I am fascinated by board games. They provide immense fun, anyone can enjoy them, they are unpredictable and best of all they are great value for money. That is why whenever I get sometime I experiment with simulating games to know them better [read Why Monopoly board game is not as random as it appears]. So, out of curiosity I have created an excel sheet that can generate bingo / housie (housey) tickets – 24 of them at a time. To get new set of tickets you would hit F9 (recalculate).
Click here to download the bingo / housie ticket generator.
Note that these are Bingo UK / India / Australia variant I am talking about, not the US 5*5 type of bingo tickets.
Read on if you want to know how this is done:
According to Wikipedia:
A typical housie/bingo ticket .., contains fifteen numbers, arranged in nine columns by three rows. Each row contains five numbers and four blank spaces. Each column contains either one, two, or very rarely three, numbers:
* The first column contains numbers from 1 to 9,
* The second column numbers from 10 to 19,
* The third 20 to 29 and so on up until the last column, which contains numbers from 80 to 90.
I have removed number “90” from the list in order to reduce some complexity in generating the tickets.
The problem is now to “generate 15 random number between 1 to 89 and fill them in 15 random spots in a grid of 3 rows by 9 columns such that each row has exactly 5 numbers”
Now I could write a function in VBA to do this, but I wanted to do this only using formulas. So I started breaking the problem.
The first challenge is to select any random 5 cells in a 9 cell row
Once we select any random 5 cells in a 9 cell row, we will fill them with bingo numbers. Now, excel has a function to generate random numbers between 1 to 9 (=round(rand()*9,0)), but this is not good for us since each time we call this function we will get a random number between 1 to 9, where as we need a 5 random numbers without repetition between 1 to 9. The function is memoryless and could repeat numbers when called 5 times.
Instead we can list all the possible “5 cells with numbers and others are empty” combinations of a 9 cells region and select a random combination every time. There are essentially 9C5 i.e. 126 ways in which you can select any 5 cells out of 9 cells (without repetition of course).
So I listed all these combinations in a table and then randomly selected one of the combinations. You can see the first five such combinations in the image below:
Selecting any five cells out of nine cells
Now I created a 3*9 region and filled the cells with 1s or 0s, “1” when the cell in bingo ticket is supposed to have a value and “0” if the cell is empty as shown below:
Next challenge is to show random values in each cell
The trick here is that first column in our 3*9 bingo ticket has any number(s) from 1 to 9, second column has any number(s) from 10 to 19 …
Again, the challenge is the numbers should not repeat, otherwise we could simply use rand()*10, rand()*10+10, rand()*10+20 ... to generate the numbers.
This time it gets even more trickier because each column can have either no values, or 1 value or 2 values or 3 values.
The ticket generation logic now looks like:
- If the column has no values in it, then we will leave all the cells in that column of bingo ticket empty
- If the column has 1 value, we will generate any random number from that column’s range of possible values (1-9, 10-19,20-29,…80-89) and place it in the cell that is supposed to have a value and leave other cells empty.
- If the column has 2 values, we will generate 2 random numbers without repetition from that column’s range of possible values and place them in cells that are supposed to have them
- If the column has 3 values, we will generate 3 random numbers without repetition from that column’s range of possible values and place them in cells that are supposed to have them
As you can see, it is easy when the column has no or 1 value in it. But when the column has 2 or 3 I used the combinations trick described earlier.
First I created all 2 number combinations and 3 number combinations. Since the numbers on Bingo ticket are always sorted from top to bottom in a column, I just had to list down 45 combinations (10C2) for 2 numbers and 120 combinations (10C3) for 3 numbers.
The rest of the details are small enough that I can leave them to your imagination. So when the ticket is generated, it looks like this:
Remember to download housie / bingo ticket generator excel sheet and print your tickets at home. Just F9 to generate new set of tickets. Un-hide the rows from 43 if you want to see how this is done.
Like this post: Join our newsletter to get more yummy Excel stuff.
26 Responses to “Get busy this weekend, with OR XOR AND [Excel Homework]”
first solution for AND
The two numbers are in A1 and B1
= SUBSTITUTE (SUBSTITUTE (A1+B1*9*9, 9, 1), 8, 0)
regards
Stef@n
next solution for OR
=1*SUBSTITUTE (A1+A2;2;1)
regards
Stef@n
last solution for XOR
=1*SUBSTITUTE (A1+A2;2;0)
regards
Stef@n
Or you could make use of the VBA logical operators!
