Generate a Baseball Pitch String

Question

Goal

Write a program or function that takes a positive integer n and randomly generate a legal series of pitches (henceforth called a Pitch string) of length n.

Input

A non-zero, positive integer n <= 100

Output

Return a random string, or list of characters, that represent a possible, valid pitch string of length n. The characters used will be:

• B - Ball. If you accumulate 4 of these, the batter is walked and finished batting.
• S - Strike. If you accumulate 3 of these, the batter is out and finished batting.
• F - Foul. Will also increase the Strike count but cannot get the batter out. I.e., you cannot have a Foul be the last pitch in a valid string. Any fouls past two strikes/fouls will not increase the Strike count (the batter already has 2 strikes at that point and a 3rd would get him out).
• H - Hit. The batter has hit a ball into play and is finished batting.

(This is simplified slightly but don't you worry about that)

Valid pitch strings are those that end in a strike-out, a walk, or a hit.

I.e., an invalid pitch string has either

• additional pitches after the 4th Ball, 3rd Strike, or Hit
• terminated before generating a 4th ball, 3rd strike, or Hit.

Rules

• Your program must be able to produce all possible results for a given input.
• Your program does not have to be uniformly random but still must follow the previous rule.
• This is code-golf.

Examples

Input => Possible Outputs
1 => [H] #Can only end with a hit
2 => [S,H], [B,H], [F,H] #Can only end with a hit
3 => [S,S,S], [F,F,S], [B,B,H], ... #Can now strike-out, otherwise must end with a hit
4 => [B,B,B,B], [S,B,S,S], [B,F,S,S], [B,B,B,H], ... #Can now be walked, struck-out, or get a hit
6 => [S,B,S,B,B,H], [F,F,F,F,F,S], ... #Can now have a full-count (3 balls, 2 strikes) before finishing 

Input => Invalid Outputs
1 => [S], [B]    #Not enough for a strike-out/walk
2 => [S,S]       #Not enough for a strike-out/walk
2 => [H,H]       #Batter has already scored a hit
3 => [S,S,F]     #Fouls will not cause a strike-out
4 => [S,S,S,H]   #Batter has already struck out
5 => [B,B,B,B,B] #Batter has already walked

Answer 1

Python 2, 128 bytes

from random import*
def g(n):
 x=i=S=0;r=''
 while(S>2)+x<3>=i-S:x=randint(0,3);r+='BFSH'[x];S+=x>0;i+=1
 return(i==n)*r or g(n)

Try it online!

Randomly generate the pitch string until the batter is done, output it if it turns out the right length, and otherwise try again from scratch.

---

Python 2, 136 bytes

from random import*
def g(n):
 B=H=F=S=0;r=''
 while(F+S<3or'S'>x)>B/4+H:x=choice('BHFS');r+=x;exec x+"+=1"
 return(len(r)==n)*r or g(n)

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Answer 2

05AB1E,  44  50 44 bytes

^{Crossed out &nbsp;44&nbsp; is no longer 44 :)}

[õ0U.µ["BFSH"3ÝΩ©è«®ĀX+U¼X2›®+3@¾X-3›~#}I¾Q#

Port of @xnor's Python 2 answer, so make sure to upvote him as well if you like this answer!
+6 bytes due to a bug-fix, and after that -6 bytes again thanks to @xnor by porting his way more efficient fix in comparison to my temporary work-around, as I was expecting. ;)

Try it online or verify some more random outputs.

Explanation:

[                # Start an infinite loop:
 õ               #  (Re)set the result-string to an empty string ""
 0U              #  (Re)set variable `X` to 0
 .µ              #  Reset the counter_variable to 0
   [             #  Start an inner infinite loop:
    "BFSH"       #   Push string "BFSH"
          3ÝΩ    #   Push a random integer in the range [0,3]
             ©   #   Store this random integer in variable `r` (without popping)
              è  #   Index it into the string "BFSH"
               « #   Append it to the result-string
    ®Ā           #   If `r` is NOT 0:
      X+U        #    Increase `X` by 1
    ¼            #   Increase the counter_variable by 1
    X2›®+        #   Calculate `X`>2 (1 if truthy; 0 if falsey) + `r`
         3@      #   Check if this is larger than or equal to 3
    ¾X-          #   Calculate counter_variable - `X`
       3›        #   Check if this is larger than 3
    ~            #   If either of the two checks above is truhy:
     #           #    Stop the inner infinite loop
   }             #  After the inner infinite loop:
    I¾Q          #  If the input and counter_variable are equal:
       #         #   Stop the outer infinite loop
                 # (and output the result-string at the top of the stack implicitly)

Answer 3

R, 148 bytes

function(n){`~`=paste0
`*`=sample
o=""
while(nchar(o)<n-1){a=c("B"[T<4],"F","S"[F<2])*1
F=F+(a>"E")
T=T+(a<"F")
o=o~a}
o~c("B"[T>3],"H","S"[F>1])*1}

Try it online!

