I'd like to calculate how many different variations of a certain amount of numbers are possible. The number of elements is variable.
Example:
I have 5 elements and each element can vary between 0 and 8. Only the first element is a bit more defined and can only vary between 1 and 8. So far I'd say I have 8*9^4 possibilities. But I have some more conditions. As soon as one of the elements gets zero the next elements should be automatically zero as well.
E.G:
6 5 4 7 8 is ok
6 3 6 8 0 is ok
3 6 7 0 5 is not possible and would turn to 3 6 7 0 0
Would somebody show me how to calculate the amount of combinations for this case and also in general, because I'd like to be able to calculate it also for 4 or 8 or 9 etc. elements. Later on I'd like to calculate this number in VBA to be able give the user a forecast how long my calculations will take.
Since once a 0 is present in the sequence, all remaining numbers in the sequence will also be 0, these are all of the possibilities: (where # below represents any digit from 1 to 8):
##### (accounts for 8^5 combinations)
####0 (accounts for 8^4 combinations)
...
#0000 (accounts for 8^1 combinations)
Therefore, the answer is (in pseudocode):
int sum = 0;
for (int x = 1; x <= 5; x++)
{
sum = sum + 8^x;
}
Or equivalently,
int prod = 0;
for (int x = 1; x <= 5; x++)
{
prod = 8*(prod+1);
}
great thank you.
Sub test()
Dim sum As Single
Dim x As Integer
For x = 1 To 6
sum = sum + 8 ^ x
Next
Debug.Print sum
End Sub
With this code I get exactly 37488. I tried also with e.g. 6 elements and it worked as well. Now I can try to estimate the calculation time
Related
I have to check how many hundreds are there in a number and translate that number to letters. For example the number 700. I have done the following code:
DATA(lv_dmbtr) = ZDS_FG-DMBTR. //Declared local variable of type DMBTR, thus DMBTR=700.
lv_dmbtr = ZDS_FG-DMBTR MOD 100. //Finding how many times 700 is in 100 via MOD and putting the value in lv_dmbtr.
IF lv_dmbtr LE 9. //The value is less or equal than 9(if larger means that the DMBTR is larger than hundreds,
e.g. 8000)
lv_hundred = lv_dmbtr / 100. // Divide the 700 with 100, taking the number 7.
lv_hundred_check = lv_hundred MOD 1. // Then taking the value of 7 into the new variable, done in case the
lv_hundred is a decimal value, e.g. 7.32.
IF lv_hundred_check > 0.
CALL FUNCTION 'SPELL_AMOUNT'
EXPORTING
amount = lv_hundred_check
* CURRENCY = ' '
* FILLER = ' '
LANGUAGE = SY-LANGU
IMPORTING
in_words = lv_hundred_string // the value is put in the new string
EXCEPTIONS
not_found = 1
too_large = 2
OTHERS = 3.
ENDIF.
Now when I debugg the code, all the variables have the value 0. Thus, lv_dmbtr, lv_hundred, lv_hundred_check all have the value 0.
May anyone of you know where the problem may be?
Thank you in advance!
Sorry for writing a lot in the code, just wanted to clarify as much as I could what I had done.
yes so I want to display the value of a specific number 700-> seven, 1400-> four.
So the basic formula to get the hundred in a number is the following: Find out how many times 100 fits completely into your number with integer division.
99 / 100 = 0
700 / 100 = 7
701 / 100 = 7
1400 / 100 = 14
1401 / 100 = 14
Now you can simply take this number MOD 10 to get the the individual hundreds.
0 MOD 10 = 0
7 MOD 10 = 7
14 MOD 10 = 4
Keep in mind that ABAP, in contrast to many other programming languages, rounds automatically. So in code this would be:
CONSTANTS lc_hundred TYPE f VALUE '100.0'.
DATA(lv_number) = 1403.
DATA(lv_hundred_count) = CONV i( floor( ( abs( lv_number ) / lc_hundred ) ) MOD 10 ).
Is there a way to give while loop a condition which makes it give an an output every ten times' executions, however it continues running after this output?
I hope I made myself clear...
Thanks!
Amy
Modulo is useful for this.
