Self numbers: Difference between revisions

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real 0m13,252s</pre>
real 0m13,252s</pre>

=={{header|Phix}}==
{{trans|Go}}
Replacing the problematic s[a+b+... line with a bit of dirty inline assembly improved performance by 90%<br>
We use a string of Y/N for the sieve to force one byte per element ('\0' and 1 would be equally valid).
<lang Phix>if machine_bits()=32 then crash("requires 64 bit") end if

function sieve()
string sv = repeat('N',2*1e9+9*9+1)
integer n = 0
for a=0 to 1 do
for b=0 to 9 do
for c=0 to 9 do
for d=0 to 9 do
for e=0 to 9 do
for f=0 to 9 do
for g=0 to 9 do
for h=0 to 9 do
for i=0 to 9 do
for j=0 to 9 do
-- n += 1
-- sv[a+b+c+d+e+f+g+h+i+j+n] = 'Y'
#ilASM{
[32] -- (allows clean compilation on 32 bit, before crash as above)
[64]
mov rax,[a]
mov r12,[b]
mov r13,[c]
mov r14,[d]
mov r15,[e]
add r12,r13
add r14,r15
add rax,r12
mov rdi,[sv]
add rax,r14
mov r12,[f]
mov r13,[g]
mov r14,[h]
mov r15,[i]
add r12,r13
shl rdi,2
mov rcx,[n]
mov r13,[j]
add r14,r15
add rax,r12
add rax,r14
add r13,rcx
add rax,r13
add rcx,1
mov byte[rdi+rax],'Y'
mov [n],rcx }
end for
end for
end for
end for
end for
end for
end for
end for
printf(1,"%d,%d\r",{a,b}) -- (show progress)
end for
end for
return sv
end function
atom t0 = time()
string sv = sieve()
printf(1,"sieve build took %s\n",{elapsed(time()-t0)})
integer count = 0
printf(1,"The first 50 self numbers are:\n")
for i=1 to length(sv) do
if sv[i]='N' then
count += 1
if count <= 50 then
printf(1,"%3d ",i-1)
if remainder(count,10)=0 then
printf(1,"\n")
end if
end if
if count == 1e8 then
printf(1,"\nThe %,dth self number is %,d (%s)\n",
{count,i-1,elapsed(time()-t0)})
exit
end if
end if
end for</lang>
{{out}}
<pre>
sieve build took 6.6s
The first 50 self numbers are:
1 3 5 7 9 20 31 42 53 64
75 86 97 108 110 121 132 143 154 165
176 187 198 209 211 222 233 244 255 266
277 288 299 310 312 323 334 345 356 367
378 389 400 411 413 424 435 446 457 468

The 100,000,000th self number is 1,022,727,208 (11.0s)
</pre>

=== generator dictionary ===
While this is dog-slow (see shocking estimate below), it is interesting to note that even by the time it generates the
10,000,000th number, it is only having to juggle a mere 27 generators.
Just a shame that we had to push over 1,000,000 generators onto that stack, and tried to push quite a few more.
Memory use is pretty stable, around a paltry 4MB.<br>
Aside: the getd_index() check is often worth trying with phix dictionaries: if there is a high probability that
the key already exists, it will yield a win, but with a low probability it will just be unhelpful overhead.
<lang Phix>integer gd = new_dict(), g = 1, pqhead = 2, n = 0

function ng(integer n)
integer r = n
while r do
n += remainder(r,10)
r = floor(r/10)
end while
return n
end function

function self(integer /*i*/)
-- note: assumes sequentially invoked (arg i unused)
integer nxt
n += 1
while n=pqhead do
while g<=pqhead do
nxt = ng(g)
if getd_index(nxt, gd)=NULL then -- (~25% gain)
setd(nxt,0,gd)
end if
g += 1
end while
integer waspqhead = pqhead
while true do
pqhead = pop_dict(gd)[1]
if pqhead!=waspqhead then exit end if
-- ?{"ding",waspqhead} -- 2, once only...
end while
nxt = ng(pqhead)
-- if getd_index(nxt, gd)=NULL then -- (~1% loss)
setd(nxt,0,gd)
-- end if
n += (n!=pqhead)
end while
return n
end function

