Stair-climbing puzzle

From Rosetta Code
Task
Stair-climbing puzzle
You are encouraged to solve this task according to the task description, using any language you may know.

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?

Here's a pseudo-code of a simple recursive solution without using variables:

func step_up()
{
    if not step() {
        step_up();
        step_up();
    }
}

Inductive proof that step_up() steps up one step, if it terminates:

  • Base case (if the step() call returns true): it stepped up one step. QED
  • Inductive case (if the step() call returns false): Assume that recursive calls to step_up() step up one step. It stepped down one step (because step() returned false), but now we step up two steps using two step_up() calls. QED


The second (tail) recursion above can be turned into an iteration, as follows:

func step_up()
{
    while not step() {
        step_up();
    }
}

ActionScript[edit]

Iterative[edit]

function stepUp()
{
var i:int = 0;
while(i < 1)
if(step())i++;
else i--;
}

Recursive[edit]

function stepUp()
{
if(!step())
{
stepUp();
stepUp();
}
}

Ada[edit]

procedure Step_Up is
begin
while not Step loop
Step_Up;
end loop;
end Step_Up;

The following is a test program simulating Step:

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;

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

Aime[edit]

Translation of: C
void step_up(void)
{
while (!step()) {
step_up();
}
}

ALGOL 68[edit]

Translation of: Ada
Works with: ALGOL 68 version Standard - no extensions to language used
Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny
Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8.8d.fc9.i386
   PROC step up = VOID:
BEGIN
WHILE NOT step DO
step up
OD
END # step up #;
The following is a test program simulating step:
 
PROC scaffolding = VOID:
BEGIN
INT level := 0;
 
PROC step = BOOL:
BEGIN
IF random > 0.5 THEN
level +:= 1;
print(("Climbed up to",level, new line));
TRUE
ELSE
level -:= 1;
print(("Fell down to",level, new line));
FALSE
FI
END # step #;
 
PROC step up = VOID:
BEGIN
WHILE NOT step DO
step up
OD
END # step up #;
 
step up
END # scaffolding #;
 
scaffolding

Sample output:

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

AutoHotkey[edit]

Recursive solution:

step_up()
{
While !step()
step_up()
}

AWK[edit]

 
function step_up() {
while (!step()) { step_up() }
}
 

BASIC[edit]

Works with: QBasic
Works with: Visual Basic

For many (most?) BASICs, STEP is a (case-insensitive) keyword, therefore the "step" function would need a different name -- in this example, "step1". (Also, for some BASICs -- notably those influenced by Microsoft's QuickBASIC -- the underscore character ("_") is invalid inside subroutine names.)

SUB stepup
IF NOT step1 THEN stepup: stepup
END SUB

See also: BBC BASIC, Liberty BASIC, PureBasic, TI-83 BASIC

BBC BASIC[edit]

Recursive solution:

      DEF PROCstepup
IF NOT FNstep PROCstepup : PROCstepup
ENDPROC

C[edit]

void step_up(void)
{
while (!step()) {
step_up();
}
}

The following uses a variable and is a bit longer, but avoids a possible stack overflow by risking a probably less likely integer overflow instead:

void step_up(void)
{
int i = 0;
 
while (i < 1) {
if (step()) {
++i;
} else {
--i;
}
}
}

C++[edit]

void step_up()
{
while (!step()) step_up();
}

The following uses a variable and is a bit longer, but avoids a possible stack overflow:

void step_up()
{
for (int i = 0; i < 1; step()? ++i : --i);
}

C#[edit]

void step_up() {
while (!step()) step_up();
}

Clojure[edit]

First, some boilerplate.

