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<h2 class="hd hd-2 unit-title">Measurements</h2>
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Measurements
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<p>
Consider the procedure where we first apply [mathjaxinline]U=e^{i(\theta /2)Y}[/mathjaxinline] to [mathjaxinline]|{\psi }\rangle[/mathjaxinline] and measure with [mathjaxinline]\{ |{0}\rangle ,|{1}\rangle \}[/mathjaxinline] basis. The preceding unitary essentially plays a role to change the measurement basis. Find the probability that measurement result is 0 for the following states. You may enter theta for [mathjaxinline]\theta[/mathjaxinline]. </p>
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<p>
[mathjaxinline]|{\psi }\rangle =|{0}\rangle[/mathjaxinline] <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_problem_Measurements_2_1" class="text-input-dynamath capa_inputtype textline">
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[mathjaxinline]|{\psi }\rangle =|{+}\rangle[/mathjaxinline] <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 2" role="group"><div id="inputtype_problem_Measurements_3_1" class="text-input-dynamath capa_inputtype textline">
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<span class="sr">unanswered</span><span class="status-icon" aria-hidden="true"/>
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<div id="display_problem_Measurements_3_1" class="equation">`{::}`</div>
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</div></div> </p>
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<li>
<p>
prepare [mathjaxinline]|{\psi }\rangle =|{0}\rangle[/mathjaxinline] with probability 1/2, prepare [mathjaxinline]|{\psi }\rangle =|{1}\rangle[/mathjaxinline] with probability 1/2. Note that the probability of getting the result 0 is [mathjaxinline]\frac{1}{2}[/mathjaxinline]Prob(getting result 0 for [mathjaxinline]|{0}\rangle[/mathjaxinline])+ [mathjaxinline]\frac{1}{2}[/mathjaxinline]Prob(getting result 0 for [mathjaxinline]|{1}\rangle[/mathjaxinline]). <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 3" role="group"><div id="inputtype_problem_Measurements_4_1" class="text-input-dynamath capa_inputtype textline">
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<div class="unanswered ">
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<span class="status unanswered" id="status_problem_Measurements_4_1" data-tooltip="Not yet answered.">
<span class="sr">unanswered</span><span class="status-icon" aria-hidden="true"/>
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<p id="answer_problem_Measurements_4_1" class="answer"/>
<div id="display_problem_Measurements_4_1" class="equation">`{::}`</div>
<textarea style="display:none" id="input_problem_Measurements_4_1_dynamath" name="input_problem_Measurements_4_1_dynamath"/>
</div>
</div></div> </p>
</li>
</ul>
<p>
One should obtain an insight from this problem how the second state is fundamentally different from the third state. Try substituting [mathjaxinline]\theta =0[/mathjaxinline] and [mathjaxinline]\theta =\pi /2[/mathjaxinline] and see how these values behave. </p>
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<h2 class="hd hd-2 unit-title">Entanglement</h2>
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Definition
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<p>
Are the following states entangled? </p>
<ul class="itemize">
<li>
<p>
<p style="display:inline">[mathjaxinline]|{00}\rangle[/mathjaxinline]&#160;&#160;&#160;</p>
<div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div class="inputtype option-input inline">
<select name="input_Definition_2_1" id="input_Definition_2_1" aria-describedby="status_Definition_2_1">
<option value="option_Definition_2_1_dummy_default">Select an option</option>
<option value="entangled"> entangled</option>
<option value="unentangled"> unentangled</option>
</select>
