0, 2, 2, 1, 6]
991 - 1000: [4, 7, 30, 25, 67, 225, 488, 0, 10, 136]
```
Or if we lose sight, for a moment, of the good advice of Donald Knuth, and fall into optimising more than is needed for the first 1000 terms, then we can define the vanEck series as a map accumulation over a range, with an array of positions as the accumulator.
```python
'''Van Eck series'''
from functools import reduce
from itertools import repeat
# vanEck :: Int -> [Int]
def vanEck(n):
'''First n terms of the vanEck sequence.'''
def go(xns, i):
(x, ns) = xns
prev = ns[x]
v = i - prev if 0 is not prev else 0
return (
(v, insert(ns, x, i)),
v
)
return [0] + mapAccumL(go)((0, list(repeat(0, n))))(
range(1, n)
)[1]
# TEST ----------------------------------------------------
# main :: IO ()
def main():
'''The last 10 of the first N vanEck terms'''
print(
fTable(main.__doc__ + ':\n')(
lambda n: 'N=' + str(n)
)(repr)(
lambda n: vanEck(n)[-10:]
)([10, 1000, 10000])
)
# FORMATTING ----------------------------------------------
# fTable :: String -> (a -> String) ->
# (b -> String) -> (a -> b) -> [a] -> String
def fTable(s):
'''Heading -> x display function -> fx display function ->
f -> xs -> tabular string.
'''
def go(xShow, fxShow, f, xs):
ys = [xShow(x) for x in xs]
w = max(map(len, ys))
return s + '\n' + '\n'.join(map(
lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),
xs, ys
))
return lambda xShow: lambda fxShow: lambda f: lambda xs: go(
xShow, fxShow, f, xs
)
# GENERIC -------------------------------------------------
# insert :: Array Int -> Int -> Int -> Array Int
def insert(xs, i, v):
'''An array updated at position i with value v.'''
xs[i] = v
return xs
# mapAccumL :: (acc -> x -> (acc, y)) -> acc -> [x] -> (acc, [y])
def mapAccumL(f):
'''A tuple of an accumulation and a list derived by a
combined map and fold,
with accumulation from left to right.
'''
def go(a, x):
tpl = f(a[0], x)
return (tpl[0], a[1] + [tpl[1]])
return lambda acc: lambda xs: (
reduce(go, xs, (acc, []))
)
# MAIN ---
if __name__ == '__main__':
main()
```
{{Out}}
```txt
The last 10 of the first N vanEck terms:
N=10 -> [0, 0, 1, 0, 2, 0, 2, 2, 1, 6]
N=1000 -> [4, 7, 30, 25, 67, 225, 488, 0, 10, 136]
N=10000 -> [7, 43, 190, 396, 2576, 3142, 0, 7, 7, 1]
```
## Racket
```racket
#lang racket
(require racket/stream)
(define (van-eck)
(define (next val n seen)
(define val1 (- n (hash-ref seen val n)))
(stream-cons val (next val1 (+ n 1) (hash-set seen val n))))
(next 0 0 (hash)))
(define (get m n s)
(stream->list
(stream-take (stream-tail s m)
(- n m))))
"First 10 terms:" (get 0 10 (van-eck))
"Terms 991 to 1000 terms:" (get 990 1000 (van-eck)) ; counting from 0
```
{{out}}
```txt
"First 10 terms:"
(0 0 1 0 2 0 2 2 1 6)
"Terms 991 to 1000 terms:"
(4 7 30 25 67 225 488 0 10 136)
```
## REXX
### using a list
This REXX version allows the specification of the ''start'' and ''end'' of the ''Van Eck'' sequence (to be displayed) as
well as the initial starting element (the default is zero).
```rexx
/*REXX pgm generates/displays the 'start ──► end' elements of the Van Eck sequence.*/
parse arg LO HI $ . /*obtain optional arguments from the CL*/
if LO=='' | LO=="," then LO= 1 /*Not specified? Then use the default.*/
if HI=='' | HI=="," then HI= 10 /* " " " " " " */
if $=='' | $=="," then $= 0 /* " " " " " " */
$$=; z= $ /*$$: old seq: $: initial value of seq*/
do HI-1; z= wordpos( reverse(z), reverse($$) ); $$= $; $= $ z
end /*HI-1*/ /*REVERSE allows backwards search in $.*/
/*stick a fork in it, we're all done. */
say 'terms ' LO " through " HI ' of the Van Eck sequence are: ' subword($,LO,HI-LO+1)
```
{{out|output|text= when using the default inputs:}}
```txt
terms 1 through 10 of the Van Eck sequence are: 0 0 1 0 2 0 2 2 1 6
```
{{out|output|text= when using the inputs of: 991 1000 }}
```txt
terms 991 through 1000 of the Van Eck sequence are: 4 7 30 25 67 225 488 0 10 136
```
{{out|output|text= when using the inputs of: 1 20 6 }}
```txt
terms 1 through 20 of the Van Eck sequence are: 6 0 0 1 0 2 0 2 2 1 6 10 0 6 3 0 3 2 9 0
```
### using a dictionary
This REXX version (which uses a dictionary) is about '''20''' times faster than using a list (in finding the previous
location of an "old" number (term).
