I want to create a poor-mans version of a password complexity checker. I determine the rough character set the password is using and its length. The search space would then be: charset ^ length. In order to compare this against a single value I want the smallest x that when used as the exponent of 2 is larger than the search space. In more mathy language I want this:
given a and b find the smallest x where a^b < 2^x;
My math sucks. Is there a quick and easy way to calculate this?
Maybe my math doesn't suck quite that much.
2^x == a^b
= define a = 2^c, c = 2loga
2^x == 2^c^b
=
2^x == 2^c*b
=
x == c*b
=
x == 2loga * b
You can solve the equation by using logarithms. Taking the logarithms on both sides yields
x > b * log(a) / log(2)
If you want to find the smallest integer such that the equation holds we can round up the right hand side. In python this can be implemented as
import math
def find_x(a, b):
return math.ceil(b * math.log(a) / math.log(2))
I'm trying to find time complexity (big O) of a recursive formula.
I tried to find a solution, you may see the formula and my solution below:
Like Brenner said, your last assumption is false. Here is why: Let's take the definition of O(n) from the Wikipedia page (using n instead of x):
f(n) = O(n) if and only if there exist constants c, n0 s.t. |f(n)| <= c |g(n)|, for alln >= n0.
We want to check if O(2^n^2) = O(2^n). Clearly, 2^n^2 is in O(2^n^2), so let's pick f(n) = 2^n^2 and check if this is in O(2^n). Put this into the above formula:
exists c, n0: 2^n^2 <= c * 2^n for all n >= n0
Let's see if we can find suitable constant values n0 and c for which the above is true, or if we can derive a contradiction to proof that it is not true:
Take the log on both sides:
log(2^n^2) <= log(c * 2 ^ n)
Simplify:
2 ^n log(2) <= log(c) + n * log(2)
Divide by log(2):
n^2 <= log(c)/log(2) * n
It's easy to see know that there is no c, n0 for which the above is true for all n >= n0, thus O(2^n^2) = O(n^2) is not a valid assumption.
The last assumption you've specified with the question mark is false! Do not make such assumptions.
The rest of the manipulations you've supplied seem to be correct. But they actually bring you nowhere.
You should have finished this exercise in the middle of your draft:
T(n) = O(T(1)^(3^log2(n)))
And that's it. That's the solution!
You could actually claim that
3^log2(n) == n^log2(3) ==~ n^1.585
and then you get:
T(n) = O(T(1)^(n^1.585))
which is somewhat similar to the manipulations you've made in the second part of the draft.
So you can also leave it like this. But you cannot mess with the exponent. Changing the value of the exponent changes the big-O classification.
I have to compute the value of this expression
(e1*cos(θ) - e2*sin(θ)) / ((cos(θ))^2 - (sin(θ))^2)
Here e1 and e2 are some complex expression.
In the case when θ approach to PI/4, then the denominator will approach to zero. But in that case e1 and e2 will also approach to same value. So at PI/4, the value of expression will be E/sqrt(2) where e1=e2=E
I can do special handing for θ=PI/4 but what about the value of θ very very near to PI/4. What is the general strategy to compute such kind of expressions
This is a bit tricky. You need to know more about how e1 and e2 behave.
First, some notation: I'll use the variable a = theta - pi/4, so that we are interested in a->0, and write e1 = E + d1, e2 = E + d2. Let F = your expression times sqrt(2)
In terms of these we have
F = ((E+d1)*(cos(a) - sin(a)) - (E+d2)*(cos(a) + sin(a))) / - sin( 2*a)
= (-(2*E+d1+d2)*sin(a) + (d1-d2)*cos(a)) / (-2*sin(a)*cos(a))
= (E+(d1+d2)/2)/cos(a) - (d1-d2)/(2*sin(a))
Assuming that d1->0 and d2->0 as a->0 the first term will tend to E.
However the second term could tend to anything, or blow up -- for example if d1=d2=sqrt(a).
We need to assume more, for example that d1-d2 has derivative D at a=0.
In this case we will have
F-> E - D/2 as a->0
To be able to compute F for values of a close to 0 we need to know even more.
