Using Z3Py, once a model has been checked for an optimization problem, is there a way to convert ArithRef expressions into values?
Such as
y = If(x > 5, 0, 0.5 * x)
Once values have been found for x, can I get the evaluated value for y, without having to calculate again based on the given values for x?
Many thanks.
You need to evaluate, but it can be done by the model for you automatically:
from z3 import *
x = Real('x')
y = If(x > 5, 0, 0.5 * x)
s = Solver()
r = s.check()
if r == sat:
m = s.model();
print("x =", m.eval(x, model_completion=True))
print("y =", m.eval(y, model_completion=True))
else:
print("Solver said:", r)
This prints:
x = 0
y = 0
Note that we used the parameter model_completion=True since there are no constraints to force x (and consequently y) to any value in this model. If you have sufficient constraints added, you wouldn't need that parameter. (Of course, having it does not hurt.)
Say I have a List of records in elm:
[ { id = 1, magnitude = 100 }
, { id = 3, magnitude = 300 }
, { id = 2, magnitude = 200 } ]
and I want to get the record with the greatest magnitude value (300). What is a good way of doing this?
The docs gives an example of using the "maximum" -method, but it uses a simple list of integers. How is it done with records?
Update based on recommendation from #robertjlooby
There is a function called maximumBy which does exactly this in elm-community/list-extra. Example:
List.Extra.maximumBy .magnitude list
Original Answer
There are a few ways to achieve this.
This first way is more concise but it involves sorting the whole list, reversing it, then taking the head.
maxOfField : (a -> comparable) -> List a -> Maybe a
maxOfField field =
List.head << List.reverse << List.sortBy field
If you want something that's more efficient and only traverses the list once, here's a more efficient version:
maxOfField : (a -> comparable) -> List a -> Maybe a
maxOfField field =
let f x acc =
case acc of
Nothing -> Just x
Just y -> if field x > field y then Just x else Just y
in List.foldr f Nothing
An example of it in use:
list =
[ { id = 1, magnitude = 100 }
, { id = 3, magnitude = 300 }
, { id = 2, magnitude = 200 } ]
main =
text <| toString <| maxOfField .magnitude list
Here is a version that uses foldl and a default record:
bigger =
let
choose x y =
if x.magnitude > y.magnitude then
x
else
y
in
List.foldl choose {id = 0, magnitude = 0} items
Sebastian's answer add an arbitrary start value which could cause a problem if all your magnitudes were negative. I would adjust to
bigger items =
case items of
[] -> []
(h :: []) -> h
(h :: tail) ->
let
choose x y =
if x.magnitude > y.magnitude then
x
else
y
in
List.foldl choose h tail
I want to check if a cubic Bezier curve is a sub-curve of another Bezier.
I think I understand basically how to do this, express the Beziers as two cubics, in x and y, then test if the cubics are scalings or translations of each other. If the scaling and translations match that tells us the curves are sub-segments of the same curve and gives us t0 prime and t1 prime of curve B in curve As space.
But I can't quite work out how to check the cubics for equivalence.
Answer based on the following comment:
Say we take a Bezier Curve, and split it up using de Casteljau's algorithm. Obviously the result is a lot of sub-curves of the original curve.The question is how to go back, and recover the t values, and the fact that the curves are part of the same curve, given only their 4 control points
Short answer: unless you have an infinite precision machine, you can't.
So we're stuck with "error threshold" testing. Given a master curve A and a "hopefully subcurve" curve B, run through the things that need to be true if B was a subcurve of A:
If B is a true subcurve then its start and end point lie on curve A. So check if that's true, within some error threshold. If they don't, then B is not a subcurve of A.
If B is a true subcurve then the derivatives at B's start and end points are the same as the derivatives for the corresponding coordinates on A. So check if that's true, within some error threshold. If they're not, B is not a subcurve of A.
If B is a true subcurve then the second derivatives at B's start an end points are the same as the second derivatives for the corresponding coordinates on A. So check if that's true, within some error threshold. If they're not, B is not a subcurve of A.
If all of these hold, we can be reasonably sure that B is a subcurve of A.
Also, since we need to come up with t values in order to check whether a point lies on A, and what derivative of A is at that point, we already know the t values that define the interval on A that maps to the full curve B.
