Using a for loop to count the digits of pi

Question

Martin 2023-10-4

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2028850-using-a-for-loop-to-count-the-digits-of-pi

评论： Sam Chak 2023-10-6

If the digits of π were random, then we would expect each of the integers

to occur with approximately equal frequency in the decimal representation of π. In this problem we are going to see if the digits of π do indeed appear to be random.

The first line of your script stores the first 100 digits of π in the vector pi_digits.

Create a vector f50 such that f50(i) is equal to the frequency with which the integer i-1 appears among the first 50 digits of π.
Replicate it for f100.

I have the following code:

pi_digits=[3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0 2 8 8 4 1 9 7 1 6 9 3 9 9 3 7 5 1 0 5 8 2 0 9 7 4 9 4 4 5 9 2 3 0 7 8 1 6 4 0 6 2 8 6 2 0 8 9 9 8 6 2 8 0 3 4 8 2 5 3 4 2 1 1 7 0 6 7];
f50=zeros(1,10);
for i=0:9
    f50(i+1)=sum(pi_digits(1:50)==i)
end
f100=zeros(1,10);
for i=0:9
    f100(i+1)=sum(pi_digits(1:100)==i)
end

the output returns the correct frequency vector, yet I get an error that it is the incorrect value.

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Dyuman Joshi 2023-10-4

@Sam, but the discussion here is for pi, not the factorial of pi :P

Sam Chak 2023-10-5

@Dyuman Joshi, I am "irrational", pun intended.

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Answer 1

Daniel 2023-10-6

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https://ww2.mathworks.cn/matlabcentral/answers/2028850-using-a-for-loop-to-count-the-digits-of-pi#answer_1327154

MA207?

need to divide f50 by 50 and f100 by 100.

apparently it wants the answer to be in decimal.

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Walter Roberson 2023-10-6

Ah, the original Question does ask about "frequency" rather than about counts, so I can see why they might normalize by the number of entries.

Sam Chak 2023-10-6

Thank you, @Daniel. I interpreted "frequency" as the number of occurrences within a given 100-digit π number. I totally overlooked that the "frequency" in question is somehow mathematically defined in statistics.

https://en.wikipedia.org/wiki/Frequency_(statistics)

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Answer 2

Sam Chak 2023-10-4

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https://ww2.mathworks.cn/matlabcentral/answers/2028850-using-a-for-loop-to-count-the-digits-of-pi#answer_1324845

在 MATLAB Online 中打开

Hi @Martin

Please try comparing the histogram with yours.

num2str(pi, 1000);

digits(100); % show 100 digits of π

numPi = vpa(pi)

numPi =

3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117068

c = char(numPi);

% Find the decimal point:

pos = strfind(c, '.')

pos = 2

% Begin to count after the decimal point

d = arrayfun(@str2num, c(pos+1:end));

% Plot the histogram for comparing with your Answer in MATLAB Grader

histogram(d, 10);

xt = linspace(0.5, 8.6, 10);

xticks(xt);

xticklabels({'0', '1', '2', '3', '4', '5', '6', '7', '8', '9'})

xlabel('Decimal Digit')

ylabel('Frequency')

title('Distribution of the digits of \pi');

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Sam Chak 2023-10-4

在 MATLAB Online 中打开

Hi @Martin

This code converts the number π into an array of integers and counts the frequency of the digits. The code also verifies the results obtained by you. Please review the question to clarify whether you want to count the frequency of the digits in the array of integers or only the decimal digits. Additionally, if I use digits(100), the last digit will be rounded up.

num2str(pi, 1000);

digits(102); % show 102 digits of π

numPi = vpa(pi)

numPi =

3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798

c = char(numPi);

c(2) = [] % remove decimal point

c = '314159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798'

% Count the digit from array of integers: 3, 1, 4, ...

d = arrayfun(@str2num, c(1:end-2)) % exclude the 101st and 102nd position in the sequence

d = 1×100

3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7

histcount100 = histcounts(d)

histcount100 = 1×10

8 8 12 12 10 8 9 8 12 13

% Plot the histogram

histogram(d, 10);

xt = linspace(0.5, 8.5, 10);

xticks(xt);

xticklabels({'0', '1', '2', '3', '4', '5', '6', '7', '8', '9'})

xlabel('Digit')

ylabel('Frequency')

title('Distribution of the digits of \pi');

Martin 2023-10-4

Thanks I appreciate it! After further review I am convinced there is an error in the solution on Matlab Grader.

Stephen23 2023-10-4

编辑：Stephen23 2023-10-4

在 MATLAB Online 中打开

Another approach using 1000 digits from https://pi2e.ch/blog/2017/03/10/pi-digits-download/

T = '31415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229489549303819644288109756659334461284756482337867831652712019091456485669234603486104543266482133936072602491412737245870066063155881748815209209628292540917153643678925903600113305305488204665213841469519415116094330572703657595919530921861173819326117931051185480744623799627495673518857527248912279381830119491298336733624406566430860213949463952247371907021798609437027705392171762931767523846748184676694051320005681271452635608277857713427577896091736371787214684409012249534301465495853710507922796892589235420199561121290219608640344181598136297747713099605187072113499999983729780499510597317328160963185950244594553469083026425223082533446850352619311881710100031378387528865875332083814206171776691473035982534904287554687311595628638823537875937519577818577805321712268066130019278766111959092164201989';

histogram(T(1:100)-'0',0:10)

Confirming a few random digits:

nnz(T(1:100)=='0')
ans = 8
nnz(T(1:100)=='6')
ans = 9
nnz(T(1:100)=='9')
ans = 13

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