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Guest Message by DevFuse
 

superprismatic

Member Since --
Offline Last Active Today, 01:32 AM
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Posts I've Made

In Topic: Bored while eating Spaghetti

08 June 2013 - 12:57 AM

Spoiler for Here are the details of my solution


In Topic: Shuffling a Deck of Cards

07 June 2013 - 06:51 PM

If a deck of 52 cards is cut exactly in half (top half and bottom half),
then the top half can be partitioned into T segments (keeping the order
of the cards) and, similarly, the bottom half can be partitioned into B
segments. If T=B then there are two ways to interleave these partitions
of cards -- dropping segments of cards alternately from T and B, beginning
with T (the first way) or beginning with B (the second way). If T=B+1,
then there is only one way to interleave the segments -- one must begin
with T. If T+1=B, one must begin with B. Any other relationship between
T and B make them impossible to interleave. We can count the number of
partitions of 26 in all possible arrangements, they are:

#partitions of 26; #of ways to make partitions of this number respecting order
1 1
2 25
3 300
4 2300
5 12650
6 53130
7 177100
8 480700
9 1081575
10 2042975
11 3268760
12 4457400
13 5200300
14 5200300
15 4457400
16 3268760
17 2042975
18 1081575
19 480700
20 177100
21 53130
22 12650
23 2300
24 300
25 25
26 1

We can now count all possible T,B pairs where T and B are at most one
apart. This gives us a grand total of 374,369,872,911,804 possible
riffle shuffles of 52 cards. Considering that 52! is the enormous
80658175170943878571660636856403766975289505440883277824000000000000,
I can't see how to show that there is some number, N, of riffle shuffles
such that one can always produce any particular permutation of the 52!
by applying at most N riffle shuffles to the identity permutation.

In Topic: Mike's Magical Scale

06 June 2013 - 01:15 AM

 

 

 



 


Spoiler for Is it?

That answer works. Is there a shorter answer, one that uses less numbers?

 

I don't think so. 5,1,1,1 also works. I think these are the only solutions

 

There are two others: 1,5,1,1 and 1,1,5,1.

 

But is there a three number solution?

 

No


In Topic: Bored while eating Spaghetti

04 June 2013 - 09:50 PM

Spoiler for I get


In Topic: Mike's Magical Scale

04 June 2013 - 09:44 PM

 



 


Spoiler for Is it?

That answer works. Is there a shorter answer, one that uses less numbers?

 

I don't think so. 5,1,1,1 also works. I think these are the only solutions

 

There are two others: 1,5,1,1 and 1,1,5,1.