atom t0 = time()
printf(1,"The first 50 self numbers are:\n")
pp(apply(tagset(50),self),{pp_IntFmt,"%3d",pp_IntCh,false})

constant limit = 10000-0-00
integer chk = 100
printf(1,"\n i n size time\n")
for i=51 to limit do
n = self(i)
if i=chk then
printf(1,"%,11d %,11d %6d %s\n",{i,n,dict_size(gd),elapsed(time()-t0)})
chk *= 10
end if
end for
printf(1,"\nEstimated time for %,d :%s\n",{1e8,elapsed((time()-t0)*1e8/limit)})</lang>
{out}
<pre>
The first 50 self numbers are:
{ 1, 3, 5, 7, 9, 20, 31, 42, 53, 64, 75, 86, 97,108,110,121,132,143,
154,165,176,187,198,209,211,222,233,244,255,266,277,288,299,310,312,323,
334,345,356,367,378,389,400,411,413,424,435,446,457,468}

i n size time
100 973 18 0.1s
1,000 10,188 13 0.2s
10,000 102,225 10 1.0s
100,000 1,022,675 20 9.3s
1,000,000 10,227,221 17 1 minute and 37s
10,000,000 102,272,662 27 16 minutes and 40s

Estimated time for 100,000,000 :2 hours, 46 minutes and 37s
</pre>
For the record, I would not be at all surprised should a translation of this beat 20 minutes (for 1e8)


=={{header|Wren}}==
=={{header|Wren}}==

Revision as of 14:29, 8 October 2020

Self numbers is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

A number n is a self number if there is no number g such that g + the sum of g's digits = n. So 18 is not a self number because 9+9=18, 43 is not a self number because 35+5+3=43.
The task is: 

 Display the first 50 self numbers;
 I believe that the 100000000th self number is 1022727208. You should either confirm or dispute my conjecture.

224036583-1 is a Mersenne prime, claimed to also be a self number. Extra credit to anyone proving it.
Wikipedia Self numbers showing some tricks especially for the number above.

F#

<lang fsharp> // Self numbers. Nigel Galloway: October 6th., 2020 let fN g=let rec fG n g=match n/10 with 0->n+g |i->fG i (g+(n%10)) in fG g g let Self=let rec Self n i g=seq{let g=g@([n..i]|>List.map fN) in yield! List.except g [n..i]; yield! Self (n+100) (i+100) (List.filter(fun n->n>i) g)} in Self 0 99 []

Self |> Seq.take 50 |> Seq.iter(printf "%d "); printfn "" printfn "\n%d" (Seq.item 99999999 Self) </lang>

Output:
1 3 5 7 9 20 31 42 53 64 75 86 97 108 110 121 132 143 154 165 176 187 198 209 211 222 233 244 255 266 277 288 299 310 312 323 334 345 356 367 378 389 400 411 413 424 435 446 457 468

1022727208

Go

Low memory

Simple-minded, uses very little memory (no sieve) but slow - over 2.5 minutes. <lang go>package main

import ( 

   "fmt"
   "time"

)

func sumDigits(n int) int { 

   sum := 0
   for n > 0 {
       sum += n % 10
       n /= 10
   }
   return sum

}

func max(x, y int) int { 

   if x > y {
       return x
   }
   return y

}

func main() { 

   st := time.Now()
   count := 0
   var selfs []int
   i := 1
   pow := 10
   digits := 1
   offset := 9
   lastSelf := 0
   for count < 1e8 {
       isSelf := true
       start := max(i-offset, 0)
       sum := sumDigits(start)
       for j := start; j < i; j++ {
           if j+sum == i {
               isSelf = false
               break
           }
           if (j+1)%10 != 0 {
               sum++
           } else {
               sum = sumDigits(j + 1)
           }
       }
       if isSelf {
           count++
           lastSelf = i
           if count <= 50 {
               selfs = append(selfs, i)
               if count == 50 {
                   fmt.Println("The first 50 self numbers are:")
                   fmt.Println(selfs)
               }
           }
       }
       i++
       if i%pow == 0 {
           pow *= 10
           digits++
           offset = digits * 9
       }
   }
   fmt.Println("\nThe 100 millionth self number is", lastSelf)
   fmt.Println("Took", time.Since(st))

