Casting out nines
You are encouraged to solve this task according to the task description, using any language you may know.
A task in three parts:
- Part 1
Write a procedure (say ) which implements Casting Out Nines as described by returning the checksum for . Demonstrate the procedure using the examples given there, or others you may consider lucky.
- Part 2
Notwithstanding past Intel microcode errors, checking computer calculations like this would not be sensible. To find a computer use for your procedure:
- Consider the statement "318682 is 101558 + 217124 and squared is 101558217124" (see: Kaprekar numbers#Casting Out Nines (fast)).
- note that has the same checksum as ();
- note that has the same checksum as () because for a Kaprekar they are made up of the same digits (sometimes with extra zeroes);
- note that this implies that for Kaprekar numbers the checksum of equals the checksum of .
Demonstrate that your procedure can be used to generate or filter a range of numbers with the property and show that this subset is a small proportion of the range and contains all the Kaprekar in the range.
- Part 3
Considering this Mathworld page, produce a efficient algorithmn based on the more mathmatical treatment of Casting Out Nines, and realizing:
- is the residual of mod ;
- the procedure can be extended to bases other than 9.
Demonstrate your algorithm by generating or filtering a range of numbers with the property and show that this subset is a small proportion of the range and contains all the Kaprekar in the range.
C++
Filter
<lang cpp>// Casting Out Nines // // Nigel Galloway. June 24th., 2012 //
- include <iostream>
int main() { int Base = 10; const int N = 2; int c1 = 0; int c2 = 0; for (int k=1; k<pow((double)Base,N); k++){ c1++; if (k%(Base-1) == (k*k)%(Base-1)){ c2++; std::cout << k << " "; } } std::cout << "\nTrying " << c2 << " numbers instead of " << c1 << " numbers saves " << 100 - ((double)c2/c1)*100 << "%" <<std::endl; return 0; }</lang>
- Produces:
1 9 10 18 19 27 28 36 37 45 46 54 55 63 64 72 73 81 82 90 91 99 Trying 22 numbers instead of 99 numbers saves 77.7778%
The kaprekar numbers in this range 1 9 45 55 and 99 are included.
Changing: "int Base = 16;" Produces:
1 6 10 15 16 21 25 30 31 36 40 45 46 51 55 60 61 66 70 75 76 81 85 90 91 96 100 105 106 111 115 120 121 126 130 135 136 141 145 150 151 156 160 165 166 171 175 180 181 186 190 195 196 201 205 210 211 216 220 225 226 231 235 240 241 246 250 255 Trying 68 numbers instead of 255 numbers saves 73.3333%
The kaprekar numbers:
- 1 is 1
- 6 is 6
- a is 10
- f is 15
- 33 is 51
- 55 is 85
- 5b is 91
- 78 is 120
- 88 is 136
- ab is 171
- cd is 205
- ff is 255
in this range are included.
Changing: "int Base = 17;" Produces:
1 16 17 32 33 48 49 64 65 80 81 96 97 112 113 128 129 144 145 160 161 176 177 19 2 193 208 209 224 225 240 241 256 257 272 273 288 Trying 36 numbers instead of 288 numbers saves 87.5%
The kaprekar numbers:
- 1 is 1
- g is 16
- 3d is 64
- d4 is 225
- gg is 288
in this range are included.
C++11 For Each Generator
<lang cpp>// Casting Out Nines Generator - Compiles with gcc4.6, MSVC 11, and CLang3 // // Nigel Galloway. June 24th., 2012 //
- include <iostream>
- include <vector>
struct ran { const int base; std::vector<int> rs; ran(const int base) : base(base) { for (int nz=0; nz<base-1; nz++) if(nz*(nz-1)%(base-1) == 0) rs.push_back(nz); } }; class co9 { private: const ran* _ran; const int _end; int _r,_x,_next; public: bool operator!=(const co9& other) const {return operator*() <= _end;} co9 begin() const {return *this;}
co9 end() const {return *this;}
int operator*() const {return _next;} co9(const int start, const int end, const ran* r) :_ran(r) ,_end(end) ,_r(1) ,_x(start/_ran->base) ,_next((_ran->base-1)*_x + _ran->rs[_r]) { while (operator*() < start) operator++(); } const co9& operator++() { const int oldr = _r; _r = ++_r%_ran->rs.size(); if (_r<oldr) _x++; _next = (_ran->base-1)*_x + _ran->rs[_r]; return *this; } };
int main() { ran r(10); for (int i : co9(1,99,&r)) { std::cout << i << ' '; } return 0; }</lang>
- Produces:
1 9 10 18 19 27 28 36 37 45 46 54 55 63 64 72 73 81 82 90 91 99
An alternative implementation for struct ran using http://rosettacode.org/wiki/Sum_digits_of_an_integer#C.2B.2B which produces the same result is: <lang cpp>struct ran { const int base; std::vector<int> rs; ran(const int base) : base(base) { for (int nz=0; nz<base-1; nz++) if(SumDigits(nz) == SumDigits(nz*nz)) rs.push_back(nz); } };</lang> Changing main: <lang cpp>int main() { ran r(16); for (int i : co9(1,255,&r)) { std::cout << i << ' '; } return 0; }</lang>
- Produces:
1 6 10 15 16 21 25 30 31 36 40 45 46 51 55 60 61 66 70 75 76 81 85 90 91 96 100 105 106 111 115 120 121 126 130 135 136 141 145 150 151 156 160 165 166 171 175 180 181 186 190 195 196 201 205 210 211 216 220 225 226 231 235 240 241 246 250 255
Changing main: <lang cpp>int main() { ran r(17); for (int i : co9(1,288,&r)) { std::cout << i << ' '; } return 0; }</lang>
- Produces:
1 16 17 32 33 48 49 64 65 80 81 96 97 112 113 128 129 144 145 160 161 176 177 192 193 208 209 224 225 240 241 256 257 272 273 288
Common Lisp
<lang lisp>;;A macro was used to ensure that the filter is inlined.