Generates the string, using conditional inclusion in the sampling datasets to ensure that the result is a possible pitch sequence.

Possibly doing rejection sampling (as xnor's python answer does) is shorter.

function(n){`~`=paste0		# alias
`*`=sample			# alias
o=""				# empty string for output
while(nchar(o)<n-1){		# do n-1 times:
a=c("B"[T<4],"F","S"[F<2])*1	# sample 1 from the string "BFS", conditionally including B or S if the ball/strike count is 3/2	
F=F+(a>"E")			# increment F (strike count) if sampled character is F or S
T=T+(a<"F")			# increment T (ball count) if sampled character is B
o=o~a}				# append a to output

o~c("B"[T>3],"H","S"[F>1])*1}	# append the sampled "BHS", conditionally including B or S if the ball/strike count is 3/2.

Random "F and S" reference that kept playing in my head every time I typed one of those letters...

Answer 4

JavaScript (SpiderMonkey), 137 bytes

f=(n,o='',i=Math.random()*4|0,r=/^(?!.*((B.*){4}|([SF].*){2}S|H).)/)=>r.test(o)?n?f(n-1,o+'BSFH'[i%4])||i<8&&f(n,o,i+1):!r.test(o+1)&&o:0

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Answer 5

Pyth, 53 bytes

u+GO?H+W<K/G\B3+W<Jl@"SF"G2\F\S\B+WqK3+WgJ2\H\S\B_UQ[

Try it online here.

This feels way too long, I think another approach may be required.

u+GO?H+W<K/G\B3+W<Jl@"SF"G2\F\S\B+WqK3+WgJ2\H\S\B_UQ[   Implicit: Q=eval(input())
                                                 _UQ    Reversed range from Q-1 to 0
u                                                   [   Reduce the above, with initial value G=[], next value as H:
                    @"SF"G                                Keep elements of G which are in "SF"
                   l                                      Length of the above
                  J                                       Store in J - this is the number of strikes and fouls so far
          /G\B                                            Count number of "B"s in G
         K                                                Store in K - this is the number of balls so far
    ?H                                                    If H is not 0 (i.e. not final pitch):
                           \F                               Start with "F" (foul is always available in non-final pitch)
                W<J       2                                 If J<2...
               +             \S                             ... append "S"
       W<K    3                                             If K<3...
      +                        \B                           ... append "B"
                                                          Else:
                                           \H               Start with "H" (hit is always available in final pitch)
                                       WgJ2                 If J >= 2...
                                      +      \S             ... append "S"
                                  WqK3                      If K == 3...
                                 +             \B           ... append "B"
   O                                                      Choose one element at random from the available options
 +G                                                       Append the above to G
                                                        Implicit print the result of the reduce operation

Answer 6

JavaScript (ES6),  107 106  99 bytes

f=(n,k=p=s=0,o='')=>p&&p>2|k-s>3|s>2&p<2?k-n?f(n):o:f(n,k+1,o+'FSBH'[p=Math.random()*4|0,s+=p<2,p])

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Commented

f = (                       // f = recursive function taking:
  n,                        //   n = requested length
  k =                       //   k = pitch counter, initialized to 0
  p =                       //   p = last pitch
  s = 0,                    //   s = sum of strikes and fouls
  o = ''                    //   o = output string
) =>                        //
  p &&                      // if the last pitch was not a foul
  p > 2 |                   // AND the last pitch was a hit
  k - s > 3 |               //     OR we have 4 balls (or 3 balls + 1 hit)
  s > 2 & p < 2 ?           //     OR more than 2 strikes or fouls, ending with a strike:
    k - n ?                 //   if k is not equal to n:
      f(n)                  //     valid series but bad timing: try again from scratch
    :                       //   else:
      o                     //     success: return o
  :                         // else:
    f(                      //   do a recursive call:
      n,                    //     n is unchanged
      k + 1,                //     increment k
      o + 'FSBH'            //     append the pitch letter to o
        [ p = Math.random() //     pick a new random pitch
              * 4 | 0,      //     in [0..3]
          s += p < 2,       //     increment s if the pitch is a foul or a strike
          p ]               //     actual index in 'FSBH'
    )                       //   end of recursive call

