Create a bingo card generator

Question

Bingo

Bingo is a numbers game where players match randomly drawn numbers to the numbers on their cards. Each bingo card is a square grid with 25 spaces, and the columns of the grid are labeled with letters such as "B", "I", "N", "G", "O". The letters help to identify the number range in each column, for example, the "B" column may contain numbers 1 to 15.

One of the spaces in the middle of the grid is often a "free" space, and is marked as such. To play the game, each player is given a bingo card with numbers on it. A caller then draws numbers randomly and announces them, and players mark the corresponding numbers on their cards. The first player to match a pre-determined pattern of marked numbers, such as a straight line, all four corners, or a full card, calls out "Bingo!" to win the game. The game continues until one player wins, or multiple players win on the same round. [1]

---

Challenge

Create a set of n bingo cards that are printed / outputted in sequence separated by one or two empty lines. The center field should be marked as free.

The output should be a 5x5 grid where columns are either separated by a single space, a tab or the grid should be fixed width, and the lines are separated by a single new line.

The randomness must be uniform, i.e. all possible permutations and combinations should be equally likely to occur. All numbers must be unique.

The numbers:

B = 1 - 15
I = 16 - 30
N = 31 - 45
G = 46 - 60
O = 61 - 75

The number n must be taken as input.

Example output for n = 3:

B I N G O
1 16 31 46 61
2 17 32 47 62
3 18 free 48 63
4 19 34 49 64
5 20 35 50 65
B I N G O
6 30 40 53 64
5 20 35 50 70
4 25 free 51 69
7 22 32 52 68
11 23 38 49 67
B I N G O
10 25 40 55 70
11 26 41 56 71
12 27 free 57 73
13 28 43 58 73
14 29 44 59 74

Yes, those were definitely randomly created.

---

_{This challenge is based on the fact that I recently had to create 100 Bingo cards. It turned out that this part of the challenge was the easy one. Making it pretty, printable and presentable was the time consuming part.}

[1] https://chat.openai.com/chat

• Stewie Griffin: Not a duplicate of this similarly named challenge.
• Shaggy: Can the "BINGO" be lowercase?
• gildux: Doesn't this challenge duplicate codegolf.stackexchange.com/questions/2631/bingo-card-generator ?
• Stewie Griffin: @gildux, No. In my opinion it's not even close. Creating a list of 5 * 15 random numbers is a whole lot different than this one. The Python answer in this one is 193 bytes while a pyhton answer in the other one is 84 bytes. That alone should show the difference, in my opinion.
• Stewie Griffin: In "my" language, MATLAB, golfing "create 5*15 random numbers" involves a completely different code than the code required to solve this challenge. Mixing numeric values with characters takes a bit of effort, and so does doing it several times without duplicating cards. I agree that in a language where replacing a numeric value with a character and adding a header to a numeric(?) array is trivial, then the differences aren't that big. In my, albeit biased, opnion, these are two different challenges that require two different approaches in some languages, and is therefore not a duplicate.

Answer 1

Vyxal J, 29 bytes

(‛6ṫ⇧Ṅ75ɾ15ẇƛÞc/o5Ẏ;∩2≬2‛λċȦV⁋¶

Try it Online!

Explained (old)

( # input times:
75ɾ15ẇ # push the range [1, 75] split into chunks of 15 items
ƛÞc/o5Ẏ; # to each: random_permutation[:5]
∩ # transpose
2≬2‛λċȦV # tos[2][2] = "free"
‛6ṫ⇧p # .prepend("BINGO")
vṄ⁋ # .map(" ".join(x)).join("\n")
¶+, # + "\n" (then print)

Answer 2

QBasic, 211 bytes

DIM b(5,5)
RANDOMIZE TIMER
INPUT n
FOR i=1TO n
ERASE b
?
?"B","I","N","G","O
FOR r=0TO 4
FOR c=0TO 4
4x=INT((c+RND)*15+1)
FOR p=0TO r
IF x=b(p,c)GOTO 4
NEXT p
IF(r=2)*(c=2)THEN?"free",ELSE?x,:b(r,c)=x
NEXT c,r,i

Try it at Archive.org! Example run:

An example run with input 3, showing 3 randomly generated bingo cards

Explanation/ungolfed

DIM b(5, 5)
RANDOMIZE TIMER
INPUT n

Declare b as a 5x5 array; we'll use this to track already-selected numbers and avoid duplicates. Seed the pseudo-random number generator using the current time. Prompt the user for the number of cards n.

