The question says
Given a two-level page table with 4-KB pages and. Assume that each level uses 10
bits. What would be the virtual address if PT1=2, PT2=3, offset=5?
The given answer is
(2×2^22)+(3×2^12)+5=8400901
I get that because the pages are size 4-KB that the PT should be multiplied by 2^12. But where does the 2^22 come from?
A clearer way to write the answer is:
2 << (10 + 12) + 3 * << 12 + 5 = 8400901
This is like:
2 << (BITS_IN_PAGE_TABLE_INDEX + BITS_IN_OFFSET) + 3 * << BITS_IN_OFFSET + 5 = 8400901
Related
Having an infinite sequence s = 1234567891011...
Let's find the number at the n position (n <= 10^18)
EX: n = 12 => 1; n = 15 => 2
import Foundation
func findNumber(n: Int) -> Character {
var i = 1
var z = ""
while i < n + 1 {
z.append(String(i))
i += 1
}
print(z)
return z[z.index(z.startIndex, offsetBy: n-1)]
}
print(findNumber(n: 12))
That's my code but when I find the number at 100.000th position, it returns an error, I thought I appended too many i to z string.
Can anyone help me, in swift language?
The problem we have here looks fairly straight forward. Take a list of all the number 1-infinity and concatenate them into a string. Then find the nth digit. Straight forward problem to understand. The issue that you are seeing though is that we do not have an infinite amount of memory nor time to be able to do this reasonably in a computer program. So we must find an alternative way around this that does not just add the numbers onto a string and then find the nth digit.
The first thing we can say is that we know what the entire list is. It will always be the same. So can we use any properties of this list to help us?
Let's call the input number n. This is the position of the digit that we want to find. Let's call the output digit d.
Well, first off, let's look at some examples.
We know all the single digit numbers are just in the same position as the number itself.
So, for n<10 ... d = n
What about for two digit numbers?
Well, we know that 10 starts at position 10. (Because there are 9 single digit numbers before it). 9 + 1 = 10
11 starts at position 12. Again, 9 single digits + one 2 digit number before it. 9 + 2 + 1 = 12
So how about, say... 25? Well that has 9 single digit numbers and 15 two digit numbers before it. So 25 starts at 9*1 + 15*2 + 1 = 40 (+ 1 as the sum gets us to the end of 24 not the start of 25).
So... 99 starts at? 9*1 + 89*2 + 1 = 188.
The we do the same for the three digit numbers...
100... 9*1 + 90*2 + 1 = 190
300... 9*1 + 90*2 + 199*3 + 1 = 787
1000...? 9*1 + 90*2 + 900*3 + 1 = 2890
OK... so now I'm seeing a pattern here that seems to need to know the number of digits in each number. Well... I can get the number of digits in a number by rounding up the log(base 10) of that number.
rounding up log base 10 of 5 = 1
rounding up log base 10 of 23 = 2
rounding up log base 10 of 99 = 2
rounding up log base 10 of 627 = 3
OK... so I think I need something like...
// in pseudo code
let lengthOfNumber = getLengthOfNumber(n)
var result = 0
for each i from 0 to lengthOfNumber - 1 {
result += 9 * 10^i * (i + 1) // this give 9*1 + 90*2 + 900*3 + ...
}
let remainder = n - 10^(lengthOfNumber - 1) // these have not been added in the loop above
result += remainder * lengthOfNumber
So, in the above pseudo code you can give it any number and it will return the position in the list that that number starts on.
This isn't the exact same as the problem you are trying to solve. And I don't want to solve it for you.
This is just a leg up on how I would go about solving it. Hopefully, this will give you some guidance on how you can take this further and solve the problem that you are trying to solve.
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I need to figure out what number base makes the expression 32 + 12 = 28 true. It looks like the second number is signed, I've tried converting to decimal and using 10's complement but haven't gotten an answer that makes sense.
If this is a positional number system, then you can replace the numerals with an expression showing how their digits multiply by the meaning of their columns.
That is,
"32 + 12 = 28"
≣ (3*R^1 + 2*R^0) + (1*R^1 + 2*R^0) = (2*R^1 + 8*R^0)
≣ (3*R + 2*1) + (1*R + 2*1) = (2*R + 8*1)
≣ (3*R + 2) + (1*R + 2) = (2*R + 8)
≣ 4*R + 4 = 2*R + 8
≣ 2*R = 4
≣ R = 2
For some radix R.
