Recaman's sequence: Difference between revisions
mNo edit summary |
Not a robot (talk | contribs) (Add BCPL) |
||
| Line 464: | Line 464: | ||
First repeated term: |
First repeated term: |
||
A( 24 ) = 42</pre> |
A( 24 ) = 42</pre> |
||
=={{header|BCPL}}== |
|||
<lang bcpl>get "libhdr" |
|||
// Generate the N'th term of the Recaman sequence |
|||
// given terms 0 to N-1. |
|||
let generate(a, n) be |
|||
a!n := n=0 -> 0, valof |
|||
$( let subterm = a!(n-1) - n |
|||
let addterm = a!(n-1) + n |
|||
if subterm <= 0 resultis addterm |
|||
for i=0 to n-1 |
|||
if a!i = subterm resultis addterm |
|||
resultis subterm |
|||
$) |
|||
let start() be |
|||
$( let a = vec 50 and n = 15 and rep = ? |
|||
writef("First %N members:*N", n) |
|||
for i = 0 to n-1 |
|||
$( generate(a, i) |
|||
writef("%N ", a!i) |
|||
$) |
|||
writef("*NFirst repeated term:*N") |
|||
rep := valof |
|||
$( generate(a, n) |
|||
for i = 0 to n-1 |
|||
if a!i = a!n resultis n |
|||
n := n + 1 |
|||
$) repeat |
|||
writef("a!%N = %N*N", rep, a!rep) |
|||
$)</lang> |
|||
{{out}} |
|||
<pre>First 15 members: |
|||
0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 |
|||
First repeated term: |
|||
a!24 = 42</pre> |
|||
=={{header|C}}== |
=={{header|C}}== |
||
Revision as of 18:14, 22 June 2021
You are encouraged to solve this task according to the task description, using any language you may know.
The Recamán's sequence generates Natural numbers.
Starting from zero, the n'th term a(n) is the previous term minus n i.e a(n) = a(n-1) - n but only if this is both positive and has not been previousely generated.
If the conditions don't hold then a(n) = a(n-1) + n.
- Task
- Generate and show here the first 15 members of the sequence.
- Find and show here, the first duplicated number in the sequence.
- Optionally: Find and show here, How many terms of the sequence are needed until all the integers 0..1000, inclusive, are generated.
- References
- A005132, The On-Line Encyclopedia of Integer Sequences.
- The Slightly Spooky Recamán Sequence, Numberphile video.
11l
<lang 11l>F recamanSucc(seen, n, r)
‘The successor for a given Recaman term,
given the set of Recaman terms seen so far.’
V back = r - n
R I 0 > back | (back C seen) {n + r} E back
F recamanUntil(p)
‘All terms of the Recaman series before the
first term for which the predicate p holds.’
V n = 1
V r = 0
V rs = [r]
V seen = Set(rs)
V blnNew = 1B
L !p(seen, n, r, blnNew)
r = recamanSucc(seen, n, r)
blnNew = r !C seen
seen.add(r)
rs.append(r)
n = 1 + n
R rs
F enumFromTo(m)
‘Integer enumeration from m to n.’ R n -> @m .< 1 + n
print("First 15 Recaman:\n "recamanUntil((seen, n, r, _) -> n == 15)) print("First duplicated Recaman:\n "recamanUntil((seen, n, r, blnNew) -> !blnNew).last) V setK = Set(enumFromTo(0)(1000)) print("Number of Recaman terms needed to generate all integers from [0..1000]:\n "(recamanUntil((seen, n, r, blnNew) -> (blnNew & r < 1001 & :setK.is_subset(seen))).len - 1))</lang>
- Output:
First 15 Recaman: [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9] First duplicated Recaman: 42 Number of Recaman terms needed to generate all integers from [0..1000]: 328002
ALGOL W
<lang algolw>begin
% calculate Recaman's sequence values %
% a hash table element - holds n, A(n) and a link to the next element with the % % same hash value % record AValue ( integer eN, eAn ; reference(AValue) eNext );
% hash modulus % integer HMOD; HMOD := 100000;
begin
reference(AValue) array hashTable ( 0 :: HMOD - 1 );
integer array A ( 0 :: 14 );
integer le1000Count, firstN, duplicateN, duplicateValue, n, An, An1, prevN;
% adds an element to the hash table, returns true if an element with value An %
% was already present, false otherwise %
% if the value was already present, its eN value is returned in prevN %
logical procedure addAValue( integer value n, An ; integer result prevN ) ;
begin
integer hash;
logical duplicate;
reference(AValue) element;
hash := An rem HMOD;
element := hashTable( hash );
duplicate := false;
while element not = null and eAn(element) not = An do element := eNext(element);
duplicate := element not = null;
if not duplicate then hashTable( hash ) := AValue( n, An, hashTable( hash ) )
else prevN := eN(element);
duplicate
end addAValue ;
% initialise the hash table %
for h := 0 until HMOD - 1 do hashTable( h ) := null;
% calculate the values of the sequence until we have found values that %
% include all numbers in 1..1000 %
% also store the first 15 values %
A( 0 ) := An1 := n := 0;
le1000Count := 0;
firstN := duplicateN := duplicateValue := -1;
while le1000Count < 1000 do begin
logical le0, duplicate;
n := n + 1;
An := An1 - n;
le0 := ( An <= 0 );
if le0 then An := An1 + n;
prevN := -1;
duplicate := addAValue( n, An, prevN );
if duplicate and not le0 then begin
An := An1 + n;
duplicate := addAValue( n, An, prevN )
end if_duplicate_and_not_le0 ;
if duplicate then begin
% the value was already present %
if firstN < 0 then begin % have the first duplicate %
firstN := n;
duplicateN := prevN;
duplicateValue := An;
end if_firstN_lt_0
end
else if An <= 1000 then le1000Count := le1000Count + 1;;
if n < 15 then A( n ) := An;
An1 := An
end while_le1000Count_lt_1000 ;
% show the first 15 values of the sequence %
write( "A( 0 .. 14 ): " );
for n := 0 until 14 do writeon( i_w := 1, A( n ) );
% positions of the first duplicate %
write( i_w := 1
, s_w := 0
, "First duplicates: "
, duplicateN
, " "
, firstN
, " ("
, duplicateValue
, ")"
);
% number of elements required to include the first 1000 integers %
write( i_w := 1, "first element to include all 1..1000: ", n )
end
end.</lang>
- Output:
A( 0 .. 14 ): 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 First duplicates: 20 24 (42) first element to include all 1..1000: 328002
APL
<lang APL>recaman←{⎕IO←0
genNext←{
R←⍵[N-1]-N←≢⍵
(R<0)∨(R∊⍵):⍵,⍵[N-1]+N
⍵,⍵[N-1]-N
}
⎕←'First 15: '
⎕←reca←(genNext⍣14),0
⎕←'First repetition: '
⎕←⊃⌽reca←genNext⍣{⍺≢∪⍺}⊢reca
⎕←'Length of sequence containing [0..1000]:'
⎕←≢reca←genNext⍣{(⍳1001)∧.∊⊂⍺}⊢reca
}</lang>
- Output:
First 15: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 First repetition: 42 Length of sequence containing [0..1000]: 328003
AppleScript
The third of these tasks probably stretches Applescript a bit beyond the point of its usefulness – it takes about 1 minute to find the result, and even that requires the use of NSMutableSet, from the Apple Foundation classes.
<lang applescript>use AppleScript version "2.4" use framework "Foundation" use scripting additions
on run
-- FIRST FIFTEEN RECAMANs ------------------------------------------------------
script term15
on |λ|(i)
15 = (i as integer)
end |λ|
end script
set strFirst15 to unwords(snd(recamanUpto(true, term15)))
set strFirstMsg to "First 15 Recamans:" & linefeed
display notification strFirstMsg & strFirst15
delay 2
-- FIRST DUPLICATE RECAMAN ----------------------------------------------------
script firstDuplicate
on |λ|(_, seen, rs)
setSize(seen) as integer is not (length of (rs as list))
end |λ|
end script
set strDuplicate to (item -1 of snd(recamanUpto(true, firstDuplicate))) as integer as string
set strDupMsg to "First duplicated Recaman:" & linefeed
display notification strDupMsg & strDuplicate
delay 2
-- NUMBER OF RECAMAN TERMS NEEDED TO GET ALL OF [0..1000]
-- (takes about a minute, depending on system)
set setK to setFromList(enumFromTo(0, 1000))
script supersetK
on |λ|(i, setR)
setK's isSubsetOfSet:(setR)
end |λ|
end script
display notification "Superset size result will take c. 1 min to find ..."
