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I wanted to ask this question: Convert to and from the factorial number system but I'm a couple of years too late!

So, instead you must convert to and from the lairotcaf backwards-factorial number base! The way it works is that the first digit is always 0, the maximum value of the next digit is 1, then 2 and so on. If a number has n digits, then each place value is given by n!/i! where i is the digit position starting at 1 on the left. The highest digit position is on the right. This means that each length uses different place values!

There is a unique shortest way to represent each number, which is what you must output for a decimal number input. If an input number starts with a 0, convert it to decimal. If it starts with any other digit, convert it to the backwards-factorial number base.

0 is both valid and the same number in both bases. In the test data no input value will use, or produce a number with a digit higher than 9. Note that the shortest representation of some numbers starts with multiple 0s.

Digit position: 1 2 3 4 5 6 Place value: 720 360 120 30 6 1 0 0 2 2 3 3 0 0 240+ 60+ 18+ 3=321

This is the shortest way to represent 321 because removing a digit makes the highest number possible only 119 (5! - 1).

Test data:

Input: Output: 0 0 1 01 002 2 3 010 011 4 5 012 0012 6 21 0121 0130 24 01000 60 01234 119 001000 120 321 002233 011235 563 0010330 987 4321 0120252 54321 000325536 654321 0010135621 0112351713 2838883 7654321 00102603574

This is so the fewest bytes wins!

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3 Answers 3

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ES6, 135 108 bytes

s=>s<"1"?[...s].reduce((t,d,i)=>+d+t*++i):eval("for(i=j=k=1;i<=s;i*=++j);for(r='';k<=j;k++)r+=s%i/(i/=k)|0")

Converting from backwards-factorial is nice and easy. Converting to it... not so nice. I found a 119-byte version that doesn't use eval:

s=>s<"1"?[...s].reduce((t,d,i)=>+d+t*++i):(i=j=k=1,g=_=>s<i||g(i*=++j),g(),[...Array(j)].map(_=>s%i/(i/=k++)|0).join``)
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Perl 6, 71 bytes

{{/^0/??.comb.reduce(* *-+^++$+*)!![R~] .polymod(+.polymod(1..*)...2)}}

Try it online!

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Jelly, 36 bytes

⁵R!>³¬TUŻ‘PƤµ³%Ḋ:ṖUŻṾ€ Jṙ1PÐƤḋV€µÇ¬?

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How it works

⁵R!>³¬TUŻ‘PƤµ³%Ḋ:ṖUŻṾ€ Aux. link (monad). Integer -> string ⁵R! [1..10]! >³¬T All indices of above <= input UŻ‘ Reverse, prepend 0, increment PƤ Product of each prefix µ Re-focus on the above ³%Ḋ:Ṗ Input % (next base value) // (own base value) UŻṾ€ Reverse, prepend zero, map each digit to string Jṙ1PÐƤḋV€µÇ¬? Main link (monad). String -> integer || Integer -> string µÇ¬? If input is string, apply the link on the left; otherwise, apply the aux. link Jṙ1 List of indices rotated once to the left PÐƤ Product of each suffix ḋV€ Convert each input char to digit, and dot product with above

Uses some tricks from my own Jelly answer for factorial base challenge.

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