What does mpn_invert_3by2 in mini-gmp do? - gmp

I really wonder the answer to this question. and I used python to calculate:
def inv(a):
return ((1 << 96) - 1) // (a << 32)
Why is python's result different from mpn_invert_limb's?
/* The 3/2 inverse is defined as
m = floor( (B^3-1) / (B u1 + u0)) - B
*/
B should be 2^32
And what is the use of mpn invert_limb?
Python code:
def inv(a):
return ((1 << 96) - 1) // (a << 32)
a = 165536
b = inv(a)
print(b & (2 ** 32 - 1))
C code:
int main()
{
mp_limb_t a = 16636;
mp_limb_t b;
b = mpn_invert_limb(a);
printf("a = %u, b = %u\n", a, b);
printf("a = %X, b = %X\n", a, b);
return 0;
}
Python output:
3522819686
C output:
a = 165536, b = 3165475657
a = 286A0, b = BCAD5349


Calling mpn_invert_limb only makes sense when your input is full-sized (has its high bit set). If the input isn't full sized then the quotient would be too big to fit in a single limb whereas in the full sized case its only 1 bit too big hence the subtraction of B in the definition.
I actually can't even run with your input of 16636, I get a division by 0 because this isn't even half a limb. Anyway, if I replace that value by a<<17 then I get a match between your Python and C. This shifting to make the top bit be set is what mini-gmp does in its usage of the function.

Related

Equation of a straight line passing through two points and distance point straight - C

I have 3 points, p1(x1,y1), p2(x2,y2) and p3(x3,y3).
I know that the equation of the straight line passing throught p1 and p2 should be obtained from (x-x1)/(x2-x1)=(y-y1)/(y2-y1), but how can I put it inside a variable?
And after that, how can I calculate the distance from p3 to this line?

first of all transform your line equation to another form
Ax + By + C = 0 // keep int mind that A^2 + B^2 != 0 (this means that A or B are unable to be equall to zero in one moment)
it will be (y1- y2)x + (x2 - x1)y + (x1y2 - x2y1) = 0;
if you have an equation of a line Ax + By + C = 0
the distance form point M(Mx, My) to your line will be
d = abs(A * Mx + B * My + C)/sqrt(A * A + B * B)
you're welcome

A line can be represented by a tuple of 3 numbers a, b and c using the form ax + by = c (or using the slope-intercept form too). So what you can do is to create a class named Line that stores 3 public members of int or float type.
You can then implement a distance function within your class using the standard point-line distance formula.
In C#, you could do something like:
class Line
{
public float a,b,c;
public float Distance(Point p)
{
return Math.Abs(a * p.X + b * p.Y + c)/Math.Sqrt(a * a + b * b)
}
}
C version should be pretty much the same.

The following code computes distance between two points.
#include <stdio.h>
#include <math.h>
int main()
{
double p1x,p1y,p2x,p2y,p1p2_distance;
//Initialize variables here
// Distance d between p1 and p2 is given by: d = sqrt((p2.x-p1.x)^2 + (p2.y-
p1.y)^2)
p1p2_distance = sqrt(pow((p2x-p1x),2)+pow((p2y-p1y),2)); // same formula can be used to calculate distance between p1,p3 and p2,p3.
printf("Distance between p1 and p2: %f \n", &p1p2_distance);
return 0;
}
On Linux compile with: gcc distance.c -o distance -lm

Check if Bezier Curve is sub-curve of another Bezier

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;
}

How can I express the term ( a < b ? 0 : 1 ) using only bitwise or arithmetic operators?

Assuming the variables a and b are 32 bit integers is there a way to compare the two and return 0 if a < b and 1 if a >= b without using a ternary nor a comparison operator?

You can do this, but it's not pretty.
The problem is, as mentioned, overflow. But you can work around it, like this:
return ((x - y) ^ ((x ^ y) & ((x - y) ^ x))) >>> 31;
For the signed version, or
return ((~x & y) | ((~x | y) & (x - y))) >>> 31;
For the unsigned version.
The >>> there is an unsigned right shift, as in Java.
Use them with x = b, y = a to make them match your function.
You can find those (and others) in Hacker's Delight, under the name "comparison predicates".