set dteStart to current date
set strSetSize to (fst(recamanUpto(false, supersetK)) - 1) as string
set dteEnd to current date
set strSetSizeMsg to "Number of Recaman terms needed to generate" & ¬
linefeed & "all integers from [0..1000]:" & linefeed
set strElapsed to "(Last result took c. " & (dteEnd - dteStart) & " seconds to find)"
display notification strSetSizeMsg & linefeed & strSetSize
-- CLEARED REFERENCE TO NSMUTABLESET -------------------------------------
set setK to missing value
-- REPORT ----------------------------------------------------------------
unlines({strFirstMsg & strFirst15, "", ¬
strDupMsg & strDuplicate, "", ¬
strSetSizeMsg & strSetSize, "", ¬
strElapsed})
end run
-- nextR :: Set Int -> Int -> Int on nextR(seen, i, n)
set bk to n - i if 0 > bk or setMember(bk, seen) then n + i else bk end if
end nextR
-- recamanUpto :: Bool -> (Int -> Set Int > [Int] -> Bool) -> (Int, [Int]) on recamanUpto(bln, p)
script recaman
property mp : mReturn(p)'s |λ|
on |λ|()
set i to 1
set r to 0
set rs to {r}
set seen to setFromList(rs)
repeat while not mp(i, seen, rs)
set r to nextR(seen, i, r)
setInsert(r, seen)
if bln then set end of rs to r
set i to i + 1
end repeat
set seen to missing value -- clear pointer to NSMutableSet
{i, rs}
end |λ|
end script
recaman's |λ|()
end recamanUpto
-- GENERIC FUNCTIONS -------------------------------------------------------
-- enumFromTo :: Int -> Int -> [Int] on enumFromTo(m, n)
if m ≤ n then
set lst to {}
repeat with i from m to n
set end of lst to i
end repeat
return lst
else
return {}
end if
end enumFromTo
-- fst :: (a, b) -> a on fst(tpl)
if class of tpl is record then |1| of tpl else item 1 of tpl end if
end fst
-- intercalateS :: String -> [String] -> String on intercalateS(sep, xs)
set {dlm, my text item delimiters} to {my text item delimiters, sep}
set s to xs as text
set my text item delimiters to dlm
return s
end intercalateS
-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: First-class m => (a -> b) -> m (a -> b) on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- NB All names of NSMutableSets should be set to *missing value* -- before the script exits. -- ( scpt files containing residual ObjC pointer values can not be saved) -- setFromList :: Ord a => [a] -> Set a on setFromList(xs)
set ca to current application ca's NSMutableSet's ¬ setWithArray:(ca's NSArray's arrayWithArray:(xs))
end setFromList
-- setMember :: Ord a => a -> Set a -> Bool on setMember(x, objcSet)
missing value is not (objcSet's member:(x))
end setMember
-- setInsert :: Ord a => a -> Set a -> Set a on setInsert(x, objcSet)
objcSet's addObject:(x) objcSet
end setInsert
-- setSize :: Set a -> Int on setSize(objcSet)
objcSet's |count|() as integer
end setSize
-- snd :: (a, b) -> b on snd(tpl)
if class of tpl is record then |2| of tpl else item 2 of tpl end if
end snd
-- unlines :: [String] -> String on unlines(xs)
set {dlm, my text item delimiters} to ¬
{my text item delimiters, linefeed}
set str to xs as text
set my text item delimiters to dlm
str
end unlines
-- unwords :: [String] -> String on unwords(xs)
intercalateS(space, xs)
end unwords</lang>
- Output:
First 15 Recamans: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 First duplicated Recaman: 42 Number of Recaman terms needed to generate all integers from [0..1000]: 328002 (Last result took c. 40 seconds to find)
AWK
<lang AWK>
- syntax: GAWK -f RECAMANS_SEQUENCE.AWK
- converted from Microsoft Small Basic
BEGIN {
found_dup = 0
n = -1
do {
n++
ap = a[n-1] + n
if (a[n-1] <= n) {
a[n] = ap
b[ap] = 1
}
else {
am = a[n-1] - n
if (b[am] == 1) {
a[n] = ap
b[ap] = 1
}
else {
a[n] = am
b[am] = 1
}
}
if (n <= 14) {
terms = sprintf("%s%s ",terms,a[n])
if (n == 14) {
printf("first %d terms: %s\n",n+1,terms)
}
}
if (!found_dup) {
if (dup[a[n]] == 1) {
printf("first duplicated term: a[%d]=%d\n",n,a[n])
found_dup = 1
}
dup[a[n]] = 1
}
if (a[n] <= 1000) {
arr[a[n]] = ""
}
} while (n <= 15 || !found_dup || length(arr) < 1001)
printf("terms needed to generate integers 0 - 1000: %d\n",n)
exit(0)
} </lang>
- Output:
first 15 terms: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 first duplicated term: a[24]=42 terms needed to generate integers 0 - 1000: 328002
BASIC
<lang BASIC>10 DEFINT A-Z: DIM A(100) 20 PRINT "First 15 terms:" 30 FOR N=0 TO 14: GOSUB 100: PRINT A(N);: NEXT 35 PRINT 40 PRINT "First repeated term:" 50 GOSUB 100 55 FOR M=0 TO N-1: IF A(M)=A(N) THEN 70 ELSE NEXT 60 N=N+1: GOTO 50 70 PRINT "A(";N;") =";A(N) 80 END 100 IF N=0 THEN A(0)=0: RETURN 110 X = A(N-1)-N: IF X<0 THEN 160 120 FOR M=0 TO N-1 130 IF A(M)=X THEN 160 140 NEXT 150 A(N)=X: RETURN 160 A(N)=A(N-1)+N: RETURN</lang>
- Output:
First 15 terms: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 First repeated term: A( 24 ) = 42
BCPL
<lang bcpl>get "libhdr"
// Generate the N'th term of the Recaman sequence // given terms 0 to N-1. let generate(a, n) be
a!n := n=0 -> 0, valof
$( let subterm = a!(n-1) - n
let addterm = a!(n-1) + n
if subterm <= 0 resultis addterm
for i=0 to n-1
if a!i = subterm resultis addterm
resultis subterm
$)
let start() be $( let a = vec 50 and n = 15 and rep = ?
writef("First %N members:*N", n)
for i = 0 to n-1
$( generate(a, i)
writef("%N ", a!i)
$)
writef("*NFirst repeated term:*N")
rep := valof
$( generate(a, n)
for i = 0 to n-1
if a!i = a!n resultis n
n := n + 1
$) repeat
writef("a!%N = %N*N", rep, a!rep)
$)</lang>
- Output:
First 15 members: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 First repeated term: a!24 = 42
C
<lang c>#include <stdio.h>
- include <stdlib.h>
- include <gmodule.h>
typedef int bool;
int main() {
int i, n, k = 0, next, *a; bool foundDup = FALSE; gboolean alreadyUsed; GHashTable* used = g_hash_table_new(g_direct_hash, g_direct_equal); GHashTable* used1000 = g_hash_table_new(g_direct_hash, g_direct_equal); a = malloc(400000 * sizeof(int)); a[0] = 0; g_hash_table_add(used, GINT_TO_POINTER(0)); g_hash_table_add(used1000, GINT_TO_POINTER(0));
for (n = 1; n <= 15 || !foundDup || k < 1001; ++n) {
next = a[n - 1] - n;
if (next < 1 || g_hash_table_contains(used, GINT_TO_POINTER(next))) {
next += 2 * n;
}
alreadyUsed = g_hash_table_contains(used, GINT_TO_POINTER(next));
a[n] = next;
if (!alreadyUsed) {
g_hash_table_add(used, GINT_TO_POINTER(next));
if (next >= 0 && next <= 1000) {
g_hash_table_add(used1000, GINT_TO_POINTER(next));
}
}
if (n == 14) {
printf("The first 15 terms of the Recaman's sequence are: ");
printf("[");
for (i = 0; i < 15; ++i) printf("%d ", a[i]);
printf("\b]\n");
}
if (!foundDup && alreadyUsed) {
printf("The first duplicated term is a[%d] = %d\n", n, next);
foundDup = TRUE;
}
k = g_hash_table_size(used1000);
if (k == 1001) {
printf("Terms up to a[%d] are needed to generate 0 to 1000\n", n);
}
}
g_hash_table_destroy(used);
g_hash_table_destroy(used1000);
free(a);
return 0;
}</lang>
- Output:
The first 15 terms of the Recaman's sequence are: [0 1 3 6 2 7 13 20 12 21 11 22 10 23 9] The first duplicated term is a[24] = 42 Terms up to a[328002] are needed to generate 0 to 1000
C#
<lang csharp>using System; using System.Collections.Generic;
namespace RecamanSequence {
class Program {
static void Main(string[] args) {
List<int> a = new List<int>() { 0 };
HashSet<int> used = new HashSet<int>() { 0 };
HashSet<int> used1000 = new HashSet<int>() { 0 };
bool foundDup = false;
int n = 1;
while (n <= 15 || !foundDup || used1000.Count < 1001) {
int next = a[n - 1] - n;
if (next < 1 || used.Contains(next)) {
next += 2 * n;
}
bool alreadyUsed = used.Contains(next);
a.Add(next);
if (!alreadyUsed) {
used.Add(next);
if (0 <= next && next <= 1000) {
used1000.Add(next);
}
}
if (n == 14) {
Console.WriteLine("The first 15 terms of the Recaman sequence are: [{0}]", string.Join(", ", a));
}
if (!foundDup && alreadyUsed) {
Console.WriteLine("The first duplicated term is a[{0}] = {1}", n, next);