- Larry Hignight. Last updated on 7/3/2012.
(defmacro kaprekar-number-filter (n &optional (base 10))
`(= (mod ,n (1- ,base)) (mod (* ,n ,n) (1- ,base))))
(defun test (&key (start 1) (stop 10000) (base 10) (collect t))
(let ((count 0)
(nums))
(loop for i from start to stop do
(when (kaprekar-number-filter i base) (if collect (push i nums)) (incf count)))
(format t "~d potential Kaprekar numbers remain (~~~$% filtered out).~%"
count (* (/ (- stop count) stop) 100))
(if collect (reverse nums))))</lang>
- Output:
CL-USER> (test :stop 99) 22 potential Kaprekar numbers remain (~77.78% filtered out). (1 9 10 18 19 27 28 36 37 45 ...) CL-USER> (test :stop 10000 :collect nil) 2223 potential Kaprekar numbers remain (~77.77% filtered out). NIL CL-USER> (test :stop 1000000 :collect nil) 222223 potential Kaprekar numbers remain (~77.78% filtered out). NIL CL-USER> (test :stop 255 :base 16) 68 potential Kaprekar numbers remain (~73.33% filtered out). (1 6 10 15 16 21 25 30 31 36 ...) CL-USER> (test :stop 288 :base 17) 36 potential Kaprekar numbers remain (~87.50% filtered out). (1 16 17 32 33 48 49 64 65 80 ...)
D
<lang d>import std.stdio, std.algorithm, std.range;
uint[] castOut(in uint base=10, in uint start=1, in uint end=999999) {
auto ran = iota(base - 1)
.filter!(x => x % (base - 1) == (x * x) % (base - 1));
auto x = start / (base - 1);
immutable y = start % (base - 1);
typeof(return) result;
while (true) {
foreach (immutable n; ran) {
immutable k = (base - 1) * x + n;
if (k < start)
continue;
if (k > end)
return result;
result ~= k;
}
x++;
}
}
void main() {
castOut(16, 1, 255).writeln; castOut(10, 1, 99).writeln; castOut(17, 1, 288).writeln;
}</lang>
- Output (some newlines added):
[1, 6, 10, 15, 16, 21, 25, 30, 31, 36, 40, 45, 46, 51, 55, 60, 61, 66, 70, 75, 76, 81, 85, 90, 91, 96, 100, 105, 106, 111, 115, 120, 121, 126, 130, 135, 136, 141, 145, 150, 151, 156, 160, 165, 166, 171, 175, 180, 181, 186, 190, 195, 196, 201, 205, 210, 211, 216, 220, 225, 226, 231, 235, 240, 241, 246, 250, 255] [1, 9, 10, 18, 19, 27, 28, 36, 37, 45, 46, 54, 55, 63, 64, 72, 73, 81, 82, 90, 91, 99] [1, 16, 17, 32, 33, 48, 49, 64, 65, 80, 81, 96, 97, 112, 113, 128, 129, 144, 145, 160, 161, 176, 177, 192, 193, 208, 209, 224, 225, 240, 241, 256, 257, 272, 273, 288]
Go
<lang go>package main
import (
"fmt" "log" "strconv"
)
// A casting out nines algorithm.
