I have program in which 4 random numbers are set to 4 buttons (one no. for each button). There are two random variables (rbvalue and loadG4). The rbvalue number is different for each button, but the loadG4 value overrides a button's value and replaces it. The idea is, none of the rbvalue numbers are meant to be equal to loadG4, and a part of the program ensures they are never equal. Here's the code:
Random GenerateG4 = new Random();
int loadG4 = GenerateG4.nextInt(10);
Random randoms1 = new Random();
final TextView number = (TextView) findViewById(R.id.number);
number.setText(""+loadG4);
for(int allrbA=0; allrbA<4; allrbA++) {
int rbvalue = randoms1.nextInt(10 - 1);
if (rbvalue==loadG4) rbvalue=9;
selectrb[allrbA].setText(""+rbvalue);
}
selectrb[rbselector].setText(""+loadG4);
The part that makes this work is:
if (rbvalue==loadG4) rbvalue=9;
Simply adding that line of code in did the job. Now out of all numbers generated from 0 to 9, there is only one of which is the value of loadG4. How did that one line do this? I always thought that after if (...) there has to be curly brackets for the actual statements e.g if (...) {System.out.println("...")} Why set rbvalue to 9? Why is there (10 -1) for the random rbvalue? What does that -1 even mean in this situation?
I'd appreciate anyone who could clear this up for me, thank you.
You can take 10 - 1 as primary school math, it equals 9, so randoms1.nextInt(10 - 1) returns a number from 0 through 8 inclusive. I don’t know why it’s not written as 9. It might be an attempt to stress that the value is 1 smaller than the 10 used in GenerateG4.nextInt(10).
When only one statement is executed when the if condition is true, the curly braces are a matter of opinion. Very many prefer to have them there; in my workplace no one gets away with leaving them out.
The if statement sets rbvalue to 9 if it was equal to loadG4. That will make sure the values differ. Since rbvalue was at most 8, we know that when they are equal, setting one of them to 9 will ensure they become different. It’s a bit subtle for my taste, I understand you needed to ask.
The code doesn’t appear to do anything to make the rbvalue numbers different for each button.
Related
I'm very (read: extremely) new to java and was going to make a table with 5 columns print out for an assignement.
The two first columns are strings, third one an int, fourth a double and fifth an int. It is three rows in total that are affected by this.
I formatted it with printf and:
System.out.printf(Locale.ENGLISH, "%s%10s%14d%20.2f%12d\n", stringOne,
stringTwo, firstInt, stupidDouble, secondInt);
and
"%s\t%s\t\t%d\t\t%2.4f\t\t%d\n" (with the format and stuff above as well ofc).
But the teacher didn't want me to "hard code" the layout since it's part of learning special commands (like \t) and thus wanted me to change it and adress it in the variable itself instead.
I've been trying everything I can think of but can't get it to skip the last decimals, which are unneccesary 0's.
I wish I could've just used the numbers as a string instead, but I need it to calculate the last int with:
static int lastInt = (int) Math.round ( stupidDouble - firstInt )
What I suspect that my teacher is looking for me to do is tabs all the way like:
"%s\t%s\t\t%d\t\t%f\t\t%d\n"
But I need to double to be e.g. 1.2345 instead of 1.234500.
If it makes any difference, it's 3 different doubles (3 different rows in the table); one with 2 decimals, one with 4 and one with 5.
It would more or less save my weekend if someone could help me with this. The simplest possible solution would be much appreciated. <3
I was wondering how to code the game, Mastermind, in Java, but with things up a notch (I want to inform the user not only how many pegs they got right or wrong, but also how many they guessed correctly in the wrong slots).
For instance, say the RNG answer of 5 digits of numbers 1-6 is:
22354
... and the user's guess is:
32624
Resulting in:
two guessed correctly (2 and 4)
two guessed partially correct (2 and 3)
one guessed incorrectly (6).
Here's my code for informing the user what they got correct:
String answer = "22354";
String guess = "32624";
int correctPegs = 0;
for (int i = 0; i < 5; i++) {
char a = answer.charAt(i);
char g = guess.charAt(i);
if (a == g) {
correctPegs++;
}
}
System.out.println(correctPegs);
How would I find the partially correct ones?