foundDup = true;
}
if (used1000.Count == 1001) {
Console.WriteLine("Terms up to a[{0}] are needed to generate 0 to 1000", n);
}
n++;
}
}
}
}</lang>
- Output:
The first 15 terms of the Recaman sequence are: [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9] The first duplicated term is a[24] = 42 Terms up to a[328002] are needed to generate 0 to 1000
C++
<lang cpp>#include <iostream>
- include <ostream>
- include <set>
- include <vector>
template<typename T> std::ostream& operator<<(std::ostream& os, const std::vector<T>& v) {
auto i = v.cbegin();
auto e = v.cend();
os << '[';
if (i != e) {
os << *i;
i = std::next(i);
}
while (i != e) {
os << ", " << *i;
i = std::next(i);
}
return os << ']';
}
int main() {
using namespace std;
vector<int> a{ 0 };
set<int> used{ 0 };
set<int> used1000{ 0 };
bool foundDup = false;
int n = 1;
while (n <= 15 || !foundDup || used1000.size() < 1001) {
int next = a[n - 1] - n;
if (next < 1 || used.find(next) != used.end()) {
next += 2 * n;
}
bool alreadyUsed = used.find(next) != used.end();
a.push_back(next);
if (!alreadyUsed) {
used.insert(next);
if (0 <= next && next <= 1000) {
used1000.insert(next);
}
}
if (n == 14) {
cout << "The first 15 terms of the Recaman sequence are: " << a << '\n';
}
if (!foundDup && alreadyUsed) {
cout << "The first duplicated term is a[" << n << "] = " << next << '\n';
foundDup = true;
}
if (used1000.size() == 1001) {
cout << "Terms up to a[" << n << "] are needed to generate 0 to 1000\n";
}
n++;
}
return 0;
}</lang>
- Output:
The first 15 terms of the Recaman sequence are: [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9] The first duplicated term is a[24] = 42 Terms up to a[328002] are needed to generate 0 to 1000
D
<lang d>import std.stdio;
void main() {
int[] a; bool[int] used; bool[int] used1000; bool foundDup;
a ~= 0; used[0] = true; used1000[0] = true;
int n = 1;
while (n <= 15 || !foundDup || used1000.length < 1001) {
int next = a[n - 1] - n;
if (next < 1 || (next in used) !is null) {
next += 2 * n;
}
bool alreadyUsed = (next in used) !is null;
a ~= next;
if (!alreadyUsed) {
used[next] = true;
if (0 <= next && next <= 1000) {
used1000[next] = true;
}
}
if (n == 14) {
writeln("The first 15 terms of the Recaman sequence are: ", a);
}
if (!foundDup && alreadyUsed) {
writefln("The first duplicated term is a[%d] = %d", n, next);
foundDup = true;
}
if (used1000.length == 1001) {
writefln("Terms up to a[%d] are needed to generate 0 to 1000", n);
}
n++;
}
}</lang>
- Output:
The first 15 terms of the Recaman sequence are: [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9] The first duplicated term is a[24] = 42 Terms up to a[328002] are needed to generate 0 to 1000
FreeBASIC
<lang freebasic>' version 26-01-2019 ' compile with: fbc -s console
Dim As UByte used() Dim As Integer sum, temp Dim As UInteger n, max, count, i
max = 1000 : ReDim used(max)
Print "The first 15 terms are 0";
For n = 0 To 14
temp = sum - n
If temp < 1 OrElse used(temp) = 1 Then
temp = sum + n
End If
If temp <= max Then used(temp) = 1
sum = temp
Print sum;
Next
sum = 0 : max = 1000 : ReDim used(max)
Print : Print
For n = 0 To 50
temp = sum - n
If temp < 1 OrElse used(temp) = 1 Then
temp = sum + n
End If
If used(temp) = 1 Then
Print "First duplicated number is a(" + Str(n) + ")"
Exit For
End If
If temp <= max Then used(temp) = 1
sum = temp
Next
sum = 0 : max = 2000000 : ReDim used(max)
Print : Print
For n = 0 To max
temp = sum - n
If temp < 1 OrElse used(temp) = 1 Then
temp = sum + n
End If
If temp <= max Then used(temp) = 1
If i = temp Then
While used(i) = 1
i += 1
If i > 1000 Then
Exit For
End If
Wend
End If
sum = temp
count += 1
Next
Print "All integers from 0 to 1000 are generated in " & count & " terms" Print
' empty keyboard buffer While Inkey <> "" : Wend Print : Print "hit any key to end program" Sleep End</lang>
- Output:
The first 15 terms are 0 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 First duplicated number is a(24) All integers from 0 to 1000 are generated in 328002 terms
FOCAL
<lang FOCAL>01.10 T "FIRST 15" 01.20 F N=0,14;D 2;T %2,A(N) 01.30 T !"FIRST REPEATED" 01.40 D 2;S Y=1 01.50 F M=0,N-1;S Y=Y*(A(M)-A(N)) 01.60 I (Y)1.7,1.8,1.7 01.70 S N=N+1;G 1.4 01.80 T A(N)," AT A(",N,")"! 01.90 Q
02.05 I (N)2.1,2.06,2.1 02.06 A(0)=0;R 02.10 S X=A(N-1)-N 02.20 I (X)2.7 02.30 S Y=1 02.40 F M=0,N-1;S Y=Y*(A(M)-X) 02.50 I (Y)2.6,2.7,2.6 02.60 S A(N)=X;R 02.70 S A(N)=A(N-1)+N</lang>
- Output:
FIRST 15= 0= 1= 3= 6= 2= 7= 13= 20= 12= 21= 11= 22= 10= 23= 9 FIRST REPEATED= 42 AT A(= 24)
Fōrmulæ
In 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 (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.
Go
<lang go>package main
import "fmt"
func main() {
a := []int{0}
used := make(map[int]bool, 1001)
used[0] = true
used1000 := make(map[int]bool, 1001)
used1000[0] = true
for n, foundDup := 1, false; n <= 15 || !foundDup || len(used1000) < 1001; n++ {
next := a[n-1] - n
if next < 1 || used[next] {
next += 2 * n
}
alreadyUsed := used[next]
a = append(a, next)
if !alreadyUsed {
used[next] = true
if next >= 0 && next <= 1000 {
used1000[next] = true
}
}
if n == 14 {
fmt.Println("The first 15 terms of the Recaman's sequence are:", a)
}
if !foundDup && alreadyUsed {
fmt.Printf("The first duplicated term is a[%d] = %d\n", n, next)
foundDup = true
}
if len(used1000) == 1001 {
fmt.Printf("Terms up to a[%d] are needed to generate 0 to 1000\n", n)
}
}
}</lang>
- Output:
The first 15 terms of the Recaman's sequence are: [0 1 3 6 2 7 13 20 12 21 11 22 10 23 9] The first duplicated term is a[24] = 42 Terms up to a[328002] are needed to generate 0 to 1000
Haskell
Recursion
A basic recursive function for the first N terms, <lang haskell>recaman :: Int -> [Int] recaman n = fst <$> reverse (go n)
where
go 0 = []
go 1 = [(0, 1)]
go x =
let xs@((r, i):_) = go (pred x)
back = r - i
in ( if 0 < back && not (any ((back ==) . fst) xs)
then back
else r + i
, succ i) :
xs
main :: IO () main = print $ recaman 15</lang>
- Output:
[0,1,3,6,2,7,13,20,12,21,11,22,10,23,9]
Conditional iteration
Or, a little more flexibly, a recamanUpto (predicate) function.
<lang haskell>import Data.Set (Set, fromList, insert, isSubsetOf, member, size) import Data.Bool (bool)
firstNRecamans :: Int -> [Int] firstNRecamans n = reverse $ recamanUpto (\(_, i, _) -> n == i)
firstDuplicateR :: Int firstDuplicateR = head $ recamanUpto (\(rs, _, set) -> size set /= length rs)
recamanSuperset :: Set Int -> [Int] recamanSuperset setInts =
tail $ recamanUpto (\(_, _, setR) -> isSubsetOf setInts setR)
recamanUpto :: (([Int], Int, Set Int) -> Bool) -> [Int] recamanUpto p = rs
where
(rs, _, _) =
until
p
(\(rs@(r:_), i, seen) ->
let n = nextR seen i r
in (n : rs, succ i, insert n seen))
([0], 1, fromList [0])
nextR :: Set Int -> Int -> Int -> Int nextR seen i r =
let back = r - i in bool back (r + i) (0 > back || member back seen)
-- TEST --------------------------------------------------------------- main :: IO () main =
(putStrLn . unlines) [ "First 15 Recamans:" , show $ firstNRecamans 15 , [] , "First duplicated Recaman:" , show firstDuplicateR , [] , "Length of Recaman series required to include [0..1000]:" , (show . length . recamanSuperset) $ fromList [0 .. 1000] ]</lang>
- Output:
First 15 Recamans: [0,1,3,6,2,7,13,20,12,21,11,22,10,23,9] First duplicated Recaman: 42 Length of Recaman series required to include [0..1000]: 328002
Lazy search over infinite lists
For a lazier solution, we could define an infinite series of Recaman sequences of growing length, starting with [0], and simply search through them for the first series of length 15, or the first to include a duplicated integer. For the third task, it would be enough to search through an infinite stream of Recaman-generated integer sets of increasing size, until we find the first that contains [0..1000] as a subset.
