consider this number:785462454105,I need an algorithm that first separates the number into groups with maximum length of three(starting from the right side) ,it would look something like:
a = 105
b = 454
c = 462
d = 785
Of course I know I can convert the number to string but I want to do this without any conversion. Also I'm not allowed to use any arrays and any special methods or class which exist in the programming language use(I'm using java but as I said I'm not allowed to use the functions).The only tools I have are loops,conditional clauses and mathematical,arithmetic and logical operators.
Also it is possible to get 454000 or 462000000 out using loops but how can I get rid of the zeros?
Note that something like 1234 should turn to:
a = 234
b = 1
It's easy to get a group of last 3 digits if you take a remainder while dividing by 1000.
785462454105 % 1000 == 105
Then you could get rid of last 3 digits dividing by 1000:
785462454105 / 1000 == 785462454
Repeat this in a loop until the number becomes zero and you're done.
The only issue left is to print leading zeros:
123045 % 1000 = 45 but we want to print 045.
Usually you'll need a separate inner loop, for example, to count decimal digits (dividing by 10 until it becomes zero) and then print number of missing zeros (it's equal to number of digits you want minus number of digits in your number).
But here it's a simple case, you could just sole it via couple of ifs:
long a = 785462454105;
while (a > 0) {
long x = a % 1000;
a /= 1000;
if (x < 10) {
System.out.print("00");
} else if (x < 100) {
System.out.print("0");
}
System.out.println(x);
}
I have a rather unorthodox homework assignment where I am to write a simple function where a double value is rounded to an integer with using only a while loop.
The main goal is to write something similar to the round function.
I made some progress where I should add or subtract a very small double value and I would eventually hit a number that will become an integer:
while(~isinteger(inumberup))
inumberup=inumberup+realmin('double');
end
However, this results in a never-ending loop. Is there a way to accomplish this task?
I'm not allowed to use round, ceil, floor, for, rem or mod for this question.
Assumption: if statements and the abs function are allowed as the list of forbidden functions does not include this.
Here's one solution. What you can do is keep subtracting the input value by 1 until you get to a point where it becomes less than 1. The number produced after this point is the fractional component of the number (i.e. if our number was 3.4, the fractional component is 0.4). You would then check to see if the fractional component, which we will call f, is less than 0.5. If it is, that means you need to round down and so you would subtract the input number with f. If the number is larger than 0.5 or equal to 0.5, you would add the input number by (1 - f) in order to go up to the next highest number. However, this only handles the case for positive values. For negative values, round in MATLAB rounds towards negative infinity, so what we ought to do is take the absolute value of the input number and do this subtraction to find the fractional part.
Once we do this, we then check to see what the fractional part is equal to, and then depending on the sign of the number, we either add or subtract accordingly. If the fractional part is less than 0.5 and if the number is positive, we need to subtract by f else we need to add by f. If the fractional part is greater than or equal to 0.5, if the number is positive we need to add by (1 - f), else we subtract by (1 - f)
Therefore, assuming that num is the input number of interest, you would do:
function out = round_hack(num)
%// Repeatedly subtract until we get a value that less than 1
%// i.e. the fractional part
%// Also make sure to take the absolute value
f = abs(num);
while f > 1
f = f - 1;
end
%// Case where we need to round down
if f < 0.5
if num > 0
out = num - f;
else
out = num + f;
end
%// Case where we need to round up
else
if num > 0
out = num + (1 - f);
else
out = num - (1 - f);
end
end
Be advised that this will be slow for larger values of num. I've also wrapped this into a function for ease of debugging. Here are a few example runs:
>> round_hack(29.1)
ans =
29
>> round_hack(29.6)
ans =
30
>> round_hack(3.4)
ans =
3
>> round_hack(3.5)
ans =
4
>> round_hack(-0.4)
ans =
0
>> round_hack(-0.6)
ans =
-1
>> round_hack(-29.7)
ans =
-30
You can check that this agrees with MATLAB's round function for the above test cases.
You can do it without loop: you can use num2str to convert the number into a string, then find the position of the . in the string and extract the string fron its beginning up to the position of the .; then you convert it back to a numebr with str2num
To round it you have to check the value of the first char (converted into a number) after the ..
r=rand*100
s=num2str(r)
idx=strfind(num2str(r),'.')
v=str2num(s(idx+1))
if(v <= 5)
rounded_val=str2num(s(1:idx-1))
else
rounded_val=str2num(s(1:idx-1))+1
end
Hope this helps.
Qapla'
I'm trying to make a program that will generate a random number, and you have to guess it by typing in the answer. The problem is that it won't match the right number as shown.
