Given N decanters (0 < N < 10) with that can hold C0 ... CN-1 liters (0 < C < 50) and a goal G liters, please determine if it is possible to reach that goal using only the following actions:

  • Fill a decanter
  • Empty a decanter
  • Pour from one decanter to another until the one being poured to is full or the one being poured from is empty

The goal amount G must be the amount of water in one of the containers at the end. You cannot have a 'output decanter'.


N: 2
C0: 5
C1: 12
G: 1
Result: Yes

N: 3
C0: 6
C1: 9
C2: 21
G: 5
Result: No

Hint: To calculate if it is possible, check to see if G is divisible by the GCD of the capacities. Also, make sure it will fit in a container.

Remember, this is , so the code with the lowest number of bytes wins.


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  • \$\begingroup\$ Related. \$\endgroup\$ Commented Sep 22, 2016 at 17:46
  • \$\begingroup\$ Is there a "output decanter"? Aka, if I have a decanter of size 1, is any capacity possible? \$\endgroup\$ Commented Sep 22, 2016 at 17:48
  • \$\begingroup\$ @MartinEnder Ahh. Fixed. \$\endgroup\$
    – Oliver Ni
    Commented Sep 22, 2016 at 17:48
  • \$\begingroup\$ @NathanMerrill There is no "output decanter." You need to be able to get it in one of the decanters given. \$\endgroup\$
    – Oliver Ni
    Commented Sep 22, 2016 at 17:49
  • 9
    \$\begingroup\$ This same challenge was being Sandboxed. \$\endgroup\$
    – xnor
    Commented Sep 22, 2016 at 17:55

11 Answers 11


Haskell, 35 bytes

l%n=n`mod`foldr1 gcd l<1&&any(>=n)l

This paper proves a result that vastly simplifies the problem. Prop 1 says that

You can achieve a goal exactly when it's both:

  • A multiple of the greatest common divisor (gcd) of the capacities,
  • At most the maximum capacity

It's clear why both of these are necessary: all amounts remain multiples of the gcd, and the goal must fit in a container. The key of the result is an algorithm to produce any goal amount that fits these conditions.

Call the operator % like [3,6,12]%9.

A 37-byte alternative:

l%n=elem n[0,foldr1 gcd l..maximum l]
  • \$\begingroup\$ I believe the goal has to fit in one of the decanters, it should be less than the largest decanter's volume (based on @Oliver's comment "You need to be able to get it in one of the decanters given."). \$\endgroup\$
    – m-chrzan
    Commented Sep 22, 2016 at 18:11
  • \$\begingroup\$ Conveniently, that's actually the definition used in the paper and I misread, so that's an easy fix. \$\endgroup\$
    – xnor
    Commented Sep 22, 2016 at 18:13

05AB1E, 9 8 9 bytes

Uses the CP-1252 encoding



          # true if
   %      # target size modulo
ZU¿       # gcd of decanter sizes
        ‹ # is smaller than
    ²X>‹  # target size is less than or equal to max decanter size

Try it online!

Saved 1 byte utilizing the less than trick from Luis Mendo's MATL answer

  • 1
    \$\begingroup\$ utilizing the less than trick ... which I learned from Dennis :-) \$\endgroup\$
    – Luis Mendo
    Commented Sep 22, 2016 at 19:17
  • \$\begingroup\$ The actual answer is still 9 bytes ;-) \$\endgroup\$ Commented Sep 22, 2016 at 19:53
  • \$\begingroup\$ @ETHproductions Oops! Seems I only updated the explanation and TIO link, not the actual code. Thanks :) \$\endgroup\$
    – Emigna
    Commented Sep 22, 2016 at 20:20

Jelly, 9 8 7 bytes

-1 byte thanks to @Dennis (use integer division, :, rather than not less than, )




Ṁ:a⁸g/ḍ - Main link: capacities, goal
Ṁ       - maximum capacity
 :      - integer division with goal (effectively not less than goal since non-0 is True)
  a     - and
   ⁸    - left argument (capacities)
    g/  - gcd reduce over list (gcd of capacities)
      ḍ - divides

MATL, 10 bytes


Try it online!

This uses @xnor's approach.

&Zd    % Take array C as input. Compute the gcd of its elements
\      % Take number G as input. Compute that number modulo the above. Call this A
&G     % Push the two inputs again: C, then G
<~a    % Gives 1 if some element of C is at least G; 0 otherwise. Call this B
<      % Gives true if A is 0 and B is 1; otherwise gives false

Excel: 43 bytes


Try it online!

How to use:
Put this formula anywhere Except A1-A10.
Then input your Decant holding volumes in cells A1:A9 (because the number of decants is fixed) and the goal in A10. cells without decants should be left blank. Wherever you put the formula will contain the result. TRUE if you can achieve the goal, FALSE if you cannot.


JavaScript (ES6), 58 bytes


Another port of @xnor's answer. Yes, I get to use reduce again!

  • 3
    \$\begingroup\$ Alternate sub-function: e=>n<=e is a visual palindrome ;) \$\endgroup\$ Commented Sep 22, 2016 at 19:50

Retina, 39 bytes


Input should be a comma-separated list of the decanters, followed by a semicolon, followed by the target volume. E.g.:


Output is 0 (falsy) or 1 (truthy).

Try it online! (The first line enables a linefeed-separated test suite.)



This just converts the input to unary. Afterwards we simply match valid inputs with a single regex:


The part inside (?>...) finds the GCD. We do this by finding the largest substring 1+ with which we can match all the decanters (allowing an optional , only after a full match of the GCD). The atomic group (the (?>...)) itself so that the regex engine doesn't backtrack to divisors of the GCD if the target volume can't be matched (otherwise group 1 will at some point be reduced to matching a single 1 and all inputs will be truthy).

Once we've found the GCD, we try to match the target volume as a multiple of it with a simple (\1+)$.

Finally, we check that the target volume not greater than the largest decanter's capacity, by ensuring that the volume can be matched inside any decanter with (?<=\3.+).


Ruby, 35 bytes


PARI/GP, 31 bytes

Pretty much straightforward. Checking the max (vecmax) is very costly, I wonder if it can be done better.


Perl, 47 bytes

Includes +2 for -ap

Run with the jar sizes on the first line of STDIN and the target jar on the second line:

decanter.pl; echo
2 5 12


#!/usr/bin/perl -p

This solution is unusual in that it processes the input line by line and outputs something for each of them. The output for the first line was carefully designed to be empty while the second line prints the solution. Two bytes are lost on () because < and > are designed to be non-associative in perl.

The regex solution is also nice but 49 bytes:

#!/usr/bin/perl -p
s/\d+/1x$&/eg;$_=/^(?>(1+)( |\1)*:)(\1+)$/&/$3./

(some parts stolen from the Retina solution)

For this one give input on STDIN as jars separated by spaces and target after a ::

decanter.pl <<< "2 5 12:1"

Hard to beat languages with a builtin gcd (21 bytes) and max (7 bytes) for this one...


Scala, 90 53 bytes

def h(g:Int,a:BigInt*)=a.max>g&&a.reduce(_ gcd _)%g<1

Works basically the same as the other answers, but scala doesn't have a built-in gcd function. Scala has a built.in gcd function, but only for BigInt.


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