// Quoting from: http://mathforum.org/library/drmath/view/55926.html /* First, for any number we can get a single digit, which I will call the "check digit," by repeatedly adding the digits. That is, we add the digits of the number, then if there is more than one digit in the result we add its digits, and so on until there is only one digit left.
...
You may notice that when you add the digits of 6395, if you just ignore the 9, and the 6+3 = 9, you still end up with 5 as your check digit. This is because any 9's make no difference in the result. That's why the process is called "casting out" nines. Also, at any step in the process, you can add digits, not just at the end: to do 8051647, I can say 8 + 5 = 13, which gives 4; plus 1 is 5, plus 6 is 11, which gives 2, plus 4 is 6, plus 7 is 13 which gives 4. I never have to work with numbers bigger than 18.
- /
// The twist is that co9Peterson returns a function to do casting out nines // in any specified base from 2 to 36. func co9Peterson(base int) (cob func(string) (byte, error), err error) {
if base < 2 || base > 36 {
return nil, fmt.Errorf("co9Peterson: %d invalid base", base)
}
// addDigits adds two digits in the specified base.
// People perfoming casting out nines by hand would usually have their
// addition facts memorized. In a program, a lookup table might be
// analogous, but we expediently use features of the programming language
// to add digits in the specified base.
addDigits := func(a, b byte) (string, error) {
ai, err := strconv.ParseInt(string(a), base, 64)
if err != nil {
return "", err
}
bi, err := strconv.ParseInt(string(b), base, 64)
if err != nil {
return "", err
}
return strconv.FormatInt(ai+bi, base), nil
}
// a '9' in the specified base. that is, the greatest digit.
s9 := strconv.FormatInt(int64(base-1), base)
b9 := s9[0]
// define result function. The result function may return an error
// if n is not a valid number in the specified base.
cob = func(n string) (r byte, err error) {
r = '0'
for i := 0; i < len(n); i++ { // for each digit of the number
d := n[i]
switch {
case d == b9: // if the digit is '9' of the base, cast it out
continue
// if the result so far is 0, the digit becomes the result
case r == '0':
r = d
continue
}
// otherwise, add the new digit to the result digit
s, err := addDigits(r, d)
if err != nil {
return 0, err
}
switch {
case s == s9: // if the sum is "9" of the base, cast it out
r = '0'
continue
// if the sum is a single digit, it becomes the result
case len(s) == 1:
r = s[0]
continue
}
// otherwise, reduce this two digit intermediate result before
// continuing.
r, err = cob(s)
if err != nil {
return 0, err
}
}
return
}
return
}
// Subset code required by task. Given a base and a range specified with // beginning and ending number in that base, return candidate Kaprekar numbers // based on the observation that k%(base-1) must equal (k*k)%(base-1). // For the % operation, rather than the language built-in operator, use // the method of casting out nines, which in fact implements %(base-1). func subset(base int, begin, end string) (s []string, err error) {
// convert begin, end to native integer types for easier iteration
begin64, err := strconv.ParseInt(begin, base, 64)
if err != nil {
return nil, fmt.Errorf("subset begin: %v", err)
}
end64, err := strconv.ParseInt(end, base, 64)
if err != nil {
return nil, fmt.Errorf("subset end: %v", err)
}
// generate casting out nines function for specified base
cob, err := co9Peterson(base)
if err != nil {
return
}
for k := begin64; k <= end64; k++ {
ks := strconv.FormatInt(k, base)
rk, err := cob(ks)
if err != nil { // assertion
panic(err) // this would indicate a bug in subset
}
rk2, err := cob(strconv.FormatInt(k*k, base))
if err != nil { // assertion
panic(err) // this would indicate a bug in subset
}
// test for candidate Kaprekar number
if rk == rk2 {
s = append(s, ks)
}
}
return
}
var testCases = []struct {
base int begin, end string kaprekar []string
}{
{10, "1", "100", []string{"1", "9", "45", "55", "99"}},
{17, "10", "gg", []string{"3d", "d4", "gg"}},
}
func main() {
for _, tc := range testCases {
fmt.Printf("\nTest case base = %d, begin = %s, end = %s:\n",
tc.base, tc.begin, tc.end)
s, err := subset(tc.base, tc.begin, tc.end)
if err != nil {
log.Fatal(err)
}
fmt.Println("Subset: ", s)
fmt.Println("Kaprekar:", tc.kaprekar)
sx := 0
for _, k := range tc.kaprekar {
for {
if sx == len(s) {
fmt.Printf("Fail:", k, "not in subset")
return
}
if s[sx] == k {
sx++
break
}
sx++
}
}
fmt.Println("Valid subset.")
