Smallest numbers: Difference between revisions
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=={{header|F_Sharp|F#}}== |
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<lang fsharp> |
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// Smallest number: Nigel Galloway. April 13th., 2021 |
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let rec fG n g=match bigint.DivRem(n,if g<10 then 10I else 100I) with (_,n) when (int n)=g->true |(n,_) when n=0I->false |(n,_)->fG n g |
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{0..50}|>Seq.iter(fun g->printf "%d " (1+({1..0x0FFFFFFF}|>Seq.map(fun n->(bigint n)**n)|>Seq.findIndex(fun n->fG n g)))); printfn "" |
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</lang> |
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{{out}} |
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<pre> |
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9 1 3 5 2 4 4 3 7 9 26 11 14 21 22 26 8 25 16 19 23 21 13 25 17 5 25 3 11 18 27 5 23 24 22 7 17 16 21 19 18 17 9 7 12 28 18 23 27 24 23 |
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Real: 00:00:00.005 |
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</pre> |
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=={{header|Factor}}== |
=={{header|Factor}}== |
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{{works with|Factor|0.99 2021-02-05}} |
{{works with|Factor|0.99 2021-02-05}} |
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Revision as of 16:40, 14 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
F#
<lang fsharp> // Smallest number: Nigel Galloway. April 13th., 2021 let rec fG n g=match bigint.DivRem(n,if g<10 then 10I else 100I) with (_,n) when (int n)=g->true |(n,_) when n=0I->false |(n,_)->fG n g {0..50}|>Seq.iter(fun g->printf "%d " (1+({1..0x0FFFFFFF}|>Seq.map(fun n->(bigint n)**n)|>Seq.findIndex(fun n->fG n g)))); printfn "" </lang>
- Output:
9 1 3 5 2 4 4 3 7 9 26 11 14 21 22 26 8 25 16 19 23 21 13 25 17 5 25 3 11 18 27 5 23 24 22 7 17 16 21 19 18 17 9 7 12 28 18 23 27 24 23 Real: 00:00:00.005
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
FreeBASIC
Reuses some code from Arbitrary-precision_integers_(included)#FreeBASIC. <lang freebasic>#Include once "gmp.bi" Dim Shared As Zstring * 100000000 outtext
Function Power(number As String,n As Uinteger) As String'automate precision
#define dp 3321921 Dim As __mpf_struct _number,FloatAnswer Dim As Ulongint ln=Len(number)*(n)*4 If ln>dp Then ln=dp mpf_init2(@FloatAnswer,ln) mpf_init2(@_number,ln) mpf_set_str(@_number,number,10) mpf_pow_ui(@Floatanswer,@_number,n) gmp_sprintf( @outtext,"%." & Str(n) & "Ff",@FloatAnswer ) Var outtxt=Trim(outtext) If Instr(outtxt,".") Then outtxt= Rtrim(outtxt,"0"):outtxt=Rtrim(outtxt,".") Return Trim(outtxt)
End Function
function is_substring( s as string, j as string ) as boolean
dim as integer nj = len(j), ns = len(s)
for i as integer = 1 to ns - nj + 1
if mid(s,i,nj) = j then return true
next i
return false
end function
dim as integer k
for i as uinteger = 0 to 50
k = 0
do
k = k + 1
loop until is_substring( Power(str(k), k), str(i) )
print k;" ";
next i</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>hasinktok(n) = for k in 1:100000 contains("$(BigInt(k)^k)", "$n") && return k 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 28 6 23 9 24 23
Pascal
made like Phix but own multiplikation to BASE 1E9 here <lang pascal>program K_pow_K; //First occurence of a numberstring with max DIGTIS digits in k^k {$IFDEF FPC}
{$MODE DELPHI}
{$Optimization ON,ALL}
{$ELSE}
{$APPTYPE CONSOLE}
{$ENDIF}
uses
sysutils;
const
LongWordDec = 1000*1000*1000; Digits = 6;
type
tMulElem = Uint32; tMul = array of tMulElem; tpMul = pUint32; tFound = Uint32;
var
