matlab precision dealing with big number
The question encountered me when I was trying to solve Project Euler #16.
The problem reads:
215 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
What is the sum of the digits of the number 21000?
I was trying to deal with it with MATLAB. Here's my code:
function [sum] = power_digit_sum_p16(pow)
sum = 0;
n = 2 ^ pow;
M = zeros(1,301);
for i = 301 : -1 : 0
a = floor(n / 10 ^ i);
M(302 - i) = a;
sum = sum + a;
n = n - 10 ^ i * a;
end
end
(Where the 301 stands for 301 digits of 2^1000)
When I ran pow_digit_sum_p16(1000), MATLAB returned 1285, which is wrong.
Then I checked the number provided by MATLAB, which is stored in matrix M, when I learned that something is wrong.
The last digit of 2^1000 should be 6 instead of 2 (it follows a 2-4-8-6-2-4-8-6 pattern)！ I don't understand what's going on here with MATLAB, but I guess the problem occurred because the number is too big, since my function works fine when pow is small.
My friend provided me with a Python solution, and it seems that python handles the big number well. Below is the code in Python:
x=2**1000
ans=0
while x>0:
ans+=(x%10)
x=x//10
print(ans)
Update: thanks to OmG's answer, I changed my code a little bit:
function [sum] = power_digit_sum_p16(pow)
sum = 0;
n = 2 ^ pow;
n = sym(n); % This line is new!
M = zeros(1,301);
for i = 301 : -1 : 0
a = floor(n / (10 ^ i));
M(302 - i) = a;
sum = sum + a;
n = n - (10 ^ i) * a;
end
end
And I got what I expected. Note that the sym function needs the Symbolic Math Toolbox.
matlab precision
edited Dec 30 '18 at 7:40
Cris Luengo
19.1k51947
asked Dec 28 '18 at 7:43
Frank LiFrank Li
83
|
show 4 more comments
The question encountered me when I was trying to solve Project Euler #16.
The problem reads:
215 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
What is the sum of the digits of the number 21000?
I was trying to deal with it with MATLAB. Here's my code:
function [sum] = power_digit_sum_p16(pow)
sum = 0;
n = 2 ^ pow;
M = zeros(1,301);
for i = 301 : -1 : 0
a = floor(n / 10 ^ i);
M(302 - i) = a;
sum = sum + a;
n = n - 10 ^ i * a;
end
end
(Where the 301 stands for 301 digits of 2^1000)
When I ran pow_digit_sum_p16(1000), MATLAB returned 1285, which is wrong.
Then I checked the number provided by MATLAB, which is stored in matrix M, when I learned that something is wrong.
The last digit of 2^1000 should be 6 instead of 2 (it follows a 2-4-8-6-2-4-8-6 pattern)！ I don't understand what's going on here with MATLAB, but I guess the problem occurred because the number is too big, since my function works fine when pow is small.
My friend provided me with a Python solution, and it seems that python handles the big number well. Below is the code in Python:
x=2**1000
ans=0
while x>0:
ans+=(x%10)
x=x//10
print(ans)
Update: thanks to OmG's answer, I changed my code a little bit:
function [sum] = power_digit_sum_p16(pow)
sum = 0;
n = 2 ^ pow;
n = sym(n); % This line is new!
M = zeros(1,301);
for i = 301 : -1 : 0
a = floor(n / (10 ^ i));
M(302 - i) = a;
sum = sum + a;
n = n - (10 ^ i) * a;
end
end
And I got what I expected. Note that the sym function needs the Symbolic Math Toolbox.
matlab precision
edited Dec 30 '18 at 7:40
Cris Luengo
19.1k51947
asked Dec 28 '18 at 7:43
Frank LiFrank Li
83
I assume it's related to the precision of the number that is related to the problem you've described above.
– Nahiyan
Dec 28 '18 at 7:45
So any idea to do to increase the precision in MATLAB computation? The fact that python handles it well make me kind of "jealous".
– Frank Li
Dec 28 '18 at 7:49
Use the repeating pattern property, read here:en.wikipedia.org/wiki/Power_of_two "The 0th through 95th powers of two"
– Mendi Barel
Dec 28 '18 at 9:56
1
Python has arbitrary-precision integers. Matlab doesn't, unless you use symbolic computations
– Luis Mendo
Dec 28 '18 at 11:02
2
Isn't the whole point of the problem to compute the sum of the digits without an explicit calculation of2^1000? Project Euler is about cunning, not about brute force.