Define the following as custom functions
Public Function BITXOR(x As Long, y As Long)
BITXOR = x Xor y
End Function
Public Function BITAND(x As Long, y As Long)
BITAND = x And y
End Function
Public Function BITOR(x As Long, y As Long)
BITOR = x Or y
End Function
and then use them such:
A B =BITOR(A,B) =BITAND(A,B) =BITXOR(A,B)
0101 0100 0101 0100 0001
an another solution for AND
=1*SUBSTITUTE (SUBSTITUTE (A1+A2;1;0);2;1)
note:
the binary numbers are in A1 and A2 !
regards
Stef@n
I was obviously playing hooky at the beach during the bit-wise math lesson – you lost me at “Understanding bit-wise operations” 🙂
After looking at the above solutions, I find my solution silly, but still:
For the following formulae,
Row 1: headers,
Row 2: OR
Row 3: AND
Row 4: XOR
Column 1: Input 1
Column 2: Input 2
Column 3: Result
OR
{=SUM(IF(MID(A2,ROW(OFFSET($A$1,0,0,LEN(A2),1)),1)+MID(B2,ROW(OFFSET($A$1,0,0,LEN(B2),1)),1)>0,1,0)*10^(LEN(A2)-ROW(OFFSET($A$1,0,0,LEN(B2),1))))}
AND
{=SUM(IF(MID(A3,ROW(OFFSET($A$1,0,0,LEN(A3),1)),1)+MID(B3,ROW(OFFSET($A$1,0,0,LEN(B3),1)),1)=2,1,0)*10^(LEN(A3)-ROW(OFFSET($A$1,0,0,LEN(B3),1))))}
XOR
{=SUM(IF(MID(A4,ROW(OFFSET($A$1,0,0,LEN(A4),1)),1)+MID(B4,ROW(OFFSET($A$1,0,0,LEN(B4),1)),1)=1,1,0)*10^(LEN(A4)-ROW(OFFSET($A$1,0,0,LEN(B4),1))))}
@Anup
Please don't consider your solution silly
Firstly, You are the 3rd person to submit an answer
Secondly, The best formula/function is the one that you know and understand.
I think I have a very tedious solution, which people won't have the patience to do except in small numbers.
I used the same problem setup as "Anup Agarwal"
AND =IF(AND(MID(B2,1,1)="1",MID(C2,1,1)="1"),1,0)&IF(AND(MID(B2,2,1)="1",MID(C2,2,1)="1"),1,0)&IF(AND(MID(B2,3,1)="1",MID(C2,3,1)="1"),1,0)&IF(AND(MID(B2,4,1)="1",MID(C2,4,1)="1"),1,0)
OR =IF(OR(MID(B3,1,1)="1",MID(C3,1,1)="1"),1,0)&IF(OR(MID(B3,2,1)="1",MID(C3,2,1)="1"),1,0)&IF(OR(MID(B3,3,1)="1",MID(C3,3,1)="1"),1,0)&IF(OR(MID(B3,4,1)="1",MID(C3,4,1)="1"),1,0)
=IF(OR(AND(MID(B4,1,1)="1",MID(C4,1,1)="0"),AND(MID(B4,1,1)="0",MID(C4,1,1)="1")),1,0)&IF(OR(AND(MID(B4,2,1)="1",MID(C4,2,1)="0"),AND(MID(B4,2,1)="0",MID(C4,2,1)="1")),1,0)&IF(OR(AND(MID(B4,3,1)="1",MID(C4,3,1)="0"),AND(MID(B4,3,1)="0",MID(C4,3,1)="1")),1,0)&IF(OR(AND(MID(B4,4,1)="1",MID(C4,4,1)="0"),AND(MID(B4,4,1)="0",MID(C4,4,1)="1")),1,0)
Sorry my last post was totally messed up
AND
=IF(AND(MID(B2,1,1)="1",MID(C2,1,1)="1"),1,0)&IF(AND(MID(B2,2,1)="1",MID(C2,2,1)="1"),1,0)&IF(AND(MID(B2,3,1)="1",MID(C2,3,1)="1"),1,0)&IF(AND(MID(B2,4,1)="1",MID(C2,4,1)="1"),1,0)
OR
=IF(OR(MID(B3,1,1)="1",MID(C3,1,1)="1"),1,0)&IF(OR(MID(B3,2,1)="1",MID(C3,2,1)="1"),1,0)&IF(OR(MID(B3,3,1)="1",MID(C3,3,1)="1"),1,0)&IF(OR(MID(B3,4,1)="1",MID(C3,4,1)="1"),1,0)
XOR
=IF(OR(AND(MID(B4,1,1)="1",MID(C4,1,1)="0"),AND(MID(B4,1,1)="0",MID(C4,1,1)="1")),1,0)&IF(OR(AND(MID(B4,2,1)="1",MID(C4,2,1)="0"),AND(MID(B4,2,1)="0",MID(C4,2,1)="1")),1,0)&IF(OR(AND(MID(B4,3,1)="1",MID(C4,3,1)="0"),AND(MID(B4,3,1)="0",MID(C4,3,1)="1")),1,0)&IF(OR(AND(MID(B4,4,1)="1",MID(C4,4,1)="0"),AND(MID(B4,4,1)="0",MID(C4,4,1)="1")),1,0)
@stefan,
I just couldn't get your solutions to work.
