Concrete mathematics : a foundation for computer science

6.1 STIRLING NUMBERS 247
We can derive a recurrence for [z] by modifying the argument we used
for {L}. Every arrangement of n objects in k cycles either puts the last object
into a cycle by itself (in [:::I wa y s ) or inserts that object into one of the [“;‘I
cycle arrangements of the first n- 1 objects. In the latter case, there are n- 1
different ways to do the insertion. (This takes some thought, but it’s not hard
to verify that there are j ways to put a new element into a j-cycle in order to
make a (j + 1)-cycle. When j = 3, for example, the cycle [A, B, C] leads to
[A, B, C, Dl , [A,B,D,Cl, o r [A,D,B,Cl
when we insert a new element D, and there are no other possibilities. Summing
over all j gives a total of n- 1 ways to insert an nth object into a cycle
decomposition of n - 1 objects.) The desired recurrence is therefore
n
[I k
= (n-l)[ni’] + [:I:], integern>O.
This is the addition-formula analog that generates Table 245.
Comparison of (6.8) and (6.3) shows that the first term on the right side is
multiplied by its upper index (n- 1) in the case of Stirling cycle numbers, but
by its lower index k in the case of Stirling subset numbers. We can therefore
perform “absorption” in terms like n[z] and k{ T}, when we do proofs by
mathematical induction.
Every permutation is equivalent to a set of cycles. For example, consider
the permutation that takes 123456789 into 384729156. We can conveniently
represent it in two rows,
123456789
384729156,
showing that 1 goes to 3 and 2 goes to 8, etc. The cycle structure comes
about because 1 goes to 3, which goes to 4, which goes to 7, which goes back
to 1; that’s the cycle [1,3,4,7]. Another cycle in this permutation is [2,8,5];
still another is [6,91. Therefore the permutation 384729156 is equivalent to
the cycle arrangement
[1,3,4,7l L&8,51 691.
If we have any permutation rr1 rrz . . . rr, of { 1,2,. . . , n}, every element is in a
unique cycle. For if we start with mu = m and look at ml = rrmor ml = rrm,,
etc., we must eventually come back to mk = TQ. (The numbers must repeat
sooner or later, and the first number to reappear must be mc because
we know the unique predecessors of the other numbers ml, ml, . . . , m-1 .)
Therefore every permutation defines a cycle arrangement. Conversely, every