Combinations with repetitions/Square digit chain
Combinations with repetitions/Square digit chain is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
Iterated digits squaring introduces RC the Project Euler Task #92. Combinations with repetitions introduce RC to the concept of generating all the combinations with repetitions of n types of things taken k at a time.
The purpose of this task is to combine these tasks as follows:
- The collections of k items will be taken from [0,1,4,9,16,25,36,49,64,81] and must be obtained using code from Combinations with repetitions. The collection of k zeroes is excluded.
- For each collection of k items determine if it translates to 1 using the rules from Iterated digits squaring
- For each collection which translates to 1 determine the number of different ways, c say, in which the k items can be uniquely ordered.
- Keep a running total of all the values of c obtained
- Answer the Project Euler Task #92 question (k=7).
- Answer the equivalent question for k=8,11,14.
- Optionally answer the question for k=17. These numbers will be larger than the basic integer type for many languages, if it is not easy to use larger numbers it is not necessary for this task.
D
// Count how many number chains for Natural Numbers < 10**K end with a value of 1.
//
import std.stdio, std.range;
const struct CombRep {
immutable uint nt, nc;
private const ulong[] combVal;
this(in uint numType, in uint numChoice) pure nothrow @safe
in {
assert(0 < numType && numType + numChoice <= 64,
"Valid only for nt + nc <= 64 (ulong bit size)");
} body {
nt = numType;
nc = numChoice;
if (nc == 0)
return;
ulong v = (1UL << (nt - 1)) - 1;
// Init to smallest number that has nt-1 bit set
// a set bit is metaphored as a _type_ seperator.
immutable limit = v << nc;
ulong[] localCombVal;
// Limit is the largest nt-1 bit set number that has nc
// zero-bit a zero-bit means a _choice_ between _type_
// seperators.
while (v <= limit) {
localCombVal ~= v;
if (v == 0)
break;
// Get next nt-1 bit number.
immutable t = (v | (v - 1)) + 1;
v = t | ((((t & -t) / (v & -v)) >> 1) - 1);
}
this.combVal = localCombVal;
}
uint length() @property const pure nothrow @safe {
return combVal.length;
}
uint[] opIndex(in uint idx) const pure nothrow @safe {
return val2set(combVal[idx]);
}
int opApply(immutable int delegate(in ref uint[]) pure nothrow @safe dg)
pure nothrow @safe {
foreach (immutable v; combVal) {
auto set = val2set(v);
if (dg(set))
break;
}
return 1;
}
private uint[] val2set(in ulong v) const pure nothrow @safe {
// Convert bit pattern to selection set
immutable uint bitLimit = nt + nc - 1;
uint typeIdx = 0;
uint[] set;
foreach (immutable bitNum; 0 .. bitLimit)
if (v & (1 << (bitLimit - bitNum - 1)))
typeIdx++;
else
set ~= typeIdx;
return set;
}
}
// For finite Random Access Range.
auto combRep(R)(R types, in uint numChoice) /*pure*/ nothrow @safe
if (hasLength!R && isRandomAccessRange!R) {
ElementType!R[][] result;
foreach (const s; CombRep(types.length, numChoice)) {
ElementType!R[] r;
foreach (immutable i; s)
r ~= types[i];
result ~= r;
}
return result;
}
void main() {
int K = 17;
ulong[] F = [1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600, 6227020800, 87178291200, 1307674368000, 20922789888000, 355687428096000];
int[] N = [0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0];
ulong z = 0;
foreach (const e; combRep([0,1,4,9,16,25,36,49,64,81], K)) {
int s = 0;
foreach (const g; e) s += g;
if (N[s] == 0) continue;
int [int] n;
foreach (const g; e) n[g] += 1;
ulong gn = F[K];
foreach (const g; n.byValue()) gn /= F[g];
z += gn;
}
writefln ("\n(k=%d) In the range 1 to %d\n%d translate to 1 and %d translate to 89\n", K, (cast (ulong) (10))^^K-1,z,(cast (ulong) (10))^^K-1-z);
}
- Output:
//(k=7) In the range 1 to 9999999 //1418853 translate to 1 and 8581146 translate to 89 //(k=8) In the range 1 to 99999999 //14255666 translate to 1 and 85744333 translate to 89 //(k=11) In the range 1 to 99999999999 //15091199356 translate to 1 and 84908800643 translate to 89 //(k=14) In the range 1 to 99999999999999 //13770853279684 translate to 1 and 86229146720315 translate to 89 //(k=17) In the range 1 to 99999999999999999 //12024696404768024 translate to 1 and 87975303595231975 translate to 89
EasyLang
func fac n .