Here's the working code.
(You can find cubic root finders quite easily)
/*
A = p3 + 3.0 * p1 - 3.0 * p2 - p0;
B = 3.0 * p0 - 6.0 * p1 + 3.0 * p2;
C = 3.0 * p1 - 3.0 * p0;
D = p0;
*/
bool CurveIsSubCurve(BezierCurve bez, BezierCurve sub, double epsilon, double *t)
{
int Nr;
double tcand[6];
int i, ii;
double ts[6], te[6];
int Ns = 0;
int Ne = 0;
Vector2 p;
/*
Take two bites at the cherry. The points may have slight errors, and a small error in x or y could represent a big error in
t. However with any luck either x or y will be close
*/
Nr = cubic_roots(bez.Ax(), bez.Bx(), bez.Cx(), bez.Dx() - sub.P0().x, tcand);
Nr += cubic_roots(bez.Ay(), bez.By(), bez.Cy(), bez.Dy() - sub.P0().y, tcand + Nr);
for(i=0;i<Nr;i++)
{
p = bez.Eval(tcand[i]);
if(fabs(p.x - sub.P0().x) < epsilon && fabs(p.y - sub.P0().y) < epsilon)
{
ts[Ns++] = tcand[i];
}
}
/* same thing of sub curve end point */
Nr = cubic_roots(bez.Ax(), bez.Bx(), bez.Cx(), bez.Dx() - sub.P3().x, tcand);
Nr += cubic_roots(bez.Ay(), bez.By(), bez.Cy(), bez.Dy() - sub.P3().y, tcand + Nr);
for(i=0;i<Nr;i++)
{
p = bez.Eval(tcand[i]);
if(fabs(p.x - sub.P3().x) < epsilon && fabs(p.y - sub.P3().y) < epsilon)
{
te[Ne++] = tcand[i];
}
}
/* do an all by all to get matches (Ns, Ne will be small, but if
we have a degenerate, i.e. a loop, the loop intersection point is
where the mother curve is quite likely to be cut, so test everything*/
for(i = 0; i < Ns; i++)
{
double s,d;
double Ax, Bx, Cx, Dx;
double Ay, By, Cy, Dy;
for(ii=0;ii<Ne;ii++)
{
s = (te[ii] - ts[i]);
d = ts[i];
/* now substitute back */
Ax = bez.Ax() *s*s*s;
Bx = bez.Ax() *2*s*s*d + bez.Ax()*s*s*d + bez.Bx()*s*s;
Cx = bez.Ax()*s*d*d + bez.Ax()*2*s*d*d + bez.Bx()*2*s*d + bez.Cx() * s;
Dx = bez.Ax() *d*d*d + bez.Bx()*d*d + bez.Cx()*d + bez.Dx();
Ay = bez.Ay() *s*s*s;
By = bez.Ay() *2*s*s*d + bez.Ay()*s*s*d + bez.By()*s*s;
Cy = bez.Ay()*s*d*d + bez.Ay()*2*s*d*d + bez.By()*2*s*d + bez.Cy() * s;
Dy = bez.Ay() *d*d*d + bez.By()*d*d + bez.Cy()*d + bez.Dy();
if(fabs(Ax - sub.Ax()) < epsilon && fabs(Bx - sub.Bx()) < epsilon &&
fabs(Cx - sub.Cx()) < epsilon && fabs(Dx - sub.Dx()) < epsilon &&
fabs(Ay - sub.Ay()) < epsilon && fabs(By - sub.By()) < epsilon &&
fabs(Cy - sub.Cy()) < epsilon && fabs(Dy - sub.Dy()) < epsilon)
{
if(t)
{
t[0] = ts[i];
t[1] = te[ii];
}
return true;
}
}
}
return false;
}
A friend and I are going back and forth with brain-teasers and I have no idea how to solve this one. My assumption is that it's possible with some bitwise operators, but not sure.
In C, with bitwise operators:
#include<stdio.h>
int add(int x, int y) {
int a, b;
do {
a = x & y;
b = x ^ y;
x = a << 1;
y = b;
} while (a);
return b;
}
int main( void ){
printf( "2 + 3 = %d", add(2,3));
return 0;
}
XOR (x ^ y) is addition without carry. (x & y) is the carry-out from each bit. (x & y) << 1 is the carry-in to each bit.