First we take the difference of the two numbers. Then we are checking the sign using the fact that numbers are rappresented as complement of two.
int Compare(int a, int b)
{
int c = a - b;
int k = (c >> 31) & 1 ^ 1;
return k;
}

SHA 256 pseuedocode?

I've been trying to work out how SHA-256 works. One thing I've been doing for other algorithms is I've worked out a sort of step by step pseudocode function for the algorithm.
I've tried to do the same for SHA256 but thus far I'm having quite a bit of trouble.
I've tried to work out how the Wikipedia diagram works but besides the text part explaining the functions I'm not sure I've got it right.
Here's what I have so far:
Input is an array 8 items long where each item is 32 bits.
Output is an array 8 items long where each item is 32 bits.
Calculate all the function boxes and store those values. I'll refer to them by function name.
Store input, right shifted by 32 bits, into output.
At this point, in the out array, E is the wrong value and A is empty
Store the function boxes.
now we need to calculate out E and out A.
note: I've replaced the modulo commands with a bitwise AND 2^(32-1)
I can't figure out how the modulus adding lines up, but I think it is like this:
Store (Input H + Ch + ( (Wt+Kt) AND 2^31 ) ) AND 2^31 As mod1
Store (sum1 + mod1) AND 2^31 as mod2
Store (d + mod2) AND 2^31 into output E
now output E is correct and all we need is output A
Store (MA + mod2) AND 2^31 as mod3
Store (sum0 + mod3) AND 2^31 into output A
output now contains the correct hash of input.
Do we return now or does this need to be run repeatedly?
Did I get all of those addition modulos right?
what are Wt and Kt?
Would this get run once, and you're done or does it need to be run a certain number of times, with the output being re-used as input?
Here's the link by the way.
http://en.wikipedia.org/wiki/SHA-2#Hash_function
Thanks alot,
Brian

Have a look at the official standard that describes the algorithm, the variables are described here: http://csrc.nist.gov/publications/fips/fips180-4/fips-180-4.pdf
(Oh, now I see I'm almost a year late with my answer, ah, never mind...)

