Python 2, 70 bytes

f=lambda n,k=0,j=0:n-1and-~f(k*[n/2,n*3+1][n%2]or f(j/99or n,1),k,j+1)

Returns 199 for one hundred or more iterations.

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Attache, 40 bytes

`-&3@`#@PeriodicSteps[CollatzSize@Max&1]

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This is a new language that I made. I wanted to get around to making a proper infix language, and this is the result: a mathematica knock-off. Hooray?

Explanation

This is a composition of a few functions. These functions are:

• PeriodicSteps[CollatzSize@Max&1] This yields a function which applies its argument until the results contain a duplicate element. This function, CollatzSize@Max&1, is applying CollatzSize to the greater of the input and 1, to avoid the invalid input 0 to CollatSize.
• `# is a quoted operator; when applied monadically in this sense, it obtains the size of its argument
• `-&3 is a bonded function, which bonds the argument 3 to the function `-, which reads as "minus 3". This is because the PeriodicSteps application yields 0s, which need to be accounted for. (It also neatly handles out-of-bounds numbers like 5, which map to -1.)

• 1
  \$\begingroup\$ Is using your own language really allowed? Cant you just create a langage for each codegolf with only uses some bytes? \$\endgroup\$

  Tweakimp

  – Tweakimp

  2018年01月08日 14:58:35 +00:00

  Commented Jan 8, 2018 at 14:58
• 2
  \$\begingroup\$ @Tweakimp Of course creating (and using) your own language is allowed. But modifying it so that a task is a single command (after the challenge is posted) is a standard loophole. \$\endgroup\$

  caird coinheringaahing

  – caird coinheringaahing ♦

  2018年01月08日 16:36:30 +00:00

  Commented Jan 8, 2018 at 16:36
• 2
  \$\begingroup\$ @Tweakimp if it makes you feel any better, I had designed this function before I saw this challenge. I am a language designer, so that's what I do. \$\endgroup\$

  Conor O'Brien

  – Conor O'Brien

  2018年01月08日 18:45:40 +00:00

  Commented Jan 8, 2018 at 18:45
• \$\begingroup\$ It was more a general question wether selfmade languages are allowed, not a negative statement that you used your own. \$\endgroup\$

  Tweakimp

  – Tweakimp

  2018年01月13日 19:20:18 +00:00

  Commented Jan 13, 2018 at 19:20

J, _{(削除) 49 (削除ここまで)} 45 bytes

-4 bytes thanks to shorter Collatz Sequence code taken from @randomra's comment here.

(2-~[:#(>&1*-:+2&|*+:+>:@-:)^:a:)^:(<101)i.1:

Outputs 101 for invalid results.

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Explanation

Unsurprisingly, this explanation has become quickly outdated. I'm going to leave it in terms of the old 49 byte answer I had, which I'm including below. If you want an update, just let me know. The way that it finds the length of the recursive sequence remains the same, I've just used a shorter Collatz Sequence method.

(1-~[:#%&2`(1+3&*)@.(2&|)^:(1&<)^:a:)^:(<101)i.1:

Finding the length of the Collatz Sequence

This section of the code is the following

(1-~[:#%&2`(1+3&*)@.(2&|)^:(1&<)^:a:)

Here's the explanation:

(1 -~ [: # %&2`(1+3&*)@.(2&|) ^: (1&<) ^: a:) Given an input n
 ^: a: Apply until convergence, collecting
 each result in an array.
 ^: (1&<) If n > 1 do the following, else
 return n.
 (2&|) Take n mod 2.
 %&2 If n mod 2 == 0, divide by 2.
 (1+3&*) If n mod 2 == 1, multiply by 3 
 and add 1.
 # Get the length of the resulting
 array.
 1 -~ Subtract 1.

Unfortunately, the apply verb (^:) when told to store results stores the initial value too, so it means we're (like always) off by one. Hence why we subtract 1.

Finding the length of the recursive sequence

(1-~[:#%&2`(1+3&*)@.(2&|)^:(1&<)^:a:) ^: (< 101) i. 1: Given an input n.
 ^: (< 101) Apply 100 times,
 collecting results
 in an array.
(1-~[:#%&2`(1+3&*)@.(2&|)^:(1&<)^:a:) Collatz sequence length.
 i. 1: Index of first 1 (returns
 101, the length of the
 array if 1 not found).

