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I have the below code:
loc_sucs(R, C, result(A, S)) :-
loc_sucs(R, C, S),
Rm is R - 1,
A \= move-north;
R = 0;
o(Rm, C);
r(Rm, C, S);
Rp is R + 1,
dim(Z, _),
A \= move-south;
R = Z;
o(Rp, C);
r(Rp, C, S);
Cp is C + 1,
dim(_, W),
A \= move-east;
C = W;
o(R, Cp);
r(R, Cp, S);
Cm is C - 1,
A \= move-west;
C = 0;
o(R, Cm);
r(R, Cm, S).
And I'm getting the singleton warning for Rm, Cm, Rp, Cp, Z and W. Why am I getting this warning if all of these variables are used more than once?
I want to implement the following function f(x,y) in Prolog
f(x,y) = a*x+b*y
where a = 1 if x > 0; a = -1 if x < 0; a = 0 if x = 0
and b = -1 if y > 0; b = 1 if y < 0; b = 0 if y = 0
For example,
f(2,-1) = 1*2 + 1*(-1) = 1
f(-2,-1) = (-1)*(-2) + (-1)*1 = 1
f(0,0) = 0*0 + 0*0 = 0
Any one can help?
How about using the following formulation?
f(X,Y,Result) :-
Result is abs(X) - abs(Y).
Let's run some queries:
?- f(0,0,0).
true.
?- f(-2,-1,1).
true.
?- f(2,-1,1).
true.
(Assuming you have a typo when defining y, y>0 not y>=0.)
You need to define a relation between the input vars and the result of the function. Prolog can then answer yes/true with substitutions or no/false.
f(X,Y,Answer):-
a_is(X,A),
b_is(Y,B),
Answer is A*X+B*Y.
a_is(X,1):-
X>0.
a_is(X,-1):-
X<0.
a_is(0,0).
b_is(Y,1):-
Y<0.
b_is(Y,-1):-
Y>0.
b_is(0,0).
Example:
?-f(2,-1,Answer).
Answer =1;
false.
Shouldn't be much more complicated than this one-liner:
f(X,Y,Z) :- Z is sign(X)*X + -sign(Y)*Y
I was making a prolog knowledge base to implement geometry rules. When testing if a rectangle had a right angle, I found two answers.
?- rect_tri(triangle(line(point(0,0),point(0,1)),line(point(0,1),point(1,0)),line(point(1,0),point(0,0)))).
true ;
false.
Here is the kwnoledge base:
point(X,Y).
line(X,Y) :- X = point(A,B), Y = point(C,D), not(X = Y).
len(X,R) :- X = line(P,Q), P = point(A,B), Q = point(C,D), not(P = Q),
R is sqrt((A - C) * (A - C) + (B - D) * (B - D)).
triangle(X,Y,Z) :- X = point(A,B), Y = point(C,D), Z = point(E,F),
not(X = Y), not(X = Z), not(Y = Z),
L1 = line(X,Y), L2 = line(X,Z), L3 = line(Y,Z),
len(L1,G), len(L2,H), len(L3,I),
G + H > I, G + I > H, H + I > G.
triangle(X,Y,Z) :- X = line(A,B), Y = line(B,C), line(A,C),
len(X,G), len(Y,H), len(Z,I),
G + H > I, G + I > H, H + I > G.
rect_tri(X) :- X = triangle(A,B,C), len(A,G), len(B,H), len(C,I),
(G is sqrt(H * H + I * I);
H is sqrt(G * G + I * I);
I is sqrt(H * H + G * G)).
When tracing, I found that the answer true comes when prolog hits the line H is sqrt(G * G + I * I), and false when it evaluates the last line.
I don't want the last evaluation to occur, because I want it to exit when a true has been found.
Daniel comment probably shows the most sensible way to solve your problem. Some other option...
in modern compilers there is the if/then/else construct:
rect_tri(X) :- X = triangle(A,B,C), len(A,G), len(B,H), len(C,I),
( G is sqrt(H * H + I * I)
-> true
; H is sqrt(G * G + I * I)
-> true
; I is sqrt(H * H + G * G)
).
