Use \pi[x...y].

diff --git a/PLAN b/PLAN
index 15b9160001f460dad5a8a618b925a7671bcc0d5b..432e52d3e12c3dd6e4ab2e83aa03250420a18ae6 100644 (file)
--- a/PLAN
+++ b/PLAN
@@ -68,7 +68,6 @@ Ranking:
 - the general perspective: is it only a technical trick?
 - ranking of permutations on general sets, relationship with integer sorting
 - mention approximation of permanent
-- use \pi[x...y]
 
 Notation:
 

diff --git a/rank.tex b/rank.tex
index 3005c42c9d6e60ee10437d84991c91e8dccb2da1..5be516dae8c24ca816ee4cd25a8bdd09dfb8dda8 100644 (file)
--- a/rank.tex
+++ b/rank.tex
@@ -79,9 +79,10 @@ RAM data structures, yields a~linear-time algorithm for lexicographic
 \nota\id{brackets}%
 We will view permutations on a~finite set $A\subseteq {\bb N}$ as ordered $\vert A\vert$-tuples
 (in other words, arrays) containing every element of~$A$ exactly once. We will
-use square brackets to index these tuples: $\pi=(\pi[1],\ldots,\pi[\vert A\vert])$.
-The corresponding lexicographic ranking and unranking functions will be denoted by~$L(\pi,A)$
-and $L^{-1}(i,A)$ respectively.
+use square brackets to index these tuples: $\pi=(\pi[1],\ldots,\pi[\vert A\vert])$,
+and sub-tuples: $\pi[i\ldots j] = (\pi[i],\ldots,\pi[j])$.
+The lexicographic ranking and unranking functions for the permutations on~$A$
+will be denoted by~$L(\pi,A)$ and $L^{-1}(i,A)$ respectively.
 
 \obs\id{permrec}%
 Let us first observe that permutations have a simple recursive structure.
@@ -89,10 +90,10 @@ If we fix the first element $\pi[1]$ of a~permutation~$\pi$ on the set~$[n]$, th
 elements $\pi[2], \ldots, \pi[n]$ form a~permutation on $[n]-\{\pi[1]\} = \{1,\ldots,\pi[1]-1,\pi[1]+1,\ldots,n\}$.
 The lexicographic order of two permutations $\pi$ and~$\pi'$ on the original set is then determined
 by $\pi[1]$ and $\pi'[1]$ and only if these elements are equal, it is decided
-by the lexicographic comparison of permutations $(\pi[2],\ldots,\pi[n])$ and
-$(\pi'[2],\ldots,\pi'[n])$. Moreover, for fixed~$\pi[1]$ all permutations on
-the smaller set occur exactly once, so the rank of $\pi$ is $(\pi[1]-1)\cdot
-(n-1)!$ plus the rank of $(\pi[2],\ldots,\pi[n])$.
+by the lexicographic comparison of permutations $\pi[2\ldots n]$ and $\pi'[2\ldots n]$.
+Moreover, for fixed~$\pi[1]$ all permutations on the smaller set occur exactly
+once, so the rank of $\pi$ is $(\pi[1]-1)\cdot (n-1)!$ plus the rank of
+$\pi[2\ldots n]$.
 
 This gives us a~reduction from (un)ranking of permutations on $[n]$ to (un)ranking
 of permutations on a $(n-1)$-element set, which suggests a straightforward
@@ -121,7 +122,7 @@ translates to the following recursive algorithms for both ranking and unranking
 \>We can call $\<Rank>(\pi,1,n,[n])$ for ranking on~$[n]$, i.e., to calculate
 $L(\pi,[n])$.
 
-\alg $\<Unrank>(j,i,n,A)$: Return an~array~$\pi$ such that $\pi[i,\ldots,n]$ is the $j$-th permutation on~$A$.
+\alg $\<Unrank>(j,i,n,A)$: Return an~array~$\pi$ such that $\pi[i\ldots n]$ is the $j$-th permutation on~$A$.
 \id{unrankalg}
 \algo
 \:If $i>n$, return $(0,\ldots,0)$.
@@ -496,7 +497,7 @@ with its $i$-th row and $j$-th column removed.
 \obs
 Let us consider a~permutation $\pi\in{\cal P}_A$ and $n=\vert A\vert$.
 When we fix the value of the element $\pi[1]$, the remaining elements form
-a~permutation $\pi'=(\pi[2],\ldots,\pi[n])$ on the set~$A'=A\setminus\{\pi[1]\}$.
+a~permutation $\pi'=\pi[2\ldots n]$ on the set~$A'=A\setminus\{\pi[1]\}$.
 The permutation~$\pi$ satisfies the restriction matrix~$M$ if and only if
 $M[1,a]=1$ for $a=R_A(\pi[1])$ and $\pi'$ satisfies a~restriction matrix~$M'=M^{1,a}$.
 This translates to the following counterparts of algorithms \ref{rankalg}