The loop keeps adding the carries until the carry is zero for all bits.
int add(int a, int b) {
const char *c=0;
return &(&c[a])[b];
}
No + right?
int add(int a, int b)
{
return -(-a) - (-b);
}
CMS's add() function is beautiful. It should not be sullied by unary negation (a non-bitwise operation, tantamount to using addition: -y==(~y)+1). So here's a subtraction function using the same bitwise-only design:
int sub(int x, int y) {
unsigned a, b;
do {
a = ~x & y;
b = x ^ y;
x = b;
y = a << 1;
} while (a);
return b;
}
Define "best". Here's a python version:
len(range(x)+range(y))
The + performs list concatenation, not addition.
Java solution with bitwise operators:
// Recursive solution
public static int addR(int x, int y) {
if (y == 0) return x;
int sum = x ^ y; //SUM of two integer is X XOR Y
int carry = (x & y) << 1; //CARRY of two integer is X AND Y
return addR(sum, carry);
}
//Iterative solution
public static int addI(int x, int y) {
while (y != 0) {
int carry = (x & y); //CARRY is AND of two bits
x = x ^ y; //SUM of two bits is X XOR Y
y = carry << 1; //shifts carry to 1 bit to calculate sum
}
return x;
}
Cheat. You could negate the number and subtract it from the first :)
Failing that, look up how a binary adder works. :)
EDIT: Ah, saw your comment after I posted.
Details of binary addition are here.
Note, this would be for an adder known as a ripple-carry adder, which works, but does not perform optimally. Most binary adders built into hardware are a form of fast adder such as a carry-look-ahead adder.
My ripple-carry adder works for both unsigned and 2's complement integers if you set carry_in to 0, and 1's complement integers if carry_in is set to 1. I also added flags to show underflow or overflow on the addition.
#define BIT_LEN 32
#define ADD_OK 0
#define ADD_UNDERFLOW 1
#define ADD_OVERFLOW 2
int ripple_add(int a, int b, char carry_in, char* flags) {
int result = 0;
int current_bit_position = 0;
char a_bit = 0, b_bit = 0, result_bit = 0;
while ((a || b) && current_bit_position < BIT_LEN) {
a_bit = a & 1;
b_bit = b & 1;
result_bit = (a_bit ^ b_bit ^ carry_in);
result |= result_bit << current_bit_position++;
carry_in = (a_bit & b_bit) | (a_bit & carry_in) | (b_bit & carry_in);
a >>= 1;
b >>= 1;
}
if (current_bit_position < BIT_LEN) {
*flags = ADD_OK;
}
else if (a_bit & b_bit & ~result_bit) {
*flags = ADD_UNDERFLOW;
}
else if (~a_bit & ~b_bit & result_bit) {
*flags = ADD_OVERFLOW;
}
else {
*flags = ADD_OK;
}
return result;
}
Go based solution
func add(a int, b int) int {
for {
carry := (a & b) << 1
a = a ^ b
b = carry
if b == 0 {
break
}
}
return a
}
same solution can be implemented in Python as follows, but there is some problem about number represent in Python, Python has more than 32 bits for integers. so we will use a mask to obtain the last 32 bits.
Eg: if we don't use mask we won't get the result for numbers (-1,1)
def add(a,b):
mask = 0xffffffff
while b & mask:
carry = a & b
a = a ^ b
b = carry << 1
return (a & mask)
Why not just incremet the first number as often, as the second number?
The reason ADD is implememted in assembler as a single instruction, rather than as some combination of bitwise operations, is that it is hard to do. You have to worry about the carries from a given low order bit to the next higher order bit. This is stuff that the machines do in hardware fast, but that even with C, you can't do in software fast.
Here's a portable one-line ternary and recursive solution.
int add(int x, int y) {
return y == 0 ? x : add(x ^ y, (x & y) << 1);
}
I saw this as problem 18.1 in the coding interview.
My python solution:
def foo(a, b):
"""iterate through a and b, count iteration via a list, check len"""
x = []
for i in range(a):
x.append(a)
for i in range(b):
x.append(b)
print len(x)
This method uses iteration, so the time complexity isn't optimal.
I believe the best way is to work at a lower level with bitwise operations.