You can solve this equation to get the value of R. If you attempt to solve it, you'll see that you get R = 2, which clearly can't be the case, since you have a digit 3 (a base-2 system would only have ... 2 digits. 0, and 1).
Thus, there is no radix R that would allows "32 + 12 = 28" to encode a valid equation.
On the face of it, it looks impossible. Adding positive integers in any base should result in a larger integer; if the second integer is negative, you end up with 20 = 28 which is clearly impossible. Considered modulo arithmetic? 32 mod 4 is 0, 12 mod 4 is 0, 28 mod 4 is 0, 0 + 0 = 0.
Is it possible for all calculations in the expression for numbers in a power to be prevented? Perhaps by pre-processing the expression or adding tellsimp rules? Or some other way?
For example, to
distrib (10 ^ 10 * (x + 1));
which produces:
1000000000 x + 1000000000
instead issued:
10 ^ 10 * x + 10 ^ 10
And similarly
factor (10 ^ 10 * x + 10 ^ 10);
returned:
10 ^ 10 * (x + 1);
Just as
factor(200);
2^3*5^2
represents power of numbers, only permanently?
Interesting question, although I don't see a good solution. Here's something I tried as an experiment, which is to display integers in factored form. I am working with Maxima 5.44.0 + SBCL.
(%i1) :lisp (defun integer-formatter (x) ($factor x))
INTEGER-FORMATTER
(%i1) :lisp (setf (get 'integer 'formatter) 'integer-formatter)
INTEGER-FORMATTER
(%i1) (x + 1000)^3;
3 3 3
(%o1) (x + 2 5 )
(%i2) 10^10*(x + 1);
2 5 2 5
(%o2) (2 5 ) (x + 1)
This is only a modification of the display; the internal representation is just a single integer.
(%i3) :lisp $%
((MTIMES SIMP) 10000000000 ((MPLUS SIMP) 1 $X))
That seems kind of clumsy, since e.g. 2^(2*5)*5^(2*5) isn't really more comprehensible than 10000000000.
A separate question is whether the arithmetic on 10^10 could be suppressed, so it actually stays as 10^10 and isn't represented internally as 10000000000. I'm pretty sure that would be difficult. Unfortunately Maxima is not too good with retracting identities which are applied, particularly with the built-in identities which are applied to perform arithmetic and other operations.
In finding the values of x and y, if (x567) + (2yx5) = (71yx) ( all in base 8) I proceeded as under.
I assumed x=abc and y=def and followed.
(abc+010 def+101 110+abc 111+101)=(111 001 def abc) //adding ()+()=() and equating LHS=RHS.
abc=111-010=101 which is 5 in base 8 and then def=001-101 which is -4
so x=5 and y=-4
Now the Question is that the answer mentioned in my book is x=4 and y=3.
Is the above method correct.If so,then what's issue here ??
you can't compare the digits beginning with the most significant digit, because you don't know the carry from the digit below. Also a digit cannot have a negative value.
You can start with the least significant digit, because there is no carry:
7 + 5 = 14
so x = 4 with a carry of 1 at the next digit.
now you can rewrite your equation to:
(4567) + (2y45) = (71y4)
now you can look at the second least significant digit (the carry in mind):
6 + 4 + 1 (carry) = 13
so y = 3, also with a carry of 1.
the whole equation is:
(4567) + (2345) = (7134)
which is true for the octal system.
I have the following predefined codes that represent an index in a binary bitmap:
0 = standard
1 = special
2 = regular
3 = late
4 = early
5 = on time
6 = generic
7 = rfu
An example value I would take as an input would be 213, which becomes 11010101 in binary. Index 0, 2, 4, 6, and 7 have their bit flipped indicating that this record is:
standard + regular + early + generic + rfu.
I am trying to figure out in perl how to take that binary data and build a string, like mentioned above with code + code + code, etc.
Any help would be greatly appreciated. Thanks.
Edit: My thoughts on how I might approach this are:
Convert decimal to binary
Find length of binary string
Using substr get the value (0 or 1) index by index
If index value = 1 then add relevant code to string
Is there a better way to go about this?
You can test bits on input from 0 to 7, and take only these that are set,
my $in = 213;
my #r = ("standard","special","regular","late","early","on time","generic","rfu");
print join " + ", #r[ grep { $in & (1 << $_) } 0 .. $#r ];
# or
# print join " + ", map { $in & (1<<$_) ? $r[$_] : () } 0 .. $#r;
output
standard + regular + early + generic + rfu