I would like to know if it is possible to bound the range of values of a universally quantified variable in Z3.
For example, let's assume that I have a variable of type Real called "time" which is used to model the time in the system.
Let's say that I have an assertion which says that the value of some unary function "func1" shall always be between 1 and 100. The function takes the input the time variable. Expressed in Z3, I have encoded the property as following:
ForAll(time, And(func1(time) >= 1, func1(time) <= 100))
Please note that I explicitly need the time variable to be universally quantified, because I want the Z3 go give me unsat if I inject property of following type:
Exists(time, func1(time) == 101)
As far as my understanding goes for Z3, all the constants have mathematical (theoretical) and not computer (practical) implementation i.e. their values are not bound (unfortunately I cannot point to the resource where I have read this at the moment). Assume that with time I model time in my systems, and according to the system constrains it cannot run for more than x hours, which I can use and say that value of time is between 0 and x*60'*60 to give the maximum execution time in seconds. I know that I can assert the allowed values for time with the following assertion:
And(time >= 0, time <= x*60*60)
but will it affect the universal quantification given in 1?
Consequently, this should lead to a situation where if property 2 is injected and for value of time I specify x*60*60 + 1, it should not be unset since the ForAll is valid only for the values of time.
but will it affect the universal quantification given in 1)?
Note that
ForAll(time, And(func1(time) >= 1, func1(time) <= 100))
treats the variable "time" as bound. The formula has the same meaning as:
ForAll(xx, And(func1(xx) >= 1, func1(xx) <= 100))
When you assert the above, the meaning is that any instantiation of xx holds (is asserted). In particular, you can instantiate the quantified variable with the free variable "time" and in particular, you can instantiate with x*60*60+1 producing the assertion:
And(func1(x*60*60+1) >= 1, func1(x*60*60+1) <= 100)
Suppose you wanted to say that
And(func1(xx) >= 1, func1(xx) <= 100))
holds for every value xx between 0 and x*60*60, then you can write:
ForAll(xx, Implies(And(xx >= 0, xx <= x*60*60), And(func1(xx) >= 1, func1(xx) <= 100)))
(unfortunately I cannot point to the resource where I have read this at the moment).
Reasonable logic text books or foundations of computer science books should explain this in depth. Z3 supports a standard first-order many-sorted logic (with background theories).
Related
I would like to assert that the most significant digit of a number is a particular value, but I don't actually know the length of the number. If it was the least significant digit, I know I could use the python mod (%) to check for it. But with an unknown number of digits, I'm unsure of how I could check this in z3.
For example, I may know that the left most digit is a 9, such as 9x, or 9xx, or 9xxx etc.
Thanks so much in advance
The generic way to do this would be to convert to a string and check that the first character matches:
from z3 import *
s = Solver()
n = Int('n')
s.add(SubString(IntToStr(n), 0, 1) == "9")
r = s.check()
if r == sat:
m = s.model()
print("n =", m[n])
else:
print("Solver said:", r)
This prints:
n = 9
Note that IntToStr expects its argument to be non-negative, so if you need to support negative numbers, you'll have to write extra code to accommodate for that. See https://smtlib.cs.uiowa.edu/theories-UnicodeStrings.shtml for details.
Aside While this will accomplish what you want in its generality, it may not be the most efficient way to encode this constraint. Since it goes through strings, the constraints generated might cause performance issues. If you have an upper limit on your number, it might be more efficient to code it explicitly. For instance, if you know your number is less than a 1000, I'd code it as (pseudocode):
n == 9 || n >= 90 && n <= 99 || n >= 900 && n <= 999
etc. until you have the range you needed covered. This would lead to much simpler constraints and perform a lot better in general. Note that this'll work even if you don't know the exact length, but have an upper bound on it. But of course, it all depends on what you are trying to achieve and what else you know about the number itself.
I am trying to create a EA in mql4, but in OrderSend function, when i use some value instead of zero it show ordersend error 130. Please help to solve this problem
Code line is
int order = OrderSend("XAUUSD",OP_SELL,0.01,Bid,3,Bid+20*0.01,tp,"",0,0,Red);
error number 130 means Invalid stops.
so that means there is a problem with the stops you set with the ordersend function.
I suggest you set it like that:
int order = OrderSend("XAUUSD",OP_SELL,0.01,Bid,3,Bid+20*Point,tp,"",0,0,Red);
so you could use Point instead of hard coding it.
and to check what is the error number means. I think you could refer to: https://book.mql4.com/appendix/errors
You should know that there exists a minimum Stop Loss Size (mSLS) given in pips. "mSLS" changes with the currency and broker. So, you need to put in the OnInit() procedure of your EA a variable to get it:
int mSLS = MarketInfo(symbol,MODE_STOPLEVEL);
The distance (in pips) from your Order Open Price (OOP) and the Stop-Loss Price (SLP) can not be smaller than mSLS value.