Objects:
2 Labels, 1 textbox, 1 Command button
My first code:
Private Sub Command1_Click()
Dim Num, Random As Integer
Label2.Caption = ""
Num = Val(Text1.Text)
Randomize (Random)
Random = Val(Label1.Caption)
Label1.Caption = Int(10 * Rnd + 1)
For Num = 1 To Num
If Num = Random Then
Label2.Caption = "you won "
Else
End If
Next
End Sub
you don't need that for loop, its checking each number up to the number you guessed.
Private Sub Command1_Click()
Dim Num, Random As Integer
Label2.Caption = ""
Num = Val(Text1.Text)
Randomize (Random)
Random = Val(Label1.Caption)
Label1.Caption = Int(10 * Rnd + 1)
If Num = Random Then
Label2.Caption = "you won "
Else
End If
End Sub
to debug it put
If Num = Random Then
Label2.Caption = "you won "
Else
Label2.Caption = "The number " & Num & " Does not equal " & Random
End If
First of all, the current code will always result in number zero being the first "randomly" generated number. Second, the formula will produce a predictable random number sequence.
The issue behind this is that computers are not smart and cannot really create random numbers, that's why you need to "seed" them with Randomize to sort of "shake" the dice and come up with a different number. But, if you randomize with the same number, it will produce exactly the same sequence of "random" numbers.
For example, if you use your code, it will always produce the following sequence of numbers: 0, 8, 7, 5 ...
That's why you need to "seed" the random number with... a random number! LOL. But how do you get a random number? Technically, you can't, but you can cheat. You can do Randomize (Timer) or Randomize, which takes Timer as a parameter and what it does is "seeds" the random number generation with a number of seconds and milliseconds elapsed since midnight. So, the only time where you will get the same sequence of random numbers if you click the button to guess the random number every day at exactly the same second and millisecond.
You can try and expand on this theory by adding day, month or year - that would expand the "seed" exponentially and you will never see repeating sequence of random numbers, but it is extremely hard to do that because once you start playing around with large seed numbers you will encounter weird issues such as if you change a very large number by 1, it would still generate the same sequence of random numbers (in my test scenario, randomizing with any number in the range from 5969992 to 5969995 will result in the same sequence of random numbers: 9, 8, 6, 6, 1). This is probably a limitation of the Randomize function itself. Personally, I don't think its worth trying to go beyond seeding with timer.
Below is your code adjusted to generate a more "random" sequence of numbers:
Dim Num, Random As Integer
Label2.Caption = ""
Num = Val(Text1.Text)
Randomize
Label1.Caption = Int(10 * Rnd + 1)
Random = Val(Label1.Caption)
If Num = Random Then
Label2.Caption = "you won "
Else
Label2.Caption = "The number " & Num & " Does not equal " & Random
End If
dim num, Random as integer
label2.caption =""
num = val(text1.text)
randomize
label1.caption = int((10 +1-1)*rnd+1)
random = val(label1.caption)
if num = random then
label2.caption ="you won"
else
label2.caption = "Try again"
end if
So I'm creating a Activation class for a VB6 project and I've run into a brain fart. I've designed how I want to generate the Serial Number for this particular product in a following way.
XXXX-XXXX-XXXX-XXXX
Each group of numbers would be representative of data that I can read if I'm aware of the matching document that allows me to understand the codes with the group of digits. So for instance the first group may represent the month that the product was sold to a customer. But I can't have all the serial numbers in January all start with the same four digits so there's some internal math that needs to be done to calculate this value. What I've landed on is this:
A B C D = digits in the first group of the serial number
(A + B) - (C + D) = #
Now # would relate to a table of Hex values that would then represent the month the product was sold. Something like...
1 - January
2 - February
3 - March
....
B - November
C - December
My question lies here - if I know I need the total to equal B(11) then how exactly can I code backwards to generate (A + B) - (C + D) = B(11)?? It's a pretty simple equation, I know - but something I've just ran into and can't seem to get started in the right direction. I'm not asking for a full work-up of code but just a push. If you have a full solution available and want to share I'm always open to learning a bit more.
I am coding in VB6 but VB.NET, C#, C++ solutions could work as well since I can just port those over relatively easily. The community help is always greatly appreciated!
There's no single solution (you have one equation with four variables). You have to pick some random numbers. Here's one that works (in Python, but you get the point):
from random import randint
X = 11 # the one you're looking for
A_plus_B = randint(X, 30)
A = randint(max(A_plus_B - 15, 0), min(A_plus_B, 15))
B = A_plus_B - A
C_plus_D = A_plus_B - X
C = randint(max(C_plus_D - 15, 0), min(C_plus_D, 15))
D = C_plus_D - C
I assume you allow hexadecimal digits; if you just want 0 to 9, replace 15 by 9 and 30 by 18.
