By Alexander Shen
That booklet does not include natural algorithms thought (like Kormen's or Skiena's book), yet difficulties (and so much of them are with solutions). each bankruptcy starts off with the straightforward challenge, by way of a few discussions of attainable ideas, and after an increasing number of hard projects ends with beautiful tough problems.
The booklet (in my opinion) is actually very functional (well, it includes a few idea, yet no longer very formal) and is worthy if 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)
Our program will consider only admissible positions. The tree of admissible positions for n = 3 Now the queens problem can be divided in two parts: (1) how to traverse all the vertices of a given tree; (2) how to represent the tree of admissible positions for the queens problem using Pascal constructs. Let us formulate the general problem of visiting all the vertices of a given tree. Imagine there is a robot that can be placed at any vertex of a tree. 1 Queens not attacking each other: position tree traversal 51 Moreover, the robot's repertoire includes tests that check whether each command can be executed: 9 is_up; 9 is_right; 9 is_down; (the last test returns True everywhere except at the root).
An-1 -b an (addition modulo 2). For n = 3, we get the following list: 000 001 011 010 110 111 101 100 It is easy to check that the transformation described (which can be applied to any sequence of n binary digits, giving another sequence of the same length) is invertible. Therefore, the list obtained contains all sequences of length n. On the other hand, adding 1 to a number in binary notation means replacement of the suffix 011... 0. This change leads to a change of exactly one digit after the transformation is applied.
If x [s] is changed from 0 to 1, we should replace 1 by 0 somewhere to keep the total number of is fixed. Therefore, it is necessary to have is on the right of x [ s ] . Conclusion: If we want to find the next bit string with k ls, we need x [s] to be the rightmost 0 that has some ls on the right. In this case we have x [ s + l ] =1 (otherwise, x [s] is not the rightmost one). Therefore, we should look for the maximal s such that x [s] =0 and x [ s + l ] =i: The term x [ s + l ] may be followed by several ls and then several 0s.
Algorithms and Programming: Problems and Solutions (Modern Birkhäuser Classics) by Alexander Shen