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Hourglass puzzle

From Rosetta Code
Hourglass puzzle is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
Task

Given two hourglasses of   4   minutes and   7   minutes,   the task is to measure   9   minutes.


Notes

Implemented as a 1-player game.


FreeBASIC[edit]

Translation of: Phyton
 
Sub Hourglass_puzzle()
Dim As Uinteger t4 = 0, limite = 1000, t7_left
While t4 < limite
t7_left = 7 - t4 Mod 7
If t7_left = 9 - 4 Then Exit While
t4 += 4
Wend
If t4 > limite Then
Print "No encontrado"
Return
End If
Print Using !"\nVoltee al mismo tiempo ambos relojes de arena y contin£e volte ndolos \n" + _
 !"cuando los relojes de arena se agoten individualmente, hasta que el \n" + _
 !"reloj de arena de 4 minutos se voltee & veces, despu‚s de lo cual el \n" + _
 !"reloj de 7 minutos se coloca inmediatamente de lado con & minutos de \n" + _
 !"arena en ‚l. \n" + _
 !"\nPuede medir 9 minutos volteando el reloj de 4 minutos una vez, luego \n" + _
 !"volteando la arena restante en el reloj de 7 minutos cuando termine \n" + _
 !"el reloj de 4 minutos."; t4/4; t7_left
End Sub
 
Hourglass_puzzle()
Sleep
 
Output:
Voltee al mismo tiempo ambos relojes de arena y continúe volteándolos
cuando los relojes de arena se agoten individualmente, hasta que el
reloj de arena de 4 minutos se voltee 4 veces, después de lo cual el
reloj de 7 minutos se coloca inmediatamente de lado con 5 minutos de
arena en él.

Puede medir 9 minutos volteando el reloj de 4 minutos una vez, luego
volteando la arena restante en el reloj de 7 minutos cuando termine
el reloj de 4 minutos.


Go[edit]

Translation of: Julia
package main
 
import (
"fmt"
"log"
)
 
func minimum(a []int) int {
min := a[0]
for i := 1; i < len(a); i++ {
if a[i] < min {
min = a[i]
}
}
return min
}
 
func sum(a []int) int {
s := 0
for _, i := range a {
s = s + i
}
return s
}
 
func hourglassFlipper(hourglasses []int, target int) (int, []int) {
flippers := make([]int, len(hourglasses))
copy(flippers, hourglasses)
var series []int
for iter := 0; iter < 10000; iter++ {
n := minimum(flippers)
series = append(series, n)
for i := 0; i < len(flippers); i++ {
flippers[i] -= n
}
for i, flipper := range flippers {
if flipper == 0 {
flippers[i] = hourglasses[i]
}
}
for start := len(series) - 1; start >= 0; start-- {
if sum(series[start:]) == target {
return start, series
}
}
}
log.Fatal("Unable to find an answer within 10,000 iterations.")
return 0, nil
}
 
func main() {
fmt.Print("Flip an hourglass every time it runs out of grains, ")
fmt.Println("and note the interval in time.")
hgs := [][]int{{4, 7}, {5, 7, 31}}
ts := []int{9, 36}
for i := 0; i < len(hgs); i++ {
start, series := hourglassFlipper(hgs[i], ts[i])
end := len(series) - 1
fmt.Println("\nSeries:", series)
fmt.Printf("Use hourglasses from indices %d to %d (inclusive) to sum ", start, end)
fmt.Println(ts[i], "using", hgs[i])
}
}
Output:
Flip an hourglass every time it runs out of grains, and note the interval in time.

Series: [4 3 1 4 2 2]
Use hourglasses from indices 2 to 5 (inclusive) to sum 9 using [4 7]

Series: [5 2 3 4 1 5 1 4 3 2 1 4 5 2 3 4 1]
Use hourglasses from indices 4 to 16 (inclusive) to sum 36 using [5 7 31]

Julia[edit]

Implemented as a game solver rather than as a game with user input.