In python using bitwise operators:
def sum_no_arithmetic_operators(x,y):
while True:
carry = x & y
x = x ^ y
y = carry << 1
if y == 0:
break
return x
Adding two integers is not that difficult; there are many examples of binary addition online.
A more challenging problem is floating point numbers! There's an example at http://pages.cs.wisc.edu/~smoler/x86text/lect.notes/arith.flpt.html
Was working on this problem myself in C# and couldn't get all test cases to pass. I then ran across this.
Here is an implementation in C# 6:
public int Sum(int a, int b) => b != 0 ? Sum(a ^ b, (a & b) << 1) : a;
Implemented in same way as we might do binary addition on paper.
int add(int x, int y)
{
int t1_set, t2_set;
int carry = 0;
int result = 0;
int mask = 0x1;
while (mask != 0) {
t1_set = x & mask;
t2_set = y & mask;
if (carry) {
if (!t1_set && !t2_set) {
carry = 0;
result |= mask;
} else if (t1_set && t2_set) {
result |= mask;
}
} else {
if ((t1_set && !t2_set) || (!t1_set && t2_set)) {
result |= mask;
} else if (t1_set && t2_set) {
carry = 1;
}
}
mask <<= 1;
}
return (result);
}
Improved for speed would be below::
int add_better (int x, int y)
{
int b1_set, b2_set;
int mask = 0x1;
int result = 0;
int carry = 0;
while (mask != 0) {
b1_set = x & mask ? 1 : 0;
b2_set = y & mask ? 1 : 0;
if ( (b1_set ^ b2_set) ^ carry)
result |= mask;
carry = (b1_set & b2_set) | (b1_set & carry) | (b2_set & carry);
mask <<= 1;
}
return (result);
}
It is my implementation on Python. It works well, when we know the number of bytes(or bits).
def summ(a, b):
#for 4 bytes(or 4*8 bits)
max_num = 0xFFFFFFFF
while a != 0:
a, b = ((a & b) << 1), (a ^ b)
if a > max_num:
b = (b&max_num)
break
return b
You can do it using bit-shifting and the AND operation.
#include <stdio.h>
int main()
{
unsigned int x = 3, y = 1, sum, carry;
sum = x ^ y; // Ex - OR x and y
carry = x & y; // AND x and y
while (carry != 0) {
carry = carry << 1; // left shift the carry
x = sum; // initialize x as sum
y = carry; // initialize y as carry
sum = x ^ y; // sum is calculated
carry = x & y; /* carry is calculated, the loop condition is
evaluated and the process is repeated until
carry is equal to 0.
*/
}
printf("%d\n", sum); // the program will print 4
return 0;
}
The most voted answer will not work if the inputs are of opposite sign. The following however will. I have cheated at one place, but only to keep the code a bit clean. Any suggestions for improvement welcome
def add(x, y):
if (x >= 0 and y >= 0) or (x < 0 and y < 0):
return _add(x, y)
else:
return __add(x, y)
def _add(x, y):
if y == 0:
return x
else:
return _add((x ^ y), ((x & y) << 1))
def __add(x, y):
if x < 0 < y:
x = _add(~x, 1)
if x > y:
diff = -sub(x, y)
else:
diff = sub(y, x)
return diff
elif y < 0 < x:
y = _add(~y, 1)
if y > x:
diff = -sub(y, x)
else:
diff = sub(y, x)
return diff
else:
raise ValueError("Invalid Input")
def sub(x, y):
if y > x:
raise ValueError('y must be less than x')
while y > 0:
b = ~x & y
x ^= y
y = b << 1
return x
Here is the solution in C++, you can find it on my github here: https://github.com/CrispenGari/Add-Without-Integers-without-operators/blob/master/main.cpp
int add(int a, int b){
while(b!=0){
int sum = a^b; // add without carrying
int carry = (a&b)<<1; // carrying without adding
a= sum;
b= carry;
}
return a;
}
// the function can be writen as follows :
int add(int a, int b){
if(b==0){
return a; // any number plus 0 = that number simple!
}
int sum = a ^ b;// adding without carrying;
int carry = (a & b)<<1; // carry, without adding
return add(sum, carry);
}
This can be done using Half Adder.
Half Adder is method to find sum of numbers with single bit.