r = 1
for i = 2 to n
r *= i
.
return r
.
fastfunc ends89 n .
repeat
s = 0
while n > 0
d = n mod 10
s += d * d
n = n div 10
.
n = s
if n = 89
return 1
.
until n = 1
.
return 0
.
items[] = [ 0 1 2 3 4 5 6 7 8 9 ]
global comb[] sum .
#
proc docomb . .
ncomb = fac len comb[]
for i = 1 to len comb[]
h = items[comb[i]]
cnt += 1
if i = len comb[] or h <> items[comb[i + 1]]
ncomb /= fac cnt
cnt = 0
.
v = v * 10 + h
.
if v > 0 and ends89 v = 1
sum += ncomb
.
.
proc combine pos val . .
if pos > len comb[]
docomb
return
.
for i = val to len items[]
comb[pos] = i
combine pos + 1 i
.
.
for h in [ 7 8 11 14 ]
len comb[] h
sum = 0
combine 1 1
print sum
.
- Output:
8581146 85744333 84908800643 86229146720315
FreeBASIC
Function endsWithOne(n As Longint) As Boolean
Dim As Longint digit, sum
Do
sum = 0
Do While n > 0
digit = n Mod 10
sum += (digit * digit)
n \= 10
Loop
If sum = 1 Then Return True
If sum = 89 Then Return False
n = sum
Loop
End Function
Dim As Integer k, n, j
Dim As Longint i, s, count1, limit
Dim ks(4) As Integer = {7, 8, 11, 14, 17}
For k = Lbound(ks) To Ubound(ks)
Dim sums(ks(k) * 81 + 1) As Longint
sums(0) = 1
sums(1) = 0
For n = 1 To ks(k)
For i = n * 81 To 1 Step -1
For j = 1 To 9
s = j * j
If s > i Then Exit For
sums(i) += sums(i - s)
Next
Next
Next
count1 = 0
For i = 1 To ks(k) * 81
If endsWithOne(i) Then count1 += sums(i)
Next
limit = 10 ^ ks(k) - 1
Print "For k ="; ks(k); " in the range 1 to"; limit
Print count1; " numbers produce 1 and"; limit - count1; " numbers produce 89"
Print
Next
Sleep
- Output:
For k = 7 in the range 1 to 9999999 1418853 numbers produce 1 and 8581146 numbers produce 89 For k = 8 in the range 1 to 99999999 14255666 numbers produce 1 and 85744333 numbers produce 89 For k = 11 in the range 1 to 99999999999 15091199356 numbers produce 1 and 84908800643 numbers produce 89 For k = 14 in the range 1 to 99999999999999 13770853279684 numbers produce 1 and 86229146720315 numbers produce 89 For k = 17 in the range 1 to 100000000000000000 12024696404768024 numbers produce 1 and 87975303595231976 numbers produce 89
Go
package main
import (
"fmt"
"math"
)
func endsWithOne(n int) bool {
sum := 0
for {
for n > 0 {
digit := n % 10
sum += digit * digit
n /= 10
}
if sum == 1 {
return true
}
if sum == 89 {
return false
}
n = sum
sum = 0
}
}
func main() {
ks := [...]int{7, 8, 11, 14, 17}
for _, k := range ks {
sums := make([]int64, k*81+1)
sums[0] = 1
sums[1] = 0
for n := 1; n <= k; n++ {
for i := n * 81; i > 0; i-- {
for j := 1; j < 10; j++ {
s := j * j
if s > i {
break
}
sums[i] += sums[i-s]
}
}
}
count1 := int64(0)
for i := 1; i <= k*81; i++ {
if endsWithOne(i) {
count1 += sums[i]
}
}
limit := int64(math.Pow10(k)) - 1
fmt.Println("For k =", k, "in the range 1 to", limit)
fmt.Println(count1, "numbers produce 1 and", limit-count1, "numbers produce 89\n")
}
}
- Output:
For k = 7 in the range 1 to 9999999 1418853 numbers produce 1 and 8581146 numbers produce 89 For k = 8 in the range 1 to 99999999 14255666 numbers produce 1 and 85744333 numbers produce 89 For k = 11 in the range 1 to 99999999999 15091199356 numbers produce 1 and 84908800643 numbers produce 89 For k = 14 in the range 1 to 99999999999999 13770853279684 numbers produce 1 and 86229146720315 numbers produce 89 For k = 17 in the range 1 to 99999999999999999 12024696404768024 numbers produce 1 and 87975303595231975 numbers produce 89
