Is there a canonical way of declaring a function by parts in Sympy? I tried
import sympy
import sympy.functions.special.delta_functions as special
sympy.init_printing()
x = sympy.symbols('x', real=True)
V = x*x * (special.Heaviside(x + 1) - special.Heaviside(x - 1)) \
+ (1 + 2*sympy.log(x)) * special.Heaviside(x - 1) \
+ (1 + 2*sympy.log(-x)) * special.Heaviside(-x - 1)
which defines a differentiable function, but
print(V.diff(x).simplify())
# Prints: (x*(x**2*(-DiracDelta(x - 1) + DiracDelta(x + 1)) - 2*x*(Heaviside(x - 1) - Heaviside(x + 1)) - (2*log(-x) + 1)*DiracDelta(x + 1) + (2*log(x) + 1)*DiracDelta(x - 1)) + 2*Heaviside(-x - 1) + 2*Heaviside(x - 1))/x
Is there a way to somehow tell Sympy to simplify DiracDelta(x - a)*f(x) to DiracDelta(x - a)*f(a)?
Piecewise-defined functions are implemented by Piecewise class. Your function would be expressed as
V = sympy.Piecewise((1 + 2*sympy.log(-x), x < -1),
(x**2, x < 1),
(1 + 2*sympy.log(x), True))
print(V.diff(x))
which prints Piecewise((2/x, x < -1), (2*x, x < 1), (2/x, True))
The (expr, cond) pairs in Piecewise are processed in the order given: the first cond that evaluates to True (if the preceding evaluated to False) causes the corresponding expr to be returned.
I have the following expression in Sympy
s = e0*a01*d1**2*u0 - e0*a01*d1**2*u1 - e0*a11*d1**2*u0 - e0*a11*d1**2*u1 + e0*d0*a00*d1*u1 + e0*d0*a01*d1*u0 + e0*d0*a10*d1*u0 - e0*d0*a11*d1*u1 + e0*d0*b0*u0 - e0*d0*b1*u1 + e0*d1*a00*d1*u0 - e0*d1*a01*d1*u1 - e0*d1*a10*d1*u1 - e0*d1*a11*d1*u0 - e0*d1*b0*u1 - e0*d1*b1*u0 - e1*a00*d1**2*u0 + e1*a00*d1**2*u1 + e1*a10*d1**2*u0 + e1*a10*d1**2*u1 - e1*d0*a00*d1*u0 + e1*d0*a01*d1*u1 + e1*d0*a10*d1*u1 + e1*d0*a11*d1*u0 + e1*d0*b0*u1 + e1*d0*b1*u0 + e1*d1*a00*d1*u1 + e1*d1*a01*d1*u0 + e1*d1*a10*d1*u0 - e1*d1*a11*d1*u1 + e1*d1*b0*u0 - e1*d1*b1*u1
So first I simpify it:
s = sympify(s,locals=T)
(T contains all these symbols in the string, that are non commutative). And I want to get the coefficient of
d1**2*u0
after "factoring" it. So I did the following:
e=sympify(d1**2*u0,locals=T)
collected_expr = collect(s,e,exact=True)
print(collected_expr)
coeff = collected_expr.coeff(e)
print(coeff)
The result of collected_expr is ok:
d1**2*u0*(e0*a01 - e0*a11 - e1*a00 + e1*a10) - e0*a01*d1**2*u1 - e0*a11*d1**2*u1 + e0*d0*a00*d1*u1 + e0*d0*a01*d1*u0 + e0*d0*a10*d1*u0 - e0*d0*a11*d1*u1 + e0*d0*b0*u0 - e0*d0*b1*u1 + e0*d1*a00*d1*u0 - e0*d1*a01*d1*u1 - e0*d1*a10*d1*u1 - e0*d1*a11*d1*u0 - e0*d1*b0*u1 - e0*d1*b1*u0 + e1*a00*d1**2*u1 + e1*a10*d1**2*u1 - e1*d0*a00*d1*u0 + e1*d0*a01*d1*u1 + e1*d0*a10*d1*u1 + e1*d0*a11*d1*u0 + e1*d0*b0*u1 + e1*d0*b1*u0 + e1*d1*a00*d1*u1 + e1*d1*a01*d1*u0 + e1*d1*a10*d1*u0 - e1*d1*a11*d1*u1 + e1*d1*b0*u0 - e1*d1*b1*u1
But coeff is not ok, as it returns 1, but I really want
e0*a01 - e0*a11 - e1*a00 + e1*a10
EDIT: I also tried
coeff = collected_expr.coeff(u0).coeff(d1).coeff(d1)
and
coeff = collected_expr.coeff(u0).coeff(d1**2)
But both things returned 0
The docstring of Expr.coeff says
When x is noncommutative, the coefficient to the left (default) or
right of x can be returned. The keyword 'right' is ignored when
x is commutative.