W_t is derived from the current block being processed while K_t is a fixed constant determined by the iteration number. The compression function is repeated 64 times for each block in SHA256. There is a specific constant K_t and a derived value W_t for each iteration 0 <= t <= 63.
I have provided my own implementation of SHA256 using Python 3.6. The tuple K contains the 64 constant values of K_t. The Sha256 function shows how the value of W_t is computed in the list W. The implementation focuses on code clarity and not high-performance.
W = 32 #Number of bits in word
M = 1 << W
FF = M - 1 #0xFFFFFFFF (for performing addition mod 2**32)
#Constants from SHA256 definition
K = (0x428a2f98, 0x71374491, 0xb5c0fbcf, 0xe9b5dba5,
0x3956c25b, 0x59f111f1, 0x923f82a4, 0xab1c5ed5,
0xd807aa98, 0x12835b01, 0x243185be, 0x550c7dc3,
0x72be5d74, 0x80deb1fe, 0x9bdc06a7, 0xc19bf174,
0xe49b69c1, 0xefbe4786, 0x0fc19dc6, 0x240ca1cc,
0x2de92c6f, 0x4a7484aa, 0x5cb0a9dc, 0x76f988da,
0x983e5152, 0xa831c66d, 0xb00327c8, 0xbf597fc7,
0xc6e00bf3, 0xd5a79147, 0x06ca6351, 0x14292967,
0x27b70a85, 0x2e1b2138, 0x4d2c6dfc, 0x53380d13,
0x650a7354, 0x766a0abb, 0x81c2c92e, 0x92722c85,
0xa2bfe8a1, 0xa81a664b, 0xc24b8b70, 0xc76c51a3,
0xd192e819, 0xd6990624, 0xf40e3585, 0x106aa070,
0x19a4c116, 0x1e376c08, 0x2748774c, 0x34b0bcb5,
0x391c0cb3, 0x4ed8aa4a, 0x5b9cca4f, 0x682e6ff3,
0x748f82ee, 0x78a5636f, 0x84c87814, 0x8cc70208,
0x90befffa, 0xa4506ceb, 0xbef9a3f7, 0xc67178f2)
#Initial values for compression function
I = (0x6a09e667, 0xbb67ae85, 0x3c6ef372, 0xa54ff53a,
0x510e527f, 0x9b05688c, 0x1f83d9ab, 0x5be0cd19)
def RR(x, b):
'''
32-bit bitwise rotate right
'''
return ((x >> b) | (x << (W - b))) & FF
def Pad(W):
'''
Pads a message and converts to byte array
'''
mdi = len(W) % 64
L = (len(W) << 3).to_bytes(8, 'big') #Binary of len(W) in bits
npad = 55 - mdi if mdi < 56 else 119 - mdi #Pad so 64 | len; add 1 block if needed
return bytes(W, 'ascii') + b'\x80' + (b'\x00' * npad) + L #64 | 1 + npad + 8 + len(W)
def Sha256CF(Wt, Kt, A, B, C, D, E, F, G, H):
'''
SHA256 Compression Function
'''
Ch = (E & F) ^ (~E & G)
Ma = (A & B) ^ (A & C) ^ (B & C) #Major
S0 = RR(A, 2) ^ RR(A, 13) ^ RR(A, 22) #Sigma_0
S1 = RR(E, 6) ^ RR(E, 11) ^ RR(E, 25) #Sigma_1
T1 = H + S1 + Ch + Wt + Kt
return (T1 + S0 + Ma) & FF, A, B, C, (D + T1) & FF, E, F, G
def Sha256(M):
'''
Performs SHA256 on an input string
M: The string to process
return: A 32 byte array of the binary digest
'''
M = Pad(M) #Pad message so that length is divisible by 64
DG = list(I) #Digest as 8 32-bit words (A-H)
for j in range(0, len(M), 64): #Iterate over message in chunks of 64
S = M[j:j + 64] #Current chunk
W = [0] * 64
W[0:16] = [int.from_bytes(S[i:i + 4], 'big') for i in range(0, 64, 4)]
for i in range(16, 64):
s0 = RR(W[i - 15], 7) ^ RR(W[i - 15], 18) ^ (W[i - 15] >> 3)
s1 = RR(W[i - 2], 17) ^ RR(W[i - 2], 19) ^ (W[i - 2] >> 10)
W[i] = (W[i - 16] + s0 + W[i-7] + s1) & FF
A, B, C, D, E, F, G, H = DG #State of the compression function
for i in range(64):
A, B, C, D, E, F, G, H = Sha256CF(W[i], K[i], A, B, C, D, E, F, G, H)
DG = [(X + Y) & FF for X, Y in zip(DG, (A, B, C, D, E, F, G, H))]
return b''.join(Di.to_bytes(4, 'big') for Di in DG) #Convert to byte array
if __name__ == "__main__":
bd = Sha256('Hello World')
print(''.join('{:02x}'.format(i) for i in bd))