function euclidean_hourglassflipper(hourglasses, target::Integer)
gcd(hourglasses) in hourglasses && !(1 in hourglasses) && throw("Hourglasses fail sanity test (not relatively prime enough)")
flippers, series = deepcopy(hourglasses), Int[]
for i in 1:typemax(target)
n = minimum(flippers)
push!(series, n)
flippers .-= n
for (i, n) in enumerate(flippers)
if n == 0
flippers[i] = hourglasses[i]
end
end
for startpoint in length(series):-1:1
if sum(series[startpoint:end]) == target
println("Series: $series")
return startpoint, length(series)
end
end
end
end
 
println("Flip an hourglass every time it runs out of grains, and note the interval in time.")
i, j = euclidean_hourglassflipper([4, 7], 9)
println("Use hourglasses from step $i to step $j (inclusive) to sum 9 using [4, 7]")
i, j = euclidean_hourglassflipper([5, 7, 31], 36)
println("Use hourglasses from step $i to step $j (inclusive) to sum 36 using [5, 7, 31]")
 
Output:
Flip an hourglass every time it runs out of grains, and note the interval in time.
Series: [4, 3, 1, 4, 2, 2]
Use hourglasses from step 3 to step 6 (inclusive) to sum 9 using [4, 7]
Series: [5, 2, 3, 4, 1, 5, 1, 4, 3, 2, 1, 4, 5, 2, 3, 4, 1]
Use hourglasses from step 5 to step 17 (inclusive) to sum 36 using [5, 7, 31]

[edit]

tested with FMSlogo

 
to bb
Make "small_capacity 4
Make "big_capacity 7
make "small 0
make "big 0
make "t 0
print "_____________decision_0_game_over
print "_________decision_1_start_timing
print "_______decision_2_flip_small
print "____decision_3_flip_big
print "__decision_4_flip_both
print "_________any_other_number________________wait
do.until [show list list :small :big :t print "your_decision_0_1_2_3_4 human_decision if :my_decision>1 [machine_computes] ] [:my_decision=0]
print list :t "minutes_passed
end
 
to human_decision
make "my_decision readword
if :my_decision=1 [print "reset_timer make "t 0]
if :my_decision=2 [print "flip_small make "small :small_capacity-:small]
if :my_decision=3 [print "flip_big make "big :big_capacity-:big]
if :my_decision=4 [print "flip_both make "small :small_capacity-:small make "big :big_capacity-:big ]
if :my_decision>4 [print "wait]
end
 
to machine_computes
ifelse :small>:big [make "my_selection :big] [make "my_selection :small]
if :small=0 [make "my_selection :big]
if :big=0 [make "my_selection :small]
make "small  :small-:my_selection
make "big  :big-:my_selection
make "t  :t+:my_selection
if :small<0 [make "small 0]
if :big<0 [make "big 0]
end
 
to zzz
;A. 7 minutes with 4- and 5-minute timers
;B. 15 minutes with 7- and 11-minute timers
;C. 14 minutes with 5- and 8-minute timers
ifelse YesNoBox [Welcome] [run / show me the code] [bb] [edall]
;A is possible: Turn both the 5 and the 4. When the 4 runs out, flip it over.Now, when the 5 runs out, start timing. The 4 will run for three more minutes, after which, you can flip it over to reach 7.
;B is possible: Turn both the 7 and the 11. When the 7 runs out, start timing. The 11 will run for 4 more minutes, after which it can be flipped to reach 15.
;C is possible: Turn both the 5 and the 8. When the 5 runs out, flip it. The 8 will then run out after 3 minutes, leaving 2 minutes in the 5. Flip the 8 then. When the 5 runs out, start timing. There are now 6 minutes left in the 8, and flipping the 8 after those 6 minutes gives 6 + 8 = 14 minutes.
end
 
Make "big 0
Make "big_capacity 5
Make "my_decision "
Make "my_selection 4
Make "small 0
Make "small_capacity 4
Make "startup [zzz]
Make "t 0
 
 
 

Nim[edit]

Translation of: Wren
import math, strutils
 
func hourglassFlipper(hourglasses: openArray[int];
target: int): tuple[start: int; series: seq[int]] =
var flippers = @hourglasses
for _ in 0..10_000:
let n = min(flippers)
result.series.add n
for i in 0..flippers.high:
dec flippers[i], n
if flippers[i] == 0: flippers[i] = hourglasses[i]
result.start = result.series.high
while result.start >= 0:
if sum(result.series[result.start..^1]) == target: return
dec result.start
raise newException(ValueError, "Unable to find an answer within 10_000 iterations.")
 
 
echo "Flip an hourglass every time it runs out of grains, "
echo "and note the interval in time."
const Tests = [(@[4, 7], 9), (@[5, 7, 31], 36)]
for test in Tests:
let
hourglasses = test[0]
target = test[1]
(start, series) = hourglassFlipper(hourglasses, target)
`end` = series.high
echo "\nSeries: ", series.join(" ")
echo "Use hourglasses from indices $1 to $2 (inclusive) to sum ".format(start, `end`),
"$1 using $2.".format(target, hourglasses.join(" "))
Output:
Flip an hourglass every time it runs out of grains, 
and note the interval in time.