}</lang>

Output:
The first 50 self numbers are:
[1 3 5 7 9 20 31 42 53 64 75 86 97 108 110 121 132 143 154 165 176 187 198 209 211 222 233 244 255 266 277 288 299 310 312 323 334 345 356 367 378 389 400 411 413 424 435 446 457 468]

The 100 millionth self number is 1022727208
Took 2m35.531949399s

Sieve based

Simple sieve, requires a lot of memory but quick - around 2 seconds.

Nested 'for's used rather than a recursive function for extra speed.

Have also incorporated Enter your username's suggestion (see Talk page) of using partial sums for each loop which improves performance by about 25%. <lang go>package main

import ( 

   "fmt"
   "time"

)

func sieve() []bool { 

   sv := make([]bool, 2*1e9+9*9 + 1)
   n := 0
   var s [8]int
   for a := 0; a < 2; a++ {
       for b := 0; b < 10; b++ {
           s[0] = a + b
           for c := 0; c < 10; c++ {
               s[1] = s[0] + c
               for d := 0; d < 10; d++ {
                   s[2] = s[1] + d
                   for e := 0; e < 10; e++ {
                       s[3] = s[2] + e
                       for f := 0; f < 10; f++ {
                           s[4] = s[3] + f
                           for g := 0; g < 10; g++ {
                               s[5] = s[4] + g
                               for h := 0; h < 10; h++ {
                                   s[6] = s[5] + h 
                                   for i := 0; i < 10; i++ {
                                       s[7] = s[6] + i
                                       for j := 0; j < 10; j++ {
                                           sv[s[7]+j+n] = true
                                           n++
                                       }
                                   }
                               }
                           }
                       }
                   }
               }
           }
       }
   }
   return sv

}

func main() { 

   st := time.Now()
   sv := sieve()
   count := 0
   fmt.Println("The first 50 self numbers are:")
   for i := 0; i < len(sv); i++ {
       if !sv[i] {
           count++
           if count <= 50 {
               fmt.Printf("%d ", i)
           }
           if count == 1e8 {
               fmt.Println("\n\nThe 100 millionth self number is", i)
               break
           }
       }
   }
   fmt.Println("Took", time.Since(st))

}</lang>

Output:
The first 50 self numbers are:
1 3 5 7 9 20 31 42 53 64 75 86 97 108 110 121 132 143 154 165 176 187 198 209 211 222 233 244 255 266 277 288 299 310 312 323 334 345 356 367 378 389 400 411 413 424 435 446 457 468 

The 100 millionth self number is 1022727208
Took 1.984969034s

Extended

Translation of: Pascal

This uses horst.h's ideas (see Talk page) to find up to the 1 billionth self number in a reasonable time and using less memory than the simple 'sieve based' approach above would have needed. <lang go>package main

import ( 

   "fmt"
   "time"

)

const MAX_COUNT = 103*1e4*1e4 + 11*9 + 1

var sv = make([]bool, MAX_COUNT+1) var digitSum = make([]int, 1e4)

func init() { 

   i := 9999
   var s, t int
   for a := 9; a >= 0; a-- {
       for b := 9; b >= 0; b-- {
           s = a + b
           for c := 9; c >= 0; c-- {
               t = s + c
               for d := 9; d >= 0; d-- {
                   digitSum[i] = t + d
                   i--
               }
           }
       }
   }

}

func sieve() { 

   n := 0
   for a := 0; a < 103; a++ {
       for b := 0; b < 1e4; b++ {
           s := digitSum[a] + digitSum[b] + n
           for c := 0; c < 1e4; c++ {
               sv[digitSum[c]+s] = true
               s++
           }
           n += 1e4
       }
   }

