If I had the following pseudocode, would I need to add anything further onto it as mentioned below. Your help is much appreciated:
**repeat
swapped = false**
for i from 1 <- N
for j <- 0 to N - 1
if a[j] > a[j + 1]
swap( a[j], a[j + 1] )
**swapped = true
end if**
**end for
until not swapped**
Are the lines I have ** REQUIRED to be in there? For example, if a question asked 'write in pseudocode for the bubble sort algorithm' would I be required to write it out fully (including the ** items) or without them is OK?
We are required to 'rope learn' the code and obviously the smaller the code the better and the easier it is to remember.
Thanks!
Sub Main()
Dim Numlist() As Integer = {23435, 1, 433, 5234}
'here you can use any numbers in any order
Dim Count As Integer = 0
Dim Swapvalue As Integer
For ii = 0 To Numlist.Length - 2
For i = 0 To Numlist.Length - 2
Count += 1
If Numlist(i) > Numlist(i + 1) Then
Swapvalue = Numlist(i)
Numlist(i) = Numlist(i + 1)
Numlist(i + 1) = Swapvalue
End If
Next
Next
Console.WriteLine("Number of comparisons: {0}", Count)
For i = 0 To Numlist.Length - 1
Console.WriteLine(Numlist(i))
Next
End Sub
How to sum 2 numbers digit by digit with pseudo code?
Note: You don't know the length of the numbers - if it has tens, hundreds, thousands...
Units should be add to units, tens to tens, hundreds to hundreds.....
If there is a value >= 10 in adding the units you need to put the value of that ten with "the tens"....
I tried
Start
Do
Add digit(x) in A to Sum(x)
Add digit(x) in B to Sum(x)
If Sum(x) > 9, then (?????)
digit(x) = digit(x+1)
while digit(x) in A and digit(x) in B is > 0
How to show the result?
I am lost with that.....
Please help!
Try this,
n = minDigit(a, b) where a and b are the numbers.
let sum be a number.
m = maxDigit(a,b)
allocate maxDigit(a,b) + 1 memory for sum
carry = 0;
for (i = 1 to n)
temp = a[i] + b[i] + carry
// reset carry
carry = 0
if (temp > 10)
carry = 1
temp = temp - 10;
sum[i] = temp
// one last step to get the leftover carry
if (digits(a) == digits(b)
sum[n + 1] = carry
return
if (digits(a) > digits(b)
toCopy = a
else
toCopy = b
for (i = n to m)
temp = toCopy[i] + carry
// reset carry
carry = 0
if (temp > 10)
carry = 1
temp = temp - 10;
sum[i] = temp
Let me know if it helps
A and B are the integers you want to sum.
Note that the while loop ends when all the three integers are equal to zero.
carry = 0
sum = 0
d = 1
while (A > 0 or B > 0 or carry > 0)
tmp = carry + A mod 10 + B mod 10
sum = sum + (tmp mod 10) * d
carry = tmp / 10
d = d * 10
A = A / 10
B = B / 10
I'm developing a program in visual basic 6.0 to display magic square. I've developed the logic, but the values are not getting displayed in the magic square. Here's the code :
Private Sub Command1_Click()
Dim limit As Integer
Dim a(100, 100) As Integer
limit = InputBox("Enter the limit")
If limit Mod 2 = 0 Then ' Rows and columns must be
MsgBox "Can't be done", vbOKCancel, "Error"
Else ' set number of rows and columns to limit
mfgsquare.Rows = limit
mfgsquare.Cols = limit
j = (n + 1) / 2
i = 1
For c = 1 To n * n
mfgsquare.TextMatrix(i, j) = c
If c Mod n = 0 Then
i = i + 1
GoTo label
End If
If i = 1 Then
i = n
Else
i = i - 1
End If
If j = n Then
j = 1
Else
j = j + 1
End If
label:
Next c
End If
End Sub
Try this:
n = InputBox("Enter the limit")
If n Mod 2 = 0 Then ' Rows and columns must be
MsgBox "Can't be done"
Else ' set number of rows and columns to limit
mfgsquare.Rows = n + 1
mfgsquare.Cols = n + 1
For i = 1 To n
For j = 1 To n
mfgsquare.TextMatrix(i, j) = n * ((i + j - 1 + Int(n / 2)) Mod n) + ((i + 2 * j - 2) Mod n) + 1
Next j
Next i
End If
I have a list of 100 items. I'd like to randomly pair these items with each other. These pairs must be unique, so there are 4950 possibilities (100 choose 2) total.
