Ludic numbers: Difference between revisions
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Ludic numbers 2000 to 2005: 21475 21481 21487 21493 21503 21511 |
Ludic numbers 2000 to 2005: 21475 21481 21487 21493 21503 21511 |
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Ludic Triplets below 250: [1 3 7] [5 7 11] [11 13 17] [23 25 29] [41 43 47] [173 175 179] [221 223 227] [233 235 239]</pre> |
Ludic Triplets below 250: [1 3 7] [5 7 11] [11 13 17] [23 25 29] [41 43 47] [173 175 179] [221 223 227] [233 235 239]</pre> |
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=={{header|FreeBASIC}}== |
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<lang freebasic>' FB 1.05.0 Win64 |
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' As it would be too expensive to actually remove elements from the array |
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' we instead set an element to 0 if it has been removed. |
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Sub ludic(n As Integer, lu() As Integer) |
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If n < 1 Then Return |
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Redim lu(1 To n) |
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lu(1) = 1 |
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If n = 1 Then Return |
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Dim As Integer count = 1, count2 |
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Dim As Integer i, j, k = 1 |
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Dim As Integer ub = n * 11 '' big enough to deal with up to 2005 ludic numbers |
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Dim a(2 To ub) As Integer |
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For i = 2 To ub : a(i) = i : Next |
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Do |
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k += 1 |
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For i = k to ub |
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If a(i) > 0 Then |
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count += 1 |
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lu(count) = a(i) |
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If n = count Then Return |
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a(i) = 0 |
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k = i |
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Exit For |
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End If |
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Next |
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count2 = 0 |
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j = k + 1 |
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While j <= ub |
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If a(j) > 0 Then |
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count2 +=1 |
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If count2 = k Then |
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a(j) = 0 |
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count2 = 0 |
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End If |
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End If |
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j += 1 |
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Wend |
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Loop |
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End Sub |
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Dim i As Integer |
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Dim lu() As Integer |
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ludic(25, lu()) |
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Print "The first 25 Ludic numbers are :" |
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For i = 1 To 25 |
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Print Using "###"; lu(i); |
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Print " "; |
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Next |
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Erase lu |
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Print |
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Dim As Integer Count = 0 |
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ludic(1000, lu()) |
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For i = 1 To 1000 |
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If lu(i) <= 1000 Then |
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count += 1 |
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Else |
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Exit For |
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End If |
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Next |
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Erase lu |
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Print |
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Print "There are"; count; " Ludic numbers <= 1000" |
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Print |
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Print "The 2000th to 2005th Ludics are :" |
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ludic(2005, lu()) |
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For i = 2000 To 2005 |
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Print lu(i); " "; |
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Next |
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Erase lu |
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Print : Print |
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Print "The Ludic triplets below 250 are : " |
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Dim As Integer j, k, ldc |
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Dim b As Boolean |
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ludic(250, lu()) |
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For i = 1 To 248 |
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ldc = lu(i) |
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If ldc >= 244 Then Exit For |
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b = False |
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For j = i + 1 To 249 |
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If lu(j) = ldc + 2 Then |
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b = True |
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k = j |
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Exit For |
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ElseIf lu(j) > ldc + 2 Then |
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Exit For |
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End If |
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Next j |
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If b = False Then Continue For |
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For j = k + 1 To 250 |
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If lu(j) = ldc + 6 Then |
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Print "("; Str(ldc); ","; ldc + 2; ","; ldc + 6; ")" |
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Exit For |
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ElseIf lu(j) > ldc + 6 Then |
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Exit For |
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End If |
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Next j |
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Next i |
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Erase lu |
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Print |
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Print "Press any key to quit" |
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Sleep </lang> |
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{{out}} |
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<pre> |
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The first 25 Ludic numbers are : |
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1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 |
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83 89 91 97 107 |
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There are 142 Ludic numbers <= 1000 |
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The 2000th to 2005th Ludics are : |
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21475 21481 21487 21493 21503 21511 |
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The Ludic triplets below 250 are : |
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(1, 3, 7) |
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(5, 7, 11) |
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(11, 13, 17) |
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(23, 25, 29) |
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(41, 43, 47) |
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(173, 175, 179) |
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(221, 223, 227) |
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(233, 235, 239) |
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</pre> |
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=={{header|Go}}== |
=={{header|Go}}== |
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Revision as of 15:08, 27 October 2016
You are encouraged to solve this task according to the task description, using any language you may know.
Ludic numbers are related to prime numbers as they are generated by a sieve quite like the Sieve of Eratosthenes is used to generate prime numbers.
The first ludic number is 1.
To generate succeeding ludic numbers create an array of increasing integers starting from 2.
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 ...
(Loop)
- Take the first member of the resultant array as the next ludic number 2.
- Remove every 2nd indexed item from the array (including the first).
234567891011121314151617181920212223242526...
- (Unrolling a few loops...)
- Take the first member of the resultant array as the next ludic number 3.
- Remove every 3rd indexed item from the array (including the first).
35 7911 131517 192123 252729 313335 373941 434547 4951...
- Take the first member of the resultant array as the next ludic number 5.
- Remove every 5th indexed item from the array (including the first).
57 11 13 171923 25 29 313537 41 43 474953 55 59 616567 71 73 77 ...
- Take the first member of the resultant array as the next ludic number 7.
- Remove every 7th indexed item from the array (including the first).
711 13 17 23 25 293137 41 43 47 53 555961 67 71 73 77 838589 91 97 ...
- ...
- Take the first member of the current array as the next ludic number L.
- Remove every Lth indexed item from the array (including the first).
- ...
- Task
- Generate and show here the first 25 ludic numbers.
- How many ludic numbers are there less than or equal to 1000?
- Show the 2000..2005th ludic numbers.
- Stretch goal
Show all triplets of ludic numbers < 250.
- A triplet is any three numbers where all three numbers are also ludic numbers.
360 Assembly
<lang 360asm>* Ludic numbers 23/04/2016 LUDICN CSECT
USING LUDICN,R15 set base register
LH R9,NMAX r9=nmax
SRA R9,1 r9=nmax/2
LA R6,2 i=2
LOOPI1 CR R6,R9 do i=2 to nmax/2
BH ELOOPI1
LA R1,LUDIC-1(R6) @ludic(i)
CLI 0(R1),X'01' if ludic(i)
BNE ELOOPJ1
SR R8,R8 n=0
LA R7,1(R6) j=i+1
LOOPJ1 CH R7,NMAX do j=i+1 to nmax
BH ELOOPJ1
LA R1,LUDIC-1(R7) @ludic(j)
CLI 0(R1),X'01' if ludic(j)
BNE NOTJ1
LA R8,1(R8) n=n+1
NOTJ1 CR R8,R6 if n=i
BNE NDIFI
LA R1,LUDIC-1(R7) @ludic(j)
MVI 0(R1),X'00' ludic(j)=false
SR R8,R8 n=0
NDIFI LA R7,1(R7) j=j+1
B LOOPJ1
ELOOPJ1 LA R6,1(R6) i=i+1
B LOOPI1
ELOOPI1 XPRNT =C'First 25 ludic numbers:',23
LA R10,BUF @buf=0
SR R8,R8 n=0
LA R6,1 i=1
LOOPI2 CH R6,NMAX do i=1 to nmax
BH ELOOPI2
LA R1,LUDIC-1(R6) @ludic(i)
CLI 0(R1),X'01' if ludic(i)
BNE NOTI2
XDECO R6,XDEC i
MVC 0(4,R10),XDEC+8 output i
LA R10,4(R10) @buf=@buf+4
LA R8,1(R8) n=n+1
LR R2,R8 n
SRDA R2,32
D R2,=F'5' r2=mod(n,5)
LTR R2,R2 if mod(n,5)=0
BNZ NOTI2
XPRNT BUF,20
LA R10,BUF @buf=0
NOTI2 EQU *
CH R8,=H'25' if n=25
BE ELOOPI2
LA R6,1(R6) i=i+1
B LOOPI2
ELOOPI2 MVC BUF(25),=C'Ludic numbers below 1000:'
SR R8,R8 n=0
LA R6,1 i=1
LOOPI3 CH R6,=H'999' do i=1 to 999
BH ELOOPI3
LA R1,LUDIC-1(R6) @ludic(i)
CLI 0(R1),X'01' if ludic(i)
BNE NOTI3
LA R8,1(R8) n=n+1
NOTI3 LA R6,1(R6) i=i+1
B LOOPI3
ELOOPI3 XDECO R8,XDEC edit n
MVC BUF+25(6),XDEC+6 output n
XPRNT BUF,31 print buffer
MVC BUF(80),=CL80'Ludic numbers 2000 to 2005:'
LA R10,BUF+28 @buf=28
SR R8,R8 n=0
LA R6,1 i=1
LOOPI4 CH R6,NMAX do i=1 to nmax
BH ELOOPI4
LA R1,LUDIC-1(R6) @ludic(i)
CLI 0(R1),X'01' if ludic(i)
BNE NOTI4
LA R8,1(R8) n=n+1
CH R8,=H'2000' if n>=2000
BL NOTI4
XDECO R6,XDEC edit i
MVC 0(6,R10),XDEC+6 output i
LA R10,6(R10) @buf=@buf+6
CH R8,=H'2005' if n=2005
BE ELOOPI4
NOTI4 LA R6,1(R6) i=i+1
B LOOPI4
ELOOPI4 XPRNT BUF,80 print buffer
XPRNT =C'Ludic triplets below 250:',25
LA R6,1 i=1
LOOPI5 CH R6,=H'243' do i=1 to 243
BH ELOOPI5
LA R1,LUDIC-1(R6) @ludic(i)
CLI 0(R1),X'01' if ludic(i)
BNE ITERI5
LA R1,LUDIC+1(R6) @ludic(i+2)
CLI 0(R1),X'01' if ludic(i+2)
BNE ITERI5
LA R1,LUDIC+5(R6) @ludic(i+6)
CLI 0(R1),X'01' if ludic(i+6)
BNE ITERI5
MVC BUF+0(1),=C'[' [
XDECO R6,XDEC edit i
MVC BUF+1(4),XDEC+8 output i
LA R2,2(R6) i+2
XDECO R2,XDEC edit i+2
MVC BUF+5(4),XDEC+8 output i+2
LA R2,6(R6) i+6
XDECO R2,XDEC edit i+6
MVC BUF+9(4),XDEC+8 output i+6
MVC BUF+13(1),=C']' ]
XPRNT BUF,14 print buffer
ITERI5 LA R6,1(R6) i=i+1
B LOOPI5
ELOOPI5 XR R15,R15 set return code
BR R14 return to caller
LTORG
BUF DS CL80 buffer XDEC DS CL12 decimal editor NMAX DC H'25000' nmax LUDIC DC 25000X'01' ludic(nmax)=true
YREGS
END LUDICN</lang>
- Output:
First 25 ludic numbers: 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107 Ludic numbers below 1000: 142 Ludic numbers 2000 to 2005: 21475 21481 21487 21493 21503 21511 Ludic triplets below 250: [ 1 3 7] [ 5 7 11] [ 11 13 17] [ 23 25 29] [ 41 43 47] [ 173 175 179] [ 221 223 227] [ 233 235 239]
ABAP
Works with NW 7.40 SP8 <lang ABAP>CLASS lcl_ludic DEFINITION CREATE PUBLIC.
PUBLIC SECTION.
TYPES: t_ludics TYPE SORTED TABLE OF i WITH UNIQUE KEY table_line.
TYPES: BEGIN OF t_triplet,
i1 TYPE i,
i2 TYPE i,
i3 TYPE i,
END OF t_triplet.
