Given this function (written in pseudocode) what's the time complexity? Trying it out i would say the time complexity is θ(n^3) since we need to traverse the tree first θ(n) then multiply the contribute of ANCESTOR which is θ(n) and the contribute of ADDTOQUEUE θ(n). Is this correct?
====================================================================
ANCESTOR does a number of operations proportionate to the depth of the node
ADDTOHEAD does a constant number of operations
ADDTOQUEUE does a number of operations proportionate to the lenght of the list
`FUNCTION(T) /* T is a tree filled with integers */
L.head = NULL /* L is a new empty linked list (of integers) */
RIC_FUNC(T.root,L)
return L
REC_FUNC(v,L)
if(v==NULL)return
if(ANCESTOR(v))
ADDTOQUEUE(L,v.info)
else
ADDHEAD(L,v.info)
REC_FUNC(v.left,L)
REC_FUNC(v.right,L)
``
Basically, you are correct: O(n^3).
But, I have a feeling (also not proveable by you pseudo-code) that ANCESTOR and ADDHEAD are on the contrary sides - which means at the first ran L is short and v is high therefor ANCESTOR will be long and ADDHEAD short and after some steps they will gets equal and from that point v is lower and L is bigger so ANCESTOR will be fast but ADDHEAD will be long.
If my assumption is correct, and the "speed" of ADDHEAD and ANCESTOR are the same complexity in different direction then you're complexity is O(n^2) (as in every node you will get: 1+n, 2+(n-1), 3+(n-2) ... which conclude each step in n+1).
Related
I am surprising to notice that it is somehow difficult to obtain a correct fit of interaction function from gam().
To be more specific, I want to estimate an additive function:
y=m_1(x)+m_2(z)+m_{12}(x,z)+u,
where m_1(x)=x^2, m_2(z)=z^2,m_{12}(x,z)=xz. The following code generate this model:
test1 <- function(x,z,sx=1,sz=1) {
#--m1(x) function
m.x<-x^2
m.x<-m.x-mean(m.x)
#--m2(z) function
m.z<-z^2
m.z<-m.z-mean(m.z)
#--m12(x,z) function
m.xz<-x*z
m.xz<-m.xz-mean(m.xz)
m<-m.x+m.z+m.xz
return(list(m=m,m.x=m.x,m.z=m.z,m.xz=m.xz))
}
n <- 1000
a=0
b=2
x <- runif(n,a,b)/20
z <- runif(n,a,b)
u <- rnorm(n,0,0.5)
model<-test1(x,z)
y <- model$m + u
So I use gam() by fitting the model as
b3 <- gam(y~ ti(x) + ti(z) + ti(x,z))
vis.gam(b3);title("tensor anova")
#---extracting basis matrix
B.f3<-model.matrix.gam(b3)
#---extracting series estimator
b3.hat<-b3$coefficients
Question: when I plot the estimated function by gam()above against its true function, I end up with
par(mfrow=c(1,3))
#---m1(x)
B.x<-B.f3[,c(2:5)]
b.x.hat<-b3.hat[c(2:5)]
plot(x,B.x%*%b.x.hat)
points(x,model$m.x,col='red')
legend('topleft',c('Estimate','True'),lty=c(1,1),col=c('black','red'))
#---m2(z)
B.z<-B.f3[,c(6:9)]
b.z.hat<-b3.hat[c(6:9)]
plot(z,B.z%*%b.z.hat)
points(z,model$m.z,col='red')
legend('topleft',c('Estimate','True'),lty=c(1,1),col=c('black','red'))
#---m12(x,z)
B.xz<-B.f3[,-c(1:9)]
b.xz.hat<-b3.hat[-c(1:9)]
plot(x,B.xz%*%b.xz.hat)
points(x,model$m.xz,col='red')
legend('topleft',c('Estimate','True'),lty=c(1,1),col=c('black','red'))
However, the function estimate of m_1(x) is largely different from x^2, and the interaction function estimate m_{12}(x,z) is also largely different from xz defined in test1 above. The results are the same if I use predict(b3).
I really can't figure it out. Can anybody help me out by explaining why the results end up with this? Greatly appreciate it!