<div class="indicator-container">
<span class="status unanswered" id="status_Definition_2_1" data-tooltip="Not yet answered.">
<span class="sr">unanswered</span><span class="status-icon" aria-hidden="true"/>
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<p class="answer" id="answer_Definition_2_1"/>
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</p>
</li>
<li>
<p>
<p style="display:inline">[mathjaxinline]\frac{1}{\sqrt{3}}|{00}\rangle +\frac{\sqrt{2}}{\sqrt{3}}|{11}\rangle[/mathjaxinline]&#160;&#160;&#160;</p>
<div class="inline" tabindex="-1" aria-label="Question 2" role="group"><div class="inputtype option-input inline">
<select name="input_Definition_3_1" id="input_Definition_3_1" aria-describedby="status_Definition_3_1">
<option value="option_Definition_3_1_dummy_default">Select an option</option>
<option value="entangled"> entangled</option>
<option value="unentangled"> unentangled</option>
</select>
<div class="indicator-container">
<span class="status unanswered" id="status_Definition_3_1" data-tooltip="Not yet answered.">
<span class="sr">unanswered</span><span class="status-icon" aria-hidden="true"/>
</span>
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<p class="answer" id="answer_Definition_3_1"/>
</div></div>
</p>
</li>
<li>
<p>
<p style="display:inline">[mathjaxinline]\left(\sqrt{\frac{5}{8}}|{0}\rangle +\sqrt{\frac{3}{8}}|{1}\rangle \right)\otimes |{1}\rangle[/mathjaxinline]&#160;&#160;&#160;</p>
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<option value="option_Definition_4_1_dummy_default">Select an option</option>
<option value="entangled"> entangled</option>
<option value="unentangled"> unentangled</option>
</select>
<div class="indicator-container">
<span class="status unanswered" id="status_Definition_4_1" data-tooltip="Not yet answered.">
<span class="sr">unanswered</span><span class="status-icon" aria-hidden="true"/>
</span>
</div>
<p class="answer" id="answer_Definition_4_1"/>
</div></div>
</p>
</li>
<li>
<p>
<p style="display:inline">[mathjaxinline]\frac{1}{2}(|{00}\rangle +|{01}\rangle +|{10}\rangle +|{11}\rangle )[/mathjaxinline]&#160;&#160;&#160;</p>
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<option value="option_Definition_5_1_dummy_default">Select an option</option>
<option value="entangled"> entangled</option>
<option value="unentangled"> unentangled</option>
</select>
<div class="indicator-container">
<span class="status unanswered" id="status_Definition_5_1" data-tooltip="Not yet answered.">
<span class="sr">unanswered</span><span class="status-icon" aria-hidden="true"/>
</span>
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<p class="answer" id="answer_Definition_5_1"/>
</div></div>
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<h3 class="hd hd-3 problem-header" id="Bell_states-problem-title" aria-describedby="block-v1:MITx+8.370.1x+1T2018+type@problem+block@Bell_states-problem-progress" tabindex="-1">
Bell states
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Let [mathjaxinline]|{\Phi ^-}\rangle =U_{CNOT(1,2)}(H\otimes I)|{10}\rangle[/mathjaxinline]. Express [mathjaxinline]|{\Phi ^-}\rangle[/mathjaxinline] using the ket notation with [mathjaxinline]\{ |{00}\rangle ,|{01}\rangle ,|{10}\rangle ,|{11}\rangle \}[/mathjaxinline] basis. Please choose the global phase so that the coefficients are real. Type in <tt class="ttfamily">|00&gt;</tt> for [mathjaxinline]|{00}\rangle[/mathjaxinline] etc. </p>
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Let [mathjaxinline]|{\Psi ^-}\rangle =(I\otimes X)|{\Phi ^-}\rangle[/mathjaxinline], and [mathjaxinline]U_{\theta }=e^{-i(\theta /2) Y}[/mathjaxinline]. Find [mathjaxinline]U_{\theta }\otimes U_{\theta }|{\Psi ^-}\rangle[/mathjaxinline] with [mathjaxinline]\{ |{0}\rangle ,|{1}\rangle \}[/mathjaxinline] basis using the ket notation with [mathjaxinline]\{ |{00}\rangle ,|{01}\rangle ,|{10}\rangle ,|{11}\rangle \}[/mathjaxinline] basis. Please choose the global phase so that the coefficients are real. Type in <tt class="ttfamily">|00&gt;</tt> for [mathjaxinline]|{00}\rangle[/mathjaxinline] etc. You may enter <tt class="ttfamily">theta</tt> for [mathjaxinline]\theta[/mathjaxinline]. </p>
