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Given a black-box function f(x) which take a value x and output true for \$p(x)\$, an unknown continious monotone function(not knowing even whether it's increasing), probable; and false otherwise. Output an infinite sequence \$a_n\$ such that \$\lim_{n\rightarrow \infty}p(a_n)=0.5\$. You can assume that the result exists.

Reasonable I/O allowed. Shortest code win.

A possible solution:

for i=1..infty
    S = [i/2] * i*i*2
    for j=0..i*i*2-1
        for k=1..i
            if f(j/i-i)
                S[j]--
    print minPos([t*t for t in S])/i-i

random code-golf

Given a black-box function f(x) which take a value x and output true for \$p(x)\$, an unknown continious monotone function(not knowing even whether it's increasing), probable; and false otherwise. Output an infinite sequence \$a_n\$ such that \$\lim_{n\rightarrow \infty}p(a_n)=0.5\$. You can assume that the result exists.

Reasonable I/O allowed. Shortest code win.

A possible solution:

for i=1..infty
    S = [i/2] * i*i*2
    for j=0..i*i*2-1
        for k=1..i
            if f(j/i-i)
                S[j]--
    print minPos([t*t for t in S])/i-i

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random code-golf

Given a black-box function f(x) which take a value x and output true for \$p(x)\$, an unknown continious monotone function(not knowing even whether it's increasing), probable; and false otherwise. Output an infinite sequence \$a_n\$ such that \$\lim_{n\rightarrow \infty}p(a_n)=0.5\$. You can assume that the result exists and \$\lim_{n\rightarrow \infty}a_n\$ lay in \$(0,1)\$.

Reasonable I/O allowed. Shortest code win.

A possible solution:

for i=1..infty
    S = [i/2] * i
    for j=0..i-1
        for k=1..i
            if f(j/i)
                S[j]--
    print minPos([t*t for t in S])/i

random code-golf

Given a black-box function f(x) which take a value x and output true for \$p(x)\$, an unknown continious monotone function(not knowing even whether it's increasing), probable; and false otherwise. Output an infinite sequence \$a_n\$ such that \$\lim_{n\rightarrow \infty}p(a_n)=0.5\$. You can assume that the result exists.

Reasonable I/O allowed. Shortest code win.

A possible solution:

for i=1..infty
    S = [i/2] * i*i*2
    for j=0..i*i*2-1
        for k=1..i
            if f(j/i-i)
                S[j]--
    print minPos([t*t for t in S])/i-i

random code-golf

Given a black-box function f(x) which take a value x and output true for \$p(x)\$, an unknown continious monotone function(not knowing even whether it's increasing), probable; and false otherwise. Output an infinite sequence \$a_n\$ such that \$\lim_{n\rightarrow \infty}p(a_n)=0.5\$. You can assume that the result exists and \$\lim_{n\rightarrow \infty}a_n\$ lay in \$(0,1)\$.

A possible solution:

for i=1..infty
    S = [i/2] * i
    for j=0..i-1
        for k=1..i
            if f(j/i)
                S[j]--
    print minPos([t*t for t in S])/i

random code-golf

Given a black-box function f(x) which take a value x and output true for \$p(x)\$, an unknown continious monotone function(not knowing even whether it's increasing), probable; and false otherwise. Output an infinite sequence \$a_n\$ such that \$\lim_{n\rightarrow \infty}p(a_n)=0.5\$. You can assume that the result exists and \$\lim_{n\rightarrow \infty}a_n\$ lay in \$(0,1)\$.

Reasonable I/O allowed. Shortest code win.

A possible solution:

for i=1..infty
    S = [i/2] * i
    for j=0..i-1
        for k=1..i
            if f(j/i)
                S[j]--
    print minPos([t*t for t in S])/i

random code-golf