Series: 4 3 1 4 2 2
Use hourglasses from indices 2 to 5 (inclusive) to sum 9 using 4 7.

Series: 5 2 3 4 1 5 1 4 3 2 1 4 5 2 3 4 1
Use hourglasses from indices 4 to 16 (inclusive) to sum 36 using 5 7 31.

Perl[edit]

Flip each hourglass when it runs out and note the time for each.

#!/usr/bin/perl
 
use strict; # https://rosettacode.org/wiki/Hourglass_puzzle
use warnings;
 
findinterval( $_, 4, 7 ) for 1 .. 20;
 
sub findinterval
{
my ($want, $hour1, $hour2) = @_;
local $_ = (('1' | ' ' x $hour1) x $hour2 | ('2' | ' ' x $hour2) x $hour1) x $want;
print /(?=\d).{$want}(?=\d)/
? "To get $want [email protected]{[$want == 1 ? '' : 's'
]}, Start at time $-[0] and End at time $+[0]\n"

: "$want is not possible\n";
}
Output:
To get 1 minute, Start at time 7 and End at time 8
To get 2 minutes, Start at time 12 and End at time 14
To get 3 minutes, Start at time 4 and End at time 7
To get 4 minutes, Start at time 0 and End at time 4
To get 5 minutes, Start at time 7 and End at time 12
To get 6 minutes, Start at time 8 and End at time 14
To get 7 minutes, Start at time 0 and End at time 7
To get 8 minutes, Start at time 0 and End at time 8
To get 9 minutes, Start at time 7 and End at time 16
To get 10 minutes, Start at time 4 and End at time 14
To get 11 minutes, Start at time 21 and End at time 32
To get 12 minutes, Start at time 0 and End at time 12
To get 13 minutes, Start at time 7 and End at time 20
To get 14 minutes, Start at time 0 and End at time 14
To get 15 minutes, Start at time 20 and End at time 35
To get 16 minutes, Start at time 0 and End at time 16
To get 17 minutes, Start at time 4 and End at time 21
To get 18 minutes, Start at time 14 and End at time 32
To get 19 minutes, Start at time 16 and End at time 35
To get 20 minutes, Start at time 0 and End at time 20

Phix[edit]

-- demo\rosetta\Hourglass_puzzle.exw
with javascript_semantics
procedure print_solution(sequence eggtimers, tries, integer target, pdx)
    sequence soln = {tries[$]}, remain
    integer n = length(eggtimers), tdx = tries[$][3], t, flipbits
    string et = ""
    for timer=1 to n do
        if timer=n then et &= " and "
        elsif timer>1 then et &= ", " end if
        et &= sprintf("%d",eggtimers[timer])
    end for
    printf(1,"\nSolution for %d minutes with %s minute eggtimers:\n",{target,et})
    while tdx do
        if tdx=pdx then soln &= 0 end if
        soln = append(soln,tries[tdx])
        tdx = tries[tdx][3]
    end while
    soln = reverse(soln[1..$-1])
    integer tp = 0, ro = 0
    sequence premain = repeat(0,n)
    for i=1 to length(soln) do
        if soln[i]=0 then
            puts(1,"start timer\n")
        else
            {remain,t,?,flipbits} = soln[i]
            sequence flip = int_to_bits(flipbits,n)
            string fs = "", lv = ""
            for timer=1 to n do
                if flip[timer] then
                    if length(fs) then fs &= " and " end if
                    fs &= sprintf("%d",eggtimers[timer])
                    if premain[timer] then
                        fs &= sprintf(" (leaving %d)",eggtimers[timer]-premain[timer])
                    end if
                end if
                if remain[timer]=0 then
                    if flip[timer] or premain[timer]!=0 then
                        ro = eggtimers[timer]
                    end if
                else
                    if length(lv) then lv &= " and " end if
                    lv &= sprintf("%d in %d",{remain[timer],eggtimers[timer]})
                end if
            end for
            lv = iff(length(lv)?" (leaving "&lv&")":"")
            printf(1,"At t=%d, flip %s, then when %d runs out%s...\n",{tp,fs,ro,lv})
            tp = t
            premain = remain
        end if
    end for
    printf(1,"At t=%d, stop timer\n",{tp})
end procedure
 