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<h2 class="hd hd-2 unit-title">Indirect measurements</h2>
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Indirect measurements
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Consider the circuit above where the meter sign denotes the measurement with [mathjaxinline]\{ |{0}\rangle , |{1}\rangle \}[/mathjaxinline] basis. </p>
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Suppose [mathjaxinline]|{\psi _ i}\rangle = a|{0}\rangle + b|{1}\rangle[/mathjaxinline] and [mathjaxinline]U=X[/mathjaxinline] where [mathjaxinline]a,b[/mathjaxinline] are real numbers satisfying [mathjaxinline]a^2 + b^2 =1[/mathjaxinline] and [mathjaxinline]X[/mathjaxinline] denotes the Pauli [mathjaxinline]X[/mathjaxinline]. Find the probability of getting result 0, and the state [mathjaxinline]|{\psi _ f}\rangle[/mathjaxinline] when the measurement result is 0. Express the probability and the state in terms of [mathjaxinline]a[/mathjaxinline] and [mathjaxinline]b[/mathjaxinline]. For expressing the state, use the ket notation with [mathjaxinline]\{ |{0}\rangle , |{1}\rangle \}[/mathjaxinline] basis. Choose the global phase so that the coefficients are real. </p>
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<p style="display:inline">probability = </p>
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<p style="display:inline">[mathjaxinline]|{\psi _ f}\rangle[/mathjaxinline] = </p>
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Suppose [mathjaxinline]U[/mathjaxinline] is a unitary having eigenvalue [mathjaxinline]e^{i\theta _1}[/mathjaxinline] with eigenstate [mathjaxinline]|{u_1}\rangle[/mathjaxinline], and eigenvalue [mathjaxinline]e^{i\theta _2}[/mathjaxinline] with eigenstate [mathjaxinline]|{u_2}\rangle[/mathjaxinline]. Also, suppose that [mathjaxinline]e^{i\theta _1}\neq e^{i\theta _2}[/mathjaxinline]. Let [mathjaxinline]|{\psi _ i}\rangle =a|{u_1}\rangle +b|{u_2}\rangle[/mathjaxinline] where [mathjaxinline]a,b[/mathjaxinline] are real numbers satisfying [mathjaxinline]a^2 + b^2 =1[/mathjaxinline]. Find the probability of getting the measurement result 0. Express the probability in terms of [mathjaxinline]a[/mathjaxinline], [mathjaxinline]\cos (\theta _1)[/mathjaxinline], and [mathjaxinline]\cos (\theta _2)[/mathjaxinline]. Input <tt class="ttfamily">cos(theta_1)</tt> for [mathjaxinline]\cos (\theta _1)[/mathjaxinline] etc. </p>
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<h3 class="hd hd-3 problem-header" id="Classical_strategy-problem-title" aria-describedby="block-v1:MITx+8.370.1x+1T2018+type@problem+block@Classical_strategy-problem-progress" tabindex="-1">
Classical strategy
</h3>
<div class="problem-progress" id="block-v1:MITx+8.370.1x+1T2018+type@problem+block@Classical_strategy-problem-progress"></div>
<div class="problem">
<div>
<p>
Consider the following game with three participants [mathjaxinline]A[/mathjaxinline], [mathjaxinline]B[/mathjaxinline] and [mathjaxinline]C[/mathjaxinline], and the referee [mathjaxinline]R[/mathjaxinline]. [mathjaxinline]R[/mathjaxinline] randomly chooses a 3 bit string from [mathjaxinline]\{ 000, 011, 101, 110\}[/mathjaxinline] and sends the first bit to [mathjaxinline]A[/mathjaxinline], the second bit to [mathjaxinline]B[/mathjaxinline], and the third bit to [mathjaxinline]C[/mathjaxinline]. We call the first bit, the second bit and the third bit that [mathjaxinline]R[/mathjaxinline] chooses [mathjaxinline]r_ A[/mathjaxinline], [mathjaxinline]r_ B[/mathjaxinline], and [mathjaxinline]r_ C[/mathjaxinline] respectively. Each of [mathjaxinline]A[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], [mathjaxinline]C[/mathjaxinline] then sends back a bit [mathjaxinline]a[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], [mathjaxinline]c[/mathjaxinline] to [mathjaxinline]R[/mathjaxinline]. [mathjaxinline]A[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], [mathjaxinline]C[/mathjaxinline] win if [mathjaxinline]r_ A \vee r_ B \vee r_ C = a \oplus b \oplus c[/mathjaxinline], and they lose otherwise. Their winning cases are summarized below. </p>
