The respective number of 'e's and 'a's in any given word is denoted by ε and α.
o (1)(a) Determine the longest common English word/words having the maximum value of ε/α
o (1)(b) Determine the longest common English word/words having the minimum value of ε/α
o (1)(c) Determine the longest common English word/words where ε=α
o (2) Extend (1) to cover the case of esoteric/archaic words.
Notes:
- The valid word/words must have at least one 'e' and at least one 'a'.
- No abbreviations/acronyms/slangs and no hyphenated words like A-Bomb, X-Ray etc.
clearvars,clc
fid=fopen('c:\words\words.txt');
longAval=0; longEval=0; longEqval=0;
longAset={}; longEset={}; longEqset={};
while ~feof(fid)
l=fgetl(fid);
alpha=length(strfind(l,'e'));
eps=length(strfind(l,'a'));
if alpha>0 && eps>0 && isequal(l,lower(l))
ratioA=alpha/eps;
ratioE=eps/alpha;
if ratioA>longAval
longAval=ratioA;
longAset={l};
else
if ratioA==longAval
longAset(length(longAset)+1)={l};
end
end
if ratioE>longEval
longEval=ratioE;
longEset={l};
else
if ratioE==longEval
longEset(length(longEset)+1)={l};
end
end
if alpha==eps
if length(l)>longEqval
longEqval=length(l);
longEqset={l};
else
if length(l)==longEqval
longEqset(length(longEqset)+1)={l};
end
end
end
end
end
fclose(fid);
disp(longAval)
disp(longAset')
disp(' ')
disp(longEval)
disp(longEset')
disp(' ')
disp(longEqval)
disp(longEqset')
finds
6 maximum alpha/eps
{'degeneratenesses' }
{'levelheadednesses' }
{'regeneratenesses' }
{'representativenesses' }
{'tenderheartednesses' }
{'unrepresentativenesses'} longest
4 (i.e., 1/4) maximum eps/alpha (min alpha/eps)
{'acanthocephalan' }
{'acanthocephalans' }
{'adenocarcinomata' }
{'agammaglobulinemia' }
{'agammaglobulinemias' }
{'amalgamate' }
{'amalgamated' }
{'amalgamates' }
{'anabaena' }
{'anabaenas' }
{'anagrammatize' }
{'anagrammatized' }
{'anagrammatizes' }
{'analemmata' }
{'anathemata' }
{'antiegalitarian' }
{'armamentaria' }
{'caravansaries' }
{'caravanserai' }
{'caravanserais' }
{'extravaganza' }
{'extravaganzas' }
{'megalomaniacal' }
{'megalomaniacally' }
{'paralanguage' }
{'paralanguages' }
{'paramagnetically' }
{'paraphernalia' }
{'paraphrasable' }
{'parliamentarian' }
{'parliamentarians' }
{'radioimmunoassayable'} tied for longest
{'radiopharmaceutical' }
{'radiopharmaceuticals'} tied for longest
{'razzamatazzes' }
{'teratocarcinomata' }
{'thalassaemia' }
{'thalassaemias' }
22 alpha = eps
{'indistinguishabilities'} tied for longest
{'overcommercializations'} tied for longest
>>
The ratios of 6 and 4 (i.e., 1/4, as specified in the problem) respectively head sets 1 and 2 and the length 22 heads the last (equality) list.
For more common words we'll try 80kwords.txt instead of words.txt:
6
{'levelheadednesses' }
{'tenderheartednesses'}
4
{'amalgamate' }
{'amalgamated' }
{'amalgamates' }
{'caravansaries' }
{'caravanserai' }
{'caravanserais' }
{'extravaganza' }
{'extravaganzas' }
{'paraphernalia' }
{'paraphernalias' }
{'parliamentarian' }
{'parliamentarians'}
21
{'straightforwardnesses'}
{'unconstitutionalities'}
or, trying the 36Kwords file:
4
{'bereavement' }
{'bereavement's' }
{'bereavements' }
{'degenerate's' }
{'degenerated' }
{'degenerates' }
{'perseverance' }
{'perseverance's' }
{'racketeered' }
{'redeemable' }
{'regenerate' }
{'regenerated' }
{'regenerates' }
{'reverberate' }
{'reverberated' }
{'reverberates' }
{'unrepresentative'}
4
{'amalgamate' }
{'amalgamated' }
{'amalgamates' }
{'paraphernalia' }
{'paraphernalia's'}
18
{'oversimplification'}
I think the winners in this case are
unrepresentative (ratio 4:1)
but unrepresentativeness (5:1) would actually be a contender, if considered common enough.
paraphernalia (ratio 1:4)
oversimplification (18 letter word for ratio=1:1)
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Posted by Charlie
on 2022-09-17 12:45:06 |
computer findings
