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.
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40 Responses to “Looking up when the data won’t co-operate (case study)”
Nice Trick.. Clever use of cell references
Here is a formula I tried to create:
=SUMPRODUCT(((NOT(ISERROR(SEARCH(L5,B4:H14))))*1),(B5:H15))
It takes care of Caveat #1 (can handle text), but Caveat #2 remains.
In situations like this, I will often use VBA to restructure the data (2 columns: dates and values) on to a new worksheet. I can then use this 'clean' source for data analysis (formula or pivot table).
=SUMPRODUCT(((NOT(ISERROR(SEARCH(L5,B4:H14))))*1),(B5:H15)) and complex formulae in general are all very well but when you come back to them in a few weeks / months time, it is not at all easy to see what they do and what the limitations are.
Hi Chandoo,
I had used this type of cell ref. various times while calculating average.
But for the situation here try below formula . Note this is an array formula and must be confirmed with Ctrl+Shift+Enter.
=SMALL(IF(MMULT((L5=B4:H14)*IF(ISNUMBER(B5:H15),B5:H15),{1;1;1;1;1;1;1}),MMULT((L5=B4:H14)*IF(ISNUMBER(B5:H15),B5:H15),{1;1;1;1;1;1;1})),1)
Regards,
Hi, I think Using SEARCH in here will create a problem say there is a text like SUN and another text SUNLIGHT both result will be added by SUMPRODUCT.
Regards,
Array option.
=SUM(IF(MOD(ROW(B4:B14),2)=MOD(ROW(B4),2),IF(B4:H14=L5,B5:H15)))
Regards
@Elias,
Nice approach.
Although not requested - the formula I suggested closes all possibilities.
Criteria: _____ Value to retrieve:
Date__________ Numeric
Date__________Textual
Textual________Numeric
Textual________Textual
While your formula copes with only the 3 first combinations.
Michael (Micky) Avidan
“Microsoft® Answer” – Wiki author & Forums Moderator
“Microsoft®” MVP – Excel (2009-2015)
ISRAEL
@Michael,
Sorry but I don’t understand your point. I believe the challenge was to return the summary of a given date. What is your really volatile formula doing that mine is not?
Regards
@Elias,
I didn't say that the challenge differs from what you just mentioned/aimed to nor that your formula doesn't provide the requested result.
Please read my previous comment again and focus on the last combination (TEXT / TEXT).
I, myself, always try to provide a global Formula that is capable to handle all sorts of data.
Michael (Micky) Avidan
“Microsoft® Answer” – Wiki author & Forums Moderator
“Microsoft®” MVP – Excel (2009-2015)
ISRAEL
@Michael,
I see your point, but you are missing the below points if you are trying to cover all sorts of data.
What happened if the lookup value does not exist?
Do you want the first, second, summary, concatenation of the values if the look value is repeated?
See they are too many possibilities to be cover with just one formula.
Regards
1) The range: B4:H15 was named: RNG.
2) The following Array Formula was "retrieved from my sleeve" and I hope it can be shorten.
3) The formula seems to take care of BOTH(!) caveats.
-------------------------------------------------------------------------------
=OFFSET(INDIRECT(ADDRESS(SMALL(IF(RNG=L5,ROW(RNG),""),1),MOD(SMALL(IF(RNG=L5,(ROW(RNG))+COLUMN(RNG)/10),1),1)*10)),1,)
-------------------------------------------------------------------------------
Michael (Micky) Avidan
“Microsoft® Answer” – Wiki author & Forums Moderator
“Microsoft®” MVP – Excel (2009-2015)
ISRAEL
@Michael,
Check what happened with your result if you type 41927 in D5.
Regards
Correct. Didn't predict that.
Will find time to work something out.
Michael (Micky) Avidan
“Microsoft® Answer” – Wiki author & Forums Moderator
“Microsoft®” MVP – Excel (2009-2015)
ISRAEL
@Elias,
Let's hope the following Array Formula "closes all open doors".