First, the problem of the above issue is not due to the package, of course. It is closely related to the identification conditions of the smooth functions. One common practice is to impose the assumptions that E(mj(.))=0 for all individual function j=1,...,d, and E(m_ij(x_i,x_j)|x_i)=E(m_ij(x_i,x_j)|x_j)=0 for i not equal to j. Those conditions require one to employ centered basis function in series estimator, which has been done already in GAM package. However, in my case above, function m(x,z)=x*z defined in test1 does not satisfy the above identification assumptions, since the integral of x*z with respect to either x or z is not zero when x and z have range from zero to two.
Furthermore, series estimator allows the individual and interaction function to be identified if one impose m(0)=0 or m(0,x_j)=m(x_i,0)=0. This can be readily achieved if we center the basis function around zero. I have tried both cases, and they work well whenever DGP satisfies the identification conditions.
I'm looking for an SSE Bitwise OR between components of same vector. (Editor's note: this is potentially an X-Y problem, see below for the real comparison logic.)
I am porting some SIMD logic from SPU intrinsics. It has an instruction
spu_orx(a)
Which according to the docs
spu_orx: OR word across d = spu_orx(a) The four word elements of
vector a are logically Ored. The result is returned in word element 0
of vector d. All other elements (1,2,3) of d are assigned a value of
zero.
How can I do that with SSE 2 - 4 involving minimum instruction? _mm_or_ps is what I got here.
UPDATE:
Here is the scenario from SPU based code:
qword res = spu_orx(spu_or(spu_fcgt(x, y), spu_fcgt(z, w)))
So it first ORs two 'greater' comparisons, then ORs its result.
Later couples of those results are ANDed to get final comparison value.
This is effectively doing (A||B||C||D||E||F||G||H) && (I||J||K||L||M||N||O||P) && ... where A..D are the 4x 32-bit elements of the fcgt(x,y) and so on.
Obviously vertical _mm_or_ps of _mm_cmp_ps results is a good way to reduce down to 1 vector, but then what? Shuffle + OR, or something else?
UPDATE 1
Regarding "but then what?"
I perform
qword res = spu_orx(spu_or(spu_fcgt(x, y), spu_fcgt(z, w)))
On SPU it goes like this:
qword aRes = si_and(res, res1);
qword aRes1 = si_and(aRes, res2);
qword aRes2 = si_and(aRes1 , res3);
return si_to_uint(aRes2 );
several times on different inputs,then AND those all into a single result,which is finally cast to integer 0 or 1 (false/true test)
SSE4.1 PTEST bool any_nonzero = !_mm_testz_si128(v,v);
That would be a good way to horizontal OR + booleanize a vector into a 0/1 integer. It will compile to multiple instructions, and ptest same,same is 2 uops on its own. But once you have the result as a scalar integer, scalar AND is even cheaper than any vector instruction, and you can branch on the result directly because it sets integer flags.
#include <immintrin.h>
bool any_nonzero_bit(__m128i v) {
return !_mm_testz_si128(v,v);
}
On Godbolt with gcc9.1 -O3 -march=nehalem:
any_nonzero(long long __vector(2)):
ptest xmm0, xmm0 # 2 uops
setne al # 1 uop with false dep on old value of RAX
ret
This is only 3 uops on Intel for a horizontal OR into a single bit in an integer register. AMD Ryzen ptest is only 1 uop so it's even better.
The only risk here is if gcc or clang creates false dependencies by not xor-zeroing eax before doing a setcc into AL. Usually gcc is pretty fanatical about spending extra uops to break false dependencies so I don't know why it doesn't here. (I did check with -march=skylake and -mtune=generic in case it was relying on Nehalem partial-register renaming for -march=nehalem. Even -march=znver1 didn't get it to xor-zero EAX before the ptest.)
It would be nice if we could avoid the _mm_or_ps and have PTEST do all the work. But even if we consider inverting the comparisons, the vertical-AND / horizontal-OR behaviour doesn't let us check something about all 8 elements of 2 vectors, or about any of those 8 elements.
e.g. Can PTEST be used to test if two registers are both zero or some other condition?
// NOT USEFUL
// 1 if all the vertical pairs AND to zero.