collect does not seem to be noncommutative-aware, however, so the factors that were on the right may collect to the left.
>>> var("A B", commutative=False)
(A, B)
>>> collect(A*B+B*A**2,B)
B*(A + A**2)
I have two univariate functions, f(x) and g(x), and I'd like to substitute g(x) = y to rewrite f(x) as some f2(y).
Here is a simple example that works:
In [240]: x = Symbol('x')
In [241]: y = Symbol('y')
In [242]: f = abs(x)**2 + 6*abs(x) + 5
In [243]: g = abs(x)
In [244]: f.subs({g: y})
Out[244]: y**2 + 6*y + 5
But now, if I try a slightly more complex example, it fails:
In [245]: h = abs(x) + 1
In [246]: f.subs({h: y})
Out[246]: Abs(x)**2 + 6*Abs(x) + 5
Is there a general approach that works for this problem?
The expression abs(x)**2 + 6*abs(x) + 5 does not actually contain abs(x) + 1 anywhere, so there is nothing to substitute for.
One can imagine changing it to abs(x)**2 + 5*(abs(x) + 1) + abs(x), with the substitution result being abs(x)**2 + 5*y + abs(x). Or maybe changing it to abs(x)**2 + 6*(abs(x) + 1) - 1, with the result being abs(x)**2 + 6*y - 1. There are other choices too. What should the result be?
There is no general approach to this task because it's not a well-defined task to begin with.
In contrast, the substitution f.subs(abs(x), y-1) is a clear instruction to replace all occurrences of abs(x) in the expression tree with y-1. It returns 6*y + (y - 1)**2 - 1.
The substitution above of abs(x) + 1 in abs(x)**2 + 6*abs(x) + 5 is a clear instruction too: to find exact occurrences of the expression abs(x) + 1 in the syntax tree of the expression abs(x)**2 + 6*abs(x) + 5, and replace those subtrees with the syntax tree of the expression abs(x) + 1. There is a caveat about heuristics though.
Aside: in addition to subs SymPy has a method .replace which supports wildcards, but I don't expect it to help here. In my experience, it is overeager to replace:
>>> a = Wild('a')
>>> b = Wild('b')
>>> f.replace(a*(abs(x) + 1) + b, a*y + b)
5*y/(Abs(x) + 1) + 6*y*Abs(x*y)/(Abs(x) + 1)**2 + (Abs(x*y)/(Abs(x) + 1))**(2*y/(Abs(x) + 1))
Eliminate a variable
There is no "eliminate" in SymPy. One can attempt to emulate it with solve by introducing another variable, e.g.,
fn = Symbol('fn')
solve([Eq(fn, f), Eq(abs(x) + 1, y)], [fn, x])
which attempts to solve for "fn" and "x", and therefore the solution for "fn" is an expression without x. If this works
In fact, it does not work with abs(); solving for something that sits inside an absolute value is not implemented in SymPy. Here is a workaround.
fn, ax = symbols('fn ax')
solve([Eq(fn, f.subs(abs(x), ax)), Eq(ax + 1, y)], [fn, ax])
This outputs [(y*(y + 4), y - 1)] where the first term is what you want; a solution for fn.