• 1
  \$\begingroup\$ If you don't mind using the header section, this will perhaps more accurately showcase your answer \$\endgroup\$

  Conor O'Brien

  – Conor O'Brien

  2018年01月08日 01:44:11 +00:00

  Commented Jan 8, 2018 at 1:44
• \$\begingroup\$ @ConorO'Brien I don't at all -- I didn't really know how to get it formatted as such (but I'll be stealing yours from now on). Thanks \$\endgroup\$

  cole

  – cole

  2018年01月08日 02:01:38 +00:00

  Commented Jan 8, 2018 at 2:01
• 1
  \$\begingroup\$ ​A​n​y​t​i​m​e​! \$\endgroup\$

  Conor O'Brien

  – Conor O'Brien

  2018年01月08日 02:06:23 +00:00

  Commented Jan 8, 2018 at 2:06
• 1
  \$\begingroup\$ 38 bytes with *i.~(<101)1&(#@}.a:2&(<*|{%~,*+1+])])] should be equivalent \$\endgroup\$

  miles

  – miles

  2018年01月09日 01:34:25 +00:00

  Commented Jan 9, 2018 at 1:34

C (gcc), 75 bytes

i,j;f(n){for(j=0;(i=n)&&j++<100;)for(n=0;i-1;++n)i=i&1?i*3+1:i/2;i=!i*j-1;}

Returns -1 for n>=100 iterations.

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C (gcc), 73 bytes

i,j;f(n){for(j=-1;(i=n)&&j++<99;)for(n=0;i-1;++n)i=i&1?i*3+1:i/2;i=!i*j;}

Returns 0 for n>=100 iterations.

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Jelly, 24 bytes

×ばつ3‘$HḂ?’пL’
Ç11?ȷ2Сi1’

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• \$\begingroup\$ 23 bytes (using ³ instead of ȷ2 is allowed). \$\endgroup\$

  Mr. Xcoder

  – Mr. Xcoder

  2018年01月08日 07:57:45 +00:00

  Commented Jan 8, 2018 at 7:57

JavaScript (ES6), 57 bytes

Returns true when it fails. Returns 0 for 1.

f=(n,k=i=0)=>n>1?f(n&1?n*3+1:n/2,k+1):k?i>99||f(k,!++i):i

Test cases

f=(n,k=i=0)=>n>1?f(n&1?n*3+1:n/2,k+1):k?i>99||f(k,!++i):i
console.log(f(2)) // 1
console.log(f(5)) // fail
console.log(f(10)) // 8
console.log(f(100)) // 13
console.log(f(1)) // 0

• \$\begingroup\$ I am sceptical if your program happens to calculate the correct result apart from overflow / inaccuracy or if rather the OP derived their results using a language with similar number implementations (I assume they did not calculate all test cases by hand). \$\endgroup\$

  Jonathan Frech

  – Jonathan Frech

  2018年01月08日 01:32:17 +00:00

  Commented Jan 8, 2018 at 1:32
• \$\begingroup\$ @JonathanFrech Indeed. It turns out both results were equally invalid. \$\endgroup\$

  Arnauld

  – Arnauld

  2018年01月08日 06:50:59 +00:00

  Commented Jan 8, 2018 at 6:50

APL (Dyalog Unicode), (削除) 39 (削除ここまで) (削除) 60 (削除ここまで) (削除) 53 (削除ここまで) (削除) 52 (削除ここまで) 49 bytes

-3 bytes thanks to @ngn

0∘{99<⍺:⋄1=⍵:0⋄1+(⍺+1)∇{1=⍵:0⋄1+∇⊃⍵⌽0 1+.5 ×ばつ⍵}⍵}

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Uses @ngn code for Collatz, but previously used @Uriel's code.

Here's the old version that didn't meet the specification:

{1=⍵:0⋄1+∇{1=×ばつ⍵⋄1+∇⍵÷2}⍵}

• \$\begingroup\$ 2|⍵:1+∇1+3×⍵⋄1+∇⍵÷2 -> 1+∇⊃⍵⌽0 1+.5 3×⍵ \$\endgroup\$

  ngn

  – ngn

  2018年01月16日 05:29:15 +00:00

  Commented Jan 16, 2018 at 5:29

Perl 6, 56 bytes

{($_,{($_,{$_%2??$_*3+1!!$_/2}...1)-1}...1).head(102)-1}

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Returns 101 for a non-terminating sequence.

Husk, 21 bytes

←1ドル↑101¡ȯ←1ドル¡?1⁄2o→*3¦2

Try it online! Returns -1 on failure, 0 on input 1.

Explanation

←1ドル↑101¡ȯ←1ドル¡?1⁄2o→*3¦2 Implicit input (a number).
 ?1⁄2o→*3¦2 Collatz function:
 ? ¦2 if divisible by 2,
 1⁄2 then halve,
 o→*3 else multiply by 3 and increment.
 ȯ←1ドル¡?1⁄2o→*3¦2 Count Collatz steps:
 ¡ iterate Collatz function and collect results in infinite list,
 1ドル get 1-based index of 1,
 ȯ← decrement.
 ¡ Iterate this function on input,
 ↑101 take first 101 values (initial value and 100 iterations),
←1ドル get index of 1 and decrement.

C (gcc), (削除) 70 (削除ここまで) 73 bytes

g(x){x=x-1?g(x%2?3*x+1:x/2)+1:0;}f(x,m){for(m=0;(x=g(x))&&100>m++;);x=m;}

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Returns 101 when number of iterations exceeds 100.