One approach is to have code like this:
if ( fabs(a) < small) { F = E-D/2 + C*a; } else { F = // normal code }
So we need to figure out what 'small' and C should be. In part this depends on what (relative) accuracy you require. The most stringent requirement would be that at a = +- small, the difference between the approximation and the normal code should be too small to represent in a double (if that's what you're using). But note that we mustn't make 'small' too small, or there is a danger that, as doubles, the normal code will evaluate to 0/0 at a += small. One approach would be to expand the numerator and denominator of F (or just the second term) as power series (to say second order), divide each by a, and then divide these series, keeping terms up to second order; the first term in this gives you C above, and the second term will allow you to estimate the error in this approximation, and hence estimate 'small'.
I am confused on how to use mathematical induction to prove Big O for a recursive function, given using its recursion relation.
Example:
The recurrence relation for recursive implementation of Towers of Hanoi is T(n) = 2T(n-1) + 1
and T(1) = 1. We claimed that this recursive method is O(n) = 2n - 1. Prove this claim using the mathematical induction.
In the case of recursion, do I always assume that n = k-1, rather than n=k? This is the assumption that the lecture notes give.
Assume f(n-1) = 2^(n-1) - 1 is true.
I understand with non-recursive mathematical induction we assume that n = k, because it is only a change of variables. Why then, is it safe to assume that n = k - 1?
One possible way: Postulate a non-recursive formula for T and proove it. After that, show that the formula you found is in the Big O you wanted.
For the proof, you may use induction, which is quick and easy in that case. To do that, you first show that your formula holds for the first value (usually 0 or 1, in your example that's 1 and trivial).
Then you show that if it holds for any number n - 1, it also holds for its successor n. For that you use the definition for T(n) (in your example that's T(n) = 2 T(n - 1) + 1): as you know that your formla holds for n - 1, you can replace occurences of T(n - 1) with your formula. In your example you then get (with formula T(n) = 2^n - 1)
T(n) = 2T(n - 1) + 1
= 2(2^(n - 1) - 1) + 1
= 2^n - 2 + 1
= 2^n + 1
As you can see, it holds for n if we assume it holds for n - 1.
Now comes the trick of induction: we showed that our formula holds for n = 1, and we showed that if it holds for any n = k - 1, it holds for k as well. That is, as we prooved it for 1, it is also proven for 2. And as it is proven for 2, it is also proven for 3. And as it is...
Thus, we do not assume that the term is true for n - 1 in our proof, we only made a statement under the assumption that it is true, then prooved our formula for one initial case, and used induction.
I have spent the last 5 hours searching for an answer. Even though I have found many answers they have not helped in any way.
What I am basically looking for is a mathematical, arithmetic only representation of the bitwise XOR operator for any 32bit unsigned integers.
Even though this sounds really simple, nobody (at least it seems so) has managed to find an answer to this question.
I hope we can brainstorm, and find a solution together.
Thanks.
XOR any numerical input between 0 and 1 including both ends
a + b - ab(1 + a + b - ab)
XOR binary input
a + b - 2ab or (a-b)²
Derivation
Basic Logical Operators
NOT = (1-x)
AND = x*y
From those operators we can get...
OR = (1-(1-a)(1-b)) = a + b - ab
Note: If a and b are mutually exclusive then their and condition will always be zero - from a Venn diagram perspective, this means there is no overlap. In that case, we could write OR = a + b, since a*b = 0 for all values of a & b.
2-Factor XOR
Defining XOR as (a OR B) AND (NOT (a AND b)):
(a OR B) --> (a + b - ab)
(NOT (a AND b)) --> (1 - ab)
AND these conditions together to get...
(a + b - ab)(1 - ab) = a + b - ab(1 + a + b - ab)
Computational Alternatives
If the input values are binary, then powers terms can be ignored to arrive at simplified computationally equivalent forms.
a + b - ab(1 + a + b - ab) = a + b - ab - a²b - ab² + a²b²
If x is binary (either 1 or 0), then we can disregard powers since 1² = 1 and 0² = 0...
a + b - ab - a²b - ab² + a²b² -- remove powers --> a + b - 2ab
XOR (binary) = a + b - 2ab
Binary also allows other equations to be computationally equivalent to the one above. For instance...
Given (a-b)² = a² + b² - 2ab
If input is binary we can ignore powers, so...
a² + b² - 2ab -- remove powers --> a + b - 2ab
Allowing us to write...
