So basically I am doing this thing and it says what is 00100110 + 01010111.
00100110 = k in binary
01010111 = X in binary
I naturally went to wikipedia and searched Alphabet. This gave me a grid where it showed me what letter corresponded to what number.
k is 11
X is 24
I added these together to get 35. I put in 35,1124,2411,13 and all of those in binary. These were ALL incorrect somehow.
I do not know what else it could be!
Characters of the alphabet are often used as a way of tagging a number so it can be easily referred to, so if the problem is not specifically asking you to guess a character out of it, the answer would be 125 in binary.
100110 = 38
1010111 = 87
100110 + 1010111 = 1111101 = 125
Related
Hello I would like to know of a quick and easy way to perform a number conversion of this binary value:
1000100000001011
to octal.
In hex I can convert fairly quickly by hand to 0xAA0B. To come up with the decimal value of this binary takes a bit more work but eventually you can arrive at 32,768 + 2,048 + 11 = 34,827.
I know the octal pattern works like 8 = 10, 9 = 11 .... 16 = 20, 17 = 21 ... 24 = 30, etc. However I am having trouble converting to octal without a large amount of effort.
Could anybody clear this up for me and perhaps provide a short-hand method that can help convert binary to octal. Programming examples are nice but I'm really looking for an explanation. Thanks
The quickest method is to break the binary number into 3-bit chunks from the right end, pad with 0's from the left as needed, then convert each chunk to an octal digit.
For example,
1000100000001011 -> 001 000 100 000 001 011 [2 0's added to the left]
-> 1 0 4 0 1 3
-> 104013
I have a BINARY number which i want to convert it into the DECIMAL and OCTAL.
(0100 1111 1011 0010)2
I know how to convert it into the decimal. But the question making me confuse. Because middle of every 4 digits there is a space "0101 1111"
can u help me how to understand this question.
Thanks
First of all, make sure that number you are converting into Decimal and Octal is actually 'Binary' and not 'Binary Coded Decimal (BCD)'. Usually when the number is grouped into 4 binary digits, it represents a BCD instead of just binary.
So, once you make sure its actually binary and not BCD, the conversion to both decimal and octal are simple steps.
For binary to octal, you group the binary number into sets of 3 digits, starting form the Least Significant Bit(LSB or right-most) to the Most Significant Bit(MSB or left-most). Add leading zeros if a group of 3 digits can not be formed at the MSB.
Now convert each group of digits from binary to octal:
(000) -> 0
(001) -> 1
.
.
(111) -> 7
Finally put the numbers together, and there you have your binary converted to octal.
Eg:-
binary - 00101101
split into groups of 2: -> 000 101 101 -> 0 5 5 - > 55
Difference between'Binary Coded Decimal' and 'Binary':
For the decimal number 1248
the binary would simply be 10011100000
However, the BCD would be -> 0001 0010 0100 1000
The spaces are not part of the number, it's just to make it easier for humans to read. Conversion from binary to octal is simple. Group the binary digits into sets of 3 (from right to left, add extra 0s to the leftmost group, then convert each group individually. Your example:
0100 1111 1011 0010 -> 100 111 110 110 010 -> 47662
The space is just for readability. Especially nice if you try to convert this to hex, because 4 binary digits make up one hex-digit.
Firstly, those spaces are for human readability; just delete them. Secondly, If this is not for a computer program, simply open up the windows calculator, go to view, and select programmer. Then chose the bin radio button and type in your number. the qword radio button should be selected. If it's for a program, I will need to know what language to help you.
To convert octal to decimal very quickly there are two methods. You can actually do the actual calculation in bitshift. In programming, you should do bitshift.
Example octal number = 147
Method one: From left to right.
Step 1: First digit is one. Take that times 8 plus 4. Got 12.
Step 2: Take 12 times 8 + 7. Got 103, and 103 is the answer.
Ultimately you can use method one to convert any base into base 10.
Method one is reading from left to right of the string. Make a result holder for calculation. When you read the first leftmost digit, you add that to a result value. Each time you read a new digit, you take the result value and multiply that by the base of the number(for octal, that would be 8), then you add the value of the new digit to the result.
Method 2, bitshift:
Octal Number: 147.
