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You are probably familiar with Word Ladders, where you attempt to transform one word into another by switching one letter at a time, with each transformation forming a new word. This puzzle is similar. The good news is that you don't have to worry about whether your intermediate strings are words. The bad news: you can only change one atomic symbol (two-letter symbols must be contiguous, in order left-to-right) into another.

The true alchemist remains frustrated, due to an inability to turn lead into anything else, but many other transformations are possible. For example, we can change the original name for element 71, LUTECIUM, into its current name with just two transmutations:

LUTECIUM (transmute CARBON to NITROGEN)
LUTENIUM (transmute NICKEL to TITANIUM)
LUTETIUM

It's pretty easy to see you cannot do it with one transmutation, so this is the shortest such chain. (Note: it is not unique. We could have done C to S, SI to TI as well; BORON/BISMUTH works too.)

Can you make the following transformations using atomic symbol transmutations?

  • WOLFRAM to TUNGSTEN
  • EKASILICON to GERMANIUM
  • KURCHATOVIUM to RUTHERFORDIUM

As benchmarks, I have solutions of 8, 9 and 10 transmutations respectively. Can you match or beat these?

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Here's the first one with

7 transmutations

 WOLFR AM -> WOLFR N
 WOL FR N -> WOL TE N
 W O LTEN -> W GA LTEN
 WG AL TEN -> WG S TEN
 W GSTEN -> TS GSTEN
 T S GSTEN -> T CN GSTEN
 T C NGSTEN -> T U NGSTEN
 

Here's the second one with

8 transmutations

 E K ASILICON -> E SR ASILICON
 ES RASILICON -> GE RASILICON
 GER AS ILICON -> GER MN ILICON
 GERM NI LICON -> GERM AC LICON
 GERMA CL ICON -> GERMA N ICON
 GERMANI CO N -> GERMANI C N
 GERMANIC N -> GERMANIC AM
 GERMANI CA M -> GERMANI U M
 

Here's the third one with

9 transmutations

 K URCHATOVIUM -> RU URCHATOVIUM
 RU U RCHATOVIUM -> RU TH RCHATOVIUM
 RUTHRCHATO V IUM -> RUTHRCHATO GD IUM
 RUTHRCHAT OG DIUM -> RUTHRCHAT ER DIUM
 RUTHRCHA TE RDIUM -> RUTHRCHA CO RDIUM
 RUTHRCH AC ORDIUM -> RUTHRCH F ORDIUM
 RUTHR C HFORDIUM -> RUTHR B HFORDIUM
 RUTH RB HFORDIUM -> RUTH ER HFORDIUM
 RUTHER HF ORDIUM -> RUTHER F ORDIUM
 

No idea if these are optimal though

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  • 4
    $\begingroup$ You beat all of mine, so you get the checkmark! We'll see if other better solutions come in. $\endgroup$ – Jeremy Dover Jul 17 at 14:53

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