I will try to explain a general algorithm I use for opening orders in my EAs, and then apply the constrain on Stop-Loss level (at step 3):
Step 1. I introduce a flag (f) for the type of operation I will open, being:
f = 1 for Buy, and
f = -1 for Sell
You know that there are mql4 constants OP_SELL=1 and OP_BUY=0 (https://docs.mql4.com/constants/tradingconstants/orderproperties).
Once I have defined f, I set my operation type variable to
int OP_TYPE = int(0.5((1+f)*OP_BUY+(1-f)*OP_SELL));
that takes value OP_TYPE=OP_BUY when f=1, while OP_TYPE=OP_SELL when f=-1.
NOTE: Regarding the color of the orders I put them in an array
color COL[2]= {clrBlue,clrRed};
then, having OP_TYPE, I set
color COLOR=COL[OP_TYPE];
Step 2. Similarly, I set the Order Open Price as
double OOP = int(0.5*((1+f)*Ask+(1-f)*Bid));
which takes value OOP=Ask when f=1, while OOP=Bid when f=-1.
Step 3. Then, given my desired Stop Loss in pips (an external POSITIVE parameter of my EA, I named sl) I make sure that sl > SLS. In other words, I check
if (sl <= mSLS) // I set my sl as the minimum allowed
{
sl = 1 + mSLS;
}
Step 4. Then I calculate the Stop-Loss Price of the order as
double SLP = OOP - f * sl * Point;
Step 5. Given my desired Take Profit in pips (an external POSITIVE parameter of my EA, I named tp) I calculate the Take-Profit Price (TPP) of the order as
double TPP = OOP + f * tp * Point;
OBSERVATION: I can not affirm, but, according to mql4 documentation, the minimum distance rule between the stop-loss limit prices and the open price also applies to the take profit limit price. In this case, a "tp" check-up needs to be done, similar to that of the sl check-up, above. that is, before calculating TPP it must be executed the control lines below
if (tp <= mSLS) // I set my tp as the minimum allowed
{
tp = 1 + mSLS;
}
Step 5. I call for order opening with a given lot size (ls) and slippage (slip) on the operating currency pair (from where I get the Ask and Bid values)
float ls = 0.01;
int slip = 3; //(pips)
int order = OrderSend(Symbol(),OP_TYPE,ls,OOP,slip,SLP,TPP,"",0,0,COLOR);
Note that with these few lines it is easy to build a function that opens orders of any type under your command, in any currency pair you are operating, without receiving error message 130, passing to the function only 3 parameters: f, sl and tp.
It is worth including in the test phase of your EA a warning when the sl is corrected for being less than the allowed, this will allow you to increase its value so that it does not violate the stop-loss minimum value rule, while you have more control about the risk of its operations. Remember that the "sl" parameter defines how much you will lose if the order fails because the asset price ended up varying too much in the opposite direction to what was expected.
I hope I could help!
Whilst the other two answers are not necessarily wrong (and I will not go over the ground they have already covered), for completeness of answers, they fail to mention that for some brokers (specifically ECN brokers) you must open your order first, without setting a stop loss or take profit. Once the order is opened, use OrderModify() to set you stop loss and/or take profit.
I work with SPSS and have difficulty finding/generating a syntax for counting cases.
I have about 120 cases and five variables. I need to know the count /proportion of cases where just one, more than one, or all of the cases have a value of 1 (dichotomous variable). Then I need to compute a new variable that shows the number / proportion of cases which include all of the aforementioned cases (also dichotomous).
For example case number one: var1=1, var2=1, var3=1, var4=0, var5=0 --> newvariable=1.
Case number two: var1=0, var2=0, var3=0, var4=0, var5=0 --> newvariable=1.
And so on...
Can anybody help me with a syntax?
Help would much appreciated!
Here we can use the sum of the variables to determine your conditions. So using a scratch variable that is the sum, we can see if it is equal to 1, more than 1 or 5 in your example.
compute #sum = SUM(var1 to var5).
compute just_one = (#sum = 1).
compute more_one = (#sum > 1).
compute all_one = (#sum = 5).
Similarly, all_one could be computed using the ANY command to evaluate if any zeroes exist, i.e. compute all_one = ANY(0,var1 to var5).. These code snippets assume that var1 to var5 are contiguous in the data frame, if not they just need to be replaced with var1,var2,var3,var4,var5 in all given instances.
You could read up on the logical function ANY in the Command Syntax Reference manual, if you negated a test for ANY with "0", then that is effectively a test for all "1"s. Use of the COUNT command would be another approach.
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.
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.