Ruby Golf on Lazy T9 Words.

[John Carter]

Assuming you are running Linux and have a list of valid English words in
the file /usr/share/dict/words or someplace or something similar.

Assume you have a cellphone with T9 predictive text input.

For those that don't, cellphones have each key overloaded to mean several
things. eg. On my Sharp GX-15
2 also means A,B,C
3 also means D,E,F
4 also means G,H,I
5 also means J,K,L
6 = M,N,O
7 = P,Q,R,S
8 = T,U,V
9 = W,X,Y,Z

In T9 mode I can type in 43556 and it only finds one word in its
dictionary that can be made with combination
{G,H,I}{D,E,F}{J,K,L}{J,K,L}{M,N,O} and that is
HELLO

Which is a lost faster than typing 4433555555666.

How ever, some words like "high", can be entered by just bouncing on one
key ... 4444.

Your mission, should you choose to accept, is to write the shortest ruby
program that will find the longest word in the dictionary that can be
produced by...
* repeatedly typing one key
* Can be produced by just using only two keys.
* Can be produced by just using only three keys.
* Can be produced by just using only four keys.
* ....


John Carter Phone : (64)(3) 358 6639
Tait Electronics Fax : (64)(3) 359 4632
PO Box 1645 Christchurch Email : (e-mail address removed)
New Zealand

Carter's Clarification of Murphy's Law.

"Things only ever go right so that they may go more spectacularly wrong later."

From this principle, all of life and physics may be deduced.

 

[wannes]

Assuming you are running Linux and have a list of valid English words in
the file /usr/share/dict/words or someplace or something similar.
Assume you have a cellphone with T9 predictive text input.
Your mission, should you choose to accept, is to write the shortest ruby
program that will find the longest word in the dictionary that can be
produced by...

If you wanted this to be a rubyquiz, you better could have send it to
James Edward Gray II. A similar quiz has allready been done in the
past. Check out http://rubyquiz.com and look around.

grtz,
wannes

 

[William James]

Assuming you are running Linux and have a list of valid English words in
the file /usr/share/dict/words or someplace or something similar.

Assume you have a cellphone with T9 predictive text input.

For those that don't, cellphones have each key overloaded to mean several
things. eg. On my Sharp GX-15
2 also means A,B,C
3 also means D,E,F
4 also means G,H,I
5 also means J,K,L
6 = M,N,O
7 = P,Q,R,S
8 = T,U,V
9 = W,X,Y,Z

In T9 mode I can type in 43556 and it only finds one word in its
dictionary that can be made with combination
{G,H,I}{D,E,F}{J,K,L}{J,K,L}{M,N,O} and that is
HELLO

Which is a lost faster than typing 4433555555666.

How ever, some words like "high", can be entered by just bouncing on one
key ... 4444.

Your mission, should you choose to accept, is to write the shortest ruby
program that will find the longest word in the dictionary that can be
produced by...
* repeatedly typing one key
* Can be produced by just using only two keys.
* Can be produced by just using only three keys.
* Can be produced by just using only four keys.
* ....

Assumes that ambiguity is disallowed; rejects words with apostrophes.

h=Hash.new(0)
a=ARGF.grep(/^\w+$/).sort_by{|x|-x.size}.map{|w|w.chomp!.downcase!
c=w.split('').map{|c|2+"abc def ghi jkl mno pqrstuv wxyz".index(c)/4}
h[c]+=1;[w,c]}.reject{|w,c|h[c]>1}
(1..8).each{|n|p [n,a.find{|w,c|n==c.uniq.size}[0]]}


[1, "deeded"]
[2, "blackjack"]
[3, "openendedness"]
[4, "prepossessingness"]
[5, "contemporaneousness"]
[6, "microspectrophotometric"]
[7, "antidisestablishmentarianism"]
[8, "microspectrophotometrically"]

 

[William James]

Assuming you are running Linux and have a list of valid English words in
the file /usr/share/dict/words or someplace or something similar.

Assume you have a cellphone with T9 predictive text input.

For those that don't, cellphones have each key overloaded to mean several
things. eg. On my Sharp GX-15
2 also means A,B,C
3 also means D,E,F
4 also means G,H,I
5 also means J,K,L
6 = M,N,O
7 = P,Q,R,S
8 = T,U,V
9 = W,X,Y,Z

In T9 mode I can type in 43556 and it only finds one word in its
dictionary that can be made with combination
{G,H,I}{D,E,F}{J,K,L}{J,K,L}{M,N,O} and that is
HELLO

Which is a lost faster than typing 4433555555666.

How ever, some words like "high", can be entered by just bouncing on one
key ... 4444.

Your mission, should you choose to accept, is to write the shortest ruby
program that will find the longest word in the dictionary that can be
produced by...
* repeatedly typing one key
* Can be produced by just using only two keys.
* Can be produced by just using only three keys.
* Can be produced by just using only four keys.
* ....

