logic

问题 I'm trying to generate all possible 'ways' from one element in matrix to another, Main conditional says that Element in matrix can be connected with other only if elements share at least one corner so in that matrix [[1,2,3] [5,4,6] [8,9,7]] 1 can be only conected with 2,4,5 but 4 is conected with all elements. Is it possible to represent that list as graph without using attract? Or maybe I can find out any easier way to do it Thanks for all answers. Ok i set forth :-) Now with predicates I

问题 I am having trouble figuring out the proper implementation of an if else statement in my application. Currently I am using the conditional operators && and || for comparison, but when my application is launched in the marketplace my code is not operating as planned. Essentially the way it should work is that if the count is greater than 100 and the application is still a trial license then ask to upgrade, otherwise if the count is less than or equal to 100 and the application is still a trial

问题 I'm trying to construct a formal proof for ((p ⇒ q) ⇒ p) ⇒ p. in Fitch. I know this is true, but how do I prove it? I can only use And Intro, And Elim, Or Inro, Or Elim, Neg Intro, Neg Elim, Impl Intro, Impl Elim, Biconditional Intro, and Biconditional Elim. 回答1: The following proof uses Klement's Fitch-style proof checker. Description of the symbols and the rules are in forallx . Links to both are below. A slightly different version is on Philosophy Stack Exchange: https://philosophy

问题 Suppose I have following structure M = (S, R, L) where S = {s0, s1, s2} is the set of possible states, R is a transition relation such that: s0 -> s1, s0 -> s2, s1 -> s0, s1 -> s2, and s2 -> s2, and L is the labeling function for each state defined by: L(s0) = {p, q}, L(s1) = {q, r}, and L(s2) = {r}. I am using notations describe in Logic in Computer Science textbook by Huth and Ryan. Clearly, from such model, we have X r is satisfied in s0 (the initial state), since r is satisfied in s1 and

问题 I've been having problems with logic and i need help addressing this issue. I come up with these types of problems all the times and i cant seem to set the variables at the right places to get the right logic to occur. basically what i want in the code is that if someone inputs the wrong (addition) answer and clicks go you should push to an array the questions and the answer. so i could get a list of the questions and answers in the array after the user finishes answering a bunch of questions

问题 I have a regex /^(?!.*([0-9])\1{2})(?=.*[0-9])(?=.*[A-Z])(?=.*[a-z])[0-9a-zA-Z]*$/ I would like to modify this to to prevent ascending or descending sequence of more than 3 characters or digits. I know this is not the most sensible solution but this is the requirement so I don't mind hard coding all the possible sequence. 回答1: My solution /^(?!.*([0-9])\1{2})(?=.*[0-9])(?=.*[A-Z])(?=.*[a-z])(?!.*0123|.*1234|.*2345|.*3456|.*4567|.*5678|.*6789|.*3210|.*4321|.*5432|.*6543|.*7654|.*8765|.*9876|.

问题 I am trying to write some futuristic card game simulator. The game itself has complicated rules and a lot of cards (~1000 total). There are few types of card (<10, types like spells, healing, moving, attacking, defending etc..). Every card belongs to exacly one type but the actions they do can be various, ex: Card1 belongs to type Healing, its action is to heal 1 Health point, Card2 also belong to type Healing, but its action is heal and ex. drop another card. Another thing is that the card

问题 Using Mustache js (a logic less templating) is there a way i can achieve switch case? this i need because a class is assigned to dom element based on the value for example: switch(tasks.Count) { case 0: element.Class = "no-tasks"; break; case 1: element.Class = "one-tasks"; break; . . . } that's the code i got now, how do i transform it to template( I Believe having methods on model being rendered is the one option) But adding methods to determine which class to use is an overkill and besides

问题 I can't see why the following logic isn't working: if (cursorCount > 1 && (!"x".equals(componentType) || !"y".equals(componentType))){ message.append("s"); } So I want to to print 's' if the cursor count is over 1 but only when the componentType does not equal x or y.. Seems to work for the cases that are y but not x interestingly. Confused.com! :) 回答1: Try if (cursorCount > 1 && !("x".equals(componentType) || "y".equals(componentType))) Equivalently you can do if (cursorCount > 1 && !"x"

问题 In my code I am getting a no such element exception when I enter a word. It does output the word correctly and the hangman, but it also crashses after doing so. What is causing this and how could I fix it? Here is the start of the error: Exception in thread "AWT-EventQueue-1" java.util.NoSuchElementException: No line found at java.util.Scanner.nextLine(Unknown Source) at Hangman.paint(Hangman.java:50) at javax.swing.RepaintManager$3.run(Unknown Source) at javax.swing.RepaintManager$3.run