20
$\begingroup$

I have devised a rule that assigns a positive integer to each English word. Here is a list of words and the integers the rule assigns them.

$$ \begin{array}{l|l} \text{Word}&\text{Integer}\\\hline \text{atmosphere}&3\\ \text{day}&2\\ \text{eclipse}&7\\ \text{geology}&4\\ \text{heavy}&3\\ \text{inflation}&1\\ \text{neanderthal}&3\\ \text{parsimoniously}&4\\ \text{span}&1\\ \text{zealous}&8\\ &&\\ &&\\ \end{array} \hspace{1cm} \begin{array}{l|l} \text{Word}&\text{Integer}\\\hline \text{a}&1\\ \text{beret}&1\\ \text{cheaters}&6\\ \text{coins}&2\\ \text{dad}&5\\ \text{fad}&2\\ \text{hectares}&6\\ \text{highways}&1\\ \text{I}&5\\ \text{laziness}&2\\ \text{lid}&1\\ \text{transportation}&2\\ \end{array}\hspace{1cm} \begin{array}{l|l} \text{Word}&\text{Integer}\\\hline \text{ab}&3\\ \text{ad}&1\\ \text{bag}&1\\ \text{banana}&5\\ \text{bann}&3\\ \text{hear}&4\\ \text{her}&3\\ \text{males}&1\\ \text{prevent}&1\\ \text{snap}&1\\ \text{thorny}&1\\ &\\ \end{array} $$

The rule can be described in a single sentence.

What is my rule?

Edit: I have chosen some additional words and computed the numbers the rule assigns them. Also, find some hints below.

Hint 1:

It is unlikely you will find a word in the dictionary that is assigned a value larger than $10$. However, the rule is defined for arbitrary strings of English letters, and it is possible write down strings with arbitrarily large output. To get an output of $11$ requires a string of at least 16 characters, and the 16 character string with output $11$ that comes first alphabetically is $\text{'dzzzzzzzzzzzzzzz'}$.

Hint 2:

It might be useful to start by figuring out what the words assigned $1$ have in common.

Hint 3:

The US coins are: penny (1¢), nickel (5¢), dime (10¢), quarter (25¢), half dollar (50¢), and dollar (100¢). This is relevant.

Hint 4:

The rule assigns anagrams the same number.

Hint 5:

The rule does not depend on any of the following:
Case, font, keyboard layout, letter frequency, word meaning

Hint 6:

The longest string that the rule assigns $1$ has one hundred letters.

$\endgroup$
  • 1
    $\begingroup$ The meaning of the words does not matter. The output is always strictly positive, never zero. $\endgroup$ – Julian Rosen May 29 '15 at 12:59
  • 1
    $\begingroup$ I got nowhere using scrabble scoring $\endgroup$ – Raystafarian May 29 '15 at 17:09
  • 1
    $\begingroup$ @TroyAndAbed No, each word determines one (and only one) value $\endgroup$ – Julian Rosen Jun 2 '15 at 15:16
  • 1
    $\begingroup$ The "I" (worth 5) is capitalized and in a serif font. Would it be 5 if it was lowercase? $\endgroup$ – Raystafarian Jun 2 '15 at 18:07
  • 2
    $\begingroup$ @Raystafarian Yes, I should have clarified: the rule does not depend on case or font $\endgroup$ – Julian Rosen Jun 3 '15 at 1:02
14
$\begingroup$

I have found the rule.

Give each letter a value. A = 1, B = 2, C = 3, and so on. Then add all the values. I have taken "snap" as an example. "snap" = 19 + 14 + 1 + 16 = 50. Then find what is the least number of coins needed to make the total, or in this case, 50 cents. You only need one coin for a total of 50 cents. Therefore, 1 is the correct answer. As listed in a hint, the US coins are used: penny (1¢), nickel (5¢), dime (10¢), quarter (25¢), half dollar (50¢), and dollar (100¢).

Another example:

cab = 3 + 1 + 2 = 6 cents = 5 cent coin + 1 cent coin = 2 coins

$\endgroup$
  • 1
    $\begingroup$ If you guys want to try this method out, use this code (JavaScript): function SOLVE(v){x="abcdefghijklmnopqrstuvwxyz".split("");v=v.split("");for(i=0,a=0;i<v.length;i++)a+=x.indexOf(v[i])+1;return c(a);}function c(a,f=Math.floor){S=f(a/100);a-=S*100;T=f(a/50);a-=T*50;U=f(a/25);a-=U*25;V=f(a/10);a-=V*10;W=f(a/5);a-=W*5;S+T+U+V+W+a;} and enter SOLVE("<your word>"). This code disregards any non-lowercase, alphabetical, English letters. $\endgroup$ – Conor O'Brien Jun 7 '15 at 23:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.