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# Hourglass puzzle

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.

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

Notes

Implemented as a 1-player game.

## Go

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

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

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

-- 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),
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

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
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

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:
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

# 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

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
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

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)
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]