Stair-climbing puzzle

Stair-climbing puzzle is a programming puzzle. It lays out a problem which Rosetta Code users are encouraged to solve, using languages and techniques they know. Multiple approaches are not discouraged, so long as the puzzle guidelines are followed. For other Puzzles, see Category:Puzzles.

From Chung-Chieh Shan (LtU):

Your stair-climbing robot has a very simple low-level API: the "step" function takes no argument and attempts to climb one step as a side effect. Unfortunately, sometimes the attempt fails and the robot clumsily falls one step instead. The "step" function detects what happens and returns a boolean flag: true on success, false on failure. Write a function "step_up" that climbs one step up [from the initial position] (by repeating "step" attempts if necessary). Assume that the robot is not already at the top of the stairs, and neither does it ever reach the bottom of the stairs. How small can you make "step_up"? Can you avoid using variables (even immutable ones) and numbers?

Ada

<lang Ada>procedure Step_Up is begin

  while not Step loop
     Step_Up;
  end loop;

end Step_Up;</lang> The following is a test program simulating Step: <lang Ada>with Ada.Numerics.Discrete_Random; with Ada.Text_IO;

procedure Scaffolding is

  package Try is new Ada.Numerics.Discrete_Random (Boolean);
  use Try;
  Dice  : Generator;
  Level : Integer := 0;

  function Step return Boolean is
  begin
     if Random (Dice) then
        Level := Level + 1;
        Ada.Text_IO.Put_Line ("Climbed up to" & Integer'Image (Level));
        return True;
     else
        Level := Level - 1;
        Ada.Text_IO.Put_Line ("Fell down to" & Integer'Image (Level));
        return False;
     end if;
  end Step;
  
  procedure Step_Up is
  begin
     while not Step loop
        Step_Up;
     end loop;
  end Step_Up;

begin

  Reset (Dice);
  Step_Up;

end Scaffolding;</lang> Sample output:

Fell down to-1
Climbed up to 0
Fell down to-1
Climbed up to 0
Fell down to-1
Fell down to-2
Climbed up to-1
Climbed up to 0
Climbed up to 1

C

The same solution of the C++ code can be used; the initial declaration of a variable inside a for loop is C99.

C++

<lang cpp>void step_up() {

 while (!step()) step_up();

}</lang>

The following uses a variable and is a bit longer, but avoids a possible stack overflow: <lang cpp>void step_up() {

 for (int i = 0; i < 1; step()? ++i : --i);

}</lang>

C#

<lang csharp>void step_up() {

   for (; step() == false; )step_up();

}</lang>

Clojure

First, some boilerplate.

<lang clojure>

the initial level

(def level (atom 41))

the probability of success

(def prob 0.5001)

(defn step

 []
 (let [success (< (rand) prob)]
   (swap! level (if success inc dec))
   success) )

</lang>

Tail-recursive

The internal recursion uses a counter; see the function documentation.

<lang clojure> (defn step-up1

 "Straightforward implementation: keep track of how many level we
  need to ascend, and stop when this count is zero."
 []
 (loop [deficit 1]
   (or (zero? deficit)

(recur (if (step) (dec deficit) (inc deficit)))) ) ) </lang>

Recursive

This satisfies Chung-chieh's challenge to avoid using numbers. Might blow the stack as p approaches 0.5.

<lang clojure> (defn step-up2

 "Non-tail-recursive. No numbers."
 []
 (if (not (step))
   (do (step-up2) ;; undo the fall

(step-up2) ;; try again )

   true))

</lang>

E

This is at first a very direct

Translation of: Clojure

The problem framework:

<lang e>var level := 41 var prob := 0.5001

def step() {

   def success := entropy.nextDouble() < prob
   level += success.pick(1, -1)
   return success

}</lang>

Counting solution:

<lang e>def stepUpCounting() {

   var deficit := 1
   while (deficit > 0) {
       deficit += step().pick(-1, 1)
   }

}</lang>

Ordinary recursive solution: <lang e>def stepUpRecur() {

   if (!step()) {
       stepUpRecur()
       stepUpRecur()
   }

}</lang>

Eventual-recursive solution. This executes on the vat queue rather than the stack, so while it has the same space usage properties as the stack-recursive version it does not use the stack which is often significantly smaller than the heap. Its return value resolves when it has completed its task.