As an example; In swift to do modulo you use the % symbol. Essentially modulo outputs the remainder of the given terms.
So;
Value 1 MODULO Value 2 outputs Remainder.
Furthermore;
6 % 2 = 0
6 % 5 = 1
6 % 4.5 = 1.5
Essentially you want every nth element to output a value, with n being the rate. You need to track how many loops of the while you have gone through already.
The code below will run through the while 1000 times, and print out every 10 times ( for a total of 100 prints of output. )
var execution : Int = 0
while ( execution != 1000 ) {
if ( execution % 10 == 0 ) {
print("output")
}
execution = execution + 1
}
Here is the same answer as given by adam but then in Labview.
I'm currently making a blackjack game for my project in school in Visual Basic.
In blackjack, when you have aces (value initially 11) their value turns to 1 when the total value of the cards is > 21. In code, this would just take away 10 for every ace
I'm stuck on this.
This is the code I have (that doesn't work):
Do While PlayerValue > 21 And counter <= noAcesPlayer
counter += 1
PlayerValue -= 10
Loop
In a senario, I have a: 2, 8, A, 8 (=29)
But since there is an Ace, and the total value is > 21, the value should have 10 subtracted from it (=19) - the above code does not do this.
Another scenario would be 10, 8, A, A (=40)
Again, the two Aces should turn into 1's, since the total value > 21, giving 20.
Any help would be greatly appreciated. :)
Here is an approach
Public Enum CardFace
None
Ace
Two
Three
Four
Five
Six
Seven
Eight
Nine
Ten
Jack
Queen
King
End Enum
This code should produce a value of twenty
Dim cards As New List(Of CardFace) From {CardFace.Ten, CardFace.Eight, CardFace.Ace, CardFace.Ace}
Dim total As Integer = 0
Dim numofAces As Integer = 0
For Each c As CardFace In cards
Debug.WriteLine(c.ToString)
If c = CardFace.Ace Then
numofAces += 1
Else
total += c
End If
Next
If numofAces > 0 Then
If total + 11 + (numofAces - 1) > 21 Then
total += numofAces
Else
total += 11 + (numofAces - 1)
End If
End If
Debug.WriteLine(total)
The correct way to build a blackjack hand is the following (in pseudo-code):
Variables: total = 0, soft-flag = false
For each card in hand:
Add card value to total. Faces are 10, aces are 1.
If the card you added was an ace, set soft-flag = true
If total < 12 and soft-flag:
Add 10 to total
Else:
set soft-flag = false
That's it. Only one loop over the cards, no extraneous variables, and you're left with the total value and a flag indicating if the total is soft.
I understand the Modulus operator in terms of the following expression:
7 % 5
This would return 2 due to the fact that 5 goes into 7 once and then gives the 2 that is left over, however my confusion comes when you reverse this statement to read:
5 % 7
This gives me the value of 5 which confuses me slightly. Although the whole of 7 doesn't go into 5, part of it does so why is there either no remainder or a remainder of positive or negative 2?
If it is calculating the value of 5 based on the fact that 7 doesn't go into 5 at all why is the remainder then not 7 instead of 5?
I feel like there is something I'm missing here in my understanding of the modulus operator.
(This explanation is only for positive numbers since it depends on the language otherwise)
Definition
The Modulus is the remainder of the euclidean division of one number by another. % is called the modulo operation.
For instance, 9 divided by 4 equals 2 but it remains 1. Here, 9 / 4 = 2 and 9 % 4 = 1.
In your example: 5 divided by 7 gives 0 but it remains 5 (5 % 7 == 5).
Calculation
The modulo operation can be calculated using this equation:
a % b = a - floor(a / b) * b
floor(a / b) represents the number of times you can divide a by b
floor(a / b) * b is the amount that was successfully shared entirely
The total (a) minus what was shared equals the remainder of the division
Applied to the last example, this gives:
5 % 7 = 5 - floor(5 / 7) * 7 = 5
Modular Arithmetic
That said, your intuition was that it could be -2 and not 5. Actually, in modular arithmetic, -2 = 5 (mod 7) because it exists k in Z such that 7k - 2 = 5.