A B SUM CARRY A & B A ^ B
0 0 0 0 0 0
0 1 1 0 0 1
1 0 1 0 0 1
1 1 0 1 0 0
We can observe here that SUM = A ^ B and CARRY = A & B
We know CARRY is always added at 1 left position from where it was
generated.
so now add ( CARRY << 1 ) in SUM, and repeat this process until we get
Carry 0.
int Addition( int a, int b)
{
if(B==0)
return A;
Addition( A ^ B, (A & B) <<1 )
}
let's add 7 (0111) and 3 (0011) answer will be 10 (1010)
A = 0100 and B = 0110
A = 0010 and B = 1000
A = 1010 and B = 0000
final answer is A.
I implemented this in Swift, I am sure someone will benefit from
var a = 3
var b = 5
var sum = 0
var carry = 0
while (b != 0) {
sum = a ^ b
carry = a & b
a = sum
b = carry << 1
}
print (sum)
You can do it iteratively or recursively. Recursive:-
public int getSum(int a, int b) {
return (b==0) ? a : getSum(a^b, (a&b)<<1);
}
Iterative:-
public int getSum(int a, int b) {
int c=0;
while(b!=0) {
c=a&b;
a=a^b;
b=c<<1;
}
return a;
}
time complexity - O(log b)
space complexity - O(1)
for further clarifications if not clear, refer leetcode or geekForGeeks explanations.
I'll interpret this question as forbidding the +,-,* operators but not ++ or -- since the question specified operator and not character (and also because that's more interesting).
A reasonable solution using the increment operator is as follows:
int add(int a, int b) {
if (b == 0)
return a;
if (b > 0)
return add(++a, --b);
else
return add(--a, ++b);
}
This function recursively nudges b towards 0, while giving a the same amount to keep the sum the same.
As an additional challenge, let's get rid of the second if block to avoid a conditional jump. This time we'll need to use some bitwise operators:
int add(int a, int b) {
if(!b)
return a;
int gt = (b > 0);
int m = -1 << (gt << 4) << (gt << 4);
return (++a & --b & 0)
| add( (~m & a--) | (m & --a),
(~m & b++) | (m & ++b)
);
}
The function trace is identical; a and b are nudged between each add call just like before.
However, some bitwise magic is employed to drop the if statement while continuing to not use +,-,*:
A mask m is set to 0xFFFFFFFF (-1 in signed decimal) if b is positive, or 0x00000000 if b is negative.
The reason for shifting the mask left by 16 twice instead a single shift left by 32 is because shifting by >= the size of the value is undefined behavior.
The final return takes a bit of thought to fully appreciate:
Consider this technique to avoid a branch when deciding between two values. Of the values, one is multiplied by the boolean while the other is multiplied by the inverse, and the results are summed like so:
double naiveFoodPrice(int ownPetBool) {
if(ownPetBool)
return 23.75;
else
return 10.50;
}
double conditionlessFoodPrice(int ownPetBool) {
double result = ownPetBool*23.75 + (!ownPetBool)*10.50;
}
This technique works great in most cases. For us, the addition operator can easily be substituted for the bitwise or | operator without changing the behavior.
The multiplication operator is also not allowed for this problem. This is the reason for our earlier mask value - a bitwise and & with the mask will achieve the same effect as multiplying by the original boolean.
The nature of the unary increment and decrement operators halts our progress.
Normally, we would easily be able to choose between an a which was incremented by 1 and an a which was decremented by 1.
However, because the increment and decrement operators modify their operand, our conditionless code will end up always performing both operations - meaning that the values of a and b will be tainted before we finish using them.
One way around this is to simply create new variables which each contain the original values of a and b, allowing a clean slate for each operation. I consider this boring, so instead we will adjust a and b in a way that does not affect the rest of the code (++a & --b & 0) in order to make full use of the differences between x++ and ++x.
We can now get both possible values for a and b, as the unary operators modifying the operands' values now works in our favor. Our techniques from earlier help us choose the correct versions of each, and we now have a working add function. :)
Python codes:
(1)
add = lambda a,b : -(-a)-(-b)
use lambda function with '-' operator
(2)
add= lambda a,b : len(list(map(lambda x:x,(i for i in range(-a,b)))))