initial_hash_values=[
'6a09e667','bb67ae85','3c6ef372','a54ff53a',
'510e527f','9b05688c','1f83d9ab','5be0cd19'
]
sha_256_constants=[
'428a2f98','71374491','b5c0fbcf','e9b5dba5',
'3956c25b','59f111f1','923f82a4','ab1c5ed5',
'd807aa98','12835b01','243185be','550c7dc3',
'72be5d74','80deb1fe','9bdc06a7','c19bf174',
'e49b69c1','efbe4786','0fc19dc6','240ca1cc',
'2de92c6f','4a7484aa','5cb0a9dc','76f988da',
'983e5152','a831c66d','b00327c8','bf597fc7',
'c6e00bf3','d5a79147','06ca6351','14292967',
'27b70a85','2e1b2138','4d2c6dfc','53380d13',
'650a7354','766a0abb','81c2c92e','92722c85',
'a2bfe8a1','a81a664b','c24b8b70','c76c51a3',
'd192e819','d6990624','f40e3585','106aa070',
'19a4c116','1e376c08','2748774c','34b0bcb5',
'391c0cb3','4ed8aa4a','5b9cca4f','682e6ff3',
'748f82ee','78a5636f','84c87814','8cc70208',
'90befffa','a4506ceb','bef9a3f7','c67178f2'
]
def bin_return(dec):
return(str(format(dec,'b')))
def bin_8bit(dec):
return(str(format(dec,'08b')))
def bin_32bit(dec):
return(str(format(dec,'032b')))
def bin_64bit(dec):
return(str(format(dec,'064b')))
def hex_return(dec):
return(str(format(dec,'x')))
def dec_return_bin(bin_string):
return(int(bin_string,2))
def dec_return_hex(hex_string):
return(int(hex_string,16))
def L_P(SET,n):
to_return=[]
j=0
k=n
while k<len(SET)+1:
to_return.append(SET[j:k])
j=k
k+=n
return(to_return)
def s_l(bit_string):
bit_list=[]
for i in range(len(bit_string)):
bit_list.append(bit_string[i])
return(bit_list)
def l_s(bit_list):
bit_string=''
for i in range(len(bit_list)):
bit_string+=bit_list[i]
return(bit_string)
def rotate_right(bit_string,n):
bit_list = s_l(bit_string)
count=0
while count <= n-1:
list_main=list(bit_list)
var_0=list_main.pop(-1)
list_main=list([var_0]+list_main)
bit_list=list(list_main)
count+=1
return(l_s(list_main))
def shift_right(bit_string,n):
bit_list=s_l(bit_string)
count=0
while count <= n-1:
bit_list.pop(-1)
count+=1
front_append=['0']*n
return(l_s(front_append+bit_list))
def mod_32_addition(input_set):
value=0
for i in range(len(input_set)):
value+=input_set[i]
mod_32 = 4294967296
return(value%mod_32)
def xor_2str(bit_string_1,bit_string_2):
xor_list=[]
for i in range(len(bit_string_1)):
if bit_string_1[i]=='0' and bit_string_2[i]=='0':
xor_list.append('0')
if bit_string_1[i]=='1' and bit_string_2[i]=='1':
xor_list.append('0')
if bit_string_1[i]=='0' and bit_string_2[i]=='1':
xor_list.append('1')
if bit_string_1[i]=='1' and bit_string_2[i]=='0':
xor_list.append('1')
return(l_s(xor_list))
def and_2str(bit_string_1,bit_string_2):
and_list=[]
for i in range(len(bit_string_1)):
if bit_string_1[i]=='1' and bit_string_2[i]=='1':
and_list.append('1')
else:
and_list.append('0')
return(l_s(and_list))
def or_2str(bit_string_1,bit_string_2):
or_list=[]
for i in range(len(bit_string_1)):
if bit_string_1[i]=='0' and bit_string_2[i]=='0':
or_list.append('0')
else:
or_list.append('1')
return(l_s(or_list))
def not_str(bit_string):
not_list=[]
for i in range(len(bit_string)):
if bit_string[i]=='0':
not_list.append('1')
else:
not_list.append('0')
return(l_s(not_list))
'''
SHA-256 Specific Functions:
'''
def Ch(x,y,z):
return(xor_2str(and_2str(x,y),and_2str(not_str(x),z)))
def Maj(x,y,z):
return(xor_2str(xor_2str(and_2str(x,y),and_2str(x,z)),and_2str(y,z)))