You may not have learned modular arithmetic, but you have probably used angles and know that -90° is the same as 270° because it is modulo 360. It's similar, it wraps! So take a circle, and say that its perimeter is 7. Then you read where is 5. And if you try with 10, it should be at 3 because 10 % 7 is 3.
Two Steps Solution.
Some of the answers here are complicated for me to understand. I will try to add one more answer in an attempt to simplify the way how to look at this.
Short Answer:
Example 1:
7 % 5 = 2
Each person should get one pizza slice.
Divide 7 slices on 5 people and every one of the 5 people will get one pizza slice and we will end up with 2 slices (remaining). 7 % 5 equals 2 is because 7 is larger than 5.
Example 2:
5 % 7 = 5
Each person should get one pizza slice
It gives 5 because 5 is less than 7. So by definition, you cannot divide whole 5items on 7 people. So the division doesn't take place at all and you end up with the same amount you started with which is 5.
Programmatic Answer:
The process is basically to ask two questions:
Example A: (7 % 5)
(Q.1) What number to multiply 5 in order to get 7?
Two Conditions: Multiplier starts from `0`. Output result should not exceed `7`.
Let's try:
Multiplier is zero 0 so, 0 x 5 = 0
Still, we are short so we add one (+1) to multiplier.
1 so, 1 x 5 = 5
We did not get 7 yet, so we add one (+1).
2 so, 2 x 5 = 10
Now we exceeded 7. So 2 is not the correct multiplier.
Let's go back one step (where we used 1) and hold in mind the result which is5. Number 5 is the key here.
(Q.2) How much do we need to add to the 5 (the number we just got from step 1) to get 7?
We deduct the two numbers: 7-5 = 2.
So the answer for: 7 % 5 is 2;
Example B: (5 % 7)
1- What number we use to multiply 7 in order to get 5?
Two Conditions: Multiplier starts from `0`. Output result and should not exceed `5`.
Let's try:
0 so, 0 x 7 = 0
We did not get 5 yet, let's try a higher number.
1 so, 1 x 7 = 7
Oh no, we exceeded 5, let's get back to the previous step where we used 0 and got the result 0.
2- How much we need to add to 0 (the number we just got from step 1) in order to reach the value of the number on the left 5?
It's clear that the number is 5. 5-0 = 5
5 % 7 = 5
Hope that helps.
As others have pointed out modulus is based on remainder system.
I think an easier way to think about modulus is what remains after a dividend (number to be divided) has been fully divided by a divisor. So if we think about 5%7, when you divide 5 by 7, 7 can go into 5 only 0 times and when you subtract 0 (7*0) from 5 (just like we learnt back in elementary school), then the remainder would be 5 ( the mod). See the illustration below.
0
______
7) 5
__-0____
5
With the same logic, -5 mod 7 will be -5 ( only 0 7s can go in -5 and -5-0*7 = -5). With the same token -5 mod -7 will also be -5.
A few more interesting cases:
5 mod (-3) = 2 i.e. 5 - (-3*-1)
(-5) mod (-3) = -2 i.e. -5 - (-3*1) = -5+3
It's just about the remainders. Let me show you how
10 % 5=0
9 % 5=4 (because the remainder of 9 when divided by 5 is 4)
8 % 5=3
7 % 5=2
6 % 5=1
5 % 5=0 (because it is fully divisible by 5)
Now we should remember one thing, mod means remainder so
4 % 5=4
but why 4?
because 5 X 0 = 0
so 0 is the nearest multiple which is less than 4
hence 4-0=4
modulus is remainders system.
So 7 % 5 = 2.
5 % 7 = 5
3 % 7 = 3
2 % 7 = 2
1 % 7 = 1
When used inside a function to determine the array index. Is it safe programming ? That is a different question. I guess.
