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Integers and Modern Applications for the Uninitiated

Sebastian Pauli

Section 2.4 The Assignment let

The instruction let \(a:=b\) assigns the value of \(b\) to \(a\text{.}\) For example, after the instruction let \(a:=5\text{,}\) the variable \(a\) has the value \(5\text{.}\) Suppose that the variable \(a\) has the value \(5\text{.}\) Then, after let \(a:=a+1\text{,}\) the variable \(a\) has the value \(6\text{.}\) First, we evaluate \(a +1\) as \(6\) because \(a = 5\text{.}\) Then we assign \(6\) to the variable \(a\text{.}\)
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In the video in Figure 2.4.1 we introduce the assignment instruction let and give the first examples.
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Figure 2.4.1. Algorithms -- The Assignment let by Matt Farmer and Stephen Steward
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The assignment instruction let can be used in two ways. We can give a value to a variable that does not yet have a value, or we can give a new value to a variable that already has a value.
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Example 2.4.2. Uses of let _ \(:=\).

We give examples for the use of the instruction let _ \(:=\text{.}\)
  1. The easiest use of let is to assign a value to a variable, for example we can use it to assign the value \(0\) to \(i\) :
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    let \(i:=0\)
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    After this instruction the value of the variable \(i\) is \(0\text{.}\)
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  2. We can also use let to assign the value of an expression to a variable. Assume that the value of the variable \(c\) is \(10\text{.}\) Consider the instruction:
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    let \(d:=c+35\)
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    First the expression on the right is evaluated. As \(c=10\) we get that \(c+35=10+35=45\text{.}\) Then the computed value 45 is assigned to \(d\text{,}\) so that the value of \(d\) after this instruction is \(45\text{.}\)
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  3. It is more confusing when the same variable shows up on both sides of the \(:=\) of a let instruction. Assume that the value of i is 5 and consider the instruction:
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    let \(i:=i+1\)
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    First the expression on the right is evaluated, we obtain \(i+1=5+1=6\text{.}\) Then the result of this computation, namely 6, is assigned to \(i\text{,}\) so that the new value of \(i\) is 6.
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The next algorithm computes the same values as Algorithm 2.2.3, namely \(c\text{,}\) \(c^2\text{,}\) \(c^3\text{,}\) and \(c^4\text{.}\)
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Now we show how Algorithm 2.4.3 works with an example.
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Example 2.4.4. Algorithm 2.4.3 with input \(c=-3\).

We follow the steps of the Algorithm 2.4.3 for the input \(c=-3\)
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Input \(c=-3\)
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  1. let \(d:=c\cdot c\text{:}\) We have \(c=-3\text{.}\) We compute \(c\cdot c=(-3)\cdot (-3)= 9\) and assign this value to the variable \(d\text{.}\) Now \(d=9\text{.}\)
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  2. let \(e:=c\cdot d\text{:}\) We have \(c=-3\) and \(d=9\text{.}\) We compute \(c\cdot d=(-3)\cdot 9=-27\) and assign this value to the variable \(e\text{.}\) Now \(e=-27\text{.}\)
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  3. let \(f:=d\cdot d\text{:}\) We have \(d=9\text{.}\) We compute \(d\cdot d=9\cdot 9=81\) and assign this value to the variable \(f\text{.}\)
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  4. return \(c\text{,}\) \(d\text{,}\) \(e\text{,}\) \(f\) : We return the values of \(c\text{,}\) \(d\text{,}\) \(e\text{,}\) and \(f\text{,}\) namely \(-3\text{,}\) \(9\text{,}\) \(-27\text{,}\) and \(81\)
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Output \(-3\text{,}\) \(9\text{,}\) \(-27\text{,}\) \(81\)
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So when the input to Algorithm 2.4.3 is \(c=-3\) the output is \(-3\text{,}\) \(9\text{,}\) \(-27\text{,}\) \(81\text{.}\)
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The input and output of Algorithm 2.2.3 and Algorithm 2.4.3 are the same, but the results are obtained in different ways. In particular the computation effort differs.
  • Algorithm 2.2.3 needs one multiplication to compute \(c^2= c\cdot c\text{,}\) two multiplications to compute \(c^3= c\cdot c\cdot c\text{,}\) and three multiplications to compute \(c^4= c\cdot c\cdot c\cdot c\) (count the multiplication symbols ‘\(\cdot\)’). This is a total of 6 multiplications.
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  • Algorithm 2.4.3 computes the same values as \(d=c\cdot c\text{,}\) \(e=c\cdot d\text{,}\) and \(f=c^2\cdot c^2\text{.}\) It only needs 3 multiplications to compute the four values \(d=c^2\text{,}\) \(e=c^3\text{,}\) and \(f=c^4\text{.}\)
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Algorithm 2.4.3 uses half as many multiplications as Algorithm 2.2.3 to compute the same values, and it uses no more operations. Therefore, Algorithm 2.4.3 is more efficient than Algorithm 2.2.3.
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Using the same strategy as in Algorithm 2.4.3, the algorithm in Example 2.4.5 computes the eighth power of its input. This is the first in a series of interactive algorithms. Click the instructions to see how they change the variable values.
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Example 2.4.5. Eighth power algorithm interactive.

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