<lang haskell>import Data.List (find, findIndex, nub) import Data.Maybe (fromJust) import Data.Set (Set, fromList, insert, isSubsetOf, member)
--- INFINITE STREAM OF RECAMAN SERIES OF GROWING LENGTH -- rSeries :: Int rSeries =
scanl
( \rs@(r : _) i ->
let back = r - i
nxt
| 0 > back || elem back rs = r + i
| otherwise = back
in nxt : rs
)
[0]
[1 ..]
INFINITE STREAM OF RECAMAN-GENERATED ---------
INTEGER SETS OF GROWING SIZE -------------
rSets :: [(Set Int, Int)] rSets =
scanl
( \(seen, r) i ->
let back = r - i
nxt
| 0 > back || member back seen = r + i
| otherwise = back
in (insert nxt seen, nxt)
)
(fromList [0], 0)
[1 ..]
TEST -------------------------
main :: IO () main = do
let setK = fromList [0 .. 1000]
(putStrLn . unlines)
[ "First 15 Recamans:",
show . reverse . fromJust $ find ((15 ==) . length) rSeries,
[],
"First duplicated Recaman:",
show . head . fromJust $ find ((/=) <$> length <*> (length . nub)) rSeries,
[],
"Length of Recaman series required to include [0..1000]:",
show . fromJust $ findIndex (\(setR, _) -> isSubsetOf setK setR) rSets
]</lang>
- Output:
First 15 Recamans: [0,1,3,6,2,7,13,20,12,21,11,22,10,23,9] First duplicated Recaman: 42 Length of Recaman series required to include [0..1000]: 328002
J
positive =: >&0
unique =: -.@:e.
condition =: (positive@:] *. unique~) ({: - #)
NB. with the agenda set by the condition, add or subtract tail with tally
recaman_term =: ({: + #)`({: - #)@.condition
NB. generate four hundred thousand terms and display the first 15
15 {. R=:(, recaman_term)^:400000]0
0 1 3 6 2 7 13 20 12 21 11 22 10 23 9
NB. plot the sequence to see why numberphile might be interested.
load'plot'
plot 470{.R
NB. binaryish search for first duplicate.
(-:&# ~.) 100 {. R
0
(-:&# ~.) 50 {. R
0
(-:&# ~.) 25 {. R
0
(-:&# ~.) 12 {. R
1
(-:&# ~.) 18 {. R
1
(-:&# ~.) 21 {. R
1
(-:&# ~.) 23 {. R
1
(-:&# ~.) 24 {. R
1
(-:&# ~.) 25 {. R
0
Let's write a binary search adverb. <lang J> average =: +/ % # NB. extra_data u Bsearch bounds NB. Bsearch returns narrowed bounds depending if u return 0 (left) or 1 (right) NB. u is called as extra_data u index NB. or as index u index NB. u is invoked as a dyad Bsearch =: 1 :'((0 1 + (u <.@:average)) { ({. , <.@:average, {:)@:])^:_' </lang>
NB. f expresses "not all [0, 1000] are in the first y members of list x"
f =: ([: -. [: *./ (i.1001) e. ~.)@:{.~
R f Bsearch 0 , #R
328002 328003
(<: 328002 328003) { R
328881 879
The sequence begins 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 . With 0 as the first term, We've learned that 25 terms are required to generate a duplicate, and that 328003 terms are needed to generate 0 through 1000 inclusively.
Java
<lang java>import java.util.ArrayList; import java.util.HashSet; import java.util.List; import java.util.Set;
public class RecamanSequence {
public static void main(String[] args) {
List<Integer> a = new ArrayList<>();
a.add(0);
Set<Integer> used = new HashSet<>();
used.add(0);
Set<Integer> used1000 = new HashSet<>();
used1000.add(0);
boolean foundDup = false;
int n = 1;
while (n <= 15 || !foundDup || used1000.size() < 1001) {
int next = a.get(n - 1) - n;
if (next < 1 || used.contains(next)) {
next += 2 * n;
}
boolean alreadyUsed = used.contains(next);
a.add(next);
if (!alreadyUsed) {
used.add(next);
if (0 <= next && next <= 1000) {
used1000.add(next);
}
}
if (n == 14) {
System.out.printf("The first 15 terms of the Recaman sequence are : %s\n", a);
}
if (!foundDup && alreadyUsed) {
System.out.printf("The first duplicate term is a[%d] = %d\n", n, next);
foundDup = true;
}
if (used1000.size() == 1001) {
System.out.printf("Terms up to a[%d] are needed to generate 0 to 1000\n", n);
}
n++;
}
}
}</lang>
- Output:
The first 15 terms of the Recaman sequence are : [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9] The first duplicate term is a[24] = 42 Terms up to a[328002] are needed to generate 0 to 1000
JavaScript
<lang javascript>(() => {
const main = () => {
console.log(
'First 15 Recaman:\n' +
recamanUpto(i => 15 === i)
);
console.log(
'\n\nFirst duplicated Recaman:\n' +
last(recamanUpto(
(_, set, rs) => set.size !== rs.length
))
);
const setK = new Set(enumFromTo(0, 1000));
console.log(
'\n\nNumber of Recaman terms needed to generate' +
'\nall integers from [0..1000]:\n' +
(recamanUpto(
(_, setR) => isSubSetOf(setK, setR)
).length - 1)
);
};
// RECAMAN --------------------------------------------
// recamanUpto :: (Int -> Set Int > [Int] -> Bool) -> [Int]
const recamanUpto = p => {
let
i = 1,
r = 0, // First term of series
rs = [r];
const seen = new Set(rs);
while (!p(i, seen, rs)) {
r = nextR(seen, i, r);
seen.add(r);
rs.push(r);
i++;
}
return rs;
}
// Next Recaman number.
// nextR :: Set Int -> Int -> Int
const nextR = (seen, i, n) => {
const back = n - i;
return (0 > back || seen.has(back)) ? (
n + i
) : back;
};
// GENERIC --------------------------------------------
// enumFromTo :: Int -> Int -> [Int]
const enumFromTo = (m, n) =>
m <= n ? iterateUntil(
x => n <= x,
x => 1 + x,
m
) : [];
// isSubsetOf :: Ord a => Set a -> Set a -> Bool
const isSubSetOf = (a, b) => {
for (let x of a) {
if (!b.has(x)) return false;
}
return true;
};
// iterateUntil :: (a -> Bool) -> (a -> a) -> a -> [a]
const iterateUntil = (p, f, x) => {
const vs = [x];
let h = x;
while (!p(h))(h = f(h), vs.push(h));
return vs;
};
// last :: [a] -> a
const last = xs =>
0 < xs.length ? xs.slice(-1)[0] : undefined;
// MAIN ------------------------------------------------ return main();
})();</lang>
- Output:
First 15 Recaman: 0,1,3,6,2,7,13,20,12,21,11,22,10,23,9 First duplicated Recaman: 42 Number of Recaman terms needed to generate all integers from [0..1000]: 328002
Julia
<lang julia>function recaman()
a = Vector{Int}([0])
used = Dict{Int, Bool}(0 => true)
used1000 = Set(0)
founddup = false
termcount = 1
while length(used1000) <= 1000
nextterm = a[termcount] - termcount
if nextterm < 1 || haskey(used, nextterm)
nextterm += termcount + termcount
end
push!(a, nextterm)
if !haskey(used, nextterm)
used[nextterm] = true
if 1 <= nextterm <= 1000
push!(used1000, nextterm)
end
elseif !founddup
println("The first duplicated term is a[$(termcount + 1)] = $nextterm.")
founddup = true
end
if termcount == 14
println("The first 15 terms of the Recaman sequence are $a")
end
termcount += 1
end
println("Terms up to $(termcount - 1) are needed to generate 0 to 1000.")
end
recaman()
</lang>
- Output:
The first 15 terms of the Recaman sequence are [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9] The first duplicated term is a[25] = 42. Terms up to 328002 are needed to generate 0 to 1000.
