By Wirth N.

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**Example text**

PROCEDURE BinaryInsertion; (* ADenS2_Sorts *) VAR i, j, m, L, R: INTEGER; x: Item; BEGIN FOR i := 1 TO n-1 DO x := a[i]; L := 0; R := i; WHILE L < R DO m := (L+R) DIV 2; IF a[m] <= x THEN L := m+1 ELSE R := m END END; FOR j := i TO R+1 BY -1 DO a[j] := a[j-1] END; a[R] := x END END BinaryInsertion Analysis of binary insertion. The insertion position is found if L = R. Thus, the search interval must in the end be of length 1; and this involves halving the interval of length i log(i) times. 44269...

Given a text T in the form of a sequence and lists of a small number of words in the form of two arrays A and B. Assume that words are short arrays of characters of a small and fixed maximum length. Write a program that transforms the text T into a text S by replacing each occurrence of a word Ai by its corresponding word Bi. 6. Compare the following three versions of the binary search with the one presented in the text. Which of the three programs are correct? Determine the relevant invariants.

2. e. the string s has the form s = s0, s1, s2, ... , sN-1 where s1 ... sN-1 are the actual characters of the string and s0 = CHR(N). This solution has the advantage that the length is directly available, and the disadvantage that the maximum length is limited to the size of the character set, that is, to 256 in the case of the ASCII set. For the subsequent search algorithm, we shall adhere to the first scheme. A string comparison then takes the form i := 0; WHILE (x[i] # 0X) & (x[i] = y[i]) DO i := i+1 END The terminating character now functions as a sentinel, the loop invariant is Aj: 0 ≤ j < i : xj = yj ≠ 0X, and the resulting condition is therefore ((xi = 0X) OR (xi ≠ yi)) & (Aj: 0 ≤ j < i : xj = yj ≠ 0X).