}

func main() { 

   st := time.Now()
   sieve()
   fmt.Println("Sieving took", time.Since(st))
   count := 0
   fmt.Println("\nThe first 50 self numbers are:")
   for i := 0; i < len(sv); i++ {
       if !sv[i] {
           count++
           if count <= 50 {
               fmt.Printf("%d ", i)
           } else {
               fmt.Println("\n\n     Index  Self number")
               break
           }
       }
   }
   count = 0
   limit := 1
   for i := 0; i < len(sv); i++ {
       if !sv[i] {
           count++
           if count == limit {
               fmt.Printf("%10d  %11d\n", count, i)
               limit *= 10
               if limit == 1e10 {
                   break
               }
           }
       }
   }
   fmt.Println("\nOverall took", time.Since(st))

}</lang>

Output:
Sieving took 8.286841692s

The first 50 self numbers are:
1 3 5 7 9 20 31 42 53 64 75 86 97 108 110 121 132 143 154 165 176 187 198 209 211 222 233 244 255 266 277 288 299 310 312 323 334 345 356 367 378 389 400 411 413 424 435 446 457 468 

     Index  Self number
         1            1
        10           64
       100          973
      1000        10188
     10000       102225
    100000      1022675
   1000000     10227221
  10000000    102272662
 100000000   1022727208
1000000000  10227272649

Overall took 14.647314803s

Julia

The code first bootstraps a sliding window of size 81 and then uses this as a sieve. Note that 81 is the window size because the sum of digits of 999,999,999 (the largest digit sum of a counting number less than 1022727208) is 81. <lang julia>gsum(i) = sum(digits(i)) + i isnonself(i) = any(x -> gsum(x) == i, i-1:-1:i-max(1, ndigits(i)*9)) const last81 = filter(isnonself, 1:5000)[1:81]

function checkselfnumbers() 

   i, selfcount = 1, 0
   while selfcount <= 100_000_000 && i <= 1022727208
       if !(i in last81)
           selfcount += 1
           if selfcount < 51
               print(i, " ")
           elseif selfcount == 51
               println()
           elseif selfcount == 100_000_000
               println(i == 1022727208 ?
                   "Yes, $i is the 100,000,000th self number." :
                   "No, instead $i is the 100,000,000th self number.")
           end
       end
       popfirst!(last81)
       push!(last81, gsum(i))
       i += 1
   end

end

checkselfnumbers()

</lang>

Output:
1 3 5 7 9 20 31 42 53 64 75 86 97 108 110 121 132 143 154 165 176 187 198 209 211 222 233 244 255 266 277 288 299 310 312 323 334 345 356 367 378 389 400 411 413 424 435 446 457 468
Yes, 1022727208 is the 100,000,000th self number.

Faster version

Translation of: Pascal

Contains tweaks peculiar to the "10 to the nth" self number. Timings include compilation times. <lang julia>const MAXCOUNT = 103 * 10000 * 10000 + 11 * 9 + 1

function dosieve!(sieve, digitsum9999) 

   n = 1
   for a in 1:103, b in 1:10000
       s = digitsum9999[a] + digitsum9999[b] + n
       for c in 1:10000
           sieve[digitsum9999[c] + s] = true
           s += 1
       end
       n += 10000
   end

end

initdigitsum() = reverse!(vec([sum(k) for k in Iterators.product(9:-1:0, 9:-1:0, 9:-1:0, 9:-1:0)]))

function findselves() 

   sieve = zeros(Bool, MAXCOUNT+1)
   println("Sieve time:")
   @time begin
       digitsum = initdigitsum()
       dosieve!(sieve, digitsum)
   end
   cnt = 1
   for i in 1:MAXCOUNT+1
       if !sieve[i]
           cnt > 50 && break
           print(i, " ")
           cnt += 1
       end
   end
   println()
   limit, cnt = 1, 0
   for i in 0:MAXCOUNT
       cnt += 1 - sieve[i + 1]
       if cnt == limit
           println(lpad(cnt, 10), lpad(i, 12))
           limit *= 10
       end
   end

end

@time findselves()