Kotlin
<lang scala>// Version 1.2.60
fun main(args: Array<String>) {
val a = mutableListOf(0)
val used = mutableSetOf(0)
val used1000 = mutableSetOf(0)
var foundDup = false
var n = 1
while (n <= 15 || !foundDup || used1000.size < 1001) {
var next = a[n - 1] - n
if (next < 1 || used.contains(next)) next += 2 * n
val alreadyUsed = used.contains(next)
a.add(next)
if (!alreadyUsed) {
used.add(next)
if (next in 0..1000) used1000.add(next)
}
if (n == 14) {
println("The first 15 terms of the Recaman's sequence are: $a")
}
if (!foundDup && alreadyUsed) {
println("The first duplicated term is a[$n] = $next")
foundDup = true
}
if (used1000.size == 1001) {
println("Terms up to a[$n] are needed to generate 0 to 1000")
}
n++
}
}</lang>
- Output:
The first 15 terms of the Recaman's sequence are: [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9] The first duplicated term is a[24] = 42 Terms up to a[328002] are needed to generate 0 to 1000
Lua
This runs out of memory determining the final part :(
<lang lua>local a = {[0]=0} local used = {[0]=true} local used1000 = {[0]=true} local foundDup = false local n = 1
while n<=15 or not foundDup or #used1000<1001 do
local nxt = a[n - 1] - n
if nxt<1 or used[nxt] ~= nil then
nxt = nxt + 2 * n
end
local alreadyUsed = used[nxt] ~= nil
table.insert(a, nxt)
if not alreadyUsed then
used[nxt] = true
if 0<=nxt and nxt<=1000 then
used1000[nxt] = true
end
end
if n==14 then
io.write("The first 15 terms of the Recaman sequence are:")
for k=0,#a do
io.write(" "..a[k])
end
print()
end
if not foundDup and alreadyUsed then
print("The first duplicated term is a["..n.."] = "..nxt)
foundDup = true
end
if #used1000 == 1001 then
print("Terms up to a["..n.."] are needed to generate 0 to 1000")
end
n = n + 1
end</lang>
- Output:
The first 15 terms of the Recaman sequence are: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 The first duplicated term is a[24] = 42 lua: not enough memory
Mathematica / Wolfram Language
<lang Mathematica>ClearAll[f] f[s_List] := Block[{a = s-1, len = Length@s},
Append[s, If[a > len && ! MemberQ[s, a - len], a - len, a + len]]]; g = Nest[f, {0}, 70]
g = Nest[f, {0}, 70]; Take[g, 15] p = Select[Tally[g], Last /* EqualTo[2]]All, 1 p = Flatten[Position[g, #]] & /@ p; TakeSmallestBy[p, Last, 1]1</lang>
- Output:
{0,1,3,6,2,7,13,20,12,21,11,22,10,23,9}
{43,42,79,78}
{21,25}
Microsoft Small Basic
Inefficency of associative array allocation in Small Basic ban to provide the optional task. <lang smallbasic>' Recaman's sequence - smallbasic - 05/08/2015
nn=15
TextWindow.WriteLine("Recaman's sequence for the first " + nn + " numbers:")
recaman()
TextWindow.WriteLine(Text.GetSubTextToEnd(recaman,2))
nn="firstdup"
recaman()
TextWindow.WriteLine("The first duplicated term is a["+n+"]="+a[n])
Sub recaman
a=""
b=""
dup=""
recaman=""
firstdup=""
If nn="firstdup" Then
nn=1000
firstdup="True"
EndIf
For n=0 To nn-1
ap=a[n-1]+n
If a[n-1]<=n Then
a[n]=ap 'a[n]=a[n-1]+n
b[ap]=1
Else
am=a[n-1]-n
If b[am]=1 Then
a[n]=ap 'a[n]=a[n-1]+n
b[ap]=1
Else
a[n]=am 'a[n]=a[n-1]-n
b[am]=1
EndIf
EndIf
If firstdup Then
If dup[a[n]]=1 Then
Goto exitsub
EndIf
dup[a[n]]=1
EndIf
recaman=recaman+","+a[n]
EndFor
exitsub:
EndSub </lang>
- Output:
Recaman's sequence for the first 15 numbers: 0,1,3,6,2,7,13,20,12,21,11,22,10,23,9 The first duplicated term is a[24]=42
Nim
<lang Nim>import sequtils, sets, strutils
iterator recaman(num: Positive = Natural.high): tuple[n, a: int; duplicate: bool] =
var a = 0
yield (0, a, false)
var known = [0].toHashSet
for n in 1..<num:
var next = a - n
if next <= 0 or next in known:
next = a + n
a = next
yield (n, a, a in known)
known.incl a
echo "First 15 numbers in Recaman’s sequence: ", toSeq(recaman(15)).mapIt(it.a).join(" ")
for (n, a, dup) in recaman():
if dup: echo "First duplicate found: a($1) = $2".format(n, a) break
var target = toSeq(0..1000).toHashSet for (n, a, dup) in recaman():
target.excl a if target.card == 0: echo "All numbers from 0 to 1000 generated after $1 terms.".format(n) break</lang>
- Output:
First 15 numbers in Recaman’s sequence: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 First duplicate found: a(24) = 42 All numbers from 0 to 1000 generated after 328002 terms.
Objeck
<lang objeck>use Collection.Generic;
class RecamanSequence {
function : Main(args : String[]) ~ Nil {
GenerateSequence();
}
function : native : GenerateSequence() ~ Nil {
a := Vector->New()<IntHolder>;
a->AddBack(0);
used := Set->New()<IntHolder>;
used->Insert(0);
used1000 := Set->New()<IntHolder>;
used1000->Insert(0);
foundDup := false;
n := 1;
while (n <= 15 | <>foundDup | used1000->Size() < 1001) {
next := a->Get(n - 1) - n;
if (next < 1 | used->Has(next)) {
next += 2 * n;
};
alreadyUsed := used->Has(next);
a->AddBack(next);
if (<>alreadyUsed) {
used->Insert(next);
if (0 <= next & next <= 1000) {
used1000->Insert(next);
};
};
if (n = 14) {
str := ToString(a);
"The first 15 terms of the Recaman sequence are : {$str}"->PrintLine();
};
if (<>foundDup & alreadyUsed) {
"The first duplicate term is a[{$n}] := {$next}"->PrintLine();
foundDup := true;
};
if (used1000->Size() = 1001) {
"Terms up to a[{$n}] are needed to generate 0 to 1000"->PrintLine();
};
n++;
};
}
function : ToString(a : Vector<IntHolder>) ~ String {
out := "[";
each(i : a) {
out += a->Get(i)->Get();
if(i + 1 < a->Size()) {
out += ',';
};
};
out += ']';
return out; }
}</lang>
- Output:
The first 15 terms of the Recaman sequence are : [0,1,3,6,2,7,13,20,12,21,11,22,10,23,9] The first duplicate term is a[24] := 42 Terms up to a[328002] are needed to generate 0 to 1000
Perl
<lang perl>use bignum;
$max = 1000; $remaining += $_ for 1..$max;
my @recamans = 0; my $previous = 0;
while ($remaining > 0) {
$term++;
my $this = $previous - $term;
$this = $previous + $term unless $this > 0 and !$seen{$this};
push @recamans, $this;
$dup = $term if !$dup and defined $seen{$this};
$remaining -= $this if $this <= $max and ! defined $seen{$this};
$seen{$this}++;
$previous = $this;
}
print "First fifteen terms of Recaman's sequence: " . join(' ', @recamans[0..14]) . "\n"; print "First duplicate at term: a[$dup]\n"; print "Range 0..1000 covered by terms up to a[$term]\n";</lang>
- Output:
First fifteen terms of Recaman's sequence: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 First duplicate at term: a[24] Range 0..1000 covered by terms up to a[328002]
Phix
<lang Phix>bool found_duplicate = false sequence used = {} -- (grows to 1,942,300 entries) integer all_used = 0 sequence a = {0} integer n = 1, next, prev = 0 while n<=15 or not found_duplicate or all_used<1000 do
next = prev - n
if next<1 or (next<=length(used) and used[next]) then
next = prev + n
end if
a &= next
integer pad = next-length(used)
bool already_used = pad<=0 and used[next]
if not already_used then
if pad>0 then used &= repeat(false,pad) end if
used[next] = true
while all_used<length(used) and used[all_used+1] do
all_used += 1
end while
end if
if length(a)=15 then
puts(1,"The first 15 terms of the Recaman sequence are: ") ?a
end if
if already_used and not found_duplicate then
printf(1,"The first duplicated term is a[%d] = %d\n", {n, next})
found_duplicate = true;
end if
if all_used>=1000 then
printf(1,"Terms up to a[%d] are needed to generate 0 to 1000\n", {n});
end if
prev = next
n += 1
end while</lang>
- Output:
The first 15 terms of the Recaman sequence are: {0,1,3,6,2,7,13,20,12,21,11,22,10,23,9}
The first duplicated term is a[24] = 42