FOR i = 1 TO n

Loop n times:

ERASE b
PRINT
PRINT "B", "I", "N", "G", "O"

Set all values in b to zero. Print a blank line followed by the B I N G O header. The commas in the PRINT statements separate the items by tabs, using a default tab stop of 14 columns.

FOR r = 0 TO 4
FOR c = 0 TO 4

Loop over rows 0 through 4; for each row, loop over columns 0 through 4.

4 x = INT((c + RND) * 15 + 1)

Label this as line number 4, which we'll use as a GOTO target later. Set x to be a random integer, greater than or equal to c*15+1, and less than (c+1)*15+1.

FOR p = 0 TO r
IF x = b(p, c) THEN GOTO 4
NEXT p

Loop over each previous row p. If x equals the value in array b at row p, column c, go back to line 4 and generate a new random number.

IF (r = 2) * (c = 2) THEN
 PRINT "free",
ELSE
 PRINT x,
 b(r, c) = x
END IF

If both r and c are 2, print the string "free"; otherwise, print x and set the appropriate value in the b array to x. In both cases, end the print statement with a tab rather than a newline.

NEXT c, r, i

Close all the FOR loops.

---

Astute readers may have noticed this doesn't print a newline at the end of each row. Because of the size of the default tab stop, it just so happens that the sixth column wraps to the next line, so printing a newline explicitly is unnecessary (and, in fact, results in unwanted blank lines). This might be bending the I/O requirements a bit. If an explicit newline must be printed, modify the B I N G O line by adding ", at the end, and add a line containing ? in between the FOR r and FOR c headers.

Answer 3

Factor, 123 bytes

[ [ "free"2 75 [1,b] 15 group [ 5 sample ] map flip
[ third set-nth ] keep "BINGO"1 group prefix simple-table. nl ] times ]

Try it online!

• [ ... ] times Do something a number of times.

"free" ! "free"
2 ! "free" 2
75 ! "free" 2 75
[1,b] ! "free" 2 { 1 2 ... 75 }
15 ! "free" 2 { 1 2 ... 75 } 15
group ! "free" 2 { { 1 2 .. 15 } { 16 17 .. 30 } ... { 61 62 .. 75 } }
[ 5 sample ] map ! "free" 2 {
 ! { 9 13 14 1 5 }
 ! { 26 24 25 27 22 }
 ! { 32 38 41 31 33 }
 ! { 54 59 55 49 57 }
 ! { 75 74 70 71 63 }
 ! }
flip ! "free" 2 {
 ! { 9 26 32 54 75 }
 ! { 13 24 38 59 74 }
 ! { 14 25 41 55 70 }
 ! { 1 27 31 49 71 }
 ! { 5 22 33 57 63 }
 ! }
[ third set-nth ] keep ! {
 ! { 9 26 32 54 75 }
 ! { 13 24 38 59 74 }
 ! { 14 25 "free" 55 70 }
 ! { 1 27 31 49 71 }
 ! { 5 22 33 57 63 }
 ! }
"BINGO" ! { ... } "BINGO"
1 group ! { ... } { "B" "I" "N" "G" "O" }
prefix ! {
 ! { "B" "I" "N" "G" "O" }
 ! { 9 26 32 54 75 }
 ! { 13 24 38 59 74 }
 ! { 14 25 "free" 55 70 }
 ! { 1 27 31 49 71 }
 ! { 5 22 33 57 63 }
 ! }
simple-table. ! output to stdout
nl ! newline

Answer 4

JavaScript (ES6), 132 bytes

f=N=>(F=n=>N?n<30?F[k=n%5]^(F[k]|=1<<(v=Math.random()*15))?(n++-17?"_BINGO"[n]||k*15-~v:'free')+` 
`[k>>2]+F(n):F(n):`
`+f(N-1):n)``

Try it online!

Commented

f = N => ( // outer function taking the number N of grids
 F = n => // inner function taking a counter n
 N ? // if there's still a grid to process:
 n < 30 ? // if n is less than 30:
 F[k = n % 5] // k is the column index in [0..4]
 ^ // if F[k] is not equal to its updated value
 ( F[k] |= 1 << ( // where the floor(v)-th bit is set
 v = // where v is
 Math.random() // randomly picked in [0,15[
 * 15 //
 )) ? // then:
 ( n++ - 17 ? // if this is not the center cell:
 "_BINGO"[n] // append either a BINGO letter (if n <= 5)
 || // or
 k * 15 - ~v // the number k * 15 + floor(v + 1)
 : // else:
 'free' // append the word 'free'
 ) + //
 ` \n`[k >> 2] + // append a line-feed if k = 4,
 // or a space otherwise
 F(n) // append the result of a recursive call to F
 : // else:
 F(n) // try again
 : // else:
 `\n` + // append a line-feed followed by
 f(N - 1) // the result of a recursive call to f
 : // else:
 n // stop the recursion
)`` // initial call to F with n zero'ish

• Shaggy: Grids need only be separated by a single newline, so you can drop the one before the recursive call to f.
• Arnauld: @Shaggy Not 100% sure about that. The challenge says separated by one or two *empty* lines.

