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Comments:
<0> But
<0> Like I said
<0> We already know how to do it
<0> Our brains are doing it when we come up with the combinations
<0> All the algorithm would do is do the same thing to come up with every combination, then pick the shortest one
<0> Or maybe compute the shortest one first
<0> Then give up on any that go longer
<1> Well...start with the biggest numbers and work down to 1
<0> Well, let's say the input is 87
<1> Like start with 10, try to come up with multiplications or addition and loop down to one
<1> Start with 10, multiply by 8, add 7
<0> Remainders are seriously involved
<1> So you could take 10, 10 is smaller than 87, multiple by 9 (too big)
<0> Maybe a recursive approach with the remainders
<0> Basically,
<1> multiply by 8
<1> You have 7 left, add 7
<0> find_an_operation_that_works(number)
<0> find_an_operation_that_works(remainder of the last operation)
<0> So like you say, 10 * 8 works
<0> Remainder is 7, so how do you account for that
<0> Same way you came up with 10*8
<0> Only now the outcome is easier
<0> That sort of makes sense to me
<2> wow, lots of numbers
<3> :(
<0> And like you said, start with the high possibles first
<3> my brain has blocked to think
<0> Wouldn't be super efficient, though
<0> The 3 7 * one wouldn't be the first choice by a long shot
<0> Maybe try the first level first
<3> 10 2 * 1 +
<3> that's the firt one
<0> No, I mean, depth
<0> Let's say we look for all the combinations that come in at or under the value, at the first depth only.
<0> So, 10 2 * works
<0> But isn't sufficient
<0> 9 2 * works
<0> Same, not sufficient
<0> 7 3 * works, and is sufficient, done.
<0> That's exactly 4 iterations
<0> Pretty slick for a number that probably has upwards of 25 representations
<0> And you might even run 10 last, since it's longer
<0> 9, 8, 7... 2, 1, 10
<0> Calculate to the first depth only, the combinations
<0> Store the combinations that work
<0> If one works and is sufficient, in this hypothetical terminology, that's your winner.
<0> So 30 would hit at 6 5 *
<3> i'm good only at computer graphic s , maths and programming is not
<3> :(
<0> It would never get to 10 3 *
<0> error, you've got to focus on figuring out what your mind is already doing.
<0> And, you have to sit down with your instructors and get help on this stuff
<0> It's not enough for people like me to give you answers
<3> i know
<3> and that's true
<0> It's fine to know, but you've got to actually do it, too
<0> Any idiot can know something
<3> :))
<0> It takes a man to saddle up and actually do it
<0> I think we have a good start on your algorithm, but I don't think you understand where we're going with it or how we got there
<0> So I'm not sure if looking at it any more will help you
<0> I mean
<0> I'm not sure if my looking at it will help you
<0> Let's try this
<0> 42
<3> 10 4 * 2 +
<0> Great
<0> How'd you come up with that
<0> :)
<0> I mean, if I show a trained monkey "42", he is NOT going to draw "10 4 * 2 +" on a chalkboard
<3> the biggest number from 1 to 10 is 10
<0> Okay?
<3> multiplied with 4 , and add 2
<0> No no
<0> That's just describing your answer.
<0> I want to know how you got there.
<0> Say that, instead of 42
<0> I give you this number:
<0> N
<0> And you come up with a representation R
<0> If I then ask you
<0> How did you get to R
<0> Your answer should be exactly the same as it was for how you got to 10 4 * 2 +
<0> Right?
<2> 'night
<0> The logic you're working through is the same
<3> yes ' i think
<0> Night, rdragon
<0> Some steps may get skipped because they don't apply to some numbers or others
<0> And you might not get to other steps
<0> But the steps are there
<0> And the goal is to sit down and take a piece of paper and play stupid
<0> Work out your answer on paper
<0> Start with "42" at the top of the piece of paper
<0> Under it, what's the very first thing you think up
<0> What's the first number
<0> 10?
<0> Put 10
<0> How do you get to 4?
<0> Explain that step to yourself, that's part of your algorithm.
<3> incremeting from 1 to 10 ..
<0> Why increment?
<0> 1's going to get you the worst case representations, by and large
<0> 1 1 + 1 + 1 + etc.
<0> And we know that 10's not a great choice because it's actually worse than 9, since it takes up two characters instead of 1
<0> 10 3 * is one character longer than 6 5 *
<0> So, the real sequence of numbers to consider would be more like:
<3> indeed
<0> 9, 8, 7... 2, 1, 10
<0> That's a little optimization in there
<0> If you start with 10, logically, it seems a little harder to get to your ideal answer right off the bat.
<0> So try 42 again
<0> With your new sequence in mind in your algorithm
<0> What new representation do you immediately come up with as the shortest one?
<3> i need to find an recursive procedure for this problem
<0> Don't jump like that
<0> One thing at a time
<0> At least until I'm satisfied that you understand these little discrete things.
<0> When I said 42
<0> You came up with 10 4 * 2 +
<0> Because you started with 10, by your own admission
<0> I talked a little about how it would make more sense to check 10 last, instead of first
<0> You sort of nodded and agreed
<0> But now try to demonstrate that you understood
<0> When I say 42 now
<0> What's the first representation you come up with
<3> 42
<3> think .. think
<3> 1 min
<3> paper and pencil
<3> 8 5 * 2 +
<0> Why not 9 4 * 8 + ?
<0> I want a serious answer to that, don't just nod and say yeah that would work too
<0> Why did you come up with 8 5 before 9 4?
<0> Your mind had a reason, even if you don't think it did
<3> i start from 1 this time
<0> Okay
<0> Wouldn't 1 give you 1 1 + 1 + 1 + etc.?
<0> And the lower numbers in general would all **** in the same way
<3> yes it can give but i's not the shortest one
<3> adding 1 by one
<3> but surley doesent work that way
<3> so
<0> So then how about 7 7 * 1 +?
<3> is biiger the 42 from the start
<0> Actually, 6 * 8 would be the shortest, but let's ignore that.
<0> Er, sorry
<0> 6 6 * 5 +
<4> Hey, do double's store in base-10 scientific notation, or base-2 scientific notation?
<0> In other words, there are a whole lot of numbers you can use
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