Terms up to a[328002] are needed to generate 0 to 1000
PHP
<lang php><?php $a = array(); array_push($a, 0);
$used = array(); array_push($used, 0);
$used1000 = array(); array_push($used1000, 0);
$foundDup = false; $n = 1;
while($n <= 15 || !$foundDup || count($used1000) < 1001) { $next = $a[$n - 1] - $n; if ($next < 1 || in_array($next, $used)) { $next += 2 * $n; } $alreadyUsed = in_array($next, $used); array_push($a, $next); if (!$alreadyUsed) { array_push($used, $next); if (0 <= $next && $next <= 1000) { array_push($used1000, $next); } } if ($n == 14) { echo "The first 15 terms of the Recaman sequence are : ["; foreach($a as $i => $v) { if ( $i == count($a) - 1) echo "$v"; else echo "$v, "; } echo "]\n"; } if (!$foundDup && $alreadyUsed) { printf("The first duplicate term is a[%d] = %d\n", $n, $next); $foundDup = true; } if (count($used1000) == 1001) { printf("Terms up to a[%d] are needed to generate 0 to 1000\n", $n); } $n++; } </lang>
- Output:
The first 15 terms of the Recaman sequence are : [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9] The first duplicate term is a[24] = 42 Terms up to a[328002] are needed to generate 0 to 1000
PureBasic
<lang PureBasic>#MAX=500000 Dim a.i(#MAX) Dim b.b(1) Dim c.b(1000) FillMemory(@c(),1000,1,#PB_Byte)
If OpenConsole() : Else : End 1 : EndIf For n_Count=0 To #MAX
If n_Count=0 a(n_Count)=0 ElseIf a(n_Count-1)-n_Count>0 And b(a(n_Count-1)-n_Count)=0 a(n_Count)=a(n_Count-1)-n_Count Else a(n_Count)=a(n_Count-1)+n_Count EndIf If ArraySize(b())<a(n_Count) : ReDim b(a(n_Count)) : EndIf If b(a(n_Count))=1 And fitD=0 : fitD=n_Count : EndIf b(a(n_Count))=1 If CompareMemory(@b(),@c(),1000) And fit1000=0 : fit1000=n_Count : Break : EndIf
Next
Print("First 15 terms: ") : For i=0 To 14 : Print(RSet(Str(a(i)),4)) : Next : PrintN("") PrintN("First duplicate term : a("+Str(fitD)+") = "+Str(a(fitD))) PrintN("Number of Recaman terms needed to generate all integers from [0..1000]: "+Str(fit1000)) Input() End</lang>
- Output:
First 15 terms: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 First duplicate term : a(24) = 42 Number of Recaman terms needed to generate all integers from [0..1000]: 328002
Python
Conditional iteration over a generator
<lang python>from itertools import islice
class Recamans():
"Recamán's sequence generator callable class"
def __init__(self):
self.a = None # Set of results so far
self.n = None # n'th term (counting from zero)
def __call__(self):
"Recamán's sequence generator"
nxt = 0
a, n = {nxt}, 0
self.a = a
self.n = n
yield nxt
while True:
an1, n = nxt, n + 1
nxt = an1 - n
if nxt < 0 or nxt in a:
nxt = an1 + n
a.add(nxt)
self.n = n
yield nxt
if __name__ == '__main__':
recamans = Recamans()
print("First fifteen members of Recamans sequence:",
list(islice(recamans(), 15)))
so_far = set()
for term in recamans():
if term in so_far:
print(f"First duplicate number in series is: a({recamans.n}) = {term}")
break
so_far.add(term)
n = 1_000
setn = set(range(n + 1)) # The target set of numbers to be covered
for _ in recamans():
if setn.issubset(recamans.a):
print(f"Range 0 ..{n} is covered by terms up to a({recamans.n})")
break</lang>
- Output:
First fifteen members of Recamans sequence: [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9] First duplicate number in series is: a(24) = 42 Range 0 ..1000 is covered by terms up to a(328002)
Parameterised query predicates
Passing different query predicates to a single more general function:
( This turns out to be c. 8X faster than than the iteration over generator approach above, on a simple start to end measure using time.time()) <lang python>Recaman sequence
- recamanUntil :: (Int -> Set Int > [Int] -> Bool) -> [Int]
def recamanUntil(p):
All terms of the Recaman series before the
first term for which the predicate p holds.
n = 1
r = 0 # First term of series
rs = [r]
seen = set(rs)
blnNew = True
while not p(seen, n, r, blnNew):
r = recamanSucc(seen, n, r)
blnNew = r not in seen
seen.add(r)
rs.append(r)
n = 1 + n
return rs
- recamanSucc :: Set Int -> Int -> Int
def recamanSucc(seen, n, r):
The successor for a given Recaman term,
given the set of Recaman terms seen so far.
back = r - n
return n + r if 0 > back or (back in seen) else back
- ------------------------- TEST -------------------------
- main :: IO ()
def main():
Test
print(
'First 15 Recaman:\r',
recamanUntil(
lambda seen, n, r, _: 15 == n
)
)
print(
'First duplicated Recaman:\r',
recamanUntil(
lambda seen, n, r, blnNew: not blnNew
)[-1]
)
setK = set(enumFromTo(0)(1000))
print(
'Number of Recaman terms needed to generate',
'all integers from [0..1000]:\r',
len(recamanUntil(
lambda seen, n, r, blnNew: (
blnNew and 1001 > r and setK.issubset(seen)
)
)) - 1
)
- ----------------------- GENERIC ------------------------
- enumFromTo :: (Int, Int) -> [Int]
def enumFromTo(m):
Integer enumeration from m to n. return lambda n: range(m, 1 + n)
if __name__ == '__main__':
main()</lang>
- Output:
First 15 Recaman: [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9] First duplicated Recaman: 42 Number of Recaman terms needed to generate all integers from [0..1000]: 328002
Alternatively, we can use query predicates in combination with the iteration of a function over a tuple,
and encapsulate the querying in the generic abstractions iterate and until.
This additional level of abstraction reduces the amount of new code that we have to write, facilitates refactoring, and turns out to have insignificant cost.
( This version is still c. 8X faster than the conditional iteration over generator version, as measured by a simple start and end test using time.time() ).
<lang python>Recaman by iteration of a function over a tuple.
from itertools import (islice)
- recamanTupleSucc :: Set Int -> (Int, Int, Bool) -> (Int, Int, Bool)
def recamanTupleSucc(seen):
The Nth in a series of Recaman tuples,
(N, previous term, boolPreviouslySeen?)
given the set of all terms seen so far.
def go(n, r, _):
back = r - n
nxt = n + r if 0 > back or (back in seen) else back
bln = nxt in seen
seen.add(nxt)
return (1 + n, nxt, bln)
return lambda tpl: go(*tpl)
- ------------------------- TEST -------------------------
- main :: IO()
def main():
First 15, and first duplicated Recaman.
f = recamanTupleSucc(set([0]))
print(
'First 15 Recaman:\n',
list(map(
snd,
take(15)(iterate(f)((1, 0, False)))
))
)
f = recamanTupleSucc(set([0]))
print(
'First duplicated Recaman:\n',
until(lambda x: x[2])(f)(
(1, 0, False)
)[1]
)
sk = set(enumFromTo(0)(1000))
sr = set([0])
f = recamanTupleSucc(sr)
print(
'Number of Recaman terms needed to generate',
'all integers from [0..1000]:\n',
until(
lambda x: not x[2] and 1001 > x[1] and sk.issubset(sr)
)(f)(
(1, 0, False)
)[0] - 1
)
- ----------------- GENERIC ABSTRACTIONS -----------------
- enumFromTo :: (Int, Int) -> [Int]
def enumFromTo(m):
Integer enumeration from m to n. return lambda n: range(m, 1 + n)
- iterate :: (a -> a) -> a -> Gen [a]
def iterate(f):
An infinite list of repeated
applications of f to x.
def go(x):
v = x
while True:
yield v
v = f(v)
return go
- snd :: (a, b) -> b
def snd(tpl):
Second component of a tuple. return tpl[1]
- take :: Int -> [a] -> [a]
- take :: Int -> String -> String
def take(n):
The prefix of xs of length n,
or xs itself if n > length xs.
return lambda xs: (
xs[0:n]
if isinstance(xs, list)
else islice(xs, n)
)
- until :: (a -> Bool) -> (a -> a) -> a -> a
def until(p):
The result of repeatedly applying f until p holds.
