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  1. PO
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    <p>I would process as below :</p> <p><strong>First you can shuffle each 4 decks (using FYK algorithm)</strong> </p> <p><strong>Then generate a 4 partitions (* I define partition below) of your 52 cards of 4 decks</strong> with the constraint of having not more than 3 element in each set of the partition :</p> <p>For example :</p> <pre><code>(1,3,0,3,2,0,1) would be a partition of 10 with this constraint (1,1,1,1,1,1,1,1,1,1) would be a partition of 10 too with this constraint </code></pre> <p><strong>Then Mix the 4 decks based on these partition.</strong></p> <p>For example if you have :</p> <pre><code>(3,2,1) (2,2,2) </code></pre> <p>you take the 3 first of deck one then 2 of deck 2 then 2 of deck one the 2 of deck 2 then 1 of deck 1 then 2 of deck 2. (okay ?)</p> <p>All partitions are not valid, so you need to add one more constraint :</p> <p>for example with this method :</p> <pre><code>(1,2,1,1,1,1,1,1) (3,3,3) </code></pre> <p>will end up having 4 elements of deck 1 at the end.</p> <p><strong>So the last partition must satisfy a constraint, I wrote a little python program to generate these partitions.</strong></p> <pre><code>from random import randint,choice def randomPartition(maxlength=float('inf'),N=52): ''' the max length is the last constraint in my expanation: you dont want l[maxlength:len(l)]) to be more than 3 in this case we are going to randomly add +1 to all element in the list that are less than 3 N is the number of element in the partition ''' l=[] # it's your result partition while(sum(l)&lt;N ): if (len(l)&gt;maxlength and sum(l[maxlength:len(l)])&gt;=3): #in the case something goes wrong availableRange=[i for i in range(len(l)) if l[i]&lt;3] #create the list of available elements to which you can add one while(sum(l)&lt;N): temp=choice(availableRange) #randomly pick element in this list l[temp]+=1 availableRange=[i for i in range(len(l)) if l[i]&lt;3] #actualize the list if availableRange==[]: #if this list gets empty your program cannot find a solution print "NO SOLUTION" break break else: temp=randint(0,min(N-sum(l),3)) # If everything goes well just add the next element in the list until you get a sum of N=52 l.append(temp) return l </code></pre> <p>Now you can generate the 4 partitions and mix the decks according to these partitions :</p> <pre><code>def generate4Partitions(): l1=randomPartition(); l2=randomPartition(); l3=randomPartition(); m=max(len(l1),len(l2),len(l3)) l4=randomPartition(m); return l1,l2,l3,l4 </code></pre> <p>This 4 partitions will always be admissible based on their definitions. </p> <p><strong>NOte:</strong></p> <p>There might be more cases where it's not admissible , for example :</p> <pre><code>(3,0,0,3) (0,3,3,0) </code></pre> <p>I guess this code needs to be modified a bit to take more constraints into account.</p> <p>But it should be easy, just to delete unwanted zeros,like this:</p> <pre><code>(3,0,3) (0,3,3,0) </code></pre> <p>Hope it's understandable and it helps</p>
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    1. POHow can I apply Fisher-Yates-Knuth to Shuffling with restrictions? Or is there any other efficient approach?
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    1. USRicky Bobby
    1. USRicky Bobby
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    1. COWhen you say "generate 4 partitions", you're actually generating more than that - you're partitioning the set into groups of at most 4.
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      1. USNick Johnson
    2. CO@NickJohnson I am generating 4 groups of at most 4 that sum up to 52. I used the word partition because I used this groups to partition create a partition of the deck of cards. (before mixing them). All the decks play a symmetrical role, so the outcome should be pretty much uniform
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      1. USRicky Bobby
    3. COIf I read your answer correctly, you're generating an undefined number of partitions of at most four elements, not just four.
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      1. USNick Johnson
 

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