# Dan Baronet, 1956 – 2016. R.I.P.

He found the shortest possible APL solution to this task:

Given a Boolean list, count the number of trailing truth values.

### Example cases

{}0

{0}0

{1}1

{0, 1, 1, 0, 0}0

{1, 1, 1, 0, 1}1

{1, 1, 0, 1, 1}2

{0, 0, 1, 1, 1}3

{1, 1, 1, 1, 1, 1}6

• Can we take the list as a string of zeros and ones? e.g. 01100? – Adnan Nov 6 '16 at 12:57
• @Adnan only if that is the most normal way for your language to represent boolean lists. – Adám Nov 6 '16 at 15:42
• Sorry for your loss. – Martin Ender Nov 6 '16 at 17:03
• @MartinEnder Thank you. It will be tough going forward. Dan taught me all I needed to know to work for Dyalog. – Adám Nov 6 '16 at 17:32
• Farewell to Dan. RIP... – Erik the Outgolfer Nov 7 '16 at 12:46

# C++17, 82 66 bytes

int f(){return 0;}int f(int H,auto... L){return(H*...*L)+f(L...);}


Uses the C++17 template parameter fold expression and essentially the same idea as Dennis. Saving 16 bytes by using Generic Variadic Lambda.

Explanation:

int f(){return 0;} //base case for empty list

int f(int H, auto... L) { //first element, variadic arguments
return (H*...*L)      //a_0*a_1*a_2*...
+ f(L...);       //+ f(a_1,a_2,...)
}


Usage:

f(1,1,0,1,1,0,1,1,1,1,1) -> 5
f() -> 0
f(1,0,1,0) -> 0


## Non competing

Albeit longer, this also works with template constants:

template <int...L> int C=0;
template <int H, int...L> int C<H,L...> = (H*...*L)+C<L...>;


Usage:

std::cout << C<1,0,1,1>  << std::endl;
std::cout << C<1,0,1,0,1>  << std::endl;
std::cout << C<1,1,1,0,1,0,1,1>  << std::endl;
std::cout << C<1,1,1,0,1,0,1,1,1,1,1,1>  << std::endl;
std::cout << C<> << std::endl;


## Pip, 8 bytes

WDQg++ii


Takes input as command-line arguments of 0 and 1 (or any truthy value). Try it online!

### Explanation

          g is cmdline args, i is 0
DQg      Dequeue item from end of g
W   ++i   While this is truthy, increment i
i  Print i


(Dequeueing from an empty list gives nil, which is falsey.)

# brainfuck, 99 bytes

Takes input like 11011. Output is a single byte/character value.

-[>+<-----]>---[>>,]<-[>+<-----]>---[-<+>>>+<<]<[->+<]>+>>+[<+[-<<+>>]<<<[-<+>>>-<<]<[->+<]>>>-]<-.


Try it online - Run with input, then click "view memory" to see the value under the pointer that was printed.

Explanation:

-[>+<-----]>---             put constant 48 (ASCII '0') at start of list
[>>,]                       receive all input, with an empty cell between each
<-[>+<-----]>---            constant 48 near end of list
[-<+>>>+<<]<[->+<]>+>>      move to right and copy right, add one to make 49 (ASCII '1')
TAPE: 48 _ i0 _ i1 _ ... in _ 49 _ 48< (pointer)

+[<+[-<<+>>]<<              LOOP. Put counter+=1 in empty cell. Move it left 2 cells.
<[-<+>>>-<<]<[->+<]>        Subtract value from value 2 cells right.
>>-]<-.                     Subtract one more. If zero, print counter-1, Else loop again.


Close, but must contain a zero, and it doesn't handle trailing zeros or an empty list. (49 bytes)

,[>>,]>>+[<+[-<<+>>]<<<[-<+>>>-<<]<[->+<]>>>-]<-.

