I’m getting quite frustrated with a very “easy” part of this
problem…reading the input! Basically, the input is given as so:
1
1 2
1 3 2
1 4 2 1
What I intend to do is read the input line by line, split the lines by
spaces and store each substring as the associated integer. Essentially I
want to end up with a two dimensional array holding the integers shown
above. I’m using the following line of code:
a = readlines.each{|i|i.split(" ").each{|j|j=j.to_i}}
However, when I do a puts I see the following:
puts a[1][1] => 32
GAH! Why is it printing the ascii character value? Do I have a logic
flaw in the way I’m reading the input? Is there a better way to read the
input? Should I just tear my hair out?
Thanks for the help…and with respect to white space, I apologize, I
was in the codegolf mindset and trying to conserve bytes. http://www.codegolf.com
Ah, then the compression is forgiven.
Also, can you explain to me the difference between map and each?
#each just evaluates the block for each element of the list (or
whatever you’re iterating over). You don’t usually use the return
value of #each.
#map (aka #collect) also evaluates the block for each element, but
returns a new list containing the return value of each evaluation.
It’s good for translating one set of values to another, like you were
doing in your code.
GAH! Why is it printing the ascii character value? Do I have a logic
flaw in the way I’m reading the input? Is there a better way to read the
input? Should I just tear my hair out?
Well firstly, because as others have mentioned, you were using #each,
so you were getting an array populated with strings:
[“1”, “1 2”, “1 3 2”, “1 4 2 1”]
^
…and the 1-index of the 1-item of the array is a space (" ") – ascii
ordinal 32.
But why doesn’t ruby give you the space – why the ordinal? Because in
1.8, indexing a string by a single integer gives you the ascii ordinal
for that character of the string (this changed in 1.9, now it gives you
the character and you have to explicitly use #ord to get the ordinal).
So, ‘a’[0] => 97. If you ever need to get around that, just make the
indexer into a range: ‘a’[0,1] => a, ‘a’[0…0] => a.
Essentially I
want to end up with a two dimensional array holding the integers shown
above.
Figure out your algorithm first.
Devin -
My first attempt at an algorithm was this:
Work backwards row by row. In each row, compare the current element with
the next, and whichever ever is larger, add this number to the element
in the same index in the next highest row. Eventually, the value in
a[0][0] will be the largest path…does this makes sense? Is it an
effective strategy?
One more hint: You probably don’t need to waste your time with map. Consider:
a=readlines.each{|i|i.split(" ").each{|j|j=j.to_i}}
That just assigns the lines you’ve read in, unchanged, to a. In the
outer each, you split lines and send the results, before you discard
them, to the inner each – where you reuse the identifier j, and then
throw its new value away. Meanwhile, all a cares about is the return
value of readlines.each, which is its receiver.
Work backwards row by row. In each row, compare the current element with
the next, and whichever ever is larger, add this number to the element
in the same index in the next highest row. Eventually, the value in
a[0][0] will be the largest path…does this makes sense? Is it an
effective strategy?
Yeah, that does seem like it’ll work. I never thought of it that way.
Think of the forwards algorithm, too. Then determine if the code you
write with the backwards is 8 bytes shorter (to make up for reverse_each
vs each). (I’ll have to look at it, myself. :P)
One more hint: You probably don’t need to waste your time with map.
Consider:
a=readlines.each{|i|i.split(" “).each{|j|j=j.to_i}}
a.reverse_each{|r|p r[0]}
vs:
readlines.reverse_each{|l|l.split(” ")[0].to_i}
Again, try it both ways and see. (Keeping in mind that if the difference
is small, then the winner’s not certain – the little optimizations
could tip it either way.)
I could suggest more things, but I don’t wanna lose my spot. It’s the
one challenge at which I was actually successful.
Devin
Wow… acting as if I’m an expert after a few little internet
challenges… Well, grain of salt and all that.
a=readlines.each{|i|i.split(" ").each{|j|j=j.to_i}}
That just assigns the lines you’ve read in, unchanged, to a.
That was a paste-o on my part. My point was that the functional means of
doing things don’t tend to golf well. Compare:
a.map{|b|b.to_i}.each{|c|}
a.each{|b|c=b.to_i}
a=readlines.each{|i|i.split(" ").each{|j|j=j.to_i}}
That just assigns the lines you’ve read in, unchanged, to a.
That was a paste-o on my part. My point was that the functional means of
doing things don’t tend to golf well. Compare:
a.map{|b|b.to_i}.each{|c|}
a.each{|b|c=b.to_i}
Doesn’t it depend what you need, though? Also, you can always do the
each-like side-effect stuff with map, but not the map-like stuff with
each.
David
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