... and for calculating how many the user guessed incorrectly, I was thinking of using basic algebra to find the remaining characters after finding the correct and partially correct ones.
I would put '0' into the strings where they match. Then for every char that is not a zero, I would look through the answer string again. If any character in the answer string matches, I would make them both zero, and increment outOfPlacePegs.
This guards against two corner cases I am assuming you do not want:
One is when you have '323' as the answer and '223' as the input. You don't want the first '2' to be recorded as out of place.
Second is when you have '223' as the answer and '442' as the input. You don't want this recorded as two numbers out of place.
I'm very new to Java and stackoverflow so I'm sorry if I seem ignorant but I wrote this program to multiply two numbers together using the Russian Peasant multiplication algorithm. The complete program includes far more operations and is hundreds of lines of code but I only included what I thought was necessary for this particular method. I have a test harness so I know that all the submethods work correctly. The problem that I'm struggling with though is the 3rd line where I'm adding factor1 to the product. The values add correctly but then when factor1 is multiplied by 2 in the 5th line then the value that was added to product in the 3rd line also gets doubled resulting in an incorrect product value. How can I make sure that when factor 1 is doubled that it doesn't carry backwards to the product term?
while (Long.parseLong(factor2.toString()) >= 1) {
if (factor2.bigIntArray[0] % 2 == 1) {
product = product.add(factor1);
}
factor1 = factor1.multiplyByTwo();
factor2 = factor2.divideByTwo();
}
I think in your method multiplyByTwo you use code
`datamember=datamember*2;`
rather than that try doing this
return new FactorClass(datamember*2);
so it doesnt change the added value.
it would be better if u could show the mulTiplyByTwo method code since that is where your actually are getting the changed value.
I have just started taking a Computer Science class online and I am quite new to Programming(a couple of week's worth of experience). I am working on an assignment, but I do not understand what a mystery method is. I have yet to find an answer that I can wrap my head around online, in my textbook, or from my professor. Any explanation using this code as an example would also be greatly appreciated!
This is the equation where I saw it in:
public static void mystery1(int n) {
System.out.print(n + " ");
if (n > 0) {
n = n - 5;
}
if (n < 0) {
n = n + 7;
} else {
n = n * 2;
}
System.out.println(n);
}
If anybody can help, that would be amazing! Thank you!
First of all, I voted your question up because I think it's a valid question for someone who is just beginning in computer programming, and I think that some people fail to understand the significance and purpose of Stack Overflow, which is to help programmers in times of need.
Secondly, I think that the couple of users that have commented on your post are on the right track. I have personally never heard of a mystery method, so I think the goal here is for you to simply figure out what the method does. In this case, the method takes a parameter for int 'n'. This means that if, at any point in the application, the 'mystery1()' method is called, an integer will have to be passed as the variable.
Let's say that a user enters the number '9'. The method would be called by the code mystery1(9). This would then run the first part of the 'if' statement, because n is greater than 0. So, n would be equal to n - 5, or 9 - 5, which is 4. (So, n=4.)
I hope my answer was somewhat helpful to you. Take care.
Your assignment is probably to figure out what this method does. More specifically, what does it print to the screen. I'll walk you through how to figure this out.
You have a function, also called a methood, called mystery1. A function is just a named block of code that you can use throughout other pieces of code. This function takes an integer argument called n. Let's assume n=12 for this example.
The first thing that happens in your function when it is called is that n is printed out via the System.out.print method. Notice that it prints a blank space after it. Notice also at the end it prints another value of n that gets assigned within the method. So the method is going to print "12 ?" without the double quotes. The question mark is what we have to figure out. The code says if n > 0 then n = n-5. Since 12 is greater than 0, n gets the new value of 7. The next if statement says if n is less than 0, n gets assigned n+7. But it is not less than zero, it is 7 at this point, so we move to the else statement. In this statement n gets multiplied by 2 which is 14. So the last statement prints 14.
So for an input value of 12 this method prints:
12 14
I hope this helps. If not, please give more detail about your assignment and what you don't understand about my explanation.
The point of this kind of exercise is that you are given a method, but they don't tell you what it does (hence the "mystery"). You are supposed to figure out what it does on your own (like "solving the mystery"). It doesn't mean that the method is special in any way.
