I am looking to find a pattern to recursively split an array to odd and even elements. I will try to describe the problem in the following:
suppose we have an array of length 16 as:
a=[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
First iteration: splitting in odd and even
[0,2,4,6,8,10,12,14]
[1,3,5,7,9,11,13,15]
which basically are
a[2i] for i=0:7
a[2i+1] for i=0:7
splitting each of these arrays into odd and even elements again we have
[0,4,8,12]
[2,6,10,14]
[1,5,9,13]
[3,7,11,15]
that similarly are
4i for i=0:3
4i+2
4i+1
4i+3
splitting again the array elements would be
[0,8]
[4,12]
.
.
[1,9]
or
8i for i=0:1
8i+4
8i+2
8i+6
8+1
8i+5
8i+3
8i+1
Arrays needed to split recursively until each array has only two elements.
One things that I noticed that the bottom half is similar to the top one and we just need to add "1" to the index terms
I was wondering how Can I find the pattern for an array with an "n" elements?
Thank you very much for your time.
assuming your number n is a power of 2 (aka 2^k):
then you will have m = n/2 = 2^(k-1) arrays with following numbers for i in {0,1}:
0: m*i+f(0)
1: m*i+f(1)
...
j: m*i+f(j)
...
m-1: m*i+f(m-1)
where f(x) is a function which takes an integer (x), transforms it into an k-1-bit binary number (b), reverses it (rb) and returns its decimal value (y).
Example for k=4 (which looks a lot like your values):
x
b
rb
f(x)=y
0
000
000
0
1
001
100
4
2
010
010
2
3
011
110
6
4
100
001
1
5
101
101
5
6
110
011
3
7
111
111
7
Related
For example, if n=9, then how many different values can be represented in 9 binary digits (bits)?
My thinking is that if I set each of those 9 bits to 1, I will make the highest number possible that those 9 digits are able to represent. Therefore, the highest value is 1 1111 1111 which equals 511 in decimal. I conclude that, therefore, 9 digits of binary can represent 511 different values.
Is my thought process correct? If not, could someone kindly explain what I'm missing? How can I generalize it to n bits?
29 = 512 values, because that's how many combinations of zeroes and ones you can have.
What those values represent however will depend on the system you are using. If it's an unsigned integer, you will have:
000000000 = 0 (min)
000000001 = 1
...
111111110 = 510
111111111 = 511 (max)
In two's complement, which is commonly used to represent integers in binary, you'll have:
000000000 = 0
000000001 = 1
...
011111110 = 254
011111111 = 255 (max)
100000000 = -256 (min) <- yay integer overflow
100000001 = -255
...
111111110 = -2
111111111 = -1
In general, with k bits you can represent 2k values. Their range will depend on the system you are using:
Unsigned: 0 to 2k-1
Signed: -2k-1 to 2k-1-1
What you're missing: Zero is a value
A better way to solve it is to start small.
Let's start with 1 bit. Which can either be 1 or 0. That's 2 values, or 10 in binary.
Now 2 bits, which can either be 00, 01, 10 or 11 That's 4 values, or 100 in binary... See the pattern?
Okay, since it already "leaked": You're missing zero, so the correct answer is 512 (511 is the greatest one, but it's 0 to 511, not 1 to 511).
By the way, an good followup exercise would be to generalize this:
How many different values can be represented in n binary digits (bits)?
Without wanting to give you the answer here is the logic.
You have 2 possible values in each digit. you have 9 of them.
like in base 10 where you have 10 different values by digit say you have 2 of them (which makes from 0 to 99) : 0 to 99 makes 100 numbers. if you do the calcul you have an exponential function
base^numberOfDigits:
10^2 = 100 ;
2^9 = 512
There's an easier way to think about this. Start with 1 bit. This can obviously represent 2 values (0 or 1). What happens when we add a bit? We can now represent twice as many values: the values we could represent before with a 0 appended and the values we could represent before with a 1 appended.
So the the number of values we can represent with n bits is just 2^n (2 to the power n)
The thing you are missing is which encoding scheme is being used. There are different ways to encode binary numbers. Look into signed number representations. For 9 bits, the ranges and the amount of numbers that can be represented will differ depending on the system used.
I have searched and found discussions and solutions to similar problems, but not quite or as complex as I'm trying to figure out.
I have an access table which consists of two columns Draw Number and Number Drawn as shown below. Draw Number is repeated 20 times, to correspond to the 20 numbers that are drawn in each particular draw.
I'm trying to figure a way to determine the most frequent occurring combination of numbers (5 numbers) for all of the draws in each of the 20 number sets. So for instance, 12341 occurs n x, 12342 occurs nx, 12343 occurs n x, etc.
