Alexander Shen's Algorithms and Programming: Problems and Solutions (Modern PDF

By Alexander Shen

ISBN-10: 0817638474

ISBN-13: 9780817638474

ISBN-10: 0817647600

ISBN-13: 9780817647605

ISBN-10: 3764338474

ISBN-13: 9783764338473

That e-book does not include natural algorithms conception (like Kormen's or Skiena's book), yet difficulties (and so much of them are with solutions). each bankruptcy starts off with the easy challenge, by way of a few discussions of attainable recommendations, and after a growing number of hard initiatives ends with lovely difficult problems.
The ebook (in my opinion) is basically very functional (well, it includes a few conception, yet now not very formal) and is precious in the event you are getting ready to the programming contests or Google/Microsoft-like interviews.

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Additional resources for Algorithms and Programming: Problems and Solutions (Modern Birkhäuser Classics)

Example text

Prove that the following program also processes all the leaves of a tree (one time each): var state: (LP, LAP); state := LP; while is_down or (state <> LAP) do begin if (state = LP) and is_up then begin I up left; end else if (state = LP) and not is up then begin I process; state := LAP; end else if (state = LAP) and is right then begin I right; state := LP; end else begin {state = LAP, not is right, is down} 1 down; end; end; Solution. The invariant relation: The value stored in the variable state is correct, that is, s t a t e = LP ~ L P i s true s t a t e ----LAP ==~ LAP is true The proof of termination: the change from LP to LAP is possible only when a vertex is processed.

1. Print all the sequences of length k composed of the numbers 1 . n. Solution. ) The first sequence in this ordering is < 1 , 1 . . . 1>; the last one is < n , n . . . n>. We use an array x [ 1 ] . x [k] to store the last sequence printed. make x [ 1 ] . . make last [l] . x := the successor of x 9 x end; Let us explain how to get the successor of x. By definition, the successor should have the same first s terms and larger ( s + l ) -th term. This is possible only if x [ s + l ] < n. To get the immediate successor, we find the maximal s with this property and increase the corresponding element by 1.

Therefore, to find the next permutation we should find the m a x i m u m k for which it is possible, that is, a k such that x[k] < x[k+l] > --. > x[n] Next we increase x [k] but keep the increase as small as possible. This means that we must find the minimal number among x [ k + l ] . x [n] that is larger than x [k]. After we exchange x [k] with the number found, we have to rearrange x [ k + l ] . x [n] to make the permutation as small as possible. To achieve this goal, we put x [ k + l ] . x [n] in increasing order.

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Algorithms and Programming: Problems and Solutions (Modern Birkhäuser Classics) by Alexander Shen


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