Combinations with repetitions/Square digit chain: Difference between revisions
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12,024,696,404,768,024 numbers produce 1 and 87,975,303,595,231,975 numbers produce 89
</pre>
Also, [[Iterated_digits_squaring#
=={{header|Raku}}==
|
Revision as of 21:25, 4 January 2022
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
<lang 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);
} </lang>
- 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
Go
<lang 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") }
}</lang>
- 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.
<lang jq># 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"</lang>
- 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
<lang 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
</lang>
- 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. <lang scala>// 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") }
}</lang>
- 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
<lang 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"</lang>
- 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
<lang 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";
}</lang>
- 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)
<lang perl6>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";
}</lang>
- 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
<lang 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" </lang>
- 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. <lang ecmascript>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")
}</lang>
- 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
<lang 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
}</lang> combosKW(k,sequence) is lazy, which, in this case, is quite a bit faster than the non-lazy version. <lang zkl>foreach K in (T(7,8,11,14,17)){ countNumberChains(K) }</lang>
- 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