// but 0 if even one vertical AND result is non-zero
_mm_testz_si128( _mm_castps_si128(_mm_cmpngt_ps(x,y)),
_mm_castps_si128(_mm_cmpngt_ps(z,w)));
I mention this only to rule it out and save you the trouble of considering this optimization idea. (#chtz suggested it in comments. Inverting the comparison is a good idea that can be useful for other ways of doing things.)
Without SSE4.1 / delaying the horizontal OR
We might be able to delay horizontal ORing / booleanizing until after combining some results from multiple vectors. This makes combining more expensive (imul or something), but saves 2 uops in the vector -> integer stage vs. PTEST.
x86 has cheap vector mask->integer bitmap with _mm_movemask_ps. Especially if you ultimately want to branch on the result, this might be a good idea. (But x86 doesn't have a || instruction that booleanizes its inputs either so you can't just & the movemask results).
One thing you can do is integer multiply movemask results: x * y is non-zero iff both inputs are non-zero. Unlike x & y which can be false for 0b0101 &0b1010for example. (Our inputs are 4-bit movemask results andunsigned` is 32-bit so we have some room before we overflow). AMD Bulldozer family has an integer multiply that isn't fully pipelined so this could be a bottleneck on old AMD CPUs. Using just 32-bit integers is also good for some low-power CPUs with slow 64-bit multiply.
This might be good if throughput is more of a bottleneck than latency, although movmskps can only run on one port.
I'm not sure if there are any cheaper integer operations that let us recover the logical-AND result later. Adding doesn't work; the result is non-zero even if only one of the inputs was non-zero. Concatenating the bits together (shift+or) is also of course like an OR if we eventually just test for any non-zero bit. We can't just bitwise AND because 2 & 1 == 0, unlike 2 && 1.
Keeping it in the vector domain
Horizontal OR of 4 elements takes multiple steps.
The obvious way is _mm_movehl_ps + OR, then another shuffle+OR. (See Fastest way to do horizontal float vector sum on x86 but replace _mm_add_ps with _mm_or_ps)
But since we don't actually need an exact bitwise-OR when our inputs are compare results, we just care if any element is non-zero. We can and should think of the vectors as integer, and look at integer instructions like 64-bit element ==. One 64-bit element covers/aliases two 32-bit elements.
__m128i cmp = _mm_castps_si128(cmpps_result); // reinterpret: zero instructions
// SSE4.1 pcmpeqq 64-bit integer elements
__m128i cmp64 = _mm_cmpeq_epi64(cmp, _mm_setzero_si128()); // -1 if both elements were zero, otherwise 0
__m128i swap = _mm_shuffle_epi32(cmp64, _MM_SHUFFLE(1,0, 3,2)); // copy and swap, no movdqa instruction needed even without AVX
__m128i bothzero = _mm_and_si128(cmp64, swap); // both halves have the full result
After this logical inversion, ORing together multiple bothzero results will give you the AND of multiple conditions you're looking for.
Alternatively, SSE4.1 _mm_minpos_epu16(cmp64) (phminposuw) will tell us in 1 uop (but 5 cycle latency) if either qword is zero. It will place either 0 or 0xFFFF in the lowest word (16 bits) of the result in this case.
If we inverted the original compares, we could use phminposuw on that (without pcmpeqq) to check if any are zero. So basically a horizontal AND across the whole vector. (Assuming that it's elements of 0 / -1). I think that's a useful result for inverted inputs. (And saves us from using _mm_xor_si128 to flip the bits).
An alternative to pcmpeqq (_mm_cmpeq_epi64) would be SSE2 psadbw against a zeroed vector to get 0 or non-zero results in the bottom of each 64-bit element. It won't be a mask, though, it's 0xFF * 8. Still, it's always that or 0 so you can still AND it. And it doesn't invert.
in fibonacci series let's assume nth fibonacci term is T. F(n)=T. but i want to write a a program that will take T as input and return n that means which term is it in the series( taken that T always will be a fibonacci number. )i want to find if there lies an efficient way to find it.
The easy way would be to simply start generating Fibonacci numbers until F(i) == T, which has a complexity of O(T) if implemented correctly (read: not recursively). This method also allows you to make sure T is a valid Fibonacci number.