jq
Works with jq (*)
Works with gojq, the Go implementation of jq
(*) For the given values of k up to and including 14, the C implementation of jq has sufficient precision, but for k==17, the unbounded precision integer arithmetic of gojq would be required. The output shown below is taken from a gojq run.
# For gojq:
def power($b): . as $in | reduce range(0;$b) as $i (1; . * $in);
def endsWithOne:
. as $start
| { n: ., sum: 0 }
| until(.stop;
until(.n <= 0;
(.n % 10) as $digit
| .sum += $digit * $digit
| .n = (.n / 10 | floor)
)
| if .sum == 1 then .stop = 1
elif .sum == 89 then .stop = 0
else .n = .sum
| .sum = 0
end )
| .stop == 1 ;
def ks: [7, 8, 11, 14, 17];
ks[] as $k
| {sums: [1,0]}
| reduce range(1; $k+1) as $n (.;
reduce range( $n*81; 0; -1) as $i (.;
.emit = false
| .j = 0
| until(.emit or (.j == 9);
.j+=1
| (.j * .j) as $s
| if ($s > $i) then .emit = true
else .sums[$i] = .sums[$i] + .sums[$i-$s]
end) ))
| .count1 = 0
| reduce range(1; 1 + $k*81) as $i (.; if $i|endsWithOne then .count1 = .count1 + .sums[$i] else . end)
| ((10|power($k)) - 1) as $limit
| "For k = \($k) in the range 1 to \($limit)",
"\(.count1) numbers produce 1 and \($limit - .count1) numbers produce 89.\n"- Output:
For k = 7 in the range 1 to 9999999 1418853 numbers produce 1 and 8581146 numbers produce 89. For k = 8 in the range 1 to 99999999 14255666 numbers produce 1 and 85744333 numbers produce 89. For k = 11 in the range 1 to 99999999999 15091199356 numbers produce 1 and 84908800643 numbers produce 89. For k = 14 in the range 1 to 99999999999999 13770853279684 numbers produce 1 and 86229146720315 numbers produce 89. For k = 17 in the range 1 to 99999999999999999 12024696404768024 numbers produce 1 and 87975303595231975 numbers produce 89.
Julia
using Combinatorics
function iterate(m::Integer)
while m != 1 && m != 89
s = 0
while m > 0 # compute sum of squares of digits
m, d = divrem(m, 10)
s += d ^ 2
end
m = s
end
return m
end
function testitersquares(numdigits)
items = [0, 1, 4, 9, 16, 25, 36, 49, 64, 81]
onecount, eightyninecount = 0, 0
for combo in with_replacement_combinations(items, numdigits)
if any(x -> x != 0, combo)
pcount = Int(factorial(length(combo)) /
prod(y -> factorial(sum(x -> x == y, combo)), unique(combo)))
if iterate(sum(combo)) == 89
eightyninecount += pcount
else
onecount += pcount
end
end
end
println("For k = $numdigits, in the range 1 to $("9" ^ numdigits),\n" *
"$onecount numbers produce 1 and $eightyninecount numbers produce 89.\n")
end
for i in [7, 8, 11, 14, 17]
testitersquares(i)
end
- Output:
For k = 2, in the range 1 to 99, 19 numbers produce 1 and 80 numbers produce 89. For k = 7, in the range 1 to 9999999, 1418853 numbers produce 1 and 8581146 numbers produce 89. For k = 8, in the range 1 to 99999999, 14255666 numbers produce 1 and 85744333 numbers produce 89. For k = 11, in the range 1 to 99999999999, 15091199356 numbers produce 1 and 84908800643 numbers produce 89. For k = 14, in the range 1 to 99999999999999, 13770853279684 numbers produce 1 and 86229146720315 numbers produce 89. For k = 17, in the range 1 to 99999999999999999, 12024696404768024 numbers produce 1 and 87975303595231975 numbers produce 89.