TYPES: t_triplets TYPE STANDARD TABLE OF t_triplet WITH EMPTY KEY.
CLASS-METHODS:
ludic_up_to
IMPORTING i_int TYPE i
RETURNING VALUE(r_ludics) TYPE t_ludics,
get_triplets
IMPORTING i_ludics TYPE t_ludics
RETURNING VALUE(r_triplets) TYPE t_triplets.
"RETURNING parameters (CallByValue) only used for readability of the demo "in "Real Life" you should use EXPORTING (CallByRef) for tables
ENDCLASS.
cl_demo_output=>begin_section( 'First 25 Ludics' ). cl_demo_output=>write( lcl_ludic=>ludic_up_to( 110 ) ).
cl_demo_output=>begin_section( 'Ludics up to 1000' ). cl_demo_output=>write( lines( lcl_ludic=>ludic_up_to( 1000 ) ) ).
cl_demo_output=>begin_section( '2000th - 2005th Ludics' ). DATA(ludics) = lcl_ludic=>ludic_up_to( 22000 ). cl_demo_output=>write( VALUE lcl_ludic=>t_ludics( FOR i = 2000 WHILE i <= 2005 ( ludics[ i ] ) ) ).
cl_demo_output=>begin_section( 'Triplets up to 250' ). cl_demo_output=>write( lcl_ludic=>get_triplets( lcl_ludic=>ludic_up_to( 250 ) ) ).
cl_demo_output=>display( ).
CLASS lcl_ludic IMPLEMENTATION.
METHOD ludic_up_to.
r_ludics = VALUE #( FOR i = 2 WHILE i <= i_int ( i ) ).
DATA(cursor) = 0.
WHILE cursor < lines( r_ludics ).
cursor = cursor + 1.
DATA(this_ludic) = r_ludics[ cursor ].
DATA(remove_cursor) = cursor + this_ludic.
WHILE remove_cursor <= lines( r_ludics ).
DELETE r_ludics INDEX remove_cursor.
remove_cursor = remove_cursor + this_ludic - 1.
ENDWHILE.
ENDWHILE.
INSERT 1 INTO TABLE r_ludics. "add one as the first Ludic number (per definition)
ENDMETHOD.
METHOD get_triplets.
DATA(i) = 0.
WHILE i < lines( i_ludics ) - 2.
i = i + 1.
DATA(this_ludic) = i_ludics[ i ].
IF line_exists( i_ludics[ table_line = this_ludic + 2 ] )
AND line_exists( i_ludics[ table_line = this_ludic + 6 ] ).
r_triplets = VALUE #(
BASE r_triplets
( i1 = i_ludics[ table_line = this_ludic ]
i2 = i_ludics[ table_line = this_ludic + 2 ]
i3 = i_ludics[ table_line = this_ludic + 6 ]
)
).
ENDIF.
ENDWHILE.
ENDMETHOD.
ENDCLASS.</lang>
- Output:
First 25 Ludics 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107 Ludics up to 1000 142 2000th - 2005th Ludics 21475 21481 21487 21493 21503 21511 Triplets up to 250 1 3 7 5 7 11 11 13 17 23 25 29 41 43 47 173 175 179 221 223 227 233 235 239
ALGOL 68
<lang algol68># find some Ludic numbers #
- sieve the Ludic numbers up to 30 000 #
INT max number = 30 000; [ 1 : max number ]INT candidates; FOR n TO UPB candidates DO candidates[ n ] := n OD; FOR n FROM 2 TO UPB candidates OVER 2 DO
IF candidates[ n ] /= 0 THEN
# have a ludic number #
INT number count := -1;
FOR remove pos FROM n TO UPB candidates DO
IF candidates[ remove pos ] /= 0 THEN
# have a number we haven't elminated yet #
number count +:= 1;
IF number count = n THEN
# this number should be removed #
candidates[ remove pos ] := 0;
number count := 0
FI
FI
OD
FI
OD;
- show some Ludic numbers and counts #
print( ( "Ludic numbers: " ) ); INT ludic count := 0; FOR n TO UPB candidates DO
IF candidates[ n ] /= 0 THEN
# have a ludic number #
ludic count +:= 1;
IF ludic count < 26 THEN
# this is one of the first few Ludic numbers #
print( ( " ", whole( n, 0 ) ) );
IF ludic count = 25 THEN
print( ( " ...", newline ) )
FI
FI;
IF ludic count = 2000 THEN
print( ( "Ludic numbers 2000-2005: ", whole( n, 0 ) ) )
ELIF ludic count > 2000 AND ludic count < 2006 THEN
print( ( " ", whole( n, 0 ) ) );
IF ludic count = 2005 THEN
print( ( newline ) )
FI
FI
FI;
IF n = 1000 THEN
# count ludic numbers up to 1000 #
print( ( "There are ", whole( ludic count, 0 ), " Ludic numbers up to 1000", newline ) )
FI
OD;
- find the Ludic triplets below 250 #
print( ( "Ludic triplets below 250:", newline ) ); FOR n TO 250 - 6 DO
IF candidates[ n ] /= 0 AND candidates[ n + 2 ] /= 0 AND candidates[ n + 6 ] /= 0 THEN
# have a triplet #
print( ( " ", whole( n, -3 ), ", ", whole( n + 2, -3 ), ", ", whole( n + 6, -3 ), newline ) )
FI
OD</lang>
- Output:
Ludic numbers: 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107 ...
There are 142 Ludic numbers up to 1000
Ludic numbers 2000-2005: 21475 21481 21487 21493 21503 21511
Ludic triplets below 250:
1, 3, 7
5, 7, 11
11, 13, 17
23, 25, 29
41, 43, 47
173, 175, 179
221, 223, 227
233, 235, 239
AutoHotkey
<lang AutoHotkey>#NoEnv SetBatchLines, -1 Ludic := LudicSieve(22000)
Loop, 25 ; the first 25 ludic numbers Task1 .= Ludic[A_Index] " "
for i, Val in Ludic ; the number of ludic numbers less than or equal to 1000 if (Val <= 1000) Task2++ else break
Loop, 6 ; the 2000..2005'th ludic numbers Task3 .= Ludic[1999 + A_Index] " "
for i, Val in Ludic { ; all triplets of ludic numbers < 250 if (Val + 6 > 249) break if (Ludic[i + 1] = Val + 2 && Ludic[i + 2] = Val + 6 || i = 1) Task4 .= "(" Val " " Val + 2 " " Val + 6 ") " }
MsgBox, % "First 25:`t`t" Task1 . "`nLudics below 1000:`t" Task2 . "`nLudic 2000 to 2005:`t" Task3 . "`nTriples below 250:`t" Task4 return
LudicSieve(Limit) { Arr := [], Ludic := [] Loop, % Limit Arr.Insert(A_Index) Ludic.Insert(Arr.Remove(1)) while Arr.MaxIndex() != 1 { Ludic.Insert(n := Arr.Remove(1)) , Removed := 0 Loop, % Arr.MaxIndex() // n { Arr.Remove(A_Index * n - Removed) , Removed++ } } Ludic.Insert(Arr[1]) return Ludic }</lang>
- Output:
First 25: 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107 Ludics below 1000: 142 Ludic 2000 to 2005: 21475 21481 21487 21493 21503 21511 Triples below 250: (1 3 7) (5 7 11) (11 13 17) (23 25 29) (41 43 47) (173 175 179) (221 223 227) (233 235 239)
C
<lang c>#include <stdio.h>
- include <stdlib.h>
typedef unsigned uint; typedef struct { uint i, v; } filt_t;
// ludics with at least so many elements and reach at least such value uint* ludic(uint min_len, uint min_val, uint *len) { uint cap, i, v, active = 1, nf = 0; filt_t *f = calloc(cap = 2, sizeof(*f)); f[1].i = 4;
for (v = 1; ; ++v) { for (i = 1; i < active && --f[i].i; i++);
if (i < active) f[i].i = f[i].v; else if (nf == f[i].i) f[i].i = f[i].v, ++active; // enable one more filter else { if (nf >= cap) f = realloc(f, sizeof(*f) * (cap*=2)); f[nf] = (filt_t){ v + nf, v }; if (++nf >= min_len && v >= min_val) break; } }
// pack the sequence into a uint[] // filt_t struct was used earlier for cache locality in loops uint *x = (void*) f; for (i = 0; i < nf; i++) x[i] = f[i].v; x = realloc(x, sizeof(*x) * nf);
*len = nf; return x; }
int find(uint *a, uint v) { uint i; for (i = 0; a[i] <= v; i++) if (v == a[i]) return 1; return 0; }
int main(void) { uint len, i, *x = ludic(2005, 1000, &len);
printf("First 25:"); for (i = 0; i < 25; i++) printf(" %u", x[i]); putchar('\n');
for (i = 0; x[i] <= 1000; i++); printf("Ludics below 1000: %u\n", i);
printf("Ludic 2000 to 2005:"); for (i = 2000; i <= 2005; i++) printf(" %u", x[i - 1]); putchar('\n');
printf("Triples below 250:"); for (i = 0; x[i] + 6 <= 250; i++) if (find(x, x[i] + 2) && find(x, x[i] + 6)) printf(" (%u %u %u)", x[i], x[i] + 2, x[i] + 6);
putchar('\n');
free(x); return 0; }</lang>
- Output:
First 25: 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107 Ludics below 1000: 142 Ludic 2000 to 2005: 21475 21481 21487 21493 21503 21511 Triples below 250: (1 3 7) (5 7 11) (11 13 17) (23 25 29) (41 43 47) (173 175 179) (221 223 227) (233 235 239)
C#
<lang csharp>using System; using System.Linq; using System.Collections.Generic;
public class Program {
public static void Main()
{
Console.WriteLine("First 25 ludic numbers:");
Console.WriteLine(string.Join(", ", LudicNumbers(150).Take(25)));
Console.WriteLine();
Console.WriteLine($"There are {LudicNumbers(1001).Count()} ludic numbers below 1000");
Console.WriteLine();
foreach (var ludic in LudicNumbers(22000).Skip(1999).Take(6)
.Select((n, i) => $"#{i+2000} = {n}")) {
Console.WriteLine(ludic);
}
Console.WriteLine();
Console.WriteLine("Triplets below 250:");
var queue = new Queue<int>(5);
foreach (int x in LudicNumbers(255)) {
if (queue.Count == 5) queue.Dequeue();
queue.Enqueue(x);
if (x - 6 < 250 && queue.Contains(x - 6) && queue.Contains(x - 4)) {
Console.WriteLine($"{x-6}, {x-4}, {x}");
}
}
}
public static IEnumerable<int> LudicNumbers(int limit) {
yield return 1;
//Like a linked list, but with value types.
//Create 2 extra entries at the start to avoid ugly index calculations
//and another at the end to avoid checking for index-out-of-bounds.