XOR (binary) = (a-b)²
Multi-Factor XOR
XOR = (1 - A*B*C...)(1 - (1-A)(1-B)(1-C)...)
Excel VBA example...
Function ArithmeticXOR(R As Range, Optional EvaluateEquation = True)
Dim AndOfNots As String
Dim AndGate As String
For Each c In R
AndOfNots = AndOfNots & "*(1-" & c.Address & ")"
AndGate = AndGate & "*" & c.Address
Next
AndOfNots = Mid(AndOfNots, 2)
AndGate = Mid(AndGate, 2)
'Now all we want is (Not(AndGate) AND Not(AndOfNots))
ArithmeticXOR = "(1 - " & AndOfNots & ")*(1 - " & AndGate & ")"
If EvaluateEquation Then
ArithmeticXOR = Application.Evaluate(xor2)
End If
End Function
Any n of k
These same methods can be extended to allow for any n number out of k conditions to qualify as true.
For instance, out of three variables a, b, and c, if you're willing to accept any two conditions, then you want a&b or a&c or b&c. This can be arithmetically modeled from the composite logic...
(a && b) || (a && c) || (b && c) ...
and applying our translations...
1 - (1-ab)(1-ac)(1-bc)...
This can be extended to any n number out of k conditions. There is a pattern of variable and exponent combinations, but this gets very long; however, you can simplify by ignoring powers for a binary context. The exact pattern is dependent on how n relates to k. For n = k-1, where k is the total number of conditions being tested, the result is as follows:
c1 + c2 + c3 ... ck - n*∏
Where c1 through ck are all n-variable combinations.
For instance, true if 3 of 4 conditions met would be
abc + abe + ace + bce - 3abce
This makes perfect logical sense since what we have is the additive OR of AND conditions minus the overlapping AND condition.
If you begin looking at n = k-2, k-3, etc. The pattern becomes more complicated because we have more overlaps to subtract out. If this is fully extended to the smallest value of n = 1, then we arrive at nothing more than a regular OR condition.
Thinking about Non-Binary Values and Fuzzy Region
The actual algebraic XOR equation a + b - ab(1 + a + b - ab) is much more complicated than the computationally equivalent binary equations like x + y - 2xy and (x-y)². Does this mean anything, and is there any value to this added complexity?
Obviously, for this to matter, you'd have to care about the decimal values outside of the discrete points (0,0), (0,1), (1,0), and (1,1). Why would this ever matter? Sometimes you want to relax the integer constraint for a discrete problem. In that case, you have to look at the premises used to convert logical operators to equations.
When it comes to translating Boolean logic into arithmetic, your basic building blocks are the AND and NOT operators, with which you can build both OR and XOR.
OR = (1-(1-a)(1-b)(1-c)...)
XOR = (1 - a*b*c...)(1 - (1-a)(1-b)(1-c)...)
So if you're thinking about the decimal region, then it's worth thinking about how we defined these operators and how they behave in that region.
Non-Binary Meaning of NOT
We expressed NOT as 1-x. Obviously, this simple equation works for binary values of 0 and 1, but the thing that's really cool about it is that it also provides the fractional or percent-wise compliment for values between 0 to 1. This is useful since NOT is also known as the Compliment in Boolean logic, and when it comes to sets, NOT refers to everything outside of the current set.
Non-Binary Meaning of AND
We expressed AND as x*y. Once again, obviously it works for 0 and 1, but its effect is a little more arbitrary for values between 0 to 1 where multiplication results in partial truths (decimal values) diminishing each other. It's possible to imagine that you would want to model truth as being averaged or accumulative in this region. For instance, if two conditions are hypothetically half true, is the AND condition only a quarter true (0.5 * 0.5), or is it entirely true (0.5 + 0.5 = 1), or does it remain half true ((0.5 + 0.5) / 2)? As it turns out, the quarter truth is actually true for conditions that are entirely discrete and the partial truth represents probability. For instance, will you flip tails (binary condition, 50% probability) both now AND again a second time? Answer is 0.5 * 0.5 = 0.25, or 25% true. Accumulation doesn't really make sense because it's basically modeling an OR condition (remember OR can be modeled by + when the AND condition is not present, so summation is characteristically OR). The average makes sense if you're looking at agreement and measurements, but it's really modeling a hybrid of AND and OR. For instance, ask 2 people to say on a scale of 1 to 10 how much do they agree with the statement "It is cold outside"? If they both say 5, then the truth of the statement "It is cold outside" is 50%.