The initial seed value is x.
def go(f):
def g(x):
v = x
while not p(v):
v = f(v)
return v
return g
return go
- MAIN ---
if __name__ == '__main__':
main()</lang>
First 15 Recaman: [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9] First duplicated Recaman: 42 Number of Recaman terms needed to generate all integers from [0..1000]: 328002
Quackery
<lang Quackery>
[ stack 0 ] is seennumbers ( --> s )
[ bit seennumbers take | seennumbers put ] is seen ( n --> )
[ dup 0 < iff
[ drop true ] done
bit seennumbers share
& 0 != ] is seen? ( n --> b )
[ 1+ bit 1 - seennumbers share over & = ] is allseen? ( n --> b )
[ stack [ ] ] is repeats ( --> s )
[ 1 seennumbers replace [] repeats replace ' [ 0 ] 1 ] is startseq ( --> [ n )
[ over -1 peek
over - dup seen? if
[ over 2 * +
dup seen? if
[ repeats take
over join
repeats put ] ]
dup seen
swap dip join
1+ ] is nextterm ( [ n --> [ n )
say "first 15 terms: "
startseq
14 times nextterm
drop echo cr
say "first duplicated term: "
startseq
[ repeats share [] = while
nextterm
again ]
drop -1 peek echo cr
say "terms needed to generate 0 to 1000: "
startseq
[ nextterm
1000 allseen? until ]
nip 1 - echo cr
</lang>
- Output:
first 15 terms: [ 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 ] first duplicated term: 42 terms needed to generate 0 to 1000: 328002
R
A bit slow because the append() function is expensive. <lang rsplus> visited <- vector('logical', 1e8)
terms <- vector('numeric')
in_a_interval <- function(v) { visitedv+1 }
add_value <- function(v) { visitedv+1 <<- TRUE terms <<- append(terms, v) }
add_value(0) step <- 1 value <- 0
founddup <- FALSE
repeat { if ((value-step>0) && (!in_a_interval(value-step))) { value <- value - step } else { value <- value + step } if (in_a_interval(value) && !founddup) { cat("The first duplicated term is a[",step,"] = ",value,"\n", sep = "") founddup <- TRUE } add_value(value) if (all(visited[1:1000])) { cat("Terms up to a[",step,"] are needed to generate 0 to 1000\n",sep = "") break } step <- step + 1 if (step == 15) { cat("The first 15 terms are :") for (aterm in terms) { cat(aterm," ", sep = "") } cat("\n") } } </lang>
- Output:
The first 15 terms are :0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 The first duplicated term is a[24] = 42 Terms up to a[328002] are needed to generate 0 to 1000
Raku
(formerly Perl 6)
<lang perl6>my @recamans = 0, {
state %seen;
state $term;
$term++;
my $this = $^previous - $term;
$this = $previous + $term unless ($this > 0) && !%seen{$this};
%seen{$this} = True;
$this
} … *;
put "First fifteen terms of Recaman's sequence: ", @recamans[^15];
say "First duplicate at term: a[{ @recamans.first({@recamans[^$_].Bag.values.max == 2})-1 }]";
my @seen; my int $i = 0; loop {
next if (my int $this = @recamans[$i++]) > 1000 or @seen[$this];
@seen[$this] = 1;
say "Range 0..1000 covered by terms up to a[{$i - 1}]" and last if ++$ == 1001;
}</lang>
- Output:
First fifteen terms of Recaman's sequence: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 First duplicate at term: a[24] Range 0..1000 covered by terms up to a[328002]
REXX
version 1
Instead of using a subroutine to perform the three tasks with one invocation, the subroutine was used three times
(once for each of the task's three requirements).
Programmer's note: the short-circuited if REXX statement (lines 14 & 15):
if z<0 then z= _ + #
else if !.z then z= _ + #
could've been replaced with:
if !.z | z<0 then z= _ + #
<lang rexx>/*REXX pgm computes a Recamán sequence up to N; the 1st dup; # terms for a range of #'s.*/ parse arg N h . /*obtain optional arguments from the CL*/ if N== | N=="," then N= 15 /*Not specified? Then use the default.*/ if h== | h=="," then h= 1000 /* " " " " " " */
say "Recamán's sequence for the first " N " numbers: " recaman(N)
say; say "The first duplicate number in the Recamán's sequence is: " recaman(0) say; say "The number of terms to complete the range 0───►"h ' is: ' recaman(-h) exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ recaman: procedure; parse arg y,,d.; $=0; !.=0; _=0; !.0=1 /*init. array and vars.*/
r= y<0; Reca= 0; hi= abs(y) /*for the 2nd invoke. */
o= y==0; if y<1 then y= 1e8 /* " " 3rd " */
do #=1 for y-1; z= _ - # /*next # might be < 0. */
if z<0 then z= _ + # /*this is faster than: */
else if !.z then z= _ + # /*if !.z | z<0 then ···*/
!.z= 1; _= z /*mark it; add to seq.*/
if r then do; if z>hi then iterate /*ignore #'s too large.*/
if d.z== then Reca= Reca + 1 /*Unique? Bump counter.*/
d.z= . /*mark # as a new low. */
if Reca>=hi then return # /*list is complete ≥ HI*/
iterate
end /* [↑] a range of #s. */
if o then do; if d.z==. then return z; d.z=.; iterate /*check if dup #.*/
end
$= $ z /*add number to $ list?*/
end /*#*/; return $ /*return the $ list. */</lang>
- output when using the default input:
Run time was about 1/6 of a second, which is over 230% times faster than version 2.
Recamán's sequence for the first 15 numbers: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 The first duplicate number in the Recamán's sequence is: 42 The number of terms to complete the range 0───►1000 is: 328002
version 2
<lang rexx>/*REXX program computes & displays the Recaman sequence */ /*improved using version 1's method for task 3 */ Call time 'R' /* Start timer */ Parse Arg n If n= Then n=15 Say 'the first' n 'elements:' recaman(n) Say ans.2 Say ans.3 Say time('E') 'seconds elapsed' Exit
recaman: Parse Arg n /* Wanted number of elements */ have.=0 /* Number not yet in sequence */ e.0=0 /* First element */ have.0=1 /* is in the sequence */ s=0 /* Sequence to be shodn */ done=0 /* turn on first duplicate switch */ d.=0 d.0=1 dn=1 /* number of elements <=1000 */
Do i=1 until dn==1001 /* Loop until all found */
ip=i-1 /* previous index */
temp=e.ip-i /* potential next element */
If temp>0 & have.temp=0 Then /* to be used */
Nop
Else /* compute the alternative */
temp=e.ip+i
e.i=temp /* Set next element */
If words(s)<n Then /* not enough in output */
s=s temp /* add the element to the output */
If temp<=1000 Then Do /* eligible for task 3 */
If d.temp=0 Then Do /* not yet encountered */
d.temp=1 /* Remember it's there */
dn=dn+1 /* count of integers<=1000 found */
End
End
If done=0 & have.temp=1 Then Do
ans.2='First duplicate ('temp') added in iteration' i,
'elapsed:' time('E') 'seconds'
done=1
End
ans.3='Element number' i 'is the last to satisfy task 3. It is' temp
Have.temp=1
End
Return s</lang>
- Output:
the first 15 elements: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 First duplicate (42) added in iteration 24 elapsed: 0 seconds Element number 328002 is the last to satisfy task 3. It is 879 7.126000 seconds elapsed
Ring
<lang ring> load "zerolib.ring"
recaman = Z(0:50) duplicate = 0 dup = [] recDuplicate = [] recnum = 0
see "working..." + nl see "the first 15 Recaman's numbers are:" + nl
for n = 1 to len(recaman) - 1
if n = 1
recaman[0] = 0
add(dup,0)
see "" + recaman[0] + " "
ok
recaman[n] = recaman[n-1] - n
if recaman[n] <= 0
recaman[n] = recaman[n-1] + n
ok
fnrec = find(dup,recaman[n])
if fnrec > 0
del(dup,fnrec)
recaman[n] = recaman[n-1] + n
add(dup,recaman[n])
else
add(dup,recaman[n])
ok
recnum = recnum + 1
if recnum < 15
see "" + recaman[n] + " "
ok
add(recDuplicate,recaman[n])
next see nl
see "the first duplicated term is a[" for n = len(recDuplicate) to 2 step -1
for m = n-1 to 1 step -1
if recDuplicate[n] = recDuplicate[m]
duplicate = recDuplicate[n]
dupnr = n
ok
next
next
see "" + dupnr + "] = " + duplicate + nl see "done..." + nl </lang>
- Output:
working... the first 15 Recaman's numbers are: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 the first duplicated term is a[24] = 42 done...
Ruby
<lang ruby>require 'set'
a = [0] used = Set[0] used1000 = Set[0] foundDup = false n = 1 while n <= 15 or not foundDup or used1000.size < 1001
nxt = a[n - 1] - n
if nxt < 1 or used === nxt then
nxt = nxt + 2 * n
end
alreadyUsed = used === nxt
a << nxt
if not alreadyUsed then
used << nxt
if nxt >= 0 and nxt <= 1000 then
used1000 << nxt
end
end
if n == 14 then
print "The first 15 terms of the Recaman's sequence are ", a, "\n"
end
if not foundDup and alreadyUsed then
print "The first duplicated term is a[", n, "] = ", nxt, "\n"
foundDup = true
end
if used1000.size == 1001 then
print "Terms up to a[", n, "] are needed to generate 0 to 1000\n"
end
n = n + 1
end</lang>
- Output:
The first 15 terms of the Recaman's sequence are [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9] The first duplicated term is a[24] = 42 Terms up to a[328002] are needed to generate 0 to 1000
Scala
- Output:
Best seen in running your browser either by ScalaFiddle (ES aka JavaScript, non JVM) or Scastie (remote JVM).