You could as well use cuts (old fashioned way, somewhat more readable here):
rect_tri(X) :- X = triangle(A,B,C), len(A,G), len(B,H), len(C,I),
( G is sqrt(H * H + I * I), !
; H is sqrt(G * G + I * I), !
; I is sqrt(H * H + G * G)
).
While converting a project from Python to C#, I found some interesting differences in the syntax families. Yet I got stuck, still unable to understand and comprehend the dissimilar behavior of comparison operator in C#.
During the course of curing this curiosity, I considered few languages of C-syntax family; C, C++, C#, Java, Javascript.. and verified the behavior. Here is how it transpired:
Let a=2, b=3, c=4, d=5;
Now, consider the following expressions:
a < a < a // returns true
c < b < a // returns true
c > b > a // returns false
a < c > b // returns false
If it was due to the right-associativity, then the following code in JavaScript shouldn't act like:
console.info(a < false); // returns false
console.info(a < a); // returns false
console.info(a < a < a); // returns true, as opposed to returning false
Here is the C/C++ version
int main(){
int a=2, b=3, c=4, d=5;
printf("%s\n","false\0true"+6*(a < a < a)); // returns true
printf("%s\n","false\0true"+6*(c < b < a)); // returns true
printf("%s\n","false\0true"+6*(c > b > a)); // returns false
printf("%s\n","false\0true"+6*(a < c > b)); // returns false
return 0;
}
Except in Python, where
a < a < a // returns false
c < b < a // returns false
c > b > a // returns true
a < c > b // returns true
Can anyone explain why C-family of languages and Python are computing the expressions differently?
Because Python uses a slightly interpretation of your input:
Formally, if a, b, c, ..., y, z are expressions and op1, op2, ..., opN are comparison operators, then a op1 b op2 c ... y opN z is equivalent to a op1 b and b op2 c and ... y opN z, except that each expression is evaluated at most once.
This means your lines will be interpreted as
a < a < a = a < a and a < a // returns false
c < b < a = c < b and b < a // returns false
c > b > a = c > b and b > a // returns true
a < c > b = a < c and c > b // returns true
In C-style languages, an comparison expression will evaluate to either false (integer value 0) or true (integer value 1). So in C it will behave like
a < a < a = (a < a) < a = 0 < a // returns true
c < b < a = (c < b) < a = 0 < a // returns true
c > b > a = (c > b) > a = 1 > a // returns false
a < c > b = (a < c) > b = 0 > b // returns false
Note that almost all languages define operators with a boolean return value, but since boolean values can be implicit converted to zero or one the proposition above is still valid:
// C++ example
struct myComparableObject{
int data;
bool operator<(const myComparableObject& o){
return data < o.data;
}
};
myComparableObject a, b;
a.data = 2;
b.data = 3;
int c = 5;
a < b; // true
a < c; // error, a cannot be converted to int / unknown operator
a.data < c; // true
a < b < c; // true, as this is equal to
// (a < b) < c = false < c = 0 < c
For example JavaScript will actually use ToNumber in order to compare two non-string objects, see [ECMAScript p78, 11.8.5 The Abstract Relational Comparison Algorithm], where ToNumber(false) is zero and ToNumber(true) === 1.
The comparison x < y, where x and y are values, produces true, false, or undefined[...]
Let px be the result of calling ToPrimitive(x, hint Number).
Let py be the result of calling ToPrimitive(y, hint Number).
If it is not the case that both Type(px) is String and Type(py) is String, then
a. Let nx be the result of calling ToNumber(px). Because px and py are primitive values evaluation
order is not important.
b. Let ny be the result of calling ToNumber(py).
c. If nx is NaN, return undefined.
d. If ny is NaN, return undefined.
e. If nx and ny are the same Number value, return false.
f. If nx is +0 and ny is -0, return false.
g. If nx is -0 and ny is +0, return false.
h. If nx is +infty, return false.
i. If ny is +infty, return true.
j. If ny is -infty, return false.
k. If nx is -infty, return true.
l. If the mathematical value of nx is less than the mathematical value of ny —note that these
mathematical values are both finite and not both zero— return true. Otherwise, return false.