def e_0(x):
return(xor_2str(xor_2str(rotate_right(x,2),rotate_right(x,13)),rotate_right(x,22)))
def e_1(x):
return(xor_2str(xor_2str(rotate_right(x,6),rotate_right(x,11)),rotate_right(x,25)))
def s_0(x):
return(xor_2str(xor_2str(rotate_right(x,7),rotate_right(x,18)),shift_right(x,3)))
def s_1(x):
return(xor_2str(xor_2str(rotate_right(x,17),rotate_right(x,19)),shift_right(x,10)))
def message_pad(bit_list):
pad_one = bit_list + '1'
pad_len = len(pad_one)
k=0
while ((pad_len+k)-448)%512 != 0:
k+=1
back_append_0 = '0'*k
back_append_1 = bin_64bit(len(bit_list))
return(pad_one+back_append_0+back_append_1)
def message_bit_return(string_input):
bit_list=[]
for i in range(len(string_input)):
bit_list.append(bin_8bit(ord(string_input[i])))
return(l_s(bit_list))
def message_pre_pro(input_string):
bit_main = message_bit_return(input_string)
return(message_pad(bit_main))
def message_parsing(input_string):
return(L_P(message_pre_pro(input_string),32))
def message_schedule(index,w_t):
new_word = bin_32bit(mod_32_addition([int(s_1(w_t[index-2]),2),int(w_t[index-7],2),int(s_0(w_t[index-15]),2),int(w_t[index-16],2)]))
return(new_word)
'''
This example of SHA_256 works for an input string >56 characters.
'''
def sha_256(input_string):
w_t=message_parsing(input_string)
a=bin_32bit(dec_return_hex(initial_hash_values[0]))
b=bin_32bit(dec_return_hex(initial_hash_values[1]))
c=bin_32bit(dec_return_hex(initial_hash_values[2]))
d=bin_32bit(dec_return_hex(initial_hash_values[3]))
e=bin_32bit(dec_return_hex(initial_hash_values[4]))
f=bin_32bit(dec_return_hex(initial_hash_values[5]))
g=bin_32bit(dec_return_hex(initial_hash_values[6]))
h=bin_32bit(dec_return_hex(initial_hash_values[7]))
for i in range(0,64):
if i <= 15:
t_1=mod_32_addition([int(h,2),int(e_1(e),2),int(Ch(e,f,g),2),int(sha_256_constants[i],16),int(w_t[i],2)])
t_2=mod_32_addition([int(e_0(a),2),int(Maj(a,b,c),2)])
h=g
g=f
f=e
e=mod_32_addition([int(d,2),t_1])
d=c
c=b
b=a
a=mod_32_addition([t_1,t_2])
a=bin_32bit(a)
e=bin_32bit(e)
if i > 15:
w_t.append(message_schedule(i,w_t))
t_1=mod_32_addition([int(h,2),int(e_1(e),2),int(Ch(e,f,g),2),int(sha_256_constants[i],16),int(w_t[i],2)])
t_2=mod_32_addition([int(e_0(a),2),int(Maj(a,b,c),2)])
h=g
g=f
f=e
e=mod_32_addition([int(d,2),t_1])
d=c
c=b
b=a
a=mod_32_addition([t_1,t_2])
a=bin_32bit(a)
e=bin_32bit(e)
hash_0 = mod_32_addition([dec_return_hex(initial_hash_values[0]),int(a,2)])
hash_1 = mod_32_addition([dec_return_hex(initial_hash_values[1]),int(b,2)])
hash_2 = mod_32_addition([dec_return_hex(initial_hash_values[2]),int(c,2)])
hash_3 = mod_32_addition([dec_return_hex(initial_hash_values[3]),int(d,2)])
hash_4 = mod_32_addition([dec_return_hex(initial_hash_values[4]),int(e,2)])
hash_5 = mod_32_addition([dec_return_hex(initial_hash_values[5]),int(f,2)])
hash_6 = mod_32_addition([dec_return_hex(initial_hash_values[6]),int(g,2)])
hash_7 = mod_32_addition([dec_return_hex(initial_hash_values[7]),int(h,2)])
final_hash = (hex_return(hash_0),
hex_return(hash_1),
hex_return(hash_2),
hex_return(hash_3),
hex_return(hash_4),
hex_return(hash_5),
hex_return(hash_6),
hex_return(hash_7))
return(final_hash)

What is the best way to add two numbers without using the + operator?

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)))))