Note: When I refer to a bridge, I mean it in the non-mathematical sense


You are on a network of islands which are connected by wooden bridges and you want to see if you can burn every bridge in the island network. However, you can only burn a bridge once you've walked over it.

Once a bridge has been burned, it has gone and you cannot go back over it. None of the islands are connected.

Since you are going on a stroll, you cannot swim.


Given a list of the bridges and which islands they connect, output a truthy value if you can burn every bridge and falsey value if you cannot.


The input will be a list of numbers. A bridge is defined as:


Where are X and Y are integers. Each bridge has an integer number which X and Y refer to.

For example,


Means that there is a bridge between the islands 2 and 6.

The input will be a list of these bridges:

2,6 8,1 8,2 6,1

You may take this a flat array, if you wish.


The numbers for the islands will always be positive integers, n, where 0 < n < 10. The island numbers will not always be contiguous.

When you walk over a bridge, you must burn it.

You will be given a maximum of eight islands. There is no limit on the number of bridges.

Island networks will always connect together. There will always be exactly one connected component.

This is an undirected network. For example, you can go either way across the bridge 1,2.


Input > Output
1,2 1,3 1,3 1,4 1,4 2,3 2,4 > Falsey
1,5 1,6 2,3 2,6 3,5 5,6 > Truthy
1,2 1,3 1,4 2,3 2,4 3,4 > Falsey
1,2 > Truthy


Shortest code in bytes wins.

  • \$\begingroup\$ Can we assume that there will not be more than one bridge between two islands (sorry, didn't notice this during review)? \$\endgroup\$
    – hyper-neutrino
    Jun 26, 2017 at 14:31
  • \$\begingroup\$ this is an undirected graph, yes? So I can walk on a (1,2) bridge from 2 to 1 \$\endgroup\$ Jun 26, 2017 at 14:33
  • \$\begingroup\$ @HyperNeutrino No, see the first example where there are two bridges between 1 and 2 and two bridges between 1 and 3 \$\endgroup\$
    – Beta Decay
    Jun 26, 2017 at 14:36
  • 1
    \$\begingroup\$ So you're just asking us to check if the graph has an Eulerian path? \$\endgroup\$
    – Not a tree
    Jun 26, 2017 at 14:42
  • 1
    \$\begingroup\$ Should "at least one connected component" say "exactly one connected component"? \$\endgroup\$ Jun 26, 2017 at 16:30

4 Answers 4


Mathematica, 27 26 bytes

Assuming that it is a connected graph.


Takes a flat list as input.


In[1]:= Tr@Mod[Last/@Tally@#,2]<3&[{1, 2, 1, 3, 1, 3, 1, 4, 1, 4, 2, 3, 2, 4}]

Out[1]= False
  • \$\begingroup\$ I'm surprised that there's no builtin for this! \$\endgroup\$ Jun 26, 2017 at 14:55
  • \$\begingroup\$ ...aand there's the built-in :D \$\endgroup\$
    – Beta Decay
    Jun 26, 2017 at 15:02
  • \$\begingroup\$ Is the FindEulerianCycle part of that answer useful? \$\endgroup\$
    – nmjcman101
    Jun 26, 2017 at 19:29

Python 2, 102 92 43 bytes (40 in Python 3)

Thanks to @Neil for reminding me that you can't have an odd number of vertices with odd degree – and saving 5 bytes.

lambda x:sum(x.count(a)%2for a in set(x))<3

Try it online!

You can cut it down to 40 bytes in Python 3 (thanks to @shooqie):

lambda x:sum(x.count(a)%2for a in{*x})<3

A graph has a Eulerian trail iff the number of vertices with odd degree is 0 or 2.

This takes the input as a flat list, as explicitly allowed by the spec.

This assumes all of the islands are connected.

  • \$\begingroup\$ Yup, BetaDecay says you must burn a bridge after walking over it, so I'll delete my comments. \$\endgroup\$
    – hyper-neutrino
    Jun 26, 2017 at 14:52
  • \$\begingroup\$ @Rod Oh yeah, true. :P \$\endgroup\$
    – hyper-neutrino
    Jun 26, 2017 at 14:52
  • \$\begingroup\$ lambda x:sum(x.count(a)%2for a in{*x})in[0,2] - you need Python 3 though \$\endgroup\$
    – shooqie
    Jun 26, 2017 at 15:43
  • \$\begingroup\$ Wow, how does that work? \$\endgroup\$
    – Beta Decay
    Jun 26, 2017 at 15:51
  • \$\begingroup\$ @BetaDecay *x "explodes" x into x1,x2,x3..., so you can pass it straight into {} which is essentially syntactic sugar for the set constructor. \$\endgroup\$ Jun 26, 2017 at 15:53

Jelly, 9 bytes


Try it online!

Port of rogaos's Python answer. If you upvote this answer, please upvote that one too. :)


Pyth, 14 11 10 bytes


Try it online!

My first Pyth program! I'm sure it could be golfed a great deal.

Thanks to @Rod for saving 1 byte by using Q implicitly.

Same general idea as my Python answer.

  • 1
    \$\begingroup\$ I guess you could reverse the check, to use the last Q implicitly -> >3sm%/Qd2{ \$\endgroup\$
    – Rod
    Jun 26, 2017 at 18:49

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