Answer 5

Bash + GNU utils, 120

for((;j++<1ドル;));{
for((;++i<75;));{
a+=" <(shuf -i$i-$[i+=14] -n5)"
}
eval paste$a|sed '1iB	I	N	G	O
3s/\w*/free/3
$a
'
}

Try it online!

Answer 6

Japt -mR, (削除) 39 (削除ここまで) 38 bytes

75õ òF Ëö¤v5Ãg2Èh2`fe`ÃÕi"BINGO"¬ m ̧·

Try it

75õ òF Ëö¤v5Ãg2Èh2`fe`ÃÕi"BINGO"¬ m ̧· :Implicit map of range [0,input)
75õ :Range [1,75]
 ò :Partitions of length
 F : 15
 Ë :Map
 ö¤ : Shuffle
 v5 : First 5 elements
 Ã :End map
 g2 :Modify the element at 0-based index 2 by
 È :Passing it through the following function
 h2 : Replace the element at 0-based index 2 with
 `fe` : Compressed string "free"
 Ã :End modification
 Õ :Transpose
 i :Prepend
 "BINGO"¬ : "BINGO", split into an array
 m :Map
 ̧ : Join with spaces
 · :Join with newlines
 :Implicit output joined with newlines

• noodle person: Came up with this independently but it's pretty close to yours: 37
• noodle person: 37 with explanation
• noodle person: Oops golfed yours to 36
• Shaggy: That won't work, @Jacob; A.ö(n) allows for duplicates to be included, hence why I'm shuffling and then grabbing the first 5 items.

Answer 7

Charcoal, (削除) 54 (削除ここまで) (削除) 52 (削除ここまで) 49 bytes

FN«≔⪪BINGO1θF25⊞θ⎇=κ12free×ばつ15%κ5...1¦16θ⟦E⪪θ5⪫κ 

Try it online! Link is to verbose version of code. Explanation: Now inspired by @Arnauld's answer.

FN«

Input n and loop that many times.

≔⪪BINGO1θ

Start with a list of BINGO as separate characters which will make up the first row.

F25

Loop 25 times.

⊞θ⎇=κ12free×ばつ15%κ5...1¦16θ

Put the word free in the middle cell, otherwise take a random element of the appropriate numeric range for the cell but subtracting all of the numbers generated so far, and push the result to the list.

⟦E⪪θ5⪫κ 

Split into rows of five cells, join the rows on spaces and output the resulting board double-spaced from the next board.

Answer 8

Pyth, 39 bytes

VQkjd"BINGO"jjL\ Cc5X%3s.SMc5S75y6"free

Try it online!

Explanation

 # implicitly assign Q = eval(input())
VQ # for N in range(Q):
 k # print a newline
 jd"BINGO" # print BINGO joined on spaces
 S75 # range(1,76)
 c5 # split into 5 equal parts
 .SM # randomly shuffle each part
 s # join the parts back together
 %3 # take every third element
 X y6"free # replace the 12th element with "free"
 c5 # split into 5 equal parts again
 C # transpose
 jL\ # join each part on whitespace, converting all elements to strings
 j # join on newlines and print

• CursorCoercer: @Arnauld See, I knew that was part of the challenge and then forgot to do it before posting. Fixed now for a measly 3 bytes. Thanks for keeping me honest :)
• Stewie Griffin: OP's note: I'll allow the single leading newline. :)

Answer 9

05AB1E, 32 bytes

F75L5äε.r5£N<i'22ドルǝ}}ø‘Èß‘Sš»,¶?

Try it online.

Explanation:

F # Loop the (implicit) input-integer amount of times:
 75L # Push a list in the range [1,75]
 5ä # Split it into five equal-sized parts
 ε # Map over each part:
 .r # Randomly shuffle the list
 5£ # Then only keep the first 5 integers
 N<i # If the index is 2:
 '2ドル '# Push dictionary string "free"
 2ǝ # Insert it at (0-based) index 2 into the quintuplet-list
 }} # Close both the if-statement and map
 ø # Zip/transpose the matrix; swapping rows/columns
 ‘Èß‘ # Push dictionary string "BINGO"
 S # Convert it to a list of characters: ["B","I","N","G","O"]
 š # Prepend it to the matrix
 » # Join each inner list by spaces; and then each string by newlines
 , # Pop and print it with trailing newline
 ¶? # And print an additional newline character

See this 05AB1E tip of mine (section How to use the dictionary?) to understand why '2ドル is "free" and ‘Èß‘ is "BINGO".