;; 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) )

Tail-recursive[edit]

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

(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)))) ) )

Recursive[edit]

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

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

Common Lisp[edit]

(defun step-up ()
(unless (step) (step-up) (step-up)))

D[edit]

The recursive version (note that "step_up" is equivalent to "step_up()" in D):

void step_up()
{
while(!step)
step_up;
}

The non-recursive version, using 1 variable:

void step_up_nr()
{
for(uint i = 0; i < 1; step ? ++i : --i) {};
}

Test program:

import std.stdio;
import std.random;
 
int position;
bool step()
{
bool r = rand() > (uint.max / 2);
if(r)
writefln("Climbed up to %d", ++position);
else
writefln("Fell down to %d", --position);
return r;
}
 
void step_up()
{
while(!step)
step_up;
}
 
void main()
{
rand_seed(0, 0); // to make it somewhat repeatable
step_up;
}

Sample output:

Fell down to -1
Fell down to -2
Fell down to -3
Fell down to -4
Climbed up to -3
Fell down to -4
Climbed up to -3
Climbed up to -2
Climbed up to -1
Climbed up to 0
Climbed up to 1

E[edit]

This is at first a very direct
Translation of: Clojure

The problem framework:

var level := 41
var prob := 0.5001
 
def step() {
def success := entropy.nextDouble() < prob
level += success.pick(1, -1)
return success
}

Counting solution:

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

Ordinary recursive solution:

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

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.

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

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

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)
}

EchoLisp[edit]

 
(define (step-up) (while (not (step)) (step-up)))
;; checking this is tail-recusive :
step-up
(#λ null (#while (#not (step)) (#lambda-tail-call)))
 
;; Experimentation (not part of the task)
;; How much step calls to climb 1000 stairs ?
;; success is the robot success probability
(define (step)
(set! STEPS (1+ STEPS)) ;; count
(< (random) SUCCESS)) ;; ->#t or #f
 
(define (climb stairs)
(when (> stairs 0) (step-up) (climb (1- stairs))))
 
(define (task (stairs 1000))
(for ((success (in-range 1 0 -5/100)))
(set! SUCCESS success)
(set! STEPS 0)
(climb stairs)
(writeln 'stairs stairs 'probability success 'steps STEPS)))
 
Output:
stairs     1000     probability     1         steps     1000    
stairs     1000     probability     19/20     steps     1062    
stairs     1000     probability     9/10      steps     1115    
stairs     1000     probability     17/20     steps     1207    
stairs     1000     probability     4/5       steps     1254    
stairs     1000     probability     3/4       steps     1305    
stairs     1000     probability     7/10      steps     1440    
stairs     1000     probability     13/20     steps     1542    
stairs     1000     probability     3/5       steps     1641    
stairs     1000     probability     11/20     steps     1865    
stairs     1000     probability     1/2       steps     2045    
stairs     1000     probability     9/20      steps     2177    
stairs     1000     probability     2/5       steps     2615    
stairs     1000     probability     7/20      steps     2769    
stairs     1000     probability     3/10      steps     3312    
stairs     1000     probability     1/4       steps     3963    
stairs     1000     probability     1/5       steps     5054    
stairs     1000     probability     3/20      steps     6573    
stairs     1000     probability     1/10      steps     9840    
stairs     1000     probability     1/20      steps     18689    

;; looks as if steps = stairs / success-probability

Elixir[edit]

Translation of: Erlang
defmodule Stair_climbing do
defp step, do: 1 == :rand.uniform(2)
 
defp step_up(true), do: :ok
defp step_up(false) do
step_up(step)
step_up(step)
end
 
def step_up, do: step_up(step)
end
 
IO.inspect Stair_climbing.step_up

Erlang[edit]

 
-module(stair).
-compile(export_all).
 
step() ->
1 == random:uniform(2).
 
step_up(true) ->
ok;
step_up(false) ->
step_up(step()),
step_up(step()).
 
step_up() ->
step_up(step()).
 

Euphoria[edit]

procedure step_up()
if not step() then
step_up()
step_up()
end if
end procedure

Forth[edit]

Recursive. May exceed return stack unless compiler optimizes tail calls.

: step-up   begin step 0= while recurse repeat ;

Counting. Avoids using a named variable.