I've been porting Sebastiano Vigna's xorshift1024* PRNG to be compatible with the standard C++11 uniform random number generator contract and noticed some strange behavior with the jump() function he provides.
According to Vigna, a call to jump() should be equivalent to 2^512 calls to next(). Therefore a series of calls to jump() and next() should be commutative. For example, assuming the generator starts in some known state,
jump();
next();
should leave the generator in the same state as
next();
jump();
since both should be equivalent to
for (bigint i = 0; i < (bigint(1) << 512) + 1; ++i)
next();
assuming bigint is some integer type with an extremely large maximum value (and assuming you are a very, very, very patient person).
Unfortunately, this doesn't work with the reference implementation Vigna provides (which I will include at the end for posterity; in case the implementation linked above changes or is taken down in the future). When testing the first two options using the following test code:
memset(s, 0xFF, sizeof(s));
p = 0;
// jump() and/or next() calls...
std::cout << p << ';';
for (int i = 0; i < 16; ++i)
std::cout << ' ' << s[i];
calling jump() before next() outputs:
1; 9726214034378009495 13187905351877324975 10033047168458208082 990371716258730972 965585206446988056 74622805968655940 11468976784638207029 3005795712504439672 6792676950637600526 9275830639065898170 6762742930827334073 16862800599087838815 13481924545051381634 16436948992084179560 6906520316916502096 12790717607058950780
while calling next() first results in:
1; 13187905351877324975 10033047168458208082 990371716258730972 965585206446988056 74622805968655940 11468976784638207029 3005795712504439672 6792676950637600526 9275830639065898170 6762742930827334073 16862800599087838815 13481924545051381634 16436948992084179560 6906520316916502096 12790717607058950780 9726214034378009495
Clearly either my understanding of what jump() is doing is wrong, or there's a bug in the jump() function, or the jump polynomial data is wrong. Vigna claims that such a jump function can be calculated for any stride of the period, but doesn't elaborate on how to calculate it (including in his paper on xorshift* generators). How can I calculate the correct jump data to verify that there's not a typo somewhere in it?
Xorshift1024* reference implementation; http://xorshift.di.unimi.it/xorshift1024star.c
/* Written in 2014-2015 by Sebastiano Vigna (vigna#acm.org)
To the extent possible under law, the author has dedicated all copyright
and related and neighboring rights to this software to the public domain
worldwide. This software is distributed without any warranty.
See <http://creativecommons.org/publicdomain/zero/1.0/>. */
#include <stdint.h>
#include <string.h>
/* This is a fast, top-quality generator. If 1024 bits of state are too
much, try a xorshift128+ generator.
The state must be seeded so that it is not everywhere zero. If you have
a 64-bit seed, we suggest to seed a splitmix64 generator and use its
output to fill s. */
uint64_t s[16];
int p;
uint64_t next(void) {
const uint64_t s0 = s[p];
uint64_t s1 = s[p = (p + 1) & 15];
s1 ^= s1 << 31; // a
s[p] = s1 ^ s0 ^ (s1 >> 11) ^ (s0 >> 30); // b,c
return s[p] * UINT64_C(1181783497276652981);
}
/* This is the jump function for the generator. It is equivalent
to 2^512 calls to next(); it can be used to generate 2^512
non-overlapping subsequences for parallel computations. */
void jump() {
static const uint64_t JUMP[] = { 0x84242f96eca9c41dULL,
0xa3c65b8776f96855ULL, 0x5b34a39f070b5837ULL, 0x4489affce4f31a1eULL,
0x2ffeeb0a48316f40ULL, 0xdc2d9891fe68c022ULL, 0x3659132bb12fea70ULL,
0xaac17d8efa43cab8ULL, 0xc4cb815590989b13ULL, 0x5ee975283d71c93bULL,
0x691548c86c1bd540ULL, 0x7910c41d10a1e6a5ULL, 0x0b5fc64563b3e2a8ULL,
0x047f7684e9fc949dULL, 0xb99181f2d8f685caULL, 0x284600e3f30e38c3ULL
};
uint64_t t[16] = { 0 };
for(int i = 0; i < sizeof JUMP / sizeof *JUMP; i++)
for(int b = 0; b < 64; b++) {
if (JUMP[i] & 1ULL << b)
for(int j = 0; j < 16; j++)
t[j] ^= s[(j + p) & 15];
next();
}
memcpy(s, t, sizeof t);
}
OK, I'm sorry but sometimes this happens (I'm the author).