Of all 4950 pairs, I'd like to have 1000 pairs randomly selected. But they key is, I'd like each item (of the 100 items) to overall appear the same amount of times (here, 20 times).
I tried to implement this with code a couple of times. And it worked fine when I tried with a lower amount of pairs chosen, but each time I try with the full 1000 pairs, I get stuck in a loop.
Does anyone have an idea for an approach? And what if I change the number of pairs I wish to select (e.g., 1500 rather than 1000 random pairs)?
My attempt (written in VBA):
Dim City1(4951) As Integer
Dim City2(4951) As Integer
Dim CityCounter(101) As Integer
Dim PairCounter(4951) As Integer
Dim i As Integer
Dim j As Integer
Dim k As Integer
i = 1
While i < 101
CityCounter(i) = 0
i = i + 1
Wend
i = 1
While i < 4951
PairCounter(i) = 0
i = i + 1
Wend
i = 1
j = 1
While j < 101
k = j + 1
While k < 101
City1(i) = j
City2(i) = k
k = k + 1
i = i + 1
Wend
j = j + 1
Wend
Dim temp As Integer
i = 1
While i < 1001
temp = Random(1,4950)
While ((PairCounter(temp) = 1) Or (CityCounter( (City1(temp)) ) = 20) Or (CityCounter( (City2(temp)) ) = 20))
temp = Random(1,4950)
Wend
PairCounter(temp) = 1
CityCounter( (City1(temp)) ) = (CityCounter( (City1(temp)) ) + 1)
CityCounter( (City2(temp)) ) = (CityCounter( (City2(temp)) ) + 1)
i = i + 1
Wend
Take a list, scramble it, and mark every two elements off as a pair. Add these pairs to a list of pairs. Ensure that list of pairs is sorted.
Scramble the list of pairs, and add each pair to a "staged" pair list. Check if it's in the list of pairs. If it's in the list of pairs, scramble and start over. If you get the entire list without any duplicates, add the staged pair list to the pair list and start this paragraph over.
Since this involves a nondeterministic step at the end I'm not sure how slow it will be, but it should work.
This is old thread, but I was looking for something similar, and finaly did it myself.
The algorithm is not 100% random (after being a bit "tired" with unsuccessfull random trials starts systematic screening of the table :) - anyway for me - "random enough") but works reasonably fast, and returns required table (unfortunalety not always, but...) usually every second or third use (look in A1 if there is your reqired number of pairs for each item).
Here is VBA code to be run in Excel environment.
Output is directed to current sheet starting from A1 cell.