• ,[>,]-[>+<-----]>---[-<<[<]>[->]>]<<[>[-<+>]<<]>. – Jo King Oct 4 '18 at 9:30
• @JoKing Go ahead and create your own answer. That doesn't resemble mine at all. – mbomb007 Oct 4 '18 at 16:28

# PHP, 38 bytes

a totally different approach

<?=strpos(strrev(join([0]+$argv)),48);  takes input from command line arguments. Save to file. [0]+$argv sets the first element (script name) to 0.
Join that without a delimiter, reverse and find the first occurence of the 0 character.

While my first solution works with any truthy and falsy values, this one obviously depends on single characters: 0 is falsy, every other character (apart from maybe the 0 byte) truthy.

• -4 Bytes for using $_GET – Jörg Hülsermann May 3 '17 at 13:32 • @JörgHülsermann That would have empty output for any list full of 1s. – Titus May 3 '17 at 14:50 • Okay now I have realize it – Jörg Hülsermann May 3 '17 at 16:48 # Pushy, 6 bytes $v;FL#


Arguments given a list on command line: $pushy truths.pshy 0,1,1,0,0. Like my binary conversion program, this takes advantage of the second stack in an interesting way. Here's how it works:  \ Implicit: Input on stack$        \ While last item is not 0:
v;      \   Move last item to auxiliary stack
FL#   \ Output length of auxiliary stack.


Note that the loop will not run if the list is empty, so the output will be 0.

• How many stacks does Pushy have? – Cows quack Dec 7 '16 at 17:09
• @KritixiLithos Two, input is automatically on stack 1 and that's where main operations take place. Stack 2 is just for storing counters / variables that need to be kept out of the way. – FlipTack Dec 7 '16 at 17:15

# PHP, 40 34 bytes

Along with $argv comes $argc ... and both are variable.
Either one of the arguments or +$argv[0]==+"-" is 0. while(+$argv[--$argc])$i++;echo$i;  takes input from command line arguments. Empty output for 0. Run with php -nr '<code>' <space separated values> unary output, 35 29 bytes while(+$argv[--$argc])echo 1;  # Brain-Flak, 30 bytes (()){{}({}{{}}<><{}>)<>([])}<>  Try it online! ## Explanation (()) #{ Start Loop. Used instead of ([]) so that the loop will run at least once on empty input. } {{} #{Remove the stack height. } ( {}{{}} #{ Add the TOS with any 1s below it (until a zero). } <><{}> #{ Remove a value on the other stack if there. } ) #{ Push the result (# of 1s just removed) to the other stack. } <> #{ Switch stacks back. } ([]) #{ Push stack height for the loop. } } #{ Once the loop is done (stack is empty)... } <> #{ Switch stacks to the most recent # of 1s removed. }  # Python2, 42 41 Byes t=input()[::-1];print len(t[:t.index(0)])  takes input as [1, 1, 0, 1, 1] • No joy with the empty list test case. – Jonathan Allan Nov 6 '16 at 15:11 # ABCR, 14 bytes c7iA)7a!xcx!Bp  Accepts the formatting [0,1,0,1,0, where any non-numeric character can be replaced with any other non-numeric character. Empty input is the empty list. Explanation: Every 7 input number i is queued up A; every "0" input number (or rather, non-"1") )7 will pop from the queue until it's empty 7a!x; before another input number is queued, a delimiter character is grabbed cx to check for the end of queue. After all the integers are grabbed, !Bp prints the length of the queue of input numbers (which will be all the trailing "1" values.) # Swift 3, 49 bytes func a(b:[Bool]){print(b.reduce(0,{$1 ?$0+1:0}))}  # Minkolang, 32 bytes 0$nI1-[1=?v0g1+1R]N.
.Ng0<


Try it online!