01010101010 + 01010101110 = 02020210120
what am i doing wrong?
@anup
...I got yours to work!
@Stephen - I get the same, but Stef@an's second solution for AND does work (at least for the test cases I used)
@ Stephen / Rich
yes , you are right ! - only this works:
OR
=1*SUBSTITUTE (A1+A2;2;1)
XOR
=1*SUBSTITUTE (A1+A2;2;0)
AND
=1*SUBSTITUTE (SUBSTITUTE (A1+A2;1;0);2;1)
@Stef@n - You're answer is really smart, I never knew about the substitute function before. Great Work!
Thx Michael 🙂
yes - it is simply easy 😉
if you add 1 and 1 - excel calculate 2
and then you have to substitute the 2 - new = 0 respectively 1
Here is a good resource for people wanting to learn binary and hexadecimal.
http://justwebware.com/bitwise/bitwise.html
Three that weren't asked for:
NOT
=SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1+A2,0,3),1,0),3,1)
EQV
=SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1+A2,0,3),2,3),1,0),3,1)
IMP
=SUBSTITUTE(SUBSTITUTE(A1+SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A2,0,3),1,0),3,1),0,1),2,0)
(was using Daniel Ferry's bitwise file to verify against)
@ Kyle
Not only takes one parameter and inverts 0 -1 and 1-0
Took out the +A2
=SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1,0,3),1,0),3,1)
Great solutions!
I'll add two:
NAND =1*SUBSTITUTE (A1+A2,2,0)
NOR=1*SUBSTITUTE(SUBSTITUTE (SUBSTITUTE(A1+A2,0,2),1,0),2,1)
This will work for binary numbers of any size (although the text format mask will have to have as many zeroes as there are digits in the longest addend)
Assume binary #s are in C35 & C36, then add and format as text in C37:
=TEXT(C36+C35,"000000000000")
-sum- = 101112211112
AND - SUBSTITUTE 0s for 1s in -sum-, then sub 1s for 2s
=SUBSTITUTE(SUBSTITUTE(C37,"1","0"),"2","1")
OR - sub 1s for 2s in -sum-
=SUBSTITUTE(C37,"2","1")
XOR - sub 0s for 2s in -sum-
=SUBSTITUTE(C37,"2","0")
Just wandered by:
AND:
=SUBSTITUTE(A1+A2,1,0)/2
Clever, Shane. I like that.
[…] post http://www.excelhero.com/blog/2010/01/5-and-3-is-1.html for examples using Sumproduct, and http://chandoo.org/wp/2011/07/29/bitwise-operations-in-excel/ for examples using Text […]
Hi Chandoo,
I am not (yet) really into bitwise calculation, but I am looking for a way to speed up my vba calculation with very big numbers. Would is ben convenient to use bitwise notation for this?
Best regards,
Ronald (the Netherlands)
p.s. love your country!
@Ronald
I'd suggest asking this in the Chandoo.org Forums
https://chandoo.org/forum/
Attach a sample file with an example of some data and describe what you want to achieve