Self-describing numbers
There are several integers numbers called "self-describing".
Integers with the property that, when digit positions are labeled 0 to N-1, the digit in each position is equal to the number of times that that digit appears in the number.
For example 2020 is a four digit self describing number.
Position "0" has value 2 and there is two 0 in the number. Position "1" has value 0 because there are not 1's in the number. Position "2" has value 2 and there is two 2. And the position "3" has value 0 and there are zero 3's.
Self-describing numbers < 100.000.000: 1210 - 2020 - 21200 - 3211000 - 42101000
- Task Description
- Write a function/routine/method/... that will check whether a given positive integer is self-describing.
- As an optional stretch goal - generate and display the set of self-describing numbers.
BASIC
<lang qbasic>Dim x, r, b, c, n, m As Integer Dim a, d As String Dim v(10), w(10) As Integer Cls For x = 1 To 5000000
a$ = ltrim$(Str$(x))
b = Len(a$)
For c = 1 To b
d$ = Mid$(a$, c, 1)
v(Val(d$)) = v(Val(d$)) + 1
w(c - 1) = Val(d$)
Next c
r = 0
For n = 0 To 10
If v(n) = w(n) Then r = r + 1
v(n) = 0
w(n) = 0
Next n
If r = 11 Then Print x; " Yes,is autodescriptive number"
Next x Print Print "End" sleep end</lang>
C
<lang c>#include <stdio.h>
- include <stdlib.h>
- include <stdbool.h>
- include <stdint.h>
- include <string.h>
bool is_self_describing(uint64_t n) {
uint8_t digits[10], i, k, d[10]; if (n == 0) return false;
memset(digits, 0, 10*sizeof(uint8_t)); memset(d, 0, 10*sizeof(uint8_t));
for(i = 0; n > 0 && i < 10; n /= 10, i++)
{
d[i] = n % 10; digits[d[i]]++;
}
if (n > 0) return false;
for(k = 0; k < i; k++)
{
if ( d[k] != digits[i - k - 1] ) return false;
}
return true;
}
int main() {
uint64_t sd[] = { 0, 1210, 1211, 2020, 2121, 21200, 3211000, 42101000 };
uint64_t i;
for(i = 0; i < sizeof(sd)/sizeof(uint64_t); i++)
printf("%llu is%s self-describing\n", sd[i], is_self_describing(sd[i]) ? "" : " NOT");
// let's find them brute-force (not a good idea...)
for(i = 521001001; i <= 9999999999LLU; i++)
{
if (is_self_describing(i)) printf("%llu\n", i);
}
return 0;
}</lang>
D
A functional version
Don't compile with -inline (DMD 2.053). <lang d>import std.stdio, std.algorithm, std.range, std.conv;
bool isSelfDescribing(long n) {
auto nu = map!q{a - '0'}(text(n));
auto f = map!((a){ return count(nu, a); })(iota(walkLength(nu)));
return equal(nu, f);
}
void main() {
writeln(filter!isSelfDescribing(iota(4_000_000)));
}</lang> Output:
[1210, 2020, 21200, 3211000]
(About 5.2 seconds run time.)
A faster version
<lang d>import std.stdio;
bool isSelfDescribing2(long n) {
if (n <= 0) return false;
__gshared static uint[10] digits, d;
digits[] = 0;
d[] = 0;
int i;
if (n < uint.max) {
uint nu = cast(uint)n;
for (i = 0; nu > 0 && i < digits.length; nu /= 10, i++) {
d[i] = cast(ubyte)(nu % 10);
digits[d[i]]++;
}
if (nu > 0)
return false;
} else {
for (i = 0; n > 0 && i < digits.length; n /= 10, i++) {
d[i] = cast(ubyte)(n % 10);
digits[d[i]]++;
}
if (n > 0)
return false;
}
foreach (k; 0 .. i)
if (d[k] != digits[i - k - 1])
return false;
return true;
}
void main() {
foreach (x; [1210, 2020, 21200, 3211000,
42101000, 521001000, 6210001000])
assert(isSelfDescribing2(x));
foreach (i; 0 .. 600_000_000)
if (isSelfDescribing2(i))
writeln(i);
}</lang> Output:
1210 2020 21200 3211000 42101000 521001000
(About 0.5 seconds run time for 4 million tests.)