I've created parameter queries which allow me to search for different number combinations from 2 to 10 numbers, and they work OK returning the number of occurrences of a combination of numbers that I input through a simple UI. But the goal is to figure out pragmatically what the optimum combination of numbers.
Hope this makes sense. And by the way, there are 36 million or so rows in the table. The para queries work quite well however; it takes just over a second to return results for each number added. So, query two numbers = 2 second wait, three numbers = 3 second wait, etc.
I've been thinking about a loop of some type but don't know how to get started? Processing time isn't an issue; can take a day if required!
This is written in VBA and has an assortment of queries, temp tables, etc to get the job done.
The text says Access, but the tags say MySql, which is it? – RBarryYoung 21 hours ago
This part confuses me: I'm trying to figure a way to determine the most frequent occurring combination of numbers (5 numbers) for all of the draws in each of the 20 number sets. So for instance, 12341 occurs n x, 12342 occurs nx, 12343 occurs n x, etc. – Newd 21 hours ago
^What do you mean five numbers? No where in your sample data do I see 12341. Please explain using the data you have, and give expected results using that data. – McAdam331 21 hours ago
drosberg - clarification:
thanks for the response. It is an Access application, but as a first-time poster Stackoverflow recommends tags?
By five numbers I mean the most frequently occurring group of five numbers (I used five as an example, could be groups of 2 to 10 numbers) which occur in each draw, where a draw consists of 20 drawn numbers from a total of 80 numbers. So the data that I posted was intended as an example. The sample provided only has 50, 51 in common. I can plug 50 and 51 into the parameter query and it will tell me that this combination occurs 60,000 times (or whatever), but perhaps 50 and 57 occurs 65,000 times.
If i was to do this manually, and assuming I'm looking for the most frequent 5 number combination I would enter the following in the parameter query: 1,2,3,4,1 group = 30,000 occurrences 1,2,3,4,2 group = 31,000 occurrences 1,2,3,4,3 group = 31,050 occurrences 1,2,3,4,4 group = 29,050 occurrences etc........... etc...........
but I would have to do this for every combination of 5 numbers that can be derived from the numbers 1 thru 80. I'm hoping to have program do the work!!
thanks
don
DRAW NUMBER NUMBER DRAWN
1 1
1 28
1 19
1 3
1 38
1 46
1 43
1 29
1 13
1 22
1 20
1 11
1 50
1 51
1 53
1 54
1 57
1 64
1 76
1 78
2 29
2 14
2 2
2 1
2 35
2 40
2 39
2 30
2 10
2 27
2 21
2 6
2 42
2 50
2 51
2 53
2 54
2 61
2 65
2 69
I wrote a post a while ago about generating permutations with and without repetition using Excel. Perhaps you can use it.
https://michiel.wordpress.com/2015/03/29/permutations-with-repetition-using-excel/
Here's how it works. I am using strings, but you can easily modify that for numbers (since you say you need 5).
You can use the MID function to grab a single char from a string, and generate permutations from it.
=MID(Pattern,MOD([N]/[P],Length)+1,1)
N revers to the column N
P refers to the horizontal row (1,4,16). You can generate these with a formula like =4^.
After putting in the code, you can make a list of all permutations in Excel and in the cell next to it generate a sql query that you can perform as well from VBA.
Example: Looking up Access database in Excel
Or find a commercial tool like http://thingiequery.com/
I don't know if there's any open source tools for it.
I'm thinking that you should consider:
Say there are 100 balls.
Setting up a table to have one row for each "Draw number" with 100 columns one for every possible number each column has type boolean.
When you look to see which draws had number 23 you just add a
WHERE Column23 = true.
For numbers 23 and 56
WHERE Column23 = true AND Column56 = true
This should massivel simplify and speed up your SQL.
You set up a table with every possible combination of numbers.
You run SQL to find the counts.
Harvey
Consider a function F[x;y] that generates a table. I also have two lists; xList:[x1;x2;x3] and yList:[y1;y2;y3]. What is the best way to do a simple comma join of F[x1;y1],F[x1;y2],F[x1;y3],F[x2;y1],..., thereby producing one large table?
You have asked for the cross product of your argument lists, so the correct answer is
raze F ./: xList cross yList
Depending on what you are doing, you might want to look into having your function operate on the entire list of x and the entire list of y and return a table, rather than on each pair and then return a list of tables which has to get razed. The performance impact can be considerable, for example see below
q)g:{x?y} //your core operation
q)//this takes each pair of x,y, performs an operation and returns a table for each
q)//which must then be flattened with raze
q)fm:{flip `x`y`res!(x;y; enlist g[x;y])}
q)//this takes all x, y at once and returns one table
q)f:{flip `x`y`res!(x;y;g'[x;y])}
q)//let's set a seed to compare answers
q)\S 1
q)\ts do[10000;rm:raze fm'[x;y]]
76 2400j
q)\S 1
q)\ts do[10000;r:f[x;y]]
22 2176j
q)rm~r
1b
Setup our example
q)f:{([] total:enlist x+y; x:enlist x; y:enlist y)}
q)x:1 2 3
q)y:4 5 6
Demonstrate F[x1;y1]
q)f[1;4]
total x y
---------
5 1 4
q)f[2;5]
total x y
---------
7 2 5
Use the multi-valent apply operator together with each' to apply to each pair of arguments.