Step 1 : 5/7 = 0.71
Step 2 : Take the left side of the decimal , so we take 0 from 0.71 and multiply by 7
0*7 = 0;
Step # : 5-0 = 5 ; Therefore , 5%7 =5
Modulus operator gives you the result in 'reduced residue system'. For example for mod 5 there are 5 integers counted: 0,1,2,3,4. In fact 19=12=5=-2=-9 (mod 7). The main difference that the answer is given by programming languages by 'reduced residue system'.
lets put it in this way:
actually Modulus operator does the same division but it does not care about the answer , it DOES CARE ABOUT reminder for example if you divide 7 to 5 ,
so , lets me take you through a simple example:
think 5 is a block, then for example we going to have 3 blocks in 15 (WITH Nothing Left) , but when that loginc comes to this kinda numbers {1,3,5,7,9,11,...} , here is where the Modulus comes out , so take that logic that i said before and apply it for 7 , so the answer gonna be that we have 1 block of 5 in 7 => with 2 reminds in our hand! that is the modulus!!!
but you were asking about 5 % 7 , right ?
so take the logic that i said , how many 7 blocks do we have in 5 ???? 0
so the modulus returns 0...
that's it ...
A novel way to find out the remainder is given below
Statement : Remainder is always constant
ex : 26 divided by 7 gives R : 5
This can be found out easily by finding the number that completely divides 26 which is closer to the
divisor and taking the difference of the both
13 is the next number after 7 that completely divides 26 because after 7 comes 8, 9, 10, 11, 12 where none of them divides 26 completely and give remainder 0.
So 13 is the closest number to 7 which divides to give remainder 0.
Now take the difference (13 ~ 7) = 5 which is the temainder.
Note: for this to work divisor should be reduced to its simplest form ex: if 14 is the divisor, 7 has to be chosen to find the closest number dividing the dividend.
As you say, the % sign is used to take the modulus (division remainder).
In w3schools' JavaScript Arithmetic page we can read in the Remainder section what I think to be a great explanation
In arithmetic, the division of two integers produces a quotient and a
remainder.
In mathematics, the result of a modulo operation is the
remainder of an arithmetic division.
So, in your specific case, when you try to divide 7 bananas into a group of 5 bananas, you're able to create 1 group of 5 (quotient) and you'll be left with 2 bananas (remainder).
If 5 bananas into a group of 7, you won't be able to and so you're left with again the 5 bananas (remainder).
Given a NxN matrix and a (row,column) position, what is a method to select a different position in a random (or pseudo-random) order, trying to avoid collisions as much as possible?
For example: consider a 5x5 matrix and start from (1,2)
0 0 0 0 0
0 0 X 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
I'm looking for a method like
(x,y) hash (x,y);
to jump to a different position in the matrix, avoiding collisions as much as possible
(do not care how to return two different values, it doesn't matter, just think of an array).
Of course, I can simply use
row = rand()%N;
column = rand()%N;
but it's not that good to avoid collisions.
I thought I could apply twice a simple hash method for both row and column and use the results as new coordinates, but I'm not sure this is a good solution.
Any ideas?
Can you determine the order of the walk before you start iterating? If your matrices are large, this approach isn't space-efficient, but it is straightforward and collision-free. I would do something like:
Generate an array of all of the coordinates. Remove the starting position from the list.
Shuffle the list (there's sample code for a Fisher-Yates shuffle here)
Use the shuffled list for your walk order.
Edit 2 & 3: A modular approach: Given s array elements, choose a prime p of form 2+3*n, p>s. For i=1 to p, use cells (iii)%p when that value is in range 1...s-1. (For row-length r, cell #c subscripts are c%r, c/r.)
Effectively, this method uses H(i) = (iii) mod p as a hash function. The reference shows that as i ranges from 1 to p, H(i) takes on each of the values from 0 to p-1, exactly one time each.
For example, with s=25 and p=29 or 47, this uses cells in following order:
p=29: 1 8 6 9 13 24 19 4 14 17 22 18 11 7 12 3 15 10 5 16 20 23 2 21 0
p=47: 1 8 17 14 24 13 15 18 7 4 10 2 6 21 3 22 9 12 11 23 5 19 16 20 0
according to bc code like
s=25;p=29;for(i=1;i<=p;++i){t=(i^3)%p; if(t<s){print " ",t}}
The text above shows the suggestion I made in Edit 2 of my answer. The text below shows my first answer.
Edit 0: (This is the suggestion to which Seamus's comment applied): A simple method to go through a vector in a "random appearing" way is to repeatedly add d (d>1) to an index. This will access all elements if d and s are coprime (where s=vector length). Note, my example below is in terms of a vector; you could do the same thing independently on the other axis of your matrix, with a different delta for it, except a problem mentioned below would occur. Note, "coprime" means that gcd(d,s)=1. If s is variable, you'd need gcd() code.