Kotlin
To achieve reasonable performance, the Kotlin entry for the Iterated digits squaring task already used a similar approach to that required by this task for k = 8.
So the following generalizes that code to deal with values of k up to 17 (which requires 64 bit integers) and to count numbers where the squared digits sum sequence eventually ends in 1 rather than 89, albeit the sum of both must of course be 10 ^ k - 1.
// version 1.1.51
fun endsWithOne(n: Int): Boolean {
var digit: Int
var sum = 0
var nn = n
while (true) {
while (nn > 0) {
digit = nn % 10
sum += digit * digit
nn /= 10
}
if (sum == 1) return true
if (sum == 89) return false
nn = sum
sum = 0
}
}
fun main(args: Array<String>) {
val ks = intArrayOf(7, 8, 11, 14, 17)
for (k in ks) {
val sums = LongArray(k * 81 + 1)
sums[0] = 1
sums[1] = 0
var s: Int
for (n in 1 .. k) {
for (i in n * 81 downTo 1) {
for (j in 1 .. 9) {
s = j * j
if (s > i) break
sums[i] += sums[i - s]
}
}
}
var count1 = 0L
for (i in 1 .. k * 81) if (endsWithOne(i)) count1 += sums[i]
val limit = Math.pow(10.0, k.toDouble()).toLong() - 1
println("For k = $k in the range 1 to $limit")
println("$count1 numbers produce 1 and ${limit - count1} numbers produce 89\n")
}
}
- Output:
For k = 7 in the range 1 to 9999999 1418853 numbers produce 1 and 8581146 numbers produce 89 For k = 8 in the range 1 to 99999999 14255666 numbers produce 1 and 85744333 numbers produce 89 For k = 11 in the range 1 to 99999999999 15091199356 numbers produce 1 and 84908800643 numbers produce 89 For k = 14 in the range 1 to 99999999999999 13770853279684 numbers produce 1 and 86229146720315 numbers produce 89 For k = 17 in the range 1 to 99999999999999999 12024696404768024 numbers produce 1 and 87975303595231975 numbers produce 89
Nim
import math, strformat
func endsWithOne(n: Natural): bool =
var n = n
while true:
var sum = 0
while n > 0:
let digit = n mod 10
sum += digit * digit
n = n div 10
if sum == 1: return true
if sum == 89: return false
n = sum
const Ks = [7, 8, 11, 14, 17]
for k in Ks:
var sums = newSeq[int64](k * 81 + 1) # Initialized to 0s.