</lang>

Output:
Sieve time:
  7.187635 seconds (2 allocations: 78.203 KiB)
1 3 5 7 9 20 31 42 53 64 75 86 97 108 110 121 132 143 154 165 176 187 198 209 211 222 233 244 255 266 277 288 299 310 312 323 334 345 356 367 378 389 400 411 413 424 435 446 457 468
         1           1
        10          64
       100         973
      1000       10188
     10000      102225
    100000     1022675
   1000000    10227221
  10000000   102272662
 100000000  1022727208
1000000000 10227272649
 16.999383 seconds (42.92 k allocations: 9.595 GiB, 0.01% gc time)

Pascal

Works with: Free Pascal


Just "sieving" with only one follower of every number

Translation of: Go

Extended to 10.23e9 <lang pascal>program selfnumb; {$IFDEF FPC} 

 {$MODE Delphi}
 {$Optimization ON,ALL}

{$IFEND} {$IFDEF DELPHI} {$APPTYPE CONSOLE} {$IFEND} uses 

 sysutils;

const 

 MAXCOUNT =103*10000*10000+11*9+ 1;

type 

 tDigitSum9999 = array[0..9999] of Uint8;
 tpDigitSum9999 = ^tDigitSum9999;

var 

 DigitSum9999 : tDigitSum9999;
 sieve : array of boolean;

procedure dosieve; var 

 pSieve : pBoolean;
 pDigitSum :tpDigitSum9999;
 n,c,b,a,s : NativeInt;

Begin 

 pSieve := @sieve[0];
 pDigitSum := @DigitSum9999[0];
 n := 0;
 for a := 0 to 102 do
   for b := 0 to 9999 do
   Begin
     s := pDigitSum^[a]+pDigitSum^[b]+n;
     for c := 0 to 9999 do
     Begin
       pSieve[pDigitSum^[c]+s] := true;
       s+=1;
     end;
     inc(n,10000);
   end;

end;

procedure InitDigitSum; var 

 i,d,c,b,a : NativeInt;

begin 

 i := 9999;
 for a := 9 downto 0 do
   for b := 9 downto 0 do
     for c := 9 downto 0 do
       for d := 9 downto 0 do
       Begin
         DigitSum9999[i] := a+b+c+d;
         dec(i);
       end;

end;

procedure OutPut(cnt,i:NativeUint); Begin 

 writeln(cnt:10,i:12);

end;

var 

 pSieve : pboolean;
 T0 : Uint64;
 i,cnt,limit,One: NativeUInt;

BEGIN 

 setlength(sieve,MAXCOUNT);
 pSieve := @sieve[0];
 T0 := GetTickCount64;
 InitDigitSum;
 dosieve;
 writeln('Sievetime : ',(GetTickCount64-T0 )/1000:8:3,' sec');
 //find first 50
 cnt := 0;
 for i := 0 to MAXCOUNT do
 Begin
   if NOT(pSieve[i]) then
   Begin
     inc(cnt);
     if cnt <= 50 then
       write(i:4)
     else
       BREAK;
   end;
 end;
 writeln;
 One := 1;
 limit := One;
 cnt := 0;
 for i := 0 to MAXCOUNT do
 Begin
   inc(cnt,One-Ord(pSieve[i]));
   if cnt = limit then
   Begin
     OutPut(cnt,i);
     limit := limit*10;
   end;
 end;

