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Darkdata
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Posted on 01-10-11 06:38:36 PM Link | Quote


Had to show a friend how to do it over the internet~

Feel free to laugh at my horrible mspaint image.

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Drag
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Posted on 01-10-11 09:26:08 PM Link | Quote
1923
Let D be any number in base 10
Let B[x] be an array of bits

1] Find the largest value of 2^N where 2^N is less than D
2] Set B[N] = 1
3] Subtract (2^N) from D
4] If D is greater than 0, go to 1

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Posted on 01-10-11 11:28:36 PM Link | Quote
I basically do it the same way as Darkdata

I had a teacher who tried to explain it some other way involving a lot of division, and I was more confused.

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Posted on 01-11-11 02:13:47 AM Link | Quote
Paulguy's Post configuration
Yeah I learned it the way darkdata said.

I even programmed a thing in qbasic that did this because it doesn't have binary ops that I could find, at least at the time. This was years ago, and it was ugly as shit.

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Posted on 01-12-11 04:01:17 AM Link | Quote
I learned how to do all of that the way Darkdata outlined. Though I don't think we went into hexadecimal too much.

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Posted on 01-12-11 04:05:10 AM Link | Quote
Drag's method is still the same thing as DD's, it's just worded differently. I learned the same thing probably because that's how we all learned it, just using a different tool to make things click. For me, it was Drag's tool.

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Lyskar
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Posted on 01-12-11 04:23:02 AM (last edited by Metal_Man88 at 01-12-11 01:23 AM) Link | Quote
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Metal_Man88's Post
Originally posted by Terra
I had a teacher who tried to explain it some other way involving a lot of division, and I was more confused.


I learned an unusual way that involves dividing by 2. Example:
DivisionResultRemainder
1234 / 26170
617 / 2 3081
308 / 21540
154 / 2770
77 / 2381
38 / 2190
19 / 291
9 / 241
4 / 220
2 / 210
1 / 201

Then you reverse the remainders to get your result, which in this case is 10011010010.

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Posted on 01-12-11 06:06:27 PM Link | Quote
00:00:00 Illumina Prevue Guide Emulator
POST #
5237		I pretty much do the conversion decimal-to-binary the same way as DD, and very similarly with hex too... other bases can be hard (such as odd-numbered ones), and you can also get octal out of this

For base 12, I do something similar but with 12s, 144s, 1728s, etc. instead of powers of 2 (replacing 10 with A and 11 with B)

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Lyskar
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Posted on 01-12-11 10:33:16 PM Link | Quote
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Metal_Man88's Post
Oh, I almost forgot. The way I learned it can be switched to any base you want by dividing by a different number.

Crazy, isn't it?

(Base 8)

DivisionResultRemainder
1234 / 81542
154 / 8192
19 / 823
2 / 802


Result: 2322 in base 8.

2 * 1 = 2
2 * 8 = 16
3 * 64 = 192
2 * 512 = 1024
--------------
= 1234

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Darkdata
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Posted on 01-12-11 10:40:53 PM Link | Quote
Originally posted by Metal_Man88
Originally posted by Terra
I had a teacher who tried to explain it some other way involving a lot of division, and I was more confused.


I learned an unusual way that involves dividing by 2. Example:
DivisionResultRemainder
1234 / 26170
617 / 2 3081
308 / 21540
154 / 2770
77 / 2381
38 / 2190
19 / 291
9 / 241
4 / 220
2 / 210
1 / 201

Then you reverse the remainders to get your result, which in this case is 10011010010.


So, I assume you round down?

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Posted on 01-13-11 12:42:59 AM Link | Quote
 
Originally posted by Darkdata
So, I assume you round down?

If you do it the way MM88 showed, yes, you truncate the division (just like you would if you did long division by hand, actually ).

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Posted on 01-13-11 04:23:21 AM Link | Quote
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Metal_Man88's Post
Yes.

My teacher worked on ballistic missiles and such. So he was able to teach the hard way by virtue of knowing his stuff--and how to teach people--something hard to come by.

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