Step 1: 1 = 1(bin) = Shift << 3 = 1000(result value)
Step 2: 4 = 100(bin) + 1000(result value) = 1100(result value)
Step 3: 1100(result value) Shift << 3 = 1100000
Step 4: 7 = 111(bin) + 1100000(result value) = 1100111
Step 5: 1100111 binary is 103 decimal.
In a programming loop, you can do something like the below, and it is lightning fast. The code is so simple that it can be converted into any programming language. Note that there isn't any error checking.
for ( int i = 0; i < length; i++ ){
c = (str.charAt(i) ^ 48);
if ( c > 7 ) return 0; // <- if char ^ 48 > 7 then that is not a valid octal number.
out = (out << 3) + c;
}
I am trying to add the following two binary numbers together, however I am unable to do so becuase im not sure whether or not this is an overflow?
110101 + 010111
The answer I get is: 1001100
Do I remove the left most 1 in the answer or do I keep it? By removing it I get 12, otherwise the answer is Not right. Am I doing something wrong?
Is this right?
The answer to this question depends upon the size of the word in bits of the system you are talking about. In an 8-bit (or higher) system, and you're doing 2's complement, the sum of
110101 + 010111 = 1001100
Is the same as:
00110101 + 00010111 = 01001100
Is: 53 + 23 = 76 with no overflow or carry out.
If it's a 7-bit system, doing 2's complement, then you have:
0110101 + 0010111 = 1001100
Which is 53 + 23 = -52. There's an overflow, but no carry out.
If it's a 6-bit system, doing 2's complement, then:
110101 + 010111 = (1)001100
Which is -11 + 23 = 12. There's no overflow but there's a carry out. Note that in a 6-bit system you can't have 1001100 technically because it's 7 bits. You would have 001100.
For reference, see The CARRY flag and OVERFLOW flag in binary arithmetic.
I am new to programming using Java and I can't use anything fancy to solve this. I assume it is some simple math but I can't seem to figure it out.
I need to determine how many units of 25 (in miles) the package is being shipped (remember that parts of 25 count as a full 25, so 75 is 3 units of 25 while 76 is 4 units of 25).
So how do I do that math to find out how many units the number the user enters will be. This is by 25.
I tried division but it wont give me two units if I do miles / 25.
Please help!
Edit
I have found online someone using (miles + 24) / 25;
But I do not understand that...
You could round it. Most computer languages have some means of rounding integers.
Many languages today also have a mod operator (often the % sign) which returns the remainder after division.
So, you can use something like
(1) int wholeUnits = miles / 25;
(2) remainder = miles % 25;
(3) if (remainder != 0) wholeUnits = wholeUnits +1;
So, (1) determines the number of whole units -- 75 would return 3, 74 would return 2, and 76 would return 3.
(2) would give you the remainder when dividing by 25. So, 75 % 25 = 0. 76 % 25 = 1. And 74 % 25 = 24.
(3) if the remainder is other than zero, you need to add one to the wholeUnits. So, 75 would have a zero remainder so it would just be 3. 74 would return 24, and thus you'd add 1 to the 2 you already have, getting 3 whole units. And 76 would return (3+1) = 4.
So, there are different ways of doing it, depending on the language you're using. The easiest thing is to figure out what the round function is and use it. But you can do it the long way with these three steps if you prefer.
Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
Write the shortest program that calculates the Frobenius number for a given set of positive numbers. The Frobenius number is the largest number that cannot be written as a sum of positive multiples of the numbers in the set.
Example: For the set of the Chicken McNuggetTM sizes [6,9,20] the Frobenius number is 43, as there is no solution for the equation a*6 + b*9 + c*20 = 43 (with a,b,c >= 0), and 43 is the largest value with this property.
It can be assumed that a Frobenius number exists for the given set. If this is not the case (e.g. for [2,4]) no particular behaviour is expected.
References:
http://en.wikipedia.org/wiki/Coin_problem
http://mathworld.wolfram.com/FrobeniusNumber.html
[Edit]
I decided to accept the GolfScript version. While the MATHEMATICA version might be considered "technically correct", it would clearly take the fun out of the competition. That said, I'm also impressed by the other solutions, especially Ruby (which was very short for a general purpose language).