Say I give you a "mystery" method like this:
public static void mystery(int n) {
System.out.println(n+1);
}
You would "solve the mystery" by telling me that this method prints out the number that comes after n. Nothing else is special here.
In the example you gave, your job would be to tell me why the method prints out 0 0 when n = 0, or 6 2 when n = 6.
I think the usage of the term "mystery method" is rather misleading, as it has clearly made you (and many, many, many others) believe that something about these methods is special and something that you need to learn about. There isn't anything special about them, and there's nothing to learn.
I think a lot of people would understand this better if instructors just said "tell me what this method does" instead of trying treat students like 5 year olds by saying "Here's a mystery method (ooh, fancy and entertaining). Can you play detective and solve the mystery for me?"
I have updated this question(found last question not clear, if you want to refer to it check out the reversion history). The current answers so far do not work because I failed to explain my question clearly(sorry, second attempt).
Goal:
Trying to take a set of numbers(pos or neg, thus needs bounds to limit growth of specific variable) and find their linear combinations that can be used to get to a specific sum. For example, to get to a sum of 10 using [2,4,5] we get:
5*2 + 0*4 + 0*5 = 10
3*2 + 1*4 + 0*5 = 10
1*2 + 2*4 + 0*5 = 10
0*2 + 0*4 + 2*5 = 10
How can I create an algo that is scalable for large number of variables and target_sums? I can write the code on my own if an algo is given, but if there's a library avail, I'm fine with any library but prefer to use java.
One idea would be to break out of the loop once you set T[z][i] to true, since you are only basically modifying T[z][i] here, and if it does become true, it won't ever be modified again.
for i = 1 to k
for z = 0 to sum:
for j = z-x_i to 0:
if(T[j][i-1]):
T[z][i]=true;
break;
EDIT2: Additionally, if I am getting it right, T[z][i] depends on the array T[z-x_i..0][i-1]. T[z+1][i] depends on T[z+1-x_i..0][i-1]. So once you know if T[z][i] is true, you only need to check one additional element (T[z+1-x_i][i-1]) to know if T[z+1][i-1] will be true.
Let's say you represent the fact whether T[z][i] was updated by a variable changed. Then, you can simply say that T[z][i] = changed && T[z-1][i]. So you should be done in two loops instead of three. This should make it much faster.
Now, to scale it - Now that T[z,i] depends only on T[z-1,i] and T[z-1-x_i,i-1], so to populate T[z,i], you do not need to wait until the whole (i-1)th column is populated. You can start working on T[z,i] as soon as the required values are populated. I can't implement it without knowing the details, but you can try this approach.
I take it this is something like unbounded knapsack? You can dispense with the loop over c entirely.
for i = 1 to k
for z = 0 to sum
T[z][i] = z >= x_i cand (T[z - x_i][i - 1] or T[z - x_i][i])
Based on the original example data you gave (linear combination of terms) and your answer to my question in the comments section (there are bounds), would a brute force approach not work?
c0x0 + c1x1 + c2x2 +...+ cnxn = SUM
I'm guessing I'm missing something important but here it is anyway:
Brute Force Divide and Conquer:
main controller generates coefficients for say, half of the terms (or however many may make sense)
it then sends each partial set of fixed coefficients to a work queue
a worker picks up a partial set of fixed coefficients and proceeds to brute force its own way through the remaining combinations
it doesn't use much memory at all as it works sequentially on each valid set of coefficients
could be optimized to ignore equivalent combinations and probably many other ways
Pseudocode for Multiprocessing
class Controller
work_queue = Queue
solution_queue = Queue
solution_sets = []
create x number of workers with access to work_queue and solution_queue
#say for 2000 terms:
for partial_set in coefficient_generator(start_term=0, end_term=999):
if worker_available(): #generate just in time
push partial set onto work_queue
while solution_queue:
add any solutions to solution_sets
#there is an efficient way to do this type of polling but I forget
class Worker
while true: #actually stops when a stop work token is received
get partial_set from the work queue
for remaining_set in coefficient_generator(start_term=1000, end_term=1999):
combine the two sets (partial_set.extend(remaining_set))
if is_solution(full_set):
push full_set onto the solution queue