Entry[] values = Enumerable.Range(0, limit + 1).Select(n => new Entry(n)).ToArray();
for (int i = 2; i < limit; i = values[i].Next) {
yield return values[i].N;
int start = i;
while (start < limit) {
Unlink(values, start);
for (int step = 0; step < i && start < limit; step++)
start = values[start].Next;
}
}
}
static void Unlink(Entry[] values, int index) {
values[values[index].Prev].Next = values[index].Next;
values[values[index].Next].Prev = values[index].Prev;
}
}
struct Entry {
public Entry(int n) : this() {
N = n;
Prev = n - 1;
Next = n + 1;
}
public int N { get; }
public int Prev { get; set; }
public int Next { get; set; }
}</lang>
- Output:
First 25 ludic numbers: 1, 2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 37, 41, 43, 47, 53, 61, 67, 71, 77, 83, 89, 91, 97, 107 There are 142 ludic numbers below 1000 #2000 = 21475 #2001 = 21481 #2002 = 21487 #2003 = 21493 #2004 = 21503 #2005 = 21511 Triplets below 250: 1, 3, 7 5, 7, 11 11, 13, 17 23, 25, 29 41, 43, 47 173, 175, 179 221, 223, 227 233, 235, 239
C++
<lang cpp>
- include <vector>
- include <iostream>
using namespace std;
class ludic { public:
void ludicList()
{
_list.push_back( 1 );
vector<int> v;
for( int x = 2; x < 22000; x++ )
v.push_back( x );
while( true )
{
vector<int>::iterator i = v.begin();
int z = *i;
_list.push_back( z );
while( true )
{
i = v.erase( i );
if( distance( i, v.end() ) <= z - 1 ) break;
advance( i, z - 1 );
}
if( v.size() < 1 ) return;
}
}
void show( int s, int e )
{
for( int x = s; x < e; x++ )
cout << _list[x] << " ";
}
void findTriplets( int e )
{
int lu, x = 0;
while( _list[x] < e )
{
lu = _list[x];
if( inList( lu + 2 ) && inList( lu + 6 ) )
cout << "(" << lu << " " << lu + 2 << " " << lu + 6 << ")\n";
x++;
}
}
int count( int e )
{
int x = 0, c = 0;
while( _list[x++] <= 1000 ) c++;
return c;
}
private:
bool inList( int lu )
{
for( int x = 0; x < 250; x++ )
if( _list[x] == lu ) return true;
return false;
}
vector<int> _list;
};
int main( int argc, char* argv[] ) {
ludic l; l.ludicList(); cout << "first 25 ludic numbers:" << "\n"; l.show( 0, 25 ); cout << "\n\nThere are " << l.count( 1000 ) << " ludic numbers <= 1000" << "\n"; cout << "\n2000 to 2005'th ludic numbers:" << "\n"; l.show( 1999, 2005 ); cout << "\n\nall triplets of ludic numbers < 250:" << "\n"; l.findTriplets( 250 ); cout << "\n\n"; return system( "pause" );
} </lang>
- Output:
first 25 ludic numbers: 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107 There are 142 ludic numbers <= 1000 2000 to 2005'th ludic numbers: 21475 21481 21487 21493 21503 21511 all triplets of ludic numbers < 250: (1 3 7) (5 7 11) (11 13 17) (23 25 29) (41 43 47) (173 175 179) (221 223 227) (233 235 239)
Clojure
<lang clojure>(defn ints-from [n]
(cons n (lazy-seq (ints-from (inc n)))))
(defn drop-nth [n seq]
(cond
(zero? n) seq
(empty? seq) []
:else (concat (take (dec n) seq) (lazy-seq (drop-nth n (drop n seq))))))
(def ludic ((fn ludic
([] (ludic 1)) ([n] (ludic n (ints-from (inc n)))) ([n [f & r]] (cons n (lazy-seq (ludic f (drop-nth f r))))))))
(defn ludic? [n] (= (first (filter (partial <= n) ludic)) n))
(print "First 25: ") (println (take 25 ludic)) (print "Count below 1000: ") (println (count (take-while (partial > 1000) ludic))) (print "2000th through 2005th: ") (println (map (partial nth ludic) (range 1999 2005))) (print "Triplets < 250: ") (println (filter (partial every? ludic?)
(for [i (range 250)] (list i (+ i 2) (+ i 6)))))</lang>
- Output:
First 25: (1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107) Count below 1000: 142 2000th through 2005th: (21475 21481 21487 21493 21503 21511) Triplets < 250: ((1 3 7) (5 7 11) (11 13 17) (23 25 29) (41 43 47) (173 175 179) (221 223 227) (233 235 239))
D
opApply Version
<lang d>struct Ludics(T) {
int opApply(int delegate(in ref T) dg) {
int result;
T[] rotor, taken = [T(1)];
result = dg(taken[0]);
if (result) return result;
for (T i = 2; ; i++) { // Shoud be stopped if T has a max.
size_t j = 0;
for (; j < rotor.length; j++)
if (!--rotor[j])
break;
if (j < rotor.length) {
rotor[j] = taken[j + 1];
} else {
result = dg(i);
if (result) return result;
taken ~= i;
rotor ~= taken[j + 1];
}
}
}
}
void main() {
import std.stdio, std.range, std.algorithm;
// std.algorithm.take can't be used here.
uint[] L;
foreach (const x; Ludics!uint())
if (L.length < 2005)
L ~= x;
else
break;
writeln("First 25 ludic primes:\n", L.take(25));
writefln("\nThere are %d ludic numbers <= 1000.",
L.until!q{ a > 1000 }.walkLength);
writeln("\n2000'th .. 2005'th ludic primes:\n", L[1999 .. 2005]);
enum m = 250;
const triplets = L.filter!(x => x + 6 < m &&
L.canFind(x + 2) && L.canFind(x + 6))
// Ugly output:
//.map!(x => tuple(x, x + 2, x + 6)).array;
.map!(x => [x, x + 2, x + 6]).array;
writefln("\nThere are %d triplets less than %d:\n%s",
triplets.length, m, triplets);
}</lang>
- Output:
First 25 ludic primes: [1, 2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 37, 41, 43, 47, 53, 61, 67, 71, 77, 83, 89, 91, 97, 107] There are 142 ludic numbers <= 1000. 2000'th .. 2005'th ludic primes: [21475, 21481, 21487, 21493, 21503, 21511] There are 8 triplets less than 250: [[1, 3, 7], [5, 7, 11], [11, 13, 17], [23, 25, 29], [41, 43, 47], [173, 175, 179], [221, 223, 227], [233, 235, 239]]
The run-time is about 0.03 seconds or less. It takes about 2.0 seconds to generate 50_000 Ludic numbers with ldc2 compiler.
Range Version
This is the same code modified to be a Range. <lang d>struct Ludics(T) {
T[] rotor, taken = [T(1)]; T i; size_t j; T front = 1; // = taken[0]; bool running = false; static immutable bool empty = false;
void popFront() pure nothrow @safe {
if (running)
goto RESUME;
else
running = true;
i = 2;
while (true) {
j = 0;
while (j < rotor.length) {
rotor[j]--;
if (!rotor[j])
break;
j++;
}
if (j < rotor.length) {
rotor[j] = taken[j + 1];
} else {
front = i;
return;
RESUME:
taken ~= i;
rotor ~= taken[j + 1];
}
i++; // Could overflow if T has a max.
}
}
}
void main() {
import std.stdio, std.range, std.algorithm, std.array;
Ludics!uint L;
writeln("First 25 ludic primes:\n", L.take(25));
writefln("\nThere are %d ludic numbers <= 1000.",
L.until!q{ a > 1000 }.walkLength);
writeln("\n2000'th .. 2005'th ludic primes:\n", L.drop(1999).take(6));
enum uint m = 250;
const few = L.until!(x => x > m).array;
const triplets = few.filter!(x => x + 6 < m && few.canFind(x + 2)
&& few.canFind(x + 6))
// Ugly output:
//.map!(x => tuple(x, x + 2, x + 6)).array;
.map!(x => [x, x + 2, x + 6]).array;
writefln("\nThere are %d triplets less than %d:\n%s",
triplets.length, m, triplets);
}</lang> The output is the same. This version is slower, it takes about 3.3 seconds to generate 50_000 Ludic numbers with ldc2 compiler.
Range Generator Version
<lang d>void main() {
import std.stdio, std.range, std.algorithm, std.concurrency;
Generator!T ludics(T)() {
return new typeof(return)({
T[] rotor, taken = [T(1)];
yield(taken[0]);
for (T i = 2; ; i++) { // Shoud be stopped if T has a max.
size_t j = 0;
for (; j < rotor.length; j++)
if (!--rotor[j])
break;
if (j < rotor.length) {
rotor[j] = taken[j + 1];
} else {
yield(i);
taken ~= i;
rotor ~= taken[j + 1];
}
}
});
}
const L = ludics!uint.take(2005).array;
writeln("First 25 ludic primes:\n", L.take(25));
writefln("\nThere are %d ludic numbers <= 1000.",
L.until!q{ a > 1000 }.walkLength);
writeln("\n2000'th .. 2005'th ludic primes:\n", L[1999 .. 2005]);
enum m = 250;
const triplets = L.filter!(x => x + 6 < m &&
L.canFind(x + 2) && L.canFind(x + 6))
// Ugly output:
//.map!(x => tuple(x, x + 2, x + 6)).array;
.map!(x => [x, x + 2, x + 6]).array;
writefln("\nThere are %d triplets less than %d:\n%s",
triplets.length, m, triplets);
}</lang> The result is the same.
Eiffel
<lang Eiffel> class LUDIC_NUMBERS
create make
feature
make (n: INTEGER) -- Initialized arrays for find_ludic_numbers. require n_positive: n > 0 local i: INTEGER do create initial.make_filled (0, 1, n - 1) create ludic_numbers.make_filled (1, 1, 1) from i := 2 until i > n loop initial.put (i, i - 1) i := i + 1 end find_ludic_numbers end
ludic_numbers: ARRAY [INTEGER]
feature {NONE}
initial: ARRAY [INTEGER]
find_ludic_numbers -- Ludic numbers in array ludic_numbers. local count: INTEGER new_array: ARRAY [INTEGER] last: INTEGER do create new_array.make_from_array (initial) last := initial.count from count := 1 until count > last loop if ludic_numbers [ludic_numbers.count] /= new_array [1] then ludic_numbers.force (new_array [1], count + 1) end new_array := delete_i_elements (new_array) count := count + 1 end end
delete_i_elements (ar: ARRAY [INTEGER]): ARRAY [INTEGER] --- Array with all multiples of 'ar[1]' deleted. require ar_not_empty: ar.count > 0 local s_array: ARRAY [INTEGER] i, k: INTEGER length: INTEGER do create s_array.make_empty length := ar.count from i := 0 k := 1 until i = length loop if (i) \\ (ar [1]) /= 0 then s_array.force (ar [i + 1], k) k := k + 1 end i := i + 1 end if s_array.count = 0 then Result := ar else Result := s_array end ensure not_empty: not Result.is_empty end
end </lang> Test: <lang Eiffel> class APPLICATION
create make
feature
make local k, count: INTEGER do create ludic.make (22000) io.put_string ("%NLudic numbers up to 25. %N") across ludic.ludic_numbers.subarray (1, 25) as ld loop io.put_string (ld.item.out + "%N") end io.put_string ("%NLudic numbers from 2000 ... 2005. %N") across ludic.ludic_numbers.subarray (2000, 2005) as ld loop io.put_string (ld.item.out + "%N") end io.put_string ("%NNumber of Ludic numbers smaller than 1000. %N") from k := 1 until ludic.ludic_numbers [k] >= 1000 loop k := k + 1 count := count + 1 end io.put_integer (count) end
ludic: LUDIC_NUMBERS
end </lang>
- Output:
Ludic numbers up to 25. 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107 Ludic numbers from 2000 ... 2005. 21475 21481 21487 21493 21503 21511 Number of Ludic numbers smaller than 1000. 142
Elixir
<lang elixir>defmodule Ludic do
def numbers(n \\ 100000) do
[h|t] = Enum.to_list(1..n)
numbers(t, [h])
end
defp numbers(list, nums) when length(list) < hd(list), do: Enum.reverse(nums, list)
defp numbers([h|_]=list, nums) do
Enum.drop_every(list, h) |> numbers([h | nums])
end
def task do
IO.puts "First 25 : #{inspect numbers(200) |> Enum.take(25)}"
IO.puts "Below 1000: #{length(numbers(1000))}"
tuple = numbers(25000) |> List.to_tuple
IO.puts "2000..2005th: #{ inspect for i <- 1999..2004, do: elem(tuple, i) }"
ludic = numbers(250)
triple = for x <- ludic, x+2 in ludic, x+6 in ludic, do: [x, x+2, x+6]
IO.puts "Triples below 250: #{inspect triple, char_lists: :as_lists}"
end
end
Ludic.task</lang>
- Output:
First 25 : [1, 2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 37, 41, 43, 47, 53, 61, 67, 71, 77, 83, 89, 91, 97, 107] Below 1000: 142 2000..2005th: [21475, 21481, 21487, 21493, 21503, 21511] Triples below 250: [[1, 3, 7], [5, 7, 11], [11, 13, 17], [23, 25, 29], [41, 43, 47], [173, 175, 179], [221, 223, 227], [233, 235, 239]]
Fortran
<lang fortran>program ludic_numbers
implicit none integer, parameter :: nmax = 25000 logical :: ludic(nmax) = .true. integer :: i, j, n
do i = 2, nmax / 2
if (ludic(i)) then
n = 0
do j = i+1, nmax
if(ludic(j)) n = n + 1
if(n == i) then
ludic(j) = .false.
