This is a "counterpart" of another puzzle, Eight coins for the fair king on Puzzling.SE.

You can read the above puzzle for the background. The details about this puzzle are as follows.

A set of 8 kinds of coins of different values are created, the king wants you to find out the maximum N such that any number of price from 0 to N can be paid with a combination no more than 8 coins and without charges.

For example, (taken from Glorfindel's answer). If a set of coins of values 1, 2, 5, 13, 34, 89, 233, 610 are given, your program should output 1596, because every number between 0 and 1596 (inclusive) can be represented by the sum of no more than 8 numbers from the given list (numbers may repeat), while 1597 cannot be represented in that way.

In a mathematical way, if the input is a set S consisting of 8 positive integers, the desired output N satisfies that for any number n between 0 and N, there exists x1, x2, x3, ..., x8 such that

$$x_1 + x_2 + ... + x_8 = n \quad\textrm{and}\quad x_1, x_2, ...,x_8 \in \{0\} \cup S$$

Your goal is to write a program, a function, or a snippet that takes 8 numbers as input, and output the maximum N as described above.


  • Flexible I/O allowed, so your program can take the input in any form that's best suitable. You may assume that the input numbers are sorted in the way that best suits your program.
    • Please state it in your answer if your program depends on input order
  • Input is a set of 8 different, positive integers (no zeros). The output is one non-negative integer.
    • In case there's no 1 in the input set, your program should output 0 because any number from 0 to 0 satisfies the requirement.
    • In the case of invalid input (the set contains zero, negative or duplicate numbers), your program can do anything.
  • Standard loopholes are forbidden.
  • Your program should run within a few minutes on a modern computer.

Test cases (mostly taken from the answers under the linked question on Puzzling):

[1, 2, 3, 4, 5, 6, 7, 8] => 64
[2, 3, 4, 5, 6, 7, 8, 9] => 0
[1, 3, 4, 5, 6, 7, 8, 9] => 72
[1, 2, 5, 13, 34, 89, 233, 610] => 1596
[1, 5, 16, 51, 130, 332, 471, 1082] => 2721
[1, 6, 20, 75, 175, 474, 756, 785] => 3356

This is a , so the shortest program or snippet in each language wins!

  • 1
    \$\begingroup\$ Nice puzzle, but I personally think that some more test cases would be helpful in order to test our submissions. \$\endgroup\$ – Mr. Xcoder Dec 25 '18 at 11:42
  • \$\begingroup\$ Wouldn't it be better to make input size a parameter? Brute force approaches will struggle with 8 \$\endgroup\$ – Luis Mendo Dec 25 '18 at 11:43
  • 1
    \$\begingroup\$ @iBug Then the usual rule is something like "submissions shoud run within a minute in a modern computer". It's fuzzy, but usually good enough, because the difference between brute force and efficient approaches is very large \$\endgroup\$ – Luis Mendo Dec 25 '18 at 11:48
  • 1
    \$\begingroup\$ Brute force is still possible with your time limit of "a few minutes". A slightly modified version of my answer runs the last test case in 1m20s on my 7 year old laptop. \$\endgroup\$ – nimi Dec 25 '18 at 12:01
  • 1
    \$\begingroup\$ @Arnauld Clarified \$\endgroup\$ – iBug Dec 26 '18 at 1:34

13 Answers 13


Python 3, 113 62 bytes

for i in[1]*3:x|={a+b for a in x for b in x}

Here x is the input as a set of ints, and i is the output.

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(Thanks: Erik the Outgolfer, Mr. Xcoder, Lynn)

  • 1
    \$\begingroup\$ 96 bytes. \$\endgroup\$ – Erik the Outgolfer Dec 25 '18 at 20:16
  • \$\begingroup\$ x=0,*x saves 1 byte. Better yet, x+=0, saves 2. \$\endgroup\$ – Mr. Xcoder Dec 25 '18 at 23:21
  • \$\begingroup\$ 78 bytes in Python 2. \$\endgroup\$ – Lynn Dec 28 '18 at 15:04

Jelly, 12 bytes


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Takes on average ~3.7 seconds to run all test cases on TIO on my phone, so surprisingly it is quite fast.