It's because a < a < a is evaluated like ((a < a) < a) which then becomes 0 < a which is true (when a >= 0)
If you run the following, the first expression changes to false.
int main(){
int a=0, b=3, c=4, d=5;
printf("%s\n","false\0true"+6*(a < a < a)); // returns false
printf("%s\n","false\0true"+6*(c < b < a)); // returns true
printf("%s\n","false\0true"+6*(c > b > a)); // returns false
printf("%s\n","false\0true"+6*(a < c > b)); // returns false
return 0;
}
Whereas in Python, a < a < a becomes a < a and a < a. So it's not comparing the result of a < a. If you add parentheses to your statements, you'll get the C-like behavior again.
(a < a) < a ## returns true
(c < b) < a ## returns true
(c > b) > a ## returns false
(a < c) > b ## returns false
Problem Hey folks. I'm looking for some advice on python performance. Some background on my problem:
Given:
A (x,y) mesh of nodes each with a value (0...255) starting at 0
A list of N input coordinates each at a specified location within the range (0...x, 0...y)
A value Z that defines the "neighborhood" in count of nodes
Increment the value of the node at the input coordinate and the node's neighbors. Neighbors beyond the mesh edge are ignored. (No wrapping)
BASE CASE: A mesh of size 1024x1024 nodes, with 400 input coordinates and a range Z of 75 nodes.
Processing should be O(x*y*Z*N). I expect x, y and Z to remain roughly around the values in the base case, but the number of input coordinates N could increase up to 100,000. My goal is to minimize processing time.
Current results Between my start and the comments below, we've got several implementations.
Running speed on my 2.26 GHz Intel Core 2 Duo with Python 2.6.1:
f1: 2.819s
f2: 1.567s
f3: 1.593s
f: 1.579s
f3b: 1.526s
f4: 0.978s
f1 is the initial naive implementation: three nested for loops.
f2 is replaces the inner for loop with a list comprehension.
f3 is based on Andrei's suggestion in the comments and replaces the outer for with map()
f is Chris's suggestion in the answers below
f3b is kriss's take on f3
f4 is Alex's contribution.
Code is included below for your perusal.
Question How can I further reduce the processing time? I'd prefer sub-1.0s for the test parameters.
Please, keep the recommendations to native Python. I know I can move to a third-party package such as numpy, but I'm trying to avoid any third party packages. Also, I've generated random input coordinates, and simplified the definition of the node value updates to keep our discussion simple. The specifics have to change slightly and are outside the scope of my question.
thanks much!
**`f1` is the initial naive implementation: three nested `for` loops.**
def f1(x,y,n,z):
rows = [[0]*x for i in xrange(y)]
for i in range(n):
inputX, inputY = (int(x*random.random()), int(y*random.random()))
topleft = (inputX - z, inputY - z)
for i in xrange(max(0, topleft[0]), min(topleft[0]+(z*2), x)):
for j in xrange(max(0, topleft[1]), min(topleft[1]+(z*2), y)):
if rows[i][j] <= 255: rows[i][j] += 1
f2 is replaces the inner for loop with a list comprehension.
def f2(x,y,n,z):
rows = [[0]*x for i in xrange(y)]
for i in range(n):
inputX, inputY = (int(x*random.random()), int(y*random.random()))
topleft = (inputX - z, inputY - z)
for i in xrange(max(0, topleft[0]), min(topleft[0]+(z*2), x)):
l = max(0, topleft[1])
r = min(topleft[1]+(z*2), y)
rows[i][l:r] = [j+(j<255) for j in rows[i][l:r]]
UPDATE: f3 is based on Andrei's suggestion in the comments and replaces the outer for with map(). My first hack at this requires several out-of-local-scope lookups, specifically recommended against by Guido: local variable lookups are much faster than global or built-in variable lookups I hardcoded all but the reference to the main data structure itself to minimize that overhead.