Answer 7

Ink, 120 119 116 117 bytes

=f(n)
->g(n,3,2)
=g(n,b,s)
~n--
{n:{~{b:b->g(n,b-1,s)}|{s:s->g(n,b,s-1)}|}f->g(n,b,s-(s>0))|{~{b:h|b}|{s:h|s}|h}}->->

Try it online!

Probably still golfable.

Ungolfed (mildly reformatted)

=f(length) // Define a stitch f, with one parameter which specifies the length of the created string. This is the intended entry point.
->g(length,3,2) // Instantly divert to g, defined below, with some extra parameters

=g(length,balls_left,strikes_left) // Define a stitch g, with three parameters.
~ length--                         // Decrement remaining length
{
    - length: // If this is not to be the last character in the string
              // randomly do one of the following:
              // 1. If balls_left is nonzero, print a b and recurse
              // 2. If strikes_left is nonzero, print an s and recurse
              // 3. Do nothing
              // If we did not divert earlier, print an f and recurse.
        {~{balls_left:b->g(length,balls_left-1,strikes_left)}|{strikes_left:s->g(length,balls_left,strikes_left-1)}|}f->g(length,balls_left,strikes_left-(strikes_left>0)) 
    - else: // Randomly do one of the following
            // 1. If a ball would result in a walk, print a b, otherwise an h.
            // 2. If a strike would result in a strikeout, print an s, otherwise an h.
            // 3. Just print an h.
            // And finally, halt.
        {~{balls_left:h|b}|{strikes_left:h|s}|h}}->->

Edits

1. Saved a byte by finishing with ->-> instead of ->END.
2. Saved three bytes by decrementing n earlier.
3. Fixed a bug that caused strikeouts in incorrect places, thanks to @veskah for spotting it (+1 byte)

Answer 8

APL (Dyalog Unicode), 77 bytes^SBCS

{'HBSF'[{⍺←⍬⋄∨/((⊃⌽⍺)=⍳3)∧1,4 3≤(⊃s),+/1↓s←⍺∘(+/=)¨1+⍳3:⍺⋄∇⍨⍺,?4}⍣{i=≢⍺}i←⍵]}

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Answer 9

Charcoal, 57 bytes

≔⁰η≔⁰ζＦ⊖Ｎ«≔‽⁺²‹ζ³ι¿›ι¹≦⊕ζ≦⊕η§SFB∨ι›η²»⊞υHＦ›η¹⊞υSＦ›ζ²⊞υB‽υ

Try it online! Link is to verbose version of code. Explanation:

≔⁰η≔⁰ζ

Start with 0 balls and 0 strikes.

Ｆ⊖Ｎ«

Loop over all of the deliveries except the last.

≔‽⁺²‹ζ³ι

If the have been fewer than three balls, then generate a random number from 0 to 2, otherwise just coinflip between 0 and 1.

¿›ι¹≦⊕ζ≦⊕η

A random value of 2 is a ball otherwise it increases the strike count.

§SFB∨ι›η²»

The values 0 to 2 map to strike, foul and ball, except that if there would be three strikes then foul is printed instead. (Four balls are excluded above.)

⊞υHＦ›η¹⊞υSＦ›ζ²⊞υB‽υ

Determine whether a strike or ball would get the batter out and choose from those or a hit as appropriate.

Answer 10

Perl 5, 122 bytes

map{$B=$S=$H=0;while($B<4&&$S<3&&!$H&&/./g){${$&}++;$S+=$&eq F&&$S<2}y///c>pos||push@a,$_}glob"{B,F,H,S}"x<>;say$a[rand@a]

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Answer 11

C (GCC) 164 145 142 bytes

-3 bytes ceilingcat

#define A(a)!i&&!r--?puts(#a),++a,--n:0;
b,s,f,h,i,r;p(n){srand(time(0));for(i=n;i--;){for(n=1;n;){r=rand()&3;b>2^!A(b)s+f>1^!A(s)!A(f)A(h)}}}

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