Combinations with repetitions/Square digit chain: Difference between revisions
(Added Kotlin) |
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//(k=17) In the range 1 to 99999999999999999 |
//(k=17) In the range 1 to 99999999999999999 |
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//12024696404768024 translate to 1 and 87975303595231975 translate to 89 |
//12024696404768024 translate to 1 and 87975303595231975 translate to 89 |
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</pre> |
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=={{header|Kotlin}}== |
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To achieve reasonable performance, the Kotlin entry for the [[Iterated digits squaring]] task used a similar approach to that required by this task for k = 8. |
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So the following generalizes that code to deal with values of k up to 17 (which requires 64 bit integers) and to identify 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. |
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<lang scala>// version 1.1.51 |
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fun endsWithOne(n: Int): Boolean { |
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var digit: Int |
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var sum = 0 |
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var nn = n |
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while (true) { |
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while (nn > 0) { |
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digit = nn % 10 |
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sum += digit * digit |
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nn /= 10 |
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} |
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if (sum == 1) return true |
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if (sum == 89) return false |
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nn = sum |
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sum = 0 |
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} |
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} |
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fun main(args: Array<String>) { |
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val ks = intArrayOf(7, 8, 11, 14, 17) |
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for (k in ks) { |
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val sums = LongArray(k * 81 + 1) |
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sums[0] = 1 |
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sums[1] = 0 |
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var s: Int |
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for (n in 1 .. k) { |
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for (i in n * 81 downTo 1) { |
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for (j in 1 .. 9) { |
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s = j * j |
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if (s > i) break |
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sums[i] += sums[i - s] |
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} |
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} |
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} |
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var count1 = 0L |
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for (i in 1 .. k * 81) if (endsWithOne(i)) count1 += sums[i] |
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val limit = Math.pow(10.0, k.toDouble()).toLong() - 1 |
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println("For k = $k in the range 1 to $limit") |
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println("$count1 numbers produce 1 and ${limit - count1} numbers produce 89\n") |
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} |
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}</lang> |
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{{out}} |
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<pre> |
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For k = 7 in the range 1 to 9999999 |
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1418853 numbers produce 1 and 8581146 numbers produce 89 |
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For k = 8 in the range 1 to 99999999 |
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14255666 numbers produce 1 and 85744333 numbers produce 89 |
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For k = 11 in the range 1 to 99999999999 |
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15091199356 numbers produce 1 and 84908800643 numbers produce 89 |
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For k = 14 in the range 1 to 99999999999999 |
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13770853279684 numbers produce 1 and 86229146720315 numbers produce 89 |
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For k = 17 in the range 1 to 99999999999999999 |
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12024696404768024 numbers produce 1 and 87975303595231975 numbers produce 89 |
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</pre> |
</pre> |
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Revision as of 15:10, 11 October 2017
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
Kotlin
To achieve reasonable performance, the Kotlin entry for the Iterated digits squaring task 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 identify 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
Phix
There is a solution to this on the Iterated_digits_squaring page
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
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