![]() Note that for the above grammar and algorithm, the probability of expanding an expression E into a terminal symbol I is only p = 1/3, while the probability to expand an expression into two further expressions is 1-p = 2/3. All Expression subclasses are required to implement an evaluate method, which works recursively on the tree structure defined by these objects and effectively implements the composite pattern. The latter two then would have a leftExpression and rightExpression. I assume you would choose to represent an expression with an interface Expression which is implemented by classes IntExpression, AddExpression and MultiplyExpression. Here's an example of the application of this algorithms: E Replace all terminal symbols I by random integers.Repeat steps 2 - 4 until only terminal symbols are left.Choose uniformly at random one of the production rules for that symbol, and apply it.Choose uniformly at random one of the non-terminal symbols.Start with E as the single symbol of the output word.Based on this definition, we can randomly build a valid arithmetic as follows: The above definition for E has three production rules. Here, E is an expression (i.e., a word of your language) and I is a terminal symbol (i.e., it's not expanded any further) representing an integer. Here's a formal description of a simplified algebraic grammar supporting only addition and multiplication: E -> I You are looking to randomly generate words (algebraic expression) from a given language (the infinite set of all syntactically correct algebraic expressions). Here's a theoretic interpretation of your problem. If my question needs any clarification, please ask in the comments. If your answer gives a solution handling this one, that's great. Those examples did not include the : operator, as I only want to manipulate ints, and this operator adds many verifications. I'm not looking for anything language-related, but for the record, I'm thinking of implementing it in Objective-C, as that's the language I'm most working with recently. If you take that in consideration in your answer, that's great. This can be done either after the expression is generated, or during its creation. with the beginning of the algorithm, or a general structure of it.Īlso note that I will have to evaluate those expressions. If you see a good way to start, I'd appreciate a lead in the right direction, e.g. I'd like to know which way you believe is the best to go, between the solutions I considered, and your own ideas. Using a crazy "for-type" loop (surely the worst).But still, I thought of a couple solutions: I can't say I tried those ideas, because I didn't really want to waste much time going in a direction that had no chance of working in the first place. I'm not even sure a single algorithm could give me the last two. But I'm having some trouble getting started with something that could create one of the hard and harder examples. Random ints separated by random operators, nothing crazy here. The easy and medium ones are pretty straight-forward. Here are some example expressions I want to generate randomly: 4 + 2 So, I decided to give it a shot and elaborate an algorithm, before coding it to test it. Yes, we will have a function with two-parameter, let’s say getRndInteger().I have this idea running around in my head, to generate and evaluate random mathematical expressions. Get the random number on the basis of minimum and maximum numbers. ![]() Var randomNumber = Math.floor(Math.random() * 10) + 1 ĥ. Since we know that normal Math.random() will give us a number between 0 to 1, we can multiply this by 10 and add 1 to get the number always greater or equals to 1. How we can always more than equal to 1 as randomly generated numbers. But most of the time, we will get the two digits numbers. We will get a random number between 0 to 99. Var randomNumber = Math.floor(Math.random() * 100) Since we have multiplied by 10, this will give the output always in a single digit.ģ. The output of the above code will vary between 0 to 9. Var randomNumber = Math.floor(Math.random() * 10) ĭocument.getElementById("random_number").innerHTML = randomNumber We can get this by extending the above function. Get the random number in the form of an integer(complete number). This will always give the result in the form of a decimal point. This number always return less than 1 as a result.ĭocument.getElementById("random_number").innerHTML = Math.random() random() function in JavaScript to deal with the random numbers. Let us see some of the examples to generate random numbers: 1. Examples of Random Number Generator in JavaScript
0 Comments
Leave a Reply. |