Answer 10

Pip, (削除) 50 (削除ここまで) 49 bytes

La{Y ZSH*,75円CH5y@2@2:"free"P_JsPBnMyH5PE"BINGO"}

Attempt This Online!

Explanation

La{...}

Given command-line argument a, loop that many times:

Y ZSH*,75円CH5
 ,75円 ; Inclusive range from 1 to 75
 CH5 ; Chop into 5 equal sections
 SH* ; Randomly shuffle each section
 Z ; Transpose
Y ; Yank that result into the y variable
y@2@2:"free" ; Set the element of y at index 2,2 to the string "free"
P_JsPBnMyH5PE"BINGO"
 yH5 ; Take the first 5 elements of y
 PE"BINGO" ; Prepend the element "BINGO"
 M ; Map to each:
 _Js ; Join on spaces
 PBn ; and push a newline to the end of the resulting string
P ; Print that list, concatenated, with a trailing newline

Answer 11

C (gcc), (削除) 262 (削除ここまで) 225 bytes

• -37 thanks to c--

a[6][5]={L"BINGO"};f(n,i,j,k,r){--n&&puts("",f(n));for(j=5;j--;)for(k=i=0;k||++i<6;k||(a[i][j]=r))for(r=j*15+rand(k=i)%15+1;--k&&a[k][j]-r;);for(i=0;i<6;i+=puts(""))for(j=0;j<5;)printf(i*j-9?i?"%d	":"%c	":"free	",a[i][j++]);}

Try it online!

Ungolfed (` used instead of literal tab character):

a[6][5]={L"BINGO"}; // bingo array, with header
f(n, // number of cards to generate
 i,j, // row and column
 k, // random number row checker
 r // random number
 ) {
 --n && puts("",f(n)); // recursively print cards, adding newline
 for(j=5;j--;) // for each column
 for(k=i=0;k||++i<6;k||(a[i][j]=r)) // for each row (redo duplicates)
 for(r=j*15+rand(k=i)%15+1;--k&&a[k][j]-r;); // detect duplicates
 for(i=0;i<6;i+=puts("")) // mark middle item as free; print board
 for(j=0;j<5;)
 printf(i*j-9?i?"%d`":"%c`":"free`",a[i][j++]);
}

• c--: it looks like you don't need to clear the array at all, because you only check for duplicates in the elements you just set
• c--: 225 bytes
• ceilingcat: 221 bytes

Answer 12

APL (Dyalog Extended), 46 bytes

Full program. Prompts for n.

('BINGO'{'free'} ̈@3@4⍤⍪(×ばつ...4)+⍤1⍉)⍤2↑5? ̈⎕5⍴15

Try it online!

⎕5⍴15 prompt for n and join that 5, then use that as dimensions for an array of 15s

5? ̈ for each 15, get 5 random numbers in the range 1 through that, without replacement

↑ mix this matrix of lists into an n-by-5-by-5 array with each row being a randome 5-of-15

(...)⍤2 on each layer:

⍉ transpose, so the 5-of-15 rows become columns

(...)+⍤1 add the following to each row:

...4 the numbers 0 through 4

×ばつ fifteen multiplied by those

'BINGO'...⍤⍪ prepend a row of the given letters, then:

...@4 at row 4:

...@3 at column 3:

{...} ̈ replace each value (though there's only one) with:

'free' the given word

• Adám: @Shaggy Lack of sleep, I guess. Should be alright now.

Answer 13

Jelly, (削除) 38 (削除ここまで) 36 bytes

75s15Ẋ€ḣ5ドルZ"ƝM»3ドル¦3ドル¦"BINGO"ṭṚK€Y7Ṅ)

Try it online!

Most of the credit goes to @cairdcoinheringaahing and @UnrelatedString for helping me in chat.

• -2 thanks to UnrelatedString

This can probably be a bit shorter, though.

