By Kulikov A. S.

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M = .. ·1 ... m = .. ·1··· m = .. ·1··· m - V4 Vs = .. ·1 .. · m V6 = .. ·1··· m - s bits .. ·1 Wk Wk Wk Wk s bits . ··1 C S C3 C2 CI s bits .. ·1 c~ C3 c~ c~ s bits···1 end marker s bits···1 right move s bits .. ·1 don't move s bits···1 left move I··· I··· I .. · I.. · I.. · I .. · I· .. 7 The information has been spread to the positions looking at work tape contents Analyzing a possible move. Assume that the corresponding block of m bits of VI are /i = om-s < i l > [i 2 ] and the corresponding m bits of Vi are Ii = om-s < jl > [h] Given the input symbol and head shift infonnation, every neighborhood of three consecutive symbols 0"10"20"3 E {17 U Q P in an instantaneous description [i 2 ] determines a set N(0"10"20"3) such that O"~ 0"~0"3 E N (0"1 0"20"3) must occupy the same neighborhood in any [h] which follows in one step from [i2]' It is easy to design a vector machine examining each neighborhood in Va ...

13. Given three binary numbers contained in registers X, Y and Z respectively, let us define: S=XEBYEBZ C = «X J\ Y) V (Y J\ Z) V (Z J\ X» i 1 Prove that S + C = X + Y + Z. Given four binary numbers W, X, Y and Z applying this method twice find S and C such that S + C = W + X + Y + Z. Use this method to demonstrate that the multiplication of nonnegative binary numbers can be performed within time o (log n), where n is the length of the result. Hint. 5. Replace U and V by Us, Uc , Vs and Vc. The addition U + V at each iteration is replaced by the carry-save operations to compute the "new" Us and Uc from the "old" Us, Uc , Vs and Vc.

A k-PRAM processor has the usual random access instructions like: Xi := constant; Xi := Xj + Xi:=Xj Xi:= Xi XXi := - l¥J; Xk; Xk; XXi; := Xj; go to 1 if X j > 0; The connection between a parent and its offspring 1 is given by the following instructions: X i·'-C'h call1(al, a2, . ,au) return (a 1, a2, ... ,au) if I returned then ml else m2 channel access call to offspring 1 return to the parent I-return test branch A Tree Machine: the k-PRAM 47 If the parent attemps to access channel register e~ or call the offspring 1 while 1 is still active the parent will be blocked at that point in the computation until offspring 1 completes its computation and returns.