n = 0
end if
end do
end if
end do
write(*, "(a)", advance = "no") "First 25 Ludic numbers: "
n = 0
do i = 1, nmax
if(ludic(i)) then
write(*, "(i0, 1x)", advance = "no") i
n = n + 1
end if
if(n == 25) exit
end do
write(*, "(/, a)", advance = "no") "Ludic numbers below 1000: "
write(*, "(i0)") count(ludic(:999))
write(*, "(a)", advance = "no") "Ludic numbers 2000 to 2005: "
n = 0
do i = 1, nmax
if(ludic(i)) then
n = n + 1
if(n >= 2000) then
write(*, "(i0, 1x)", advance = "no") i
if(n == 2005) exit
end if
end if
end do
write(*, "(/, a)", advance = "no") "Ludic Triplets below 250: "
do i = 1, 243
if(ludic(i) .and. ludic(i+2) .and. ludic(i+6)) then
write(*, "(a, 2(i0, 1x), i0, a, 1x)", advance = "no") "[", i, i+2, i+6, "]"
end if
end do
end program</lang> Output:
First 25 Ludic numbers: 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107 Ludic numbers below 1000: 142 Ludic numbers 2000 to 2005: 21475 21481 21487 21493 21503 21511 Ludic Triplets below 250: [1 3 7] [5 7 11] [11 13 17] [23 25 29] [41 43 47] [173 175 179] [221 223 227] [233 235 239]
FreeBASIC
<lang freebasic>' FB 1.05.0 Win64
' As it would be too expensive to actually remove elements from the array ' we instead set an element to 0 if it has been removed.
Sub ludic(n As Integer, lu() As Integer)
If n < 1 Then Return Redim lu(1 To n) lu(1) = 1 If n = 1 Then Return Dim As Integer count = 1, count2 Dim As Integer i, j, k = 1 Dim As Integer ub = n * 11 big enough to deal with up to 2005 ludic numbers Dim a(2 To ub) As Integer For i = 2 To ub : a(i) = i : Next
Do
k += 1
For i = k to ub
If a(i) > 0 Then
count += 1
lu(count) = a(i)
If n = count Then Return
a(i) = 0
k = i
Exit For
End If
Next
count2 = 0
j = k + 1
While j <= ub
If a(j) > 0 Then
count2 +=1
If count2 = k Then
a(j) = 0
count2 = 0
End If
End If
j += 1
Wend
Loop
End Sub
Dim i As Integer Dim lu() As Integer ludic(25, lu()) Print "The first 25 Ludic numbers are :" For i = 1 To 25
Print Using "###"; lu(i); Print " ";
Next Erase lu Print
Dim As Integer Count = 0 ludic(1000, lu()) For i = 1 To 1000
If lu(i) <= 1000 Then count += 1 Else Exit For End If
Next Erase lu Print Print "There are"; count; " Ludic numbers <= 1000" Print
Print "The 2000th to 2005th Ludics are :" ludic(2005, lu()) For i = 2000 To 2005
Print lu(i); " ";
Next Erase lu Print : Print
Print "The Ludic triplets below 250 are : " Dim As Integer j, k, ldc Dim b As Boolean ludic(250, lu()) For i = 1 To 248
ldc = lu(i)
If ldc >= 244 Then Exit For
b = False
For j = i + 1 To 249
If lu(j) = ldc + 2 Then
b = True
k = j
Exit For
ElseIf lu(j) > ldc + 2 Then
Exit For
End If
Next j
If b = False Then Continue For
For j = k + 1 To 250
If lu(j) = ldc + 6 Then
Print "("; Str(ldc); ","; ldc + 2; ","; ldc + 6; ")"
Exit For
ElseIf lu(j) > ldc + 6 Then
Exit For
End If
Next j
Next i Erase lu Print
Print "Press any key to quit" Sleep </lang>
- Output:
The first 25 Ludic numbers are : 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107 There are 142 Ludic numbers <= 1000 The 2000th to 2005th Ludics are : 21475 21481 21487 21493 21503 21511 The Ludic triplets below 250 are : (1, 3, 7) (5, 7, 11) (11, 13, 17) (23, 25, 29) (41, 43, 47) (173, 175, 179) (221, 223, 227) (233, 235, 239)
Go
<lang go>package main
import "fmt"
// Ludic returns a slice of Ludic numbers stopping after // either n entries or when max is exceeded. // Either argument may be <=0 to disable that limit. func Ludic(n int, max int) []uint32 { const maxInt32 = 1<<31 - 1 // i.e. math.MaxInt32 if max > 0 && n < 0 { n = maxInt32 } if n < 1 { return nil } if max < 0 { max = maxInt32 } sieve := make([]uint32, 10760) // XXX big enough for 2005 Ludics sieve[0] = 1 sieve[1] = 2 if n > 2 { // We start with even numbers already removed for i, j := 2, uint32(3); i < len(sieve); i, j = i+1, j+2 { sieve[i] = j } // We leave the Ludic numbers in place, // k is the index of the next Ludic for k := 2; k < n; k++ { l := int(sieve[k]) if l >= max { n = k break } i := l l-- // last is the last valid index last := k + i - 1 for j := k + i + 1; j < len(sieve); i, j = i+1, j+1 { last = k + i sieve[last] = sieve[j] if i%l == 0 { j++ } } // Truncate down to only the valid entries if last < len(sieve)-1 { sieve = sieve[:last+1] } } } if n > len(sieve) { panic("program error") // should never happen } return sieve[:n] }
func has(x []uint32, v uint32) bool { for i := 0; i < len(x) && x[i] <= v; i++ { if x[i] == v { return true } } return false }
func main() { // Ludic() is so quick we just call it repeatedly fmt.Println("First 25:", Ludic(25, -1)) fmt.Println("Numner of Ludics below 1000:", len(Ludic(-1, 1000))) fmt.Println("Ludic 2000 to 2005:", Ludic(2005, -1)[1999:])
fmt.Print("Tripples below 250:") x := Ludic(-1, 250) for i, v := range x[:len(x)-2] { if has(x[i+1:], v+2) && has(x[i+2:], v+6) { fmt.Printf(", (%d %d %d)", v, v+2, v+6) } } fmt.Println() }</lang> Run in Go Playground.
- Output:
First 25: [1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107] Numner of Ludics below 1000: 142 Ludic 2000 to 2005: [21475 21481 21487 21493 21503 21511] Tripples below 250:, (1 3 7), (5 7 11), (11 13 17), (23 25 29), (41 43 47), (173 175 179), (221 223 227), (233 235 239)
Haskell
<lang haskell>import Data.List (unfoldr, genericSplitAt)
ludic :: [Integer] ludic = 1 : unfoldr (\xs@(x:_) -> Just (x, dropEvery x xs)) [2..] where
dropEvery n = concat . map tail . unfoldr (Just . genericSplitAt n)
main :: IO () main = do
print $ take 25 $ ludic print $ length $ takeWhile (<= 1000) $ ludic print $ take 6 $ drop 1999 $ ludic -- haven't done triplets task yet</lang>
- Output:
[1,2,3,5,7,11,13,17,23,25,29,37,41,43,47,53,61,67,71,77,83,89,91,97,107] 142 [21475,21481,21487,21493,21503,21511]
The filter for dropping every n-th number can be delayed until it's needed, which speeds up the generator, more so when a longer sequence is taken. <lang haskell>ludic = 1:2 : f 3 [3..] [(4,2)] where f n (x:xs) yy@((i,y):ys) | n == i = f n (dropEvery y xs) ys | otherwise = x : f (1+n) xs (yy ++ [(n+x, x)])
dropEvery n s = a ++ dropEvery n (tail b) where (a,b) = splitAt (n-1) s
main = print $ ludic !! 10000</lang>
Icon and Unicon
This is inefficient, but was fun to code as a cascade of filters. Works in both languages. <lang unicon>global num, cascade, sieve, nfilter
procedure main(A)
lds := ludic(2005) # All we need for the four tasks.
every writes("First 25:" | (" "||!lds)\25 | "\n")
every (n := 0) +:= (!lds < 1000, 1)
write("There are ",n," Ludic numbers < 1000.")
