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| − | <h2>Base Case</h2>
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| − | The base case <math>(z = 0)</math> is true since it returns zero<br>
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| − | :<code>if <math>z = 0</math> then return(0)</code>
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| − | <h2>Assumptions</h2><br/>
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| − | multiply(y,z) gives the correct answer where:
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| − | * z <= n
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| − | * c >= 2
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| − | * y >= 0<br />
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| − | <br />
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| − | We will also assume the following. A brief proof will follow:
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| − | * <math>\left \lfloor z / c \right \rfloor c + (z\,\bmod\,c) = z</math>
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| − | <br/>
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| − | <h2>Proof</h2><br />
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| − | Where z = n+1, c >= 2, y >= 0<br/>
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| − | <br />
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| − | First we break the result of multiply into two parts:<br />
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| − | <br/>
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| − | A = <math>multiply(cy, \left \lfloor [n+1]/c \right \rfloor)</math><br/>
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| − | B = <math>y([n+1]\,\bmod\,c)</math><br/>
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| − | <br/>
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| − | <math>multiply(y,z) = A + B</math><br/>
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| − | <br/>
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| − | <br/>
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| − | Now, because c >= 2 we know that <math>\left \lfloor (n+1) / c \right \rfloor < (n+1)</math>.<br />
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| − | Based on that, we know that the call to multiply in "A" returns the correct result.<br />
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| − | <br />
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| − | <math>A = multiply(cy, \left \lfloor [n+1]/c \right \rfloor) = cy * \left \lfloor [n+1] / c \right \rfloor</math><br />
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| − | <br />
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| − | So now let's transform A into something more useful:<br />
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| − | <br />
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| − | <math>A = cy * \left \lfloor [n+1] / c \right \rfloor = y * \left \lfloor z / c \right \rfloor c</math> <br/>
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| − | (Note: We transformed n+1 back into z for simplicity)<br />
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| − | <br />
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| − | Based on our earlier assumption, we can transform this further:<br />
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| − | <br />
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| − | <math> \left \lfloor z / c \right \rfloor c + (z \,\bmod\, c) = z</math><br />
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| − | <math> \left \lfloor z / c \right \rfloor c = z - (z \,\bmod\, c)</math><br/>
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| − | <br />
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| − | <math>A = y * \left \lfloor z / c \right \rfloor c = y (z - z \,\bmod\, c) = yz - y(z \,\bmod\, c)</math><br />
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| − | <br />
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| − | And now we add B back into the mix:<br />
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| − | <br />
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| − | <math>A + B = yz - y(z \,\bmod\, c) + y(z \,\bmod\, c) = yz</math><br />
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| − | <br />
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| − | <br/>
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| − | <h2>Subproof</h2>
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| − | <b>Show that <math>\left \lfloor z/c \right \rfloor c + z\,\bmod\,c = z</math> where c >= 2</b><br/>
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| − | <br/>
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| − | We can prove this statement using a general example.<br/>
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| − | <br/>
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| − | First, assume that <math>z\,\bmod\,c = a.</math><br/>
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| − | <br/>
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| − | Now, due to the definition of floor, we know the following:<br/>
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| − | <br/>
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| − | <math>(z-a) / c = \left \lfloor z / c \right \rfloor</math><br/>
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| − | <br/>
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| − | This is because the remainder can't possibly be taken into account during a "floor" operation.<br/>
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| − | <br/>
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| − | Based on that, we can restate the equation as:<br/>
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| − | <br/>
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| − | <math>(z-a) / c * c + a = (z - a) + a = z</math><br/>
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| − | <br/>
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| − | | |
| − | [[introduction-TADM2E|Back to ''Introduction ...'']]
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