### Explanation

0                             pushes 0 (this is the number that will keep track of the number of 1s)
• return+$i; needs no space. – Titus Dec 7 '16 at 16:08 Clojure, 39 36 bytes #(count(take-while{1 1}(reverse %)))  Input as integers [1 0 1 1 0 1 1 1], anything other than 1 is falsy. {1 1} is a hash-map with key 1 and value 1 which can also used as a function. For example ({1 999} 1) is 999 (truthy) whereas ({1 999} 0) is nil (falsy). Original, based on booleans: #(count(take-while(fn[i]i)(reverse %)))  Amazing how (fn[i]i) is shorter than identity and also doesn't force you to put a space after take-while. Must take a list or vector of booleans, integers can be converted like this: (map #(= 1 %) [1 0 1 1 0 1 1 1]) # PHP, 33 Bytes <?=strspn(strrev(join($_GET)),1);


Online Version

strspn

# Pari/GP, 30 bytes

p->valuation(Pol(p)*(x-1)+1,x)


Converts the list to a polynomial, say p, then finds the valuation of p*(x-1)+1 with respect to x, i.e., the minimal degree of its nonzero terms.

For example, if we take [1, 0, 1, 1, 1] as input, then the polynomial p is x^4 + x^2 + x + 1, and p*(x-1)+1 is x^5 - x^4 + x^3, whose valuation is 3.

Try it online!

• Well, that is truly an original method. +1 – Adám Jul 13 '17 at 5:47

# Japt-hx, 3 bytes

i ô


Try it online!

Japt's flag combinations rock.

### How it works

Ui ô

Ui  Insert undefined at index 0
ô   Split at falsy items

-h  Take last element
-x  Sum


If we didn't have to handle the special case [], we could get away with 1 byte ô, winning over APL.

• Half the job is done by flags‽ – Adám Jun 1 '18 at 8:22

# brainfuck, 33 bytes

,[+++>-[<->-----],]<[>[-<+>]<<]>.


Try it online!

Takes input via a string like 101011, and then outputs via byte value. I've added some code in the footer to add 48 to the value to output a digit.

 ,[  Loop over input
+++>-[<->-----]  Subtract 48 from each value to form a tape of 0s and 1s
,]
<[>[-<+>]<<]   Add up all the trailing ones
>.             And print the value


# Taxi, 15801576 1564 bytes

-4 bytes because of a less complex route to the Crime Lab (string equality checker). Also, you don't run out of gas for a sufficiently large number of truthy values anymore!

-12 bytes by getting rid of the quotes.

Go to the Post Office:w 1 l 1 r 1 l.
Pickup a passenger going to Chop Suey.
Go to Chop Suey:n 1 r 1 l 4 r 1 l.
[B]
Switch to plan C if no one is waiting.
Pickup a passenger going to Cyclone.
Go to Cyclone:n 1 l 3 l.
Pickup a passenger going to Narrow Path Park.
Pickup a passenger going to Joyless Park.
Go to Zoom Zoom:n.
Go to Narrow Path Park:w 1 l 1 l 1 r.
Go to Joyless Park:e 1 r 3 l.
Go to Chop Suey:w 1 r 1 r 1 l.
Switch to plan B.
[C]
0 is waiting at Starchild Numerology.
Go to Starchild Numerology:n 1 l 3 l 3 l 2 r.
Pickup a passenger going to Sunny Skies Park.
Go to Sunny Skies Park:w 1 r.
Go to Narrow Path Park:n 1 r 1 r 1 l 1 r.
[D]
Switch to plan F if no one is waiting.
Pickup a passenger going to Crime Lab.
1 is waiting at Writer's Depot.
Go to Writer's Depot:w 1 l 1 r 2 l.
Pickup a passenger going to Crime Lab.
Go to Zoom Zoom:n.
Go to Crime Lab:w 1 l 2 r.
Switch to plan E if no one is waiting.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery:s.
Pickup a passenger going to Addition Alley.
Go to Sunny Skies Park:n 1 l 1 l 1 r.
Pickup a passenger going to Addition Alley.
Go to Addition Alley:n 1 r 1 r 1 r.
Pickup a passenger going to Sunny Skies Park.
Go to Sunny Skies Park:n 1 l 1 l 1 l.
Go to Narrow Path Park:n 1 r 1 r 1 l 1 r.
Switch to plan D.
[E]
Go to Narrow Path Park:n 5 l.
[F]
Go to Sunny Skies Park:w 1 l 1 r 2 l 1 l.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery:s 1 l 1 r 1 r.
Pickup a passenger going to Post Office.
Go to Post Office:n 1 l 1 r.
Go to Taxi Garage:n 1 r 1 l 1 r.