sums[0] = 1
for n in 1..k:
for i in countdown(n * 81, 1):
for j in 1..9:
let s = j * j
if s > i: break
sums[i] += sums[i - s]
var count1 = 0i64
for i in 1..k*81:
if i.endsWithOne(): count1 += sums[i]
let limit = 10^k - 1
echo &"For k = {k} in the range 1 to {limit}"
echo &"{count1} numbers produce 1 and {limit - count1} numbers produce 89\n"
- Output:
For k = 7 in the range 1 to 9999999 1418853 numbers produce 1 and 8581146 numbers produce 89 For k = 8 in the range 1 to 99999999 14255666 numbers produce 1 and 85744333 numbers produce 89 For k = 11 in the range 1 to 99999999999 15091199356 numbers produce 1 and 84908800643 numbers produce 89 For k = 14 in the range 1 to 99999999999999 13770853279684 numbers produce 1 and 86229146720315 numbers produce 89 For k = 17 in the range 1 to 99999999999999999 12024696404768024 numbers produce 1 and 87975303595231975 numbers produce 89
Perl
use strict;
use warnings;
use feature 'say';
use Math::AnyNum qw(:overload);
sub comma { reverse ((reverse shift) =~ s/(.{3})/$1,/gr) =~ s/^,//r }
sub endsWithOne {
my($n) = @_;
my $digit;
my $sum = 0;
my $nn = $n;
while () {
while ($nn > 0) {
$digit = $nn % 10;
$sum += $digit**2;
$nn = int $nn / 10;
}
return 1 if $sum == 1;
return 0 if $sum == 89;
$nn = $sum;
$sum = 0;
}
}
my @ks = <7 8 11 14 17>;
for my $k (@ks) {
my @sums = <1 0>;
my $s;
for my $n (1 .. $k) {
for my $i (reverse 1 .. $n*81) {
for my $j (1 .. 9) {
no warnings 'uninitialized';
last if ($s = $j**2) > $i;
$sums[$i] += $sums[$i-$s];
}
}
}
my $count1 = 0;
for my $i (1 .. $k*81) { $count1 += $sums[$i] if endsWithOne($i) }
my $limit = 10**$k - 1;
say "For k = $k in the range 1 to " . comma $limit;
say comma($count1) . ' numbers produce 1 and ' . comma($limit-$count1) . " numbers produce 89\n";
}
- Output:
For k = 7 in the range 1 to 9,999,999 1,418,853 numbers produce 1 and 8,581,146 numbers produce 89 For k = 8 in the range 1 to 99,999,999 14,255,666 numbers produce 1 and 85,744,333 numbers produce 89 For k = 11 in the range 1 to 99,999,999,999 15,091,199,356 numbers produce 1 and 84,908,800,643 numbers produce 89 For k = 14 in the range 1 to 99,999,999,999,999 13,770,853,279,684 numbers produce 1 and 86,229,146,720,315 numbers produce 89 For k = 17 in the range 1 to 99,999,999,999,999,999 12,024,696,404,768,024 numbers produce 1 and 87,975,303,595,231,975 numbers produce 89
Phix
with javascript_semantics include mpfr.e function endsWithOne(integer n) integer total = 0 while true do while n>0 do integer digit = remainder(n,10) total += digit * digit n = floor(n/10) end while if total==1 then return true end if if total==89 then return false end if n = total total = 0 end while end function constant ks = {7, 8, 11, 14, 17} mpz count1 = mpz_init(0), si = mpz_init(), limit = mpz_init() for ki=1 to length(ks) do integer k = ks[ki] sequence sums = repeat(0,k*81+1) sums[1] = 1 sums[2] = 0 for n=1 to k do for i=n*81+1 to 2 by -1 do for j=1 to 9 do integer s = j * j if s>i-1 then exit end if sums[i] += sums[i-s] end for end for end for mpz_set_si(count1,0) for i=1 to k*81 do if endsWithOne(i) then mpz_set_d(si,sums[i+1]) mpz_add(count1,count1,si) end if end for mpz_ui_pow_ui(limit, 10, k) mpz_sub_si(limit,limit,1) mpz_sub(si,limit,count1) string l = mpz_get_str(limit,comma_fill:=true), c = mpz_get_str(count1,comma_fill:=true), s = mpz_get_str(si,comma_fill:=true) printf(1,"For k = %d in the range 1 to %s\n",{k,l}) printf(1,"%s numbers produce 1 and %s numbers produce 89\n\n",{c,s}) end for
- Output:
For k = 7 in the range 1 to 9,999,999 1,418,853 numbers produce 1 and 8,581,146 numbers produce 89 For k = 8 in the range 1 to 99,999,999 14,255,666 numbers produce 1 and 85,744,333 numbers produce 89 For k = 11 in the range 1 to 99,999,999,999 15,091,199,356 numbers produce 1 and 84,908,800,643 numbers produce 89 For k = 14 in the range 1 to 99,999,999,999,999 13,770,853,279,684 numbers produce 1 and 86,229,146,720,315 numbers produce 89 For k = 17 in the range 1 to 99,999,999,999,999,999 12,024,696,404,768,024 numbers produce 1 and 87,975,303,595,231,975 numbers produce 89
Also, Iterated_digits_squaring#Phix produces some of the same numbers (just not so high).