Haskell
<lang Haskell>import Data.Char
count :: Int -> [Int] -> Int count x = length . filter (x ==)
isSelfDescribing :: Integer -> Bool isSelfDescribing n =
nu == f where
nu = map digitToInt (show n)
f = map (\a -> count a nu) [0 .. ((length nu)-1)]
main = do
let tests = [1210, 2020, 21200, 3211000,
42101000, 521001000, 6210001000]
print $ map isSelfDescribing tests
print $ filter isSelfDescribing [0 .. 4000000]</lang>
Output:
[True,True,True,True,True,True,True] [1210,2020,21200,3211000]
J
<lang j> NB. background material:
digits=: 10 #.inv ] digits 2020
2 0 2 0
(,~ i.@#@digits)2020
0 1 2 3 2020
(digits ,~ i.@#@digits)2020
0 1 2 3 2 0 2 0
(,~ i.@#)&digits 2020
0 1 2 3 2 0 2 0
_1 + #/.~@(,~ i.@#)&digits 2020
2 0 2 0
NB. task item 1:
(digits -: _1 + #/.~@(,~ i.@#)&digits) 2020
1
(digits -: _1 + #/.~@(,~ i.@#)&digits) 1210
1
(digits -: _1 + #/.~@(,~ i.@#)&digits) 21200
1
(digits -: _1 + #/.~@(,~ i.@#)&digits) 3211000
1
(digits -: _1 + #/.~@(,~ i.@#)&digits) 43101000
0
(digits -: _1 + #/.~@(,~ i.@#)&digits) 42101000
1
NB. task item 2:
I. (digits -: _1 + #/.~@(,~ i.@#)&digits)"0 i.1e6 NB. task item 2
1210 2020 21200</lang>
Logo
<lang logo>TO XX BT MAKE "AA (ARRAY 10 0) MAKE "BB (ARRAY 10 0) FOR [Z 0 9][SETITEM :Z :AA "0 SETITEM :Z :BB "0 ]
FOR [A 1 50000][
MAKE "B COUNT :A
MAKE "Y 0
MAKE "X 0
MAKE "R 0
MAKE "J 0
MAKE "K 0
FOR [C 1 :B][MAKE "D ITEM :C :A
SETITEM :C - 1 :AA :D
MAKE "X ITEM :D :BB
MAKE "Y :X + 1
SETITEM :D :BB :Y
MAKE "R 0]
FOR [Z 0 9][MAKE "J ITEM :Z :AA
MAKE "K ITEM :Z :BB
IF :J = :K [MAKE "R :R + 1]]
IF :R = 10 [PR :A] FOR [Z 0 9][SETITEM :Z :AA "0 SETITEM :Z :BB "0 ]] PR [END] END</lang>
PicoLisp
<lang PicoLisp>(de selfDescribing (N)
(not
(find '((D I) (<> D (cnt = N (circ I))))
(setq N (mapcar format (chop N)))
(range 0 (length N)) ) ) )</lang>
Output:
: (filter selfDescribing (range 1 4000000)) -> (1210 2020 21200 3211000)
Prolog
Works with SWI-Prolog and library clpfd written by Markus Triska. <lang Prolog>:- use_module(library(clpfd)).
self_describling :- forall(between(1, 10, I), (findall(N, self_describling(I,N), L), format('Len ~w, Numbers ~w~n', [I, L]))).
% search of the self_describling numbers of a given len self_describling(Len, N) :- length(L, Len), Len1 is Len - 1, L = [H|T],
% the first figure is greater than 0 H in 1..Len1,
% there is a least to figures so the number of these figures % is at most Len - 2 Len2 is Len - 2, T ins 0..Len2,
% the sum of the figures is equal to the len of the number sum(L, #=, Len),
% There is at least one figure corresponding to the number of zeros H1 #= H+1, element(H1, L, V), V #> 0,
% create the list label(L),
% test the list msort(L, LNS), packList(LNS,LNP), numlist(0, Len1, NumList), verif(LNP,NumList, L),
% list is OK, create the number maplist(atom_number, LA, L), number_chars(N, LA).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% testing a number (not use in this program)
self_describling(N) :-
number_chars(N, L),
maplist(atom_number, L, LN),
msort(LN, LNS),
packList(LNS,LNP), !,
length(L, Len),
Len1 is Len - 1,
numlist(0, Len1, NumList),
verif(LNP,NumList, LN).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% verif(PackList, Order_of_Numeral, Numeral_of_the_nuber_to_test)
% Packlist is of the form [[Number_of_Numeral, Order_of_Numeral]|_]
% Test succeed when
% All lists are empty verif([], [], []).
% Packlist is empty and all lasting numerals are 0 verif([], [_N|S], [0|T]) :- verif([], S, T).
% Number of numerals N is V verif([[V, N]|R], [N|S], [V|T]) :- verif(R, S, T).
% Number of numerals N is 0 verif([[V, N1]|R], [N|S], [0|T]) :- N #< N1, verif([[V,N1]|R], S, T).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % ?- packList([a,a,a,b,c,c,c,d,d,e], L). % L = [[3,a],[1,b],[3,c],[2,d],[1,e]] . % ?- packList(R, [[3,a],[1,b],[3,c],[2,d],[1,e]]). % R = [a,a,a,b,c,c,c,d,d,e] . % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% packList([],[]).
packList([X],1,X) :- !.
packList([X|Rest],[XRun|Packed]):-
run(X,Rest, XRun,RRest), packList(RRest,Packed).
run(Var,[],[1, Var],[]).
run(Var,[Var|LRest],[N1, Var],RRest):-
N #> 0, N1 #= N + 1, run(Var,LRest,[N, Var],RRest).
run(Var,[Other|RRest], [1, Var],[Other|RRest]):-
dif(Var,Other).</lang>
Output
?- self_describling. Len 1, Numbers [] Len 2, Numbers [] Len 3, Numbers [] Len 4, Numbers [1210,2020] Len 5, Numbers [21200] Len 6, Numbers [] Len 7, Numbers [3211000] Len 8, Numbers [42101000] Len 9, Numbers [521001000] Len 10, Numbers [6210001000] true.
