Please, can anyone help me get the general formula of the sequence below.
when n=3 the number of arrows is 2
when n=4 the number of arrows is 4
when n=5 the number of arrows is 8
when n=6 the number of arrows is 12
when n=7 the number of arrows is 18
when n=8 the number of arrows is 24
when n=9 the number of arrows is 32
when n=10 the number of arrows is 40 etc
the n start from 3 upward.
Thank you
I'm not seeing an immediate way to combine it into a single equation, but it looks like if you split it between odd and evens there are two basic formulas. Maybe this will help you get on the right track:
Even: (n^2/2) - n
Odd: [(n^2 + 1)/2] - n
Edit
I got the idea for the combined formula from this Stack Exchange https://meta.stackexchange.com/questions/76902/how-can-i-write-math-formula-in-a-post
Here is the formula:
Note the absolute value around the (-1)^n - 1. If you don't have that, it will give you the -1 instead of +1 for every odd value.
Related
i am trying to read a csv file and my code is as follows
param=csvRead("C:\Users\USER\Dropbox\VOA-BK code\assets\Iris.csv",",","%i",'double',[],[],[1 2 3 4]); //reads number of clusters and features
data=csvRead("C:\Users\USER\Dropbox\VOA-BK code\assets\Iris.csv",",","%f",'double',[],[],[3 1 19 4]); //reads the values
numft=param(1,1);//save number of features
numcl=param(2,1);//save number of clusters
data_pts=0;
data_pts = max(size(data, "r"));//checks how many number of rows
disp(data(numft-3:data_pts,:));//print all data points (I added -3 otherwise it displays only 15 rows)
disp(numft);//print features
disp(data_pts);//print features
disp(param);
endfunction
below is the values that i am trying to read
features,4,,
clusters,3,,
5.1,3.5,1.4,0.2
4.9,3,1.4,0.2
4.7,3.2,1.3,0.2
4.6,3.1,1.5,0.2
5,3.6,1.4,0.2
7,3.2,4.7,1.4
6.4,3.2,4.5,1.5
6.9,3.1,4.9,1.5
5.5,2.3,4,1.3
6.5,2.8,4.6,1.5
5.7,2.8,4.5,1.3
6.3,3.3,6,2.5
5.8,2.7,5.1,1.9
7.1,3,5.9,2.1
6.3,2.9,5.6,1.8
6.5,3,5.8,2.2
7.6,3,6.6,2.1
I do not know why the code only displays 15 rows instead of 17. The only time it displays the correct matrix is when i put -3 in numft but with that, the number of columns would be 1. I am so confused. Is there a better way to read the values?
In the csvRead call in the first line of your script the boundaries of the region to read is incorrect, it should be corrected like this:
param=csvRead("C:\Users\USER\Dropbox\VOA-BK code\assets\Iris.csv",",","%i",'double',[],[],[1 2 2 2]);
Lets say for example we have the number 12345.
This sums to 15 when you add 1 + 2 + 3 + 4 + 5, which sums to 6 when you add 1 + 5.
My question is, what would the time complexity be for a repetitive adding algorithm like this be? This process is happens until there is only a single digit left.
I know that for any given number, the # of digits is approximately ln(n). Im thinking that this means that the big o would look something like (ln(n))^k, for some k. However, I am not confident because each time you sum, the number of digits gets smaller (first summed 5 digits, then only 2).
How would I go about figuring this out?
I need a code to generate only random EVEN numbers 2-100
There is tutorials on the web that generate random numbers but they're odd and even.
Please understand i just need even numbers to generate.
1, generate numbers 1-50
2, multiply all the numbers by 2
all numbers multiplied by 2 are even
This will work:
NSInteger evenNumber = (arc4random_uniform(50) + 1) * 2;
arc4random_uniform(50) will give results in the range 0 - 49. Adding 1 gives a value in the range 1 - 50. Multiplying by two gives you even numbers in the range 2 - 100.
The following is the problem from Interviewstreet I am not getting any help from their site, so asking a question here. I am not interested in an algorithm/solution, but I did not understand the solution given by them as an example for their second input. Can anybody please help me to understand the second Input and Output as specified in the problem statement.
Circle Summation (30 Points)
There are N children sitting along a circle, numbered 1,2,...,N clockwise. The ith child has a piece of paper with number ai written on it. They play the following game:
In the first round, the child numbered x adds to his number the sum of the numbers of his neighbors.
In the second round, the child next in clockwise order adds to his number the sum of the numbers of his neighbors, and so on.
The game ends after M rounds have been played.
Input:
The first line contains T, the number of test cases. T cases follow. The first line for a test case contains two space seperated integers N and M. The next line contains N integers, the ith number being ai.
Output:
For each test case, output N lines each having N integers. The jth integer on the ith line contains the number that the jth child ends up with if the game starts with child i playing the first round. Output a blank line after each test case except the last one. Since the numbers can be really huge, output them modulo 1000000007.
Constraints:
1 <= T <= 15
3 <= N <= 50
1 <= M <= 10^9
1 <= ai <= 10^9
Sample Input:
2
5 1
10 20 30 40 50
3 4
1 2 1
Sample Output:
80 20 30 40 50
10 60 30 40 50
10 20 90 40 50
10 20 30 120 50
10 20 30 40 100
23 7 12
11 21 6
7 13 24
Here is an explanation of the second test case. I will use a notation (a, b, c) meaning that child one has number a, child two has number b and child three has number c. In the beginning, the position is always (1,2,1).
