BUGFIX-66

Bug 57: Backward Over The Edge

This is a data structure for backtrack programming. It maintains a set of integers in [0, n).
NewResidue(n) returns a set containing integers 0, 1, ..., n-1. Slice returns all integers in the set.
If 0 <= x < n is not in the set: Contains(x) returns false; Remove(x) returns false with no other effect.
If 0 <= x < n is in the set: Contains(x) returns true; Remove(x) removes x from the set and returns true.
Undo reverts the effect of the most recent call to Remove that returned true and has not yet been reverted.
There is a stack of removed integers. Remove(x) removes x from the set and pushes x onto the stack.
If the stack is not empty, Undo pops x off the stack, inserts x into the set, and returns true.
If the stack is empty, Undo returns false with no other effect.

Fix The Tiny Bug In This Go Code:

type Residue struct { elems []int count int split int } func NewResidue(n int) *Residue { elems := make([]int, n+n) for i := 0; i < n; i++ { elems[0+i] = i elems[n+i] = i } return &Residue{ elems: elems, count: n, split: n, } } func (r *Residue) Slice() []int { return r.elems[:r.count] } func (r *Residue) Contains(x int) bool { return r.elems[r.split+x] < r.count } func (r *Residue) Remove(x int) bool { at := r.elems[r.split+x] if at >= r.count { return false } if r.count--; at < r.count { y := r.elems[r.count] r.elems[r.count] = x r.elems[at] = y r.elems[r.split+x] = r.count r.elems[r.split+y] = at } return true } func (r *Residue) Undo() bool { if r.count == r.split { return false } r.split++ return true }

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