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== Mojibake and misspellings ==
(As the task description is fairly difficult to read as it stands, I took a shot at revising it slightly, correcting misspellings and translating the erroneous Windows1252>utf8 mojibake. Note that this is just my interpretation as I understood it. I am not the task author. Feel free to edit / correct anything I have gotten wrong.)
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You need a random arrangement of a deck of cards, you are sick of lame ways of doing this.
This task is a supercool way of doing this using factorial base numbers.
The first 25 factorial base numbers in increasing order are:
0.0.0, 0.0.1, 0.1.0, 0.1.1, 0.2.0, 0.2.1, 1.0.0, 1.0.1, 1.1.0, 1.1.1, 1.2.0, 1.2.1, 2.0.0, 2.0.1, 2.1.0, 2.1.1, 2.2.0, 2.2.1, 3.0.0, 3.0.1, 3.1.0, 3.1.1, 3.2.0, 3.2.1, 1.0.0.0
Observe that the least significant digit is base 2, the next; base 3, and so on. In general, an '''n'''digit factorial base number will use the digits '''0..n''' in base '''n+1''' for each place '''1..n''' (from least to most significant.)
It is simple to produce a 1to1 mapping between an '''n''' digit factorial base number and permutations of an '''n+1''' element array:
0.0.0 > 0123 0.0.1 > 0132 0.1.0 > 0213 0.1.1 > 0231 0.2.0 > 0312 0.2.1 > 0321 1.0.0 > 1023 1.0.1 > 1032 1.1.0 > 1203 1.1.1 > 1230 1.2.0 > 1302 1.2.1 > 1320 2.0.0 > 2013 2.0.1 > 2031 2.1.0 > 2103 2.1.1 > 2130 2.2.0 > 2301 2.2.1 > 2310 3.0.0 > 3012 3.0.1 > 3021 3.1.0 > 3102 3.1.1 > 3120 3.2.0 > 3201 3.2.1 > 3210
The following pseudocode demonstrates the procedure to generate the mapping:
Starting with '''m=0''' and '''Ω''', an array of elements to be permutated, for each digit '''g''' starting with the '''most''' significant digit in the factorial base number.
 If '''g''' is greater than zero, rotate the elements from '''m''' to '''m+g''' in '''Ω''' (see example)
 Increment '''m''' and repeat the first step using the next most significant digit until the factorial base number is exhausted.
For example: using the factorial base number '''2.0.1''' and '''Ω''' = '''0 1 2 3''' where place 0 in both is the most significant (leftmost) digit/element.
 Step 1: '''m=0 g=2'''; Rotate places 0 through 2. '''0 1 2 3''' becomes '''2 0 1 3'''
 Step 2: '''m=1 g=0'''; No action.
 Step 3: '''m=2 g=1'''; Rotate places 2 through 3. '''2 0 1 3''' becomes '''2 0 3 1'''
;Task:

'''Part 1:''' Use your function to recreate the permutation table of 3 digit factorial base numbers as above. Show the output here.

'''Part 2:''' Use your function to generate all permutations of 11 digits. Count them, don't display them, and compare this method with the method in RCs [[Permutations]] task.

'''Part 3:''' Use your function to create the corresponding permutations of the following two random 51 digit factorial base numbers:
39.49.7.47.29.30.2.12.10.3.29.37.33.17.12.31.29.34.17.25.2.4.25.4.1.14.20.6.21.18.1.1.1.4.0.5.15.12.4.3.10.10.9.1.6.5.5.3.0.0.0 51.48.16.22.3.0.19.34.29.1.36.30.12.32.12.29.30.26.14.21.8.12.1.3.10.4.7.17.6.21.8.12.15.15.13.15.7.3.12.11.9.5.5.6.6.3.4.0.3.2.1
Use the following shoe of cards as your '''Ω''' array:
A♠K♠Q♠J♠10♠9♠8♠7♠6♠5♠4♠3♠2♠A♥K♥Q♥J♥10♥9♥8♥7♥6♥5♥4♥3♥2♥A♦K♦Q♦J♦10♦9♦8♦7♦6♦5♦4♦3♦2♦A♣K♣Q♣J♣10♣9♣8♣7♣6♣5♣4♣3♣2♣
 '''Part 4:''' Finally, create your own 51 digit factorial base number and produce the corresponding permutation of the above shoe.
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::Thanks[[User:Nigel GallowayNigel Galloway]] ([[User talk:Nigel Gallowaytalk]]) 11:21, 10 December 2018 (UTC)
== Why not using the inverse of [http://rosettacode.org/wiki/Knuth_shuffle Knuth_shuffle] ==
this has the same factorial base number, but one has to do only length1 swaps instead of rotation. It ist easy to create a map between source and goal arrangement.
source 1.3.2.0 with value,indices (0,31,02,13,1)
goal 0,1,2,3 with value,indices (0,01,12,23,3)
start index = 0
first swap 0 <> 1 => index 0 <> 3 > update Value,Indices for swapped values >
(Swap Index = 3 out of [0..3])
source 0.3.2.1 with value,indices (0,01,32,23,1)
index++ => 1
swap 3 <> 1 => index 1 <> 3
(Swap Index = 3 out of [index..3])
source 0.1.2.3 with value,indices (0,01,12,23,3)
index++ => 2
index 2 = index of value > no swap;
source 0.1.2.3 with value,indices (0,01,12,23,3)
index++ => 3
index 3 = index of value > no swap;
source 0.1.2.3 with value,indices (0,01,12,23,3)
I see that the advantage of the rotation is the lexicographical order of generating the permutations, but what is it good for? :see [[First perfect square in base N with N unique digits#Using Factorial base numbers indexing permutations of a collection]][[User:Nigel GallowayNigel Galloway]] ([[User talk:Nigel Gallowaytalk]]) 18:34, 30 May 2019 (UTC)