}
}</lang>
- Output:
Test case base = 10, begin = 1, end = 100: Subset: [1 9 10 18 19 27 28 36 37 45 46 54 55 63 64 72 73 81 82 90 91 99 100] Kaprekar: [1 9 45 55 99] Valid subset. Test case base = 17, begin = 10, end = gg: Subset: [10 1f 1g 2e 2f 3d 3e 4c 4d 5b 5c 6a 6b 79 7a 88 89 97 98 a6 a7 b5 b6 c4 c5 d3 d4 e2 e3 f1 f2 g0 g1 gg] Kaprekar: [3d d4 gg] Valid subset.
J
This is an implementation of: "given two numbers which mark the beginning and end of a range of integers, and another number which denotes an integer base, return numbers from within the range where the number is equal (modulo the base minus 1) to its square". At the time of this writing, this task is a draft task and this description does not precisely match the task description on this page. Eventually, either the task description will change to match this implementation (which means this paragraph should be removed) or the task description will change to conflict with this implementation (so this entire section should be re-written). <lang J>castout=: 1 :0
[: (#~ ] =&((m-1)&|) *:) <. + [: i. (+*)@-~
)</lang> Example use: <lang J> 0 (10 castout) 100 0 1 9 10 18 19 27 28 36 37 45 46 54 55 63 64 72 73 81 82 90 91 99 100</lang> Alternate implementation: <lang J>castout=: 1 :0
[: (#~ 0 = (m-1) | 0 _1 1&p.) <. + [: i. (+*)@-~
)</lang> Note that about half of the code here is the code that implements "range of numbers". If we factor that out, and represent the desired values directly the code becomes much simpler: <lang J> (#~ 0=9|0 _1 1&p.) i.101 0 1 9 10 18 19 27 28 36 37 45 46 54 55 63 64 72 73 81 82 90 91 99 100
(#~ ] =&(9&|) *:) i. 101
0 1 9 10 18 19 27 28 36 37 45 46 54 55 63 64 72 73 81 82 90 91 99 100</lang> And, of course, we can name parts of these expressions. For example: <lang J> (#~ ] =&(co9=: 9&|) *:) i. 101 0 1 9 10 18 19 27 28 36 37 45 46 54 55 63 64 72 73 81 82 90 91 99 100</lang> Or, if you prefer: <lang J>co9=: 9&|
(#~ ] =&co9 *:) i. 101
0 1 9 10 18 19 27 28 36 37 45 46 54 55 63 64 72 73 81 82 90 91 99 100</lang>
jq
In the following, the filter is_kaprekar as defined at Kaprekar_numbers#jq is used. Since it is only defined for decimals, this section is correspondingly restricted.
Definition of co9: <lang jq>def co9:
def digits: tostring | explode | map(. - 48); # "0" is 48 if . == 9 then 0 elif 0 <= . and . <= 8 then . else digits | add | co9 end;</lang>
For convenience, we also define a function to check whether co9(i) equals co9(i*i) for a given integer, i: <lang jq>def co9_equals_co9_squared: co9 == ((.*.)|co9);</lang>
Example:
Integers in 1 .. 100 satisfying co9(i) == co9(i*i):
<lang jq>[range (1;101) | select( co9_equals_co9_squared )</lang> produces:
[1,9,10,18,19,27,28,36,37,45,46,54,55,63,64,72,73,81,82,90,91,99,100]
Verification:
One way to verify that the Kaprekar numbers satisfy the co9_equals_co9_squared condition is by inspection. For the range 1..100 considered above, we have:
<lang jq>[ range(1;101) | select(is_kaprekar) ]</lang>
[1,9,45,55,99]
To check the condition programmatically for a given range of integers, we can define a function which will emit any exceptions, e.g. <lang jq>def verify:
range(1; .) | select(is_kaprekar and (co9_equals_co9_squared | not));</lang>
For example, running (1000 | verify) produces an empty stream.
Proportion of integers in 1 .. n satisfying the mod (b-1) condition:
For a given base, "b", the following function computes the proportion of integers, i, in 1 .. n such that i % (b-1) == (i*i) % (b-1):
<lang jq>def proportion(base):
def count(stream): reduce stream as $i (0; . + 1); . as $n | (base - 1) as $b | count( range(1; 1+$n) | select( . % $b == (.*.) % $b) ) / $n ;</lang>
For example:
(10, 100, 1000, 10000, 100000) | proportion(16)
produces: <lang sh>0.3 0.27 0.267 0.2667 0.26667</lang>
Julia
<lang Julia>co9(x) = x == 9 ? 0 :
1<=x<=8 ? x :
co9(sum(digits(x)))</lang>
iskaprekar is defined in the task Kaprekar_numbers#Julia.
