Hamming code is one of the Computer Science\/Telecommunication classics.<\/p>\n
In this article, we’ll revisit the topic and implement a stateless Hamming(7,4) encoder using Kotlin.<\/p>\n
Our communication channels and data storages are error-prone<\/a> – bits can flip due to various things like electric\/magnetic interferences, background radiation, or just because of the low quality of materials used.<\/p>\n Since the neutron flux<\/a> is ~300 higher at around 10km altitude, a particular attention is necessary when dealing with systems operating at high altitudes – the case study of the\u00a0Cassini-Huygens<\/a> proves it – in space, a number of reported errors was over four times bigger than on earth, hence the need for efficient error correction.<\/p>\n Richard Hamming<\/a>‘s Code is one of the solutions to the problem. It’s a perfect code<\/em> (at least, according to Hamming’s definition) which can expose\u00a0and correct errors in transmitted messages.<\/p>\n Simply put, it adds metadata to the message (in the form of parity bits) that can be used for validation and correction of errors in messages.<\/p>\n I bet you already wondered what did (7,4) in “Hamming(7,4)” mean.<\/p>\n Simply put, N and M in “Hamming(N, M)” represent\u00a0the block length and the message size – so, (7,4) means that it\u00a0encodes four bits into seven bits by adding three additional parity bits<\/strong> – as simple as that.<\/p>\n This particular version can detect and correct single-bit errors, and detect (but not correct) double-bit errors.<\/p>\n In the Hamming’s codeword, parity bits always occupy all indexes that are powers of two (if we use 1-based-indexing).<\/p>\n So, if our initial message is 1111,\u00a0<\/em>the codeword will look somewhat like [][]1[]111<\/em> – with three parity bits\u00a0for us to fill in.<\/p>\n If we want to calculate the n-th parity bit, we start on its position in a codeword, we take n elements, skip n elements, take n elements, skip n elements… and so on.\u00a0If the number of taken ones is odd, we set the parity bit to one, otherwise zero.<\/p>\n In our case:<\/p>\n And that’s all – the codeword is 1111111<\/em>.<\/p>\n In this case, it might be tempting to think that every sequence containing only ones will be encoded to another sequence comprising only ones… but that’s not the case… but every message containing only zeros will always be encoded to zeros exclusively.<\/p>\n First things first, we can leverage Type Driven Development for making our life easier when working with Strings representing raw and encoded messages:<\/p>\n Using this approach, it’ll be slightly harder to mix them up.<\/p>\n We’ll need a method for calculating the encoded codeword size for a given message. In this case, we simply find the lowest number of parity pairs that can cover the given message:<\/p>\n Next, we’ll need a method for calculating parity and data bits at given indexes for a given message:<\/p>\n Where parityIndicesSequence()<\/em> is defined as:<\/p>\n Now, we can put it all together to form the actual solution, which simply is simply going through the whole codeword and filling it with parity bits and actual data:<\/p>\n Note that isPowerOfTwo()<\/em> is our custom extension function and is not available out-of-the-box in Kotlin:<\/p>\n The interesting thing is that the whole computation can be inlined to a single Goliath sequence:<\/p>\n Not the most readable version, but interesting to have a look.