Again - it has nothing to do with your formula which works fine as long as the 3 first mentioned combinations are concerned.
-------------------------------------------------------------------------------
=INDEX(RNG,LARGE(IF(RNG=L5,MOD(ROW(RNG)-1,2)*(ROW(RNG)),""),1)-2,(MOD(SMALL(IF(RNG=L5,(ROW(RNG))+COLUMN(RNG)/10),1),1)*10)-1)
-------------------------------------------------------------------------------
Michael (Micky) Avidan
“Microsoft® Answer” – Wiki author & Forums Moderator
“Microsoft®” MVP – Excel (2009-2015)
ISRAEL
Ok, if you insist. The following will cover all the scenarios you listed. However, I’ll never recommend/use such of formula.
Defined names:
rDat = $B$4:$H$15
rRow =ROW(rDat)-MIN(ROW(rDat))+1
rCol =COLUMN(rDat)-MIN(COLUMN(rDat))+1
rInc =MOD(rRow,2)=MOD(MIN(rRow),2)
L6=INDEX(rDat,MAX(IF(rInc,IF(rDat=L5,rRow)))+1,MAX(IF(rInc,IF(rDat=L5,rCol))))
Array Enter
Regards
@Michael,
unfortunately, your array formula still seems to return wrong results (eg 3-Nov).
If data are organized like in the example, ie. looks like a calendar, the INDEX formula seems quite simple:
=INDEX($B$4:$H$15,ROUNDDOWN((L5-B4)/7,0)*2+2,MOD((L5-B4),7)+1)
Yours is effectively the same as what I just came up with, and I believe this is the optimal answer to this particular problem.
My solution, before I saw yours:
=OFFSET(B5,QUOTIENT(L5-B4,7)*2,MOD(L5-B4,7))
OFFSET will work for an arbitrary list size, but INDEX might be easier to read.
QUOTIENT does the round and division in a single step.
If there's an improvement over Elias's solution then I for one can't see it.
Perhaps a non-CSE version which would also mean that only two references (B4:H14 and B5:H15), as opposed to three (B4, B4:B14 and and B5:H15), would require manually amending should the data range change, i.e.:
=SUMPRODUCT((ISEVEN(ROW(B4:H14)-MIN(ROW(B4:H14)))*(B4:H14=L5)*B5:H15))
I suppose we could make it a single, uniform range reference:
=SUMPRODUCT((ISEVEN(ROW(B4:H14)-MIN(ROW(B4:H14)))*(B4:H14=L5)*OFFSET(B4:H14,1,,,)))
which might be more appropriate should we e.g. wish to use a Defined Name for our range, i.e.:
=SUMPRODUCT((ISEVEN(ROW(Rng)-MIN(ROW(Rng)))*(Rng=L5)*OFFSET(Rng,1,,,)))
though whether that compensates for the extra, volatile function call is something to be debated.
Regards
I have tried something and then my Excel workbooks got shut down. Maybe that was too much?
Anyway here is what I've tried:
=SUMPRODUCT(INDEX(B5:H15;IF(ISEVEN(ROW(B5:H15));ROW(B5:B15)-ROW(B5)+1);{1\2\3\4\5\6\7}))
Guess that was wrong? Would this approach work anyway?
Looking forward to learn something from you Excel Experts.
Sorry, I haven't took notice of XOR LX's answer. I guess that's kind of what I was looking for.
@Michael Avidan
As it stands that is not a very rigorous construction.
You say "I, myself, always try to provide a global Formula that is capable to handle all sorts of data", which is a wonderful philosophy, but isn't it at least as important that we ensure that our formulas are independent of the row and column references of the data range in question, so that, should that range change, we do not have to re-work our solution?
What happens with your formula, for example, if RNG is instead re-located one row down, from B4:H15 to B5:H16?