- Output:
julia> show(filter(x->co9(x)==co9(x^2), 1:100)) [1,9,10,18,19,27,28,36,37,45,46,54,55,63,64,72,73,81,82,90,91,99,100] julia> show(filter(iskaprekar, 1:100)) [1,9,45,55,99] julia> show(filter(x->x%15 == (x^2)%15, 1:100)) # base 16 [1,6,10,15,16,21,25,30,31,36,40,45,46,51,55,60,61,66,70,75,76,81,85,90,91,96,100]
Nim
<lang nim>import sequtils
iterator castOut(base = 10, start = 1, ending = 999_999) =
var ran: seq[int] = @[]
for y in 0 .. <base-1:
if y mod (base - 1) == (y*y) mod (base - 1):
ran.add(y)
var x = start div (base - 1) var y = start mod (base - 1)
block outer:
while true:
for n in ran:
let k = (base - 1) * x + n
if k < start:
continue
if k > ending:
break outer
yield k
inc x
echo toSeq(castOut(base=16, start=1, ending=255))</lang>
- Output:
@[1, 6, 10, 15, 16, 21, 25, 30, 31, 36, 40, 45, 46, 51, 55, 60, 61, 66, 70, 75, 76, 81, 85, 90, 91, 96, 100, 105, 106, 111, 115, 120, 121, 126, 130, 135, 136, 141, 145, 150, 151, 156, 160, 165, 166, 171, 175, 180, 181, 186, 190, 195, 196, 201, 205, 210, 211, 216, 220, 225, 226, 231, 235, 240, 241, 246, 250, 255]
Objeck
<lang objeck>class CastingNines {
function : Main(args : String[]) ~ Nil {
base := 10;
N := 2;
c1 := 0;
c2 := 0;
for (k:=1; k<base->As(Float)->Power(N->As(Float)); k+=1;){
c1+=1;
if (k%(base-1) = (k*k)%(base-1)){
c2+=1;
IO.Console->Print(k)->Print(" ");
};
};
IO.Console->Print("\nTrying ")->Print(c2)->Print(" numbers instead of ")
->Print(c1)->Print(" numbers saves ")->Print(100 - (c2->As(Float)/c1
->As(Float)*100))->PrintLine("%");
}
}</lang>
- Output:
1 9 10 18 19 27 28 36 37 45 46 54 55 63 64 72 73 81 82 90 91 99 Trying 22 numbers instead of 99 numbers saves 77.7777778%
PARI/GP
<lang parigp>{base=10; N=2; c1=c2=0; for(k=1,base^N-1,
c1++; if (k%(base-1) == k^2%(base-1), c2++; print1(k" ") );
); print("\nTrying "c2" numbers instead of "c1" numbers saves " 100.-(c2/c1)*100 "%")} </lang>
- Produces:
1 9 10 18 19 27 28 36 37 45 46 54 55 63 64 72 73 81 82 90 91 99 Trying 22 numbers instead of 99 numbers saves 77.77777777777777777777777778%
Changing to: "base = 16;" produces:
1 6 10 15 16 21 25 30 31 36 40 45 46 51 55 60 61 66 70 75 76 81 85 90 91 96 100 105 106 111 115 120 121 126 130 135 136 141 145 150 151 156 160 165 166 171 175 180 181 186 190 195 196 201 205 210 211 216 220 225 226 231 235 240 241 246 250 255 Trying 68 numbers instead of 255 numbers saves 73.33333333333333333333333333%
Perl
<lang perl>sub co9 { # Follows the simple procedure asked for in Part 1
my $n = shift; return $n if $n < 10; my $sum = 0; $sum += $_ for split(//,$n); co9($sum);
}
sub showadd {
my($n,$m) = @_;
print "( $n [",co9($n),"] + $m [",co9($m),"] ) [",co9(co9($n)+co9($m)),"]",
" = ", $n+$m," [",co9($n+$m),"]\n";
}
sub co9filter {
my $base = shift;
die unless $base >= 2;
my($beg, $end, $basem1) = (1, $base*$base-1, $base-1);
my @list = grep { $_ % $basem1 == $_*$_ % $basem1 } $beg .. $end;
($end, scalar(@list), @list);
}
print "Part 1: Create a simple filter and demonstrate using simple example.\n"; showadd(6395, 1259);
print "\nPart 2: Use this to filter a range with co9(k) == co9(k^2).\n"; print join(" ", grep { co9($_) == co9($_*$_) } 1..99), "\n";
print "\nPart 3: Use efficient method on range.\n"; for my $base (10, 17) {
my($N, $n, @l) = co9filter($base);
printf "[@l]\nIn base %d, trying %d numbers instead of %d saves %.4f%%\n\n",
$base, $n, $N, 100-($n/$N)*100;
}</lang>
- Output:
