# Hourglass puzzle

From Rosetta Code

*is a*

**Hourglass puzzle****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 hourglass of 4 minutes and 7 minutes, the task is to measure 9 minutes.

- Notes

Implemented as a 1-player game.

## Go[edit]

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]

## 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

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}} -- {{remain,t,link,flip}}

while tdx<=length(tries) do

for flipbits=1 to power(2,n)-1 do

{remain,t} = 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

## Logo[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

## 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]

/*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]

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]