Assumes that ambiguity is disallowed; rejects words with apostrophes.

h=Hash.new(0)
a=ARGF.grep(/^\w+$/).sort_by{|x|-x.size}.map{|w|w.chomp!.downcase!
c=w.split('').map{|c|2+"abc def ghi jkl mno pqrstuv wxyz".index(c)/4}
h[c]+=1;[w,c]}.reject{|w,c|h[c]>1}
(1..8).each{|n|p [n,a.find{|w,c|n==c.uniq.size}[0]]}


[1, "deeded"]
[2, "blackjack"]
[3, "openendedness"]
[4, "prepossessingness"]
[5, "contemporaneousness"]
[6, "microspectrophotometric"]
[7, "antidisestablishmentarianism"]
[8, "microspectrophotometrically"]

Shorter:

h=Hash.new(0)
a=ARGF.grep(/^\w+$/).sort_by{|x|-x.size}.map{|w|w.chomp!.downcase!
c=w.split('').map{|c|2+"abc def ghi jkl mno pqrstuv wxyz".index(c)/4}
h[c]+=1;[w,c]}
(1..8).each{|n|p [n,a.find{|w,c|1==h[c]&&n==c.uniq.size}[0]]}

 

[John Carter]

If you wanted this to be a rubyquiz, you better could have send it to
James Edward Gray II.

Didn't even occur to me. Just thought it was a nice simple problem to
which I personally wanted the answer. Since I thought it would be fun, I
wondered if anyone else would have fun with it. RubyQuiz must have been
somewhere in my subconscious since I borrowed his "wish to accept" tag
line.

A similar quiz has allready been done in the
past. Check out http://rubyquiz.com and look around.

There was a quiz to design a "better than multitap" algorithm, not given
T9, what is the longest word with shortest taps.

I started crafting a solution to my problem when I got lost with wheels in
wheels in wheels and stopped.

I then went for a walk to buy milk and had the notion that regexes would
be good....
=========================================================
a=Hash.new('')
IO.read('/usr/share/dict/words').scan(/[a-z]+/i){|m|
x={}

m.upcase.tr( 'ABCDEFGHIJKLMNOPQRSTUVWXYZ',
'22233344455566677778889999').each_byte{|c|
x[c]=c
}
z=x.keys.size
i=a[z]
a[z]=m if i.size < m.size
}
p a
=======================================================
But than gave just the one example, so I extended it to give me all words
of the maximum length..
=============================================================
# I love the flexibility of the Ruby hash constructor. I use that
# feature all the time.
a=Hash.new([''])

# Hah! I didn't even know about ARGF. Ah well, I learn something
# new every day!

# I should of used \w instead of [a-z], that would have been shorter.
IO.read('/usr/share/dict/words').scan(/[a-z]+/i){|m|
x={}

# I chose upcase instead of downcase, saved me 2 chars. ;-)
# Hmm, perhaps I should have done the tr before the scan.
m.upcase.tr( 'ABCDEFGHIJKLMNOPQRSTUVWXYZ',
'22233344455566677778889999').each_byte{|c|
x[c]=c
}

# Somehow I sure if I was smarter there would be a shorter way of
# finding the number of uniq chars in a string. Especially
# as I could make those chars whatever I like.

z=x.keys.size

i=a[z]
if i[0].size < m.size
a[z]=[m]
elsif i[0].size == m.size
# Now this is actually quite subtle.
# Who says ruby doesn't have pointers?
# I have another fun problem in my head that relates to that.
i << m
end
}
a.keys.sort.each{|k| puts "#{k}\n\t#{a[k].sort.uniq.join("\n\t")}\n"}

The result is...

1
deeded
2
blackball
blackjack
depressed
depresses
preferred
redressed
redresses
repressed
represses
3
shepherdesses
4
presumptuousness
5
transubstantiation
6
contradistinctions
disenfranchisement
disproportionating
hypersensitivities
misinterpretations
misrepresentations
7
counterrevolutionaries
electroencephalographs
8
counterrevolutionary
uncharacteristically

It is interesting that the 8 key words are shorter than the 7.

John Carter

The Cybernetic Entomologist - (e-mail address removed)

http://geocities.yahoo.com/cy_ent

I'm becoming less and less convinced of humans as rational beings.
I suspect we are merely meme collectors, and the reason meme is only
kept on to help count our change.


John Carter Phone : (64)(3) 358 6639
Tait Electronics Fax : (64)(3) 359 4632
PO Box 1645 Christchurch Email : (e-mail address removed)
New Zealand

Carter's Clarification of Murphy's Law.

"Things only ever go right so that they may go more spectacularly wrong later."

From this principle, all of life and physics may be deduced.

 

[John Carter]

So having learnt from the other posts I get down to...

a=Hash.new([''])
ARGF.read.upcase.scan(/\w+/){|m|

z = m.tr( 'A-Z',
'22233344455566677778889999').split('').sort.uniq.size

i=a[z]
if (j=i[0].size) < n = m.size
a[z]=[m]
elsif j == n
i << m
end
}
a.keys.sort.each{|k| puts "#{k}\n\t#{a[k].sort.uniq.join("\n\t")}\n"}

1
DEEDED
2
BLACKBALL
BLACKJACK
DEPRESSED
DEPRESSES
PREFERRED
REDRESSED
REDRESSES
REPRESSED
REPRESSES
3
SHEPHERDESSES
4
PRESUMPTUOUSNESS
5
TRANSUBSTANTIATION
6
CONTRADISTINCTIONS
DISENFRANCHISEMENT
DISPROPORTIONATING
HYPERSENSITIVITIES
MISINTERPRETATIONS
MISREPRESENTATIONS
7
COUNTERREVOLUTIONARIES
ELECTROENCEPHALOGRAPHS
8
COUNTERREVOLUTIONARY
UNCHARACTERISTICALLY




John Carter Phone : (64)(3) 358 6639
Tait Electronics Fax : (64)(3) 359 4632
PO Box 1645 Christchurch Email : (e-mail address removed)
New Zealand

Carter's Clarification of Murphy's Law.

"Things only ever go right so that they may go more spectacularly wrong later."

From this principle, all of life and physics may be deduced.