Part 1: Create a simple filter and demonstrate using simple example. ( 6395 [5] + 1259 [8] ) [4] = 7654 [4] Part 2: Use this to filter a range with co9(k) == co9(k^2). 1 9 10 18 19 27 28 36 37 45 46 54 55 63 64 72 73 81 82 90 91 99 Part 3: Use efficient method on range. [1 9 10 18 19 27 28 36 37 45 46 54 55 63 64 72 73 81 82 90 91 99] In base 10, trying 22 numbers instead of 99 saves 77.7778% [1 16 17 32 33 48 49 64 65 80 81 96 97 112 113 128 129 144 145 160 161 176 177 192 193 208 209 224 225 240 241 256 257 272 273 288] In base 17, trying 36 numbers instead of 288 saves 87.5000%
Perl 6
<lang perl6>sub cast-out(\BASE = 10, \MIN = 1, \MAX = BASE**2 - 1) {
my \B9 = BASE - 1;
my @ran = ($_ if $_ % B9 == $_**2 % B9 for ^B9);
my $x = MIN div B9;
gather loop {
for @ran -> \n {
my \k = B9 * $x + n;
take k if k >= MIN;
}
$x++;
} ...^ * > MAX;
}
say cast-out; say cast-out 16; say cast-out 17;</lang>
- Output:
1 9 10 18 19 27 28 36 37 45 46 54 55 63 64 72 73 81 82 90 91 99 1 6 10 15 16 21 25 30 31 36 40 45 46 51 55 60 61 66 70 75 76 81 85 90 91 96 100 105 106 111 115 120 121 126 130 135 136 141 145 150 151 156 160 165 166 171 175 180 181 186 190 195 196 201 205 210 211 216 220 225 226 231 235 240 241 246 250 255 1 16 17 32 33 48 49 64 65 80 81 96 97 112 113 128 129 144 145 160 161 176 177 192 193 208 209 224 225 240 241 256 257 272 273 288
PicoLisp
<lang PicoLisp>(de kaprekar (N)
(let L (cons 0 (chop (* N N)))
(for ((I . R) (cdr L) R (cdr R))
(NIL (gt0 (format R)))
(T (= N (+ @ (format (head I L)))) N) ) ) )
(de co9 (N)
(until
(> 9
(setq N
(sum
'((N) (unless (= "9" N) (format N)))
(chop N) ) ) ) )
N )
(println 'Part1:) (println
(=
(co9 (+ 6395 1259))
(co9 (+ (co9 6395) (co9 1259))) ) )
(println 'Part2:) (println
(filter
'((N) (= (co9 N) (co9 (* N N))))
(range 1 100) ) )
(println
(filter kaprekar (range 1 100)) )
(println 'Part3 '- 'base17:) (println
(filter
'((N) (= (% N 16) (% (* N N) 16)))
(range 1 100) ) )
(bye)</lang>
- Output:
Part1: T Part2: (1 9 10 18 19 27 28 36 37 45 46 54 55 63 64 72 73 81 82 90 91 99 100) (1 9 45 55 99) Part3 - base17: (1 16 17 32 33 48 49 64 65 80 81 96 97)
Python
This works slightly differently, generating the "wierd" (as defined by Counting Out Nines) numbers which may be Kaprekar, rather than filtering all numbers in a range. <lang Python># Casting out Nines
- Nigel Galloway: June 27th., 2012,
def CastOut(Base=10, Start=1, End=999999):
ran = [y for y in range(Base-1) if y%(Base-1) == (y*y)%(Base-1)]
x,y = divmod(Start, Base-1)
while True:
for n in ran:
k = (Base-1)*x + n
if k < Start:
continue
if k > End:
return
yield k
x += 1
for V in CastOut(Base=16,Start=1,End=255):
print(V, end=' ')</lang>
Produces:
1 6 10 15 16 21 25 30 31 36 40 45 46 51 55 60 61 66 70 75 76 81 85 90 91 96 100 105 106 111 115 120 121 126 130 135 136 141 145 150 151 156 160 165 166 171 175 180 181 186 190 195 196 201 205 210 211 216 220 225 226 231 235 240 241 246 250 255
CastOut(Base=10,Start=1,End=99) produces:
1 9 10 18 19 27 28 36 37 45 46 54 55 63 64 72 73 81 82 90 91 99
CastOut(Base=17,Start=1,End=288) produces:
1 16 17 32 33 48 49 64 65 80 81 96 97 112 113 128 129 144 145 160 161 176 177 192 193 208 209 224 225 240 241 256 257 272 273 288
Racket
<lang racket>#lang racket (require math)
(define (digits n)
(map (compose1 string->number string)
(string->list (number->string n))))
(define (cast-out-nines n)
(with-modulus 9
(for/fold ([sum 0]) ([d (digits n)])
(mod+ sum d))))</lang>
REXX