œċⱮ8Ẏ§ṢQJƑƤS     Monadic link / Full program.
  Ɱ8             Promote 8 to [1 ... 8] and for each value k:
œċ                    Generate all combinations of k elements from the list.
    Ẏ§           Tighten, then sum. Flatten to a 2D list then sum each.
      ṢQ         Sort the result and remove equal entries.
        JƑƤ      For each prefix of this list, return 1 if it is equal to its length range, 0 otherwise.
           S     Finally, sum the result (counts the 1's which is equivalent to what is being asked).

Haskell, 56 50 bytes

g c=[x|x<-[1..],all((/=x).sum)$mapM(0:)$c<$c]!!0-1

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A brute force approach. Add 0 to the list of coins and try all combinations of 8 picks. Find the first number n that is not equal to the sum of any of the picks and return n-1.

Takes about 5m30s for [1, 2, 5, 13, 34, 89, 233, 610] on my 7 year old laptop hardware.

Edit: -6 bytes thanks to @Ørjan Johansen

An even shorter version (-2 bytes, again thanks to @Ørjan Johansen) is

Haskell, 48 bytes

g c=[x|x<-[1..],all((/=x).sum)$mapM(:0:c)c]!!0-1

but it uses significantly more memory and runs into heavy paging on my machine and does not finish "within a few minutes".

  • 1
    \$\begingroup\$ You can use mapM(0:)$c<$c. (In fact mapM(:0:c)c should work, but times out on TIO for the given test case.) \$\endgroup\$ – Ørjan Johansen Dec 25 '18 at 20:06

Jelly, 9 bytes


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How it works

Żœċ8§ḟ’$Ṃ  Main link. Argument: A (array)

Ż          Prepend a 0 to A.
 œċ8       Take all combinations of length 8, with repetitions.
    §      Take the sum of each combination.
       $   Combine the two links to the left into a monadic chain.
      ’      Decrement all sums.
     ḟ       Filterfalse; keep only sums that do not appear in the decremented sums.
        Ṃ  Take the minimum.
  • 2
    \$\begingroup\$ Żṗ8§ḟ’$Ṃ saves one byte, but I'm not sure if 8.5 minutes counts as a few. \$\endgroup\$ – Dennis Dec 26 '18 at 4:02

Wolfram Language (Mathematica), 53 bytes


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JavaScript (ES6),  100 88 80  76 bytes

This is essentially a brute-force search, but enhanced with pruning to speed it up. The average execution time for the test cases is close to 1 second on TIO.

Assumes that the input array is sorted from highest to lowest.


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a =>                      // a[] = input array
  [...Array(a[0] * 9)]    // create an array of 9 * max(a) entries
  .findIndex(             // find the position of the first truthy result
    g = (i = 8, s) =>     // g = recursive function taking a counter i, initialized to 8
                          //     and a sum s, initialized to the position in the above array
      s * i > 0 ?         //   if s is positive and i is not equal to 0:
        a.every(x =>      //     for each value x in a[]:
          g(i - 1, s - x) //       do a recursive call with i - 1 and s - x
        )                 //     end of every()
      :                   //   else:
        s                 //     yield s (s = 0 means success and makes findIndex go on)
  ) - 1                   // end of findIndex(); decrement the result

Python 2, 145 bytes

from itertools import*

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Pari/GP, 57 bytes


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  • \$\begingroup\$ is this using generating function? \$\endgroup\$ – don bright Jan 2 '19 at 0:03
  • 1
    \$\begingroup\$ @donbright Yes. \$\endgroup\$ – alephalpha Jan 2 '19 at 7:18
  • 1
    \$\begingroup\$ that is awesome.. one of the few answers not brute-forcing the solution. alot of languages probably dont have built in polynomial symbolic features. Pari GP is cool. \$\endgroup\$ – don bright Jan 2 '19 at 7:36

Python 2, 125 115 111 bytes

lambda c:sum(i==j for i,j in enumerate(sorted(set(map(sum,product([0]+c,repeat=8))))))-1
from itertools import*

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Expects a list of integers as input.