<\/p>\n We can verify that the implementation works as expected by leveraging JUnit5 and Parameterized Tests:<\/p>\n … and by using a home-made property testing:<\/p>\n The most important property of this implementation is statelessness – it could be achieved by making sure that we’re using only pure functions and avoiding shared mutable state – all necessary data is always passed explicitly as input parameters and not held in any form of internal state.<\/p>\n Unfortunately, it results in some repetition and performance overhead that could’ve been avoided if we’re just modifying one mutable list and passing it around… but now we can utilize our resources wiser by parallelizing the whole operation – which should result in a performance improvement.<\/p>\n Without running the code that’s just wishful thinking so let’s do that.<\/p>\n We can parallelize the operation (naively) using Java 8’s parallel streams:<\/p>\n To not give the sequential implementation an unfair advantage (no toList()<\/em> conversion so far), we’ll need to change the implementation slightly:<\/p>\n And now, we can perform some benchmarking using JMH (message.size == 10_000):<\/p>\n As we can see, we can notice a major performance improvement in favor of the parallelized implementation – of course; results might drastically change because of various factors so do not think that we’ve found a silver bullet – they do not exist.<\/p>\n For example, here’re the results for encoding a very short message (message.size == 10)):<\/p>\n In this case, the overhead of splitting the operation among multiple threads makes the parallelized implementation perform eight times slower(sic!).<\/p>\n Here’s the full table for the reference:<\/p>\n We saw how to implement a thread-safe Hamming(7,4)<\/a> encoder using Kotlin and what parallelization can potentially give us.<\/p>\n In the second part of the article, we’ll implement a Hamming decoder and see how we can correct single-bit errors and detect double-bit ones.<\/p>\nA Brief Explanation<\/h3>\n
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Encoding<\/h2>\n
data class EncodedString(val value: String)\n\ndata class BinaryString(val value: String)<\/pre>\n
fun codewordSize(msgLength: Int) = generateSequence(2) { it + 1 }\n .first { r -> msgLength + r + 1 <= (1 shl r) } + msgLength<\/pre>\nfun getParityBit(codeWordIndex: Int, msg: BinaryString) =\n parityIndicesSequence(codeWordIndex, codewordSize(msg.value.length))\n .map { getDataBit(it, msg).toInt() }\n .reduce { a, b -> a xor b }\n .toString()\n\nfun getDataBit(ind: Int, input: BinaryString) = input\n .value[ind - Integer.toBinaryString(ind).length].toString()<\/pre>\nfun parityIndicesSequence(start: Int, endEx: Int) = generateSequence(start) { it + 1 }\n .take(endEx - start)\n .filterIndexed { i, _ -> i % ((2 * (start + 1))) < start + 1 }\n .drop(1) \/\/ ignore the parity bit<\/pre>\noverride fun encode(input: BinaryString): EncodedString {\n fun toHammingCodeValue(it: Int, input: BinaryString) =\n when ((it + 1).isPowerOfTwo()) {\n true -> hammingHelper.getParityBit(it, input)\n false -> hammingHelper.getDataBit(it, input)\n }\n\n return hammingHelper.getHammingCodewordIndices(input.value.length)\n .map { toHammingCodeValue(it, input) }\n .joinToString(\"\")\n .let(::EncodedString)\n}<\/pre>\ninternal fun Int.isPowerOfTwo() = this != 0 && this and this - 1 == 0<\/pre>\n
Inlined<\/h4>\n