: step-up   -1 begin step if 1+ else 1- then ?dup 0= until ;

Fortran[edit]

Translation of: C++
Works with: Fortran version 90 and later
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

Go[edit]

38 bytes, no variables, no numbers.

func step_up(){for !step(){step_up()}}

Groovy[edit]

 
class Stair_climbing{
static void main(String[] args){
}
static def step_up(){
while not step(){
step_up();
}
}
 
}
 

Haskell[edit]

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:

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

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

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 ++ "."

Icon and Unicon[edit]

Icon (and Unicon) don't have boolean values. Instead expressions (all expressions) either succeed or they fail. Control structures respond to this success or failure. Assuming that step() is implemented in Icon (or Unicon) and fails only when the implementation in another language returns false, then:

procedure step_up()
return step() | (step_up(),step_up())
end

You can subtract a few more characters (and multiply the difficulty of understanding) with:

procedure step_up()
(|not step(), step_up())
end

J[edit]

Solution (Tacit):

step         =: 0.6 > ?@0:
attemptClimb =: [: <:`>:@.step 0:
isNotUpOne =: -.@(+/@])
 
step_up=: (] , attemptClimb)^:isNotUpOne^:_

Note that 0: is not a number but a verb (function) that returns the number zero irrespective of its argument(s). And, arguably, infinity is not any specific number. And, finally, step is presumed to pre-exist in the task description. Therefore the above solution for step_up could validly be said to meet the restrictions of no variables or numbers.

J's verbs (functions) always take an argument. J programmers use verbs which ignore their arguments (e.g. step and attemptClimb) to serve as verbs which "do not take an argument".

Solution (Explicit):

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.
)

Example usage:

   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


Another approach might be:

keepTrying=: (, {: - _1 ^ step)^:({. >: {:)^:_

Here, the argument is the number of the starting step and the result is a list of the numbers of each visited step including the initial and final steps. For example:

   keepTrying 2
2 1 0 1 2 3
keepTrying 3
3 2 3 2 3 2 3 4
keepTrying 4
4 5

Java[edit]

Translation of: C++
public void stepUp() {
while (!step()) stepUp();
}

The following uses a variable and is a bit longer, but avoids a possible stack overflow:

public void stepUp(){
for (int i = 0; i < 1; step() ? ++i : --i);
}

jq[edit]

Works with: jq version 1.4

Since jq is a purely functional language, we need to keep track of time explicitly. This can be done using a clock that ticks the time:

def tick: .+1;

To model the robot's success and failure, we shall assume a sufficiently large array of 0/1 values is available. To avoid problems with modeling infinite time, we will pad the array with 1s if necessary.

def random: [0, 0, 0, 1, 0, 1, 1, 0];

"step" returns true or false based on the current time (the input) and the value of "random":

def step:
random as $r
| if . >= ($r|length) then true else ($r[.] == 1) end ;

We can now define step_up:

def step_up:
if step then tick
else tick | step_up | step_up
end;

Now we can start the simulation at time 0; step_up will then emit the number of "step" attempts that have been made to achieve success:

0 | step_up
Output:
$ jq -n -f stair-climbing_puzzle.jq
11

Tail Call Optimization[edit]

To take advantage of jq's TCO (available in versions of jq after the release of Version 1.4), the step_up function must be tail-recursive and have arity 0. This can be achieved by providing [time, goal] as the input as follows:

def tco_step_up:
.[0] as $time | .[1] as $goal
| if $goal == 0 then $time
else
if $time|step then $goal - 1 else $goal + 1 end
| [ ($time|tick), .] | tco_step_up
end ;

The simulation can then be started as follows:

[0,1] | tco_step_up

Liberty BASIC[edit]

'This demo will try to get the robot to step up
'Run it several times to see the differences; sometimes the robot falls
'quite a ways before making it to the next step up, but sometimes he makes it
'on the first try
 
result = Stepp.Up()
 