q)raze .'[f;flip (x;y)]
total x y
---------
5 1 4
7 2 5
9 3 6
Another way to achieve it using each-both :
x: 1 2 3
y: 4 5 6
f:{x+y}
f2:{ a:flip x cross y ; f'[a 0;a 1] }
f2[x;y]
5j, 6j, 7j, 6j, 7j, 8j, 7j, 8j, 9j
Assume a simple machine uses 4 bits to represent it instruction set. How many different instruction can this machine have? How many instruction could it have if eight bit are used? How many if 16 bits are used?
Sorry with the homework theory.. I didnt know how else to put it.. thanks
A bit can have two values: 0 or 1.
How many unique values are there of no bits? Just one. I'd show it here, but I don't know how to show no bits.
How many unique values are there of one bit? Two: 0 1
How many unique values are there of two bits? Four: 00 01 10 11
How many unique values are there of three bits? Eight: 000 001 010 011 100 101 110 111
Notice anything? Each time you add another bit, you double the number of values. You can represent that with this recursive formula:
unique_values(0) -> 1
unique_values(Bits) -> 2 * unique_values(Bits - 1)
This happens to be a recursive definition of "two to the power of," which can also be represented in this non-recursive formula:
unique_values = 2 ^ bits # ^ is exponentiation
Now you can compute the number of unique values that can be held by any number of bits, without having to count them all out. How many unique values can four bits hold? Two to the fourth power, which is 2 * 2 * 2 * 2 which is 16.
It's 2 to the power "bits". So
4 bits = 16 instructions
8 bits = 256 instructions
16 bits = 65536 instructions
You can have 2 raised to the power of the number of bits (since each bit can be 1 or zero). E.g. for the 4 bit computer: 2^4 = 16.
For example, if n=9, then how many different values can be represented in 9 binary digits (bits)?
My thinking is that if I set each of those 9 bits to 1, I will make the highest number possible that those 9 digits are able to represent. Therefore, the highest value is 1 1111 1111 which equals 511 in decimal. I conclude that, therefore, 9 digits of binary can represent 511 different values.
Is my thought process correct? If not, could someone kindly explain what I'm missing? How can I generalize it to n bits?
29 = 512 values, because that's how many combinations of zeroes and ones you can have.
What those values represent however will depend on the system you are using. If it's an unsigned integer, you will have:
000000000 = 0 (min)
000000001 = 1
...
111111110 = 510
111111111 = 511 (max)
In two's complement, which is commonly used to represent integers in binary, you'll have:
000000000 = 0
000000001 = 1
...
011111110 = 254
011111111 = 255 (max)
100000000 = -256 (min) <- yay integer overflow
100000001 = -255
...
111111110 = -2
111111111 = -1
In general, with k bits you can represent 2k values. Their range will depend on the system you are using:
Unsigned: 0 to 2k-1
Signed: -2k-1 to 2k-1-1
What you're missing: Zero is a value
A better way to solve it is to start small.
Let's start with 1 bit. Which can either be 1 or 0. That's 2 values, or 10 in binary.
Now 2 bits, which can either be 00, 01, 10 or 11 That's 4 values, or 100 in binary... See the pattern?
Okay, since it already "leaked": You're missing zero, so the correct answer is 512 (511 is the greatest one, but it's 0 to 511, not 1 to 511).
By the way, an good followup exercise would be to generalize this:
How many different values can be represented in n binary digits (bits)?
Without wanting to give you the answer here is the logic.
You have 2 possible values in each digit. you have 9 of them.
like in base 10 where you have 10 different values by digit say you have 2 of them (which makes from 0 to 99) : 0 to 99 makes 100 numbers. if you do the calcul you have an exponential function
base^numberOfDigits:
10^2 = 100 ;
2^9 = 512
There's an easier way to think about this. Start with 1 bit. This can obviously represent 2 values (0 or 1). What happens when we add a bit? We can now represent twice as many values: the values we could represent before with a 0 appended and the values we could represent before with a 1 appended.
So the the number of values we can represent with n bits is just 2^n (2 to the power n)
The thing you are missing is which encoding scheme is being used. There are different ways to encode binary numbers. Look into signed number representations. For 9 bits, the ranges and the amount of numbers that can be represented will differ depending on the system used.