procedure solve(sequence eggtimers, integer target)
    integer n = length(eggtimers), tdx = 1, t, pdx
    sequence remain = repeat(0,n),
             tries = {{remain,0,0,0}} -- Template:Remain,t,link,flip
    while tdx<=length(tries) do
        for flipbits=1 to power(2,n)-1 do
            {remain,t} = deep_copy(tries[tdx])
            sequence flip = int_to_bits(flipbits,n)
            for timer=1 to n do
                if flip[timer] then
                    remain[timer] = eggtimers[timer]-remain[timer]
                end if
            end for
            integer mr = min(filter(remain,">",0))
            remain = sq_max(sq_sub(remain,mr),0)
            mr += t
            tries = append(tries,{remain,mr,tdx,flipbits})
            pdx = tdx
            while pdx do
                mr -= tries[pdx][2]
                if mr>=target then
                    if mr>target then exit end if
                    print_solution(eggtimers, tries, target, pdx)
                    return
                end if  
                mr += tries[pdx][2]
                pdx = tries[pdx][3]
            end while
        end for
        tdx += 1
        -- totally arbitrary sanity crash:
        if length(tries)>20000 then crash("no solution") end if
    end while
end procedure
solve({4,7},9)
solve({4,7},15)
solve({5,7,31},36) -- (slightly better output than Julia, I think...)
solve({4,5},7)   -- (logo solution stops timer at t=12, this manages t=11)
solve({7,11},15) -- (logo solution stops timer at t=22, this manages t=15)
solve({5,8},14)  -- (logo solution stops timer at t=24, this manages t=19)
Output:
Solution for 9 minutes with 4 and 7 minute eggtimers:
start timer
At t=0, flip 4 and 7, then when 4 runs out (leaving 3 in 7)...
At t=4, flip 4, then when 7 runs out (leaving 1 in 4)...
At t=7, flip 7, then when 4 runs out (leaving 6 in 7)...
At t=8, flip 7 (leaving 1), then when 7 runs out...
At t=9, stop timer

Solution for 15 minutes with 4 and 7 minute eggtimers:
start timer
At t=0, flip 4, then when 4 runs out...
At t=4, flip 4, then when 4 runs out...
At t=8, flip 7, then when 7 runs out...
At t=15, stop timer

Solution for 36 minutes with 5, 7 and 31 minute eggtimers:
start timer
At t=0, flip 5, then when 5 runs out...
At t=5, flip 31, then when 31 runs out...
At t=36, stop timer

Solution for 7 minutes with 4 and 5 minute eggtimers:
At t=0, flip 4 and 5, then when 4 runs out (leaving 1 in 5)...
start timer
At t=4, flip 4, then when 5 runs out (leaving 3 in 4)...
At t=5, flip 4 (leaving 1), then when 4 runs out...
At t=6, flip 5, then when 5 runs out...
At t=11, stop timer

Solution for 15 minutes with 7 and 11 minute eggtimers:
start timer
At t=0, flip 7 and 11, then when 7 runs out (leaving 4 in 11)...
At t=7, flip 7, then when 11 runs out (leaving 3 in 7)...
At t=11, flip 7 (leaving 4), then when 7 runs out...
At t=15, stop timer

Solution for 14 minutes with 5 and 8 minute eggtimers:
At t=0, flip 5 and 8, then when 5 runs out (leaving 3 in 8)...
start timer
At t=5, flip 5, then when 8 runs out (leaving 2 in 5)...
At t=8, flip 5 (leaving 3), then when 5 runs out...
At t=11, flip 8, then when 8 runs out...
At t=19, stop timer

Python[edit]

There isn't much of a task description as I write this, but, here goes...

def hourglass_puzzle():
t4 = 0
while t4 < 10_000:
t7_left = 7 - t4 % 7
if t7_left == 9 - 4:
break
t4 += 4
else:
print('Not found')
return
print(f"""
Turn over both hour glasses at the same time and continue flipping them each
when they individually run down until the 4 hour glass is flipped {t4//4} times,
wherupon the 7 hour glass is immediately placed on its side with {t7_left} hours
of sand in it.
You can measure 9 hours by flipping the 4 hour glass once, then
flipping the remaining sand in the 7 hour glass when the 4 hour glass ends.
"""
)
 
hourglass_puzzle()
Output:
Turn over both hour glasses at the same time and continue flipping them each
when they individually run down until the 4 hour glass is flipped 4 times,
wherupon the 7 hour glass is immediately placed on its side with 5 hours 
of sand in it.
You can measure 9 hours by flipping the 4 hour glass once, then
flipping the remaining sand in the 7 hour glass when the 4 hour glass ends.

