i want a code to define the prompted number by user is prime or not . since it's an assignment
i'm not allowed to use ' isprime ' predefined code .
the following approach was not useful :
N = input( 'please enter a positive enteger value = ' ) ;
Quotient = floor(N - (mod(N,2)./2)) ;
for i = 1 : Quotient
if mod(N,i ) == 0
fprintf(' your prompted number is not prime ' ) ;
if mod(N,i) ~= 0
fprintf(' your prompted number is prime ' ) ;
end
end
end
for example if i enter a prime number like 13 it results in this :
your prompted number is prime
but if i enter a Non-prime num like 12 it repeats the ' your prompted number is prime ' message for 10 times .
for i = 1 : Quotient
if mod(N,i ) == 0
That will give you every number since x mod 1 is always zero. In other words, the remainder (when you divide any positive integer by one) is zero, since all of them divide perfectly.
You need to start at 2 rather than 1.
In addition, once you've found out the number is not prime, you should stop the loop since there's no possibility of it becoming prime again after that :-) And, for efficiency, you only need to go up to the square root of the number since, if it has a factor above that, you would have already found the equivalent factor below that.
The pseudo-code for such a beast would be:
set isprime to true
set chkval to 2
while chkval * chkval <= number:
if number mod chkval is zero:
set isprime to false
exit while
end if
increment chkval
end while
if isprime:
say number, " is prime"
else:
say number, " is composite"
Try to find factors and as soon as you find one you know it's not prime:
prime = true
for f = 2:ceil(sqrt(N)) %// Start from 2 as prime numbers DO have 1 as a factor. Anything larger than sqrt(N) will have to have a corresponding factor smaller than this so there is no point checking them
if mod(N,f) == 0
prime = false;
break;
end
end
There are 2 problems with your code. First, as already explained by paxdiablo, you need to start your loop from 2. Secondly you have nested your if statements, and since they are mutually exclusive conditions, the inner condition will never trigger.
Related
I need to determine whether the following Hash Functions Set is universal or not:
Let U be the set of the keys - {000, 001, 002, 003, ... ,999} - all the numbers between 0 and 999 padded with 0 in the beginning where needed. Let n = 10 and 1 < a < 9 ,an integer between 1 and 9. We denote by ha(x) the rightmost digit of the number a*x.
For example, h2(123) = 6, because, 2 * 123 = 246.
We also denote H = {h1, h2, h3, ... ,h9} as our set of hash functions.
Is H is universal? prove.
I know I need to calculate the probability for collision of 2 different keys and check if it's smaller or equal to 1/n (which is 1/10), so I tried to separate into cases - if a is odd or even, because when a is even the last digit of a*x will be 0/2/4/6/8, else it could be anything. But it didn't help me so much as I'm stuck on it.
Would be very glad for some help here.
I encountered the following in a leetcode article to determine whether an integer is a palindrome
Now let's think about how to revert the last half of the number. For number 1221, if we do 1221 % 10, we get the last digit 1, to get the second to the last digit, we need to remove the last digit from 1221, we could do so by dividing it by 10, 1221 / 10 = 122. Then we can get the last digit again by doing a modulus by 10, 122 % 10 = 2, and if we multiply the last digit by 10 and add the second last digit, 1 * 10 + 2 = 12, it gives us the reverted number we want. Continuing this process would give us the reverted number with more digits.
Now the question is, how do we know that we've reached the half of the number?
Since we divided the number by 10, and multiplied the reversed number by 10, when the original number is less than the reversed number, it means we've processed half of the number digits.
Can someone explain the last two sentences please?! Thank you!
Here is the enclosed C# code:
public class Solution {
public bool IsPalindrome(int x) {
// Special cases:
// As discussed above, when x < 0, x is not a palindrome.
// Also if the last digit of the number is 0, in order to be a palindrome,
// the first digit of the number also needs to be 0.
// Only 0 satisfy this property.
if(x < 0 || (x % 10 == 0 && x != 0)) {
return false;
}
int revertedNumber = 0;
while(x > revertedNumber) {
revertedNumber = revertedNumber * 10 + x % 10;
x /= 10;
}
// When the length is an odd number, we can get rid of the middle digit by revertedNumber/10
// For example when the input is 12321, at the end of the while loop we get x = 12, revertedNumber = 123,
// since the middle digit doesn't matter in palidrome(it will always equal to itself), we can simply get rid of it.
return x == revertedNumber || x == revertedNumber/10;
}
}
The reason is due to the original given input, x will be decreasing by 1 digit while the reverted string increases by 1 digit at the same time. The process keeps on going until x is less than or equal to the reverted string. Hence due to the change length changes, when it terminates, we would approximately reach half of the string.