OK - pen and paper is always the solution... so here goes...
Attempting to find what values should be for (A + B) - (C + D) to equal a certain number called X. First I know that I want HEX values so that limits me to 0-F or 0-15. From there I need a better starting place so I'll generate a random number that will represent the total of (A + B), we'll call this Y, but not be lower than value X. Then subtract from that number Y value of X to determine that value that will represent (C + D), which we'll call Z. Use similar logic to break down Y and Z into two numbers each that can represent (A + B) = Y and (C + D) = Z. After it's all said and done I should have a good randomization of creating 4 numbers that when plugged into my equation will return a suitable result.
Just had to get past the brain fart.
This may seem a little hackish, and it may not take you where you're trying to go. However it should produce a wider range of values for your key strings:
Option Explicit
Private Function MonthString(ByVal MonthNum As Integer) As String
'MonthNum: January=1, ... December=12. Altered to base 0
'value for use internally.
Dim lngdigits As Long
MonthNum = MonthNum - 1
lngdigits = (Rnd() * &H10000) - MonthNum
MonthString = Right$("000" & Hex$(lngdigits + (MonthNum - lngdigits Mod 12)), 4)
End Function
Private Function MonthRecov(ByVal MonthString As String) As Integer
'Value returned is base 1, i.e. 1=January.
MonthRecov = CInt(CLng("&H" & MonthString) Mod 12) + 1
End Function
Private Sub Form_Load()
Dim intMonth As Integer
Dim strMonth As String
Dim intMonthRecov As Integer
Dim J As Integer
Randomize
For intMonth = 1 To 12
For J = 1 To 2
strMonth = MonthString(intMonth)
intMonthRecov = MonthRecov(strMonth)
Debug.Print intMonth, strMonth, intMonthRecov, Hex$(intMonthRecov)
Next
Next
End Sub
Not sure how best to explain it, other than using an example...
Imagine having a client with 10 outstanding invoices, and one day they provide you with a cheque, but do not tell you which invoices it's for.
What would be the best way to return all the possible combination of values which can produce the required total?
My current thinking is a kind of brute force method, which involves using a self-calling function that runs though all the possibilities (see current version).
For example, with 3 numbers, there are 15 ways to add them together:
A
A + B
A + B + C
A + C
A + C + B
B
B + A
B + A + C
B + C
B + C + A
C
C + A
C + A + B
C + B
C + B + A
Which, if you remove the duplicates, give you 7 unique ways to add them together:
A
A + B
A + B + C
A + C
B
B + C
C
However, this kind of falls apart after you have:
15 numbers (32,767 possibilities / ~2 seconds to calculate)
16 numbers (65,535 possibilities / ~6 seconds to calculate)
17 numbers (131,071 possibilities / ~9 seconds to calculate)
18 numbers (262,143 possibilities / ~20 seconds to calculate)
Where, I would like this function to handle at least 100 numbers.
So, any ideas on how to improve it? (in any language)
This is a pretty common variation of the subset sum problem, and it is indeed quite hard. The section on the Pseudo-polynomial time dynamic programming solution on the page linked is what you're after.
This is strictly for the number of possibilities and does not consider overlap. I am unsure what you want.
Consider the states that any single value could be at one time - it could either be included or excluded. That is two different states so the number of different states for all n items will be 2^n. However there is one state that is not wanted; that state is when none of the numbers are included.
And thus, for any n numbers, the number of combinations is equal to 2^n-1.
def setNumbers(n): return 2**n-1
print(setNumbers(15))
These findings are very closely related to combinations and permutations.
Instead, though, I think you may be after telling whether given a set of values any combination of them sum to a value k. For this Bill the Lizard pointed you in the right direction.
Following from that, and bearing in mind I haven't read the whole Wikipedia article, I propose this algorithm in Python:
def combs(arr):
r = set()
for i in range(len(arr)):
v = arr[i]
new = set()
new.add(v)
for a in r: new.add(a+v)
r |= new
return r
def subsetSum(arr, val):
middle = len(arr)//2
seta = combs(arr[:middle])
setb = combs(arr[middle:])
for a in seta:
if (val-a) in setb:
return True
return False
print(subsetSum([2, 3, 5, 8, 9], 8))
Basically the algorithm works as this:
Splits the list into 2 lists of approximately half the length. [O(n)]
Finds the set of subset sums. [O(2n/2 n)]
Loops through the first set of up to 2floor(n/2)-1 values seeing if the another value in the second set would total to k. [O(2n/2 n)]
So I think overall it runs in O(2n/2 n) - still pretty slow but much better.