Non-Binary Values in Summary
The take away from this look at non-binary values is that we can capture actual logic in our choice of operators and construct equations from the ground up, but we have to keep in mind numerical behavior. We are used to thinking about logic as discrete (binary) and computer processing as discrete, but non-binary logic is becoming more and more common and can help make problems that are difficult with discrete logic easier/possible to solve. You'll need to give thought to how values interact in this region and how to translate them into something meaningful.
"mathematical, arithmetic only representation" are not correct terms anyway. What you are looking for is a function which goes from IxI to I (domain of integer numbers).
Which restrictions would you like to have on this function? Only linear algebra? (+ , - , * , /) then it's impossible to emulate the XOR operator.
If instead you accept some non-linear operators like Max() Sgn() etc, you can emulate the XOR operator with some "simpler" operators.
Given that (a-b)(a-b) quite obviously computes xor for a single bit, you could construct a function with the floor or mod arithmetic operators to split the bits out, then xor them, then sum to recombine. (a-b)(a-b) = a2 -2·a·b + b2 so one bit of xor gives a polynomial with 3 terms.
Without floor or mod, the different bits interfere with each other, so you're stuck with looking at a solution which is a polynomial interpolation treating the input a,b as a single value: a xor b = g(a · 232 + b)
The polynomial has 264-1 terms, though will be symmetric in a and b as xor is commutative so you only have to calculate half of the coefficients. I don't have the space to write it out for you.
I wasn't able to find any solution for 32-bit unsigned integers but I've found some solutions for 2-bit integers which I was trying to use in my Prolog program.
One of my solutions (which uses exponentiation and modulo) is described in this StackOverflow question and the others (some without exponentiation, pure algebra) can be found in this code repository on Github: see different xor0 and o_xor0 implementations.
The nicest xor represention for 2-bit uints seems to be: xor(A,B) = (A + B*((-1)^A)) mod 4.
Solution with +,-,*,/ expressed as Excel formula (where cells from A2 to A5 and cells from B1 to E1 contain numbers 0-4) to be inserted in cells from A2 to E5:
(1-$A2)*(2-$A2)*(3-$A2)*($A2+B$1)/6 - $A2*(1-$A2)*(3-$A2)*($A2+B$1)/2 + $A2*(1-$A2)*(2-$A2)*($A2-B$1)/6 + $A2*(2-$A2)*(3-$A2)*($A2-B$1)/2 - B$1*(1-B$1)*(3-B$1)*$A2*(3-$A2)*(6-4*$A2)/2 + B$1*(1-B$1)*(2-B$1)*$A2*($A2-3)*(6-4*$A2)/6
It may be possible to adapt and optimize this solution for 32-bit unsigned integers. It's complicated and it uses logarithms but seems to be the most universal one as it can be used on any integer number. Additionaly, you'll have to check if it really works for all number combinations.
I do realize that this is sort of an old topic, but the question is worth answering and yes, this is possible using an algorithm. And rather than go into great detail about how it works, I'll just demonstrate with a simple example (written in C):
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
typedef unsigned long
number;
number XOR(number a, number b)
{
number
result = 0,
/*
The following calculation just gives us the highest power of
two (and thus the most significant bit) for this data type.
*/
power = pow(2, (sizeof(number) * 8) - 1);
/*
Loop until no more bits are left to test...
*/
while(power != 0)
{
result *= 2;
/*
The != comparison works just like the XOR operation.
*/
if((power > a) != (power > b))
result += 1;
a %= power;
b %= power;
power /= 2;
}
return result;
}
int main()
{
srand(time(0));
for(;;)
{
number
a = rand(),
b = rand();
printf("a = %lu\n", a);
printf("b = %lu\n", b);
printf("a ^ b = %lu\n", a ^ b);
printf("XOR(a, b) = %lu\n", XOR(a, b));
getchar();
}
}
I think this relation might help in answering your question
A + B = (A XOR B ) + 2*(A.B)
(a-b)*(a-b) is the right answer. the only one? I guess so!