Pot_N_str : AnsiString; Str_Found : array of tFound; FirstMissing :NativeInt; T0 : INt64;
procedure Out_Results(number,found:NativeInt); var
i : NativeInt;
Begin
writeln;
writeln(#10,'Found: ',found,' at ',number,' with ',length(Pot_N_str),
' digits in Time used ',(GetTickCount64-T0)/1000:8:3,' secs');
writeln ;
writeln(' 0 1 2 3 4 5 6 7 8 9');
write(' |__________________________________________________');
For i := 0 to 99 do//decLimit-1 do
begin
if i MOD 10 = 0 then
Begin
writeln;
write((i DIV 10)*10:10,'|');
end;
number := Str_Found[i]-1;
if number > 0 then
write(number:5);
end;
writeln;
end;
procedure Mul_12(var Mul1,Mul2:tMul); //Mul2 = Mul1*Mul2; var
TmpMul : tMul; carry, n,prod: Uint64; lmt1,lmt2,i,j : NativeInt;
begin
lmt1 := High(MUl1);
lmt2 := High(Mul2);
setlength(TmpMul,lmt1+lmt2+3);
For i := 0 to lmt1 do
Begin
carry := 0;
n := Mul1[i];
For j := 0 to lmt2 do
Begin
prod := n*Mul2[j]+TmpMul[i+j]+carry;
carry := prod DIV LongWordDec;
TmpMul[i+j]:=prod-carry*LongWordDec;
end;
TmpMul[i+lmt2+1] += carry;
end;
Mul2 := TmpMul;
setlength(TmpMul,0);
i := High(Mul2);
while (i>=1) AND (Mul2[i]=0) do
dec(i);
setlength(Mul2,i+1);
end;
procedure ConvToStr(var s:Ansistring;const Mul:tMul;i:NativeInt); var
s9: string[9]; pS : pChar; j,k : NativeInt;
begin // i := High(MUL);
j := (i+1)*9;
setlength(s,j+1);
pS := pChar(s);
// fill complete with '0'
fillchar(pS[0],j,'0');
str(Mul[i],S9);
j := length(s9);
move(s9[1],pS[0],j);
k := j;
dec(i);
If i >= 0 then
repeat
str(Mul[i],S9);// no leading '0'
j := length(s9);
inc(k,9);
//move to the right place, leading '0' is already there
move(s9[1],pS[k-j],j);
dec(i);
until i<0;
setlength(s,k);
end;
function CheckOneString(const s:Ansistring;pow:NativeInt):NativeInt; //check every possible number from one to DIGITS digits var
i,k,lmt,num : NativeInt;
begin
result := 0;
lmt := length(s);
For i := 1 to lmt do
Begin
k := i;
num := 0;
repeat
num := num*10+ Ord(s[k])-Ord('0');
IF (num >= FirstMissing) AND (str_Found[num] = 0) then
begin
str_Found[num]:= pow+1;
// commatize only once. reference counted string
inc(result);
if num =FirstMissing then
Begin
while str_Found[FirstMissing] <> 0 do
inc(FirstMissing);
end;
end;
inc(k)
until (k>lmt) or (k-i >DIGITS-1);
end;
end;
var
MulErg,Square :tMUl; number,i,j,found,decLimit: Int32;
Begin
T0 := GetTickCount64; decLimit := 1; For i := 1 to digits do decLimit *= 10; setlength(Str_Found,decLimit);
found := 0; FirstMissing := 0; number := 1; repeat setlength(MulErg,1); MulErg[0] := 1; setlength(Square,1); Square[0]:= number;
If number AND 1 <> 0 then
MulErg[0] := number;
j := 2;
while j <= number do
Begin
Mul_12(Square,Square);
If number AND J <> 0 then
Mul_12(Square,MulErg);
j:= j*2;
end;
ConvToStr(Pot_N_str,MulErg,High(MulErg));
inc(found,CheckOneString(Pot_N_str,number));
inc(number);
if number AND 511 = 0 then
write(#13,number:7,' with ',length(Pot_N_str), ' digits.Found ',found);
until found >=decLimit;
Out_Results(number,found);
end. </lang>
- Output:
TIO.RUN for 6 Digits
512 with 1385 digits.Found 334811
1024 with 3080 digits.Found 777542
1536 with 4891 digits.Found 968756
2048 with 6778 digits.Found 998285
2560 with 8722 digits.Found 999959
3072 with 10710 digits.Found 999999
Found: 1000000 at 3173 with 11107 digits in Time used 2.719 secs
0 1 2 3 4 5 6 7 8 9
|__________________________________________________
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 13 18 11 7 14 4 18 14 13
60| 19 11 25 17 17 6 6 8 14 27
70| 11 26 8 16 9 13 17 8 15 19
80| 14 21 7 21 16 11 17 9 17 9
90| 15 12 13 15 27 16 18 19 21 23
... at home for 7 Digits
only calc k^k for 1..9604
Found: 0 at 9604 with 0 digits in Time used 45.700 secs
with ConvToStr
Found: 0 at 9604 with 38244 digits in Time used 46.406 secs
with ConvToStr and CheckOneString
Found: 10000000 at 9604 with 38244 digits in Time used 52.222 secs