<lang e>def stepUpEventualRecur() {

   if (!step()) {
       return when (stepUpEventualRecur <- (),
                    stepUpEventualRecur <- ()) -> {} 
   }

}</lang>

Fully eventual counting solution. This would be appropriate for controlling an actual robot, where the climb operation is non-instantaneous (and therefore eventual): <lang e>def stepUpEventual() {

   # This general structure (tail-recursive def{if{when}}) is rather common
   # and probably ought to be defined in a library.
 
   def loop(deficit) {
       if (deficit > 0) {
           return when (def success := step()) -> {
               loop(deficit + success.pick(-1, 1))
           }
       }
   }
   return loop(1)

}</lang>

Fortran

Translation of: C++

Works with: Fortran version 90 and later

<lang fortran>module StairRobot

 implicit none

contains

 logical function step()
   ! try to climb up and return true or false
   step = .true.     ! to avoid compiler warning
 end function step

 recursive subroutine step_up_rec
   do while ( .not. step() )
      call step_up_rec
   end do
 end subroutine step_up_rec

 subroutine step_up_iter
   integer :: i = 0
   do while ( i < 1 )
      if ( step() ) then
         i = i + 1
      else
         i = i - 1
      end if
   end do
 end subroutine step_up_iter

end module StairRobot</lang>

Haskell

In Haskell, stateful computation is only allowed in a monad. Then suppose we have a monad Robot with an action step :: Robot Bool. We can implement stepUp like this:

<lang haskell>stepUp :: Robot () stepUp = untilM step stepUp

untilM :: Monad m => m Bool -> m () -> m () untilM test action = do

   result <- test
   if result then return () else action >> untilM test action</lang>

Here's an example implementation of Robot and step, as well as a main with which to test stepUp.

<lang haskell>import Control.Monad.State import System.Random (StdGen, getStdGen, random)

type Robot = State (Int, StdGen)

step :: Robot Bool step = do

   (i, g) <- get
   let (succeeded, g') = random g
   put (if succeeded then i + 1 else i - 1, g')
   return succeeded

startingPos = 20 :: Int

main = do

   g <- getStdGen
   putStrLn $ "The robot is at step #" ++ show startingPos ++ "."
   let (endingPos, _) = execState stepUp (startingPos, g)
   putStrLn $ "The robot is at step #" ++ show endingPos ++ "."</lang>

J

Solution (Tacit): <lang j>step =: 0.6 > ?@0: attemptClimb =: [: <:`>:@.step 0: isNotUpOne =: -.@(+/@])

step_up=: (] , attemptClimb)^:isNotUpOne^:_</lang> Note that 0: is not a number but a verb (function) that returns the number zero irrespective of its argument(s). Therefore the above solution for step_up meets the restrictions of no variables or numbers.

J's verbs (functions) always take an argument i.e. there is no such thing as a niladic verb. Verbs that ignore their arguments (e.g. step and attemptClimb) achieve the same effect.

Solution (Explicit): <lang j>step_upX=: monad define NB. iterative

 while. -. +/y do. y=. y , _1 1 {~ step 0 end.

)

step_upR=: monad define NB. recursive (stack overflow possible!)

  while. -. step do. step_upR end.

)</lang>

Example usage: <lang j> step_up NB. output is sequence of falls & climbs required to climb one step. _1 1 _1 _1 1 1 1

  +/\ _1 1 _1 _1 1 1 1  NB. running sum of output (current step relative to start)

_1 0 _1 _2 _1 0 1

  +/\ step_up         NB. another example

_1 _2 _3 _2 _3 _2 _1 _2 _3 _4 _3 _2 _3 _2 _3 _2 _3 _2 _1 _2 _1 _2 _1 0 1</lang>

Java

Translation of: C++

<lang java>public void stepUp() {

 while (!step()) stepUp();

}</lang> The following uses a variable and is a bit longer, but avoids a possible stack overflow: <lang cpp>public void stepUp(){

 for (int i = 0; i < 1; step() ? ++i : --i);

}</lang>

OCaml

<lang ocaml>let rec step_up() =

 while not(step()) do
   step_up()
 done

</lang>

Perl

<lang perl>sub step_up { step_up until step; }</lang>

Perl 6

<lang perl6>sub step_up { step_up until step; }</lang>

Python

Iterative

<lang python>def step_up1()

 "Straightforward implementation: keep track of how many level we
  need to ascend, and stop when this count is zero."
 deficit = 1
 while deficit > 0:
   if step():
     deficit -= 1
   else:
     deficit += 1</lang>

Recursive

This satisfies Chung-chieh's challenge to avoid using numbers. Might blow the stack as p approaches 0.5.

<lang python>def step_up2():

 "No numbers."
 while not step():
   step_up2() # undo the fall</lang>

Smalltalk

Works with: GNU Smalltalk

The following uses a block closure and the recursive solution which consumes stack until successful. <lang smalltalk>Smalltalk at: #stepUp put: 0. stepUp := [ [ step value ] whileFalse: [ stepUp value ] ]. </lang>

Tcl

The setup (note that level and steps are not used elsewhere, but are great for testing…) <lang tcl>set level 41 set prob 0.5001 proc step {} {

   global level prob steps
   incr steps
   if {rand() < $prob} {

incr level 1 return 1

   } else {

incr level -1 return 0

   }

}</lang>

Iterative Solution

All iterative solutions require a counter variable, but at least we can avoid any literal digits... <lang tcl>proc step-up-iter {} {

   for {incr d} {$d} {incr d} {

incr d [set s -[step]]; incr d $s

   }

}</lang>

Recursive Solution

This is the simplest possible recursive solution: <lang tcl>proc step-up-rec {} {

   while {![step]} step-up-rec

}</lang>