If T is guaranteed to be a valid Fibonacci number, you can use approximation rules:
Formula
It looks complicated, but it's not. The point is: from a certain point on, the ratio of F(i+1)/F(i) becomes a constant value. Since we're not generating Fibonacci Numbers but are merely finding the "index", we can drop most of it and just realize the following:
breakpoint := f(T)
Any f(i) where i > T = f(i-1)*Ratio = f(T) * Ratio^(i-T)
We can get the reverse by simply taking Log(N, R), R being Ratio. By adjusting for the inaccuracy for early numbers, we don't even have to select a breakpoint (if you do: it's ~ correct for i > 17).
The Ratio is, approximately, 1.618034. Taking the log(1.618034) of 6765 (= F(20)), we get a value of 18.3277. The accuracy remains the same for any higher Fibonacci numbers, so simply rounding down and adding 2 gives us the exact Fibonacci "rank" (provided that F(1) = F(2) = 1).
The first step is to implement fib numbers in a non-recursive way such as
fib1=0;fib2=1;
for(i=startIndex;i<stopIndex;i++)
{
if(fib1<fib2)
{
fib1+=fib2;
if(fib1=T) return i;
if(fib1>T) return -1;
}
else
{
fib2+=fib1;
if(fib2=T) return i;
if(fib2>t) return -1;
}
}
Here startIndex would be set to 3 stopIndex would be set to 10000 or so. To cut down in the iteration, you can also select 2 seed number that are sequential fib numbers further down the sequence. startIndex is then set to the next index and do the computation with an appropriate adjustment to the stopIndex. I would suggest breaking the sequence up in several section depending on machine performance and the maximum expected input to minimize the run time.
Suppose you've got a single linked list of size N, and you want to perform an operation on every element, beginning at the end.
I've come up with the following pseudocode:
while N > 0
Current = LinkedList
for 0 to N
Current = Current.tail
end
Operation(Current.head)
N := N-1
end
Now I've got to determine which Big-O this algorithm is.
Supposing that Operation() is O(1), I think it's something like this:
N + (N-1) + (N-2) + ... + (N-(N-1)) + 1
But I'm not sure what Big-O that actually is. I think it is definitely smaller than O(N^2), but I don't think you can say its O(N) either ...
Your equation is basically that of the triangular numbers, and sums to N(N+1)/2. I'll leave you to determine the O() from that!
A quicker way to do this is to construct a new list that is the reverse of the original list, and then perform the operations on that.
Your algorithm is O(n^2) as you suggest in your post. You can do it in O(n), though.
It's important to remember that Big-O notation is an upper bound on the algorithm's time complexity.
1+2+3+...+n = n*(n+1)/2 = 0.5*n^2+O(n)
This is O(n^2), and O(n^2) is tight, i.e. there is no lower runtime order that'd contain your runtime.
A faster algorithm that works from front-to-back could have O(n) instead of O(n^2)
Your runtime analysis is correct, the runtime is 1 + 2 + ... + N which is a sum of the arithmetic progression and therefore = (N²-N) / 2.
Is there any input that SHA-1 will compute to a hex value of fourty-zeros, i.e. "0000000000000000000000000000000000000000"?
Yes, it's just incredibly unlikely. I.e. one in 2^160, or 0.00000000000000000000000000000000000000000000006842277657836021%.
Also, becuase SHA1 is cryptographically strong, it would also be computationally unfeasible (at least with current computer technology -- all bets are off for emergent technologies such as quantum computing) to find out what data would result in an all-zero hash until it occurred in practice. If you really must use the "0" hash as a sentinel be sure to include an appropriate assertion (that you did not just hash input data to your "zero" hash sentinel) that survives into production. It is a failure condition your code will permanently need to check for. WARNING: Your code will permanently be broken if it does.
Depending on your situation (if your logic can cope with handling the empty string as a special case in order to forbid it from input) you could use the SHA1 hash ('da39a3ee5e6b4b0d3255bfef95601890afd80709') of the empty string. Also possible is using the hash for any string not in your input domain such as sha1('a') if your input has numeric-only as an invariant. If the input is preprocessed to add any regular decoration then a hash of something without the decoration would work as well (eg: sha1('abc') if your inputs like 'foo' are decorated with quotes to something like '"foo"').