Sounds like a bin packing problem. Those are NP-complete, i.e. it's nearly impossible to find a perfect solution for large problem sets. But you can get pretty close using heuristics, which are probably applicable to your problem even if it's not strictly a bin packing problem.
This is a variant on a similar problem.
But you can solve this by creating a counter with n bits. Where n is the amount of numbers. Then you count from 000 to 111 (n 1's) and for each number a 1 is equivalent to an available number:
001 = A
010 = B
011 = A+B
100 = C
101 = A+C
110 = B+C
111 = A+B+C
(But that was not the question, ah well I leave it as a target).
It's not strictly a bin packing problem. It's a what combination of values could have produced another value.
It's more like the change making problem, which has a bunch of papers detailing how to solve it. Google pointed me here: http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.57.3243
I don't know how often it would work in practice as there are many exceptions to this oversimplified case, but here's a thought:
In a perfect world, the invoices are going to be paid up to a certain point. People will pay A, or A+B, or A+B+C, but not A+C - if they've received invoice C then they've received invoice B already. In the perfect world the problem is not to find to a combination, it's to find a point along a line.
Rather than brute forcing every combination of invoice totals, you could iterate through the outstanding invoices in order of date issued, and simply add each invoice amount to a running total which you compare with the target figure.
Back in the real world, it's a trivially quick check you can do before launching into the heavy number-crunching, or chasing them up. Any hits it gets are a bonus :)
Here is an optimized Object-Oriented version of the exact integer solution to the Subset Sums problem(Horowitz, Sahni 1974). On my laptop (which is nothing special) this vb.net Class solves 1900 subset sums a second (for 20 items):
Option Explicit On
Public Class SubsetSum
'Class to solve exact integer Subset Sum problems'
''
' 06-sep-09 RBarryYoung Created.'
Dim Power2() As Integer = {1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32764}
Public ForceMatch As Boolean
Public watch As New Stopwatch
Public w0 As Integer, w1 As Integer, w1a As Integer, w2 As Integer, w3 As Integer, w4 As Integer
Public Function SolveMany(ByVal ItemCount As Integer, ByVal Range As Integer, ByVal Iterations As Integer) As Integer
' Solve many subset sum problems in sequence.'
''
' 06-sep-09 RBarryYoung Created.'
Dim TotalFound As Integer
Dim Items() As Integer
ReDim Items(ItemCount - 1)
'First create our list of selectable items:'
Randomize()
For item As Integer = 0 To Items.GetUpperBound(0)
Items(item) = Rnd() * Range
Next
For iteration As Integer = 1 To Iterations
Dim TargetSum As Integer
If ForceMatch Then
'Use a random value but make sure that it can be matched:'
' First, make a random bitmask to use:'
Dim bits As Integer = Rnd() * (2 ^ (Items.GetUpperBound(0) + 1) - 1)
' Now enumerate the bits and match them to the Items:'
Dim sum As Integer = 0
For b As Integer = 0 To Items.GetUpperBound(0)
'build the sum from the corresponding items:'
If b < 16 Then
If Power2(b) = (bits And Power2(b)) Then
sum = sum + Items(b)
End If
Else
If Power2(b - 15) * Power2(15) = (bits And (Power2(b - 15) * Power2(15))) Then
sum = sum + Items(b)
End If
End If
Next
TargetSum = sum
Else
'Use a completely random Target Sum (low chance of matching): (Range / 2^ItemCount)'
TargetSum = ((Rnd() * Range / 4) + Range * (3.0 / 8.0)) * ItemCount
End If
'Now see if there is a match'
If SolveOne(TargetSum, ItemCount, Range, Items) Then TotalFound += 1
Next
Return TotalFound
End Function
Public Function SolveOne(ByVal TargetSum As Integer, ByVal ItemCount As Integer _
, ByVal Range As Integer, ByRef Items() As Integer) As Boolean
' Solve a single Subset Sum problem: determine if the TargetSum can be made from'
'the integer items.'