Raku
(formerly Perl 6)
use v6;
sub endsWithOne($n --> Bool) {
my $digit;
my $sum = 0;
my $nn = $n;
loop {
while ($nn > 0) {
$digit = $nn % 10;
$sum += $digit²;
$nn = $nn div 10;
}
($sum == 1) and return True;
($sum == 89) and return False;
$nn = $sum;
$sum = 0;
}
}
my @ks = (7, 8, 11, 14, 17);
for @ks -> $k {
my @sums is default(0) = 1,0;
my $s;
for (1 .. $k) -> $n {
for ($n*81 ... 1) -> $i {
for (1 .. 9) -> $j {
$s = $j²;
if ($s > $i) { last };
@sums[$i] += @sums[$i-$s];
}
}
}
my $count1 = 0;
for (1 .. $k*81) -> $i { if (endsWithOne($i)) {$count1 += @sums[$i]} }
my $limit = 10**$k - 1;
say "For k = $k in the range 1 to $limit";
say "$count1 numbers produce 1 and ",$limit-$count1," numbers produce 89";
}
- Output:
For k = 7 in the range 1 to 9999999 1418853 numbers produce 1 and 8581146 numbers produce 89 For k = 8 in the range 1 to 99999999 14255666 numbers produce 1 and 85744333 numbers produce 89 For k = 11 in the range 1 to 99999999999 15091199356 numbers produce 1 and 84908800643 numbers produce 89 For k = 14 in the range 1 to 99999999999999 13770853279684 numbers produce 1 and 86229146720315 numbers produce 89 For k = 17 in the range 1 to 99999999999999999 12024696404768024 numbers produce 1 and 87975303595231975 numbers produce 89
Ruby
# Count how many number chains for Natural Numbers < 10**K end with a value of 1.
#
# Nigel_Galloway
# August 26th., 2014.
K = 17
F = Array.new(K+1){|n| n==0?1:(1..n).inject(:*)} #Some small factorials
g = -> n, gn=[n,0], res=0 { while gn[0]>0
gn = gn[0].divmod(10)
res += gn[1]**2
end
return res==89?0:res
}
#An array: N[n]==1 means that n translates to 1, 0 means that it does not.
N = (G=Array.new(K*81+1){|n| n==0? 0:(i=g.call(n))==89 ? 0:i}).collect{|n| while n>1 do n = G[n] end; n }
z = 0 #Running count of numbers translating to 1
(0..9).collect{|n| n**2}.repeated_combination(K).each{|n| #Iterate over unique digit combinations
next if N[n.inject(:+)] == 0 #Count only ones
nn = Hash.new{0} #Determine how many numbers this digit combination corresponds to
n.each{|n| nn[n] += 1} #and
z += nn.values.inject(F[K]){|gn,n| gn/F[n]} #Add to the count of numbers terminating in 1
}
puts "\nk=(#{K}) in the range 1 to #{10**K-1}\n#{z} numbers produce 1 and #{10**K-1-z} numbers produce 89"
- Output:
#(k=7) in the range 1 to 9999999 #1418853 numbers produce 1 and 8581146 numbers produce 89 #(k=8) in the range 1 to 99999999 #14255666 numbers produce 1 and 85744333 numbers produce 89 #(k=11) in the range 1 to 99999999999 #15091199356 numbers produce 1 and 84908800643 numbers produce 89 #(k=14) in the range 1 to 99999999999999 #13770853279684 numbers produce 1 and 86229146720315 numbers produce 89 #(k=17) in the range 1 to 99999999999999999 #12024696404768024 numbers produce 1 and 87975303595231975 numbers produce 89
Wren
As Wren doesn't have 64 bit integers, it is necessary to use BigInt here to process k = 17.