– High Performance Mark
Dec 28 '18 at 11:15
|
show 4 more comments
The question encountered me when I was trying to solve Project Euler #16.
The problem reads:
215 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
What is the sum of the digits of the number 21000?
I was trying to deal with it with MATLAB. Here's my code:
function [sum] = power_digit_sum_p16(pow)
sum = 0;
n = 2 ^ pow;
M = zeros(1,301);
for i = 301 : -1 : 0
a = floor(n / 10 ^ i);
M(302 - i) = a;
sum = sum + a;
n = n - 10 ^ i * a;
end
end
(Where the 301 stands for 301 digits of 2^1000)
When I ran pow_digit_sum_p16(1000), MATLAB returned 1285, which is wrong.
Then I checked the number provided by MATLAB, which is stored in matrix M, when I learned that something is wrong.
The last digit of 2^1000 should be 6 instead of 2 (it follows a 2-4-8-6-2-4-8-6 pattern)！ I don't understand what's going on here with MATLAB, but I guess the problem occurred because the number is too big, since my function works fine when pow is small.
My friend provided me with a Python solution, and it seems that python handles the big number well. Below is the code in Python:
x=2**1000
ans=0
while x>0:
ans+=(x%10)
x=x//10
print(ans)
Update: thanks to OmG's answer, I changed my code a little bit:
function [sum] = power_digit_sum_p16(pow)
sum = 0;
n = 2 ^ pow;
n = sym(n); % This line is new!
M = zeros(1,301);
for i = 301 : -1 : 0
a = floor(n / (10 ^ i));
M(302 - i) = a;
sum = sum + a;
n = n - (10 ^ i) * a;
end
end
And I got what I expected. Note that the sym function needs the Symbolic Math Toolbox.
matlab precision
edited Dec 30 '18 at 7:40
Cris Luengo
19.1k51947
asked Dec 28 '18 at 7:43
Frank LiFrank Li
83
The question encountered me when I was trying to solve Project Euler #16.
The problem reads:
215 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
What is the sum of the digits of the number 21000?
I was trying to deal with it with MATLAB. Here's my code:
function [sum] = power_digit_sum_p16(pow)
sum = 0;
n = 2 ^ pow;
M = zeros(1,301);
for i = 301 : -1 : 0
a = floor(n / 10 ^ i);
M(302 - i) = a;
sum = sum + a;
n = n - 10 ^ i * a;
end
end
(Where the 301 stands for 301 digits of 2^1000)
When I ran pow_digit_sum_p16(1000), MATLAB returned 1285, which is wrong.
Then I checked the number provided by MATLAB, which is stored in matrix M, when I learned that something is wrong.
The last digit of 2^1000 should be 6 instead of 2 (it follows a 2-4-8-6-2-4-8-6 pattern)！ I don't understand what's going on here with MATLAB, but I guess the problem occurred because the number is too big, since my function works fine when pow is small.
My friend provided me with a Python solution, and it seems that python handles the big number well. Below is the code in Python:
x=2**1000
ans=0
while x>0:
ans+=(x%10)
x=x//10
print(ans)
Update: thanks to OmG's answer, I changed my code a little bit:
function [sum] = power_digit_sum_p16(pow)
sum = 0;
n = 2 ^ pow;
n = sym(n); % This line is new!
M = zeros(1,301);
for i = 301 : -1 : 0
a = floor(n / (10 ^ i));
M(302 - i) = a;
sum = sum + a;
n = n - (10 ^ i) * a;
end
end
And I got what I expected. Note that the sym function needs the Symbolic Math Toolbox.
matlab precision
matlab precision
edited Dec 30 '18 at 7:40
Cris Luengo
19.1k51947
asked Dec 28 '18 at 7:43
Frank LiFrank Li
83
edited Dec 30 '18 at 7:40
Cris Luengo
19.1k51947
asked Dec 28 '18 at 7:43
Frank LiFrank Li
83
edited Dec 30 '18 at 7:40
Cris Luengo
19.1k51947
edited Dec 30 '18 at 7:40
Cris Luengo
19.1k51947
edited Dec 30 '18 at 7:40
Cris Luengo
19.1k51947
19.1k51947
asked Dec 28 '18 at 7:43
Frank LiFrank Li
83
asked Dec 28 '18 at 7:43
Frank LiFrank Li
83
asked Dec 28 '18 at 7:43
Frank LiFrank Li
83
83
I assume it's related to the precision of the number that is related to the problem you've described above.