rows = [[0]*x for i in xrange(y)]
def f3(x,y,n,z):
inputs = [(int(x*random.random()), int(y*random.random())) for i in range(n)]
rows = map(g, inputs)
def g(input):
inputX, inputY = input
topleft = (inputX - 75, inputY - 75)
for i in xrange(max(0, topleft[0]), min(topleft[0]+(75*2), 1024)):
l = max(0, topleft[1])
r = min(topleft[1]+(75*2), 1024)
rows[i][l:r] = [j+(j<255) for j in rows[i][l:r]]
UPDATE3: ChristopeD also pointed out a couple improvements.
def f(x,y,n,z):
rows = [[0] * y for i in xrange(x)]
rn = random.random
for i in xrange(n):
topleft = (int(x*rn()) - z, int(y*rn()) - z)
l = max(0, topleft[1])
r = min(topleft[1]+(z*2), y)
for u in xrange(max(0, topleft[0]), min(topleft[0]+(z*2), x)):
rows[u][l:r] = [j+(j<255) for j in rows[u][l:r]]
UPDATE4: kriss added a few improvements to f3, replacing min/max with the new ternary operator syntax.
def f3b(x,y,n,z):
rn = random.random
rows = [g1(x, y, z) for x, y in [(int(x*rn()), int(y*rn())) for i in xrange(n)]]
def g1(x, y, z):
l = y - z if y - z > 0 else 0
r = y + z if y + z < 1024 else 1024
for i in xrange(x - z if x - z > 0 else 0, x + z if x + z < 1024 else 1024 ):
rows[i][l:r] = [j+(j<255) for j in rows[i][l:r]]
UPDATE5: Alex weighed in with his substantive revision, adding a separate map() operation to cap the values at 255 and removing all non-local-scope lookups. The perf differences are non-trivial.
def f4(x,y,n,z):
rows = [[0]*y for i in range(x)]
rr = random.randrange
inc = (1).__add__
sat = (0xff).__and__
for i in range(n):
inputX, inputY = rr(x), rr(y)
b = max(0, inputX - z)
t = min(inputX + z, x)
l = max(0, inputY - z)
r = min(inputY + z, y)
for i in range(b, t):
rows[i][l:r] = map(inc, rows[i][l:r])
for i in range(x):
rows[i] = map(sat, rows[i])
Also, since we all seem to be hacking around with variations, here's my test harness to compare speeds: (improved by ChristopheD)
def timing(f,x,y,z,n):
fn = "%s(%d,%d,%d,%d)" % (f.__name__, x, y, z, n)
ctx = "from __main__ import %s" % f.__name__
results = timeit.Timer(fn, ctx).timeit(10)
return "%4.4s: %.3f" % (f.__name__, results / 10.0)
if __name__ == "__main__":
print timing(f, 1024, 1024, 400, 75)
#add more here.
On my (slow-ish;-) first-day Macbook Air, 1.6GHz Core 2 Duo, system Python 2.5 on MacOSX 10.5, after saving your code in op.py I see the following timings:
$ python -mtimeit -s'import op' 'op.f1()'
10 loops, best of 3: 5.58 sec per loop
$ python -mtimeit -s'import op' 'op.f2()'
10 loops, best of 3: 3.15 sec per loop
So, my machine is slower than yours by a factor of a bit more than 1.9.
The fastest code I have for this task is:
def f3(x=x,y=y,n=n,z=z):
rows = [[0]*y for i in range(x)]
rr = random.randrange
inc = (1).__add__
sat = (0xff).__and__
for i in range(n):
inputX, inputY = rr(x), rr(y)
b = max(0, inputX - z)
t = min(inputX + z, x)
l = max(0, inputY - z)
r = min(inputY + z, y)
for i in range(b, t):
rows[i][l:r] = map(inc, rows[i][l:r])
for i in range(x):
rows[i] = map(sat, rows[i])
which times as:
$ python -mtimeit -s'import op' 'op.f3()'
10 loops, best of 3: 3 sec per loop
so, a very modest speedup, projecting to more than 1.5 seconds on your machine - well above the 1.0 you're aiming for:-(.