Example: Say s is 10. gcd(s,x) is 1 for x in {1,3,7,9} and is not 1 for x in {2,4,5,6,8,10}. Suppose we choose d=7, and start with i=0. i will take on values 0, 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, which modulo 10 is 0, 7, 4, 1, 8, 5, 2, 9, 6, 3, 0.
Edit 1 & 3: Unfortunately this will have a problem in the two-axis case; for example, if you use d=7 for x axis, and e=3 for y-axis, while the first 21 hits will be distinct, it will then continue repeating the same 21 hits. To address this, treat the whole matrix as a vector, use d with gcd(d,s)=1, and convert cell numbers to subscripts as above.
If you just want to iterate through the matrix, what is wrong with row++; if (row == N) {row = 0; column++}?
If you iterate through the row and the column independently, and each cycles back to the beginning after N steps, then the (row, column) pair will interate through only N of the N^2 cells of the matrix.
If you want to iterate through all of the cells of the matrix in pseudo-random order, you could look at questions here on random permutations.
This is a companion answer to address a question about my previous answer: How to find an appropriate prime p >= s (where s = the number of matrix elements) to use in the hash function H(i) = (i*i*i) mod p.
We need to find a prime of form 3n+2, where n is any odd integer such that 3*n+2 >= s. Note that n odd gives 3n+2 = 3(2k+1)+2 = 6k+5 where k need not be odd. In the example code below, p = 5+6*(s/6); initializes p to be a number of form 6k+5, and p += 6; maintains p in this form.
The code below shows that half-a-dozen lines of code are enough for the calculation. Timings are shown after the code, which is reasonably fast: 12 us at s=half a million, 200 us at s=half a billion, where us denotes microseconds.
// timing how long to find primes of form 2+3*n by division
// jiw 20 Sep 2011
#include <stdlib.h>
#include <stdio.h>
#include <sys/time.h>
double ttime(double base) {
struct timeval tod;
gettimeofday(&tod, NULL);
return tod.tv_sec + tod.tv_usec/1e6 - base;
}
int main(int argc, char *argv[]) {
int d, s, p, par=0;
double t0=ttime(0);
++par; s=5000; if (argc > par) s = atoi(argv[par]);
p = 5+6*(s/6);
while (1) {
for (d=3; d*d<p; d+=2)
if (p%d==0) break;
if (d*d >= p) break;
p += 6;
}
printf ("p = %d after %.6f seconds\n", p, ttime(t0));
return 0;
}
Timing results on 2.5GHz Athlon 5200+:
qili ~/px > for i in 0 00 000 0000 00000 000000; do ./divide-timing 500$i; done
p = 5003 after 0.000008 seconds
p = 50021 after 0.000010 seconds
p = 500009 after 0.000012 seconds
p = 5000081 after 0.000031 seconds
p = 50000021 after 0.000072 seconds
p = 500000003 after 0.000200 seconds
qili ~/px > factor 5003 50021 500009 5000081 50000021 500000003
5003: 5003
50021: 50021
500009: 500009
5000081: 5000081
50000021: 50000021
500000003: 500000003
Update 1 Of course, timing is not determinate (ie, can vary substantially depending on the value of s, other processes on machine, etc); for example:
qili ~/px > time for i in 000 004 010 058 070 094 100 118 184; do ./divide-timing 500000$i; done
p = 500000003 after 0.000201 seconds
p = 500000009 after 0.000201 seconds
p = 500000057 after 0.000235 seconds
p = 500000069 after 0.000394 seconds
p = 500000093 after 0.000200 seconds
p = 500000099 after 0.000201 seconds
p = 500000117 after 0.000201 seconds
p = 500000183 after 0.000211 seconds
p = 500000201 after 0.000223 seconds
real 0m0.011s
user 0m0.002s
sys 0m0.004s
Consider using a double hash function to get a better distribution inside the matrix,
but given that you cannot avoid colisions, what I suggest is to use an array of sentinels
and mark the positions you visit, this way you are sure you get to visit a cell once.