every writes("2000th through 2005th: " | (lds[2000 to 20005]||" ") | "\n")
writes("Triplets:")
every (250 > (x := !lds)) & (250 > (x+2 = !lds)) & (250 > (x+6 = !lds)) do
writes(" [",x,",",x+2,",",x+6,"]")
write()
end
procedure ludic(limit)
candidates := create seq(2)
put(cascade := [], create {
repeat {
report(l := num, limit)
put(cascade, create (cnt:=0, repeat ((cnt+:=1)%l=0, @sieve) | @@nfilter))
cascade[-2] :=: cascade[-1] # keep this sink as the last filter
@sieve
}
})
sieve := create while num := @candidates do @@(nfilter := create !cascade)
report(1, limit)
return @sieve
end
procedure report(ludic, limit)
static count, lds
initial {count := 0; lds := []}
if (count +:= 1) > limit then lds@&main
put(lds, ludic)
end</lang>
Output:
->ludic First 25: 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107 There are 142 Ludic numbers < 1000. 2000th through 20005th: 21475 21481 21487 21493 21503 21511 Triplets: [1,3,7] [5,7,11] [11,13,17] [23,25,29] [41,43,47] [173,175,179] [221,223,227] [233,235,239] ->
J
Solution (naive / brute force):<lang j> ludic =: _1 |.!.1 [: {."1 [: (#~ 0 ~: {. | i.@#)^:a: 2 + i.</lang> Examples:<lang j> # ludic 110 NB. 110 is sufficient to generate 25 Ludic numbers 25
ludic 110 NB. First 25 Ludic numbers
1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107
#ludic 1000 NB. 142 Ludic numbers <= 1000
142
# ludic 22000 NB. 22000 is sufficient to generate > 2005 Ludic numbers
2042
(2000+i.6) { ludic 22000 NB. Ludic numbers 2000-2005
21481 21487 21493 21503 21511 21523
0 2 6 (] (*./ .e.~ # |:@]) +/) ludic 250 NB. Ludic triplets <= 250 1 3 7 5 7 11 11 13 17 23 25 29 41 43 47
173 175 179 221 223 227 233 235 239</lang>
Java
This example uses pre-calculated ranges for the first and third task items (noted in comments). <lang java5>import java.util.ArrayList; import java.util.List;
public class Ludic{ public static List<Integer> ludicUpTo(int n){ List<Integer> ludics = new ArrayList<Integer>(n); for(int i = 1; i <= n; i++){ //fill the initial list ludics.add(i); }
//start at index 1 because the first ludic number is 1 and we don't remove anything for it for(int cursor = 1; cursor < ludics.size(); cursor++){ int thisLudic = ludics.get(cursor); //the first item in the list is a ludic number int removeCursor = cursor + thisLudic; //start removing that many items later while(removeCursor < ludics.size()){ ludics.remove(removeCursor); //remove the next item removeCursor = removeCursor + thisLudic - 1; //move the removal cursor up as many spaces as we need to //then back one to make up for the item we just removed } } return ludics; }
public static List<List<Integer>> getTriplets(List<Integer> ludics){ List<List<Integer>> triplets = new ArrayList<List<Integer>>(); for(int i = 0; i < ludics.size() - 2; i++){ //only need to check up to the third to last item int thisLudic = ludics.get(i); if(ludics.contains(thisLudic + 2) && ludics.contains(thisLudic + 6)){ List<Integer> triplet = new ArrayList<Integer>(3); triplet.add(thisLudic); triplet.add(thisLudic + 2); triplet.add(thisLudic + 6); triplets.add(triplet); } } return triplets; }
public static void main(String[] srgs){ System.out.println("First 25 Ludics: " + ludicUpTo(110)); //110 will get us 25 numbers System.out.println("Ludics up to 1000: " + ludicUpTo(1000).size()); System.out.println("2000th - 2005th Ludics: " + ludicUpTo(22000).subList(1999, 2005)); //22000 will get us 2005 numbers System.out.println("Triplets up to 250: " + getTriplets(ludicUpTo(250))); } }</lang>
- Output:
First 25 Ludics: [1, 2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 37, 41, 43, 47, 53, 61, 67, 71, 77, 83, 89, 91, 97, 107] Ludics up to 1000: 142 2000th - 2005th Ludics: [21475, 21481, 21487, 21493, 21503, 21511] Triplets up to 250: [[1, 3, 7], [5, 7, 11], [11, 13, 17], [23, 25, 29], [41, 43, 47], [173, 175, 179], [221, 223, 227], [233, 235, 239]]
Julia
<lang Julia> function ludic_filter{T<:Integer}(n::T)
0 < n || throw(DomainError())
slud = trues(n)
for i in 2:(n-1)
slud[i] || continue
x = 0
for j in (i+1):n
slud[j] || continue
x += 1
x %= i
x == 0 || continue
slud[j] = false
end
end
return slud
end
ludlen = 10^5 slud = ludic_filter(ludlen) ludics = collect(1:ludlen)[slud]
n = 25 println("Generate and show here the first ", n, " ludic numbers.") print(" ") crwid = 76 wid = 0 for i in 1:(n-1)
s = @sprintf "%d, " ludics[i]
wid += length(s)
if crwid < wid
print("\n ")
wid = 0
end
print(s)
end println(ludics[n])
n = 10^3 println() println("How many ludic numbers are there less than or equal to ", n, "?") println(" ", sum(slud[1:n]))
lo = 2000 hi = lo+5 println() println("Show the ", lo, "..", hi, "'th ludic numbers.") for i in lo:hi
println(" Ludic(", i, ") = ", ludics[i])
end
n = 250 println() println("Show all triplets of ludic numbers < ", n) for i = 1:n-7
slud[i] || continue
j = i+2
slud[j] || continue
k = i+6
slud[k] || continue
println(" ", i, ", ", j, ", ", k)
end </lang>
- Output:
Generate and show here the first 25 ludic numbers.
1, 2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 37, 41, 43, 47, 53, 61, 67, 71, 77,
83, 89, 91, 97, 107
How many ludic numbers are there less than or equal to 1000?
142
Show the 2000..2005'th ludic numbers.
Ludic(2000) = 21475
Ludic(2001) = 21481
Ludic(2002) = 21487
Ludic(2003) = 21493
Ludic(2004) = 21503
Ludic(2005) = 21511
Show all triplets of ludic numbers < 250
1, 3, 7
5, 7, 11
11, 13, 17
23, 25, 29
41, 43, 47
173, 175, 179
221, 223, 227
233, 235, 239
Lua
<lang Lua>-- Return table of ludic numbers below limit function ludics (limit)
local ludList, numList, index = {1}, {}
for n = 2, limit do table.insert(numList, n) end
while #numList > 0 do
index = numList[1]
table.insert(ludList, index)
for key = #numList, 1, -1 do
if key % index == 1 then table.remove(numList, key) end
end
end
return ludList
end
-- Return true if n is found in t or false otherwise function foundIn (t, n)
for k, v in pairs(t) do
if v == n then return true end
end
return false
end
-- Display msg followed by all values in t function show (msg, t)
io.write(msg)
for _, v in pairs(t) do io.write(" " .. v) end
print("\n")
end
-- Main procedure local first25, under1k, inRange, tripList, triplets = {}, 0, {}, {}, {} for k, v in pairs(ludics(30000)) do
if k <= 25 then table.insert(first25, v) end if v <= 1000 then under1k = under1k + 1 end if k >= 2000 and k <= 2005 then table.insert(inRange, v) end if v < 250 then table.insert(tripList, v) end
end for _, x in pairs(tripList) do
if foundIn(tripList, x + 2) and foundIn(tripList, x + 6) then
table.insert(triplets, "\n{" .. x .. "," .. x+2 .. "," .. x+6 .. "}")
end
end show("First 25:", first25) print(under1k .. " are less than or equal to 1000\n") show("2000th to 2005th:", inRange) show("Triplets:", triplets)</lang>
- Output:
First 25: 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107
142 are less than or equal to 1000
2000th to 2005th: 21475 21481 21487 21493 21503 21511
Triplets:
{1,3,7}
{5,7,11}
{11,13,17}
{23,25,29}
{41,43,47}
{173,175,179}
{221,223,227}
{233,235,239}
Mathematica
<lang Mathematica>n=10^5; Ludic={1}; seq=Range[2,n]; ClearAll[DoStep] DoStep[seq:{f_,___}]:=Module[{out=seq},
AppendTo[Ludic,f]; out;;;;f=Sequence[]; out
] Nest[DoStep,seq,2500];</lang>
- Output:
Ludic[[;; 25]]
LengthWhile[Ludic, # < 1000 &]
Ludic[[2000 ;; 2005]]
Select[Subsets[Select[Ludic, # < 250 &], {3}], Differences[#] == {2, 4} &]
{1, 2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 37, 41, 43, 47, 53, 61, 67, 71, 77, 83, 89, 91, 97, 107}
142
{21475, 21481, 21487, 21493, 21503, 21511}
{{1, 3, 7}, {5, 7, 11}, {11, 13, 17}, {23, 25, 29}, {41, 43, 47}, {173, 175, 179}, {221, 223, 227}, {233, 235, 239}}
Oforth
<lang Oforth>: ludic(n) | ludics l p |
ListBuffer newSize(n) seqFrom(2, n) over addAll ->l ListBuffer newSize(n) dup add(1) dup ->ludics
while(l notEmpty) [
l removeFirst dup ludics add ->p
l size p / p * while(dup 1 > ) [ dup l removeAt drop p - ] drop
] ;
- ludics
| l i |
ludic(22000) ->l "First 25 : " print l left(25) println "Below 1000 : " print l filter(#[ 1000 < ]) size println "2000 to 2005 : " print l extract(2000, 2005) println
250 loop: i [
l include(i) ifFalse: [ continue ]
l include(i 2 +) ifFalse: [ continue ]
l include(i 6 +) ifFalse: [ continue ]
i print ", " print i 2 + print ", " print i 6 + println
] ;</lang>
- Output:
First 25 : [1, 2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 37, 41, 43, 47, 53, 61, 67, 71, 77, 83, 89, 91, 97, 107] Below 1000 : 142 2000 to 2005 : [21475, 21481, 21487, 21493, 21503, 21511] 1, 3, 7 5, 7, 11 11, 13, 17 23, 25, 29 41, 43, 47 173, 175, 179 221, 223, 227 233, 235, 239
PARI/GP
Version #1. Creating vector of ludic numbers' flags, where the index of each flag=1 is the ludic number.
<lang parigp> \\ Creating Vlf - Vector of ludic numbers' flags, \\ where the index of each flag=1 is the ludic number. \\ 2/28/16 aev ludic(maxn)={my(Vlf=vector(maxn,z,1),n2=maxn\2,k,j1); for(i=2,n2,
if(Vlf[i], k=0; j1=i+1;
for(j=j1,maxn, if(Vlf[j], k++); if(k==i, Vlf[j]=0; k=0))
);
);
return(Vlf); }
{ \\ Required tests: my(Vr,L=List(),k=0,maxn=25000); Vr=ludic(maxn); print("The first 25 Ludic numbers: "); for(i=1,maxn, if(Vr[i]==1, k++; print1(i," "); if(k==25, break))); print("");print(""); k=0; for(i=1,999, if(Vr[i]==1, k++)); print("Ludic numbers below 1000: ",k); print(""); k=0; print("Ludic numbers 2000 to 2005: "); for(i=1,maxn, if(Vr[i]==1, k++; if(k>=2000&&k<=2005, listput(L,i)); if(k>2005, break))); for(i=1,6, print1(L[i]," ")); print(""); print(""); print("Ludic Triplets below 250: "); for(i=1,250, if(Vr[i]&&Vr[i+2]&&Vr[i+6], print1("(",i," ",i+2," ",i+6,") "))); } </lang>
- Output:
The first 25 Ludic numbers: 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107 Ludic numbers below 1000: 142 Ludic numbers 2000 to 2005: 21475 21481 21487 21493 21503 21511 Ludic Triplets below 250: (1 3 7) (5 7 11) (11 13 17) (23 25 29) (41 43 47) (173 175 179) (221 223 227) (233 235 239)
Version #2. Creating vector of ludic numbers.
Upgraded script from A003309 to meet task requirements.
