## Conjugate of a partition

**Authors:** Jean-Christophe Filliâtre

**Topics:** Ghost code

**Tools:** Why3

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(* Conjugate of a partition. Author: Jean-Christophe Filliatre (CNRS) A partition of an integer n is a way of writing n as a sum of positive integers. For instance 7 = 3 + 2 + 2 + 1. Such a partition can be displayed as a Ferrer diagram: 3 * * * 2 * * 2 * * 1 * The conjugate of that a partition is another partition of 7, obtained by flipping the diagram above along its main diagonal. We get 4 * * * * 3 * * * 1 * Equivalently, this is the number of cells in each column of the original Ferrer diagram: 4 3 1 3 * * * 2 * * 2 * * 1 * The following program computes the conjugate of a partition. A partition is represented as an array of integers, sorted in non-increasing order, with a least a 0 at the end. This was inspired by this article: Franck Butelle, Florent Hivert, Micaela Mayero, and Frédéric Toumazet. Formal Proof of SCHUR Conjugate Function. In Proceedings of Calculemus 2010, pages 158-171. Springer-Verlag LNAI, 2010. *) module Conjugate use import int.Int use import ref.Refint use import array.Array predicate is_partition (a: array int) = (* at least one element *) a.length > 0 && (* sorted in non-increasing order *) (forall i j: int. 0 <= i <= j < a.length -> a[i] >= a[j]) && (* at least one 0 sentinel *) a[a.length - 1] = 0 (* values in a[0..n[ are > v, and a[n] <= v *) predicate numofgt (a: array int) (n: int) (v: int) = 0 <= n < a.length && (forall j: int. 0 <= j < n -> v < a[j]) && v >= a[n] predicate is_conjugate (a b: array int) = b.length > a[0] && forall j: int. 0 <= j < b.length -> numofgt a b[j] j let conjugate (a: array int) : array int requires { is_partition a } ensures { is_conjugate a result } = let b = Array.make (a[0] + 1) 0 in let partc = ref 0 in while a[!partc] <> 0 do invariant { 0 <= !partc < a.length } invariant { forall j: int. a[!partc] <= j < b.length -> numofgt a b[j] j } variant { a.length - !partc } 'L: let ghost start = !partc in let edge = a[!partc] in incr partc; while a[!partc] = edge do invariant { start <= !partc < a.length } invariant { forall j: int. start <= j < !partc -> a[j] = edge } variant { a.length - !partc } incr partc done; for i = a[!partc] to edge - 1 do invariant { forall j: int. edge <= j < b.length -> b[j] = (at b 'L)[j] } invariant { forall j: int. a[!partc] <= j < i -> b[j] = !partc } b[i] <- !partc done done; assert { a[!partc] = 0 }; b end module Test use import int.Int use import array.Array use import Conjugate let test () ensures { result.length >= 4 } = let a = make 5 0 in a[0] <- 3; a[1] <- 2; a[2] <- 2; a[3] <- 1; conjugate a exception BenchFailure let bench () raises { BenchFailure -> true } = let a = test () in if a[0] <> 4 then raise BenchFailure; if a[1] <> 3 then raise BenchFailure; if a[2] <> 1 then raise BenchFailure; if a[3] <> 0 then raise BenchFailure; a end (* Original C code from SCHUR Note that arrays are one-based (that code was translated from Pascal code where arrays were one-based) #define MAX 100 void conjgte (int A[MAX], int B[MAX]) { int i, partc = 1, edge = 0; while (A[partc] != 0) { edge = A[partc]; do partc = partc + 1; while (A[partc] == edge); for (i = A[partc] + 1; i <= edge; i++) B[i] = partc - 1; } } *)

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