Using any set of items that can be "valued" sequentially - meaning you
can identify one of the items as the smallest value or 1, the next
item as 2, the next as 3, etc. - a rising sequence is when the items
appear in order as you go from one end to the next. For example, if
you have the following set of items:
1, 2, 3, 4, 5, 6, 7, 8, 9
and we agree that the left is the beginning (or top) of the list, then
you would find the item that equals 1 and begin moving up the list to
the right until you get to the end of the list; those numbers are
considered a rising sequence. Once you get to the end of the list (to
the last item on the right), you would repeat the same process from
the beginning of the list (the left) until you find the first value
that has not been used in a rising (that hasn't already been used).
This continues until all numbers are used.
OK... let's try this. From the set above, there is ONE rising sequence
because we start with 1 and count up until we get to the end. In this
case, all numbers have been used at this point so we don't need to
look for another one.
Now, let's say we "shuffle" these numbers/cards and then lay them out
from left to right (top card goes on far left and cards are laid out
to the right, like in Solitaire), and we get the following order:
1, 2, 6, 7, 3, 8, 4, 5, 9
To find the first sequential rising, find the lowest value, 1, and
count sequentially to the right until you get to the end)
I can do 1, 2, 3, 4, 5 before I get to the end. NOW, I start from
the left and find the lowest number that I didn't already use: I've
used 1 & 2, so I start with 6 and can count 6, 7, 8, 9, which uses up
all of the numbers and gives us TWO sequential risings for this list.
One last example:
2, 6, 3, 1, 9, 5, 4, 7, 8
How many sequential risings are in this list?
1st sequential rising = 1
2nd sequential rising = 2, 3, 4
3rd sequential rising = 6, 7, 8
4th sequential rising = 9
5th sequential rising = 5
There are FIVE. Hopefully the concept of a "rising sequence" is a
little more clear now.
To help understand this concept in terms of card shuffling, take 10
sequential cards from a deck -- preferably A, 2, 3, 4, 5, 6, 7, 8, 9,
and 10 -- and make the ACE = 1. Stack them in order from Ace to Ten so
that the ten is on the bottom. Now, if you shuffle the cards using a
traditional "rifle" shuffle, you will split the stack in half (an
equal split is not required to make this work) and then intersperse
the cards from each hand. Example, I split the deck and have A, 2, 3,
4 in my right hand and 5, 6, 7, 8, 9, and 10 in my left hand. When I
shuffle using the regular method, no matter how bad I am at making
sure I intersperse one card at a time from each hand, you will find
that the maximum number of sequential risings will be two for the
first shuffle. Remember to lay them out from left to right and then
count the risings.
This concept gets more complex when you increase the number of cards
and shuffles and ESPECIALLY when you begin with a deck that has the
highest value on top. This leads to the mathematic formula: For all
positive integers n and a, Qa(w) = (n+a-r/n)/an with r = r(w) the
number of rising sequences in w.
Hope this helps; it's not an easy one to explain! |