With a simple C-coded extensions, exte.c...:
#include "Python.h"
static PyObject*
dopoint(PyObject* self, PyObject* args)
{
int x, y, z, px, py;
int b, t, l, r;
int i, j;
PyObject* rows;
if(!PyArg_ParseTuple(args, "iiiiiO",
&x, &y, &z, &px, &py, &rows
))
return 0;
b = px - z;
if (b < 0) b = 0;
t = px + z;
if (t > x) t = x;
l = py - z;
if (l < 0) l = 0;
r = py + z;
if (r > y) r = y;
for(i = b; i < t; ++i) {
PyObject* row = PyList_GetItem(rows, i);
for(j = l; j < r; ++j) {
PyObject* pyitem = PyList_GetItem(row, j);
long item = PyInt_AsLong(pyitem);
if (item < 255) {
PyObject* newitem = PyInt_FromLong(item + 1);
PyList_SetItem(row, j, newitem);
}
}
}
Py_RETURN_NONE;
}
static PyMethodDef exteMethods[] = {
{"dopoint", dopoint, METH_VARARGS, "process a point"},
{0}
};
void
initexte()
{
Py_InitModule("exte", exteMethods);
}
(note: I haven't checked it carefully -- I think it doesn't leak memory due to the correct interplay of reference stealing and borrowing, but it should be code inspected very carefully before being put in production;-), we could do
import exte
def f4(x=x,y=y,n=n,z=z):
rows = [[0]*y for i in range(x)]
rr = random.randrange
for i in range(n):
inputX, inputY = rr(x), rr(y)
exte.dopoint(x, y, z, inputX, inputY, rows)
and the timing
$ python -mtimeit -s'import op' 'op.f4()'
10 loops, best of 3: 345 msec per loop
shows an acceleration of 8-9 times, which should put you in the ballpark you desire. I've seen a comment saying you don't want any third-party extension, but, well, this tiny extension you could make entirely your own;-). ((Not sure what licensing conditions apply to code on Stack Overflow, but I'll be glad to re-release this under the Apache 2 license or the like, if you need that;-)).
1. A (smaller) speedup could definitely be the initialization of your rows...
Replace
rows = []
for i in range(x):
rows.append([0 for i in xrange(y)])
with
rows = [[0] * y for i in xrange(x)]
2. You can also avoid some lookups by moving random.random out of the loops (saves a little).
3. EDIT: after corrections -- you could arrive at something like this:
def f(x,y,n,z):
rows = [[0] * y for i in xrange(x)]
rn = random.random
for i in xrange(n):
topleft = (int(x*rn()) - z, int(y*rn()) - z)
l = max(0, topleft[1])
r = min(topleft[1]+(z*2), y)
for u in xrange(max(0, topleft[0]), min(topleft[0]+(z*2), x)):
rows[u][l:r] = [j+(j<255) for j in rows[u][l:r]]
EDIT: some new timings with timeit (10 runs) -- seems this provides only minor speedups:
import timeit
print timeit.Timer("f1(1024,1024,400,75)", "from __main__ import f1").timeit(10)
print timeit.Timer("f2(1024,1024,400,75)", "from __main__ import f2").timeit(10)
print timeit.Timer("f(1024,1024,400,75)", "from __main__ import f3").timeit(10)
f1 21.1669280529
f2 12.9376120567
f 11.1249599457
in your f3 rewrite, g can be simplified. (Can also be applied to f4)
You have the following code inside a for loop.
l = max(0, topleft[1])
r = min(topleft[1]+(75*2), 1024)
However, it appears that those values never change inside the for loop. So calculate them once, outside the loop instead.
Based on your f3 version I played with the code. As l and r are constants you can avoid to compute them in g1 loop. Also using new ternary if instead of min and max seems to be consistently faster. Also simplified expression with topleft. On my system it appears to be about 20% faster using with the code below.
def f3b(x,y,n,z):
rows = [g1(x, y, z) for x, y in [(int(x*random.random()), int(y*random.random())) for i in range(n)]]
def g1(x, y, z):
l = y - z if y - z > 0 else 0
r = y + z if y + z < 1024 else 1024
for i in xrange(x - z if x - z > 0 else 0, x + z if x + z < 1024 else 1024 ):
rows[i][l:r] = [j+(j<255) for j in rows[i][l:r]]
You can create your own Python module in C, and control the performance as you want:
http://docs.python.org/extending/