END.</lang>

Output:
 time ./selfnumb
Sievetime :    6.579 sec
   1   3   5   7   9  20  31  42  53  64  75  86  97 108 110 121 132 143 154 165 176 187 198 209 211 222 233 244 255 266 277 288 299 310 312 323 334 345 356 367 378 389 400 411 413 424 435 446 457 468
         1           1
        10          64
       100         973
      1000       10188
     10000      102225
    100000     1022675
   1000000    10227221
  10000000   102272662
 100000000  1022727208
1000000000 10227272649

real  0m13,252s

Phix

Translation of: Go

Replacing the problematic s[a+b+... line with a bit of dirty inline assembly improved performance by 90%
We use a string of Y/N for the sieve to force one byte per element ('\0' and 1 would be equally valid). <lang Phix>if machine_bits()=32 then crash("requires 64 bit") end if

function sieve() 

   string sv = repeat('N',2*1e9+9*9+1)
   integer n = 0
   for a=0 to 1 do
       for b=0 to 9 do
           for c=0 to 9 do
               for d=0 to 9 do                  
                  for e=0 to 9 do
                       for f=0 to 9 do                         
                           for g=0 to 9 do
                               for h=0 to 9 do
                                   for i=0 to 9 do
                                       for j=0 to 9 do

-- n += 1 -- sv[a+b+c+d+e+f+g+h+i+j+n] = 'Y'

  1. ilASM{
 [32]  -- (allows clean compilation on 32 bit, before crash as above)
 [64]
   mov rax,[a]
   mov r12,[b]
   mov r13,[c]
   mov r14,[d]
   mov r15,[e]
   add r12,r13
   add r14,r15
   add rax,r12 
   mov rdi,[sv]
   add rax,r14 
   mov r12,[f]
   mov r13,[g]
   mov r14,[h]
   mov r15,[i]
   add r12,r13
   shl rdi,2
   mov rcx,[n]
   mov r13,[j]
   add r14,r15
   add rax,r12 
   add rax,r14 
   add r13,rcx
   add rax,r13 
   add rcx,1
   mov byte[rdi+rax],'Y'
   mov [n],rcx }
                                       end for
                                   end for                                  
                               end for
                           end for  
                       end for
                   end for
               end for
           end for
           printf(1,"%d,%d\r",{a,b}) -- (show progress)
       end for
   end for
   return sv

end function

atom t0 = time() string sv = sieve() printf(1,"sieve build took %s\n",{elapsed(time()-t0)}) integer count = 0 printf(1,"The first 50 self numbers are:\n") for i=1 to length(sv) do 

   if sv[i]='N' then
       count += 1
       if count <= 50 then
           printf(1,"%3d ",i-1)
           if remainder(count,10)=0 then
               printf(1,"\n")
           end if
       end if
       if count == 1e8 then
           printf(1,"\nThe %,dth self number is %,d (%s)\n",
                    {count,i-1,elapsed(time()-t0)})
           exit
       end if
   end if

end for</lang>

Output:
sieve build took 6.6s
The first 50 self numbers are:
  1   3   5   7   9  20  31  42  53  64
 75  86  97 108 110 121 132 143 154 165
176 187 198 209 211 222 233 244 255 266
277 288 299 310 312 323 334 345 356 367
378 389 400 411 413 424 435 446 457 468

The 100,000,000th self number is 1,022,727,208 (11.0s)

generator dictionary

While this is dog-slow (see shocking estimate below), it is interesting to note that even by the time it generates the 10,000,000th number, it is only having to juggle a mere 27 generators. Just a shame that we had to push over 1,000,000 generators onto that stack, and tried to push quite a few more. Memory use is pretty stable, around a paltry 4MB.
Aside: the getd_index() check is often worth trying with phix dictionaries: if there is a high probability that the key already exists, it will yield a win, but with a low probability it will just be unhelpful overhead. <lang Phix>integer gd = new_dict(), g = 1, pqhead = 2, n = 0

function ng(integer n) 

   integer r = n
   while r do
       n += remainder(r,10)
       r = floor(r/10)
   end while
   return n

end function

function self(integer /*i*/) -- note: assumes sequentially invoked (arg i unused) integer nxt 

   n += 1
   while n=pqhead do
       while g<=pqhead do
           nxt = ng(g)
           if getd_index(nxt, gd)=NULL then -- (~25% gain)
               setd(nxt,0,gd)
           end if
           g += 1
       end while
       integer waspqhead = pqhead
       while true do
           pqhead = pop_dict(gd)[1]
           if pqhead!=waspqhead then exit end if