If the first child is the first to sum its neighbours, the table goes through the following situations (I will put an asterisk in front of the child that just added its two neighbouring numbers):
(1,2,1)->(*4,2,1)->(4,*7,1)->(4,7,*12)->(*23,7,12)
If the second child is the first to move:
(1,2,1)->(1,*4,1)->(1,4,*6)->(*11,4,6)->(11,*21,6)
And last if the third child is first to move:
(1,2,1)->(1,2,*4)->(*7,2,4)->(7,*13,4)->(7,13,*24)
And as you notice the output to the second case are exactly the three triples computed that way.
Hope that helps.
I'm trying to emulate a function in SQL that a client has produced in Excel. In effect, they have a unique, 10-digit numeric value (VARCHAR) as the primary key in one of their enterprise database systems. Within another database, they require a unique, 5-digit alphanumeric identifier. They want that 5-digit alphanumeric value to be a representation of the 10-digit number. So what they did in excel was to split the 10-digit number into pairs, then convert each of those pairs into a hexadecimal value, then stitch them back together.
The EXCEL equation is:
=IF(VALUE(MID(A2,1,4))>0,DEC2HEX(VALUE(MID(A2,3,2)))&DEC2HEX(VALUE(MID(A2,5,2)))&DEC2HEX(VALUE(MID(A2,7,2)))&DEC2HEX(VALUE(MID(A2,9,2))),DEC2HEX(VALUE(MID(A2,5,2)))&DEC2HEX(VALUE(MID(A2,7,2)))&DEC2HEX((VALUE(MID(A2,9,2)))))
I need the SQL equivalent of this. Of course, should someone out there know a better way to accomplish their goal of "a 5-digit alphanumeric identifier" based off the 10-digit number, I'm all ears.
ADDED 8/2/2011
First of all, thank you to everyone for the replies. Nice to see folks willing to help and even enjoying it! Based on all the responses, I'm apt to tell my client they're intent is sound, only their method is off kilter. I'd also like to recommend a solution. So the challenge remains, just modified slightly:
CHALLENGE: Within SQL, take a 10 digit, unique NUMERIC string and represent it ALPHANUMERICALLY in as few characters as possible. The resulting string must also be unique.
Note that the first 3-4 characters in the 10-digit string are likely to be zeros, and that they could be stripped to shorten the resulting alphanumeric string. Not required, but perhaps helpful.
This problem is inherently impossible. You have a 10 digit numeric value that you want to convert to a 5 digit alphanumeric value. Since there are 10 numeric characters, this means that there are 10^10 = 10 000 000 000 unique values for your 10 digit number. Since there are 36 alphanumeric characters (26 letters + 10 numbers), there are 36^5 = 60 466 176 unique values for your 5 digit number. You cannot map a set of 10 billion elements into a set with around 60 million.
Now, lets take a closer look at what your client's code is doing:
So what they did in excel was to split the 10-digit number into pairs, then convert each of those pairs into a hexadecimal value, then stitch them back together.
This isn't 100% accurate. The excel code never uses the first 2 digits, but performs this operation on the remaining 8. There are two main problems with this algorithm which may not be intuitively obvious:
Two 10 digit numbers can map to the same 5 digit number. Consider the numbers 1000000117 and 1000001701. The last four digits of 1000000117 get mapped to 1 11, where the last four digits of 1000001701 get mapped to 11 1. This causes both to map to 00111.
The 5 digit number may not even end up being 5 digits! For example, 1000001616 gets mapped to 001010.
So, what is a possible solution? Well, if you don't care if that 5 digit number is unique or not, in MySQL you can use something like:
hex(<NUMERIC VALUE> % 0xFFFFF)
The log of 10^10 base 2 is 33.219280948874
> return math.log(10 ^ 10) / math.log(2)
33.219280948874
> = 2 ^ 33.21928
9999993422.9114
So, it takes 34 bits to represent this number. In hex this will take 34/4 = 8.5 characters, much more than 5.
> return math.log(10 ^ 10) / math.log(16)
8.3048202372184
The Excel macro is ignoring the first 4 (or 6) characters of the 10 character string.
You could try encoding in base 36 instead of 16. This will get you to 7 characters or less.
> return math.log(10 ^ 10) / math.log(36)
6.4254860446923
The popular base 64 encoding will get you to 6 characters
> return math.log(10 ^ 10) / math.log(64)
5.5365468248123
Even Ascii85 encoding won't get you down to 5.
> return math.log(10 ^ 10) / math.log(85)
5.1829075929158
You need base 100 to get to 5 characters
> return math.log(10 ^ 10) / math.log(100)
5
There aren't 100 printable ASCII characters, so this is not going to work, as zkhr explained as well, unless you're willing to go beyond ASCII.
I found your question interesting (although I don't claim to know the answer) - I googled a bit for you out of interest and found this which may help you http://dpatrickcaldwell.blogspot.com/2009/05/converting-decimal-to-hexadecimal-with.html