<center>
<table class="tabular" cellspacing="0" style="table-layout:auto">
<tr>
<td style="text-align:center; border-right:1px solid black; border:none">
[mathjaxinline]R[/mathjaxinline]'s bits </td>
<td style="text-align:center; border:none">
[mathjaxinline]a\oplus b \oplus c[/mathjaxinline] </td>
</tr>
<tr>
<td style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border:none">
000 </td>
<td style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border:none">
0 </td>
</tr>
<tr>
<td style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border:none">
011 </td>
<td style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border:none">
1 </td>
</tr>
<tr>
<td style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border:none">
101 </td>
<td style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border:none">
1 </td>
</tr>
<tr>
<td style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border:none">
110 </td>
<td style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border:none">
1 </td>
</tr>
</table>
</center>
<p>
For simplicity, we will only consider deterministic strategies in this problem. A deterministic strategy refers to the correspondence between [mathjaxinline]R[/mathjaxinline]'s bits and the choice of [mathjaxinline]a[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], [mathjaxinline]c[/mathjaxinline] depending on the [mathjaxinline]R[/mathjaxinline]'s value. Namely, [mathjaxinline]A[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], [mathjaxinline]C[/mathjaxinline] choose a strategy [mathjaxinline]a_0[/mathjaxinline], [mathjaxinline]a_1[/mathjaxinline], [mathjaxinline]b_0[/mathjaxinline], [mathjaxinline]b_1[/mathjaxinline], [mathjaxinline]c_0[/mathjaxinline], [mathjaxinline]c_1[/mathjaxinline] where [mathjaxinline]a_0[/mathjaxinline] is the value that [mathjaxinline]A[/mathjaxinline] will send back when [mathjaxinline]r_ A = 0[/mathjaxinline] etc. </p>
<p>
Let us define [mathjaxinline]S(i,j,k)\equiv a_ i \oplus b_ j \oplus c_ k[/mathjaxinline]. </p>
<ul class="itemize">
<li>
<p>
If [mathjaxinline]A[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], [mathjaxinline]C[/mathjaxinline] could always win the game, what must the value of [mathjaxinline]S(0,0,0) \oplus S(0,1,1) \oplus S(1,0,1) \oplus S(1,1,0)[/mathjaxinline] be? <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_Classical_strategy_2_1" class="text-input-dynamath capa_inputtype textline">
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</li>
<li>
<p>
What is the actual value of [mathjaxinline]S(0,0,0) \oplus S(0,1,1) \oplus S(1,0,1) \oplus S(1,1,0)[/mathjaxinline]? <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 2" role="group"><div id="inputtype_Classical_strategy_3_1" class="text-input-dynamath capa_inputtype textline">
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</div></div> </p>
</li>
<li>
<p>
In this setting, what is the maximum winning probability? </p>
<p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 3" role="group"><div id="inputtype_Classical_strategy_4_1" class="text-input-dynamath capa_inputtype textline">
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<h3 class="hd hd-3 problem-header" id="Quantum_strategy-problem-title" aria-describedby="block-v1:MITx+8.370.1x+1T2018+type@problem+block@Quantum_strategy-problem-progress" tabindex="-1">
Quantum strategy
</h3>
<div class="problem-progress" id="block-v1:MITx+8.370.1x+1T2018+type@problem+block@Quantum_strategy-problem-progress"></div>