Let's visit a few examples with positive numbers to understand the process. I would write (x,y) as the (original number, reverted string). The third example is purposely designed to show that it need not be exactly half though but the code would still work.
The first example is 1221, where there are even number of digits. It will go from (1221, 0) to (122, 1) to (12, 12), at this stage, the two terms are equal and hence the process terminates and we can conclude that it is a palindrome.
The next example is 1223, where there are even number of digits. It will go from (1223, 0) to (122, 3) to (12, 32), at this stage, the termination condition holds and hence the process terminates and we can conclude that it is not a palindrome.
Now, the third example is 1211,then the sequence is (1211,0), (121, 1), (12,11), (1,112), after which we terminates from the string and it would conclude that it is not a palindrome
Now, let's make the number consists of odd number of digits:
For 12321. It will go from (12321, 0) to (1232, 1) to (123, 12) to (12, 123) and at this point, the condition breaks. We then divide the reverted string by 10 and we end up with (12,12) and we can conclude that it is a palindrome.
For 12323. It will go from (12323, 0) to (1232, 3) to (123, 32) to (12, 323) and at this point, the condition breaks. We then divide the reverted string by 10 and we end up with (12,32) and we can conclude that it is not a palindrome.
For 12311. It will go from (12311, 0) to (1231, 1) to (123, 11) to (12, 113) and at this point, the condition breaks. We then divide the reverted string by 10 and we end up with (12,11) and we can conclude that it is not a palindrome.
I hope these examples would help you to understand what the post mean.
I am using scratch. I acquire two values from the user and have to find the numbers divisible by 2 & 3 between those values . How can I count those numbers without using arrays ( just by using basic operations) ?
If you only need to count those numbers, arrays are not needed. Just iterate through the range and count:
Here's what you can do (do mind me, I am not good at creating questions and variables) ...
when flag clicked
ask (starting number is?) and wait
set (startrange) to (answer)
ask (ending number is?) and wait
set (endrange) to (answer)
set (counter) to (startrange)
set (divisibleby2) to (0)
set (divisibleby3) to (0)
set (divisibleby6) to (0)
repeat until counter = endrange
if (counter) mod 6 = 0
change (divisibleby6) by (1)
else
if (counter) mod 3 = 0
change (divisibleby3) by (1)
else
if (counter) mod 2 = 0
change (divisibleby2) by (1)
say (join (The number of numbers from the two inputs that are divisible by 2, 3 is) ((divisibleby2) + ((divisibleby3) + (divisibleby6)))
So, why is a divisibleby6 variable needed? It is because some numbers are divisible by 2 and 3 simultaneously, which means, the number would be recorded twice if the code was altered. However, if you want this to happen, this would be the code for you:
hen flag clicked
ask (starting number is?) and wait
set (startrange) to (answer)
ask (ending number is?) and wait
set (endrange) to (answer)
set (counter) to (startrange)
set (divisibleby2) to (0)
set (divisibleby3) to (0)
repeat until counter = endrange
if (counter) mod 3 = 0
change (divisibleby3) by (1)
if (counter) mod 2 = 0
change (divisibleby2) by (1)
say (join (The number of numbers from the two inputs that are divisible by 2, 3 is) ((divisibleby2) + (divisibleby3))
Given an arbitrarily long number, how can I output its double? I know how to multiply small numbers together as long as the result is <10, but what about larger integers like 32984335, and doubling something like that? I don't know the right way to handle something like this.
This is the algorithm you need to implement:
Start the current count with 0;
Multiply the current count by ten: this can be achieved by dupping 10 times, and then adding all dupes together;
Read a digit;
If it's null proceed to 8;
Convert it to an actual number: this can be achieved by subtracting 48;
Add it to the current count;
Proceed to 2;
Duplicate the current count;
Adding the dupes together;
Divide by ten using repeated subtraction; keep quotient and remainder;
Grab the remainder;
Make it a digit (add 48);
Print it;
Grab the quotient from 10;
If it's not zero, goto 10;
The end.
All these steps consists of basic brainfuck idioms, so it should be easy to implement.
Here's a start. It will multiply a byte of input, but I think you can build off it to make it work for any number. Basically, you take in a number, and store the number to multiply by (2) in the next pointer. You loop decrementing the first number, and then nest a loop decrementing the second number; in each iteration of the inner loop, you increment the pointer to the right of your second operand. This is your result.
