1. The main result.
Consider a stack of books, containing both white and black books.
Suppose that we want to sort them out, putting the white books on the right, and the black books on the left (fig. 1). This will be done by a finite sequence of elementary transpositions. In other words, if we have a stack of all black books of lengthafollowed by a stack of all white books of lengthb, we are allowed to reverse their order at the cost ofa1b. We are interested in a lower bound on the total cost of the rearrangement.
Assume that initially the white and black books are highly mixed. By this we mean that inside every segment of lengthe one can find at least ke white books and at leastkeblack books, for a givenk]0 , 1[. We want to show that, as eK0, the cost of sorting the books grows like NlogeN.
We reformulate the problem in more precise terms. Consider the set F of all piecewise constant functions f : [ 0 , 1 ]O]0 , 1(. We think of Bu]x; f(x)41( as the set of black books. We say that g is obtained from fby an elementary transpositionif the following holds. For some y[ 0 , 1[ and a,bD0, one has
f(x)4./
´ 1 0
if yExEy1a,
if y1aExEy1a1b,
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E-mail: bressanHsissa.it
Figure 1
g(x)4
. /
´
0 1 f(x)
if yExEy1b,
if y1bExEy1b1a, if x[y,y1a1b] .
In this case, we define thecost of the transposition asa1b. By a rear- rangement of the function f we mean a sequence f0,f1,R,fnF such that
(i) f04f,
(ii) fn is nondecreasing,
(iii) each fj is obtained from fj21 by an elementary transposi- tion.
The function fn is then characterized as
fn(x)4 . /
´ 0 1
if 0ExE12j,
if 12jExE1 , ju
0 1
f(s)ds. (1.1)
G(f0,R,fn)FCkNlogeN (1.3)
for some constant Ck independent of f,e.
PROOF. For each s[ 0 , k], define V(s) as the minimum cost of any sequence of transpositionsf0,f1,R,fnsuch thatf04fandfnf1 on some interval with length Fs. By (1.1)-(1.2) it follows
G(f0,R,fn)FV( 12j)FV(k/2 ) , (1.4)
V(s)Fs2e, (1.5)
V(s)F min
0EsEsV(s2s)1V(s)1k2s1( 12k2)s sDe. (1.6)
The last inequality is due to the fact that, in order to create a stack of white books of lengthFs, we need first to form a stack of white books with length s2s, removing toward the right all the black books that were initially located within them. The amount of such black books can be estimated as
Fk((s2s)2( 12k)e)Fk2(s2s)
for s2sFe. Then we need to form a second stack of white books of length s, and finally join together the two stacks, switching the second white stack with the black stack in the middle, at the cost ofs1k2(s2 2s). Notice that the possible choices2sEe, not considered in the previ- ous argument, certainly does not attain the minimum. Indeed, its cost is
FV(s2e)1( 12k2)(s2e) .
Using (1.5)-(1.6) one finds
V(s)F( 11k2)s22e.
Continuing by induction, for every integer n we obtain V(s)F( 11nk2)s22ne.
For each eD0 we now define n(e) as the largest integer n such that ( 11nk2)k
2 F2n11e. (1.7)
Of course, this implies
G(f0,R,fn)FV(k/2 )F k3 4 n(e) .
By (1.7), an elementary estimate yieldsn(e)FCk8NlogeN, for some con- stant Ck8. This proves the lemma.
2. A conjecture on mixing vector fields.
It is interesting to speculate whether a continuous version of the above lemma can be true, for mixing vector fields on a compact manifold.
To fix the ideas, consider the two-dimensional torusKuR2/Z2, with co- ordinates x4(x1,x2)[ 0 , 1[3[ 0 , 1[ . Consider the set (fig. 2)
Au](x1,x2); 0Gx2G1 /2(%K.
Figure 2
from constant. Indeed, it should steer nearby points to quite different lo- cations. The previous lemma thus motivates the following
CONJECTURE 1. Let f4f(t,x) be a smooth vector field on K, and assume that the associated flow Ct is nearly incompressible so that, for some k8 D0,
k8meas (V)Gmeas (Ct(V))G 1 k8
meas (V) , (2.2)
for all V%K and t[ 0 , 1 ]. Then there exists a constant C depending only on k,k8 such that, if f mixes the set A up to scale e, then
0 1
K
N˜xfNdx dtFCNlogeN. (2.3)
Reversing time, the above conjecture says that, in order to rearrange the setA8uC×(A), one needs a vector field whose cost (measured by the total variation) grows logarithmically withe. A closely related conjecture can be stated as follows.
CONJECTURE 2. Consider two setsX,Y%Rnwith unit measure. As- sume that they aree-close, in the sense that there exists a measure pre- serving transformation W:XOY such that
Nx2W(x)N Ge for all xX.
Let f4f(t,x) be a smooth vector field, whose flow (t,x)OCt(x) is nearly incompressible and eventually separatesXfromY. More precise-
ly we assume that (2.2) holds and that, at some time TD0, NCT(x)2CT(y)N F1 for all xX, yY.
Then the total variation of f must be large:
0 T
Rn
N˜xfNdx dtFCNlogeN. (2.4)
R E F E R E N C E S
[1] A. BRESSAN, An ill posed Cauchy problem for a hyperbolic system in two space dimensions, Rend. Sem. Mat. Univ. Padova, in this volume p. 103.
Manoscritto pervenuto in redazione il 3 dicembre 2002.