When a formula is reliant upon the addition/subtraction of certain constants within the formula, which themselves are necessarily dependent upon the specific rows/columns in which the data lies at any given time (e.g. the -1 in MOD(ROW(RNG)-1,2)), then that formula is not a very flexible one.
Hence the reason for my choice of a slightly longer construction:
ROW(B4:H14)-MIN(ROW(B4:H14))
which ensures that this part of the calculation is not dependent upon the precise location of the data range within the worksheet, and so will give correct answers even if that range is re-located.
Regards
{=OFFSET(B4,MAX((B4:H15=L5)*ISODD(ROW(1:12))*ROW(1:12)),MAX((B4:H15=L5)*ISODD(ROW(1:12))*COLUMN(A:G))-1)}
Non-array formula:
=INDEX(B4:H15,SUMPRODUCT((B4:H15=L5)*(ROW(B4:H15)-ROW(B4)+1)*ISODD(ROW(B4:H15)-ROW(B4)+1))+1,SUMPRODUCT((B4:H15=L5)*(COLUMN(B4:H15)-COLUMN(B4)+1)*ISODD(ROW(B4:H15)-ROW(B4)+1)))
Using one range (B4:H15), one reference (B4), one lookup value (L5) and no INDIRECT or OFFSET.
My trial with defined names:
DateRange
=$B$4:$H$4,$B$6:$H$6,$B$8:$H$8,$B$10:$H$10,$B$12:$H$12,$B$14:$H$14
Position
=RANK('lookup problem'!$L$5,DateRange,1)
L6
=OFFSET(B4,ROUNDUP(Position/7,0)*2- 1,IF(MOD(Position,7)=0,6,MOD(Position,7)-1))
I'd probably just run with something like:
=SUMPRODUCT((B4:H14=L5)*(MOD(ROW(B4:H14),2)=MOD(ROW(B4),2))*B5:H15)
...which is basically the same as Elias' but without the IFs
The opposite of elegant but it works...
=INDEX(B4:H15,IFERROR(MATCH(L5,B4:B14,0),0)+IFERROR(MATCH(L5,C4:C14,0),0)+IFERROR(MATCH(L5,D4:D14,0),0)+IFERROR(MATCH(L5,E4:E14,0),0)+IFERROR(MATCH(L5,F4:F14,0),0)+IFERROR(MATCH(L5,G4:G14,0),0)+IFERROR(MATCH(L5,H4:H14,0),0)+1,IFERROR(MATCH(L5,B4:H4,0),0)+IFERROR(MATCH(L5,B6:H6,0),0)+IFERROR(MATCH(L5,B8:H8,0),0)+IFERROR(MATCH(L5,B10:H10,0),0)+IFERROR(MATCH(L5,B12:H12,0),0)+IFERROR(MATCH(L5,B14:H14,0),0))
=INDEX(B4:H15,
IFERROR(MATCH(L5,B4:B14,0),0)+
IFERROR(MATCH(L5,C4:C14,0),0)+
IFERROR(MATCH(L5,D4:D14,0),0)+
IFERROR(MATCH(L5,E4:E14,0),0)+
IFERROR(MATCH(L5,F4:F14,0),0)+
IFERROR(MATCH(L5,G4:G14,0),0)+
IFERROR(MATCH(L5,H4:H14,0),0)+1,
IFERROR(MATCH(L5,B4:H4,0),0)+
IFERROR(MATCH(L5,B6:H6,0),0)+
IFERROR(MATCH(L5,B8:H8,0),0)+
IFERROR(MATCH(L5,B10:H10,0),0)+
IFERROR(MATCH(L5,B12:H12,0),0)+
IFERROR(MATCH(L5,B14:H14,0),0))
Named Range
rownum = SUMPRODUCT(('lookup problem'!$B$4:$H$14='lookup problem'!$L$5)*ROW('lookup problem'!$B$4:$H$14)*ISEVEN(ROW('lookup problem'!$B$4:$H$14)))
Formula
=OFFSET($A$1,rownum,MATCH(L5,INDIRECT("$B"&rownum&":$H"&rownum),0))
How about SUM(IF(B4:H14=L5,B5:H15)) with array..it should work
Sorry, Chandoo, you can't find stuff this way in every possible scenario.