Raku[edit]

# 20201230 Raku programming solution
 
my @hourglasses = 4, 7;
my $target = 9;
my @output = [];
my %elapsed = 0 => 1 ;
my $done = False ;
 
for 1 ..-> $t {
my $flip-happened = False;
for @hourglasses -> $hg {
unless $t % $hg {
%elapsed{$t} = 1 unless %elapsed{$t};
with @output[$t] { $_ ~= "\t, flip hourglass $hg " } else {
$_ = "At time t = $t , flip hourglass $hg" }
$flip-happened = True
}
}
if $flip-happened {
for %elapsed.keys.sort -> $t1 {
if ($t - $t1) == $target {
@output[$t1] ~= "\tbegin = 0";
@output[$t] ~= "\tend = $target";
$done = True
}
%elapsed{$t} = 1 unless %elapsed{$t} ;
}
}
last if $done
}
 
.say if .defined for @output
Output:
At time t = 4 , flip hourglass 4
At time t = 7 , flip hourglass 7        begin = 0
At time t = 8 , flip hourglass 4
At time t = 12 , flip hourglass 4
At time t = 14 , flip hourglass 7
At time t = 16 , flip hourglass 4       end   = 9

REXX[edit]

Translation of: Python
/*REXX program determines  if there is a  solution  to measure  9 minutes  using a      */
/*──────────────────────────────────── four and seven minute sandglasses. */
t4= 0
mx= 10000
do t4=0 by 4 to mx
t7_left= 7 - t4 % 7
if t7_left==9-4 then leave
end /*t4*/
say
if t4>mx then do
say 'Not found.'
exit 4
end
 
say "Turn over both sandglasses (at the same time) and continue"
say "flipping them each when the sandglasses individually run down"
say "until the four-minute glass is flipped " t4%4 ' times,'
say "whereupon the seven-minute glass is immediately placed on its"
say "side with " t7_left ' minutes of sand in it.'
say
say "You can measure 9 minutes by flipping the four-minute glass"
say "once, then flipping the remaining sand in the seven-minute"
say "glass when the four-minute glass ends."
say
exit 0
output   when using the internal default input:
Turn over both sandglasses  (at the same time)  and continue
flipping them each when the sandglasses individually run down
until the four-minute glass is flipped  4  times,
whereupon the seven-minute glass is immediately placed on its
side with  5  minutes of sand in it.

You can measure 9 minutes by flipping the four-minute glass
once,  then flipping the remaining sand in the seven-minute
glass when the four-minute glass ends.

Wren[edit]

Translation of: Julia
Library: Wren-math
import "/math" for Nums
 
var hourglassFlipper = Fn.new { |hourglasses, target|
var flippers = hourglasses.toList
var series = []
for (iter in 0...10000) {
var n = Nums.min(flippers)
series.add(n)
for (i in 0...flippers.count) flippers[i] = flippers[i] - n
var i = 0
for (flipper in flippers) {
if (flipper == 0) flippers[i] = hourglasses[i]
i = i + 1
}
for (start in series.count-1..0) {
if (Nums.sum(series[start..-1]) == target) return [start, series]
}
}
Fiber.abort("Unable to find an answer within 10,000 iterations.")
}
 
System.write("Flip an hourglass every time it runs out of grains, ")
System.print("and note the interval in time.")
var tests = [ [[4, 7], 9], [[5, 7, 31], 36] ]
for (test in tests) {
var hourglasses = test[0]
var target = test[1]
var res = hourglassFlipper.call(hourglasses, target)
var start = res[0]
var series = res[1]
var end = series.count - 1
System.print("\nSeries: %(series)")
System.write("Use hourglasses from indices %(start) to %(end) (inclusive) to sum ")
System.print("%(target) using %(hourglasses)")
}
Output:
Flip an hourglass every time it runs out of grains, and note the interval in time.

Series: [4, 3, 1, 4, 2, 2]
Use hourglasses from indices 2 to 5 (inclusive) to sum 9 using [4, 7]

Series: [5, 2, 3, 4, 1, 5, 1, 4, 3, 2, 1, 4, 5, 2, 3, 4, 1]
Use hourglasses from indices 4 to 16 (inclusive) to sum 36 using [5, 7, 31]