Originally the function had two memcpy(). Then I realised then a circular copy was needed. But I replaced just the first memcpy(). Stupid, stupid, stupid. All files on the site have been fixed. The arXiv copy is undergoing update. See http://xorshift.di.unimi.it/xorshift1024star.c
Incidentally: I didn't "publish" anything wrong in the scientific sense, as the jump() function is not part of the ACM Trans. Math. Soft. paper—it just has been added few weeks ago on the site and on the arXiv/WWW version. The fast publication path of the web and arXiv means that, sometimes, one distributes unpolished papers. I can only thank the reporter for reporting this bug (OK, technically StackOverflow is not reporting bugs, but I got an email, too).
Unfortunately, the unit test I had did not consider the case p ≠ 0. My main concern was that the correctness of the computed polynomial. The function, as noted above, is correct when p = 0.
As for the computation: to each generator corresponds a characteristic polynomial P(x). The jump polynomial for k is just x^k mod P(x). I use fermat to compute such powers, and then I have some scripts generating the C code.
Of course I can't test 2^512, but since my generation code works perfectly from 2 to 2^30 (the range you can easily test), I'm confident it works at 2^512, too. It's just fermat computing x^{2^512} instead of x^{2^30}. But independent verifications are more than welcome.
I have code working only for powers of the form x^{2^t}. This is what I need to compute useful jump functions. Computing polynomials modulo P(x) is not difficult, so one could conceivably have a completely generic jump function for any value, but frankly I find this totally overkill.
If anybody is interested in getting other jump polynomials, I can provide the scripts. They will be part, as it happens for all other code, of the next xorshift distribution, but I need to complete the documentation before giving them out.
For the record, the characteristic polynomial of xorshift1024* is x^1024 + x^974 + x^973 + x^972 + x^971 + x^966 + x^965 + x^964 + x^963 + x^960 + x^958 + x^957 + x^956 + x^955 + x^950 + x^949 + x^948 + x^947 + x^942 + x^941 + x^940 + x^939 + x^934 + x^933 + x^932 + x^931 + x^926 + x^925 + x^923 + x^922 + x^920 + x^917 + x^916 + x^915 + x^908 + x^906 + x^904 + x^902 + x^890 + x^886 + x^873 + x^870 + x^857 + x^856 + x^846 + x^845 + x^844 + x^843 + x^841 + x^840 + x^837 + x^835 + x^830 + x^828 + x^825 + x^824 + x^820 + x^816 + x^814 + x^813 + x^811 + x^810 + x^803 + x^798 + x^797 + x^790 + x^788 + x^787 + x^786 + x^783 + x^774 + x^772 + x^771 + x^770 + x^769 + x^768 + x^767 + x^765 + x^760 + x^758 + x^753 + x^749 + x^747 + x^746 + x^743 + x^741 + x^740 + x^738 + x^737 + x^736 + x^735 + x^728 + x^726 + x^723 + x^722 + x^721 + x^720 + x^718 + x^716 + x^715 + x^714 + x^710 + x^709 + x^707 + x^694 + x^687 + x^686 + x^685 + x^684 + x^679 + x^678 + x^677 + x^674 + x^670 + x^669 + x^667 + x^666 + x^665 + x^663 + x^658 + x^655 + x^651 + x^639 + x^638 + x^635 + x^634 + x^632 + x^630 + x^623 + x^621 + x^618 + x^617 + x^616 + x^615 + x^614 + x^613 + x^609 + x^606 + x^604 + x^601 + x^600 + x^598 + x^597 + x^596 + x^594 + x^593 + x^592 + x^590 + x^589 + x^588 + x^584 + x^583 + x^582 + x^581 + x^579 + x^577 + x^575 + x^573 + x^572 + x^571 + x^569 + x^567 + x^565 + x^564 + x^563 + x^561 + x^559 + x^557 + x^556 + x^553 + x^552 + x^550 + x^544 + x^543 + x^542 + x^541 + x^537 + x^534 + x^532 + x^530 + x^528 + x^526 + x^523 + x^521 + x^520 + x^518 + x^516 + x^515 + x^512 + x^511 + x^510 + x^508 + x^507 + x^506 + x^505 + x^504 + x^502 + x^501 + x^499 + x^497 + x^494 + x^493 + x^492 + x^491 + x^490 + x^487 + x^485 + x^483 + x^482 + x^480 + x^479 + x^477 + x^476 + x^475 + x^473 + x^469 + x^468 + x^465 + x^463 + x^461 + x^460 + x^459 + x^458 + x^455 + x^453 + x^451 + x^448 + x^447 + x^446 + x^445 + x^443 + x^438 + x^437 + x^431 + x^430 + x^429 + x^428 + x^423 + x^417 + x^416 + x^415 + x^414 + x^412 + x^410 + x^409 + x^408 + x^400 + x^398 + x^396 + x^395 + x^391 + x^390 + x^386 + x^385 + x^381 + x^380 + x^378 + x^375 + x^373 + x^372 + x^369 + x^368 + x^365 + x^360 + x^358 + x^357 + x^354 + x^350 + x^348 + x^346 + x^345 + x^344 + x^343 + x^342 + x^340 + x^338 + x^337 + x^336 + x^335 + x^333 + x^332 + x^325 + x^323 + x^318 + x^315 + x^313 + x^309 + x^308 + x^305 + x^303 + x^302 + x^300 + x^294 + x^290 + x^281 + x^279 + x^276 + x^275 + x^273 + x^272 + x^267 + x^263 + x^262 + x^261 + x^260 + x^258 + x^257 + x^256 + x^249 + x^248 + x^243 + x^242 + x^240 + x^238 + x^236 + x^233 + x^232 + x^230 + x^228 + x^225 + x^216 + x^214 + x^212 + x^210 + x^208 + x^206 + x^205 + x^200 + x^197 + x^196 + x^184 + x^180 + x^176 + x^175 + x^174 + x^173 + x^168 + x^167 + x^166 + x^157 + x^155 + x^153 + x^152 + x^151 + x^150 + x^144 + x^143 + x^136 + x^135 + x^125 + x^121 + x^111 + x^109 + x^107 + x^105 + x^92 + x^90 + x^79 + x^78 + x^77 + x^76 + x^60 + 1
tldr: I'm pretty sure there's a bug in the original code:
The memcpy in jump() must consider the p rotation too.
The author didn't test nearly as much as appropriate before publishing a paper...
Long version:
One next() call changes only one of the 16 s array elements, the one with index p. p starts at 0, gets increased each next() call, and after 15 it becomes 0 again. Let's call s[p] the "current" array element. Another (slower) possibility for implementing next() would be that the current element is always the first one, there is no p, and instead of incrementing p the whole s array is rotated (ie. the first element moves to the last position and the previous second element becomes the first).
Independent of the current p value, 16 calls to next() should result in the same p value as before, ie. the whole cycle is done and the current element is the same position as before the 16 calls. jump() should do 2^512 next(), 2^512 is a multiple of 16, so with one jump, the p value before and after it should be the same.