```rexx
/*REXX pgm generates/displays the 'start ──► end' elements of the Van Eck sequence.*/
parse arg LO HI $ . /*obtain optional arguments from the CL*/
if LO=='' | LO=="," then LO= 1 /*Not specified? Then use the default.*/
if HI=='' | HI=="," then HI= 10 /* " " " " " " */
if $=='' | $=="," then $= 0 /* " " " " " " */
x=$; @.=. /*$: the Van Eck sequence as a list. */
do #=1 for HI /*X: is the last term being examined. */
if @.x==. then do; @.x= #; $= $ 0; x= 0; end /* a new term.*/
else do; z= # - @.x; $= $ z; @.x= #; x= z; end /*an old term.*/
end /*#*/ /*Z: the new term being added to list.*/
/*stick a fork in it, we're all done. */
say 'terms ' LO " through " HI ' of the Van Eck sequence are: ' subword($,LO,HI-LO+1)
```
{{out|output|text= is identical to the 1st REXX version.}}
## Ruby
```ruby
van_eck = Enumerator.new do |y|
ar = [0]
loop do
y << (term = ar.last) # yield
ar << (ar.count(term)==1 ? 0 : ar.size - 1 - ar[0..-2].rindex(term))
end
end
ve = van_eck.take(1000)
p ve.first(10), ve.last(10)
```
{{out}}
```txt
[0, 0, 1, 0, 2, 0, 2, 2, 1, 6]
[4, 7, 30, 25, 67, 225, 488, 0, 10, 136]
```
## Scala
```scala
object VanEck extends App {
def vanEck(n: Int): List[Int] = {
def vanEck(values: List[Int]): List[Int] =
if (values.size < n)
vanEck(math.max(0, values.indexOf(values.head, 1)) :: values)
else
values
vanEck(List(0)).reverse
}
val vanEck1000 = vanEck(1000)
println(s"The first 10 terms are ${vanEck1000.take(10)}.")
println(s"Terms 991 to 1000 are ${vanEck1000.drop(990)}.")
}
```
{{out}}
```txt
The first 10 terms are List(0, 0, 1, 0, 2, 0, 2, 2, 1, 6).
Terms 991 to 1000 are List(4, 7, 30, 25, 67, 225, 488, 0, 10, 136).
```
## Sidef
```ruby
func van_eck(n) {
var seen = Hash()
var seq = [0]
var prev = seq[-1]
for k in (1 ..^ n) {
seq << (seen.has(prev) ? (k - seen{prev}) : 0)
seen{prev} = k
prev = seq[-1]
}
seq
}
say van_eck(10)
say van_eck(1000).slice(991-1, 1000-1)
```
{{out}}
```txt
[0, 0, 1, 0, 2, 0, 2, 2, 1, 6]
[4, 7, 30, 25, 67, 225, 488, 0, 10, 136]
```
## zkl
{{trans|Perl6}}
```zkl
fcn vanEck(startAt=0){ // --> iterator
(startAt).walker(*).tweak(fcn(n,seen,rprev){
prev,t := rprev.value, n - seen.find(prev,n);
seen[prev] = n;
rprev.set(t);
t
}.fp1(Dictionary(),Ref(startAt))).push(startAt)
}
```
```zkl
foreach n in (9){
ve:=vanEck(n);
println("The first ten terms of the Van Eck (%d) sequence are:".fmt(n));
println("\t",ve.walk(10).concat(","));
println(" Terms 991 to 1000 of the sequence are:");
println("\t",ve.drop(990-10).walk(10).concat(","));
}
```
{{out}}
The first ten terms of the Van Eck (0) sequence are:
0,0,1,0,2,0,2,2,1,6
Terms 991 to 1000 of the sequence are:
4,7,30,25,67,225,488,0,10,136
The first ten terms of the Van Eck (1) sequence are:
1,0,0,1,3,0,3,2,0,3
Terms 991 to 1000 of the sequence are:
0,6,53,114,302,0,5,9,22,71
The first ten terms of the Van Eck (2) sequence are:
2,0,0,1,0,2,5,0,3,0
Terms 991 to 1000 of the sequence are:
8,92,186,0,5,19,41,413,0,5
The first ten terms of the Van Eck (3) sequence are:
3,0,0,1,0,2,0,2,2,1
Terms 991 to 1000 of the sequence are:
5,5,1,17,192,0,6,34,38,179
The first ten terms of the Van Eck (4) sequence are:
4,0,0,1,0,2,0,2,2,1
Terms 991 to 1000 of the sequence are:
33,410,0,6,149,0,3,267,0,3
The first ten terms of the Van Eck (5) sequence are:
5,0,0,1,0,2,0,2,2,1
Terms 991 to 1000 of the sequence are:
60,459,0,7,13,243,0,4,10,211
The first ten terms of the Van Eck (6) sequence are:
6,0,0,1,0,2,0,2,2,1
Terms 991 to 1000 of the sequence are:
6,19,11,59,292,0,6,6,1,12
The first ten terms of the Van Eck (7) sequence are:
7,0,0,1,0,2,0,2,2,1
Terms 991 to 1000 of the sequence are:
11,7,2,7,2,2,1,34,24,238
The first ten terms of the Van Eck (8) sequence are:
8,0,0,1,0,2,0,2,2,1
Terms 991 to 1000 of the sequence are:
16,183,0,6,21,10,249,0,5,48
```