I don't think so.
There is no easy way to show why it's not possible. If there was, then this would itself be the basis of an algorithm to find collisions.
Longer analysis:
The preprocessing makes sure that there is always at least one 1 bit in the input.
The loop over w[i] will leave the original stream alone, so there is at least one 1 bit in the input (words 0 to 15). Even with clever design of the bit patterns, at least some of the values from 0 to 15 must be non-zero since the loop doesn't affect them.
Note: leftrotate is circular, so no 1 bits will get lost.
In the main loop, it's easy to see that the factor k is never zero, so temp can't be zero for the reason that all operands on the right hand side are zero (k never is).
This leaves us with the question whether you can create a bit pattern for which (a leftrotate 5) + f + e + k + w[i] returns 0 by overflowing the sum. For this, we need to find values for w[i] such that w[i] = 0 - ((a leftrotate 5) + f + e + k)
This is possible for the first 16 values of w[i] since you have full control over them. But the words 16 to 79 are again created by xoring the first 16 values.
So the next step could be to unroll the loops and create a system of linear equations. I'll leave that as an exercise to the reader ;-) The system is interesting since we have a loop that creates additional equations until we end up with a stable result.
Basically, the algorithm was chosen in such a way that you can create individual 0 words by selecting input patterns but these effects are countered by xoring the input patterns to create the 64 other inputs.
Just an example: To make temp 0, we have
a = h0 = 0x67452301
f = (b and c) or ((not b) and d)
= (h1 and h2) or ((not h1) and h3)
= (0xEFCDAB89 & 0x98BADCFE) | (~0x98BADCFE & 0x10325476)
= 0x98badcfe
e = 0xC3D2E1F0
k = 0x5A827999
which gives us w[0] = 0x9fb498b3, etc. This value is then used in the words 16, 19, 22, 24-25, 27-28, 30-79.
Word 1, similarly, is used in words 1, 17, 20, 23, 25-26, 28-29, 31-79.
As you can see, there is a lot of overlap. If you calculate the input value that would give you a 0 result, that value influences at last 32 other input values.
The post by Aaron is incorrect. It is getting hung up on the internals of the SHA1 computation while ignoring what happens at the end of the round function.
Specifically, see the pseudo-code from Wikipedia. At the end of the round, the following computation is done:
h0 = h0 + a
h1 = h1 + b
h2 = h2 + c
h3 = h3 + d
h4 = h4 + e
So an all 0 output can happen if h0 == -a, h1 == -b, h2 == -c, h3 == -d, and h4 == -e going into this last section, where the computations are mod 2^32.
To answer your question: nobody knows whether there exists an input that produces all zero outputs, but cryptographers expect that there are based upon the simple argument provided by daf.
Without any knowledge of SHA-1 internals, I don't see why any particular value should be impossible (unless explicitly stated in the description of the algorithm). An all-zero value is no more or less probable than any other specific value.
Contrary to all of the current answers here, nobody knows that. There's a big difference between a probability estimation and a proof.
But you can safely assume it won't happen. In fact, you can safely assume that just about ANY value won't be the result (assuming it wasn't obtained through some SHA-1-like procedures). You can assume this as long as SHA-1 is secure (it actually isn't anymore, at least theoretically).
People doesn't seem realize just how improbable it is (if all humanity focused all of it's current resources on finding a zero hash by bruteforcing, it would take about xxx... ages of the current universe to crack it).
If you know the function is safe, it's not wrong to assume it won't happen. That may change in the future, so assume some malicious inputs could give that value (e.g. don't erase user's HDD if you find a zero hash).
If anyone still thinks it's not "clean" or something, I can tell you that nothing is guaranteed in the real world, because of quantum mechanics. You assume you can't walk through a solid wall just because of an insanely low probability.
[I'm done with this site... My first answer here, I tried to write a nice answer, but all I see is a bunch of downvoting morons who are wrong and can't even tell the reason why are they doing it. Your community really disappointed me. I'll still use this site, but only passively]
Contrary to all answers here, the answer is simply No.
The hash value always contains bits set to 1.