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Euler circuit
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</XXXXXX><p>So, to XXXXXXXXX the type XX the XXXXX XXXXX i.e. XXXXXXX it's an Euler path or Euler XXXXXXX, we'll start with XXXX XXXXXXXXX moving XXXX XXX edges only once and XXXXXXXXXXX XXXXXXX the XXXXXXXX XXX ending XXXX is same or XXX.</p><p>XX'll XXXX the XXXX in order as XXXXXXX:</p><p>X(Start) - X - 6- 3 - X - 7 - XX - XX - 9 - XX - 12 - 8 - XX - 14 - XX - XX - XX - 22 - XX - XX - 25 - XX - 17 - XX - 5 - X (end)</p><figure><img src="https://schoolsolver.XX.amazonaws.XXX/XXXXX/XXXXXXXXXXX/XXXXXXXXXXXXXXX.XXX" data-s3="XXXXX://schoolsolver.XX.amazonaws.XXX/XXXXX/attachments/XXXXXXXXXXXXXXX.png"></figure><p>Since, XXX XXXXXXXX and finishing XXXXXX are XXXX, XXX XXXXX is an <strong>XXXXX XXXXXXX.</XXXXXX></p><p>X XXXXX XXX been XXXXXXXX to XXXX XXXXXXXXXXXX description of XXX path formation XX the question.</p><p><XX></p><XX>Brute XXXXX XXXXXXXXX for finding XXXXXXXX Circuits</h4> <ol><li>XXXX of XXX possible XXXXXXXX circuits, </li><li>XXXXXXXXX the XXXXXX of XXXX XXXXXXX XXXXX in XXXX 1 </li><li>XXXX the XXXXXXX XXXX has XXX<b>smallest</b>XXXXXX.</li></ol><h4>So, why it is XXXX XXXXXX?</h4><p>This XXXXXX is<b>inefficient</b>, i.e., takes a XXX XX time. The XXXXXX is that if XX XXXX a XXXXXXXX graph, XXXX N vertices XXXX XXXXX XXX (N-1)! XXXXXXXX to list, calculate the weight, XXX XXXX select XXX smallest XXXX. XXXX XX XX XXX this XXXX number of (N-X)! by half, XXXXX XXX N XX XXXXX XX XX, the time it takes even XXX XXXXXXX computers of our day XX<b>XXXXX-XXXXX</b>is longer XXXX the age of our XXXXXXXX.Thus, XX XXXXXXXX XXXXX XXXXX XXXXX XXXXXX for XXXXXX having XXXXXXXX greater XXXX X.</p>">
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