<div class="problem">
<div>
<p>
Now, assume that [mathjaxinline]A[/mathjaxinline], [mathjaxinline]B[/mathjaxinline] and [mathjaxinline]C[/mathjaxinline] share the following quantum state </p>
<table id="a0000000032" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]|{\psi }\rangle =\frac{1}{2}(|{000}\rangle -|{011}\rangle -|{101}\rangle -|{110}\rangle )[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(1.27)</td>
</tr>
</table>
<p>
Each of [mathjaxinline]A[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], [mathjaxinline]C[/mathjaxinline] measures his(her) own qubit with [mathjaxinline]\{ |{0}\rangle , |{1}\rangle \}[/mathjaxinline] basis if the bit sent from [mathjaxinline]R[/mathjaxinline] is 0, and sends back the measurement result to [mathjaxinline]R[/mathjaxinline]. If the bit sent from [mathjaxinline]R[/mathjaxinline] is 1, they apply the Hadamard gate in their own system and measure it with [mathjaxinline]\{ |{0}\rangle ,|{1}\rangle \}[/mathjaxinline] basis, and send back the measurement results to [mathjaxinline]R[/mathjaxinline]. For instance, if [mathjaxinline]r_ A=1, r_ B=1, r_ C =0[/mathjaxinline], [mathjaxinline]A[/mathjaxinline] and [mathjaxinline]B[/mathjaxinline] apply [mathjaxinline]H[/mathjaxinline] on their system where the total state becomes [mathjaxinline](H\otimes H \otimes I)|{\psi }\rangle[/mathjaxinline]. They then measure it with [mathjaxinline]\{ |{0}\rangle , |{1}\rangle \}[/mathjaxinline] basis. </p>
<ul class="itemize">
<li>
<p>
Suppose [mathjaxinline]R[/mathjaxinline] sends [mathjaxinline]r_ A=0, r_ B=1, r_ C=1[/mathjaxinline]. What is the state right before the measurement? Express it with the ket notation with [mathjaxinline]\{ |{0}\rangle , |{1}\rangle \}[/mathjaxinline] basis. Choose the global phase so that the coefficients are real. </p>
<p>
<p style="display:inline">[mathjaxinline]|{\psi _ f}\rangle[/mathjaxinline] = </p>
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<p>
What is the winning probability in this setting? </p>
<p>
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var mke = function(x){ return lb + x + rb; }
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<h2 class="hd hd-2 unit-title">Measurement with general basis</h2>
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Measurement with general basis
</h3>
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<p>
Recall that [mathjaxinline]\vec{n}\cdot \vec{\sigma }=n_ x X + n_ y Y + n_ z Z[/mathjaxinline] is an observable having eigenvalues +1 and -1. Note that [mathjaxinline]\vec{n}[/mathjaxinline] is a real unit vector. Find the probability of getting the result +1 for a measurement of [mathjaxinline]\vec{n}\cdot \vec{\sigma }[/mathjaxinline] on the state [mathjaxinline]|{0}\rangle[/mathjaxinline]. Input <tt class="ttfamily">n_z</tt> for [mathjaxinline]n_ z[/mathjaxinline] etc. </p>
<p>
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vid = vuid[5];
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vid = vuid[2];
}
var mfnpre = vid.split("_video",1)[0];
var mfnid = mfnpre; // no periods
mfnpre = mfnpre.replace('8_370', '8.370'); // periods in gh filename
var lb = String.fromCharCode(60);
var rb = String.fromCharCode(62);
var mke = function(x){ return lb + x + rb; }
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console.log("html = ", html);
vblock.after(html)
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var cindex = Number(cst.data('index'));
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gurl = "https://github.com/mitocw/content-mit-8370x-subtitles/blob/master/";
}
gurl += mfnpre + ".txt#L" + String(cindex + 10 + 1);
console.log("going to ", gurl);
window.open(gurl, "MITx 8.370x subtitle source");
});
});
}
try{
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}
catch(err){
console.log(err);
}
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}
catch(err){
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