Function Stepp.Up()
While Not(Stepp())
result = Stepp.Up()
Wend
End Function
 
Function Stepp()
Stepp = Int((Rnd(1) * 2))
If Stepp Then
Print "Robot stepped up"
Else
Print "Robot fell down"
End If
End Function

[edit]

Recursive.

to step.up
if not step [step.up step.up]
end

Constant space (fully tail-recursive).

to step.up [:n 1]
if :n=0 [stop]
(step.up ifelse step [:n-1] [:n+1])
end

Lua[edit]

 
function step_up()
while not step() do step_up() end
end
 

Mathematica[edit]

StepUp[] := If[!Step[], StepUp[]; StepUp[]]

MATLAB[edit]

function step_up()
while ~step()
step_up();
end

Nim[edit]

proc stepUp1 =
var deficit = 1
while deficit > 0:
if step():
dec deficit
else:
inc deficit
 
proc stepUp2 =
while not step():
stepUp2()

OCaml[edit]

let rec step_up() =
while not(step()) do
step_up()
done
;;

Oz[edit]

Recursive solution:

proc {StepUp}
if {Not {Step}} then
{StepUp} %% make up for the fall
{StepUp} %% repeat original attempt
end
end

Might lead to a stack overflow because the first call to StepUp is not in tail position.

Iterative solution:

proc {StepUp}
Level = {NewCell 0}
in
for until:@Level == 1 do
if {Step} then Level := @Level + 1
else Level := @Level - 1
end
end
end
 

Oz has arbitrary large integers. So if the robot is very unlucky, the contents of the Level variable will fill up all the memory and the program will fail. I believe this problem needs infinite memory to be solved for all cases.

PARI/GP[edit]

step_up()=while(!step(),step_up())

Pascal[edit]

Recursive solution:

procedure stepUp;
begin
while not step do
stepUp;
end;

Perl[edit]

sub step_up { step_up until step; }

Perl 6[edit]

sub step_up { step_up until step; }

Phix[edit]

procedure step_up()
while not step() do step_up() end while
end procedure

PicoLisp[edit]

(de stepUp ()
(until (step1) # ('step1', because 'step' is a system function)
(stepUp) ) )

PowerShell[edit]

function StepUp
{
If ( -not ( Step ) )
{
StepUp
StepUp
}
}
 
# Step simulator for testing
function Step
{
If ( Get-Random 0,1 )
{
$Success = $True
Write-Verbose "Up one step"
}
Else
{
$Success = $False
Write-Verbose "Fell one step"
}
return $Success
}
 
# Test
$VerbosePreference = 'Continue'
StepUp
Output:
VERBOSE: Fell one step
VERBOSE: Fell one step
VERBOSE: Up one step
VERBOSE: Fell one step
VERBOSE: Fell one step
VERBOSE: Up one step
VERBOSE: Up one step
VERBOSE: Up one step
VERBOSE: Fell one step
VERBOSE: Up one step
VERBOSE: Up one step

PureBasic[edit]

Iterative version using one variable.

Procedure step_up()
Protected i
Repeat: If _step(): i + 1: Else: i - 1: EndIf: Until i = 1
EndProcedure

Recursive version. Stack may overflow as probability of a fall approaches or exceeds 50%.

Procedure step_up()
While Not _step()
step_up()
Wend
EndProcedure

Demonstration program.

Global level
 
Procedure _step()
If Random(1) ;equal chance of stepping up or falling down
level + 1
PrintN("Climbed up to " + Str(level))
ProcedureReturn #True
Else
level - 1
PrintN("Fell down to " + Str(level))
ProcedureReturn #False
EndIf
EndProcedure
 
;recursive
Procedure step_up()
While Not _step()
step_up()
Wend
EndProcedure
 
If OpenConsole()
PrintN("Begin at level: " + Str(level))
step_up()
PrintN("*** Now at level: " + Str(level))
 
Print(#CRLF$ + #CRLF$ + "Press ENTER to exit")
Input()
CloseConsole()
EndIf

Sample output:

Begin at level: 0
Fell down to -1
Climbed up to 0
Fell down to -1
Climbed up to 0
Climbed up to 1
*** Now at level: 1

Python[edit]

Iterative[edit]

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

Recursive[edit]

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

def step_up2():
"No numbers."
while not step():
step_up2() # undo the fall

R[edit]

The step() function described would not be idiomatic R, since it would require using the global assignment operator to get the side effect.

step <- function() {
success <- runif(1) > p
## Requires that the "robot" is a variable named "level"
level <<- level - 1 + (2 * success)
success
}

Recursive Solution[edit]

stepUp <- function() {
while(! step()) {
stepUp()
}
}

Iterative Solution[edit]

stepUpIter <- function() {
i <- 0
while ( ! i) {
i <- i - 1 + (2 * step())
}
}

Example output:

> p <- 0.25
> level <- 1
> print(level)
[1] 1
> stepUp()
> print(level)
[1] 2
> stepUpIter()
> print(level)
[1] 3

Racket[edit]

#lang racket
(define p 0.5001)
(define (step)
(> p (random)))
 
(define (step-up n)
(cond ((zero? n) 'done)
((step) (step-up (sub1 n)))
(else (step-up (add1 n)))))
 
(step-up 1)

REBOL[edit]

rebol [
Title: "Stair Climber"
Date: 2009-12-14
Author: oofoe
URL: http://rosettacode.org/wiki/Stair_Climbing
]

 
random/seed now
 
step: does [random/only reduce [yes no]]
 
; Iterative solution with symbol stack. No numbers, draws a nifty
; diagram of number of steps to go. This is intended more to
; demonstrate a correct solution:
 
step_up: func [/steps s] [
either not steps [
print "Starting up..."
step_up/steps copy [|]
][
while [not empty? s][
print [" Steps left:" s]
either step [remove s][append s '|]
]
]
]
 
step_up print ["Success!" crlf]
 
; Recursive solution. No numbers, no variables. "R" means a recover
; step, "+" means a step up.
 
step_upr: does [if not step [prin "R " step_upr prin "+ " step_upr]]
 
step_upr print ["Success!" crlf]
 
; Small recursive solution, no monitoring:
 
step_upt: does [if not step [step_upt step_upt]]
 
step_upt print "Success!"

Output:

Starting up...
    Steps left: |
    Steps left: | |
    Steps left: |
Success!

R R + R R R R R R R + + R + + + + + + R + R R + R + R + R + + + Success!

Success!

REXX[edit]

step_up:           do  while \step();   call step_up
end
return

Ring[edit]

 
stepup()
 
func stepup
n = 0
while n < 1
if stp() n=n+1 else n= n-1 ok
see n + nl
end
 
func stp
return 0
 

Run BASIC[edit]

 
result = stepUp()
 
Function stepUp()
While Not(stepp())
result = stepUp()
Wend
End Function
 
Function stepp()
stepp = int((Rnd(1) * 2))
print "Robot stepped "+word$("up down",stepp+1)
End Function
 

Ruby[edit]

def step_up
start_position = $position
step until ($position == start_position + 1)
end
 
# assumptions about the step function:
# - it maintains the current position of the robot "as a side effect"
# - the robot is equally likely to step back as to step up
def step
if rand < 0.5
$position -= 1
p "fall (#$position)" if $DEBUG
return false
else
$position += 1
p "rise (#$position)" if $DEBUG
return true
end
end
 
$position = 0
step_up

Sample run:

$ ruby -d stair.climbing.rb
"fall (-1)"
"rise (0)"
"fall (-1)"
"rise (0)"
"fall (-1)"
"fall (-2)"
"rise (-1)"
"rise (0)"
"rise (1)"

SAS[edit]

 
%macro step();
%sysfunc(round(%sysfunc(ranuni(0))))
%mend step;
 
%macro step_up();
 