• 1
  \$\begingroup\$ Welcome to PPCG! This answer isn't quite valid, because all function submissions need to be reusable. I think you can fix this by inserting m=0 into your f (probably even making use of the currently empty for intiailiser to save one a ;). \$\endgroup\$

  Martin Ender

  – Martin Ender

  2018年01月09日 12:37:51 +00:00

  Commented Jan 9, 2018 at 12:37

Clean, (削除) 146 (削除ここまで) ... 86 bytes

-11 bytes thanks to Ørjan Johansen

import StdEnv
?f l n=hd[u\1円<-iterate f n&u<-l]

?(?(\b|isOdd b=3*b+1=b/2)[0..])[0..99]

As a partial function literal.

Try it online!

Aborts with hd of [] if the number of iterations exceeds 100.
Exits with Heap Full for inputs above ~2^23 unless you specify a larger heap size.

• 1
  \$\begingroup\$ I'm starting to understand some Clean syntax (as it differs from Haskell) from your answers... you can shorten that with a helper function j f l n=hd[u\1円<-iterate f n&u<-l]. \$\endgroup\$

  Ørjan Johansen

  – Ørjan Johansen

  2018年01月10日 00:42:13 +00:00

  Commented Jan 10, 2018 at 0:42
• \$\begingroup\$ @ØrjanJohansen Thanks! \$\endgroup\$

  Οurous

  – Οurous

  2018年01月10日 00:56:43 +00:00

  Commented Jan 10, 2018 at 0:56
• \$\begingroup\$ You don't need the \a=...a part, it curries. (Or eta reduces.) \$\endgroup\$

  Ørjan Johansen

  – Ørjan Johansen

  2018年01月10日 01:16:40 +00:00

  Commented Jan 10, 2018 at 1:16
• \$\begingroup\$ @ØrjanJohansen oh, missed that, thanks! \$\endgroup\$

  Οurous

  – Οurous

  2018年01月10日 01:41:49 +00:00

  Commented Jan 10, 2018 at 1:41

Python 2, (削除) 99 (削除ここまで) (削除) 98 (削除ここまで) 97 bytes

• Saved a byte by using c and t or f instead of t if c else f.
• Saved a byte by outputting -1 instead of f or 'f' for non-halting inputs.

exec"f,F="+"lambda n,i=0:n<2and i or %s"*2%("f([n/2,3*n+1][n%2],-~i),","i>99and-1or F(f(n),-~i)")

Try it online!

BiwaScheme, 151 chars

(define(f n i s)(if(= s 0) 'F(if(= n 0)i(f(letrec((c(lambda(m k)(if(= m 1)k(c(if(=(mod m 2)0)(/ m 2)(+(* m 3)1))(+ k 1))))))(c n 0))(+ i 1)(- s 1)))))

You can try it here.

R, (削除) 119 (削除ここまで) 107 bytes

Partially uses Jarko Dubbeldam's collatz code from here. Returns 0 for>100 iterations (failure).

pryr::f(x,{N=n=0
while(x-1){while(x-1){x=`if`(x%%2,3*x+1,x/2);n=n+1}
x=n
n=0
N=N+1
if(N==100)return(0)}
N})

Try it online!

APL NARS, 115 bytes, 63 chars

{d←0⋄{⍵=1:d⋄99<d+←1: ̄1⋄∇{c←0⋄{1=⍵:c⋄c×ばつ⍵⋄∇⍵÷2}⍵}⍵}⍵}

Probably using loops it would be more clear... There are 4 functions, 2 nested and ricorsive, and the first only for define and initialize to =0, the variable d, seen from the 2th function as a global variable counter.

q←{c←0⋄{1=⍵:c⋄c×ばつ⍵⋄∇⍵÷2}⍵}

This 3th function, would be the function that return how many call there are for resolve the Collatz conjecture for its arg

{⍵=1:d⋄99<d+←1: ̄1⋄∇q⍵}

This is the 2th function, if has its arg =1,stop its recursion and return d the number of time it is called itself-1; else if itself is called more than 99times stop its recursion and return -1(fail) else calculate Collatz conjecture for its arg, and call itself for the Collatz sequence length value. For me even if all this seems run could be a big problem if is defined a global variable and one variable in a function of the same name, when the programmer sees it as just a local variable.

 f←{d←0⋄{⍵=1:d⋄99<d+←1: ̄1⋄∇{c←0⋄{1=⍵:c⋄c×ばつ⍵⋄∇⍵÷2}⍵}⍵}⍵} 
 f 2
1
 f 3
5
 f 5
 ̄1
 f 10
8
 f 100
13
 f 12313
7
 f 1
0

(Emacs, Common, ...) Lisp, 105 bytes

Returns t for iterations> 100

(defun f(n k c)(or(> c 100)(if(= n 1)(if(= k 0)c(f k 0(1+ c)))(f(if(oddp
n)(+ n n n 1)(/ n 2))(1+ k)c))))

Expanded:

(defun f (n k c)
 (or (> c 100)
 (if (= n 1)
 (if (= k 0) c
 (f k 0 (1+ c)))
 (f (if (oddp n) (+ n n n 1) (/ n 2))
 (1+ k) c))))
(f (read) 0 0)

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