<lang rexx>/*REXX program to demonstrate casting-out-nines (with Kaprekar numbers).*/ parse arg start stop base . /*get the user args (if any). */ if base== then base=10 /*use base ten if not specified. */ if start== then do; start=1; stop=1000; end /*use the defaults?*/ if stop== then stop=start /*not specified? Use the default*/ numbers=castOut(start,stop,base) /*generate a list of numbers. */ @cast_out='cast-out-'||(base-1) "test" /*construct a text shortcut. */ say 'For' start "through" stop', the following passed the' @cast_out":" say numbers; say /*display the list of numbers. */ q=stop-start+1 /*Q is the range of #s in list.*/ p=words(numbers) /*P is the number of #s in list.*/ pc=format(p/q*100,,2)/1 || '%' /*calculate the percentage (%). */ say 'For' q "numbers," p 'passed the' @cast_out "("pc') for base' base"." if base\==10 then exit /*if radix ¬ ten, then stop pgm. */ Kaps=kaprekar(start,stop) /*generate a list of Kaprekar #s.*/ say; say 'The Kaprekar numbers in the same range are:' Kaps say
do i=1 for words(Kaps); x=word(Kaps,i) /*verify 'em in list. */
if wordpos(x,numbers)\==0 then iterate /*OK so far ... */
say 'Kaprekar number' x "isn't in the numbers list." /*oops!*/
exit /*go spank the coder. */
end
say 'All Kaprekar numbers are in the' @cast_out "numbers list." /*ok*/ exit /*stick a fork in it, we're done.*/ /*──────────────────────────────────CASTOUT subroutine──────────────────*/ castOut: procedure; parse arg low,high,radix; $=
high=word(high low,1) /*use default of LOW for HIGH.*/
radix=word(radix 10,1) /*use default ot 10 for RADIX.*/ niner=radix-1 /*a fast way to use RADIX - 1. */
do j=low to high /*traipse through the arg range. */
if j//niner==j*j//niner then $=$ j /*pass the test?*/
end /*j*/
return strip($) /*strip leading blank from result*/ /*──────────────────────────────────Kaprekar subroutine─────────────────*/ kaprekar: procedure; parse arg L,H; #=0; $=; if L<=1 then $=1 numeric digits max(10,2*length(H**2)) /*insure enough digits for H². */
do j=max(2,L) to H; s=j*j /*slow way to find Kaprekar #'s. */
do k=1 for length(s)%2
if j==left(s,k)+substr(s,k+1) then $=$ j /*found a Kaprekar #.*/
end /*k*/
end /*j*/
return strip($) /*return all the Kaprekar numbers*/</lang>
- Output:
when using the default input
For 1 through 1000, the following passed the cast-out-9 test: 1 9 10 18 19 27 28 36 37 45 46 54 55 63 64 72 73 81 82 90 91 99 100 108 109 117 118 126 127 135 136 144 145 153 154 162 163 171 172 180 181 189 190 198 199 207 208 216 217 225 226 234 235 243 244 252 253 261 262 270 271 279 280 288 289 297 298 306 307 315 316 324 325 333 334 342 343 351 352 360 361 369 370 378 379 387 388 396 397 405 406 414 415 423 424 432 433 441 442 450 451 459 460 468 469 477 478 486 487 495 496 504 505 513 514 522 523 531 532 540 541 549 550 558 559 567 568 576 577 585 586 594 595 603 604 612 613 621 622 630 631 639 640 648 649 657 658 666 667 675 676 684 685 693 694 702 703 711 712 720 721 729 730 738 739 747 748 756 757 765 766 774 775 783 784 792 793 801 802 810 811 819 820 828 829 837 838 846 847 855 856 864 865 873 874 882 883 891 892 900 901 909 910 918 919 927 928 936 937 945 946 954 955 963 964 972 973 981 982 990 991 999 1000 For 1000 numbers, 223 passed the cast-out-9 test (22.3%) for base 10. The Kaprekar numbers in the same range are: 1 9 45 55 99 297 703 999 All Kaprekar numbers are in the cast-out-9 test numbers list.
- Output:
when using the input of
For 1 through 256, the following passed the cast-out-15 test: 1 6 10 15 16 21 25 30 31 36 40 45 46 51 55 60 61 66 70 75 76 81 85 90 91 96 100 105 106 111 115 120 121 126 130 135 136 141 145 150 151 156 160 165 166 171 175 180 181 186 190 195 196 201 205 210 211 216 220 225 226 231 235 240 241 246 250 255 256 For 256 numbers, 69 passed the cast-out-15 test (26.95%) for base 16.