--  ?{"ding",waspqhead} -- 2, once only... 

       end while
       nxt = ng(pqhead)

-- if getd_index(nxt, gd)=NULL then -- (~1% loss) 

           setd(nxt,0,gd)

-- end if 

       n += (n!=pqhead)
   end while
   return n

end function

atom t0 = time() printf(1,"The first 50 self numbers are:\n") pp(apply(tagset(50),self),{pp_IntFmt,"%3d",pp_IntCh,false})

constant limit = 10000-0-00 integer chk = 100 printf(1,"\n i n size time\n") for i=51 to limit do 

   n = self(i)
   if i=chk then
       printf(1,"%,11d %,11d %6d %s\n",{i,n,dict_size(gd),elapsed(time()-t0)})
       chk *= 10
   end if

end for printf(1,"\nEstimated time for %,d :%s\n",{1e8,elapsed((time()-t0)*1e8/limit)})</lang> {out} 

The first 50 self numbers are:
{  1,  3,  5,  7,  9, 20, 31, 42, 53, 64, 75, 86, 97,108,110,121,132,143,
 154,165,176,187,198,209,211,222,233,244,255,266,277,288,299,310,312,323,
 334,345,356,367,378,389,400,411,413,424,435,446,457,468}

          i           n   size time
        100         973     18 0.1s
      1,000      10,188     13 0.2s
     10,000     102,225     10 1.0s
    100,000   1,022,675     20 9.3s
  1,000,000  10,227,221     17 1 minute and 37s
 10,000,000 102,272,662     27 16 minutes and 40s

Estimated time for 100,000,000 :2 hours, 46 minutes and 37s

For the record, I would not be at all surprised should a translation of this beat 20 minutes (for 1e8)

Wren

Translation of: Go

Just the sieve based version as the low memory version would take too long to run in Wren.

Note that you need a lot of memory to run this as Bools in Wren require 8 bytes of storage compared to 1 byte in Go.

Unsurprisingly, very slow compared to the Go version as Wren is interpreted and uses floating point arithmetic for all numerical work. <lang ecmascript> var sieve = Fn.new { 

   var sv = List.filled(2*1e9+9*9+1, false)
   var n = 0
   var s = [0] * 8
   for (a in 0..1) {
       for (b in 0..9) {
           s[0] = a + b
           for (c in 0..9) {
               s[1] = s[0] + c
               for (d in 0..9) { 
                  s[2] = s[1] + d                   
                  for (e in 0..9) {
                       s[3] = s[2] + e
                       for (f in 0..9) {
                           s[4] = s[3] + f                           
                           for (g in 0..9) {
                               s[5] = s[4] + g
                               for (h in 0..9) {
                                   s[6] = s[5] + h
                                   for (i in 0..9) {
                                       s[7] = s[6] + i
                                       for (j in 0..9) {                                           
                                          sv[s[7] + j + n] = true
                                          n = n + 1
                                       }
                                   }                                    
                               }
                           }  
                       }
                   }
               }
           }
       }
   }
   return sv

}

var st = System.clock var sv = sieve.call() var count = 0 System.print("The first 50 self numbers are:") for (i in 0...sv.count) { 

   if (!sv[i]) {
       count = count + 1
       if (count <= 50) System.write("%(i) ")
       if (count == 1e8) {
           System.print("\n\nThe 100 millionth self number is %(i)")
           break
       }
   }

} System.print("Took %(System.clock-st) seconds.")</lang>

Output:
The first 50 self numbers are:
1 3 5 7 9 20 31 42 53 64 75 86 97 108 110 121 132 143 154 165 176 187 198 209 211 222 233 244 255 266 277 288 299 310 312 323 334 345 356 367 378 389 400 411 413 424 435 446 457 468 

The 100 millionth self number is 1022727208
Took 222.789713 seconds.
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