override fun encode(input: BinaryString) = generateSequence(0) { it + 1 }\n .take(generateSequence(2) { it + 1 }\n .first { r -> input.value.length + r + 1 <= (1 shl r) } + input.value.length)\n .map {\n when ((it + 1).isPowerOfTwo()) {\n true -> generateSequence(it) { it + 1 }\n .take(generateSequence(2) { it + 1 }\n .first { r -> input.value.length + r + 1 <= (1 shl r) } + input.value.length - it)\n .filterIndexed { i, _ -> i % ((2 * (it + 1))) < it + 1 }\n .drop(1)\n .map {\n input\n .value[it - Integer.toBinaryString(it).length].toString().toInt()\n }\n .reduce { a, b -> a xor b }\n .toString()\n false -> input\n .value[it - Integer.toBinaryString(it).length].toString()\n }\n }\n .joinToString(\"\")\n .let(::EncodedString)<\/pre>\nIn Action<\/h3>\n
@ParameterizedTest(name = \"{0} should be encoded to {1}\")\n@CsvSource(\n \"1,111\",\n \"01,10011\",\n \"11,01111\",\n \"1001000,00110010000\",\n \"1100001,10111001001\",\n \"1101101,11101010101\",\n \"1101001,01101011001\",\n \"1101110,01101010110\",\n \"1100111,01111001111\",\n \"0100000,10011000000\",\n \"1100011,11111000011\",\n \"1101111,10101011111\",\n \"1100100,11111001100\",\n \"1100101,00111000101\",\n \"10011010,011100101010\")\nfun shouldEncode(first: String, second: String) {\n assertThat(sut.encode(BinaryString(first)))\n .isEqualTo(EncodedString(second))\n}<\/pre>\n@Test\n@DisplayName(\"should always encode zeros to zeros\")\nfun shouldEncodeZeros() {\n generateSequence(\"0\") { it + \"0\" }\n .take(1000)\n .map { sut.encode(BinaryString(it)).value }\n .forEach {\n assertThat(it).doesNotContain(\"1\")\n }\n}<\/pre>\nGoing Parallel<\/span><\/h2>\n
override fun encode(input: BinaryString) = hammingHelper.getHammingCodewordIndices(input.value.length)\n .toList().parallelStream()\n .map { toHammingCodeValue(it, input) }\n .reduce(\"\") { t, u -> t + u }\n .let(::EncodedString)<\/pre>\noverride fun encode(input: BinaryString) = hammingHelper.getHammingCodewordIndices(input.value.length)\n .toList().stream() \/\/ to be fair.\n .map { toHammingCodeValue(it, input) }\n .reduce(\"\") { t, u -> t + u }\n .let(::EncodedString)<\/pre>\nResult \"com.pivovarit.hamming.benchmarks.SimpleBenchmark.parallel\":\n 3.690 \u00b1(99.9%) 0.018 ms\/op [Average]\n (min, avg, max) = (3.524, 3.690, 3.974), stdev = 0.076\n CI (99.9%): [3.672, 3.708] (assumes normal distribution)\n\nResult \"com.pivovarit.hamming.benchmarks.SimpleBenchmark.sequential\":\n 10.877 \u00b1(99.9%) 0.097 ms\/op [Average]\n (min, avg, max) = (10.482, 10.877, 13.498), stdev = 0.410\n CI (99.9%): [10.780, 10.974] (assumes normal distribution)\n\n\n# Run complete. Total time: 00:15:14\n\nBenchmark Mode Cnt Score Error Units\nSimpleBenchmark.parallel avgt 200 3.690<\/strong> \u00b1 0.018 ms\/op<\/strong>\nSimpleBenchmark.sequential avgt 200 10.877<\/strong> \u00b1 0.097 ms\/op<\/strong><\/pre>\n
Benchmark Mode Cnt Score Error Units\nSimpleBenchmark.parallel avgt 200 0.024<\/strong> \u00b1 0.001 ms\/op<\/strong>\nSimpleBenchmark.sequential avgt 200 0.003<\/strong> \u00b1 0.001 ms\/op<\/strong><\/pre>\n
Benchmark (messageSize) Mode Cnt Score Error Units\nBenchmark.parallel 10 avgt 200 0.022 \u00b1 0.001 ms\/op\nBenchmark.sequential 10 avgt 200 0.003 \u00b1 0.001 ms\/op\n \nBenchmark.parallel 100 avgt 200 0.038 \u00b1 0.001 ms\/op\nBenchmark.sequential 100 avgt 200 0.031 \u00b1 0.001 ms\/op\n\nBenchmark.parallel 1000 avgt 200 0.273 \u00b1 0.011 ms\/op \nBenchmark.sequential 1000 avgt 200 0.470 \u00b1 0.008 ms\/op\n\nBenchmark.parallel 10000 avgt 200 3.731 \u00b1 0.047 ms\/op\nBenchmark.sequential 10000 avgt 200 12.425 \u00b1 0.336 ms\/op\n<\/pre>\n
Conclusion<\/h2>\n