Hourglass puzzle: Difference between revisions
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glass when the four-minute glass ends. |
glass when the four-minute glass ends. |
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</pre> |
</pre> |
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=={{header|Vlang}}== |
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{{trans|go}} |
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<lang vlang>import arrays {sum, min} |
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fn hourglass_flipper(hourglasses []int, target int) (int, []int) { |
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mut flippers := hourglasses.clone() |
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mut series := []int{} |
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for _ in 0..10000 { |
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n := min<int>(flippers) or {flippers[0]} |
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series << n |
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for i in 0..flippers.len { |
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flippers[i] -= n |
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} |
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for i, flipper in flippers { |
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if flipper == 0 { |
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flippers[i] = hourglasses[i] |
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} |
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} |
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for start := series.len - 1; start >= 0; start-- { |
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if sum<int>(series[start..]) or {-1} == target { |
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return start, series |
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} |
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} |
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} |
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return 0, []int{} |
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} |
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fn main() { |
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print("Flip an hourglass every time it runs out of grains, ") |
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println("and note the interval in time.") |
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hgs := [[4, 7], [5, 7, 31]] |
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ts := [9, 36] |
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for i in 0..hgs.len { |
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start, series := hourglass_flipper(hgs[i], ts[i]) |
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end := series.len - 1 |
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println("\nSeries: $series") |
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print("Use hourglasses from indices $start to $end (inclusive) to sum ") |
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println("${ts[i]} using ${hgs[i]}") |
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} |
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}</lang> |
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{{out}} |
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<pre>Same as Go entry</pre> |
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=={{header|Wren}}== |
=={{header|Wren}}== |
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Revision as of 16:59, 12 May 2022
- Task
Given two hourglasses of 4 minutes and 7 minutes, the task is to measure 9 minutes.
- Notes
Implemented as a 1-player game.
11l
<lang 11l>V t4 = 0 L t4 < 10'000
V t7_left = 7 - t4 % 7
I t7_left == 9 - 4
print(|‘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 #. times,
wherupon the 7 hour glass is immediately placed on its side with #. 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.
’.format(t4 I/ 4, t7_left))
L.break
t4 += 4
L.was_no_break
print(‘Not found’)</lang>
- 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.
AWK
<lang AWK>
- syntax: GAWK -f HOURGLASS_PUZZLE.AWK
BEGIN {
limit = 100
t4 = 0
while (t4 < limit) {
t7_left = 7 - t4 % 7
if (t7_left == 9 - 4) {
break
}
t4 += 4
}
if (t4 > limit) {
printf("Unable to find an answer within %d iterations\n",limit)
exit(1)
}
str = sprintf("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 %d times, " \
"wherupon the 7 hour glass is immediately placed on its side with %d minutes " \
"of sand in it. " \
"You can measure 9 minutes by flipping the 4 hour glass once, then " \
"flipping the remaining sand in the 7 hour glass when the 4 hour glass ends.",t4/4,t7_left)
fold(str)
exit(0)
} function fold(rec, chars_printed,indx,text) {
line_length = 80
while (1) {
indx = match(rec," ")
if (indx == 0) {
printf("%s\n",rec)
break
}
text = substr(rec,1,indx)
printf("%s",text)
rec = substr(rec,RSTART+1)
chars_printed += length(text)
if (chars_printed > line_length) {
printf("\n")
chars_printed = 0
}
}
} </lang>
- 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 minutes of sand in it. You can measure 9 minutes by flipping the 4 hour glass once, then flipping the remaining sand in the 7 hour glass when the 4 hour glass ends.
FreeBASIC
<lang freebasic> 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 </lang>
- 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
<lang go>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])
}
}</lang>
- 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. <lang julia>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]")
</lang>
- 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]
Logo
tested with FMSlogo <lang logo> 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
</lang>
Nim
<lang Nim>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(" "))</lang>
- 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
Flip each hourglass when it runs out and note the time for each. <lang perl>#!/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 $want minute@{[$want == 1 ? : 's'
]}, Start at time $-[0] and End at time $+[0]\n"
: "$want is not possible\n";
}</lang>
- 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 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
There isn't much of a task description as I write this, but, here goes...
<lang python>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()</lang>
- 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
<lang perl6># 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</lang>
- 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
<lang rexx>/*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</lang>
- 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.
Vlang
<lang vlang>import arrays {sum, min}
fn hourglass_flipper(hourglasses []int, target int) (int, []int) {
mut flippers := hourglasses.clone()
mut series := []int{}
for _ in 0..10000 {
n := min<int>(flippers) or {flippers[0]}
series << n
for i in 0..flippers.len {
flippers[i] -= n
}
for i, flipper in flippers {
if flipper == 0 {
flippers[i] = hourglasses[i]
}
}
for start := series.len - 1; start >= 0; start-- {
if sum<int>(series[start..]) or {-1} == target {
return start, series
}
}
}
return 0, []int{}
}
fn main() {
print("Flip an hourglass every time it runs out of grains, ")
println("and note the interval in time.")
hgs := [[4, 7], [5, 7, 31]]
ts := [9, 36]
for i in 0..hgs.len {
start, series := hourglass_flipper(hgs[i], ts[i])
end := series.len - 1
println("\nSeries: $series")
print("Use hourglasses from indices $start to $end (inclusive) to sum ")
println("${ts[i]} using ${hgs[i]}")
}
}</lang>
- Output:
Same as Go entry
Wren
<lang ecmascript>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)")
}</lang>
- 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]