, take input to ptr 0
[
- decrement first operand (input)
>++ number to multiply by (2) at ptr 1
[
>+ result in ptr 2
<- decrement second operand (2)
]
<
]
>> move to result
Here is my BF code: http://ideone.com/2Y9pk8
->>+>,+
[
-----------
[
-->++++++[-<------>]<
[->+<[->+<[->+<[->+<[>----<-<+>[->+<]]]]]]>
[-<++>]<+
>,+
[
-----------
[->+<]
]
>[-<+>]<
]
<[<]
>-[<++++++++[->++++++<]>.[-]]
>[<++++++++[->++++++<]>-.[-]>]
++++++++++.[-]
<+[-<+]
->>+>,+
]
It reads each number in each line until EOF, and multiply all numbers by two..
Here is the code for multiplying a number by 2.
,[>++<-]>.
Hope this helps.
How can I compare two numbers with an inequality? (greater than or less than)
I want to compare single digits
For example
1 2
5 3
9 2
etc.
This is the best way to compare two numbers.Why because, if you are intelligent enough, you can use the same code in bigger programs.It's highly portable.
Assume we have two numbers a,b.
we have two blocks : if( a>=b ) and else,
Hope its enough.
0 1 0 a b 0
Make the array like this. And point to the (4) i.e. point to the a
+>+< This is for managing if a=0 and b=0
[->-[>]<<] This is a magic loop. if a is the one which
reaches 0 first (a<b),then pointer will be at(4).
Else it will be at (3)
<[-
// BLOCK (a>=b)
//You are at (2) and do whatever you want and come back to (2).
//Its a must
]
<[-<
// BLOCK(a<b)
//You are at (1) and do whatever you want and come back to (1).
//Its a must
]
It will not affect the following program code as both the code blocks will end up in (1) You can do further coding assuming that pointer will reach (1)
Please remove the documentation if you copy the code. Because code contains some valid brainfuck symbols like < . , etc.
Once you know which is the distance between the two numbers you should or decrement both of them in the same loop iteration and then check both for being zero: you will understand which one is the smaller.
Eg:
+++++ > +++ < [->-< check is first is zero, then second]
(this is just to give you a hint, you will have to take care about equal numbers and similar issues.
I was thinking about this too, and while I'm sure this isn't the best solution, at least it can answer the question of which number is larger =)
The program asks for two characters, outputs '<' if the first is smaller, '>' if it is larger, and '=' if they are equal. After outputting one char, the program halts by asking for additional input.
+>,>,<<[>-[>>>]<[>>-[>++++++++++[->++++++<]>.,]++++++++++[->++++++<]>+.,]<-[>>>]<<[>>>++++++++++[->++++++<]>++.,]<<<]
Hopefully somewhat clearer:
+ init (0) to 1
>, read (1)
>, read (2)
<<[ loop forever
>-[>>>] decrement (1) going to (4) if (1) != 0
<[ goto (0) == 1 if (1) reached 0 (otherwise goto (3))
>>-[>++++++++++[->++++++<]>.,] decrement (2) printing lessthan if larger than 0
++++++++++[->++++++<]>+., if (2) == 0 print '='
]
<-[>>>] decrement (2) going to (5) if (2) != 0
<<[ goto (0) == 1 if (2) reached 0 (otherwise goto (3))
>>>++++++++++[->++++++<]>++., print largerthan since (2) reached 0 first
]
<<< goto(0)
]
I made a solution, that gives you back a boolean and the pointer always at the same point.
This is how it looks like at the beginning:
0 0 0 a b 0 0
p
And these are the two possible outputs:
0 0 0 0 0 1 0 #true
p
0 0 0 0 0 0 0 #false
p
The code:
>>>>
[ # while cell != 0
- # decrement a
[ # if a != 0
>- # decrement b
[ # if b != 0
< # go left
<-< # undo the finally-block;
] # finally-block
<[-]> # clear a
>+> # res = 1; move to end-position
<<< # undo the finally-block
] # finally-block
>[-]>> # clear b; res = 0; move to end-position
] #
minified version:
>>>>[-[>-[< <-<]<[-]>>+><<<]>[-]>>]
Given two numbers A and B, the following code will print A if A is greater than B, B if B is greater than A and C if both are equal.
>>>>>>>>>++++++[>+++++++++++<-]>[>+>+>+<<<-]>+>->
<<<<<<<<<<<,>,<
[->-<[>]<<]>>>[>>]>>>>>>>>.
No such thing exists in BF. The > and < in BF move the pointer to the right and to the left, respectively.