<lang parigp> \\ Creating Vl - Vector of ludic numbers. \\ 2/28/16 aev ludic2(maxn)={my(Vw=vector(maxn, x, x+1),Vl=Vec([1]),vwn=#Vw,i); while(vwn>0, i=Vw[1]; Vl=concat(Vl,[i]);
Vw=vector((vwn*(i-1))\i,x,Vw[(x*i+i-2)\(i-1)]); vwn=#Vw
); return(Vl); } { \\ Required tests: my(Vr,L=List(),k=0,maxn=22000,vrs,vi); Vr=ludic2(maxn); vrs=#Vr; print("The first 25 Ludic numbers: "); for(i=1,25, print1(Vr[i]," ")); print("");print(""); k=0; for(i=1,vrs, if(Vr[i]<1000, k++, break)); print("Ludic numbers below 1000: ",k); print(""); k=0; print("Ludic numbers 2000 to 2005: "); for(i=2000,2005, print1(Vr[i]," ")); print("");print(""); print("Ludic Triplets below 250: "); for(i=1,vrs, vi=Vr[i]; if(i==1,print1("(",vi," ",vi+2," ",vi+6,") "); next); if(vi+6<250,if(Vr[i+1]==vi+2&&Vr[i+2]==vi+6, print1("(",vi," ",vi+2," ",vi+6,") ")))); } </lang>
- Output:
The first 25 Ludic numbers: 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107 Ludic numbers below 1000: 142 Ludic numbers 2000 to 2005: 21475 21481 21487 21493 21503 21511 Ludic Triplets below 250: (1 3 7) (5 7 11) (11 13 17) (23 25 29) (41 43 47) (173 175 179) (221 223 227) (233 235 239)
Pascal
Inspired by "rotors" of perl 6 . Runtime nearly quadratic: maxLudicCnt = 10000 -> 0.03 s =>maxLudicCnt= 100000 -> 3 s <lang pascal>program lucid; {$IFDEF FPC}
{$MODE objFPC} // useful for x64
{$ENDIF}
const
//66164 -> last < 1000*1000; maxLudicCnt = 2005;//must be > 1
type
tDelta = record
dNum,
dCnt : LongInt;
end;
tpDelta = ^tDelta; tLudicList = array of tDelta;
tArrdelta =array[0..0] of tDelta; tpLl = ^tArrdelta;
function isLudic(plL:tpLl;maxIdx:nativeInt):boolean; var
i, cn : NativeInt;
Begin
//check if n is 'hit' by a prior ludic number
For i := 1 to maxIdx do
with plL^[i] do
Begin
//Mask read modify write reread
//dec(dCnt);IF dCnt= 0
cn := dCnt;
IF cn = 1 then
Begin
dcnt := dNum;
isLudic := false;
EXIT;
end;
dcnt := cn-1;
end;
isLudic := true;
end;
procedure CreateLudicList(var Ll:tLudicList); var
plL : tpLl; n,LudicCnt : NativeUint;
begin
// special case 1 n := 1; Ll[0].dNum := 1;
plL := @Ll[0];
LudicCnt := 0;
repeat
inc(n);
If isLudic(plL,LudicCnt ) then
Begin
inc(LudicCnt);
with plL^[LudicCnt] do
Begin
dNum := n;
dCnt := n;
end;
IF (LudicCnt >= High(LL)) then
BREAK;
end;
until false;
end;
procedure firstN(var Ll:tLudicList;cnt: NativeUint); var
i : NativeInt;
Begin
writeln('First ',cnt,' ludic numbers:');
For i := 0 to cnt-2 do
write(Ll[i].dNum,',');
writeln(Ll[cnt-1].dNum);
end;
procedure triples(var Ll:tLudicList;max: NativeUint); var
i, chk : NativeUint;
Begin
// special case 1,3,7
writeln('Ludic triples below ',max);
write('(',ll[0].dNum,',',ll[2].dNum,',',ll[4].dNum,') ');
For i := 1 to High(Ll) do
Begin
chk := ll[i].dNum;
If chk> max then
break;
If (ll[i+2].dNum = chk+6) AND (ll[i+1].dNum = chk+2) then
write('(',ll[i].dNum,',',ll[i+1].dNum,',',ll[i+2].dNum,') ');
end;
writeln;
writeln;
end;
procedure LastLucid(var Ll:tLudicList;start,cnt: NativeUint); var
limit,i : NativeUint;
Begin
dec(start); limit := high(Ll); IF cnt >= limit then cnt := limit; if start+cnt >limit then start := limit-cnt; writeln(Start+1,'.th to ',Start+cnt+1,'.th ludic number'); For i := 0 to cnt-1 do write(Ll[i+start].dNum,','); writeln(Ll[start+cnt].dNum); writeln;
end;
function CountLudic(var Ll:tLudicList;Limit: NativeUint):NativeUint; var
i,res : NativeUint;
Begin
res := 0;
For i := 0 to High(Ll) do begin
IF Ll[i].dnum <= Limit then
inc(res)
else
BREAK;
CountLudic:= res;
end;
end; var
LudicList : tLudicList;
BEGIN
setlength(LudicList,maxLudicCnt);
CreateLudicList(LudicList);
firstN(LudicList,25);
writeln('There are ',CountLudic(LudicList,1000),' ludic numbers below 1000');
LastLucid(LudicList,2000,5);
LastLucid(LudicList,maxLudicCnt,5);
triples(LudicList,250);//all-> (LudicList,LudicList[High(LudicList)].dNum);
END. { 1,2,3,5,7,11,13,17,23,25,29,37,41,43,47,53,61,67,71,77,83,89,91,97,107 There are 142 ludic numbers below 1000 2000.th to 2005.th ludic number 21475,21481,21487,21493,21503,21511
99995.th to 100000.th ludic number 1561243,1561291,1561301,1561307,1561313,1561333
Ludic triples below 250 (1,3,7) (5,7,11) (11,13,17) (23,25,29) (41,43,47) (173,175,179) (221,223,227) (233,235,239)
real 0m2.921s}</lang>
- Output:
First 25 ludic numbers: 1,2,3,5,7,11,13,17,23,25,29,37,41,43,47,53,61,67,71,77,83,89,91,97,107 There are 142 ludic numbers below 1000 2000.th to 2005.th ludic number 21481,21487,21493,21503,21511 Ludic triples below 250 (1,3,7) (5,7,11) (11,13,17) (23,25,29) (41,43,47) (173,175,179) (221,223,227) (233,235,239)
Perl
The "ludic" subroutine caches the longest generated sequence so far. It also generates the candidates only if no candidates remain. <lang perl>#!/usr/bin/perl use warnings; use strict; use feature qw{ say };
{ my @ludic = (1);
my $max = 3; my @candidates;
sub sieve {
my $l = shift;
for (my $i = 0; $i <= $#candidates; $i += $l) {
splice @candidates, $i, 1;
}
}
sub ludic {
my ($type, $n) = @_;
die "Arg0 Type must be 'count' or 'max'\n"
unless grep $_ eq $type, qw( count max );
die "Arg1 Number must be > 0\n" if 0 >= $n;
return (@ludic[ 0 .. $n - 1 ]) if 'count' eq $type and @ludic >= $n;
return (grep $_ <= $n, @ludic) if 'max' eq $type and $ludic[-1] >= $n;
while (1) {
if (@candidates) {
last if ('max' eq $type and $candidates[0] > $n)
or ($n == @ludic);
push @ludic, $candidates[0];
sieve($ludic[-1] - 1);
} else {
$max *= 2;
@candidates = 2 .. $max;
for my $l (@ludic) {
sieve($l - 1) unless 1 == $l;
}
}
}
return (@ludic)
}
}
my @triplet; my %ludic; undef @ludic{ ludic(max => 250) }; for my $i (keys %ludic) {
push @triplet, $i if exists $ludic{ $i + 2 } and exists $ludic { $i + 6 };
}
say 'First 25: ', join ' ', ludic(count => 25); say 'Count < 1000: ', scalar ludic(max => 1000); say '2000..2005th: ', join ' ', (ludic(count => 2005))[1999 .. 2004]; say 'triplets < 250: ', join ' ',
map { '(' . join(' ',$_, $_ + 2, $_ + 6) . ')' }
sort { $a <=> $b } @triplet;</lang>
- Output:
First 25: 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107 Count < 1000: 142 2000..2005th: 21475 21481 21487 21493 21503 21511 triplets < 250: (1 3 7) (5 7 11) (11 13 17) (23 25 29) (41 43 47) (173 175 179) (221 223 227) (233 235 239)
Perl 6
This implementation has no arbitrary upper limit, since it can keep adding new rotors on the fly. It just gets slower and slower instead... :-) <lang perl6>constant @ludic = gather {
my @taken = take 1;
my @rotor;
for 2..* -> $i {
loop (my $j = 0; $j < @rotor; $j++) {
--@rotor[$j] or last;
}
if $j < @rotor {
@rotor[$j] = @taken[$j+1];
}
else {
push @taken, take $i;
push @rotor, @taken[$j+1];
}
}
}
say @ludic[^25]; say "Number of Ludic numbers <= 1000: ", +(@ludic ...^ * > 1000); say "Ludic numbers 2000..2005: ", @ludic[1999..2004];
my \l250 = set @ludic ...^ * > 250; say "Ludic triples < 250: ", gather
for l250.keys.sort -> $a {
my $b = $a + 2;
my $c = $a + 6;
take "<$a $b $c>" if $b ∈ l250 and $c ∈ l250;
}</lang>
- Output:
(1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107) Number of Ludic numbers <= 1000: 142 Ludic numbers 2000..2005: (21475 21481 21487 21493 21503 21511) Ludic triples < 250: (<1 3 7> <5 7 11> <11 13 17> <23 25 29> <41 43 47> <173 175 179> <221 223 227> <233 235 239>)
PicoLisp
<lang PicoLisp>(de drop (Lst)
(let N (car Lst)
(make
(for (I . X) (cdr Lst)
(unless (=0 (% I N)) (link X)) ) ) ) )
(de comb (M Lst)
(cond
((=0 M) '(NIL))
((not Lst))
(T
(conc
(mapcar
'((Y) (cons (car Lst) Y))
(comb (dec M) (cdr Lst)) )
(comb M (cdr Lst)) ) ) ) )
(de ludic (N)
(let Ludic (range 1 100000)
(make
(link (pop 'Ludic))
(do (dec N)
(link (car Ludic))
(setq Ludic (drop Ludic)) ) ) ) )
(let L (ludic 2005)
(println (head 25 L))
(println (cnt '((X) (< X 1000)) L))
(println (tail 6 L))
(println
(filter
'((Lst)
(and
(= (+ 2 (car Lst)) (cadr Lst))
(= (+ 6 (car Lst)) (caddr Lst)) ) )
(comb
3
(filter '((X) (< X 250)) L) ) ) ) )
(bye)</lang>
- Output:
(1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107) 142 (21475 21481 21487 21493 21503 21511)
((1 3 7) (5 7 11) (11 13 17) (23 25 29) (41 43 47) (173 175 179) (221 223 227) (233 235 239))
PL/I
<lang PL/I>Ludic_numbers: procedure options (main); /* 18 April 2014 */
declare V(2:22000) fixed, L(2200) fixed; declare (step, i, j, k, n) fixed binary;
Ludic: procedure;
n = hbound(V,1); k = 1; L(1) = 1; do i = 2 to n; V(i) = i; end;
do forever;
k = k + 1; L(k), step = V(2);
do i = 2 to n by step;
V(i) = 0;
end;
call compress;
if L(k) >= 21511 then leave;
end;
put skip list ('The first 25 Ludic numbers are:');
put skip edit ( (L(i) do i = 1 to 25) ) (F(4));
k = 0;
do i = 1 by 1;
if L(i) < 1000 then k = k + 1; else leave;
end;
put skip list ('There are ' || trim(k) || ' Ludic numbers < 1000');
put skip list ('Six Ludic numbers from the 2000-th:');
put skip edit ( (L(i) do i = 2000 to 2005) ) (f(7));
/* Triples are values of the form x, x+2, x+6 */
put skip list ('Triples are:');
put skip;
i = 1;
put edit ('(', L(1), L(3), L(5), ') ' ) (A, 3 F(4), A);
do i = 1 by 1 while (L(i+2) <= 250);
if (L(i) = L(i+1) - 2) & (L(i) = L(i+2) - 6) then
put edit ('(', L(i), L(i+1), L(i+2), ') ' ) (A, 3 F(4), A);
end;
compress: procedure;
j = 2;
do i = 2 to n;
if V(i) ^= 0 then do; V(j) = V(i); j = j + 1; end;
end;
n = j-1;
end compress;
end Ludic;
call Ludic;
end Ludic_numbers;</lang> Output:
The first 25 Ludic numbers are: 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107 There are 142 Ludic numbers < 1000 Six Ludic numbers from the 2000-th: 21475 21481 21487 21493 21503 21511 Triples are: ( 1 3 7) ( 5 7 11) ( 11 13 17) ( 23 25 29) ( 41 43 47) ( 173 175 179) ( 221 223 227) ( 233 235 239)
PL/SQL
<lang plsql>SET SERVEROUTPUT ON DECLARE