PureBasic
<lang PureBasic>Procedure isSelfDescribing(x.q)
;returns 1 if number is self-describing, otherwise it returns 0
Protected digitCount, digit, i, digitSum
Dim digitTally(10)
Dim digitprediction(10)
If x <= 0
ProcedureReturn 0 ;number must be positive and non-zero
EndIf
While x > 0 And i < 10
digit = x % 10
digitSum + digit
If digitSum > 10
ProcedureReturn 0 ;sum of digits' values exceeds maximum possible
EndIf
digitprediction(i) = digit
digitTally(digit) + 1
x / 10
i + 1
Wend
digitCount = i - 1
If digitSum < digitCount Or x > 0
ProcedureReturn 0 ;sum of digits' values is too small or number has more than 10 digits
EndIf
For i = 0 To digitCount
If digitTally(i) <> digitprediction(digitCount - i)
ProcedureReturn 0 ;number is not self-describing
EndIf
Next
ProcedureReturn 1 ;number is self-describing
EndProcedure
Procedure displayAll()
Protected i, j, t
PrintN("Starting search for all self-describing numbers..." + #CRLF$)
For j = 0 To 9
PrintN(#CRLF$ + "Searching possibilites " + Str(j * 1000000000) + " -> " + Str((j + 1) * 1000000000 - 1)+ "...")
t = ElapsedMilliseconds()
For i = 0 To 999999999
If isSelfDescribing(j * 1000000000 + i)
PrintN(Str(j * 1000000000 + i))
EndIf
Next
PrintN("Time to search this range of possibilities: " + Str((ElapsedMilliseconds() - t) / 1000) + "s.")
Next
PrintN(#CRLF$ + "Search complete.")
EndProcedure
If OpenConsole()
DataSection
Data.q 1210, 2020, 21200, 3211000, 42101000, 521001000, 6210001000, 3214314
EndDataSection
Define i, x.q
For i = 1 To 8
Read.q x
Print(Str(x) + " is ")
If Not isSelfDescribing(x)
Print("not ")
EndIf
PrintN("selfdescribing.")
Next
PrintN(#CRLF$)
displayAll()
Print(#CRLF$ + #CRLF$ + "Press ENTER to exit"): Input()
CloseConsole()
EndIf</lang> Sample output:
1210 is selfdescribing. 2020 is selfdescribing. 21200 is selfdescribing. 3211000 is selfdescribing. 42101000 is selfdescribing. 521001000 is selfdescribing. 6210001000 is selfdescribing. 3214314 is not selfdescribing. Starting search for all self-describing numbers... Searching possibilites 0 -> 999999999... 1210 2020 21200 3211000 42101000 521001000 Time to search this range of possibilities: 615s. Searching possibilites 1000000000 -> 1999999999... Time to search this range of possibilities: 614s. Searching possibilites 2000000000 -> 2999999999... Time to search this range of possibilities: 628s. Searching possibilites 3000000000 -> 3999999999... Time to search this range of possibilities: 631s. Searching possibilites 4000000000 -> 4999999999... Time to search this range of possibilities: 630s. Searching possibilites 5000000000 -> 5999999999... Time to search this range of possibilities: 628s. Searching possibilites 6000000000 -> 6999999999... 6210001000 Time to search this range of possibilities: 629s. Searching possibilites 7000000000 -> 7999999999... Time to search this range of possibilities: 631s. Searching possibilites 8000000000 -> 8999999999... Time to search this range of possibilities: 629s. Searching possibilites 9000000000 -> 9999999999... Time to search this range of possibilities: 629s. Search complete.
Python
<lang python>>>> def isSelfDescribing(n): s = str(n) return all(s.count(str(i)) == int(ch) for i, ch in enumerate(s))
>>> [x for x in range(4000000) if isSelfDescribing(x)] [1210, 2020, 21200, 3211000] >>> [(x, isSelfDescribing(x)) for x in (1210, 2020, 21200, 3211000, 42101000, 521001000, 6210001000)] [(1210, True), (2020, True), (21200, True), (3211000, True), (42101000, True), (521001000, True), (6210001000, True)]</lang>
Tcl
<lang tcl>package require Tcl 8.5 proc isSelfDescribing num {
set digits [split $num ""]
set len [llength $digits]
set count [lrepeat $len 0]
foreach d $digits {
if {$d >= $len} {return false} lset count $d [expr {[lindex $count $d] + 1}]
}
foreach d $digits c $count {if {$c != $d} {return false}}
return true
}
for {set i 0} {$i < 100000000} {incr i} {
if {[isSelfDescribing $i]} {puts $i}
}</lang>