In the image you can see twelve mutation chambers (A to L). Each alien on the left side of the aliens goes through the three mutation chambers of its right side. The alien on the right side is the result after the three mutations. So, the alien left of chamber A goes through A, B and C. The result is the one right of C.

Same goes for the aliens op top. Every alien on the top of the image goes through the four mutation chambers below it. The one above A goes through A, D, G and J. The alien below chamber J is the result.

Every chamber applies exactly one mutation (make a head square, ...).

These mutations can occur:

  • Add antennas
  • Remove antennas
  • Add arms
  • Remove arms
  • Add legs
  • Remove legs
  • Make square head
  • Make round head
  • Make square body
  • Make round body
  • Walk upright
  • Walk on 4 legs (2 arms and 2 legs)

Each of these mutations can occur as many times as you want. It is not said that every mutation described above will occur for sure. This means that multiple chambers can remove leggs and that adding arms is done nowhere (e.g.).

enter image description here

Can you tell me which room provides which mutation? If you can tell me, tell me in spoiler tags!

Source: Raadselweb.net (translated from Dutch to English by me)

  • 2
    $\begingroup$ Nice puzzle and +1 for the graphics. Erik was way too fast for me, but I'll puzzle it out myself without his spoilers... ;c) $\endgroup$
    – BmyGuest
    Dec 5, 2014 at 14:42
  • 1
    $\begingroup$ I'm hopping on the "nice puzzle" wagon here. Very fun to work out, nice original puzzle :-) Would puzzle again $\endgroup$
    – Joe
    Dec 5, 2014 at 14:48
  • $\begingroup$ Thanks for your kind words! I'll keep them coming on a regular basis :). Admitting, I didn't create it myself, I found it somewhere and thought it was a nice puzzle :). $\endgroup$ Dec 5, 2014 at 14:55
  • $\begingroup$ It might be worth taking a quick look at a policy on plagiarism just to make sure you're not falling foul of it :-) $\endgroup$
    – Joe
    Dec 5, 2014 at 15:04
  • 1
    $\begingroup$ @HemantAgarwal at the time I wrote the comment (6 years ago... shudder) it was, but the author quickly edited it to include the source. It's currently fine, to the best of my knowledge $\endgroup$
    – Joe
    Apr 1, 2021 at 7:50

3 Answers 3


Think I got it...

A = make head square
B = add legs
C = make body square
D = remove arms
E = add antennae
F = walk upright
G = walk on all fours
H = make head round
I = remove antennae
J = add arms
K = walk upright
L = make head square

  • 5
    $\begingroup$ To use line breaks inside spoiler tags, use <br>. $\endgroup$ Dec 5, 2014 at 14:16
  • 1
    $\begingroup$ Thanks. I was wondering why I couldn't get it look properly. $\endgroup$
    – Erik
    Dec 5, 2014 at 14:16
  • 2
    $\begingroup$ I think you may have 'H' and 'K' backwards. $\endgroup$
    – YoungJohn
    Dec 5, 2014 at 15:44
  • $\begingroup$ You were the first one, congratz! You can swap H and K, both are correct! $\endgroup$ Dec 7, 2014 at 23:17
  • $\begingroup$ This could be improved by adding an explanation of how you reached your answer. Notice a later answer has a score three times higher, for this exact reason. $\endgroup$ Dec 8, 2014 at 6:44

As I haven't been the first to present a solution, I thought I might at least provide a detailed deduction route here. ( I did not look at the solution first. )

To solve this puzzle, I first make a reduction of information, moving form the (nice!) pictures to pure information. I use the following abbreviations:

+ added property
- removed property
0 unchanged property

A Antennae 
H Hands (arms)
L Legs
S Square Head
B Square Body
U Upright Position

By simply comparing left & right as well as top & bottom one gets:

enter image description here

The next step is to count which mutations actually appear, giving us:

Once of each: +A -A +H -H +L -S +B -U
Twice of each: +S +U

Note that twice means that the same mutation appears twice in either the rows or twice in the columns.
Summing those mutations, we get 12, which is good, as we have 12 cells to fill.

Now lets start assigning mutations to cells:

Start with the modifications which appear once in total, and also appear once in either the rows or columns. Their cross-section has to be the mutation cell:
enter image description here
By this, we have not yet assigned the single: +H -H -S -U

Now we look at the obvious double-counted mutations:

+S +U appear each twice but we see already that for the first row only a single cell is left over for +S. For the second row a net +U can only be achieved if there is only a positive and no negative mutation in that row. This immediately gives:
enter image description here

Next we can expand the logic:

In the third column 4 mutations have to happen but only one cell isn't yet assigned. It has to be +S. But now the last row needs to be neutral in S, which we can only achieve by a -S in the same row. This has to be in the 2nd column, because the net mutation for columns is +S in the first and -S in the second column, respectively. This gives us:
enter image description here

We can use the same arguments again for the upright (U) property:

In the third row there has to be +U and -U, and there is only one possibility to fit this:
enter image description here

And now there is only one possibility to fill in the rest for the final solution:

enter image description here

  • $\begingroup$ I believe this is a much better answer as things stand right now. Excellent explanation $\endgroup$
    – Joe
    Dec 5, 2014 at 17:09

A [head-square]
B [add-legs]
C [body-square]
D [remove arms]
E [add-ant]
F [walk-up]
G [walk-down]
H [walk-up]
I [remove-ant]
J [add-arms]
K [head-round]
L [head-square]

I noticed that my answer is different from BmyGuest; maybe there are multiple solutions?

ed: comparing the two answers:

BmyGuest     me 

+S +L +B   +S +L +B  

+A +A +U   -H +A +U  

-U +U -A   -U +U -A  

+H -S +S   +H -S +S  

Actually, I think that D must be -H (remove arms/hands)

I was much less methodical.

I found A, B, and C by process of elimination. Then I listed all the possiblities for E, F, H, I, K, and L. Then I was able to find E and I, followed by the rest through guessing and checking.

  • $\begingroup$ You're of course right, I mistyped the last entry. (The logic stays the same, but my typing was wrong. There mustn't be two A+. I lol edit, but thanks again. $\endgroup$
    – BmyGuest
    Dec 5, 2014 at 19:46

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