Input is in the form of a string of 0's and 1's (for example: 11101).

Try it online!

Taxi is an esolang in which all programming is done by picking up and dropping off passengers at various stops in the fictional town of Townsburg. Of course, your taxicab will sometimes run out of gas, so you also need to visit gas stations every so often, and pay using the credits you receive as fare.

I had to do some strange management of fuel in this program. Specifically, I have a loop at the beginning (between plans B and C) which pushes each character of the input to Narrow Path Park (which is a LIFO queue, or a stack) by going back and forth between Narrow Path Park and Chop Suey (which split the input string into characters in the first place). However, for sufficiently large inputs, this can cause me to run out of gas. Simply going to a gas station every iteration is insufficient, because I'm not earning enough money from actually bringing passengers to their destinations. The best way I've figured out to earn enough money to make going to a gas station every iteration "worth it" is bringing each new passenger to Cyclone (which clones them), and taking one of those clones to Narrow Path Park while leaving one forever trapped at Joyless Park (which is a FIFO queue, but for the purposes of this program is a dumping ground).

(Note: Bringing passengers to Riverview Bridge is a good way to get rid of them, but they don't pay you - because they always seem to fall off the bridge into the river before they can pay - so I couldn't do this either.)

# Braingolf, 7 bytes

!?&gGL|


Try it online!

## Explanation

!?&gGL|  Implicit input from commandline args
!?       If last item on stack is > 0..
&g     ..Combine all items into single number (1, 1, 0, 1 becomes 1101)
G    ..Split into digit runs (1101 becomes 11, 0, 1)
L   ..Pop last item and push length of item (111 becomes 3)
|  Endif
Implicit output of last item on stack, either number of trailing ones, or zero

• You use 3 chars for the conditional. Could you maybe multiply by the parity instead? – Adám Jul 13 '17 at 9:46
• @Adám that would still be 3 bytes, 2 bytes to get the parity (2%) and one byte to multiply (*) – Skidsdev Jul 13 '17 at 9:49

# Backhand, 17 bytes

v I^: ]|{]1$|{O @  Try it online! ### Explanation: v I^: ]|{]1$|{O @
v                  Decrease step value to 2
I                Get input as a number
: ]|{          Reflect if not EOF
^               Increase step value to 3
v                  Decrease to 2 and repeat the loop
]         The leftover EOF (-1) is now our counter. Increment it to 0
$|{ Reflect if the top value is truthy |{ 1 Reflect and repeat ] Increment the counter for every truthy value O @ Output the final state of the counter  # Ohm v2, 5 bytes Ö⁾ì]*  RLE encodes the string, takes the last element, flattens, and multiplies Try it online! # Mathematica, 36 bytes Length[#/.({___,0,x___/;x==1}:>{x})]&  Not the shortest, but short enough where I was proud to submit it :) # V, 7 bytes Ó1*0 ø1  Try it online! Takes input as a string of 0 and 1. ## Explanation Ó1*0 Remove all occurences of any number of ones followed by a zero ø1 Count the number of remaining ones  # MBASIC, 112 bytes 1 INPUT B$:T=0:FOR I=LEN(B$) TO 1 STEP -1:C$=MID$(B$,I,1):IF C$="0" THEN 4 2 IF C$="1" THEN T=T+1
3 NEXT
4 PRINT T


Just wanted to see if I could do it.

Explanation

Input is a string of 1's and 0's. String is traversed from right to left. If the current digit is a 0, bail out and print the total. If the digit is a 1, increment the total and continue to loop.

Output

? 01100
0

? 11011
2

? 11101
1

? 111111
6