How many squares are drawn if foo(0,0,5,1) is called? Assume that the function drawSquare(double x, double y, double halfLen) draws a square witha side length that is double of halfLen, centered at (x, y). For example, drawSquare(0,0,5) drawsa square centered at (0,0) with each side of length 10.Consider the following function foo defined as:public static void foo(double x, double y, double len, int n) {if (n < 1) return;drawSquare (x, y, len);foo(x-len/2, y-len/2, len/2,n-1);foo(x-len/2, y+len/2, len/2,n-1);foo(x+len/2, y+len/2, len/2,n-1);foo(x+len/2, y-len/2, len/2,n-1);}

Tree is a widespread abstract data form simulating a structure of a hierarchical tree with root values and children's sub-trees with a nudity of parent, as a collection of connected nodes. A tree data structure can be repetitiously specified as the node set (starting from a rooted node), where each of the nodes is a value data structure, along with a reference list (the "children") that does not repeat references and no references to root. Since the function invokation only depends on attribute ‘n’ which is last parameter in function foo. Using Tree Method: foo(0,0,5,1) Number of square draw=Number of invocation of draw()=1

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How many squares are drawn if foo(0,0,5,1) is called?

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Question

How many squares are drawn if foo(0,0,5,1) is called?

Assume that the function drawSquare(double x, double y, double halfLen) draws a square with
a side length that is double of halfLen, centered at (x, y). For example, drawSquare(0,0,5) draws
a square centered at (0,0) with each side of length 10.
Consider the following function foo defined as:
public static void foo(double x, double y, double len, int n) {
if (n < 1) return;
drawSquare (x, y, len);
foo(x-len/2, y-len/2, len/2,n-1);
foo(x-len/2, y+len/2, len/2,n-1);
foo(x+len/2, y+len/2, len/2,n-1);
foo(x+len/2, y-len/2, len/2,n-1);
}

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