Stair-climbing puzzle
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 (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?
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>
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
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>
OCaml
<lang ocaml>let rec step_up() =
while not(step()) do step_up() done
- </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>
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>