import "./big" for BigInt
var endsWithOne = Fn.new { |n|
var sum = 0
while (true) {
while (n > 0) {
var digit = n % 10
sum = sum + digit * digit
n = (n/10).floor
}
if (sum == 1) return true
if (sum == 89) return false
n = sum
sum = 0
}
}
var ks = [7, 8, 11, 14, 17]
for (k in ks) {
var sums = List.filled(k * 81 + 1, 0)
sums[0] = 1
sums[1] = 0
for (n in 1..k) {
for (i in n*81..1) {
for (j in 1..9) {
var s = j * j
if (s > i) break
sums[i] = sums[i] + sums[i-s]
}
}
}
var count1 = BigInt.zero
for (i in 1..k*81) if (endsWithOne.call(i)) count1 = count1 + sums[i]
var limit = BigInt.ten.pow(k) - 1
System.print("For k = %(k) in the range 1 to %(limit)")
System.print("%(count1) numbers produce 1 and %(limit - count1) numbers produce 89\n")
}
- Output:
For k = 7 in the range 1 to 9999999 1418853 numbers produce 1 and 8581146 numbers produce 89 For k = 8 in the range 1 to 99999999 14255666 numbers produce 1 and 85744333 numbers produce 89 For k = 11 in the range 1 to 99999999999 15091199356 numbers produce 1 and 84908800643 numbers produce 89 For k = 14 in the range 1 to 99999999999999 13770853279684 numbers produce 1 and 86229146720315 numbers produce 89 For k = 17 in the range 1 to 99999999999999999 12024696404768024 numbers produce 1 and 87975303595231975 numbers produce 89
zkl
fcn countNumberChains(K){
F:=(K+1).pump(List,fcn(n){ (1).reduce(n,'*,1) }); #Some small factorials
g:=fcn(n){
gn,res:=L(n,0),0;
while(gn[0]>0){ gn=gn[0].divr(10); res+=gn[1].pow(2); }
if(res==89) 0 else res
};
#An array: N[n]==1 means that n translates to 1, 0 means that it does not.
n,G:=K*81+1,n.pump(List,g);
N:=n.pump(List,'wrap(n){ n=g(n); while(n>1){ n=G[n] } n });
z:=([0..9].pump(List,fcn(n){ n*n }):Utils.Helpers.combosKW(K,_)) #combos of (0,1,4,9,16,25,36,49,64,81)
.reduce('wrap(z,ds){ #Iterate over unique digit combinations
if(N[ds.sum(0)]==0) return(z); #Count only ones
nn:=Dictionary(); #Determine how many numbers this digit combination corresponds to
ds.pump(Void,nn.incV); #and (eg (0,0,0,0,0,1,9)-->(0:5, 1:1, 9:1)
z + nn.values.reduce( #Add to the count of numbers terminating in 1
'wrap(gn,n){ gn/F[n] },F[K]);
},0);
println("\nk=(%d) in the range 1 to %,d".fmt(K,(10).pow(K)-1));
println("%,d numbers produce 1 and %,d numbers produce 89".fmt(z,(10).pow(K)-1-z));
z
}combosKW(k,sequence) is lazy, which, in this case, is quite a bit faster than the non-lazy version.
foreach K in (T(7,8,11,14,17)){ countNumberChains(K) }- Output:
k=(7) in the range 1 to 9,999,999 1,418,853 numbers produce 1 and 8,581,146 numbers produce 89 k=(8) in the range 1 to 99,999,999 14,255,666 numbers produce 1 and 85,744,333 numbers produce 89 k=(11) in the range 1 to 99,999,999,999 15,091,199,356 numbers produce 1 and 84,908,800,643 numbers produce 89 k=(14) in the range 1 to 99,999,999,999,999 13,770,853,279,684 numbers produce 1 and 86,229,146,720,315 numbers produce 89 k=(17) in the range 1 to 99,999,999,999,999,999 12,024,696,404,768,024 numbers produce 1 and 87,975,303,595,231,975 numbers produce 89