Smallest numbers: Difference between revisions
(Added Algol 68) |
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16 17 9 7 12 28 6 23 9 24 |
16 17 9 7 12 28 6 23 9 24 |
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23 |
23 |
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</pre> |
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=={{header|Julia}}== |
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<lang julia>function hasinktok(n, limit=1000) |
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nlen = ndigits(n) |
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for k in 1:limit |
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d = digits(Int128(k)^k) |
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for j in 1:length(d)-nlen+1 |
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evalpoly(10, d[j:j+nlen-1]) == n && return k |
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end |
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end |
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error("Could not find a valid k where k <= $limit and k^k contains $n") |
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end |
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foreach(p -> print(rpad(p[2], 4), p[1] % 17 == 0 ? "\n" : ""), enumerate(map(hasinktok, 0:50))) |
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</lang>{{out}} |
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<pre> |
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9 1 3 5 2 4 4 3 7 9 10 11 5 19 22 26 8 |
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17 16 19 9 8 13 7 17 4 17 3 11 18 13 5 23 17 |
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18 7 17 15 9 18 16 17 9 7 12 35 6 23 9 24 23 |
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</pre> |
</pre> |
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Revision as of 18:36, 11 April 2021
- Task
Smallest number k > 0 such that the decimal expansion of k^k contains n, where n < 51
ALGOL 68
Uses ALGOL 68G's LOMG LONG INT which provides large integers (the default precision is sufficient for the task). Also uses the ALGOL 68G string in string procedure. <lang algol68>BEGIN # find the smallest k such that the decimal representation of k^k contains n for 0 <= n <= 50 #
# start with powers up to 20^20, if this proves insufficient, the kk array will be extended #
FLEX[ 1 : 20 ]STRING kk;
FOR k TO UPB kk DO kk[ k ] := whole( LONG LONG INT( k ) ^ k, 0 ) OD;
# find the numbers #
FOR i FROM 0 TO 50 DO
STRING n = whole( i, 0 );
BOOL try again := TRUE;
WHILE try again DO
try again := FALSE;
BOOL found := FALSE;
FOR k FROM LWB kk TO UPB kk WHILE NOT found DO
IF string in string( n, NIL, kk[ k ] ) THEN
found := TRUE;
print( ( " ", whole( k, -3 ) ) )
FI
OD;
IF NOT found THEN
# haven't got enough k^k values - get some more #
kk := HEAP[ 1 : UPB kk * 2 ]STRING;
FOR k TO UPB kk DO kk[ k ] := whole( LONG LONG INT( k ) ^ k, 0 ) OD;
try again := TRUE
FI
OD;
IF i MOD 10 = 9 THEN print( ( newline ) ) FI
OD
END</lang>
- Output:
9 1 3 5 2 4 4 3 7 9 10 11 5 19 22 26 8 17 16 19 9 8 13 7 17 4 17 3 11 18 13 5 23 17 18 7 17 15 9 18 16 17 9 7 12 28 6 23 9 24 23
Factor
<lang factor>USING: formatting grouping io kernel lists lists.lazy math.functions present sequences ;
- smallest ( m -- n )
present 1 lfrom [ dup ^ present subseq? ] with lfilter car ;
51 <iota> [ smallest ] map 10 group [ [ "%3d" printf ] each nl ] each</lang>
- Output:
9 1 3 5 2 4 4 3 7 9 10 11 5 19 22 26 8 17 16 19 9 8 13 7 17 4 17 3 11 18 13 5 23 17 18 7 17 15 9 18 16 17 9 7 12 28 6 23 9 24 23
Julia
<lang julia>function hasinktok(n, limit=1000)
nlen = ndigits(n)
for k in 1:limit
d = digits(Int128(k)^k)
for j in 1:length(d)-nlen+1
evalpoly(10, d[j:j+nlen-1]) == n && return k
end
end
error("Could not find a valid k where k <= $limit and k^k contains $n")
end
foreach(p -> print(rpad(p[2], 4), p[1] % 17 == 0 ? "\n" : ""), enumerate(map(hasinktok, 0:50)))
</lang>
- Output:
9 1 3 5 2 4 4 3 7 9 10 11 5 19 22 26 8 17 16 19 9 8 13 7 17 4 17 3 11 18 13 5 23 17 18 7 17 15 9 18 16 17 9 7 12 35 6 23 9 24 23
Phix
Native numbers won't cope (14^14 exceeds a 64-bit float, 17^17 an 80-bit one), so instead of gmp I've gone with string math again. (Related recent tasks: here and here)
constant lim = 51 -- (tested to 1,000,000) atom t0 = time(), t1 = t0+1 sequence res = repeat(0,lim) integer found = 0, k = 1 while found<lim do string kk = "1" for i=1 to k do integer carry = 0 for j=length(kk) to 1 by -1 do integer digit = (kk[j]-'0')*k+carry kk[j] = remainder(digit,10)+'0' carry = floor(digit/10) end for while carry do kk = remainder(carry,10)+'0' & kk carry = floor(carry/10) end while end for for i=1 to length(kk) do integer digit = 0, j = i while j<=length(kk) and digit<=lim do digit = digit*10+kk[j]-'0' if digit<lim and res[digit+1]=0 then res[digit+1] = sprintf("%2d",k) found += 1 end if j += 1 end while end for if platform()!=JS and time()>t1 then progress("found %,d/%,d, at %d^%d which has %,d digits (%s)", {found,lim,k,k,length(kk),elapsed(time()-t0)}) t1 = time()+1 end if k += 1 end while puts(1,join_by(shorten(res,"",30),1,10))
- Output:
9 1 3 5 2 4 4 3 7 9 10 11 5 19 22 26 8 17 16 19 9 8 13 7 17 4 17 3 11 18 13 5 23 17 18 7 17 15 9 18 16 17 9 7 12 28 6 23 9 24 23
Testing to 1,000,000 took 12mins 35s.
