Self numbers\Phix
My humiliatingly bad prior attempts at this task:
Replacing the problematic sv[a+b+... line with a bit of dirty inline assembly improved performance by 90%
(Of course you lose bounds checking, type checking, negative subscripts, fraction handling, and all that jazz.)
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) -- (1.86GB)
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>
- 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 10,000,000 generators onto that stack, and tried to push quite a few more.
Memory use is pretty low, around ~4MB.
[unlike the above, this is perfectly happy on both 32 and 64 bit]
Long story short: this works much the same as a prime sieve, in which you only need to eliminate multiples
of previous primes. Here, you only need to eliminate digital additions of previous safe numbers.
Only after writing this did I understand how to write a sliding sieve, which it turns out is a much better idea (see below).
Still, this is pretty interesting and quite neat.
<lang Phix>integer gd = new_dict(), gdhead = 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)
n += 1
while n=gdhead do
gdhead = pop_dict(gd)[1]
setd(ng(gdhead),0,gd)
n += (n!=gdhead)
end while
setd(ng(n),0,gd)
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 = 10000000 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>
- 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}
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)
sliding sieve
Mid-speed, perhaps helped by a slightly smarter way of calculating/updating the digit sums
Similar to some other entries, we (only) need a sieve of 9*9, +1 here as I test an entry after the slide.
<lang Phix>--sequence sieve = repeat(0,82), -- (~25% slower)
sequence sieve = repeat(0,8192),
digits = repeat(0,10)
integer offset = 0,
digit_sum = 0,
n = 0
procedure next_self()
while true do
n += 1
for i=length(digits) to 1 by -1 do
integer d = digits[i]
if d!=9 then
digits[i] = d+1
digit_sum += 1
exit
end if
digits[i] = 0
digit_sum -= 9
end for
integer k = n+digit_sum-offset
if k>length(sieve) then
integer j = 1
for i=n-offset to length(sieve) do
sieve[j] = sieve[i]
j += 1
end for
sieve[j..$] = 0
offset = n-1
k = digit_sum+1
end if
sieve[k] = 1
if sieve[n-offset]=0 then exit end if
end while
-- (result in n)
end procedure
constant limit = 100000000 atom t0 = time() printf(1,"The first 50 self numbers are:\n") integer chk = 100 for i=1 to limit do
next_self()
if i<=50 then
printf(1," %3d%s",{n,iff(mod(i,25)=0?"\n":"")})
elsif i=chk then
if chk=100 then
printf(1,"\n i n time\n")
end if
printf(1,"%,12d %,13d %s\n",{i,n,elapsed(time()-t0)})
chk *= 10
end if
end for printf(1,"\nEstimated time for %,d :%s\n",{1e9,elapsed((time()-t0)*1e9/limit)})</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
i n time
100 973 0.1s
1,000 10,188 0.1s
10,000 102,225 0.1s
100,000 1,022,675 0.1s
1,000,000 10,227,221 0.5s
10,000,000 102,272,662 4.5s
100,000,000 1,022,727,208 44.5s
Estimated time for 1,000,000,000 :7 minutes and 25s