Option Explicit
Public generalmax%, oldgeneralmax%, generalmin%, alloweddiff%, i&
Public outtable() As Integer
Const maxpair = 100, upperlimit = 20
Sub generate_random_unique_pairs()
'by Kaper 2015.02 for stackoverflow.com/questions/14884975
Dim x%, y%, counter%
Randomize
ReDim outtable(1 To maxpair + 1, 1 To maxpair + 1)
Range("A1").Resize(maxpair + 1, maxpair + 1).ClearContents
alloweddiff = 1
Do
i = i + 1
If counter > (0.5 * upperlimit) Then 'try some systematic approach
For x = 1 To maxpair - 1 ' top-left or:' To 1 Step -1 ' bottom-right
For y = x + 1 To maxpair
Call test_and_fill(x, y, counter)
Next y
Next x
If counter > 0 Then
alloweddiff = alloweddiff + 1
counter = 0
End If
End If
' mostly used - random mode
x = WorksheetFunction.RandBetween(1, maxpair - 1)
y = WorksheetFunction.RandBetween(x + 1, maxpair)
counter = counter + 1
Call test_and_fill(x, y, counter)
If counter = 0 Then alloweddiff = WorksheetFunction.Max(alloweddiff, 1)
If i > (2.5 * upperlimit) Then Exit Do
Loop Until generalmin = upperlimit
Range("A1").Resize(maxpair + 1, maxpair + 1).Value = outtable
Range("A1").Value = generalmin
Application.StatusBar = ""
End Sub
Sub test_and_fill(x%, y%, ByRef counter%)
Dim temprowx%, temprowy%, tempcolx%, tempcoly%, tempmax%, j%
tempcolx = outtable(1, x + 1)
tempcoly = outtable(1, y + 1)
temprowx = outtable(x + 1, 1)
temprowy = outtable(y + 1, 1)
tempmax = 1+ WorksheetFunction.Max(tempcolx, tempcoly, temprowx, temprowy)
If tempmax <= (generalmin + alloweddiff) And tempmax <= upperlimit And outtable(y + 1, x + 1) = 0 Then
counter = 0
outtable(y + 1, x + 1) = 1
outtable(x + 1, y + 1) = 1
outtable(x + 1, 1) = 1 + outtable(x + 1, 1)
outtable(y + 1, 1) = 1 + outtable(y + 1, 1)
outtable(1, x + 1) = 1 + outtable(1, x + 1)
outtable(1, y + 1) = 1 + outtable(1, y + 1)
generalmax = WorksheetFunction.Max(generalmax, outtable(x + 1, 1), outtable(y + 1, 1), outtable(1, x + 1), outtable(1, y + 1))
generalmin = outtable(x + 1, 1)
For j = 1 To maxpair
If outtable(j + 1, 1) < generalmin Then generalmin = outtable(j + 1, 1)
If outtable(1, j + 1) < generalmin Then generalmin = outtable(1, j + 1)
Next j
If generalmax > oldgeneralmax Then
oldgeneralmax = generalmax
Application.StatusBar = "Working on pairs " & generalmax & "Total progress (non-linear): " & Format(1# * generalmax / upperlimit, "0%")
End If
alloweddiff = alloweddiff - 1
i = 0
End If
End Sub
Have an array appeared[] which keeps track of how many times each item already appeared in answer. Let's say each element has to appear k times. Iterate over the array, and while current element has its appeared value less than k, choose a random pair for it from that element who also have appeared less than k times. Add that pair to answer and increase appearance count for both.
create a 2-dimensional 100*100 matrix of booleans, all False
of these 10K booleans, set 1K of them to true, with the following constraints:
the diagonal should stay empty
no row or column should have more than 20 true values
at the end, every row and column should have 20 True values.
Now, there is the X=Y diagonal symmetry. Just add the following constraints:
the triangle at one side of the diagonal should stay empty
in the above constraints, the restrictions for rows&columns should be combined/added
what is the algorithm (or rather formula) which, for each pair integers i and j with j >= i, gives an integer k = k(i,j) such that
k(0,0) = 0
k(i,j2) = k(i,j1)+1 for j2 = j1 + 1
k(i,0) = k(i-1,i-1) + 1 , i >= 1
holds?
In other words, if you fill up the left-lower part of matrix row by row from left to right with the natural numbers, starting at 0, how can you compute the value of a cell given the index of its row i and the column index j <= i?
Thank you very much!
proof of Alleo answer:
first write your second formula from j to 1
k(i,j)= k(i,j-1) + 1
k(i,j-1) = k(i,j-2) + 1
...
k(i,1) = k(i,0) + 1
sum up these formulas you get :
k(i,j) = k(i,0) + 1+1 ..+1 = k(i,0) + j (1)
now from your 3rd formula:
k(i,0) = k(i-1,i-1) + 1
using (1) :
k(i-1,i-1) = k(i-1,0) + i-1
then
k(i,0) = k(i-1,0) + i
then since k(0,0) = 0
k(i,0) = sum(p for p=0 to i) = i*(i+1)/2 (2)
then
(1) & (2) => k(i,j) = i*(i+1)/2 + j
This is i*(i+1)/2 + j. You are welcome to check