9216 with 36533 digits.Found 9999997
gmp-version
<lang pascal>program K_pow_K_gmp; //First occurence of a numberstring with max DIGTIS digits in k^k {$IFDEF FPC}
{$MODE DELPHI}
{$Optimization ON,ALL}
{$ELSE}
{$APPTYPE CONSOLE}
{$ENDIF}
uses
sysutils,gmp;
const
LongWordDec = 1000*1000*1000;
Digits = 7;
var
Pot_N_str : AnsiString; Str_Found : array of Uint32; FirstMissing :NativeInt; T0 : INt64;
procedure Out_Results(number,found:NativeInt); var
i : NativeInt;
Begin
writeln;
writeln(#10,'Found: ',found,' at ',number,' with ',length(pChar(Pot_N_str)),
' digits in Time used ',(GetTickCount64-T0)/1000:8:3,' secs');
writeln ;
writeln(' 0 1 2 3 4 5 6 7 8 9');
write(' |__________________________________________________');
For i := 0 to 99 do//decLimit-1 do
begin
if i MOD 10 = 0 then
Begin
writeln;
write((i DIV 10)*10:10,'|');
end;
number := Str_Found[i]-1;
if number > 0 then
write(number:5);
end;
writeln;
end;
function CheckOneString(const s:Ansistring;lmt,pow:NativeInt):NativeInt; //check every possible number from one to DIGITS digits var
i,k,num : NativeInt;
begin
result := 0;
For i := 1 to lmt do
Begin
k := i;
num := 0;
repeat
num := num*10+ Ord(s[k])-Ord('0');
IF (num >= FirstMissing) AND (str_Found[num] = 0) then
begin
str_Found[num]:= pow+1;
inc(result);
if num =FirstMissing then
Begin
while str_Found[FirstMissing] <> 0 do
inc(FirstMissing);
end;
end;
inc(k)
until (k>lmt) or (k-i >DIGITS-1);
end;
end;
var
zkk: mpz_t; number,i,found,lenS,decLimit: Int32;
Begin
T0 := GetTickCount64; mpz_init(zkk);
decLimit := 1; For i := 1 to digits do decLimit *= 10; setlength(Str_Found,decLimit);
//calc digits for max number := 10000 number:= 10000; i := trunc(number*ln(number)/ln(10))+5; setlength(Pot_N_str,i);
found := 0; FirstMissing := 0; number := 1; lenS :=1; repeat mpz_ui_pow_ui(zkk,number,number); mpz_get_str(pChar(Pot_N_str),10,zkk); while Pot_N_str[lenS] <> #0 do inc(lenS);
// lenS := length(pChar(Pot_N_str));
inc(found,CheckOneString(Pot_N_str,lenS,number));
inc(number);
if number AND 511 = 0 then
write(#13,number:7,' with ',lenS, ' digits.Found ',found);
until number>9604;// found >=decLimit;
Out_Results(number,found);
end.</lang>
- Output:
TIO.RUN for 7 digits same as above
512 with 1386 digits.Found 608645
1024 with 3081 digits.Found 1952296
...
Found: 10000000 at 9604 with 38244 digits in Time used 13.538 secs
//only mpz_ui_pow_ui(zkk,number,number); takes <0.5s up to 9604 with string conversion 3.3s
Perl
<lang perl>use strict; use warnings; use feature 'say'; use List::Util 'first'; use Math::AnyNum 'ipow';
sub smallest { first { ipow($_,$_) =~ /$_[0]/ } 1..1e4 } say join ' ', map { smallest($_) } 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 28 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 "bignumber.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 = 50 limit2 = 30
for n = 0 to limit1
strn = string(n)
for m = 1 to limit2
powm = pow(m,m)
ind = substr(powm,strn)
if ind > 0
exit
ok
next
row = row + 1
see "" + m + " "
if row%10 = 0
see nl
ok
next
see nl + "done..." + nl
func pow(num1,num2)
num1 = string(num1)
num2 = string(num2)
return FuncPower(num1,num2)
</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 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 done...
Wren
<lang ecmascript>import "/big" for BigInt import "/seq" for Lst import "/fmt" for Fmt
var res = [] for (n in 0..50) {
var k = 1
while (true) {
var s = BigInt.new(k).pow(k).toString
if (s.contains(n.toString)) {
res.add(k)
break
}
k = k + 1
}
} System.print("The smallest positive integers K where K ^ K contains N (0..50) are:") for (chunk in Lst.chunks(res, 17)) Fmt.print("$2d", chunk)</lang>
- Output:
The smallest positive integers K where K ^ K contains N (0..50) are: 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