Seed7
<lang seed7>$ include "seed7_05.s7i";
const func bitset: castOut (in integer: base, in integer: start, in integer: ending) is func
result
var bitset: casted is {};
local
var bitset: ran is {};
var integer: x is 0;
var integer: n is 0;
var integer: k is 0;
var boolean: finished is FALSE;
begin
for x range 0 to base - 2 do
if x rem pred(base) = x ** 2 rem pred(base) then
incl(ran, x);
end if;
end for;
x := start div pred(base);
repeat
for n range ran until finished do
k := pred(base) * x + n;
if k >= start then
if k > ending then
finished := TRUE;
else
incl(casted, k);
end if;
end if;
end for;
incr(x);
until finished;
end func;
const proc: main is func
begin writeln(castOut(16, 1, 255)); end func;</lang>
- Output:
{1, 6, 10, 15, 16, 21, 25, 30, 31, 36, 40, 45, 46, 51, 55, 60, 61, 66, 70, 75, 76, 81, 85, 90, 91, 96, 100, 105, 106, 111, 115, 120, 121, 126, 130, 135, 136, 141, 145, 150, 151, 156, 160, 165, 166, 171, 175, 180, 181, 186, 190, 195, 196, 201, 205, 210, 211, 216, 220, 225, 226, 231, 235, 240, 241, 246, 250, 255}
Tcl
<lang tcl>proc co9 {x} {
while {[string length $x] > 1} {
set x [tcl::mathop::+ {*}[split $x ""]]
} return $x
}
- Extended to the general case
proc coBase {x {base 10}} {
while {$x >= $base} {
for {set digits {}} {$x} {set x [expr {$x / $base}]} { lappend digits [expr {$x % $base}] } set x [tcl::mathop::+ {*}$digits]
} return $x
}
- Simple helper
proc percent {part whole} {format "%.2f%%" [expr {($whole - $part) * 100.0 / $whole}]}
puts "In base 10..." set satisfying {} for {set i 1} {$i < 100} {incr i} {
if {[co9 $i] == [co9 [expr {$i*$i}]]} {
lappend satisfying $i
}
} puts $satisfying puts "Trying [llength $satisfying] numbers instead of 99 numbers saves [percent [llength $satisfying] 99]"
puts "In base 16..." set satisfying {} for {set i 1} {$i < 256} {incr i} {
if {[coBase $i 16] == [coBase [expr {$i*$i}] 16]} {
lappend satisfying $i
}
} puts $satisfying puts "Trying [llength $satisfying] numbers instead of 255 numbers saves [percent [llength $satisfying] 255]"</lang>
- Output:
With some newlines inserted…
In base 10... 1 9 10 18 19 27 28 36 37 45 46 54 55 63 64 72 73 81 82 90 91 99 Trying 22 numbers instead of 99 numbers saves 77.78% In base 16... 1 6 10 15 16 21 25 30 31 36 40 45 46 51 55 60 61 66 70 75 76 81 85 90 91 96 100 105 106 111 115 120 121 126 130 135 136 141 145 150 151 156 160 165 166 171 175 180 181 186 190 195 196 201 205 210 211 216 220 225 226 231 235 240 241 246 250 255 Trying 68 numbers instead of 255 numbers saves 73.33%
zkl
<lang zkl>fcn castOut(base=10, start=1, end=999999){
base-=1;
ran := (0).filter(base,'wrap(n){ n%base == (n*n)%base });
reg result=Sink(List);
foreach a in ([start/base ..]){
foreach b in (ran){
k := base*a + b;
if (k < start) continue; if (k > end) return(result.close()); result.write(k);
} }
}</lang> <lang zkl>castOut(16, 1, 255).toString(*).println("\n-----"); castOut(10, 1, 99).toString(*).println("\n-----"); castOut(17, 1, 288).toString(*).println();</lang>
- Output:
L(1,6,10,15,16,21,25,30,31,36,40,45,46,51,55,60,61,66,70,75, 76,81,85,90,91,96,100,105,106,111,115,120,121,126,130,135,136, 141,145,150,151,156,160,165,166,171,175,180,181,186,190,195,196, 201,205,210,211,216,220,225,226,231,235,240,241,246,250,255) ----- L(1,9,10,18,19,27,28,36,37,45,46,54,55,63,64,72,73,81,82,90,91,99) ----- L(1,16,17,32,33,48,49,64,65,80,81,96,97,112,113,128, 129,144,145,160,161,176,177,192,193,208,209,224,225,240, 241,256,257,272,273,288)