Rosetta Code

Inconsummate numbers in base 10: Difference between revisions

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Inconsummate numbers in base 10 (view source)

Revision as of 08:49, 29 September 2022

157 bytes removed ,  1 year ago
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→‎{{header|Wren}}: Changes to preamble and made second version more compact.
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{{libheader|Wren-fmt}}
It appears to be more than enough to calculate ratios for all numbers up to 999,999 (which only takes about 0.4 seconds on my machine) to be sure of finding the 1,000th inconsummate number.
 
In fact it still finds the 1,000th number if you limit the search to the first 249,999 (but not 248,999) numbers which may be where the first magic number of '250' comes from in the Pascal/Phix entries.
<syntaxhighlight lang="ecmascript">import "./math" for Int
import "./fmt" for Fmt
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Alternatively and more generally:
{{trans|Python}}
...though much quicker as I'm using lower figures for the second component of the minDigitSums tuples.
Though I think the Python version is in fact wrong for the 100,000th number since if you enumerate up to 10,000 you get the 10,000th inconsummate number to be 42,171 rather than 59,853.
 
The problem seems to be that the minimum divisor for (say) 6 digit numbers is not 999999/54 = 18518 but 109999/37 = 2972. I've corrected for that in the following translation.
<syntaxhighlight lang="ecmascript">import "./math" for Int, Nums
import "./fmt" for Fmt
 
var generateInconsummate = FnFiber.new { |maxWanted|
var minDigitSums = (2..14).map { |i| [10.pow(i), ((10.pow(i-2) * 11 - 1) / (9 * i - 17)).floor] }
var limit = Nums.min(minDigitSums.where { |p| p[1] > maxWanted }.map { |p| p[0] })
var arr = List.filled(limit, 0)
arr[0] = 1
for (dividend in 1...limit) {
var ds = Int.digitSum(dividend)
var quo = (dividend/ds).floor
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}
 
var gi = Fiber.new(generateInconsummate)
var incons = List.filled(50, 0)
var incons1k
var incons10k
var incons100k
System.print("First 50 inconsummate numbers in base 10:")
 
for (i in 1..100000) {
var j = gigenerateInconsummate.call(100000)
if (i <= 50) {
incons[i-1] = j
if (i == 50) Fmt.tprint("$3d", incons, 10)▼
} else if (i == 1000) {
incons1kFmt.print("\nOne =thousandth $,d", j)
} else if (i == 10000) {
incons10kFmt.print("Ten =thousandth $,d", j)
} else if (i == 100000) {
incons100kFmt.print("100 =thousandth $,d", j)
}
}</syntaxhighlight>
}
▲Fmt.tprint("$3d", incons, 10)
Fmt.print("\nOne thousandth $,d", incons1k)
Fmt.print("Ten thousandth $,d", incons10k)
Fmt.print("100 thousandth $,d", incons100k)</syntaxhighlight>
 
{{out}}
PureFox
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