logic

问题 I've encountered some obj-c code and I'm wondering if there's a way to simplify it: #if ( A && !(B || C)) || ( B || C ) is this the same as? #if ( A || B || C ) If not, is there another way to formulate it that would be easier to read? [edit] I tried the truth table before asking the question, but thought I had to be missing something because I doubted that Foundation.framework/Foundation.h would employ this more complex form. Is there a good reason for it? Here's the original code (from

问题 The idea behind Data.Constraint.Forall, as I understand it, is to use coercion in the implementation, but ensure safety using the type system. I have two questions regarding the latter. Why do we need two skolem variables — A and B? I would imagine that if a constraint is satisfied by an «unknown» type, then it is polymorphic. How does the second type give more safety? Why are these types called skolem variables? I thought that skolemnization is used to remove existential quantification, and

问题 The idea behind Data.Constraint.Forall, as I understand it, is to use coercion in the implementation, but ensure safety using the type system. I have two questions regarding the latter. Why do we need two skolem variables — A and B? I would imagine that if a constraint is satisfied by an «unknown» type, then it is polymorphic. How does the second type give more safety? Why are these types called skolem variables? I thought that skolemnization is used to remove existential quantification, and

问题 We are building a survey engine for our internal use. I would like to know how to persist the question branching logic into the database? Any body done this before or any ideas on the schema for the database? If the user responses with an answer, we need to skip to the next questions based on the logic added to the questions Each question can have multiple logic added to it. For eg: Question: Is it Sunny, Raining or Cloudy? Answer: Raining. The next question should be based on the previous

问题 I'm new but have managed to learn a lot and create a pretty awesome (I hope) app that's near completion. One of my last tasks is to create a PDF of dynamically generated user data. It has been the most frustrating part of this whole process as there is no real modern clear cut template or guide. Apple's documentation isn't very descriptive (and some parts I don't understand) and the Q/A here on stack and examples on Google all seem very case specific. I thought I almost had it by using a

问题 I've been able to create a proof that shows the maximum total nodes in a tree is equal to n = 2^(h+1) - 1 and logically I know that the height of a binary tree is log n (can draw it out to see) but I'm having trouble constructing a formal proof to show that a tree with n leaves has "at least" log n. Every proof I've come across or been able to put together always deals with perfect binary trees, but I need something for any situation. Any tips to lead me in the right direction? 回答1: Lemma :

问题 I have a couple time_tables in this array. There are four time_tables that are related to each other in a linear way by their start_location - end_location and start_date - end_date . When the first time_table ends, the other time_table starts, and so on. My code: arr = [ { name: 01, start_date: '2014-04-24 22:03:00', start_location: 'A', end_date: '2014-04-24 22:10:00', end_location: 'B' }, { name: 05, start_date: '2014-04-24 22:10:00', start_location: 'C', end_date: '2014-04-24 23:10:00',

问题 I was playing around on HackerEarth and I came across this issue. What I try to do is to compare the strings and check if they have the same characters or not. var string = "" while let thing = readLine() { string += thing + " " } var arrayStr = string.split(separator: " ").map{String(($0))} var firstString = [String]() var secondString = [String]() var cas = arrayStr[0] for i in 1..<arrayStr.count { if i % 2 != 0 { firstString.append(String(arrayStr[i])) } else { secondString.append(String

问题 I have a list of 75200 words. I need to give a 'unique' id to each word, and the length of each id could be 3 letters or less. I can use numbers, letters or even symbols but the max length is 3. Below is my code. import java.io.*; import java.util.*; public class HashCreator { private Map completedWordMap; private String [] simpleLetters = {"a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"}; private String[] symbols = {"!","@","#","$","%",

问题 I have a list of 75200 words. I need to give a 'unique' id to each word, and the length of each id could be 3 letters or less. I can use numbers, letters or even symbols but the max length is 3. Below is my code. import java.io.*; import java.util.*; public class HashCreator { private Map completedWordMap; private String [] simpleLetters = {"a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"}; private String[] symbols = {"!","@","#","$","%",