What if 2014-10-01 sales would equal 41.927 ? Which is serial number for 2014-10-15 ? SUMIF would fail to retrive correct answer. And your example data suggest that such number is possible in your table.
It's better not to search through dates and numbers at the same time.
If I'd solve a problem like this, it'd reformat table first so I get one column with dates and the other with numbers.
In this case, formula to form date column would be:
=INDIRECT(ADDRESS((INT((ROW()-4)/COUNT($B$4:$H$4))+1)*2+2;MOD(ROW()-4;COUNT($B$4:$H$4))+2;4;1))
and numbers would be the same formula with sight adjustment (+3 instead of +2 at the end of first argument):
=INDIRECT(ADDRESS((INT((ROW()-4)/COUNT($B$4:$H$4))+1)*2+3;MOD(ROW()-4;COUNT($B$4:$H$4))+2;4;1))
And now you got two columns that you can safely use for searching!
Oops, sorry, you actually mentioned that it doesn't work if number=date! I missed that part 🙁
={OFFSET(A1,SUM((B4:H14=L5)*ROW((B4:H14))),SUM((B4:H14=L5)*COLUMN((B4:H14)))-1)}
Works for all data... the solution I got for indirect looks little lengthy
I want to count last 20 records of a person, whose marks is greater than 2 and grade "manager". ....
Assume A1 has got names (James, John...etc...)
A2 "Manager"
A3 "2"
Someone please reply
I want to count last 20 records of a person, whose marks is greater than "2" and grade "Manager"
Assume A1 "geroge" A2 "Michael" A3 "George" etc...name can found anywhere in the rows
B1 "Manager" B2" clerk"
C1 "2" C2, "4"
please reply
Simplest I can come up with. No limitations for either 1 or 2. This does assume dates are an ordered list with 7 per row, and 2 rows per set. Assuming this is always true this will work for an arbitrary long list of dates.
=OFFSET(B5,QUOTIENT($L$5-$B$4,7)*2,MOD($L$5-$B$4,7))
@Marc,
Nice approach - however, as there are no "Negative Dates" - try:
=OFFSET(B5,INT(L5-B4)/7)*2,MOD(L5-B4,7))
——————————————————————————-
Michael (Micky) Avidan
“Microsoft® Answer” – Wiki author & Forums Moderator
“Microsoft®” MVP – Excel (2009-2015)
ISRAEL
=OFFSET(B4,ROUNDUP((L5-41911+1)/7,0)*2-1,MOD(L5-41911,7))
B4 has been used as reference cell for OFFSET().
FOR ROWS:
ROUNDUP(....,0) gives the integer value of a division. In case of presence of a remainder, ROUNDUP will add 1 to the Quotient.
As opposed to ROUNDUP(), the INT() or QUOTIENT() functions eliminate the remainder.
41911 = 01-Sept-2014, the first date in the data.
*2 has been used because there are 2 columns per set of data.
/7 has been used because there are 7 columns per set of data.
For columns
MOD(L5-41911,7))
Vijaykumar Shetye,
Panaji, Goa, India
This is how i did it
{=INDEX(B4:H15, MAX((L5=B4:H15)*ROW(B4:H15))-2, MAX((L5=B4:H15)*COLUMN(B4:H15))-1 )}
Here's my solution:
=INDEX(B4:H15,MATCH(1,MMULT(--(B4:H15=L5),TRANSPOSE(COLUMN(B4:H15)^0)),0)+1,MATCH(1,MMULT(TRANSPOSE(--(B4:H15=L5)),ROW(B4:H15)^0),0))
Sorry, forgot to mention Ctrl Shift Enter is needed.