– Nahiyan
Dec 28 '18 at 7:45
So any idea to do to increase the precision in MATLAB computation? The fact that python handles it well make me kind of "jealous".
– Frank Li
Dec 28 '18 at 7:49
Use the repeating pattern property, read here:en.wikipedia.org/wiki/Power_of_two "The 0th through 95th powers of two"
– Mendi Barel
Dec 28 '18 at 9:56
1
Python has arbitrary-precision integers. Matlab doesn't, unless you use symbolic computations
– Luis Mendo
Dec 28 '18 at 11:02
2
Isn't the whole point of the problem to compute the sum of the digits without an explicit calculation of2^1000? Project Euler is about cunning, not about brute force.
– High Performance Mark
Dec 28 '18 at 11:15
|
show 4 more comments
I assume it's related to the precision of the number that is related to the problem you've described above.
– Nahiyan
Dec 28 '18 at 7:45
So any idea to do to increase the precision in MATLAB computation? The fact that python handles it well make me kind of "jealous".
– Frank Li
Dec 28 '18 at 7:49
Use the repeating pattern property, read here:en.wikipedia.org/wiki/Power_of_two "The 0th through 95th powers of two"
– Mendi Barel
Dec 28 '18 at 9:56
1
Python has arbitrary-precision integers. Matlab doesn't, unless you use symbolic computations
– Luis Mendo
Dec 28 '18 at 11:02
2
Isn't the whole point of the problem to compute the sum of the digits without an explicit calculation of2^1000? Project Euler is about cunning, not about brute force.
– High Performance Mark
Dec 28 '18 at 11:15
I assume it's related to the precision of the number that is related to the problem you've described above.
– Nahiyan
Dec 28 '18 at 7:45
I assume it's related to the precision of the number that is related to the problem you've described above.
– Nahiyan
Dec 28 '18 at 7:45
So any idea to do to increase the precision in MATLAB computation? The fact that python handles it well make me kind of "jealous".
– Frank Li
Dec 28 '18 at 7:49
So any idea to do to increase the precision in MATLAB computation? The fact that python handles it well make me kind of "jealous".
– Frank Li
Dec 28 '18 at 7:49
Use the repeating pattern property, read here:en.wikipedia.org/wiki/Power_of_two "The 0th through 95th powers of two"
– Mendi Barel
Dec 28 '18 at 9:56
Use the repeating pattern property, read here:en.wikipedia.org/wiki/Power_of_two "The 0th through 95th powers of two"
– Mendi Barel
Dec 28 '18 at 9:56
1
1
Python has arbitrary-precision integers. Matlab doesn't, unless you use symbolic computations
– Luis Mendo
Dec 28 '18 at 11:02
Python has arbitrary-precision integers. Matlab doesn't, unless you use symbolic computations
– Luis Mendo
Dec 28 '18 at 11:02
2
2
Isn't the whole point of the problem to compute the sum of the digits without an explicit calculation of
2^1000 ? Project Euler is about cunning, not about brute force.– High Performance Mark
Dec 28 '18 at 11:15
Isn't the whole point of the problem to compute the sum of the digits without an explicit calculation of
2^1000 ? Project Euler is about cunning, not about brute force.– High Performance Mark
Dec 28 '18 at 11:15
|
show 4 more comments
1 Answer
1
active
oldest
votes
A solution is using symbolic tools. For example for finding the true value for the remaining for 2^1000 over 10 you can do it likes the following:
x = sym('2^1000');
reminder = fix(double(mod(x,10)))
The answer is:
reminder =
6
For 100 and 1000 is 67 and 376 respectively.
answered Dec 28 '18 at 13:14
OmGOmG
7,89352643
The only thing wrong with this answer is that it is not an answer to the question asked.
– High Performance Mark
Dec 28 '18 at 18:12
Your is the answer I wanted, thank you!
– Frank Li
Dec 29 '18 at 16:32
Something to improve in your answer, 1.solutio--->solution(a typo) 2. x=sym(2^1000), do not make it a string.
– Frank Li
Dec 29 '18 at 20:32
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
A solution is using symbolic tools. For example for finding the true value for the remaining for 2^1000 over 10 you can do it likes the following:
x = sym('2^1000');
reminder = fix(double(mod(x,10)))
The answer is:
reminder =
6
For 100 and 1000 is 67 and 376 respectively.
answered Dec 28 '18 at 13:14
OmGOmG
7,89352643
The only thing wrong with this answer is that it is not an answer to the question asked.
