# Part 1: Reduction

We have seen how to map a function across a vector of data, with the return value of each function call placed into a vector of results. For example, you summed together two lists of numbers using map using code such as this.

#include "part1.h"

int sum(int x, int y)
{
return x + y;
}

int main(int argc, char **argv)
{
auto a = std::vector<int>( { 1, 2, 3, 4, 5 } );
auto b = std::vector<int>( { 6, 7, 8, 9, 10 } );

auto results = map( sum, a, b );

print_vector(results);

return 0;
}

This returns a vector of results. However, what if we want to sum every value in the returned vector of results to form a single value? We could write the code by hand, e.g. create a new C++ file called reduce.cpp and copy into it the following code;

#include "part1.h"

using namespace part1;

int sum(int x, int y)
{
return x + y;
}

int main(int argc, char **argv)
{
auto a = std::vector<int>( { 1, 2, 3, 4, 5 } );
auto b = std::vector<int>( { 6, 7, 8, 9, 10 } );

auto results = map( sum, a, b );

int total = 0;

for (int result : results)
{
total += result;
}

std::cout << "Total = " << total << std::endl;

return 0;
}

Compile and run this code using

g++ --std=c++14 -Iinclude reduce.cpp -o reduce
./reduce

You should see printed out

Total = 55

This process of summing a vector of numbers into a total is an example of reduction. The vector of numbers has been reduced into a total by adding each value onto a running total. Reduction is the complement to mapping, and as such, in include/part1.h there is defined a pair of reduce functions, which look like this;

template<class FUNC, class T>
T reduce(FUNC func, const std::vector<T> &values)
{
if (values.empty())
{
return T();
}
else
{
T result = values[0];

for (size_t i=1; i<values.size(); ++i)
{
result = func(result, values[i]);
}

return result;
}
}

template<class FUNC, class T>
T reduce(FUNC func, const std::vector<T> &values, const T &initial)
{
if (values.empty())
{
return initial;
}
else
{
T result = initial;

for (const T &value : values)
{
result = func(result, value);
}

return result;
}
}

You can use the reduce function to reduce the above vector of results by creating editing your reduce.cpp file and changing the main function to read;

int main(int argc, char **argv)
{
auto a = std::vector<int>( { 1, 2, 3, 4, 5 } );
auto b = std::vector<int>( { 6, 7, 8, 9, 10 } );

auto results = map( sum, a, b );

auto total = reduce( sum, results );

std::cout << "Total = " << total << std::endl;

return 0;
}

Compile and run using the commands;

g++ --std=c++14 -Iinclude reduce.cpp -o reduce
./reduce

You should see printed out

Total = 55

The reduce functions takes two required arguments and one additional, optional argument:

1. The reduction function used to reduce a pair of arguments to a single result, e.g. sum takes two arguments and returns the sum of those arguments. This can be any function that accepts two arguments and returns a single result.

2. The vector of values to be reduced.

3. An (optional) initial value that is used as the first value for the reduction.

For example, edit your reduce.cpp to contain;

#include "part1.h"

using namespace part1;

int sum(int x, int y)
{
return x + y;
}

int main(int argc, char **argv)
{
auto a = std::vector<int>( {1, 2, 3, 4, 5} );

auto total = reduce( sum, a, 10 );

std::cout << total << std::endl;
}

Compile and run using the commands;

g++ --std=c++14 -Iinclude reduce.cpp -o reduce
./reduce

You should see that the total is 25. Why do you think the answer is 25?

Our reduce applies the reduction function (in this case sum) cumalatively from left to right along the items of a vector. If an initial value is supplied then this is used as the first value. Otherwise, the first value is the result of the reduction function applied to the first two items in the vector. In the above case, reduce performed;

1. total = 10
2. total = sum(total,1)
3. total = sum(total,2)
4. total = sum(total,3)
5. total = sum(total,4)
6. total = sum(total,5)

The result is thus 25, i.e. (((((10+1)+2)+3)+4)+5).

The reduction function can be any function that accepts two arguments and returns a single value. For example, let’s now use reduce to calculate the product of all of the values in the list. To do this, we need to create a new function that will take in two arguments and return their product.

Edit your reduce.cpp file to contain;

#include "part1.h"

using namespace part1;

int multiply(int x, int y)
{
return x * y;
}

int main(int argc, char **argv)
{
auto a = std::vector<int>( { 1, 2, 3, 4, 5 } );

auto result = reduce( multiply, a );

std::cout << result << std::endl;

return 0;
}

Compile and run using

g++ --std=c++14 -Iinclude reduce.cpp -o reduce
./reduce

You should see that the product is 120. Is this what you expected? In this case, reduce performed;

1. total = multiply(1, 2)
2. total = multiply(total, 3)
3. total = multiply(total, 4)
4. total = multiply(total, 5)

i.e. it set total equal to ((((1x2)x3)x4)x5) = 120.

Note that the reduction function is not limited to just numbers. You can write a reduction function to reduce any types of object together. For example, we could use reduce to join together some strings. Change your reduce.cpp file to contain;

#include "part1.h"

using namespace part1;

std::string join_strings(std::string x, std::string y)
{
return x + " " + y;
}

int main(int argc, char **argv)
{
auto a = std::vector<std::string>( { "cat", "dog", "mouse", "fish" } );

auto result = reduce( join_strings, a );

std::cout << result << std::endl;
}

Compile and run using

g++ --std=c++14 -Iinclude reduce.cpp -o reduce
./reduce

You should see the result cat dog mouse fish.

## Exercise

Modify your countlines.cpp program so that, in addition to printing out the total number of lines in each Shakespeare play, it also uses reduce to print out the total number of lines in all Shakespeare plays.

If you get stuck or want some inspiration, a possible answer is given here.

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