c_limit CONSTANT PLS_INTEGER := 25000; TYPE t_nums IS TABLE OF PLS_INTEGER INDEX BY PLS_INTEGER; v_nums t_nums; v_ludic t_nums; v_count_ludic PLS_INTEGER; v_count_pos PLS_INTEGER; v_pos PLS_INTEGER; v_next_ludic PLS_INTEGER;
FUNCTION is_ludic(p_num PLS_INTEGER) RETURN BOOLEAN IS
BEGIN
FOR i IN 1..v_ludic.COUNT LOOP
EXIT WHEN v_ludic(i) > p_num;
IF v_ludic(i) = p_num THEN
RETURN TRUE;
END IF;
END LOOP;
RETURN FALSE;
END;
BEGIN
FOR i IN 1..c_limit LOOP v_nums(i) := i; END LOOP;
v_count_ludic := 1; v_next_ludic := 1; v_ludic(v_count_ludic) := v_next_ludic; v_nums.DELETE(1);
WHILE v_nums.COUNT > 0 LOOP
v_pos := v_nums.FIRST;
v_next_ludic := v_nums(v_pos);
v_count_ludic := v_count_ludic + 1;
v_ludic(v_count_ludic) := v_next_ludic;
v_count_pos := 0;
WHILE v_pos IS NOT NULL LOOP
IF MOD(v_count_pos, v_next_ludic) = 0 THEN
v_nums.DELETE(v_pos);
END IF;
v_pos := v_nums.NEXT(v_pos);
v_count_pos := v_count_pos + 1;
END LOOP;
END LOOP;
dbms_output.put_line('Generate and show here the first 25 ludic numbers.');
FOR i IN 1..25 LOOP
dbms_output.put(v_ludic(i) || ' ');
END LOOP;
dbms_output.put_line();
dbms_output.put_line('How many ludic numbers are there less than or equal to 1000?');
v_count_ludic := 0;
FOR i IN 1..v_ludic.COUNT LOOP
EXIT WHEN v_ludic(i) > 1000;
v_count_ludic := v_count_ludic + 1;
END LOOP;
dbms_output.put_line(v_count_ludic);
dbms_output.put_line('Show the 2000..2005th ludic numbers.');
FOR i IN 2000..2005 LOOP
dbms_output.put(v_ludic(i) || ' ');
END LOOP;
dbms_output.put_line();
dbms_output.put_line('A triplet is any three numbers x, x + 2, x + 6 where all three numbers are also ludic numbers.');
dbms_output.put_line('Show all triplets of ludic numbers < 250 (Stretch goal)');
FOR i IN 1..v_ludic.COUNT LOOP
EXIT WHEN (v_ludic(i)+6) >= 250;
IF is_ludic(v_ludic(i)+2) AND is_ludic(v_ludic(i)+6) THEN
dbms_output.put_line(v_ludic(i) || ', ' || (v_ludic(i)+2) || ', ' || (v_ludic(i)+6));
END IF;
END LOOP;
END; / </lang>
- Output:
Generate and show here the first 25 ludic numbers. 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107 How many ludic numbers are there less than or equal to 1000? 142 Show the 2000..2005'th ludic numbers. 21475 21481 21487 21493 21503 21511 A triplet is any three numbers x, x + 2, x + 6 where all three numbers are also ludic numbers. Show all triplets of ludic numbers < 250 (Stretch goal) 1, 3, 7 5, 7, 11 11, 13, 17 23, 25, 29 41, 43, 47 173, 175, 179 221, 223, 227 233, 235, 239
PowerShell
<lang PowerShell>
- Start with a pool large enough to meet the requirements
$Pool = [System.Collections.ArrayList]( 2..22000 )
- Start with 1, because it's grandfathered in
$Ludic = @( 1 )
- While the size of the pool is still larger than the next Ludic number...
While ( $Pool.Count -gt $Pool[0] )
{
# Add the next Ludic number to the list
$Ludic += $Pool[0]
# Remove from the pool all entries whose index is a multiple of the next Ludic number
[math]::Truncate( ( $Pool.Count - 1 )/ $Pool[0])..0 | ForEach { $Pool.RemoveAt( $_ * $Pool[0] ) }
}
- Add the rest of the numbers in the pool to the list of Ludic numbers
$Ludic += $Pool.ToArray() </lang> <lang PowerShell>
- Display the first 25 Ludic numbers
$Ludic[0..24] -join ", "
- Display the count of all Ludic numbers under 1000
$Ludic.Where{ $_ -le 1000 }.Count
- Display the 2000th through the 2005th Ludic number
$Ludic[1999..2004] -join ", "
- Display all Ludic triplets less than 250
$TripletStart = $Ludic.Where{ $_ -lt 244 -and ( $_ + 2 ) -in $Ludic -and ( $_ + 6 ) -in $Ludic } $TripletStart.ForEach{ $_, ( $_ + 2 ), ( $_ + 6 ) -join ", " } </lang>
- Output:
1, 2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 37, 41, 43, 47, 53, 61, 67, 71, 77, 83, 89, 91, 97, 107 142 21475, 21481, 21487, 21493, 21503, 21511 1, 3, 7 5, 7, 11 11, 13, 17 23, 25, 29 41, 43, 47 173, 175, 179 221, 223, 227 233, 235, 239
Python
Python: Fast
<lang python>def ludic(nmax=100000):
yield 1
lst = list(range(2, nmax + 1))
while lst:
yield lst[0]
del lst[::lst[0]]
ludics = [l for l in ludic()]
print('First 25 ludic primes:') print(ludics[:25]) print("\nThere are %i ludic numbers <= 1000"
% sum(1 for l in ludics if l <= 1000))
print("\n2000'th..2005'th ludic primes:") print(ludics[2000-1: 2005])
n = 250 triplets = [(x, x+2, x+6)
for x in ludics
if x+6 < n and x+2 in ludics and x+6 in ludics]
print('\nThere are %i triplets less than %i:\n %r'
% (len(triplets), n, triplets))</lang>
- Output:
First 25 ludic primes: [1, 2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 37, 41, 43, 47, 53, 61, 67, 71, 77, 83, 89, 91, 97, 107] There are 142 ludic numbers <= 1000 2000'th..2005'th ludic primes: [21475, 21481, 21487, 21493, 21503, 21511] There are 8 triplets less than 250: [(1, 3, 7), (5, 7, 11), (11, 13, 17), (23, 25, 29), (41, 43, 47), (173, 175, 179), (221, 223, 227), (233, 235, 239)]
Python: No set maximum
The following version of function ludic will return ludic numbers until reaching system limits. It is less efficient than the fast version as all lucid numbers so far are cached; on exhausting the current lst a new list of twice the size is created and the previous deletions applied before continuing. <lang python>def ludic(nmax=64):
yield 1
taken = []
while True:
lst, nmax = list(range(2, nmax + 1)), nmax * 2
for t in taken:
del lst[::t]
while lst:
t = lst[0]
taken.append(t)
yield t
del lst[::t]</lang>
Output is the same as for the fast version.
Racket
<lang racket>#lang racket (define lucid-sieve-size 25000) ; this should be enough to do me! (define lucid?
(let ((lucid-bytes-sieve
(delay
(define sieve-bytes (make-bytes lucid-sieve-size 1))
(bytes-set! sieve-bytes 0 0) ; not a lucid number
(define (sieve-pass L)
(let loop ((idx (add1 L)) (skip (sub1 L)))
(cond
[(= idx lucid-sieve-size)
(for/first ((rv (in-range (add1 L) lucid-sieve-size))
#:unless (zero? (bytes-ref sieve-bytes rv))) rv)]
[(zero? (bytes-ref sieve-bytes idx))
(loop (add1 idx) skip)]
[(= skip 0)
(bytes-set! sieve-bytes idx 0)
(loop (add1 idx) (sub1 L))]
[else (loop (add1 idx) (sub1 skip))])))
(let loop ((l 2))
(when l (loop (sieve-pass l))))
sieve-bytes)))
(λ (n) (= 1 (bytes-ref (force lucid-bytes-sieve) n)))))
(define (dnl . things) (for-each displayln things))
(dnl
"Generate and show here the first 25 ludic numbers." (for/list ((_ 25) (l (sequence-filter lucid? (in-naturals)))) l) "How many ludic numbers are there less than or equal to 1000?" (for/sum ((n 1001) #:when (lucid? n)) 1) "Show the 2000..2005'th ludic numbers." (for/list ((i 2006) (l (sequence-filter lucid? (in-naturals))) #:when (>= i 2000)) l) #<<EOS
A triplet is any three numbers x, x + 2, x + 6 where all three numbers are also ludic numbers. Show all triplets of ludic numbers < 250 (Stretch goal) EOS
(for/list ((x (in-range 250)) #:when (and (lucid? x) (lucid? (+ x 2)) (lucid? (+ x 6)))) (list x (+ x 2) (+ x 6))))</lang>
- Output:
Generate and show here the first 25 ludic numbers. (1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107) How many ludic numbers are there less than or equal to 1000? 142 Show the 2000..2005'th ludic numbers. (21481 21487 21493 21503 21511 21523) A triplet is any three numbers x, x + 2, x + 6 where all three numbers are also ludic numbers. Show all triplets of ludic numbers < 250 (Stretch goal) ((1 3 7) (5 7 11) (11 13 17) (23 25 29) (41 43 47) (173 175 179) (221 223 227) (233 235 239)) cpu time: 18753 real time: 18766 gc time: 80
REXX
<lang rexx>/*REXX program displays (a range of) ludic numbers, or a count of when a range is used.*/ parse arg N count bot top triples . /*obtain optional arguments from the CL*/ if N== | N=="," then N=25 /*Not specified? Then use the default.*/ if count== | count=="," then count=1000 /* " " " " " " */ if bot== | bot=="," then bot=2000 /* " " " " " " */ if top== | top=="," then top=2005 /* " " " " " " */ if triples== | triples=="," then triples=250-1 /* " " " " " " */ say 'The first ' N " ludic numbers: " ludic(n) /*display title for what's coming next.*/ say say "There are " words(ludic(-count)) ' ludic numbers from 1───►'count " (inclusive)." say say "The " bot ' to ' top " ludic numbers are: " ludic(bot,top) $=ludic(-triples) 0 0; #=0; @= say
do j=1 for words($); _=word($,j) /*it is known that ludic _ exists. */
if wordpos(_+2, $)==0 | wordpos(_+6, $)==0 then iterate /*Not triple? Skip it.*/
#=#+1; @=@ '◄'_ _+2 _+6"► " /*bump the triple counter, and ··· */
end /*j*/ /* [↑] append the found triple ──► @ */
if @== then say 'From 1──►'triples", no triples found."
else say 'From 1──►'triples", " # ' triples found:' @
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ ludic: procedure; parse arg m,h; am=abs(m); if h\== then am=h; $=1 2; yes=m>0 | h\== @= /*$≡ludic numbers superset; @≡sequence*/
do j=3 by 2 to am*max(1, 15*yes) /*construct an initial list of numbers.*/
@=@ j /* [↓] construct a ludic sequence. */
end /*j*/ /* [↑] high limit: approx or exact. */
@=@' ' /*append a blank to the number sequence*/
do while words(@)\==0; f=word(@,1) /* [↓] examine the first word. */
$=$ f /*append this first word to the list. */
do d=1 by f while d<=words(@) /*use 1st number, elide all occurrences*/
y=word(@,d) /*obtain the Yth word of @ string.*/
@=changestr(' 'y" ", @, ' . ') /*delete the number in the sequence. */
end /*d*/ /* [↑] done eliding the "1st" number. */
@=translate(@, , .) /*translate periods (dots) to blanks. */
end /*while*/ /* [↑] done eliding ludic numbers. */
if h== then return subword($, 1, am) /*return a range of ludic numbers. */
return subword($, am, h-m+1) /*return a section of a range. */</lang>
Some older REXXes don't have a changestr BIF, so one is included here ──► CHANGESTR.REX.