gmp version
constant lim = 51 -- (tested to 1,000,000) include mpfr.e mpz zkk = mpz_init() atom t0 = time(), t1 = t0+1 sequence res = repeat(0,lim) integer found = 0, k = 1 while found<lim do mpz_ui_pow_ui(zkk,k,k) string kk = mpz_get_str(zkk) for i=1 to length(kk) do integer digit = 0, j = i while j<=length(kk) and digit<=lim do digit = digit*10+kk[j]-'0' if digit<lim and res[digit+1]=0 then res[digit+1] = sprintf("%2d",k) found += 1 end if j += 1 end while end for if platform()!=JS and time()>t1 then progress("found %,d/%,d, at %d^%d which has %,d digits (%s)", {found,lim,k,k,length(kk),elapsed(time()-t0)}) t1 = time()+1 end if k += 1 end while puts(1,join_by(shorten(res,"",30),1,10))
Same results, but nearly 30 times faster, finishing the 1,000,000 test in just 26.6s
Raku
<lang perl6>sub smallest ( $n ) {
state @powers = , |map { $_ ** $_ }, 1 .. *;
return @powers.first: :k, *.contains($n);
}
.say for (^51).map(&smallest).batch(10)».fmt('%2d');</lang>
- Output:
( 9 1 3 5 2 4 4 3 7 9) (10 11 5 19 22 26 8 17 16 19) ( 9 8 13 7 17 4 17 3 11 18) (13 5 23 17 18 7 17 15 9 18) (16 17 9 7 12 28 6 23 9 24) (23)
REXX
<lang rexx>/*REXX pgm finds the smallest positive integer K where K**K contains N, N < 51 */ numeric digits 200 /*ensure enough decimal digs for k**k */ parse arg hi cols . /*obtain optional argument from the CL.*/ if hi== | hi=="," then hi= 51 /*Not specified? Then use the default.*/ if cols== | cols=="," then cols= 10 /* " " " " " " */ w= 6 /*width of a number in any column. */ @spiKK=' smallest positive integer K where K**K contains N, 0 ≤ N < ' commas(hi) say ' N │'center(@spiKK, 5 + cols*(w+1) ) /*display the title of the output. */ say '─────┼'center("" , 5 + cols*(w+1), '─') /* " " separator " " " */ $=; idx= 0 /*define $ output list; index to 0.*/
do j=0 for hi; n= j + 1 /*look for a power of 6 that contains N*/
do k=1 until pos(j, k**k)>0 /*calculate a bunch of powers (K**K). */
end /*k*/
c= commas(k) /*maybe add commas to the powe of six. */
$= $ right(c, max(w, length(c) ) ) /*add a K (power) ──► list, allow big#*/
if n//cols\==0 then iterate /*have we populated a line of output? */
say center(idx, 5)'│'substr($, 2); $= /*display what we have so far (cols). */
idx= idx + cols /*bump the index count for the output*/
end /*j*/
if $\== then say center(idx, 5)"│"substr($,2) /*possible display any residual output.*/ say '─────┴'center("" , 5 + cols*(w+1), '─') /* " " separator " " " */ exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ commas: parse arg ?; do jc=length(?)-3 to 1 by -3; ?=insert(',', ?, jc); end; return ?</lang>
- output when using the default inputs:
N │ smallest positive integer K where K**K contains N, 0 ≤ N < 51 ─────┼─────────────────────────────────────────────────────────────────────────── 0 │ 9 1 3 5 2 4 4 3 7 9 10 │ 10 11 5 19 22 26 8 17 16 19 20 │ 9 8 13 7 17 4 17 3 11 18 30 │ 13 5 23 17 18 7 17 15 9 18 40 │ 16 17 9 7 12 28 6 23 9 24 50 │ 23 ─────┴───────────────────────────────────────────────────────────────────────────
Ring
<lang ring> load "stdlib.ring"
decimals(0) see "working..." + nl see "Smallest number k > 0 such that the decimal expansion of k^k contains n are:" + nl
row = 0 limit1 = 49 limit2 = 30
for n = 0 to limit1
strn = string(n)
for m = 1 to limit2
powm = pow(m,m)
strm = string(powm)
ind = substr(strm,strn)
if ind > 0
exit
ok
next
row = row + 1
see "" + m + " "
if row%10 = 0
see nl
ok
next
see "done..." + nl </lang>
- Output:
working... Smallest number k > 0 such that the decimal expansion of k^k contains n are: 9 1 3 5 2 4 4 3 7 9 10 11 5 19 21 18 8 25 16 19 9 8 13 7 17 4 17 3 11 18 13 5 19 17 18 7 17 15 9 15 16 18 9 7 12 25 6 23 9 23 done...