You probably noticed already that your two different results are only rotated one time, ie. one solution is "9726214034378009495 somethingelse" and one is "somethingelse 9726214034378009495"
...because you did one next() before/after the jump() and jump() can't handle p other than 0.
If you'd test it with 16 next() (or 32 or 0 or ...) before/after jump() instead of one, the two results are equal. The reason is, within jump, while for the s array the current element / p is handled as it is in next(), the t array is semantically rotated so that the current element is always the first one (t[j] ^= s[(j + p) & 15];). Then, right before the function terminates, memcpy(s, t, sizeof t); copies the new values from t back to s without considering the rotation at all. Just replace the memcpy with a proper loop including the p offset, then it should be fine.
(Well, but that doesn't mean jump() is really the same as 2^512 next(). But at least it could be.)
As Vigna himself said, that was actually a bug.
While working on a Java implementation, I found, if not mistaken, a small improvement on the correct implementation:
If you update t array also circularly from p to p-1, then you can just memcpy it back to the state and it will work correctly.
Moreover, the loop updating t gets tighter, as you do not need to add p + j every time. For instance:
int j = p;
do {
t[j] ^= s[j];
++j;
j &= 15;
} while (j != p);
Ok, as bcrist correctly noted, the previous code is wrong, as p changes for each bit in JUMP array. The best alternative I come up with is the following:
void jump() {
static const uint64_t JUMP[] = { 0x84242f96eca9c41dULL,
0xa3c65b8776f96855ULL, 0x5b34a39f070b5837ULL, 0x4489affce4f31a1eULL,
0x2ffeeb0a48316f40ULL, 0xdc2d9891fe68c022ULL, 0x3659132bb12fea70ULL,
0xaac17d8efa43cab8ULL, 0xc4cb815590989b13ULL, 0x5ee975283d71c93bULL,
0x691548c86c1bd540ULL, 0x7910c41d10a1e6a5ULL, 0x0b5fc64563b3e2a8ULL,
0x047f7684e9fc949dULL, 0xb99181f2d8f685caULL, 0x284600e3f30e38c3ULL
};
uint64_t t[16] = { 0 };
const int base = p;
int j = base;
for(int i = 0; i < sizeof JUMP / sizeof *JUMP; i++)
for(int b = 0; b < 64; b++) {
if (JUMP[i] & 1ULL << b) {
int k = p;
do {
t[j++] ^= s[k++];
j &= 15;
k &= 15;
} while (j != base);
}
next();
}
memcpy(s, t, sizeof t);
}
As p will have its original value in the end, this should work.
Not very sure whether it is actually an improvement in performance, as I am trading one addition for an increment and a bitwise AND.
I think it will not be slower, even if increment is as expensive as addition, due to the lack of data dependency between j and k updates. Hopefully, it may be slightly faster.
Opinions / corrections are more than welcome.
I am wanting to evaluate the expression, (an + bn + cn) % 1000000003 , in C++. I a getting overflow errors when n is very large. Can someone help me with this ? More specifically a = q + 1, b = - 2 * q and c = q - 1. I have been following the function outlined in this
Can I break (an + bn + cn) % 1000000003 into (an) % 1000000003 + (bn) % 100000003 + (cn) % 1000000003 or something similar ?
Also I cannot use anything more than unsigned long long int
You can distribute your modulo. Mathematically, this will be sound:
( ((a^n)%1000000003) + ((b^n)%100000003) + ((c^n)%1000000003) ) % 1000000003;
This will prevent you from having to compute numbers that are out of bounds, allowing you to choose larger values for n.
Proof.
Just be sure to use pow in the math.h module:
( ((pow(a, n))%1000000003)
+ ((pow(b, n))%100000003)
+ ((pow(c, n))%1000000003) ) % 1000000003;