gcd method

[PvtBillPilgrim]

I'm very new to Java and I was looking for some help on this particular problem.

I need to create a method that takes a String as a parameter and returns "true" if the parameter is the letter Y or the letter N (in either upper or lower case), or false otherwise.

Most of the methods we've been working on are numbers (integers mostly) not strings, so this one confuses me quite a bit. How do you make the computer look for a certain character like y, Y, n, N. I don't want to declare them as variables, I just want Java to see them as letters.

I have:
public static boolean isYorN(String str)
{
boolean character;
switch (character)
{
case 'y':
character = true;
return true;
break;

case 'Y':
character = true;
return true;
break;

case 'n':
character = true;
return true;
break;

case 'N':
character = true;
return true;
break;

default:
character = false;
return false;

}
}
}

I think a switch case is the best method, but I'm missing something about naming because all my compiling errors are telling me that I have incompatible types (i.e. found: boolean, required: int).

Thank you in advance for any help.

[sicarie]

I believe the errors are coming from the fact that you are using 'Y', 'y', 'N', and 'n' as the cases, when you tell the switch statement (swtich (character)) that it should expect a boolean - either a true or a false.

I would recommend using the .equals() function in the String class. That returns a boolean, so you can resolve it as being either true or false (use the fact that it will return as either true or false).

(And in future posts, please use [code ] and [/code ] around your code - only without the spaces at the end, it really helps with readability. Thanks!)

[nomad]

I believe the errors are coming from the fact that you are using 'Y', 'y', 'N', and 'n' as the cases, when you tell the switch statement (swtich (character)) that it should expect a boolean - either a true or a false.

I would recommend using the .equals() function in the String class. That returns a boolean, so you can resolve it as being either true or false (use the fact that it will return as either true or false).

(And in future posts, please use [code ] and [/code ] around your code - only without the spaces at the end, it really helps with readability. Thanks!)

Can PvtBillPilgrim use scanner and do this
if(kbd.equalsIg noreCase("Y"));

then use the switch or if statement
...

[PvtBillPilgrim]

gcd method

I am a beginner programmer in Java. Just started learning the past few weeks.

I'm taking a class and need to create a method that finds the greatest common divisor of two integers. I can assume that both are positive, but I cannot use the Euclidean Algorithm.

I'm sort of lost, but I think I'm on the right track, although this may look confusing:

public static int gcd(int a, int c)
{
int gcd;
int attempt;
if (a > c)
{
 if (a % c == 0)
{
gcd = c;
return gcd;
}
else
{
for (attempt = c; attempt = 1; attempt--)
do
{
if (c % attempt == 0 && a % attempt == 0)
{
gcd = attempt;
return attempt;
}
}
while (c % attempt != 0 || a % attempt == 0);
}
}

And then I basically copied the code with an else statement if c > a.

I'm sure there are a lot of mistakes. Could someone just point them out and offer a simpler way of doing it (possibly without using the Euclidean Algorithm)?

Thanks.
Michael

[JosAH]

Hint: for any two (positive) numbers a and c the gcd is the largest number in the
range [1 ... min(a, c) ]. Only a single for loop could do the job. There's no need
to duplicate code for a < c or c < a and there's also no need for two nested loops.

kind regards,

Jos

[r035198x]

....I think a switch case is the best method...

I'm afraid you're wrong there and that's where you missed it. Just an if else will do it.

big hint:
[CODE=java] if(str.equalsIg noreCase("y")) ....[/CODE]

[rsrinivasan]

I am a beginner programmer in Java. Just started learning the past few weeks.

I'm taking a class and need to create a method that finds the greatest common divisor of two integers. I can assume that both are positive, but I cannot use the Euclidean Algorithm.

I'm sort of lost, but I think I'm on the right track, although this may look confusing:

public static int gcd(int a, int c)
{
int gcd;
int attempt;
if (a > c)
{
 if (a % c == 0)
{
gcd = c;
return gcd;
}
else
{
for (attempt = c; attempt = 1; attempt--)
do
{
if (c % attempt == 0 && a % attempt == 0)
{
gcd = attempt;
return attempt;
}
}
while (c % attempt != 0 || a % attempt == 0);
}
}

And then I basically copied the code with an else statement if c > a.

I'm sure there are a lot of mistakes. Could someone just point them out and offer a simpler way of doing it (possibly without using the Euclidean Algorithm)?

Thanks.
Michael

Hi,
I modified ur function to find GCD of two numbers.
Try this function repl ur feedback

[edit] Don't supply full spoonfed code; it's not allowed in this forum. Thanks;

Thanks,

Srinivas r.

[PvtBillPilgrim]

OK. I think this is a lot better. It compiles and makes sense to me at least. However, it doesn't work properly for more difficult integers like 20, 30. Can anyone spot the problem (as I said nothing wrong with compiling, just doesn't work logically). As I said before, I can't use the Euclidean Algorithm, so I basically had to write the simplest program to find gcd.

public static int gcd(int a, int b)
{
int gcd;
if (a > b && a % b == 0)
{
gcd = b;
return gcd;
}
else if (b > a && b % a == 0)
{
gcd = a;
return gcd;
}
else if (a > b && a % b != 0)
{
int count = b;
do
{
count = count - 1;
}
while (b % count != 0 && a % count != 0);
gcd = count;
return gcd;
}
else if (b > a && b % a != 0)
{
int count = a;
do
{
count = count - 1;
}
while (b % count != 0 && a % count != 0);
gcd = count;
return gcd;
}
else
{
gcd = a;
return gcd;
}
}

[blazedaces]

I'm not 100% sure about the forum guidelines, but I'm pretty sure you should post this as a new thread/topic because it is indeed a ... new... topic. Anyways, may I ask why you're using subtraction and not mod? What do you mean you can't use the euclidean algorithm?

Ok, you're subtracting by one each time... why? Why not subtract by intervals of the smaller number (original euclidean algorithm) or use the mod method?

You realize there's a Math.max(int a, int b) method?

I don't know what is logically wrong in your code, but you don't have a good base of an idea, try rewriting and also tell us what exactly outputs when you have higher integers.

-blazed

[astolpho]

I'm afraid you're wrong there and that's where you missed it. Just an if else will do it.

big hint:
[CODE=java] if(str.equalsIg noreCase("y")) ....[/CODE]

perhaps simply

return str.equalsIgnor eCase("y") || str.equalsIgnor eCase("n");

[prometheuzz]

OK. I think this is a lot better. It compiles and makes sense to me at least. However, it doesn't work properly for more difficult integers like 20, 30. Can anyone spot the problem (as I said nothing wrong with compiling, just doesn't work logically). As I said before, I can't use the Euclidean Algorithm, so I basically had to write the simplest program to find gcd.

Oh man, why so complicated? Here's some pseudo code to do it (non-Euclidean-style, why not Euclidean, btw?):

public static int gcd(int a, int b) {

  counter <- the minimum value between 'a' and 'b'

  WHILE counter is more than 0
	IF 'counter' divides 'a' AND 'counter' divides 'b'
	  break out of this loop
	END IF
	decrease 'counter'
  END WHILE

  return 'counter'
}

Good luck.

[JosAH]

Oh man, why so complicated? Here's some pseudo code to do it (non-Euclidean-style, why not Euclidean, btw?):

Yup, exactly what I suggest in the thread where the sequel of this
thread belongs ;-)

kind regards,

Jos

ps. I just merged the two threads.