# an anonymous function
lambda c:
                                                          # get all length-8 combinations of values, from (0,0,0,0,0,0,0,0) to (8,8,8,8,8,8,8,8)
                                                          # zero is added to ensure that combinations of fewer than 8 coins are represented Ex:(1,0,0,0,0,0,0,0)
                                                  # for each combination, sum the values
                                       # get unique values, then sort them smallest to largest
             # for each index, value pair, return if the index is equal to the value
             i==j for i,j in enumerate(.............................................)
         # in Python arithmetic, False is 0 and True is 1. So, count how many items match their index.
         # Since zero was added to the list, there will always be one extra match (0==0). So offset by one.
from itertools import*

Perl6, 65 63 41 bytes (39 37 chars)

{@_=(0,|@_)X+(0,|@_)for ^3;($_ if $_==$++for @_.sort.unique)-1}

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This is an anonymous block that gets passed its data as an array. The (0,|@_) is a quick way to add a 0 to @_, and even though it's done twice, it's still a bit shorter than @_.push: 0; which would then need spaces after the _. This is a brute force approach that cheeses a bit on the fact that it's 8 combinations. After cross adding, an anonymous list is created for sequential values. With math operators, lists evaluate to their length, so the -1 pulls double duty: accounting for the 0 and coercing to Int.

This can take its sweet time, but by changing one or both (0,|@_) to (0,|@_.unique) before the first for it can be sped up considerably. That adds +7 (runtime <60s) or +14 (runtime <10s) to the score if you feel the first is too slow (I did this for the linked code to avoid timeouts after 60 seconds).

Edit: JoKing in the comments improved it (same idea, cross add, then return the last consecutive result) to an astonishing 39 chars (41 bytes):

{(@_=@_ X+0,|@_)xx 3;first *+1∉@_,^∞}

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The final tabulation doesn't need a 0, saving a few bytes by only needing to add 0 in once. The xx 3 mimics the for loop (still cheeses on the coins being a power of 2). The first sub returns the first number in the infinite list 0..* (^Inf is possible too, but doesn't save space) whose +1 is not a member of the cross added list. Like mine, it's slow, so add +7 for a unique after the first equals if you feel it's too slow for guidelines.

  • 1
    \$\begingroup\$ 48 bytes. Technically, the unique is not needed, but it speeds it up a lot \$\endgroup\$ – Jo King Dec 28 '18 at 7:14
  • \$\begingroup\$ @JoKing nice, I don't know why I didn't think about using xx. I knew there had to be a way to do the final tabulation in a much shorter way using set functions, but my brain wasn't working. \$\endgroup\$ – user0721090601 Dec 28 '18 at 15:17
  • \$\begingroup\$ The xx 1 should be xx 3 \$\endgroup\$ – Jo King Mar 29 '19 at 0:56
  • \$\begingroup\$ @JoKing fixed. Also I realized two chars (but no bytes) can be saved by using ^∞ \$\endgroup\$ – user0721090601 Mar 29 '19 at 1:01
  • \$\begingroup\$ Actually, you can save some bytes with (1...*∉@_)-1 instead of using first, (which I realise is the same method I used here) \$\endgroup\$ – Jo King Mar 29 '19 at 1:05

JavaScript (Node.js), 171 145 115 bytes

f=(s,n=3)=>n?f(s=new Set(a=[0,...s]),n-1,a.map(m=>a.map(n=>s.add(m+n)))):Math.min(...[...s].filter(m=>!s.has(m+1)))

Try it online! Port of @Mark's Python 3 answer. 108 bytes in Firefox 30-57:

f=(s,n=3)=>n?f(new Set((for(n of s=[0,...s])for(m of s)n+m)),n-1):Math.min(...[...s].filter(m=>!s.has(m+1)))

Wolfram Language (Mathematica), 46 bytes


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Brute force approach: checks integers counting upward until it reaches a value that can't be paid for in 8 coins. Very, very slow (tio times out), but I'm fairly sure the condition is correct.


Clean, 161 bytes

import StdEnv,Data.List
$l=:[1:_]#k=sort(nub(map sum(iter 8(concatMap(\[h:t]=[[e,h:t]\\e<-[0:l]|e>=h]))[[0]])))
=length(takeWhile((>=)1)(zipWith(-)(tl k)k))

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