<lang Scala>import scala.collection.mutable
object RecamansSequence extends App {
val (a, used) = (mutable.ArrayBuffer[Int](0), mutable.BitSet()) var (foundDup, hop, nUsed1000) = (false, 1, 0)
while (nUsed1000 < 1000) {
val _next = a(hop - 1) - hop
val next = if (_next < 1 || used.contains(_next)) _next + 2 * hop else _next
val alreadyUsed = used.contains(next)
a += next
if (!alreadyUsed) {
used.add(next)
if (next <= 1000) nUsed1000 += 1
}
if (!foundDup && alreadyUsed) {
println(s"The first duplicate term is a($hop) = $next")
foundDup = true
}
if (nUsed1000 == 1000)
println(s"Terms up to $hop are needed to generate 0 to 1000")
hop += 1 }
println(s"The first 15 terms of the Recaman sequence are : ${a.take(15)}")
}</lang>
Sidef
<lang ruby>func recamans_generator() {
var term = 0 var prev = 0 var seen = Hash()
{
var this = (prev - term)
if ((this <= 0) || seen{this}) {
this = (prev + term)
}
prev = this
seen{this} = true
term++
this
}
}
with (recamans_generator()) { |r|
say ("First 15 terms of the Recaman's sequence: ", 15.of { r.run }.join(', '))
}
with (recamans_generator()) {|r|
var seen = Hash()
Inf.times {|i|
var n = r.run
if (seen{n}) {
say "First duplicate term in the series is a(#{i}) = #{n}"
break
}
seen{n} = true
}
}
with (recamans_generator()) {|r|
var seen = Hash()
Inf.times {|i|
var n = r.run
if ((n <= 1000) && (seen{n} := true) && (seen.len == 1001)) {
say "Terms up to a(#{i}) are needed to generate 0 to 1000"
break
}
}
}</lang>
- Output:
First 15 terms of the Recaman's sequence: 0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9 First duplicate term in the series is a(24) = 42 Terms up to a(328002) are needed to generate 0 to 1000
uBasic/4tH
<lang basic>a = 0 ' the first one is free ;-) Print "First 15 numbers:"
For i = 1 Step 1 ' start loop
If i<16 Then Print a, ' print first 15 numbers b = Iif ((a-i<1) + (Func(_Peek(Max(0, a-i)))), a+i, a-i) If Func(_Peek(b)) Then Print "\nFirst repetition: ";b : Break Proc _Set(Set(a, b)) ' set bit in bitmap
Next End ' terminate program
' bitmap functions
_Set Param(1) : Let @(a@/32) = Func(_Poke(a@/32, a@%32)) : Return _Poke Param(2) : Return (Or(@(a@), Shl(1, b@))) _Peek Param(1) : Return (And(@(a@/32), Shl(1, a@%32))>0) </lang>
- Output:
First 15 numbers: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 First repetition: 42 0 OK, 0:459
VBScript
To run in console mode with cscript. <lang vb>' Recaman's sequence - vbscript - 04/08/2015 nx=15 h=1000 Wscript.StdOut.WriteLine "Recaman's sequence for the first " & nx & " numbers:" Wscript.StdOut.WriteLine recaman("seq",nx) Wscript.StdOut.WriteLine "The first duplicate number is: " & recaman("firstdup",0) Wscript.StdOut.WriteLine "The number of terms to complete the range 0--->"& h &" is: "& recaman("numterm",h) Wscript.StdOut.Write vbCrlf&".../...": zz=Wscript.StdIn.ReadLine()
function recaman(op,nn) Dim b,d,h Set b = CreateObject("Scripting.Dictionary") Set d = CreateObject("Scripting.Dictionary")
list="0" : firstdup=0
if op="firstdup" then nn=1000 : firstdup=1 end if if op="numterm" then h=nn : nn=10000000 : numterm=1 end if ax=0 'a(0)=0 b.Add 0,1 'b(0)=1 s=0 for n=1 to nn-1
an=ax-n
if an<=0 then an=ax+n elseif b.Exists(an) then an=ax+n end if ax=an 'a(n)=an if not b.Exists(an) then b.Add an,1 'b(an)=1 if op="seq" then list=list&" "&an end if if firstdup then if d.Exists(an) then recaman="a("&n&")="&an exit function else d.Add an,1 'd(an)=1 end if end if if numterm then if an<=h then if not d.Exists(an) then s=s+1 d.Add an,1 'd(an)=1 end if if s>=h then recaman=n exit function end if end if end if next 'n recaman=list end function 'recaman</lang>
- Output:
Recaman's sequence for the first 15 numbers: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 The first duplicate number is: a(24)=42 The number of terms to complete the range 0--->1000 is: 328002
Visual Basic .NET
<lang vbnet>Imports System Imports System.Collections.Generic
Module Module1
Sub Main(ByVal args As String())
Dim a As List(Of Integer) = New List(Of Integer)() From { 0 },
used As HashSet(Of Integer) = New HashSet(Of Integer)() From { 0 },
used1000 As HashSet(Of Integer) = used.ToHashSet(),
foundDup As Boolean = False
For n As Integer = 1 to Integer.MaxValue
Dim nv As Integer = a(n - 1) - n
If nv < 1 OrElse used.Contains(nv) Then nv += 2 * n
Dim alreadyUsed As Boolean = used.Contains(nv) : a.Add(nv)
If Not alreadyUsed Then used.Add(nv) : If nv > 0 AndAlso nv <= 1000 Then used1000.Add(nv)
If Not foundDup Then
If a.Count = 15 Then _
Console.WriteLine("The first 15 terms of the Recamán sequence are: ({0})", String.Join(", ", a))
If alreadyUsed Then _
Console.WriteLine("The first duplicated term is a({0}) = {1}", n, nv) : foundDup = True
End If
If used1000.Count = 1001 Then _
Console.WriteLine("Terms up to a({0}) are needed to generate 0 to 1000", n) : Exit For
Next
End Sub
End Module</lang>
- Output:
The first 15 terms of the Recamán sequence are: (0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9) The first duplicated term is a(24) = 42 Terms up to a(328002) are needed to generate 0 to 1000
Wren
<lang ecmascript>var a = [0] var used = { 0: true } var used1000 = { 0: true } var foundDup = false var n = 1 while (n <= 15 || !foundDup || used1000.count < 1001) {
var next = a[n-1] - n
if (next < 1 || used[next]) next = next + 2*n
var alreadyUsed = used[next]
a.add(next)
if (!alreadyUsed) {
used[next] = true
if (next >= 0 && next <= 1000) used1000[next] = true
}
if (n == 14) System.print("The first 15 terms of the Recaman's sequence are:\n%(a)")
if (!foundDup && alreadyUsed) {
System.print("The first duplicated term is a[%(n)] = %(next)")
foundDup = true
}
if (used1000.count == 1001) {
System.print("Terms up to a[%(n)] are needed to generate 0 to 1000")
}
n = n + 1
}</lang>
- Output:
The first 15 terms of the Recaman's sequence are: [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9] The first duplicated term is a[24] = 42 Terms up to a[328002] are needed to generate 0 to 1000
zkl
<lang zkl>fcn recamanW{ // -->iterator -->(n,a,True if a is a dup)
Walker.tweak(fcn(rn,rp,d){
n,p,a := rn.value, rp.value, p - n;
if(a<=0 or d.find(a)) a+=2*n;
d.incV(a); rp.set(a);
return(rn.inc(),a,d[a]>1);
}.fp(Ref(0),Ref(0),Dictionary()) )
}</lang> <lang zkl>print("First 15 members of Recaman's sequence: "); recamanW().walk(15).apply("get",1).println();
n,a := recamanW().filter1("get",2); // ie filter(a[n].dup) println("First duplicate number in series is: a(%d) = %d".fmt(n,a));
rw,ns,n,a,dup := recamanW(),1000,0,0,0; do{ n,a,dup=rw.next(); if(not dup and a<1000) ns-=1; }while(ns); println("Range 0..1000 is covered by terms up to a(%,d)".fmt(n));</lang>
- Output:
First 15 members of Recamans sequence: L(0,1,3,6,2,7,13,20,12,21,11,22,10,23,9) First duplicate number in series is: a(24) = 42 Range 0..1000 is covered by terms up to a(328,002)
- Programming Tasks
- Solutions by Programming Task
- 11l
- ALGOL W
- APL
- AppleScript
- AWK
- BASIC
- BCPL
- C
- GLib
- C sharp
- C++
- D
- FreeBASIC
- FOCAL
- Fōrmulæ
- Go
- Haskell
- J
- Java
- JavaScript
- Julia
- Kotlin
- Lua
- Mathematica
- Wolfram Language
- Microsoft Small Basic
- Nim
- Objeck
- Perl
- Phix
- PHP
- PureBasic
- Python
- Quackery
- R
- Raku
- REXX
- Ring
- Ruby
- Scala
- Sidef
- UBasic/4tH
- VBScript
- Visual Basic .NET
- Wren
- Zkl