%if not %step %then %do;
%put Step Down;
%step_up;
%step_up;
%end;
%else %put Step Up;
 
%mend step_up;
 
%step_up;
 

Sample Output:

Step Down
Step Down
Step Down
Step Up
Step Down
Step Down
Step Up
Step Up
Step Up
Step Up
Step Up

Scala[edit]

Simple recursive solution:

def stepUp { while (! step) stepUp }

Non-recursive solution which almost gets away with not having named variables:

def stepUp {
def rec: List[Boolean] => Boolean = step :: (_: List[Boolean]) match {
case true :: Nil => true
case true :: false :: rest => rec(rest)
case other => rec(other)
}
rec(Nil)
}

Scheme[edit]

(define (step-up n-steps)
(cond ((zero? n-steps) 'done)
((step) (step-up (- n-steps 1)))
(else (step-up (+ n-steps 1)))))

Seed7[edit]

const proc: step_up is func
begin
while not doStep do
step_up;
end while;
end func;

Sidef[edit]

func step_up() {
while (!step()) {
step_up();
}
}

Smalltalk[edit]

Works with: GNU Smalltalk

The following uses a block closure and the recursive solution which consumes stack until successful.

Smalltalk at: #stepUp put: 0.
stepUp := [ [ step value ] whileFalse: [ stepUp value ] ].

Standard ML[edit]

 
(*
* val step : unit -> bool
* This is a stub for a function which returns true if successfully climb a step or false otherwise.
*)
fun step() = true
 
(*
* val step_up : unit -> bool
*)
fun step_up() = step() orelse (step_up() andalso step_up())
 

Swift[edit]

func step_up() {
while !step() {
step_up()
}
}

The following uses a variable and is a bit longer, but avoids a possible stack overflow:

func step_up() {
for var i = 0; i < 1; step()? ++i : --i { }
}

Tcl[edit]

The setup (note that level and steps are not used elsewhere, but are great for testing…)

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
}
}

Iterative Solution[edit]

All iterative solutions require a counter variable, but at least we can avoid any literal digits...

proc step-up-iter {} {
for {incr d} {$d} {incr d} {
incr d [set s -[step]]; incr d $s
}
}

Recursive Solution[edit]

This is the simplest possible recursive solution:

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

TI-83 BASIC[edit]

TI-83 BASIC doesn't have functions (only subroutines), so a variable must be used as the return value for prgmSTEP. Set A to 1 before calling to display the offset from the stair it was called from (store in D). Set B to 1 to pause after each attempt. C is the return variable for prgmSTEP. D is the stair it is on (only used for display, and not used if A isn't 1).

prgmSTEP:

If rand>.5:Then
0→C
Disp "FALL"
If A=1:Then
D-1→D
Disp D
End
Else
1→C
Disp "CLIMB"
If A=1:Then
D+1→D
Disp D
End
End
If B=1
Pause

prgmSTEPUP:

prgmSTEP
While C=0
prgmSTEPUP
prgmSTEP
End

XPL0[edit]

proc Step_up;           \Iterative version
int I;
[I:= 0;
while I<1 do
if Step then I:= I+1
else I:= I-1;
];
 
proc Step_up; \Recursive version
while not Step do Step_up;

zkl[edit]

According to Leon P Smith in the referenced LTU thread:

This version consumes stack space proportional to the number of steps it needs to go up, as opposed to the number of steps it attempts. This is a substantial difference if the probability of success is only somewhat greater than or equal to 1/2.

fcn step{  } // add code to return Bool
fcn stepUp{ while(not step()){ self.fcn() } }

You could also use "stepUp" instead of self.fcn, self.fcn seems a little clearer and makes it easier to refactor.

An example step function:

Translation of: D
var position=0;
fcn step(){ //-->0|1
r:=(0).random(2); // 0 or 1
if(r) println("Climbed up to ",position+=1);
else println("Fell down to ", position-=1);
r
}
stepUp();
Output:
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