'first split the items into two half-lists: [O(n)]'
Dim H1() As Integer, H2() As Integer
Dim hu1 As Integer, hu2 As Integer
If ItemCount Mod 2 = 0 Then
'even is easy:'
hu1 = (ItemCount / 2) - 1 : hu2 = (ItemCount / 2) - 1
ReDim H1((ItemCount / 2) - 1), H2((ItemCount / 2) - 1)
Else
'odd is a little harder, give the first half the extra item:'
hu1 = ((ItemCount + 1) / 2) - 1 : hu2 = ((ItemCount - 1) / 2) - 1
ReDim H1(hu1), H2(hu2)
End If
For i As Integer = 0 To ItemCount - 1 Step 2
H1(i / 2) = Items(i)
'make sure that H2 doesnt run over on the last item of an odd-numbered list:'
If (i + 1) <= ItemCount - 1 Then
H2(i / 2) = Items(i + 1)
End If
Next
'Now generate all of the sums for each half-list: [O( 2^(n/2) * n )] **(this is the slowest step)'
Dim S1() As Integer, S2() As Integer
Dim sum1 As Integer, sum2 As Integer
Dim su1 As Integer = 2 ^ (hu1 + 1) - 1, su2 As Integer = 2 ^ (hu2 + 1) - 1
ReDim S1(su1), S2(su2)
For i As Integer = 0 To su1
' Use the binary bitmask of our enumerator(i) to select items to use in our candidate sums:'
sum1 = 0 : sum2 = 0
For b As Integer = 0 To hu1
If 0 < (i And Power2(b)) Then
sum1 += H1(b)
If i <= su2 Then sum2 += H2(b)
End If
Next
S1(i) = sum1
If i <= su2 Then S2(i) = sum2
Next
'Sort both lists: [O( 2^(n/2) * n )] **(this is the 2nd slowest step)'
Array.Sort(S1)
Array.Sort(S2)
' Start the first half-sums from lowest to highest,'
'and the second half sums from highest to lowest.'
Dim i1 As Integer = 0, i2 As Integer = su2
' Now do a merge-match on the lists (but reversing S2) and looking to '
'match their sum to the target sum: [O( 2^(n/2) )]'
Dim sum As Integer
Do While i1 <= su1 And i2 >= 0
sum = S1(i1) + S2(i2)
If sum < TargetSum Then
'if the Sum is too low, then we need to increase the ascending side (S1):'
i1 += 1
ElseIf sum > TargetSum Then
'if the Sum is too high, then we need to decrease the descending side (S2):'
i2 -= 1
Else
'Sums match:'
Return True
End If
Loop
'if we got here, then there are no matches to the TargetSum'
Return False
End Function
End Class
Here is the Forms code to go along with it:
Public Class frmSubsetSum
Dim ssm As New SubsetSum
Private Sub btnGo_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles btnGo.Click
Dim Total As Integer
Dim datStart As Date, datEnd As Date
Dim Iterations As Integer, Range As Integer, NumberCount As Integer
Iterations = CInt(txtIterations.Text)
Range = CInt(txtRange.Text)
NumberCount = CInt(txtNumberCount.Text)
ssm.ForceMatch = chkForceMatch.Checked
datStart = Now
Total = ssm.SolveMany(NumberCount, Range, Iterations)
datEnd = Now()
lblStart.Text = datStart.TimeOfDay.ToString
lblEnd.Text = datEnd.TimeOfDay.ToString
lblRate.Text = Format(Iterations / (datEnd - datStart).TotalMilliseconds * 1000, "####0.0")
ListBox1.Items.Insert(0, "Found " & Total.ToString & " Matches out of " & Iterations.ToString & " tries.")
ListBox1.Items.Insert(1, "Tics 0:" & ssm.w0 _
& " 1:" & Format(ssm.w1 - ssm.w0, "###,###,##0") _
& " 1a:" & Format(ssm.w1a - ssm.w1, "###,###,##0") _
& " 2:" & Format(ssm.w2 - ssm.w1a, "###,###,##0") _
& " 3:" & Format(ssm.w3 - ssm.w2, "###,###,##0") _
& " 4:" & Format(ssm.w4 - ssm.w3, "###,###,##0") _
& ", tics/sec = " & Stopwatch.Frequency)
End Sub
End Class
Let me know if you have any questions.
For the record, here is some fairly simple Java code that uses recursion to solve this problem. It is optimised for simplicity rather than performance, although with 100 elements it seems to be quite fast. With 1000 elements it takes dramatically longer, so if you are processing larger amounts of data you could better use a more sophisticated algorithm.
public static List<Double> getMatchingAmounts(Double goal, List<Double> amounts) {
List<Double> remaining = new ArrayList<Double>(amounts);
for (final Double amount : amounts) {
if (amount > goal) {
continue;
} else if (amount.equals(goal)) {
return new ArrayList<Double>(){{ add(amount); }};
}
remaining.remove(amount);
List<Double> matchingAmounts = getMatchingAmounts(goal - amount, remaining);
if (matchingAmounts != null) {
matchingAmounts.add(amount);
return matchingAmounts;
}
}
return null;
}