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Algorithm 2.1 Exchanging The Values Of Two Variables

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Learning objectives​

  • Explain the main ideas and vocabulary in Algorithm 2.1 Exchanging The Values Of Two Variables.
  • Work through the source examples for Algorithm 2.1 Exchanging The Values Of Two Variables without depending on raw chunk order.
  • Use Algorithm 2.1 Exchanging The Values Of Two Variables as selective reference when learner modules point back to How To Solve It By Computers.

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  • module-05-problem-solving

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Source-of-truth note​

This unit is anchored to How To Solve It By Computers and the source chapter "Algorithm 2.1 Exchanging The Values Of Two Variables". Use external resources only to clarify, extend, or modernize details without replacing the chapter's conceptual spine.

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Source provenance​

  • Primary source: How To Solve It By Computers
  • Source chapter: Algorithm 2.1 Exchanging The Values Of Two Variables
  • Raw source file: 018-algorithm-2-1-exchanging-the-values-of-two-variables.md

Merged source​

Algorithm 2.1 Exchanging The Values Of Two Variables​

id: '018-algorithm-2-1-exchanging-the-values-of-two-variables' book: "How to Solve It by Computer" source_file: "../how-to-solve-it-by-computer-r-g-dromey-for-unit-1.pdf" ocr_source: "../ocr-extracted.txt" chunk_index: 18 chunk_total: 138 title: "Algorithm 2.1 EXCHANGING THE VALUES OF TWO VARIABLES" part: "" chapter: "" pages: start: 64 end: 68 words: 828​

Algorithm 2.1 EXCHANGING THE VALUES OF TWO VARIABLES

Algorithm 2.1 EXCHANGING THE VALUES OF TWO VARIABLES​

Problem

Given two variables, a and b, exchange the values assigned to them.

ueveuupruuell'

The problem of interchanging the values associated with two variables

involves a very fundamental mechanism that occurs in many sorting and data

manipulation algorithms. To define the problem more clearly we will

examine a specific example.

Consider that the variables a and b are assigned values as outlined

Starting configuration

463 721

This means that memory cell or variable a contains the value 721, and

contents of a with 463, and the contents of b with 721. In other words we

want to end up with the configuration below:

Target configuration

a

721 463

To change the value of a variable we can use the assignment operator.

Because we want a to assume the value currently belonging to b, and b the

helnno;no tn n we nerhnnc the evohnnoe with the follnwino b assignments:

(1)

(2)

where " is the assignment operator. In (1) "

causes the value stored

in memory cell b to be copied into memory cell a.

Let us work through these two steps to make sure they have the desired

effect.

qtnrted ennfionrntinn with We the

then nfter eyecutinn of nqqionment have the we

The assignment (1) has changed the value of a but has left the value of b

untoucncu. cnccK111g With oui colli1guEdL1011 wc see iliac a nas

assumed the value 463 as required. So far so good! We must also check on b.

When the nqqiønment cten (2) i e h: = n iq made after executing sten (1 we

end up with:

In executing steo (2) a is not changed while b takes on the value that currentlv

VVV up UUVD

represent the solution we are seeking. The problem arises because in making

the assignment:

b

we have lost the value that originally belonged to a (i.e. 721 has been lost). It

stated more carefully as:

old value of b; new value of a

old value of a new value of b

What we have done with our present proposal is to make the assignment

new value of b new value of a

In other words when we execute step (2) we are not using the value a that will

make things work correctly--because a has already changed.

To solve this exchange problem we need to find a way of not destroying

"the old value of a" when we make the assignment

b a.

A way to do this is to introduce a temporary variable t and copy the original

value of a into this variable before executing step (1). The steps to do this

are:

b a.

After these two steps we have

b a

We are better off than last time because now we still have the old value of a

stored in t. It is this value that we need for assignment to b. We can therefore

make the assignment

t

After execution of this step we nave:

a

Rechecking with our target configuration we see that a and b have now been

interchanged as required.

The exchange nrocednre he outlined can now

Alaorithm descriDtion

  1. Save the original value of a in t.

  2. Assign to a the original value of b.

  3. Assign to b the original value of a that is stored in t.

The exchange mechanism as a programming tool is most usefully

implemented as a Drocedure that acceots two variables and returns their

Pascal implementation

procedure exchange (var a,b: integer);

var. ameger, begin {save the original value of a then exchange a and b} {assert:

a =aOAb --

b:=t (assert:

a =bOAb end

Notes on design

  1. The use of an intermediate temporary variable allows the exchange of

two variables to proceed correctly.

VI" pnaDIL..,VD at a KA y a a v a 1

always assumes the value dictated by the most recent assignment made

to that variable.

  1. Working through the mechanism with a particular example can be a

useful way of detecting design faults.

A more common application of exchange involves two array elements 4.

(e.g. a[i] and a[j]). The steps for such an exchange are:

e-- nrn.

a[j];

t

Applications

Sorting algorithms.

Supplementary problems

2.1.1 Given two glasses marked A and B. Glass A is full of raspberry drink​

and glass B is full of lemonade. Suggest a way of exchanging the

onntentQ nf A R

2.1.2 Design an algorithm that makes the following exchanges:​

The arrows indicate that b is to assume the value of a, c the value of

b, and so on.

2.1.3 Design an algorithm that makes the following exchanges:​

Given two variables of integer type a and b, exhange their values 2.1.4

without using a third temporary variable.

2.1.5 What happens when the arguments given to the procedure exhange​

are i and a[i]?