– High Performance Mark
Dec 28 '18 at 18:12
Your is the answer I wanted, thank you!
– Frank Li
Dec 29 '18 at 16:32
Something to improve in your answer, 1.solutio--->solution(a typo) 2. x=sym(2^1000), do not make it a string.
– Frank Li
Dec 29 '18 at 20:32
add a comment |
A solution is using symbolic tools. For example for finding the true value for the remaining for 2^1000 over 10 you can do it likes the following:
x = sym('2^1000');
reminder = fix(double(mod(x,10)))
The answer is:
reminder =
6
For 100 and 1000 is 67 and 376 respectively.
answered Dec 28 '18 at 13:14
OmGOmG
7,89352643
The only thing wrong with this answer is that it is not an answer to the question asked.
– High Performance Mark
Dec 28 '18 at 18:12
Your is the answer I wanted, thank you!
– Frank Li
Dec 29 '18 at 16:32
Something to improve in your answer, 1.solutio--->solution(a typo) 2. x=sym(2^1000), do not make it a string.
– Frank Li
Dec 29 '18 at 20:32
add a comment |
A solution is using symbolic tools. For example for finding the true value for the remaining for 2^1000 over 10 you can do it likes the following:
x = sym('2^1000');
reminder = fix(double(mod(x,10)))
The answer is:
reminder =
6
For 100 and 1000 is 67 and 376 respectively.
answered Dec 28 '18 at 13:14
OmGOmG
7,89352643
A solution is using symbolic tools. For example for finding the true value for the remaining for 2^1000 over 10 you can do it likes the following:
x = sym('2^1000');
reminder = fix(double(mod(x,10)))
The answer is:
reminder =
6
For 100 and 1000 is 67 and 376 respectively.
answered Dec 28 '18 at 13:14
OmGOmG
7,89352643
edited Dec 30 '18 at 6:50
answered Dec 28 '18 at 13:14
OmGOmG
7,89352643
answered Dec 28 '18 at 13:14
OmGOmG
7,89352643
answered Dec 28 '18 at 13:14
OmGOmG
7,89352643
7,89352643
The only thing wrong with this answer is that it is not an answer to the question asked.
– High Performance Mark
Dec 28 '18 at 18:12
Your is the answer I wanted, thank you!
– Frank Li
Dec 29 '18 at 16:32
Something to improve in your answer, 1.solutio--->solution(a typo) 2. x=sym(2^1000), do not make it a string.
– Frank Li
Dec 29 '18 at 20:32
add a comment |
The only thing wrong with this answer is that it is not an answer to the question asked.
– High Performance Mark
Dec 28 '18 at 18:12
Your is the answer I wanted, thank you!
– Frank Li
Dec 29 '18 at 16:32
Something to improve in your answer, 1.solutio--->solution(a typo) 2. x=sym(2^1000), do not make it a string.
– Frank Li
Dec 29 '18 at 20:32
The only thing wrong with this answer is that it is not an answer to the question asked.
– High Performance Mark
Dec 28 '18 at 18:12
The only thing wrong with this answer is that it is not an answer to the question asked.
– High Performance Mark
Dec 28 '18 at 18:12
Your is the answer I wanted, thank you!
– Frank Li
Dec 29 '18 at 16:32
Your is the answer I wanted, thank you!
– Frank Li
Dec 29 '18 at 16:32
Something to improve in your answer, 1.solutio--->solution(a typo) 2. x=sym(2^1000), do not make it a string.
– Frank Li
Dec 29 '18 at 20:32
Something to improve in your answer, 1.solutio--->solution(a typo) 2. x=sym(2^1000), do not make it a string.
– Frank Li
Dec 29 '18 at 20:32
add a comment |
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I assume it's related to the precision of the number that is related to the problem you've described above.
– Nahiyan
Dec 28 '18 at 7:45
So any idea to do to increase the precision in MATLAB computation? The fact that python handles it well make me kind of "jealous".
– Frank Li
Dec 28 '18 at 7:49
Use the repeating pattern property, read here:en.wikipedia.org/wiki/Power_of_two "The 0th through 95th powers of two"
– Mendi Barel
Dec 28 '18 at 9:56
1
Python has arbitrary-precision integers. Matlab doesn't, unless you use symbolic computations
– Luis Mendo
Dec 28 '18 at 11:02
2
Isn't the whole point of the problem to compute the sum of the digits without an explicit calculation of
2^1000? Project Euler is about cunning, not about brute force.– High Performance Mark
Dec 28 '18 at 11:15