output using the default values for input:
The first 25 ludic numbers: 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107 There are 142 ludic numbers from 1───►1000 (inclusive). The 2000 to 2005 ludic numbers are: 21475 21481 21487 21493 21503 21511 From 1──►249, 8 triples found: ◄1 3 7► ◄5 7 11► ◄11 13 17► ◄23 25 29► ◄41 43 47► ◄173 175 179► ◄221 223 227► ◄233 235 239►
Ruby
<lang ruby>def ludic(nmax=100000)
Enumerator.new do |y|
y << 1
ary = *2..nmax
until ary.empty?
y << (n = ary.first)
(0...ary.size).step(n){|i| ary[i] = nil}
ary.compact!
end
end
end
puts "First 25 Ludic numbers:", ludic.first(25).to_s
puts "Ludics below 1000:", ludic(1000).count
puts "Ludic numbers 2000 to 2005:", ludic.first(2005).last(6).to_s
ludics = ludic(250).to_a puts "Ludic triples below 250:",
ludics.select{|x| ludics.include?(x+2) and ludics.include?(x+6)}.map{|x| [x, x+2, x+6]}.to_s</lang>
- Output:
First 25 Ludic numbers: [1, 2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 37, 41, 43, 47, 53, 61, 67, 71, 77, 83, 89, 91, 97, 107] Ludics below 1000: 142 Ludic numbers 2000 to 2005: [21475, 21481, 21487, 21493, 21503, 21511] Ludic triples below 250: [[1, 3, 7], [5, 7, 11], [11, 13, 17], [23, 25, 29], [41, 43, 47], [173, 175, 179], [221, 223, 227], [233, 235, 239]]
Seed7
<lang seed7>$ include "seed7_05.s7i";
const func set of integer: ludicNumbers (in integer: n) is func
result
var set of integer: ludicNumbers is {1};
local
var set of integer: sieve is EMPTY_SET;
var integer: ludicNumber is 0;
var integer: number is 0;
var integer: count is 0;
begin
sieve := {2 .. n};
while sieve <> EMPTY_SET do
ludicNumber := min(sieve);
incl(ludicNumbers, ludicNumber);
count := 0;
for number range sieve do
if count rem ludicNumber = 0 then
excl(sieve, number);
end if;
incr(count);
end for;
end while;
end func;
const integer: limit is 22000; const set of integer: ludicNumbers is ludicNumbers(limit);
const proc: main is func
local
var integer: number is 0;
var integer: count is 0;
begin
write("First 25:");
for number range ludicNumbers until count = 25 do
write(" " <& number);
incr(count);
end for;
writeln;
count := 0;
for number range ludicNumbers until number > 1000 do
incr(count);
end for;
writeln("Ludics below 1000: " <& count);
write("Ludic 2000 to 2005:");
count := 0;
for number range ludicNumbers until count >= 2005 do
incr(count);
if count >= 2000 then
write(" " <& number);
end if;
end for;
writeln;
write("Triples below 250:");
for number range ludicNumbers until number > 250 do
if number + 2 in ludicNumbers and number + 6 in ludicNumbers then
write(" (" <& number <& ", " <& number + 2 <& ", " <& number + 6 <& ")");
end if;
end for;
writeln;
end func;</lang>
- Output:
First 25: 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107 Ludics below 1000: 142 Ludic 2000 to 2005: 21475 21481 21487 21493 21503 21511 Triples below 250: (1, 3, 7) (5, 7, 11) (11, 13, 17) (23, 25, 29) (41, 43, 47) (173, 175, 179) (221, 223, 227) (233, 235, 239)
Sidef
<lang ruby>func ludics_upto(nmax=100000) {
Enumerator({ |collect|
collect(1)
var arr = @(2..nmax)
while (arr) {
collect(var n = arr[0])
arr.range.by(n).each {|i| arr[i] = nil}
arr.compact!
}
})
}
func ludics_first(n) {
ludics_upto(n * n.log2).first(n)
}
say("First 25 Ludic numbers: ", ludics_first(25).join(' ')) say("Ludics below 1000: ", ludics_upto(1000).len) say("Ludic numbers 2000 to 2005: ", ludics_first(2005).last(6).join(' '))
var a = ludics_upto(250).to_a say("Ludic triples below 250: ", a.grep{|x| a.contains_all([x+2, x+6]) } \
.map {|x| '(' + [x, x+2, x+6].join(' ') + ')' } \
.join(' '))</lang>
- Output:
First 25 Ludic numbers: 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107 Ludics below 1000: 142 Ludic numbers 2000 to 2005: 21475 21481 21487 21493 21503 21511 Ludic triples below 250: (1 3 7) (5 7 11) (11 13 17) (23 25 29) (41 43 47) (173 175 179) (221 223 227) (233 235 239)
Tcl
The limit on the number of values generated is the depth of stack; this can be set to arbitrarily deep to go as far as you want. Provided you are prepared to wait for the values to be generated. <lang tcl>package require Tcl 8.6
proc ludic n {
global ludicList ludicGenerator
for {} {[llength $ludicList] <= $n} {lappend ludicList $i} {
set i [$ludicGenerator] set ludicGenerator [coroutine L_$i apply {{gen k} { yield [info coroutine] while true { set val [$gen] if {[incr i] == $k} {set i 0} else {yield $val} } }} $ludicGenerator $i]
} return [lindex $ludicList $n]
}
- Bootstrap the generator sequence
set ludicList [list 1] set ludicGenerator [coroutine L_1 apply {{} {
set n 1
yield [info coroutine]
while true {yield [incr n]}
}}]
- Default of 1000 is not enough
interp recursionlimit {} 5000
for {set i 0;set l {}} {$i < 25} {incr i} {lappend l [ludic $i]} puts "first25: [join $l ,]"
for {set i 0} {[ludic $i] <= 1000} {incr i} {} puts "below=1000: $i"
for {set i 1999;set l {}} {$i < 2005} {incr i} {lappend l [ludic $i]} puts "2000-2005: [join $l ,]"
for {set i 0} {[ludic $i] < 256} {incr i} {set isl([ludic $i]) $i} for {set i 1;set l {}} {$i < 250} {incr i} {
if {[info exists isl($i)] && [info exists isl([expr {$i+2}])] && [info exists isl([expr {$i+6}])]} {
lappend l ($i,[expr {$i+2}],[expr {$i+6}])
}
} puts "triplets: [join $l ,]"</lang>
- Output:
first25: 1,2,3,5,7,11,13,17,23,25,29,37,41,43,47,53,61,67,71,77,83,89,91,97,107 below=1000: 142 2000-2005: 21475,21481,21487,21493,21503,21511 triplets: (1,3,7),(5,7,11),(11,13,17),(23,25,29),(41,43,47),(173,175,179),(221,223,227),(233,235,239)
VBScript
<lang vb> Set list = CreateObject("System.Collections.Arraylist") Set ludic = CreateObject("System.Collections.Arraylist")
'populate the list For i = 1 To 25000 list.Add i Next
'set 1 as the first ludic number ludic.Add list(0) list.RemoveAt(0)
'variable to count ludic numbers <= 1000 up_to_1k = 1
'determine the succeeding ludic numbers For j = 2 To 2005 If list.Count > 0 Then If list(0) <= 1000 Then up_to_1k = up_to_1k + 1 End If ludic.Add list(0) Else Exit For End If increment = list(0) - 1 n = 0 Do While n <= list.Count - 1 list.RemoveAt(n) n = n + increment Loop Next
'the first 25 ludics WScript.StdOut.WriteLine "First 25 Ludic Numbers:" For k = 0 To 24 If k < 24 Then WScript.StdOut.Write ludic(k) & ", " Else WScript.StdOut.Write ludic(k) End If Next WScript.StdOut.WriteBlankLines(2)
'the number of ludics up to 1000 WScript.StdOut.WriteLine "Ludics up to 1000: " WScript.StdOut.WriteLine up_to_1k WScript.StdOut.WriteBlankLines(1)
'2000th - 2005th ludics WScript.StdOut.WriteLine "The 2000th - 2005th Ludic Numbers:" For k = 1999 To 2004 If k < 2004 Then WScript.StdOut.Write ludic(k) & ", " Else WScript.StdOut.Write ludic(k) End If Next WScript.StdOut.WriteBlankLines(2)
'triplets up to 250: x, x+2, and x+6 WScript.StdOut.WriteLine "Ludic Triplets up to 250: " triplets = "" k = 0 Do While ludic(k) + 6 <= 250 x2 = ludic(k) + 2 x6 = ludic(k) + 6 If ludic.IndexOf(x2,1) > 0 And ludic.IndexOf(x6,1) > 0 Then triplets = triplets & ludic(k) & ", " & x2 & ", " & x6 & vbCrLf End If k = k + 1 Loop WScript.StdOut.WriteLine triplets </lang>
- Output:
First 25 Ludic Numbers: 1, 2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 37, 41, 43, 47, 53, 61, 67, 71, 77, 83, 89, 91, 97, 107 Ludics up to 1000: 142 The 2000th - 2005th Ludic Numbers: 21475, 21481, 21487, 21493, 21503, 21511 Ludic Triplets up to 250: 1, 3, 7 5, 7, 11 11, 13, 17 23, 25, 29 41, 43, 47 173, 175, 179 221, 223, 227 233, 235, 239
zkl
This solution builds an iterator with filters, one for each Ludic number, each extending the previous filter. A "master" iterator sits at the top and provides the interface. When the next Ludic number is requested, the next odd number sent down the list of filters and if it makes to the end, it is the next Ludic number. A new filter is then attached [to the iterator] with a starting index of 1 and which indexes to strike. <lang zkl>fcn dropNth(n,seq){
seq.tweak(fcn(n,skipper,idx){ if(0==idx.inc()%skipper) Void.Skip else n }
.fp1(n,Ref(1))) // skip every nth number of previous sequence } fcn ludic{ //-->Walker
Walker(fcn(rw){ w:=rw.value; n:=w.next(); rw.set(dropNth(n,w)); n }
.fp(Ref([3..*,2]))) // odd numbers starting at 3
.push(1,2); // first two Ludic numbers
}</lang> <lang zkl>ludic().walk(25).toString(*).println(); ludic().reduce(fcn(sum,n){ if(n<1000) return(sum+1); return(Void.Stop,sum); },0).println(); ludic().drop(1999).walk(6).println(); // Ludic's between 2000 & 2005
ls:=ludic().filter(fcn(n){ (n<250) and True or Void.Stop }); // Ludic's < 250 ls.filter('wrap(n){ ls.holds(n+2) and ls.holds(n+6) }).apply(fcn(n){ T(n,n+2,n+6) }).println();</lang>
- Output:
L(1,2,3,5,7,11,13,17,23,25,29,37,41,43,47,53,61,67,71,77,83,89,91,97,107) 142 L(21475,21481,21487,21493,21503,21511) L(L(1,3,7),L(5,7,11),L(11,13,17),L(23,25,29),L(41,43,47),L(173,175,179),L(221,223,227),L(233,235,239))