Explanation

75 # 75 (implicit range)
 s15 # Split into 15 parts
 Ẋ€ # Random permutation of each
ḣ5ドル # First 5 elements of each
 Z # Transpose
 "ƝM» # Replace ↓ with "free"
3ドル¦3ドル¦ # The third item in the third item (1-indexed)
"BINGO" # Push "BINGO"
 ṭṚ # Append and reverse (i.e. prepend)
K€ # Join each by spaces
 Y7 # Join by newlines and append a newline
 Ṅ # Print
 ) # Repeat input number of times

Answer 14

Python 3, 151 bytes

from random import *
print("B I N G O")
x=lambda n:n*15+randint(0,15)
[print(*[*("free"if(i==2 and j==2)else x(j)for j in range(5))])for i in range(5)]

Answer 15

Raku, 94 bytes

{join "
",({S:13th/\d+/free/}((<B I N G O>,|[Z] (1..75).rotor(15)».pick(5)).join("
"))xx$_)}

• (1..75).rotor(15) splits the range 1-75 into five subranges of size 15: 1-15, 16-30, etc.
• ».pick(5) selects five distinct random elements from each of those subranges.
• [Z] reduces that list of lists of random choices with the Zip operator. It effectively transposes the matrix, so the 1-15 selections are in the first column instead of the first row.
• <B I N G O>,|... prepends the header row to the lists of numbers.
• .join("\n") joins the rows with newlines, producing a Bingo board as a single string. The actual code uses a single newline character between the quotes to save a byte.
• S:13th/\d+/free/ changes the thirteenth occurrence of a number in that string to free.
• xx $_ replicates the preceding process a number of times given by the input argument $_, producing a list of board strings.
• join "\n\n" joins those boards together with two newlines between them.

Answer 16

Python 3, (削除) 198 166 (削除ここまで) 157 bytes

lambda n:exec('m=["BINGO",*map(list,zip(*[sample(range(i,i+15),5)for i in range(1,75,15)])),""];m[3][2]="free";[print(*i)for i in m];'*n)
from random import*

Try it online!

(ATO isn't working right now so I switched to TIO)

• -32 thanks to tsh

This can probably be golfed a bit more.

• Iuri Guilherme: typo at line 2, need space before or
• The Thonnu: @IuriGuilherme it's not a typo - Python will let you remove the space after a number in most cases.
• Iuri Guilherme: not in python3.11 anymore
• gildux: So change to say it's Python prior 3.11 or add the space...
• tsh: 166 bytes: lambda n:exec('m=["BINGO",*map(list,zip(*[sample(range(i,i+15),5)for i in range(1,75,15)])),""];m[3][2]="FREE";[print(*i,sep="\t")for i in m];'*n)

Answer 17

Zsh +coreutils, 105 bytes

repeat 1ドル {A=(B I N G O `(for i ({1..75..15})shuf -i$i-$[i+14] -n5)|rs -t 5`)
A[18]=free;<<<$A|rs 6;echo}

Try it online!

Answer 18

Kotlin 2.0+, 188 bytes

fun b(x:Int)=repeat(x){println("\nB I N G O");val c=List(5){(it*15+1..it*15+15).shuffled().take(5)};repeat(5){i->println((0..4).joinToString(" "){if(i==2&&it==2)"free"else"${c[it][i]}"})}}

Online Test

I take advantage of shuffled() to randomize numbers but not have duplicates, also saves an import of Kotlin's random; aside from that I just print the columns of the 2d vector c but print free if I get to the center of the 2d vector which is c[2][2]

Answer 19

APL(NARS), 202 chars

next←1
r←n rndu ×ばつ⍳(w<n)∨(n≤0)∨(w≥1E9)∨w≤0⋄r←⍬
next←2147483648∣12345+ne×ばつ1103515245
t←1+w∣⌊ne×ばつ⍳0<n-←1
f←{⍪{M←⊃(×ばつ0..4)+{5 rndu ⍵} ×ばつ⍵⍴1⋄M[3;3]←⊂'free'⋄⍉'BINGO',M} ̈⍵⍴5}

//7+13+36+38+41+67 rndu would return n random numbers in 1..w without repetition, or ∅ if n>w. Reference is the function rand in the book "The C Language" Brian W. Kernignan, Dennis M. Ritchie. This seems ok even if I m not 100% sure.

test:

 f 3
 B I N G O 
 8 26 35 48 74 
 1 17 31 60 67 
12 21 free 47 75 
 7 20 44 58 61 
13 28 42 51 71 
 
 B I N G O 
 1 28 40 57 64 
10 23 45 55 62 
 9 20 free 52 72 
 3 29 39 48 70 
13 30 34 46 73 
 
 B I N G O 
10 20 39 59 67 
14 29 42 58 63 
12 23 free 60 68 
 5 25 41 51 72 
 3 21 31 52 66 
 f 1
 B I N G O 
15 30 34 50 73 
 7 16 44 46 72 
13 24 free 47 67 
 3 18 38 58 69 
10 26 35 52 66