Error Detection and Correction: Small Applications of Exclusive-Or

Transcription

1 Error Detection and Correction: Small Applications of Exclusive-Or Greg Plaxton Theory in Programming Practice, Fall 2005 Department of Computer Science University of Texas at Austin

2 Exclusive-Or (XOR, ) For Booleans p and q, p q is true if exactly one of p and q is true, and false otherwise When applied to bits, we treat 0 as false and 1 as true Here are some basic properties of Commutativity: x y = y x Associativity: (x y) z = x (y z) Zero and complementation: x 0 = x, x 1 = x Inverse: x x = 0, x x = 1

3 Applying to Words Often we will be working with binary strings of a particular length, which we call words Let x and y be two words We interpret x y as the binary string of length k with ith bit equal to the XOR of the ith bits of x and y, for all i The value of the ith output bit only depends on the values of the ith input bits Because of this, many properties of on words follow easily from the corresponding property of on bits Example: Associativity, commutativity of on words Similar comments apply to other binary operators such as and

4 Applying to Nonnegative Integers Convert each operand to its binary representation Pad the shorter operand with leading zeros Compute the output word as on the previous slide Convert the output word to the corresponding nonnegative integer Similar comments apply to other binary operators such as and

5 Outline Small applications of : Bit selection; toggling; exchange; storage for doubly-linked lists; Hamming distance See the course packet for a detailed discussion of each of these applications Here we will sketch alternate proofs based on the following two-step approach: First, we give a proof for the special case of words of length 1 (i.e., bits) Then, we argue that the proof for the general case (i.e., words of arbitrary length) follows

6 Selecting a Bit Suppose we are given three Booleans (bits) x, y, and u We d like to compute a Boolean w that is equal to x if u is false and to y if u is true We can set w to ((x y) u) x)

7 Forming a Word by Selecting Bits from Two Given Words Suppose we are given three words x, y, and u We d like to compute the word w with ith bit equal to that of x if the ith bit of u is 0, and to that of y if the ith bit of u is 1, for all i Since and are bitwise operators, we can set w to ((x y) u) x) as on the previous slide

8 Toggling a Boolean Variable We d like to write a single assignment statement S that toggles the value of a Boolean variable x between two given Boolean values p and q If x = p then S should set x to q If x = q then S should set x to p Solution: Initialize an auxiliary variable t to p q and define S as x := x t

9 Word Toggling We d like to write a single assignment statement S that toggles the value of a word x between two given words p and q If x = p then S should set x to q If x = q then S should set x to p Since is a bitwise operator, we can solve the problem as on the previous slide

10 Exchanging the Values of Two Boolean Variables We d like to swap the values of two Boolean variables x and y without using a temporary variable Solution: Set x to x y, then y to x y, and then x to x y

11 Exchanging the Values of Two Words We d like to swap the values of two words x and y without using a temporary variable Since is a bitwise operator, we can solve the problem as on the previous slide

12 Storage for Doubly-Linked Lists Normally each node of a doubly-linked list has two pointers, a left pointer and a right pointer Can we get by with just one? The trick: Just keep the XOR of the left and right pointers Works if you always arrive at a node from either the left or right neighbor (or via the head or tail pointer) Won t work if your application involves other pointers outside the list data structure that point to nodes in the list But remember Dijkstra s advice: Avoid clever tricks like the plague!

13 Hamming Distance The Hamming distance between two words x and y is the number of 1s in x y Equal to the number of bit flips needed to transform x into y (or vice versa) Claim: For any nonnegative integer k, Hamming distance defines a metric space over the set of all words of length k Before attempting to prove this claim, we ll need to look at the definition of a metric space

14 Metric Space A distance function over a set S is a function from S S to the reals A distance function d defined on a set S is metric if it satisfies the following properties for all x, y, and z in S Nonnegativity: d(x, y) 0 Distinctness: d(x, y) = 0 iff x = y Symmetry: d(x, y) = d(y, x) Triangle inequality: d(x, y) + d(y, z) d(x, z) A metric space is a set with an associated metric distance function Numerous applications give rise to computational problems defined over metric spaces Example: Automatic clustering of web pages

15 An Observation Concerning Metric Spaces Suppose that (S, d) is a metric space Let k be a nonnegative integer and let d (x, y) = 1 i k d(x i, y i ) for all x = (x 1,..., x k ) and y = (y 1,..., y k ) in S k Observation (not hard to prove; try it!): (S k, d ) is a metric space How can we use this observation to prove that Hamming distance defines a metric space over the set of all words of a given length?

16 Hamming Distance Defines a Metric Space Claim: For any nonnegative integer k, Hamming distance defines a metric space over the set of all words of length k By the lemma stated on the previous slide, the claim follows if we can establish that Hamming distance defines a metric space over {0, 1}, i.e., over the set of all words of length 1 This is a straightforward exercise