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- Path: sparky!uunet!ogicse!uwm.edu!linac!att!princeton!fish.Princeton.EDU!lhjensen
- From: lhjensen@fish.Princeton.EDU (Leif Jensen)
- Newsgroups: sci.math
- Subject: Re: Another chocolate fish problem.
- Message-ID: <1992Oct16.053009.24883@Princeton.EDU>
- Date: 16 Oct 92 05:30:09 GMT
- Article-I.D.: Princeto.1992Oct16.053009.24883
- References: <Bw3E8p.FEn@cantua.canterbury.ac.nz>
- Sender: news@Princeton.EDU (USENET News System)
- Organization: Princeton University
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- Originator: news@nimaster
- Nntp-Posting-Host: fish.princeton.edu
-
- OK, since my last proof was so popular, here is another. First a rehash
- of the problem.
-
- u(n) is the number of distinct ordered partitions of a set of n elements
- into sets of one or two elements. v(n) is the number of distinct ordered
- partitions of a set of n elements into sets of two elements or more.
- Prove that u(n) = v(n+2) for integers n >= 1.
-
- We denote ordered partitions by a sequence of integers abcd... in which
- the first digit represents the order of the first set of the partition,
- the second the order of the second set of the partition, etc.
-
- We define an algorithm to convert ordered partitions of n into ordered
- partitions of n+2 and an algorithm to convert the other way. Each
- builds up a partition by saying "output x". These xi's are to be collected
- and when set in order x1x2x3... represent the resulting partition.
-
- Algorithm 1: operates on ordered partitions into sets of 1 or 2.
-
- 1) Remove the leftmost group of n 1's followed by a 2 from the input
- partition and output n+2.
- 2) Goto 1.
- 3) At this point only n 1's remain in the input. Output n+2.
-
- It is easy to see that this algorithm maps every ordered partition of n
- into sets of 1 or 2 to a unique ordered partition of n+2 into sets of
- 2 or more. Thus we have u(n) <= v(n+2).
-
- Algorithm 2: operates on ordered partitions into sets of 2 or more.
-
- 1) If the input partition has only one element n, repeatedly output 1
- (n-2) times and stop execution of the algorithm.
- 2) Remove the leftmost element n from the input partition, then
- repeatedly output 1 (n-2) times, and then output 2
- 3) Goto 1
-
- It is easy to see that this algorithm maps every ordered partition of n
- into sets of 2 or more to a unique ordered partition of n-2 into sets of
- 1 or 2. Thus we have v(n) <= u(n-2)
-
- We can now conclude that u(n) = v(n+2) for all n >= 1.
-
- OK, so which proof do you prefer? Are they both correct? This second
- one gives absolutely no insight into the problem. Is that good or
- bad? Do you like defining algorithms as here in proof 2 to solve
- math problems or does it remind you too much of Computer Science?
-
- All comments are welcome.
-
- --
- Leif Jensen
- lhjensen@phoenix.princeton.edu
-