This is in the spirit of the What is a Word/Phrase™ series started by JLee with a special brand of Phrase™ and Word™ puzzles.

If a word has a certain property, I call it a Shifting Word™.

You can use the examples below to find the property:

Shifting™ Not Shifting™
ace jack
army navy
commute common
effort stress
egg apple
grumpy sad
influx income
Knox fort
moth bug
obey allow
solo alone
wonky wobbly
wrens crows
yolk white

Here's a CSV version:


Can you tell me the rule?

Hint 1:

Most of the Shifting Words™ are short. In fact, no English word (even an invented one!) may contain more than 7 letters. However, other languages like French, Russian, Thai etc. may possibly have longer words.

  • 1
    $\begingroup$ Is the capitalization of Knox important, other than just because it's a proper noun? $\endgroup$ – Lukas Rotter Jan 13 at 13:35
  • $\begingroup$ @LukasRotter In this case it doesn't matter. $\endgroup$ – trolley813 Jan 13 at 16:25
  • $\begingroup$ The only way (direction) I've found that makes sense is rot13(fbzrguvat gb qb jvgu NFPVV ovg rapbqvat, orpnhfr nyy Ratyvfu yrggref unir bayl 7 ovgf, jurernf Serapu, Ehffvna naq Gunv nyy unir nqqvgvbany yrggref gung qba'g snyy jvguva gur 7-ovg NFPVV pbqr enatr). But I'm stuck after this point. I'm sitting duck waiting for Hint 2 now. $\endgroup$ – iBug Jan 15 at 19:28
  • $\begingroup$ rot13(Nqqvat gb vOht'f pbzzrag, fuvsgvat jbeqf nyy fgneg jvgu na bqq punenpgre (v.r. pune -> NFPVV % 2 == 1). Hafuvsgvat jbeqf pbhyq fgneg jvgu rvgure na bqq be rira punenpgre. V gubhtug guvf jnf na vaqvpngvba gung gur jubyr jbeq unf gb or pbairegrq gb n fvatyr NFPVV yrggre, naq gur 1fg yrggre qrgrezvarf gur 6gu (0-vaqrkrq) ovg bs vg, 2aq -> 5gu, rgp..., fvapr gur 6gu ovg zhfg arprffnevyl or frg va beqre gb unir gur cbgragvny gb perngr nal ratyvfu yrggre jvguva NFPVV. Unq ab yhpx svaqvat n uneq ehyr, gubhtu, rvgure) $\endgroup$ – Lukas Rotter Jan 16 at 11:39

The rule for a Shifting word is

When converting the word to binary (with ASCII), the ith bit of each letter is set, where i is the current 0-based letter index within the word. So for the first letter, the 0th bit must be set. For the second, the 1st, and so on. Note that bit #0 is the rightmost bit.

Another way of thinking about this is: Write out the letters in binary, one letter each line. Then draw a diagonal starting from the top right down to the bottom left. If the diagonal only intersects with 1's, it's a shifting word. Otherwise it isn't

enter image description here

The reason they're called shifting words is because

The operation described above can also be done via ascending bit-wise right shifts, and then checking if the 0th bit of the shifted value is 1 (or if the number itself is even/odd). E.g. 1101 1011 1001 1111 would be unshifting, because (1101 >> 0) % 2 = 1, (1011 >> 1) % 2 = 1, (1001 >> 2) % 2 = 0, (1111 >> 3) % 2 = 1

Some remarks

Just by looking at the words, we can see that the first letter of each shifting word is (and must be) an "odd" letter, meaning it has an odd index within ASCII (or the alphabet, for that matter). The 0th bit of an odd number is always 1 by definition, because (anOddNumber mod 2 = 1) Unshifting words can either have an odd or even first letter. Furthermore, as noted in hint #1, there is no shifting english string of characters with a length greater than 7. This is because all english characters in ASCII only have 7 bits (meaning the imagined 7th bit would always be 0)

  • $\begingroup$ Great! You found it. $\endgroup$ – trolley813 Jan 16 at 16:13

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