Mathematica 0 chars (or 19 chars counting the invoke command)
Invoke wtih
FrobeniusNumber[{a,b,c,...}]
Example
In[3]:= FrobeniusNumber[{6, 9, 20}]
Out[3]= 43
Is it a record? :)
Ruby 100 86 80 chars
(newline not needed)
Invoke with frob.rb 6 9 20
a=$*.map &:to_i;
p ((1..eval(a*"*")).map{|i|a<<i if(a&a.map{|v|i-v})[0];i}-a)[-1]
Works just like the Perl solution (except better:). $* is an array of command line strings; a is the same array as ints, which is then used to collect all the numbers which can be made; eval(a*"*") is the product, the max number to check.
In Ruby 1.9, you can save one additional character in by replacing "*" with ?*.
Edit: Shortened to 86 using Symbol#to_proc in $*.map, inlining m and shortening its calculation by folding the array.
Edit 2: Replaced .times with .map, traded .to_a for ;i.
Mathematica PROGRAM - 28 chars
Well, this is a REAL (unnecessary) program. As the other Mathematica entry shows clearly, you can compute the answer without writing a program ... but here it is
f[x__]:=FrobeniusNumber[{x}]
Invoke with
f[6, 9, 20]
43
GolfScript 47/42 chars
Faster solution (47).
~:+{0+{.1<{$}{1=}if|}/.!1):1\{:X}*+0=-X<}do];X(
Slow solution (42). Checks all values up to the product of every number in the set...
~:+{*}*{0+{.1<{$}{1=}if|}/1):1;}*]-1%.0?>,
Sample I/O:
$ echo "[6 9 20]"|golfscript frobenius.gs
43
$ echo "[60 90 2011]"|golfscript frobenius.gs
58349
Haskell 155 chars
The function f does the work and expects the list to be sorted. For example f [6,9,20] = 43
b x n=sequence$replicate n[0..x]
f a=last$filter(not.(flip elem)(map(sum.zipWith(*)a)(b u(length a))))[1..u] where
h=head a
l=last a
u=h*l-h-l
P.S. since that's my first code golf submission I'm not sure how to handle input, what are the rules?
C#, 360 characters
using System;using System.Linq;class a{static void Main(string[]b)
{var c=(b.Select(d=>int.Parse(d))).ToArray();int e=c[0]*c[1];a:--e;
var f=c.Length;var g=new int[f];g[f-1]=1;int h=1;for(;;){int i=0;for
(int j=0;j<f;j++)i+=c[j]*g[j];if(i==e){goto a;}if(i<e){g[f-1]++;h=1;}
else{if(h>=f){Console.Write(e);return;}for(int k=f-1;k>=f-h;k--)
g[k]=0;g[f-h-1]++;h++;}}}}
I'm sure there's a shorter C# solution than this, but this is what I came up with.
This is a complete program that takes the values as command-line parameters and outputs the result to the screen.
Perl 105 107 110 119 122 127 152 158 characters
Latest edit: Compound assignment is good for you!
$h{0}=$t=1;$t*=$_ for#ARGV;for$x(1..$t){$h{$x}=grep$h{$x-$_},#ARGV}#b=grep!$h{$_},1..$t;print pop#b,"\n"
Explanation:
$t = 1;
$t *= $_ foreach(#ARGV);
Set $t to the product of all of the input numbers. This is our upper limit.
foreach $x (1..$t)
{
$h{$x} = grep {$_ == $x || $h{$x-$_} } #ARGV;
}
For each number from 1 to $t: If it's one of the input numbers, mark it using the %h hash; otherwise, if there is a marked entry from further back (difference being anything in the input), mark this entry. All marked entries are non-candidates for Frobenius numbers.
#b=grep{!$h{$_}}(1..$t);
Extract all UNMARKED entries. These are Frobenius candidates...
print pop #b, "\n"
...and the last of these, the highest, is our Frobenius number.
Haskell 153 chars
A different take on a Haskell solution. I'm a rank novice at Haskell, so I'd be surprised if this couldn't be shortened.
m(x:a)(y:b)
|x==y=x:m a b
|x<y=x:m(y:b)a
|True=y:m(x:a)b
f d=l!!s-1where
l=0:foldl1 m[map(n+)l|n<-d]
g=minimum d
s=until(\n->l!!(n+g)-l!!n==g)(+1)0
Call it with, e.g., f [9,6,20].
FrobeniusScript 5 characters
solve
Sadly there does not yet exist any compiler/interpreter for this language.
No params, the interpreter will handle that:
$ echo solve > myProgram
$ frobeniusScript myProgram
6
9
20
^D
Your answer is: 43
$ exit