Let's substitute each letter of a given word by the value of its ordinal count in the ABC: A=>1, B==>2 .. Z==>26.
Evaluate the total for that word and call it f(k), k being the number of letters in the word.
Let MW(K) be a word for which f(k) is minimal.
Example: assuming the word CAB generates the lowest f(3) then MW(3) is CAB and its f(3)=6.
Question: What triplet of common English words will generate the lowest
f(7)+f(9)+ f(11)?
(In reply to
computer solution by Charlie)
C+A+B+B+A+G+E=3+1+2+2+1+7+5=21
A+B+A+N+D+O+N+E+D=1+2+1+14+4+15+14+5+4=60
A+B+B+R+E+V+I+A+T+E+D=1+2+2+18+5+22+9+1+20+5+4=89
B+E+D+A+B+B+L+E+D=2+5+4+1+2+2+12+5+4=37
A+B+R+A+C+A+D+A+B+R+A=1+2+18+1+3+1+4+1+2+18+1=52
Therefore, cabbage, abandoned, abbreviated adds up to 21+60+89=170. The best solution is cabbage, bedabbled, abracadabra, which adds up to 21+37+52=110.
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Posted by Math Man
on 2017-10-22 09:19:05 |
re: computer solution
