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user21820
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P=NP $\mathbf{P} = \mathbf{NP}$, What'swhat's the problem?

Using "..." instead of \cdots causes a conspicuous asymmetry. \mod instead of \bmod puts too much white space to its left, inappropriate in this context. And other copy-editing.
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Let's take the problem of the backpack: $A_1,... ,A_n$$A_1,\ldots ,A_n$ the weights that are integers, and we want to know if we can achieve a total weight of $V$.

We take $I=\dfrac{1}{2\pi}\int_0^{2\pi} \exp(-iVt)\times (1+\exp(iA_1t))\times...\times(1+\exp(iA_nt))dt$.$$I=\dfrac{1}{2\pi}\int_0^{2\pi} \exp(-iVt)\times (1+\exp(iA_1t))\times\cdots\times(1+\exp(iA_nt)) \, dt.$$

The question then becomes what $I=0$ or $I\geq 1$.

Can we not approach the value of $I$ with the method of Monte Carlo, or others?

Why are these approaches not succeeding?

Remark : it's not difficult to find a good approximation of $B\times t \mod 2\pi$$B\times t \bmod 2\pi$, where $B$ is a big integer and $t\in [0,2\pi]$, because we know excellent approximation of $\pi$.

Let's take the problem of the backpack: $A_1,... ,A_n$ the weights that are integers, and we want to know if we can achieve a total weight of $V$.

We take $I=\dfrac{1}{2\pi}\int_0^{2\pi} \exp(-iVt)\times (1+\exp(iA_1t))\times...\times(1+\exp(iA_nt))dt$.

The question then becomes what $I=0$ or $I\geq 1$.

Can we not approach the value of $I$ with the method of Monte Carlo, or others?

Why are these approaches not succeeding?

Remark : it's not difficult to find a good approximation of $B\times t \mod 2\pi$, where $B$ is a big integer and $t\in [0,2\pi]$, because we know excellent approximation of $\pi$.

Let's take the problem of the backpack: $A_1,\ldots ,A_n$ the weights that are integers, and we want to know if we can achieve a total weight of $V$.

We take $$I=\dfrac{1}{2\pi}\int_0^{2\pi} \exp(-iVt)\times (1+\exp(iA_1t))\times\cdots\times(1+\exp(iA_nt)) \, dt.$$

The question then becomes what $I=0$ or $I\geq 1$.

Can we not approach the value of $I$ with the method of Monte Carlo, or others?

Why are these approaches not succeeding?

Remark : it's not difficult to find a good approximation of $B\times t \bmod 2\pi$, where $B$ is a big integer and $t\in [0,2\pi]$, because we know excellent approximation of $\pi$.

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Dattier
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Let's take the problem of the backpack: $A_1,... ,A_n$ the weights that are integers, and we want to know if we can achieve a total weight of $V$.

We take $I=\dfrac{1}{2\pi}\int_0^{2\pi} \exp(-iVt)\times (1+\exp(iA_1t))\times...\times(1+\exp(iA_nt))dt$.

The question then becomes what $I=0$ or $I=1$$I\geq 1$.

Can we not approach the value of $I$ with the method of Monte Carlo, or others?

Why are these approaches not succeeding?

Remark : it's not difficult to find a good approximation of $B\times t \mod 2\pi$, where $B$ is a big integer and $t\in [0,2\pi]$, because we know excellent approximation of $\pi$.

Let's take the problem of the backpack: $A_1,... ,A_n$ the weights that are integers, and we want to know if we can achieve a total weight of $V$.

We take $I=\dfrac{1}{2\pi}\int_0^{2\pi} \exp(-iVt)\times (1+\exp(iA_1t))\times...\times(1+\exp(iA_nt))dt$.

The question then becomes what $I=0$ or $I=1$.

Can we not approach the value of $I$ with the method of Monte Carlo, or others?

Why are these approaches not succeeding?

Remark : it's not difficult to find a good approximation of $B\times t \mod 2\pi$, where $B$ is a big integer and $t\in [0,2\pi]$, because we know excellent approximation of $\pi$.

Let's take the problem of the backpack: $A_1,... ,A_n$ the weights that are integers, and we want to know if we can achieve a total weight of $V$.

We take $I=\dfrac{1}{2\pi}\int_0^{2\pi} \exp(-iVt)\times (1+\exp(iA_1t))\times...\times(1+\exp(iA_nt))dt$.

The question then becomes what $I=0$ or $I\geq 1$.

Can we not approach the value of $I$ with the method of Monte Carlo, or others?

Why are these approaches not succeeding?

Remark : it's not difficult to find a good approximation of $B\times t \mod 2\pi$, where $B$ is a big integer and $t\in [0,2\pi]$, because we know excellent approximation of $\pi$.

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Dattier
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